strength of materials - priodeep.weebly.com · machines, the book also covers the subject of...
TRANSCRIPT
STRENGTH OF MATERIALS A COURSE FOR STUDENTS
B Y
P E T E R BLACK
P E R G A M O N P R E S S
OXFORD · LONDON · E D I N B U R G H · NEW YORK
TORONTO · PARIS · F R A N K F U R T
Pergamon Press Ltd. , Headington Hill Hall, Oxford 4 & 5 Fitzroy Square, London W . l
Pergamon Press (Scotland) Ltd. , 2 & 3 Teviot Place, Edinburgh I
Pergamon Press Inc. , 44-01 21st Street, Long Island City, New York 11101
Pergamon of Canada, Ltd. , 6 Adelaide Street East , Toronto, Ontario
Pergamon Press S.A.R.L., 24 rue des Écoles, Paris 5e
Pergamon Press GmbH, Kaiserstrasse 75, Frankfurt-am-Main
Copyright © 1966
Pergamon Press Ltd .
First edition 1966
Library of Congress Catalog Card No. 66-16364
2532/66
PREFACE
T H E contents of this book fulfil the requirements of those studying
for Higher National Certificates and Diplomas in Mechanical
Engineering and will be useful to students of Civil and Structural
Engineering. Together with i ts companion volume Mechanics of
Machines, the book also covers the subject of Applied Mechanics
for those aiming a t Higher National Diploma in Elect r ica l Engineer-
ing.
Only theory essential to simple stress analysis has been given but
this is complete as no steps in the mathemat ics have been omitted.
Discussion of assumptions made and of l imitations of theory as
derived, has been kept to a minimum as i t is felt tha t this can best
and safely be left to the lecturer .
No a t tempt has been made to deal with the test ing of materials,
other than in simple tension. The student will normally gain
experience of the various standard tests in the laboratory, where
also he will undertake pract ical work designed to verify—where
possible— the theoret ical subject mat te r here presented.
The scope of this book is not such as to include a discussion of
the properties of materials, which topic is inseparable from the
subject of Engineering Metallurgy.
A range of worked examples of graduated difficulty has been
included a t each stage to amplify the t ex t , the la ter ones in each
group being intended to give the student some idea of the sort of
question he will meet in his examination paper. The Council of
the Insti tution of Mechanical Engineers and the Senate of the Uni-
versity of London are thanked for their permission to reproduce
questions from past examination papers.
P E T E R B L A C K
l a SM ix
SYMBOLS
The following symbols are used in this book and conform with
B S . 1 9 9 1 :
L Length , span
A Area, cross-section
V Volume
W Weight , load (Note : W\g for mass)
w Weight per unit length, intensi ty of loading
R Radius of curvature
F Force , load, shear force
M Bending moment
Τ Twisting moment, turning moment, torque, periodic t ime
Ε Modulus of elastici ty
G Modulus of rigidity
Κ B u l k modulus
J Polar 2nd Moment of section (torsion constant)
I Moment of Iner t ia , 2nd Moment of area (bending constant)
Ζ Sect ion modulus, Maximum deflection of beam
U S t ra in energy
y Distance from neutral axis, deflection a t any point of beam
r Radius, distance from polar axis , amplitude
d Diameter , depth
k Radius of gyrat ion
ρ P i t ch , pressure
w In tens i ty of loading
χ Displacement, change in length
ν L inear veloci ty
ω Angular veloci ty
/ Linear acceleration
oc Angular accelerat ion
g Gravitat ional acceleration
t T ime
η F requency
l a * XI
xii S Y M B O L S
λ Stiffness, force per unit deflection
π Circumference/diameter rat io
μ Coefficient of friction
η Efficiency
ρ Dens i ty
δ Deflection
σ Poisson's R a t i o
q, fs Shear stress
e Direct strain
φ Shear strain, pressure angle
0 Angle of twist, angle turned through
Σ Sum of
Derived Units
Abbre-
Quant i ty viated Fu l l form
form
Linear velocity ft/s feet per second
Linear acceleration f t / s2
ft/s per second Angular veloci ty rad/s radians per second
Angular acceleration rad/s2
rad/s per second
Stress lbf/in2
pounds force per square inch
Bending moment and) Ibf in pounds force inches
torque J tonf in tons force inches
Densi ty lbf/in3
pounds force per cubic inch
Strain, kinet ic and 1 in Ibf inch pounds force
potential energy J in tonf inch tons force
ft Ibf foot pounds force
Load per unit length lbf/ft pounds force per foot Spring stiffness lbf/in pounds force per inch
Moment of Iner t ia orl Ibf ft s2
pounds force feet seconds squared
2nd Moment of Mass) slug f t2
slug feet squared (slugs = Ihi/g)
2nd Moment of Area in4
inches to the fourth
Section Modulus i n3
inches cubed
Area in2
inches squared
C H A P T E R I
SIMPLE STRESS, STRAIN AND STRAIN ENERGY
Load
Where an engineering component is concerned, this may consist
of forces due to one or more of the following :
1. A stat ionary (dead) load,
2. A change in veloci ty (inertia force),
3 . Ro ta t ion (centrifugal force),
4 . Fr ic t ion ,
5 . Bending,
6. Twisting (Torsion),
7. A change in temperature.
Stress
The application of the load causes a deformation of the compo-
nent which induces an equal and opposite resisting force in the
material, the intensi ty of which is referred to as stress.
LLLLL / / / / / / / /
I ι X
T
F (load)
(Section) A
L (length)
Τ7ΤΤ77Τ7Ύ777777Ύ7
(b) Compressive (fc)
Plan
A (area of 'section)
"I
Li pi
777Π7777Τ777777777777Γ/
(a) Tensile (f+) FIG. 1
1
(c) Shear (fs or q) (L = Arm of shear couple)
2 S T R E N G T H O F M A T E R I A L S
Thus, Stress = Load per unit area (usully lbf/in2)
or / = FjA for simple cases as shown above.
Strain
This is a measure of the deformation produced in the component
b y the load and is denoted b y e. F o r tensile and compressive loads,
the strain is defined as the change in length per unit length, i.e.,
χ
Note tha t this is a rat io and therefore has no units. F o r shear loads
e = x\L = φ radians, since χ is very small.
The effects of shear will be discussed la ter in greater detail .
Elasticity
A material is elastic as long as the strain disappears with the
removal of the load. A limiting value of load exists beyond which
this does not happen and the corresponding stress is called the
Elastic limit
FIG. 2 FIG. 3
Ehstic Limit. After passing this the residual strain is referred to as a Permanent Set.
I n the case of ferrous materials (e.g. steel and iron), the deforma-tion χ is, up to the elastic limit, proportional to the load (Fig. 2) and this is known as Hooke's Law. However, for most non-ferrous
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 3
metals (e.g. aluminium) the l imit of proportionality occurs before
the elastic l imit.
Now, Stress / = F\A
. ι e
= a constant χ — . χ
B u t F fx is the slope of the graph in Fig . 2 and is constant
within the elastic l imit so t ha t (below the elastic l imit) the ratio
fje is also constant and is clearly the slope of the graph obtained
by plotting stress against strain (Fig. 3 ) . This constant is called the
Modulus of Elasticity (Young's Modulus) and is denoted b y E. I t s
value depends on the material .
Thus , Ε = — e
or Stress = Ε χ Strain.
and Strain e = x/L,
-LI—
L F
A χ
— = / χ — i t follows tha t , if i t were possible
(which, within the elastic limit, i t is not ) to continue loading until
χ = L, then Ε = f. Thus Ε would be the stress required to produce
an elastic extension equal to the original length. F o r steel this
stress would be of the order of 30 χ 1 06 lbf/in
2 and this is the value
of Ε for this material .
/ fL Note: Since change in length χ = eL and e = — , .'. χ = .
Rigidity
When equal and opposite forces are applied as in F ig . 1 (c) so
as to induce a s tate of shear (i.e. forces are parallel and not co-
axial) the resulting couple produces a change in shape in the
material. This is not the case with materials under simple tension
or compression, where the shape remains the same notwith-
4 S T R E N G T H O F M A T E R I A L S
standing the deformation due to the load. The resistance offered
by the material to such change in shape is dependent on what is
called the rigidity of the material .
Clearly the material may fail (as shown in F ig . 4 ) a t any one of
an infinite number of sections such as X X of area abed (= A).
As with simple tension and compression, the ratio stress/strain
F
FIG. 4
is constant provided the elastic l imit is not exceeded, and this
constant is called the Modulus of Rigidity and denoted by G. I t s
value is a measure of the abil i ty of a material to retain i ts shape
under load and, as will be shown later, is dependent on the value
of E. Thus, Shear stress = Modulus of rigidity χ Shear strain,
F or fs = Gcp where fs = —,
A
and φ = angle of shear.
Average values (in lbf/ in2) of Ε and G for some common materials
are given in the following table :
Material Ε G
Plain carbon steel (0-1-0-5% C) 29-7 χ 106
11-6 χ 106
Aluminium and its base alloys 10-3 χ 106
3-7 χ 106
Hard drawn copper wire 18-0 χ 106
6-5 χ 106
Cold drawn brass 14-3 χ 106
5-2 χ 106
Cast iron 16-5 χ 106
β·4 χ 10e
Rubber 750 approx. 240 approx.
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 5
The Tensile Test
As already stated, extension is proportional to load up to the
limit of proportionality, which, in the case of ferrous metals ,
coincides for pract ical purposes with the elastic limit. When the
extension in a steel tes t piece (see relevant Bri t i sh Standard) is
measured under load and the loading is continued until fracture
occurs, the complete F-x graph has the form shown in F ig . 5
Beyond the elastic l imit ( E . L . , F ig . 5) the rate of extension
increases until a point called the Yield Point ( Y . P. , F ig . 5) is
reached a t which occurs a sudden extension with no increase in load. Fur ther loading results in an extension a t a greatly increased rate until a point of maximum load (Ì, F ig . 5) is reached. Around this point the material begins to extend locally without further load increase, and a " n e c k " is formed a t the midpoint of which fracture finally occurs.
Since / = FjA and e = x/L, the f - e graph is exac t ly similar to the F-x graph up to the point where the neck begins to form, i.e. up to the point beyond which the Sect ion A is no longer constant but diminishing. Thus, although the actual load falls after the point M, due to the reduction in A, the net effect is a continued stress increase as shown in F ig . 6. I f A is assumed constant , i.e. the nominal stress is plotted, the dot ted position of F ig . 6 is obtained.
\ Elastic zone
FIG. 5
ι \ V Uniform Local strain strain
FIG. 6
6 S T R E N G T H O F M A T E R I A L S
Proof Stress
Since most metals exhibit no yield point, i.e. no sudden extension
a t any load, some criterion other than yield stress must be used in
the design of non-ferrous components. The quant i ty used is defined
as the stress which will induce in the tes t piece an arbitrary pre-
determined permanent strain, (set) usually 0-001, i.e. the permanent
increase in length after subjection to such stress is OOOIL where L
is the tes t or " g a u g e " length. (See relevant Br i t i sh Standard for
tes t specimens).
This stress is then know as the 0-1 per cent Proof stress and the
corresponding load as the 0-1 per cent Proof load.
FIG. 7
The proof load may be obtained by drawing on the F-x graph (Fig. 7) a straight line parallel to the straight part of the graph from a point χ = 0 0 0 1 L from the origin, on the #-axis, and projecting horizontally (as shown) from its intersection Ρ with the graph.
The 0-1 per cent Proof stress is then the load obtained as above divided b y the test-piece section. I t is usually impossible to fix the position of the limit of proportionality (L.O.P) with any accuracy and the determination of proof stress eliminates the need to t ry .
Alternatively, the proof stress may be obtained directly from the / — e graph by drawing the parallel line from a strain of
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 7
•* 6 i n - — * • 12 in — 6 in •
1-25 in dia. lO in dia. 1*5 in dia
Solution
F o r the L H end
F o r the centre portion
F o r the R H end
To ta l reduction
FIG. 8
A1 = 0-785 χ 1-252 = 1-227 in
2.
11
/a
1-227 8-97 tonf/in
2.
0-785 χ 1-02 = 0-785 in
2.
11
0-785 = 14-0 tonf/in
2.
0-785 χ 1-52 = 1-769 in
2.
11
1-769 6-22 tonf/in
2.
X — X^ -\- #2 4~ i
Ε Ε Ε '
1 = — [(8-97 χ 6) + (14 χ 12)
Ε
+ (6-22 χ β ) ] ,
(53-8 + 168 + 37 -3 ) ,
= 0-0207 in.
12500
259
12500
0-1 per cent, i.e. of 0-001. The 0-2 per cent and 0-5 per cent Proof
stresses, also in common use, may be found in the same way.
Note: As a rule, the stress in a component carrying a non-fluctuating tensile load should not exceed two thirds of the specified proof stress.
E X A M P L E . Es t ima te the to ta l reduction in length of the strut
shown when carrying an axial compressive load of 11 tonf. Assume
Ε = 12,500 tonf/in2.
8 S T R E N G T H O F M A T E R I A L S
E X A M P L E . A steel tube is secured in a socket by a number of
radial steel pins 0 - 2 5 in. dia. as shown. I f the axial load is such as
to induce a tensile stress of 2 0 tonf/in2 in the tube, find the number
of pins required to limit the shear stress in them to 8 tonf/in2.
FIG. 9
Solution
Tube section = ^ ( 1 - 8 82 - 1 - 7 9
2) = 0 - 2 5 9 in
2.
Load on tube ftAt = 2 0 χ 0 - 2 5 9 - 5 - 1 8 tonf.
Pin section = x 0 - 2 52 = 0 - 0 9 8 in
2.
4
.'. Permissible shear load per pin = fpAp = 8 χ 0 - 0 9 8 = 0 - 7 8 tonf.
.'. Required No. of pins = -77^3-, 0 * 7 8
= 6 - 6 5 , say, 7 .
Note: Such joints usually fail due to excessive bearing stress, i.e. the holes elongate.
E X A M P L E . TWO vert ical steel wires each 0 - 0 5 in. dia. and 4 8 in.
long support a light rod over a span of 1 2 in. A concentrated load
of 6 0 lbf is then applied vertically downwards a t a point 4 in. from
one of the wires.
Calculate: 1 . the load on each wire,
2 . the stress in each wire,
3. the extension in each wire,
4 . the angular displacement of the rod from the
horizontal, assuming i t does not bend.
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y
Solution
Moments about L H end give
F2 χ 12 = 60 χ 4 .
.·. F2 = 201b f ,
i.e. F1=40lbi.
| * - I 2 i n - * |
111(11 III 11(11111 III III .
0-05 in_ dia.
Fx Stress in L H wire f, = —-, 1 1
A 40
0-785 χ 0 0 52
= 20 ,350 lbf/in2.
Since J1. = - ^ - , Λ / 2 = 10,175 lbf / in
2.
, 4 in
Extens ion in L H wire χ Ε '
20,350 χ 48
30 χ 1 06 3
: 0 - 0 3 2 6 i n .
4 8 in
6 0 l b f
FIG. 10
The extension in the R H wire will be half this amount,
i.e. x2 = 0-0163 in ,
.*. Difference in level of ends == x1 — x2,
= 0 0 1 6 3 in .
Angle required r
_ 0 0 1 6 3
~~ —
Ϊ 2 '
= 0-00136 radian,
180 = 0-00136 χ
π
χ 6 0 , = 4-67 min .
0 - 0 1 6 3 in
12 in
F I G . 1 1
9
10 S T R E N G T H O F M A T E R I A L S
Example. Three struts 2 in. dia. and 24 in. long are arranged in
line and equidistant and support a rigid load of 55 tonf. I f Ε for
the material is 12,000 tonf/in2, determine the proportion of the
load carried by each when the centre strut sinks 0-01 in.
Load level
2 4 in 23-99 in
Β A
FIG.12
1
Solution
Area of strut section = 0-785 χ 22,
= 3-14 in2.
Load = Stress χ Section,
= Ε χ Strain χ Sect ion.
.·. Load on A = 12,000 ( ^ 9 ) 3 1 4
>
= 1570 a;.
χ + 0-01 Load on Β and C = 2 [ 12,000
\ 24 = 3140 χ + 31-4 .
x + O-OI
3-14 ,
/ . 55 = 1570α: + 3140a; + 31 -4 .
4170a; = 23 -58 .
23-58
4710 '
= 0-005 in .
.·. Load on A = 1570 χ 0 -005,
- 7-85 t o n f .
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 1 1
Solution
Steel section As = 0 - 7 8 5 χ 0 - 1 0 52,
= 0 - 0 0 8 6 6 i n2.
Bronze section Ab = 6 ( 0 - 7 8 5 χ 0 - 0 7 82) ,
= 0 - 0 4 5 1 i n2.
τ j -ο ë ç 5 5 - 7 - 8 5 4 7 - 1 5 00 _ , , .*. Load on  and C = = — - — = 2 3 - 5 7 tont each.
Δ Δ
E X A M P L E . A steel wire 0 - 1 0 5 in. dia. is covered by s ix bronze
wires each 0 - 0 9 8 in. dia. I f the working stress in the bronze is
4 tonf/in2, calculate :
(a) the strength of the combination,
(b) the equivalent tensile modulus.
Es = 2 8 χ 1 06 lbf/ in
2 and Eh = 1 2 χ 1 0
6 lbf/in
2. (L .U.)
Common tensile strain = Λ) /s ç r
ºÃ = -Åà and / b
= 4 tonf/in2.
.*. Stress in steel 9 - 3 3 tonf/ in2.
FIG. 1 3
12 S T R E N G T H O F M A T E R T A L S
Common strain =
Since stress
Tota l load
Tota l section
or Evq,
4 χ 2240
12 χ 1 06 '
0-000747.
Ε χ Strain,
Equivalent Ε χ Common Strain
ZW ι χ — ΣΑ
585 1
(0-00866 + 0-0451)
585
0-0538 χ 0-000747 '
14-58 χ ΙΟ6 lbf/in
2.
0 -000747 '
Strain Energy
This is the work done in the elastic straining of a material , whether under tension, torsion, bending or a combination of one or more of these. Such work (mechanical energy) is stored in the strained material and may be released either under controlled conditions (as in an unwinding clock spring) or suddenly, in which case the result is a vibration. The amount of such strain energy (sometimes referred to as Resilience) is a measure of the abil i ty of a material to resist shock without permanent deformation.
There are three ways of applying a load.
Permissible load = /S^4S + fbAh,
= (9-33 χ 0-00866) + (4 χ 0 -0451) ,
= 0-0808 + 0-1804,
= 0-261 tonf ,
= 585 lbf .
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 13
I f / and Ε are in lbf/in2 and V is in i n
3 then U is in in lbf. The
maximum possible value of U is tha t obtaining a t the elastic limit,
i.e. if / E L = elastic l imit stress, then
/ 2
tfmax = -wk- Pe r
unit volume.
2. Suddenly
If , from being held jus t in contac t with the collar, the load is suddenly released, collar and load travel beyond the equilibrium position by an amount equal to the s tat ic deflection x, i.e. the to ta l instantaneous extension is 2x. The maximum instantaneous strain (and hence instantaneous stress) is thus twice the strain corresponding to a gradual loading.
To produce a s tat ic extension of 2x clearly would require a gradually applied load of twice as much, i.e. of 2 I F in which case the graph would be as in Fig . 16, the mean load being W.
1. Gradually
The load starts from zero and increases uniformly up to i ts
final value, e.g. liquid being poured into a tank.
Suppose a load W is held jus t in contac t with the collar Fig . 14
and released gradually. The rod will extend in proportion (like a
spring) and the graph of load against extension will be a straight
line through the origin as in F ig . 15. The final value χ is called the
Static Deflection and when this has been at tained, the load is in the
Equilibrium Position. Then Tensile strain energy U = Average force χ Extens ion
W = —— χ ( = shaded area under
graph), fA fL
= ——- χ —— where 2 Ε
f = max. stress induced,
f2
f2
.*. U = —— χ Volume.
14 S T R E N G T H O F M A T E R I A L S
Section A
W
Γ I
-J- ΖΖ1^-Γ, Equili I I posit
-Collar
librium ion
FIG. 1 4
FIG. 1 5 F I G . 1 6
Ι
F I G . 1 7
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 15
3. With Shock
I f the load W is permitted to fall freely through a height h
before striking the collar, then the energy absorbed will be tha t due
to a sudden load application plus the
kinet ic energy gained during the fall.
Thus U = Wx + K . E . (where
K . E . = P . E . = Wh)
= Wx + Wh.
This is clearly the to ta l loss in poten-
t ia l energy of the load W. Now, maxi-
mum force in rod = fA where / = max .
stress, induced ;
Mean force in rod
Hence,
Equat ing,
U = fA
JA. 2
χ χ.
ÎA.1L 2 Ε
'///////////////.
t L . Τ -
α; = Wx + Wh and χ
W-^- + Wh,
Fia. 18
IL Ε
i .e.
Then U = shaded area of F ig . 16,
= W χ 2x,
= 2Wx,
= 4 (Area under first graph).
Since the full value of the load is in fact suddenly applied, the
actual graph is the horizontal straight line of F ig . 17 the area under
which is clearly 2Wx. Thus b y applying a load suddenly instead
of gradually, the instantaneous deflection, strain and stress are
doubled and the instantaneous strain energy is multiplied b y four.
16 S T R E N G T H O F M A T E R I A L S
Transposing gives - ^ r /2 τ τ " / — Wh = 0 ,
which is a quadratic equation enabling the maximum stress / to be found.
The behaviour of the rod in the foregoing cases of gradual, sudden
and shock application of the load is comparable with tha t of the
spring shown in Fig . 19.
FIG. 1 9
(a) Spring unloaded.
(b) Load released gradually from no-load position X X . The load
remains stat ionary in the stat ical ly deflected (equilibrium)
position.
(c) Load released suddenly from no-load position X X . The load
makes oscillations of which the amplitude (i.e. maximum
displacement on either side of the equilibrium position) is
equal to the s tat ic deflection x.
(d) Load released suddenly from a point above the no-load posi-
t ion X X . Since the spring is initially compressed, the load
already possesses kinet ic energy a t the ins tant of passing
position X X so tha t the behaviour of the spring is similar to
tha t of the rod subjected to shock loading. The amplitude of
the resulting oscillation is thus increased by some amount y
(Fig. 19(d)) .
Note: For a full treatment of vibrations see Mechanics of Machines by the same author.
No load position X-L-
Equilibrium position —
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 1 7
E X A M P L E . A ship of 8 0 0 ton is checked by two 4 in. dia. steel
hawsers when moving a t 0 - 5 f t /s . Assume tha t the kinet ic energy
of the vessel is all converted into strain energy and est imate the
length of hawser required to l imit the tensile stress in the steel to
6 tonf/in2. Take Ε = 1 3 , 4 0 0 tonf/in
2.
Solution
1 W K . E . = — — v 2,
2 g
8 0 0 χ 0 - 5 *
2 χ 3 2 - 2
3 - 1 f t / tonf
/2 Strain energy = - — per unit volume,
IE
χ F (where V = hawser volume in i n3) ,
2 χ 1 3 , 4 0 0
0 - 0 0 1 3 4 V inch tonf.
3 1 χ 1 2 Λ Volume of hawser required =
0 - 0 0 1 3 4 '
2 7 , 7 0 0 in3.
2 7 , 7 0 0
L
= ( 0 - 7 8 5 χ 42) L where
L = length in inches,
2 7 , 7 0 0
2 ( 0 , 7 8 5 χ 42)
1 1 0 5 in,
9 2 ft.
E X A M P L E . A steel rod 1 2 in. long 1 - 2 5 in. dia. has 7 in. of i ts length machined down to 1 in. dia. I f an axial blow induces a stress of 9 - 6 tonf/in
2 in the th icker part, calculate :
(a) the stress induced in the thinner part, (b) the force in the rod, (c) the strain energy of each part in inch tonf,
(d) the to ta l strain energy in inch lbf.
Ε = 1 3 , 5 0 0 tonf/in2.
18 S T R E N G T H O F M A T E R I A L S
7 in
in dia.-
1-25 in ο ι α -
Ι 5 in
Blow Solution
Force = Stress χ Sect ion,
. ' . ί = / Λ = / Λ
/« = / i-
- 9-6
l2
1-252
f, = 9-6 -A,
l-O2 '
= 15 tonf / in2,
.·. F χ 15 χ 0-785 χ l2,
= 11-87 t o n f .
'////////////
FIG. 20
Strain energy of th icker part
Q-62
(0-785 χ l - 2 5 2) 5 ,
A . x F
2 £ 1
2 χ 13,500
= 0-0285 inch/ tonf .
Strain energy of thinner part U2 = x F 2 ,
1 52
(0-785 χ l - 02) 7 ,
i n ^ e
1-5 in
ill 4 in
6 in
FIG. 21
2 χ 13,500
= 0-0458 inch tonf .
To ta l strain energy
U=U1 + U%
= (0-0285 + 0-0458) 2 2 4 0 ,
= 166-5 inch lbf .
E X A M P L E . TWO round bars of
the same mater ial have the dimen-
sions shown. I f Β receives an axia l
Ψ blow inducing a stress of 5 tonf/in2,
find the max . stress induced b y a
similar blow on A.
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 1 9
Ε
ü a = A ( 4 x i } + A ( 6 x i , 5 )
(where f1 = stress in reduced part) ,
2ft ±5
Ε +
li
- 1 ( 2 / ? + 4-5/1) and / 2 χ 1-5 = U χ 1-0
(since force is the same, i.e. / 2 = - j ^ - ,
= - 1 ( 2 / ? + 4 · 5 /!
i i Γ 1 ' " 1 - 5 «
= 4 ( 2 / ! + 2 / ? ) '
- A / 2
B u t J 7 A = C 7 n . Since the blow is the same.
. 4
/ l2 1 25
· · ,2 1 25
OT o r
.'. / x = 5-58 tonf/ in2
E X A M P L E . A 1 in. dia. bolt has an effective length of 1 0 in. I f the nut is t ightened until the strain energy is 5 in lbf, calculate :
(a) the stress induced, (b) the extension produced, (c) the force in the bolt . Take Ε = 3 0 χ 1 0 6 lbf/in2.
Solution
Effective volume = 0 -785 χ l 2 χ 1 0 ,
- 7-85 i n 3 .
Solution
U b = 2Ë
X V o l u m e'
52
= - ^ ( i o x i ) ,
* ^ - i n c h tonf .
20 S T R E N G T H O F M A T E R I A L S
Strain energy U 2E
χ Volume.
5 = /2
2 χ 30 χ 1 06
5 χ 2 χ 30 χ 1 06
7-85,
Extens ion
7-85
. / - 6-2 χ 1 03 = 6200 lbf/in
2.
jL 6200 χ 10
38-2 χ 106,
Ε 30 χ 1 06
Force in bolt = Stress χ Sec t ion ,
= 6200 χ 0-785 χ l2,
= 48601bf .
0-002 in .
E X A M P L E . I n the arrangement shown, the upper support may
be taken as rigid. Ε = 13,200 tonf/in2.
Es t ima te :
(a) the extension required to
induce a tensile stress of
5 tonf/in2 in the rod,
(b) the height h from which the
annular weight must be re-
leased to induce this stress
on impact .
Assume no energy to be lost in
deforming the stop.
Solution
Required extension χ = ,
5(120)
~ 13,200 '
= 0-0454 in .
///////////////////////
FIG. 22
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 2 1
Loss in Poten t ia l energy = Average force χ Extens ion .
- Μ -
0 · 2 5 ( Α χ 0 - 0 4 5 4 )
h + 0 - 0 4 5 4
— χ 0 - 0 4 5 4
5 0 0 4 5 4 Λ^ Ο Κ
F χ
Ί > 2 5 - Χ 0 , 7 85 Χ 1 ,1
= 0 - 3 5 6 ,
.·. h = 0 - 3 5 6 - 0 - 0 4 5 ,
= 0 - 3 1 in (approx.) .
E X A M P L E . A certain big-end bolt is 1 - 2 5 in. dia. from head to
nut and when the nut is t ightened, the strain energy is 3 in. lbf.
Es t imate the increase in strain energy brought about b y reducing
the dia. to 0 - 1 in. over three quarters of the length, assuming the
tension in the bolt to remain the same.
Solution
i ( 0 - 7 8 5 x l - 2 5z) = F -
r 1-25 in
i
Ι-ΦΟΊΗ
M Υ ι Ο l +
I
FIG. 2 3
Force = Stress χ Sec t ion .
.·. F = ( 0 - 7 8 5 χ 1 - 2 52) / = ( 0 - 7 8 5 χ 1 - 0
2) f±
(where f± = stress in reduced part) ,
• · h h Q2 ,
= 1 - 5 6 / .
Strain energy = — ~ χ Volume .
2 SM
22 S T R E N G T H O F M A T E R I A L S
1. For original state
3
Ε
2E (0-785 χ 1 - 2 5
2) L ,
3 x 2
0-785
4-9
1-252 '
2. For modified state
JL 2E
U =IL (0-785 χ 1·252) 0 · 25£ + (0-785 x Ι Ο
2) 0-75 Ζ,,
Ε χ 0 1 5 3 +
2E
2E
(0-785 χ 0-75 L),
= (0-153 + 0-715) , Ε
= 4-9 χ 0-868,
= 4-25 inch lbf .
Increase = 4-25 — 3-0,
= 1-25 inch lbf .
ν//////////////,
E X A M P L E . A load of 0-5 ton is to fall
0-25 in. on to a collar a t the lower end of
a vert ical t ie 72 in. long. Es t imate the
diameter d required to l imit the strain in
the t ie to 0-0005 given tha t Ε = 12,500
tonf /in2.
72 in
Solution
Strain e = = 0 0 0 0 5 ,
72
0 0 0 0 5 χ 7 2 ,
0-036 in .
Stress / = Ε χ St ra in ,
12,500 χ 2240 χ 0-0005,
14,000 lbf/in2.
0-5 ton
I : L 0-25in
FIG. 2 4
or
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 23
Strain energy = K . E . of mass + work done during sudden applica-tion.
,\ΙΑ·χ = (0-5 χ 2240) 0-25 + (0-5 χ 2240)a; , where χ = 0-036, Δ
14,000.4 0-036 - 280 + 4 0 - 3 ,
= 320 -3 .
_ 320-3 χ 2
~ 14,000 χ 0 0 3 6 '
= 1-27 i n2.
.·. 0-785 d2 = 1-27,
d2 = 1-62,
.·. d = 1-27 in .
E X A M P L E . A load of 2000 lb is being lowered a t a steady rate of
2 f t / s a t the end of a steel cable of 1-0 i n2 section. I f the pulley
j ams when the load is 30 ft below i t , es t imate the additional stress
induced in the cable by the stoppage taking Ε = 30 x 1 06 lbf/in
2.
Neglect the weight of the cable and assume the pulley axis not to
be deflected.
Solution
Stress before jamming occurs
Load
Section
2000
1-0 ' 3 0 f t
= 2000 lbf/in2.
On jamming
Strain energy of cable = K . E . of load -1- work done during sudden x -loading.
• JL ·· 2E
Cable volume
•v2 + Wx,
-1-0 in2 section
2 0 0 0 Ibf
v = 2 f t / s
2*
FIG. 2 5
24 S T R E N G T H O F M A T E R I A L S
where χ
^ χ 30 χ 12 χ 1-0 =
/ 30 χ 12 \
\ 3 0 x 1 06/ ' '
0-000012 / (where / - max. stress
induced).
Ί 9ΑΠΠ Ί
12
1 2000 •
¥X
" 3 2 ¥X 2
"
and /
* * 2(30 χ 106)
+ 2000 (0-000012 / )
(Note tha t the kinet ic energy must be in inch lbf)
i.e. 6 /2 x 1 0 '
6 = 1490 + 0 - 0 2 4 / ,
or /2 = 4 0 0 0 / + (248 χ 10
6)
/ 1 06\
(multiplying b y — I .
.·. /2 - 4 0 0 0 / - (248 χ 1 0
6) = 0 ,
4000 ± V[(16 x 106) + (4 χ 248 χ 10
6) ]
2
= 2000 + 15,880, taking the positive root,
= 17,880 lbf/in2.
B u t stress prior to jamming = 2000 lbf/in2.
.·. Increase due to jamming = 15,880 lbf/in2.
9
E X A M P L E . Two wire ropes each — in. dia. and weighing 16
0-85 lb/ft support a hoist weighing 0-5 ton, the ropes being 80 ft
long when the hoist is a t ground level. Es t ima te the number of
packing cases each weighing 50 lb which m a y safely be onloaded
assuming a permissible stress of 15,000 lbf/in2 and allowing for the
fact t ha t each case may be dropped from a height of 1 ft on to the
platform. T a k e Ε = 30 x 1 0 e lbf/ in 2 and neglect the instantaneous
extension due to impact .
Solution Γ / 9 \
2
R o p e section = 2 0-785 I—J
- 0-496 in2.
S I M P L E S T R E S S . S T R A I N A N D S T R A I N E N E R G Y 25
Weight of ropes = 2(0-85 χ 8 0 ) ,
= 68 χ 2 ,
= 136 lbf .
To ta l dead load
= 136 + (0-5 x 2 2 4 0 ) ,
= 1256 lbf .
Stress due to this
1256 8 0 ft
0-496 = 2525 lbf/in
2.
Energy absorbed in fall =
Wh = —— χ Volume, where lté
• 1-0 f t
/ = max . stress induced due to fall.
Wh χ 2E
FIG. 2 6
/ 2 = Vol
where Volume = 0-496(80 χ 12)
= 477 in3,
(50 χ 12) (2 χ 30 χ 106)
477
= 0-755 χ 108,
/ = 0-867 χ 104,
= 8670 lbf/in2.
To ta l stress due to initial dead load plus impact of one case
= 2525 + 8 6 7 0 ,
- 11,195 lbf/in2.
Stress available for remaining packing cases = 15,000 — 11 ,195 ,
= 3805 lbf/in2.
Load required to induce this = 3805 χ 0-496,
- 1885 lb f .
26 S T R E N G T H O F M A T E R I A L S
.*. Number of 50 lb cases = 1885
50 '
= 37-7 say 3 7 .
Hence, to ta l number of cases = 37 + 1,
= 3 8 .
E X A M P L E . The cylinder shown which
weighs 6500 lbf is released from a height
h above the spring, the stiffnes of which
is 3000 lbf/in. I f the maximum stress
induced in the rod is 9000 lbf/in2, find :
(a) the extension in the rod (x),
(b) the reduction in spring length (y),
(c) the value of h.
Assume no losses and tha t the rod
anchorage is immovable.
Solution
Extens ion in rod iL " Ε
9
9000 χ 100
30 χ ΙΟ6 '
= 0-03 in .
Strain energy of rod Ur JL 2E
100 in.
x Volume
9 0 0 02
- 2 in dia.
1— η —
1
• ι
FIG. 27
2(30 χ 106)
418 inch lb f .
92 100
I f W = gradually applied load to produce χ and y,
W
then, = ~2~ x>
W i.e. 418 = — x 0 0 3 ,
W
2
418 0 0 3
27,900 lbf .
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 27
Reduct ion in spring length y = Stiffness
27 ,900
3000 '
= 9-3 in .
Strain energy of spring Us
W
27 ,900 χ 9-3
= 129,500 inch lbf .
To ta l work done = Î 7 R + Z7 S,
.·. 6500(A + y + χ) = 418 + 129 ,500 ,
· · · ' ·« · » ^ · .·. A = 19-9 - 9 -33 ,
= 10-57 i n .
Clearly, in comparison with Us the strain energy of the rod is negligible.
Simple Load Shared by Two Materials—Compound Column
Consider the short strut composed of a steel rod and copper sleeve and let the dot ted line (Fig. 28) represent the loaded posi-t ion.
total load
nnnnnin //////////fin/m in π ninr/m//////// Steel . section As
v Copper section Ac
FIG. 2 8
W
2 8 S T R E N G T H O F M A T E R I A L S
I f Es and Ec are the values of Young 's Modulus for s teel and
copper, t hen : χ f f
Common strain = —— = - ~ == . L Es Ec Ε
Stress in copper / c = χ / s , where / s = Stress in steel.
To ta l load W = Ws + Wc,
Ε = fsAs + / s X y X i t .
Es and Et: are known while As and Ac can be calculated from given
dimensions. Hence, if W is known, / s can be found from the above
equation. / L
Then, reduction in length χ = .
Since, Load = Ε χ Strain χ Section,
Load on steel Ws = Es χ — χ As, Ε
and Load on copper Wc = W — Ws.
E X A M P L E . A short steel tube 4 in. inside dia. and 0 - 5 in. th ick is
loosely surrounded by a brass tube of the same length and thick-
ness. Together they carry an axial thrust of 0 - 5 tonf. I f the com-
mon length is 3 * 5 in., es t imate :
(a) The rat io of the compressive stresses in each,
(b) the stress in each,
(c) the load on each, (d) the common reduction in length.
Es = 3 0 x 1 0E, Eb = 1 1 - 8 x 1 0
E lbf/in
2.
Solution
Steel section As = 0 - 7 8 5 ( 5 2 - 4 2 ) = 7 - 0 7 in 2.
Brass section Ah = 0 - 7 8 5 ( 6 2 - 5 2 ) = 8 - 6 4 in 2. x f* = / b
L Es Eh
fh Eh 1 1 - 8
Common strain
= 0 - 3 9 3 .
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y
FIG. 2 9
i.e.
, \ Load on steel
Load on brass
1120
10-47 '
107 lbf/in2.
107 χ 0 -393 ,
42 lbf/in2.
Ws = fsAs = 107 χ 7-07 - 755 lbf.
Wh = fbAh = 42 χ 8-64 = 365 lbf.
Common reduction in length χ = -ψς— , E«
107 χ 3-5
30 χ ΙΟ6 0-0000125 in.
2a SM
Tota l load W = Ws + WH,
Λ 1120 =fsAs +fhAh,
= fsAs + 0-393 fsAh,
= / s (7-07 + 3 - 4 ) ,
29
30 S T R E N G T H O F M A T E R I A L S
Tempvraturo Stresses
A change in temperature produces a change in dimension, which,
if prevented sets up a stress. F o r a given material , linear expansion
is proportional to rise in temperature and to length.
Hence, if oc = change in length per degree per unit length,
change in length χ = ocTL,
where Τ temperature change, L = original length,
oc = expansion coefficient.
F o r a rise in temperature, χ is an increase, and if this is prevented
then this is equivalent to compressing a bar of length (L -|- x) back
to a length L, χ
i.e. Strain = — , JU
Stress / = Ε χ s t rain,
χ i.e. / = Ε χ — where χ = aTL.
Ε
E X A M P L E . A steel pipe is 6 0 ft long a t 2 0 ° C . E s t i m a t e :
(a) the elongation due to a rise in temperature to 9 0 ° C , (b) the stress induced by the prevention of this elongation.
T a k e oc = 0 - 0 0 0 0 1 2 / ° C . Ε = 3 0 χ 1 06 lbf/in
2.
Solution
Temperature rise
Elongation
Τ = 9 0 - 2 0 ,
= 7 0 ° C .
ocTL,
0 - 0 0 0 0 1 2 χ 7 0 χ 6 0 ,
0 - 0 5 0 4 f t ,
0 - 6 2 4 in.
Stress / = Ε χ S t ra in ,
oc TL Ε χ L '
- Ε x ocT,
= 3 0 χ 1 0E χ 0 0 0 0 0 1 2 χ 7 0 ,
= 2 5 , 0 0 0 lbf/in2. (approx.)
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 31
E X A M P L E . A copper cylinder 6 in. outside dia. has walls 1 in.
th ick. E n d plates are stayed by a 1 in dia. bar of steel which
passes through the cylinder and is jus t t ight a t 12°F . Es t ima te the
stress in cylinder and s tay when steam a t atmospheric pressure is
admitted. Assume the end plates do not distort.
occ = 0 -00001/°F, E, - 15-7 χ 1 06 lbf/in
2,
« s = O0000062 / °F , Es - 30 χ 1 06 lbf/in
2.
Solution
Steel section = 0-785 i n2
Copper section = 0-785 ( 62 - 4
2) ,
= 15-70 in2.
- - * | a cT L I - -
ν///////////////////λ {
1 1
1 J
1 J
__J ^ Common change
I in length
I 12 °F
212 °F
F i g . 3 0
Compression in copper = occTL — χ,
.'. Strain in copper
Extens ion in steel
.*. Strain in steel
- (0-00001 χ 200 L) - a- =
= 0-002 - 4 " · Jj = χ - otsTL,
= χ - (0-0000062 χ 200 L)
= χ - 0 0 0 1 2 4 L,
--=4- - 0 0 0 1 2 4 .
0-002 L - x.
32 S T R E N G T H O F M A T E R I A L S
Now, Load = Ε χ Strain χ Sect ion and the loads in steel and copper are equal.
.·. - 0 -00124J 30 χ ΙΟ6 χ 0-785 = 15-7 χ ΙΟ
6
0-002 - j " ) 15-7,
± - 0-00124) - 1 5
·? X
f X 1 5
'7
L J 30 Χ 10° X 0-785
Χ (0-002 - 4" Ε
= 10-5 0-002 - 4-Ε
4- - 0-00124 - 0-021 - 10-5 4" L Ε
11-5 4 · = 0 0 2 2 2 . L
χ
Τ 0 0 0 1 9 3
Stress in steel = 30 χ ΙΟ6 (0-00193 - 0-00124) = 20 ,800 lbf/in
2.
Stress in copper = 15-7 χ 1 06 (0-002 - 0-00193) = 1000 lbf/in
2.
E X A M P L E . The body of a small condenser is made from 8 in.
outside dia. brass tube. The end plates are also brass and are held
in place by four long | in. dia. steel bolts . I f these are t ightened
equally a t 5 2 ° F so tha t each carries 1000 lbf, what will they carry
a t the operating temperature of 212°F? The tube is \ in. th ick .
ocs = 0 -0000062/°F , Es = 30 χ 1 06 lbf/in
2,
ah = 0 -0000095 / °F , Eh= 9 χ 1 06 lbf / in
2.
Solution 9π
Steel Section - 4(0-785 χ 0-3752) = — in
2.
64 15JT
Brass Section = 0-785(8 2 - 7 2) = - j - i n 2 .
Temperature rise = 212 - 52 = 160°F .
Strain in steel es = = -= oc/1 Ε Ε
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 33
y 4 - -g- in dia. steel
Common extension
- a hT L -
7 in 8 in I
1
. . . . . .
X j
FIG. 3 1
Hence, 9 χ 1 06
(0-0000095 χ 160) 15π
L
9 χ 15
- (0-0000062 χ 160
00152 - — 1j
L
9π
64 '
30 χ 9 / χ
= 30 χ 1 06
n i . r - 0-000992) ()4 \ /v /
whence, — - 0-00146. Ε
Strain in brass e b - "b T L
* - ocbT - * L L
The loads on steel and brass are equal and Load = Ε χ Strain χ Section.
34 S T R E N G T H O F M A T E R I A L S
Stress in steel /s = Εs χ Strain in s teel ,
= 30 χ 1 06
= 30 χ 1 06 [0-00146
= 14,100 lbf / in2.
(0-0000062 + 160) ] ,
Load /Bo l t -Stress χ Section
• 4 . 1 0 0 x | x |
1000 ,
+ 1000 1560 + 1000
= 2 5 6 0 l b f .
E X A M P L E . The two side members of a water cooler are of 7 χ 2 in.
box section aluminium 0-25 in. th ick and 2 4 in. long. Between them
are 256 vert ical copper tubes 0-25 in. outside dia. 0-125 in. bore,
of the same length. Assuming tha t
(a) assembly was carried out at 20°C,
(b) the headers are rigid,
(c) the tubes do not buckle,
(d) the side members remain cold,
calculate, for an operating temperature of 90°C,
(a) the stress in the tubes,
(b) the increase in height of the cooler.
F o r copper ac = 17-1 χ 10"e/ °C and Ec = 15 χ 1 0
6 lbf/in
2.
F o r aluminium Ea = 12 χ 1 06 lbf/in
2.
Solution
Aluminium section Α.. = 2 (3 χ 2) 2
x¥
: 2(6 - 3-75) ,
4-5 in2.
Copper section Ar = 256[0-785(0-25 + 0-125) (0-25 - 0-125) ,
256 χ 0-785 χ 0-375 χ 0 1 2 5 ,
9-43 in2.
Temperature rise Τ = 90 — 2 0 ,
= 70°C.
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y
Compression in copper = occ TL — E.t
0 - 2 5 in -
0 · 2 5 ί η 0 1 2 5 in 1
Tubes-
Side member^
FIG. 3 2
occ TL
.*. Strain in copper = EiX oc'T
.*. Stress in copper / c = Ec χ Strain = Ec
and / { Α = / Λ ,
Ε.Λ
"•'-•it\
Extens ion in copper when free to expand = occTL.
Actual extension in copper = extension in aluminium =
35
36 S T R E N G T H O F M A T E R I A L S
i.e.
Hence,
and
f f A
< f 9
'4 3
2 1 /
.·. / c = 15 X 1 06
/ c
17-1
1 0e
15(1195 - 0 175 / c )
χ 70 -2 1 / c
12 x 1 06
15 + 0-175 / c = 1195, whence / c = 4950 lbf/in
2
/ a = 4950 χ 2-1,
= 10,400 lbf/in2.
fsL Extension in aluminium (i.e. in height of cooler) = ,
10400 χ 24
12 x 1 06, '
0-0208 in.
Examples
1. The force required to punch a hole 1-2 in. dia. through a steel plate 0-75 in. thick is found to be 51 tonf. Estimate the shear strength of the ma-terial (18 tonf/in
2).
2. The section of a steel strip is 3-0 χ 0-25 in. and that of a similar copper strip is 3-0 χ 0-125 in. If the two strips are superimposed and rivetted together, estimate the stress in each component when the combination resists a load of 6 tonf applied so that the two sections are in pure tension. Assume Es = 13,000 tonf/in
2 and Ec = 4200 tonf/in
2 (5-17 tonf and 0-83 tonf).
3. Compare the strain energy of a 12 in. length of rod 1-25 in. dia. with that obtainable by reducing 7 in. of the rod to 1-0 in. dia., assuming the maximum tensile stress to be the same in each case (1: 3-45).
4. Two alloy plates are rivetted together at 10°C using copper rivets. Neg-lect the compressive strain in the plates and estimate the stress increase in the rivets resulting from a rise in temperature to 21 °C. Assume Ec — 18 χ 10
6 lbf/in
2, <xc = 0-000005/°C, and a a = 0-000007/°C (1150 lbf/in
2).
5. As a result of a tensile test the following figures were obtained : Load (lbf χ 1000) 3 4 5 6 7 8 9 Extension (inch χ 10
4) 2-4 4-95 7-90 10-70 13-50 16-45 19-10
Load 10 11 12 13 14 15 16 17 18 19 Extension 21-95 25-00 28-00 31-00 34-28 37-10 40-60 44-90 54-00 87-00 Gauge length 2 in. Maximum load 21,000 lbf.
S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 37
Original dia. 0-5615 in. Load to fracture 14,000 lbf. Fracture dia. 0-375 in. Length after fracture 2-41 in. Plot the graph and find :
(a) Ε (29-5 χ 106 lbf/in
2),
(b) 0-1 per cent proof stress (74,600 lbf/in2),
(c) Percentage elongation (20-5 per cent).
C H A P T E R I I
BEAMS I—BENDING MOMENT
Moment of a Force Producing Bending
The moment of a force about a point is the product of the force
and the perpendicular distance of the point from i ts line of action.
Such a moment, when applied to a component (shaft, beam strut)
in such a way as to result in bending, is referred to as a Bending
Moment and denoted by M. The units are usually lbf/in or tonf/in,
i.e. product of force and length of arm.
Convention: To the left of any section, an anti-clockwise moment is considered
An example of this is a beam built into a wall as shown in
Fig . 33 . This is called a Cantilever.
negative and vice versa.
Beam Rigidly Supported at One End,
With Concentrated Load at the Other
w L
χ X
!x
Ο χ
-moxT x =-WL
M
A
3 8
F i g . 3 3
B E A M S I — B E N D I N G M O M E N T 39
F o r any section X X ,
Bending moment Mx = Moment of W,
= -Wx.
Thus M is a function of .τ, i.e., i t is proportional to x, the distance
from the origin O. I t s graph is therefore the straight line OA.
At the support, where χ = L,
Μιη:ικ = -WL.
The bending moment causes tensile stress (and therefore an
increase in length) in the upper layers of the beam, and compressive
stress in the lower layers. At some intermediate point there will be
a plane of zero stress and i t will be shown later tha t the stress
(tensile or compressive) is proportional to the distance from this
" n e u t r a l " plane.
Cantilever With Several Concentrated Loads
The value of M a t any section will be the sum of the moments
of all forces to the left of the section. Thus, taking an anticlockwise
moment as negative to the left of a section (Fig. 3 4 ) :
I f A = -W^L, ~L2)
MB = -W^L, - £8) - WS(L2 - £,)
Mc = -W.L, - W2L2 - W3L3
and Mc is clearly the maximum value.
The graph of M is thus a series of straight lines as shown.
E X A M P L E . F ind the bending moment a t each of the points A,
Β and C in the system shown in Fig. 35 and draw the graph of M.
Solution
Μ\ = - 0 - 5 ( 1 0 - 5 ) ,
= 2-5 tonf f t .
MB = - 0 - 5 ( 1 0 - 2) - 1(5 - 2 ) ,
= - 4 - 3 ,
= - 7 - 0 tonf f t .
Mc - - ( 0 - 5 χ 10) - (1 χ 5) - (2 χ 2)
^ - 1 4 - 0 tonf f t .
S T R E N G T H O F M A T E R I A L S 40
i
FIG. 34
14 F r o . 35
B E A M S I — B E N D I N G M O M E N T 41
Simple Span With Central Load
Clearly, load per support = W/2. The bending moment a t any
section is the ne t effect of all forces acting on t ha t par t of the beam
to the left of the section, clockwise direction being taken as
positive.
Thus for the L H half of the beam M is proportional to the
distance from the L H support and i ts graph is the straight line OP.
FIG. 3 6
At any section X X to the left of the load,
W L
Μ„ = —r-x where χ < — . x 2 2
42 S T R E N G T H O F M A T E R I A L S
A t the centre where L
w
w L
Μ = ΎΧΎ WL
and this is clearly the maximum value.
At any section Y Y to the right of the load :
- w r L]
Wy
2 Wy + — ,
WL Wy
2 2
W -y) where
L y > τ·
At the R H support when y = L, M — 0
Thus for the R H half of the beam M is proportional to the
distance from the R H support and i ts graph is the straight line
P Q .
Simply Supported Beam With Several Concentrated Loads
The reactions must first be found b y taking moments about point 0 (Fig. 3 7 ) :
Thus,
R2L = WXLX + W2L2 + WsLz
from which R« can be found.
Bt = (W, + W2+ W3) R2. Then
Then,
MA = R ^ .
Mc = R,L3 - WX(LZ - Lx) - W2(LS - Lz).
The graph of M against L will thus be a series of s traight lines as
shown.
B E A M S I — B E N D I N G M O M E N T
W, W,
43
L 3
M
FIG. 37
E X A M P L E . Calculate the reactions and the value of M a t points
A, B , C and D, Fig. 38 . Draw the graph of the lat ter .
Solution Moments about 0 give :
WR2 = (6 χ 2) + (8 X 4) + (10 χ 12 ) ,
= 12 -I- 32 + 120 .
164
16 = 10-25 tonf ,
R1 = (6 + 8 + 10) - 10-25,
- 24 - 10-25,
= 13-75 tonf .
44 S T R E N G T H O F M A T E R I A L S
6 8 10 tonf
(13-75) (10-25)
Ö j
FIG. 3 8
Beam With Loaded Ends Overhanging Supports
The diagram represents a wagon axle in which the springs rest
on the ends and carry W each.
I t follows tha t the wheels, which support the axle a t points
distant a from each end, also carry W each.
MA = (13-65 χ 2) = 27-50 tonf ft.
MB = (13-75 χ 4) - (6 χ 2 ) ,
= 55 - 12 ,
= 43 tonf f t .
M(] - (13-75 χ 12) - (6 χ 10) - (8 χ 8 ) ,
- 165 - 60 - 6 4 ,
= 41 tonf f t .
Hence,
B E A M S I — B E N D I N G M O M E N T 45
FIG. 3 9
FIG. 4 0
W
46 S T R E N G T H O F M A T E R I A L S
At any section X X ,
Mx = — Wx, where χ < a.
Hence M is proportional to x, i.e. i t is zero at the ends and has a
value a t the wheels of — Wa.
At any section Y Y , My = -Wy + W(y - a),
= -Wy + Wy - Wa,
= - Wa.
Since a is a constant, the bending moment between the wheels is
also constant and the graph of M is as shown (Fig. 39) :
I f the loads are unequal and/or if the supports are not symmetric-
ally spaced, i.e. c > a then the graph of M is modified as shown,
and the value between the supports is constant only if Wxa = W2c.
The supports themselves may thus share the load unequally and
the load on each must first be found by taking moments about one
of them.
General Case
F o r the system of loads and supports given, (Fig. 41) Moments
about A give
Εφ -ι- Wxa - W2(d - a) -|- Ws{d - a + c)
w, w2 w3 J
•4 d
A c
Β
b * r*— 0 *
R, R 2
FIG. 41
from which R2 and hence Rx can be calculated.
B E A M S I — B E N D I N G M O M E N T 47
Then,
MA = - W > .
MB = -Wxd + B1(d - a).
Mc = - Wx(a + b) + Rxb - W2(e - c)
( = — W3c al ternatively).
FIG. 4 2
Thus, as shown in F ig . 42 , the introduction of W2 has reduced
the negative bending moment between the supports. I f W2 is
made sufficiently large, the bending moment between the supports
may become positive as shown in F ig . 4 3 .
M
01
FIG. 4 3
I n general, when there is no change in load between two sections
of a beam, the bending moment changes uniformly.
48 S T R E N G T H O F M A T E R I A L S
E X A M P L E . Determine the bending moment a t points A , B , C
and D in the system shown and draw the bending moment graph
to scale.
2 tonf
-10 f t -
A
H O f t - - 1 0 f t -
D
-5fH
FIG. 44
Solution
Moments about A give
( i ? 2 x 20) + (2 χ 5) = (4 χ 5) + (6 χ 15) + (2 χ 2 5 ) .
20R2 + 10 = 2 0 + 90 + 5 0 .
20 + 90 + 50 - 10 Ra =
20 7-5 tonf .
.'. R1 = (2 + 4 + 6 - j - 2) — 7-5,
= 14 - 7-5,
- 6.5 tonf .
B E A M S I — B E N D I N G M O M E N T 49
Hence, Mx = - ( 2 χ 5 ) ,
- - 10 tonf f t .
MB = - ( 2 χ 10) + (6-5 χ 5 ) ,
= - 2 0 + 32 -5 ,
= + 12-5 tonf f t .
Mc = - ( 2 χ 20) + (6-5 χ 15) - (4 χ 10 ) ,
= - 4 0 + 97-5 - 4 0 ,
= + 17-5 tonf f t .
ΜΏ = - ( 2 χ 25) + (6-5 χ 20)
- ( 4 χ 15) - (6 χ 5 ) ,
= - 5 0 + 130 - 60 - 3 0 , = - 1 0 tonf ft [ = al ternatively ( - ( 2 x 5 ) ] .
E X A M P L E . Determine the bending moment a t points A, B , C
and D and draw to scale the graph of bending moment.
4 2 5
8 f t - 9 f t
A Β C D
2 0 f t
R. R 2
M
0^
- 2 0
FIG. 45
50 S T R E N G T H O F M A T E R I A L S
Cantilever With Uniform Load
The only force acting to the left of the section X X (Fig. 46 ) is
the weight of the portion x, (i.e. wx) and this acts a t a distance
x\2.
Solution
Moments about A give
(R2 x 20) + (4 χ 5) = (2 χ 3) + (5 χ 12) + (3 χ 1 8 ) ,
20R2 + 2 0 - 6 + 60 + 5 4 ,
_ 6 + 60 -I- 54 - 20 / V
20 '
= 5 tonf.
. \ Ä 1 = ( 4 + 2 + 5 + 3 ) - 5 ,
= 9 t o n f .
Hence, MA = - ( 4 x 5 ) ,
- - 2 0 tonf f t .
Mn = - ( 4 χ 8) + (9 χ 3 ) ,
= - 3 2 + 2 7 ,
= - 5 tonf f t .
Mc - - ( 4 x 17) + (9 χ 12) - (2 χ 9 ) ,
= - 6 8 + 108 - 1 8 ,
= + 22 tonf f t .
MD = - ( 4 χ 23) + (9 χ 18) - (2 χ 15) - (5 χ 6 ) ,
= 92 + 162 - 30 - 3 0 ,
- + 1 0 tonf f t .
B E A M S I — B E N D I N G M O M E N T 51
Hence, i f x = — wx taking an anticlockwise moment as negative.
W 2
The graph of M against χ is thus a parabola as shown. At the
support when χ = L,
w i i i n ; ix = — and wL = W (the to ta l load)
W L
: . J f . „ : , V = —
I
Για. 46
Simply Supported Beam With Uniform Load
Tota l load = wL,
.'. Load/Support =
52 S T R E N G T H O F M A T E R I A L S
At any (Fig. 4 7 ) ,
section X X
wL X X
WX X —-
wL
w/unit length
The graph of M is thus
a parabola. The value of
χ which makes M a max-
imum can be found b y JL differentiating the above
and equating to zero : FIG. 4 7
dx = — wx = 0 for a max. ,
i.e. wx =
χ =
wL
"2
wL
ΊΓ _L
Τ
Hence, Mn
wL L
wL2 wL
2
w (L\2
4
wL2
8
and wL = W (the to ta l load)
FIG. 4 8
B E A M S I — B E N D I N G M O M E N T 53
w/unit length
Fia. 49
and, from Fig . 51 :
Mb= —w[a 2
b δ2
w ί 0 Λ δ b2\ w Ί w Ίη
= -T(a2 + 2aJ+Tj + Y a b + T b ' w „ w . w w , w ,„
= - ^ r «2 --z-ab --ζ-V + s-eb + — δ
2,
wa
FIG. 50
Thus the maximum value of M between the supports is less w
(numerically) than tha t a t a support b y an amount — b2 and the graph of M is as shown in F ig . 52 . 3 SM
Uniformly Loaded Beam With Simple Supports Not at the Ends
Suppose the beam to be supported a t points A and C (Fig. 49)
such tha t a > 0·5δ and let point Β denote the centre. Then, from
Fig . 5 0 :
M a = — wa ^- = Mc, Δι
54 S T R E N G T H O F M A T E R I A L S
U—α =0-5 tM
FIG. 5 3
Clearly, if a < 0-56, Mh becomes positive. B e s t use is made of
the beam when
i.e. when
or when
w w w
Yai + Tb = Ô α ' W 1,2 2
— b2 = wa
2,
8
or when
The graph of M is then as shown in F ig . 54 .
J FIG. 5 2
w w F o r Mh to be zero : 0 = - — a
2 + — b
2,
Ζ o
i.e. a2 = ——
4
or a = 0·5δ ^ =
The graph of M is then as shown in Fig . 53 .
B E A M S 1 — B E N D I N G M O M E N T 55
I f the supports Rx and R2 are not symmetrically spaced, i.e. if
c > a (Fig. 55) then Rx Φ R2 and the graph of M is as shown in
Fig. 56 .
M
- V — F
* I 0 Mo
a*0-35b 1* b J
FIG. 54
Ρ Q
Deflected shape of beam (exaggerated )
FIG. 57
3 *
FIG. 56
FIG. 55
56 S T R E N G T H O F M A T E R I A L S
Contraflexure
At points Ρ and Q (Figs. 55 and 57) the bending moment is zero
so tha t a t these points the beam is straight. Also, since M changes
sign a t these points, the curvature also changes sign, i.e. the profile
changes from convex to concave. F o r this reason points such as Ρ
and Q are called Points of Contraflexure.
E X A M P L E . A beam is simply supported and loaded as shown in
Fig . 58 .
F ind the value of W required to make Rx and R2 equal. Calculate,
for this value of W the bending moment a t points A and Β and a t
two points 1-825 ft and 7-50 ft respectively to the right of point A.
Sketch the graph of M using these values.
Solution
Moments about point A give :
(W X 4) + (R2 χ 12) = (6 χ 11) + (12 χ 1 ) - ^ ,
12R2 = 66 + 72 - 4 I F , fu f ρ
138 - m so tha t iio = — . 2 12 Moments about C give :
(W χ 16) + ( 6 x l ) | (12 χ l ) ^ - = R1 χ 1 2 ,
\mx = 6 + 72 + \m, 78 + 161Γ
K l ~ Î 2 * 78 + 16Tf 138 - 4W
Since RX=R2, — = — ,
20 W = 60, i.e. W = 3 tonf.
FIG. 58
B E A M S I — B E N D I N G M O M E N T 57
Hence, ιΔ
M A — . (3 χ 4 ) = - 1 2 tonf ft.
11 MB = - ( 3 χ 15) + (10-5 χ 11) - (11 χ 1 ) — = + 1 0 tonf ft.
Δ At a point 1-825 ft to the right of point A,
M = - 3(4 + 1-825) + (10-5 χ 1-825) - (1-825 χ 1)
= - 1 7 - 4 7 5 + 19-2 - 1-68,
= 0 (nearly).
At a point 7-50 ft to the right of point A :
7-5 M = - 3 (4 + 7-5) + (10-5 χ 7-5) - (7-5 χ 1) — ,
Δ
= - 3 4 - 5 + 79 - 2 8 ,
= + 1 6 - 5 tonf ft,
1-825
FIG. 59
The graph of M is then as shown.
Examples I I
1. Find the maximum bending moment in a timber joist which is simply supported over a span of 20 ft and loaded with 1 ton at the centre and 2 tons at a point 4 ft from one end. Neglect the weight of the joist itself (9-0 tonf ft).
58 S T R E N G T H O F M A T E R I A L S
2. An R.S.J, weighing 133 lb/ft rests on supports 40 ft apart. Calculate the maximum bending moment due to its own weight. (26,400 lbf ft).
3. A beam 22 ft long is simply supported at the LH end and at a point 2 ft from the RH end. At a point 4 ft from the LH end there is a concentrated load of 5 tonf and from this point to the RH end there is a distributed load of 1-0 tonf ft. Draw to scale the bending moment graph and find from it the position and magnitude of the greatest bending moment (55-2 tonf ft at 9-3 ft from the LH end).
4. ABODE is a beam supported at Β and D 10 ft apart. At points A and C and Ε are concentrated loads of 1-5, 2-0 and 1-0 tonf respectively while from Β to D there is a uniform load of 0-5 tonf/ft. The remaining dimensions are : AB = 2 ft, BC = 5 ft, DE = 3 ft. Determine graphically the value of the maximum bending moment (8*25 tonf ft).
5. A joist is simply supported over a span of 40 ft and carries a wall weighing 1-75 tonf/ft from the LH end to the midpoint. In addition there are concentrated loads of 5 tonf at points 6 ft from each support. Calculate the load on each support, draw the graph of bending moment to scale and from it find the maximum value of M and its position (31-25 and 13-75 tonf. 228 tonf ft at 15 ft from the LH end).
C H A P T E R I I I
1st AND 2nd MOMENTS
Centroid
This is the point in a body a t which i ts mass may be assumed
concentrated. I n the case of a lamina i t is the point a t which the
area may be assumed concentrated.
1st Moment of Area
F o r an element of area αλ (F ig . 60) distant xx from an axis Y Y
this is defined as the product axxx.
Total area Α «Σα
FIG. 6 0
The 1st Moment of the whole figure will be the sum of the 1st
Moments of all such elements, v iz : axxx + #2^2 + W + a n
d so
on and this is written Σαχ.
59
60 S T R E N G T H O F M A T E R I A L S
Datum
I*—2-0 in—H I
3Oin
Iy
FIG. 6 1
Solution
From symmetry the centroid will be on the axis Y Y . The 1st
Moment of the shaded area is the difference between tha t of the
to ta l area and tha t of the hole.
Thus, 1st Moment of area about Υ Υ Σα χ
= α
ιχι +
α2χ2 + a
sxz
+ and so on.
I f the to ta l area A is assumed concentrated a t the centroid G a t
χ from Y Y , then the to ta l 1st Moment will be the product Ax.
Hence, Ax = Σαχ,
_ . J „ „ , T r Tr J ^ a x 1st Moment of area i.e. Distance of G from 7 7 £ = — - — = .
A Area
Note tha t if the axis Y Y passes through G, then
X = 0 ,
i.e. Ax = 0 ,
. · . 2 > * = o.
Hence the 1st Moment of any figure about an axis through i ts
centroid is zero.
E X A M P L E . F ind the position of the centroid of the box-section
shown.
1 s t A N D 2 n d M O M E N T S 61
10
V
FIG. 6 2
The two rectangular areas comprising the section are (2 χ 10) and
(8 χ 2) with centroids a t 5 and 11 in respectively from Y Y .
3 a SM
Original area as = 3 x 4 -5 ,
= 13-5 in2 a t G s = 2-25 in. from X X .
Area of hole ah = 2-0 χ 2 -0 ,
= 4-0 in2 a t G h - 2-50 in. from X X .
Shaded area A = 13-5 — 4-0 ,
= 9-5 in2 a t £ in. from X X .
.·. 9-5 χ χ = (13-5 χ 2-25) - (4 χ 2 -5 ) .
30-4 - 1 0 0
* = 9-5 -
= 2-145 in from the given datum, i.e. X X .
Example. F ind the position of the centroid for the given Τ
section.
Solution
From symmetry the centroid will be on the axis X X .
62 S T R E N G T H O F M A T E R I A L S
(2 χ 1 0 ) 5 + (8 χ 2 ) 1 1
(2 χ 10) + (8 + 2)
100 + 176
36
= 7-68 in. from Y Y .
Area under a Graph —1st Moment
The area enclosed between the graph of y = f(x) and the
x-axis can be found by integration as follows.
FIG. 6 3
Area of element = y dx b
Tota l area A = j y dx n.
Hence,
1st Moment of element about Y Y = y dx χ χ by definition,
= yx dx.
b .'. To ta l 1st Moment about Y Y = J yx dx and this is equal to
a
Ax where A is the to ta l area and X the distance of the centroid from Y Y .
1 s t A N D 2 n d M O M E N T S 63
Hence.
J yxdx a
b
J y dx
1st Moment about Y Y
Area
y Again, 1st Moment of element about X X = y dx ~ - (since the
centroid of the element is a t —from X X ) , Δ
y2 * = — A r .
Γ y2
.'. To ta l 1st Moment about X X =^= j ~- dx and this is equal to a
Ay where y is the distance of the centroid G from X X .
Hence, y = • b
j ydx
1st Moment about X X
Area
The above are general expressions and may be applied to particular cases.
Quadrant of a Circle
Since y2 r
2 - x
2, the
equation of the graph is
y = (r2 — x
2)
11" where r is a
constant.
F rom Fig . 64,
Area of element = y dx.
Fi rs t Moment about X X
A y
dx
3 η*
FIG. 6 4
6 4 S T R E N G T H O F M A T E R I A L S
.'. To ta l 1st Moment of quadrant = -^Jy2 dx,
r
= — J (R2 — x2) dx,
1 ~2
r2x — X3
Ύ 1 1 ^3 r
3\
~2 Τ —
Y Area of quadrant A = — π r
2
and Ay = 1st Moment about X X ,
3 nr2
4r
~3π from X X .
The centroid of the quadrant is, from symmetry, also a t a distance
of 4r/37r from Y Y .
Semi Circle
This is composed of two quadrants, the centroid s G,7 of which
lie symmetrically on either side of Y Y a t a distance 4r/3jr from
both axes. Clearly the centroid Gs of the semicircle lies on Y Y a t a
distance of 4r/37r from X X .
Y
•Β' - Ι
\ 4r
\ 3TT
1 I 0
Y Λ 4 r .
3ττ _
FIG. 65
1 s t A N D 2 n d M O M E N T S 65
W To ta l mass
g
As before, the 1st Moment of any solid about an axis through G is zero.
Cone : 1st Moment and Position of Centroid
F o r the slice of thickness dx (Fig. 66) : χ
Volume = ny2 dx where, from similar triangles, y = — r,
Tir2 Tir
2
= —rr- χ2 dx and W = — - r - ρ x
2 dx.
h2 h
2
1st Moment about YY = i^f x
2 dx\ χ x = x
zdx,
\ gw I Λ Tota l 1st Moment
= f x3 dx
2r> \ nr
2Q
W gh
2 h
nr*q x*
gh2
nr2q h*
gh2 ' 4 '
nr2h
2q
1st Moment of Mass
F o r an element of mass wjg distant χ from an axis Y Y this is w
defined as the product — x. The 1st Moment of the whole body ^ w
will be the sum of the 1st Moments of all such elements, i.e. Σ — W î w\
g
I f the to ta l mass — = Σ — is assumed concentrated a t the 9 \ gl centroid G a t a; from Y Y , then the to ta l 1st Moment will also be
W _ the product — x.
Wx _ wx Hence, = λ ,
g g
v . W
\ ' ft q 1st Moment of mass
i.e. χ =
2nd Moment of Area
F o r an element of area ax (Fig. 67) distant x1 from an axis Y Y
this is defined as the product axx\.
The 2nd Moment of the whole figure will be the sum of the 2nd
Moments of all such elements, viz. : αλχ\ + a2x\ + and so on
and this is written Σαχ2- I t is denoted b y / . Thus, 2nd Moment of area about Y Y
Iu = Σαχ2 = aiXl + a2
X2 + a3Xl + an(^ SO 0n*
I f Κ is a point in the figure distant ky from Y Y such tha t
Aky = I}J where A = to ta l area,
then **-V-T-
66 S T R E N G T H OF M A T E R I A L S
B u t To ta l 1st Moment =
From similar A's.,(t)r
Equating,
Alternatively, h from base of cone.
1 s t A N D 2 n d M O M E N T S 67
Total area A
FIG. 6 7
Parallel Axes Theorem
F o r the irregular plane area of F ig . 68,
Iy = Σαχ2 ° y definition,
= ΣίΦ + r)«],
= ΣίΦ2 + + r2)\,
= Σ{βχ2 + 2xar + a r
2) ,
= ^ ( α ά1) + Σ(^) + Σ(*?2),
= χ2Σα + 2 x i ; ( a r ) + 2/(« r 2) (taking out the
constants) .
B u t the 3rd term Σ(ατ2) = 2nd Moment of area about an
axis through G parallel to Y Y ,
= h-And in the 2nd term Σ(
αν)
= 1 ^ Moment of Area about an
axis through G parallel to Y Y ,
= 0 . This term is therefore zero.
The distance ky is called the Radius of Gyration about Y Y .
Similarly, kx = .
68 S T R E N G T H O F M A T E R I A L S
Hence,
Similarly,
ig + Ax2.
l x = Ig+ Ay*
Total area Α=Σα
FIG. 6 8
Thus, if the the value of / about an axis through the centroid G
is known, the value of / about any other parallel axis Y Y can be
found by adding the product Ax2 where χ is the distance between
the two axes .
Rectangle—2nd Moment about an Edge
In Fig . 69, area of element = b ay.
2nd Moment about X X = b ay x y2,
= by2 ay.
b
I
1 dy
Τ
FIG. 6 9
And in the 1st term Σα =
A (the to ta l area).
1 s t A N D 2 n d M O M E N T S
d 69
Tota l 2nd Moment Ix = j by2 dy,
= b
• (in4 if b and d are in inches.)
B u t Ah\ = Ix, i.e. h\ = - j - where ^4 =
ta3 J _
" 3 X
bd 9
d2
, from which kx can be found.
Rectangle—2nd Moment about an Axis Through the Centroid
I n Fig . 70, area of element = b dy
2nd Moment about X X = b dy x y2,
= by2dy.
.'. 2nd Moment of shaded part above X X
= / by2 dy.
0
F I G . 7 0
70 S T R E N G T H O F M A T E R I A L S
The total 2nd Moment will be twice this,
d/2
i.e.
B u t Alq. = Ix, i.e.
Ix = 2fby*dy9
- 2b
- — 3
d_
d/2
0
bd*
H T *
" A '
6d3
where A
1
6d,
12 X bd
= —— , from which &r can be found. 12
Note that , from the Parallel axes theorem, considering the axis AA,
/ a A (since ti/2 is the distance between AA and X X ) ,
bdä
12
bd*
bd d*
(as found above).
FIG. 71
1 s t A N D 2 n d M O M E N T S 7 1
E X A M P L E . Find, for the section shown (Fig. 7 1 ) , the values of
IxIyIa and Iz in in4. F ind also the corresponding values of k.
Solution
Area of section A = bd,
= 1 6 χ 6 ,
= 9 6 in2.
_ 16 χ 63
~ 12 '
- 288 in4,
i.e. k, 1-73 in.
_ 6 χ 1 63
~ Ï 2
= 2047 in*.
• / :
2 2 0 47 21 3
i.e. Α.. = 4-62 in.
= Ix + ^ 4 ( 3 )2 (from the parallel axes theorem),
= 2 8 8 -f- ( 9 6 χ 9 ) ,
- 2 8 8 + 8 6 4 ,
= 1 1 5 2 in4,
1 1 52 12 ,
" a
96
i.e. ku = 3-47 in.
72 S T R E N G T H O F M A T E R I A L S
Alternatively,
_6*_
3 '
= 12 (as before).
Iz = Ale2 = IX + A{5)
2 (from the parallel
axes theorem),
= 288 + (96 χ 2 5 ) ,
= 288 + 2 4 0 0 ,
= 2688 i n4,
2688 Β =
Je,
96 '
2 8 ,
5-28 in .
E X A M P L E . Show tha t the 2nd Moment of area of a rectangle of
sides b and d about a side of length b is bd2/3. F ind the 2nd Moment
of area of the section shown about
(1) the axis A B
(2) an axis through the
centroid parallel to A B
Solution
Area of element
= bay.
2nd Moment about A B
= b dy χ y*,
= by2 dy.
*. ^ A B = / by2 dy,
- b
ό
1 s t A N D 2 n d M O M E N T S 73
F o r the section given.
• 8 χ 63
9 Γ 3 χ 4
31
3 — Δι ί 3 J
y =
= 576 - 1 2 8 ,
= 448 in4.
(8 χ 2) 5 + (2 χ 4) 2
(8 χ 2) + (2 χ 4 )
80 + 16
~ 16 + 8 '
_ 9β
" 2 4 '
= 4 inch, i.e. G lies in the edge of the flange
8 χ 23 \ / 2 χ 4
3
FIG. 73
/ 8 χ 23 \ / 2 χ 4
3 \
_ j>4 128
" Τ " +
" 3 ~ '
192
3 9
= 64 in4.
Alternatively, by the parallel axes theorem,
I G = / A B - (24 x 42),
= 448 - 3 8 4 ,
= 64 in4.
Area Under a Graph—2nd Moment about Y Y
The values of I and k for the area enclosed between the graph of
y = f(x) and the #-axis can be found as follows :
74 S T R E N G T H O F M A T E R I A L S
y = f U )
Shaded area A=J ydx
FIG. 74
2nd Moment of element about Y Y = y dx χ x2 by definition,
= yx2 dx.
b
/ . Tota l 2nd Moment about Y Y , Iy = f yx2 dx
and Akff = Iy, where A j y dx a
b jyx
2 dx
h y
J y dx
from which ky can be found.
Arm under a Graph —2nd Moment about XX
In Fig . 74, the whole of the element is a t the same distance χ
from Y Y and the 2nd Moment is simply y dx χ x2. Since this is
not true for the axis X X the 2nd Moment of the strip about X X must first be found.
Considering an element of length dh distant h from X X and forming part of a strip of length y and uniform width dx (Fig. 75) :
Area = dh dx.
2nd Moment about X X = dh dx x h2,
= dxh2dh.
1 s t A N D 2 n d M O M E N T S
y
ο = dx
da;
And -4 A^ftrip — - ^ s t r i p *
dh :
^ s t r i p ^ ^i p
where A = y dx,
dx_ 3 1
3 V X
ydx9
x
- - | - about X X . dx
FIG. 75
75
The area enclosed between the graph of y = f(x) and the a:-axis
(Fig. 74) can be divided into a series of elements of length y for
each of which the radius of gyration is y2/3 (from above) .
Hence, for each strip, 2nd Moment about X X = Area χ k2,
y dx χ 3 '
y dx.
Tota l 2nd Moment
And
-dx.
Ak2 = Ix where A — j ydx,
f i d *
from which kr can be found.
j ydx
Tota l 2nd Moment
Λίπρ = / dxh2dh and da; is constant ,
h3
76 S T R E N G T H O F M A T E R I A L S
Common Sections
F o r the channel and I sections shown
jTN A = — (2nd Moment about N.A. of shaded area.)
N.A.
FIG. 76
FIG. 77
E X A M P L E . F ind for the section shown, the values of
(a) Ix,
( b ) / „ , (C) J « , (d)L.
-7 in-|Y
V///////////Ä
I in -
- 5 in-
6 in
V7777/)Ç7777A
4ziin
lOin 12in
Solution
(a) / 7 χ 1 2
3 \ / 6 χ 1 0
3 \
I 12 ) " \ 12 )9
12
= 1008 - 5 0 0 ,
= 508 in4.
1 s t A N D 2 n d M O M E N T S 77
Λ / 1 x 73\ / Ι Ο χ 1 » \
( b ) J
- = 2
( - Ï 2 ~ )+
( - ^ 2 — )
= 58 in4.
(c) F rom the parallel axes theorem : Ia = Ix + (Area) 62,
= 508 + [ ( 7 x 1 ) 2 +
+ (10 χ 1 ) ] 3 6 ,
= 508 + 864,
- 1372 in4.
(d) Similarly, Iz = Iy + (Area) 52,
= 58 + (24 χ 25 ) ,
= 58 + 600,
= 658 in4.
Perpendicular Axes Theorem
F o r the element of area a (Fig. 78) forming part of the shaded area in plane Y O X , by definition, 2nd Moment about X X = ay
2
and, for the whole figure
h = Zw2-Similarly, Iy = Σαχ2>
F I G . 7 8
78 S T R E N G T H O F M A T E R I A L S
and I. = 2Jaz2> where ZZ is normal to the
plane Y O X ,
= ΣΜν2 + χ2)λ (Pythagoras)
= Σ(αν2 + αχ2),
- ι τ + //7.
iVote: For a thin lamina, the above relation is approximately true for the 2nd Moments of mass about the three mutually perpendicular axes. The relation does not hold good for other three-dimensional solids, for which Iz must be found separately (see under 2nd Moment of mass p. 80 et seq.
Circular Section—2nd Moment of Area
F o r the circular element of radius r and thickness dr (Fig. 79) ,
Area = Length χ Thickness,
= 2nr χ dr.
FIG. 79
Since every part of the element is a t the same distance r from the
axis ZZ, b y definition :
2nd Moment of area about ZZ = 2nr dr χ r2,
= 2nr'6 dr.
1 s t A N D 2 n d M O M E N T S 79
2nd Moment of whole Iz = j 2nrz dr,
ο - 2π
ο R
I υ
i.e. / , = £ = D4 (where Z>= 2 Ä ) .
» Δ ΟΔ
This is called the Polar 2nd Moment and denoted (usually) by J.
nD* ... I f D is in inches then J =
32
Since J z = ^4&2, where kz = radius of gyration about ZZ,
/m k\ = -γ-, where ^4 = π ^2,
nRi 1
2 π Α2 '
R2 D
2
" or
τ -
From the perpendicular axes theorem Iz = Iv - f Iy and clearly
r L Hence, l x = γ ,
64
iü2 D
2
I t follows tha t Η = — - or — -4 16
PZcme Surface—Graphical Method for Determination of Position of
Centroid and value of Ig
F o r the irregular plane area A in F ig . 80 , let G be distant y
from any axis X X .
Construction
Draw any axis P P parallel to X X and distant d from it . Draw, on the figure, any line B C parallel to X X . Pro jec t B D and C E . J o i n D and Ε to any point F in X X . Mark the intersections Bx and C 2.
R
80 S T R E N G T H O F M A T E R I A L S
Projec t B1D1 and CJ£V J o i n D x and Ej^ to point Γ . Mark the
intersections B 2 and C 2. Repea t for a series of parallel lines such
as B C using the same point F . Then the area A x enclosed b y a
smooth curve connecting points such as B 2 and C 2 is known as the
1st Derived Area and can be measured with a planimeter.
χ F I G . 8 0
The area A 2 enclosed by a similar smooth curve connecting
points such as B 2 and C 2 is known as the 2nd Derived Area and can
be similarly measured.
Then, 1st Moment of area about X X , Ay = Axd,
A,d
(χ may be found similarly).
And 2nd Moment of area about X X , Ix = A2d2.
From the parallel axes theorem, Ig = Ix — Ay2.
2nd Moment of Mass
F o r an element of mass wjg distant xt from an axis Y Y this is defined as the product (wjg) x\. The 2nd Moment of the whole body will be the sum of the 2nd Moments of all such elements
1 s t A N D 2 n d M O M E N T S 81
viz (wjg) x\ + (w2/g) x\ + (wjg) %l and so on and this is written
Yj (wig) x2. Although, like the 2nd Moment of area, i t is denoted
by I7J i t is usually called the Moment of Iner t ia . (See also Chapt.
I , Mechanics of Machines by the same author.)
I f Κ is a point in the body distant ky from Y Y such tha t W
— k2 = Iy, where W = to ta l weight,
then,
Similarly,
= Radius of gyration about Y Y .
= Radius of gyration about X X .
F o r the circular element of radius r and thickness dx (Fig. 81) ,
Weight = Volume χ Densi ty,
= nr2 dx χ ρ.
Mass = —dx.
Sphere—2nd Moment of Mass about a Diameter
FIG. 8 1
82 S T R E N G T H O F M A T E R I A L S
dx χ 2 '
πρ
πρ
πρ
R2
r4 dx and r
2
(R2 - x
2)
2dx,
(R* - 2R2x
2 + x*) dx.
Since χ varies from zero to R, integration of the above between these
limits will give the 2nd Moment of the R H hemisphere.
Hence, 2nd Moment of whole sphere
η
I_ 2 JR %
^ - ( / T> 4 - 2Rh
2 + A
4) dx,
ο
πρ
(J R*x - 2R
2-
9 \ 3 5 / '
I-
πρ b
_8πρΡ^_
Rb,
W Since / . = — k
2, k
2 I. 4
—f—, where W = —nR3p,
w/g 3
8 π ρ #5
3ί7 log 4:πR
3ρ'
2 W (2R2\
= 4 *a
» Hence 1. 5
Cone—Polar 2nd Moment of Mass
F o r the circular element of radius r and thickness dx (Fig. 82) ,
Weight = Volume χ Density,
= πτ2 dxρ.
Polar 2nd Moment = Mass χ k2,
που* Γ Λ 2nd Moment of whole 7~ (or J) = I x* dx
* v ' 2gh* J
h
0
nqR* χ h
2gh* J 0
9
2gh* X
5 '
~10g~'
1 s t AND 2 n d M O M E N T S 83
Mass = — ax. 9
Polar 2nd Moment = Mass χ k2,
nr2g r
2
9 2
7t0 X = - r ^ - r
4 do: and from similar triangles r = —R,
2g h
πρ ΙχΒγ Ί
= - f — α ;4 dx.
2gh*
FIG. 8 2
84 S T R E N G T H O F M A T E R I A L S
W/g ' 3
πΚ4&ρ 3gr
3iü2
I c T '
Hence, I 5 = — ^ — J .
Cylinder—Polar 2nd Moment of Mass
F o r the circular element of radius r, radial thickness dr and length
£ (Fig. 83) ,
FIG. 8 3
Weight = Volume χ Densi ty,
= (2nr dr χ L) ρ.
2πΖ/ρ Mass χ rd r slugs.
Since every part of the element is a t the same distance r from the
axis ZZ, b y definition : 2nLq
2nd Moment of mass about ZZ 9
2πΣρ
r dr χ r2,
rz dr.
W
Since Iz
= k2
z
, where kz
= polar radius of gyration,
. - . i P - i - , where W -
1 s t A N D 2 n d M O M E N T S 85
.'. 2nd Moment of whole ΙΎ = %πΣρ Γ 3 ^
ο 2nLQ | r
4|
ß
2ττΧρ i ?4
x —-
2g
w Since = ~ ^ z > where &z = radius of gyration about ZZ,
Λ kl = -^η-, where W = π #2£ ρ (total weight),
nR^Lq g
2g nR2Lq
9
_ R 2
~ ΊΓ' W R
2
Thus, for a disk, / = — k2, where k
2 = ——. Note tha t if the
9 2 Si + R\
cylinder is hollow (e.g. flywheel rim) then k2 — , where
R2 and R1 are the inner and outer radii respectively.
4 SM
R
h-FIG. 8 4
86 S T R E N G T H O F M A T E R I A L S
E X A M P L E . Four holes each 4 in. dia. are bored in a steel disk
20 in. dia. 4 in. thick. Their polar axes are 5 in. from and parallel
to tha t of the disk. Take the density of steel as 0-283 lb / in3 and
calculate (in slug f t2) the polar moment of the bored disk, Fig . 84 .
Sohlt ion
Init ial weight of disk - (0-785 χ 2 02 χ 4) 0-283,
- 356 lbf.
k d ~ 2 I 12
356 1 / 1 0 \2
.*. Polar moment before drilling Id = ^ ^ · — I — 1 ,
= 3-84 slug f t2.
Weight removed/hole = (0-785 χ 42 χ 4 ) 0 - 2 8 3 ,
- 14-2 lbf.
« - I i
Polar moment/hole Ih = 14-2 1
32-2 2 12
= 0-0061 slug f t2.
Polar moment of hole about disk axis Ihl = Ih
14-2 / 5 \2
Required
32-2 \ 1 2 /
(from parallel axis theorem),
- 0-0061 Η 0-0764,
- 0-0825 slug f t2.
J = 3-84 - (4 χ 0-0825) = 3-52 slug f t2.
E X A M P L E . Show from first principles that , for a uniform cylinder
of weight W the polar moment of inert ia is given by / = W/Sg
χ (D\ + Dl), where Ώχ and D2 are the outside and inside dia-
meters respectively. Hence, determine the value of / for a single
cylinder diesel engine flywheel consisting of a disk 4-625 in. thick,
20-5 in. outside dia. and 2-625 in. inside dia. Take ρ == 0-28 lb/ in3.
Es t imate the K . E . stored a t 12000 rev/min.
Solution
Mass of elemental ring =
1 s t A N D 2 n d M O M E N T S 87
, where ρ = density.
F I G . 8 5
2jrr dr to 2nd Moment about ZZ = χ r
2 2 π ί ρ
χ r3d r .
2 π ί ρ In
/r
3 dr ,
Jit
2ntQ I R \ - R\
^.(Rl-RlURf+Bl)
and n(R\ -R\)tQ=W,
W D — {D\ + Dl), substituting — for R. Sg 2
I n the given case, W = 0-785 (20-52 - 2 -625
2) 4-625 χ 0-28,
= 4 2 0 lbf.
/ = ^ 4 2
? ^ ^ χ 4 f (2° -
2 52 +
2*
6 2 5 2) > 8 χ 32-2 1 2
2
0 0 1 1 3 (420 + 6-9) ,
4-84 slug f t2.
4*
[2nr dr] to
At 1200 rev/min,
S T R E N G T H O F M A T E R I A L S
ω = 60
125-5 rad/s.
K . E . =4· /ω2' Δ
4-84 χ 125-5
2,
= 38200 ft lbf.
Cylinder—2nd Moment about an Axis Through the Centroid
/ / / / / / / / / /
te
.Disk
1
FIG. 86
As already shown, for a circular section
64 and ky le"
Hence, for the infinitely thin disk of thickness da; and diameter D
(Fig. 86) distant χ from an axis through the centroid G,
2nd Moment of mass about Y Y , Iy = Mass χ k*
^ 4 Jo 16
2π χ 1200
88
1 s t A N D 2 n d M O M E N T S 89
F rom the parallel axis theorem,
2nd Moment about the axis through G,
(Mass χ . τ2) .
Hence, for the disk
B \ 4 / g lb \ 4
,2
- (πΌ2 9
~ \ 4 ~ 7
D2
— +x*)dx.
.·. F o r the whole cylinder h = 2
j ( χ 2 )2
~ ) (j£ + χ2
) ά χ
'
ο πΌ
2ρ
2g
Ό2 χ
16
L/2
|ο
and πΌ
2
χ 7>ρ
πΌ2ρ ι D
2L L*
2g \ 32 +
"24
π £2 Σρ ID
2 L
2
~ ι ~ χ ι γ 1 ί 6 " + Ί 2 "
IF (the to ta l weight).
W
g
D2 L
2
l ö " +
~Ï2~
W Since / „ = kl, where &σ =
8 g
ë
of suspension i t follows tha t k2
radius of gyration about the axis D
2 L
2
16 + 12 D2
Note tha t if L is small relative to D, then k\ ~ —- ~ k2
b 16
J
E X A M P L E . A rectangular solid of length L, depth d and breadth b
is suspended from a single point with i ts length and breadth hori-zontal. Show from first principles tha t the radius of gyration about the point of suspension is given by
- V ' b
2 + L
2
12
90 S T R E N G T H O F M A T E R I A L S
Solution
F o r L and b to be horizontal, the axis of suspension must pass
through the centroid G. As already shown, for a rectangular section
Iy = — and kl = — .
Hence, for the infinitely thin rectangular slice of thickness da;
distant χ from an axis through the centroid G,
A Y
/ S
I 1 1
/
L dx * 2 -
.Slice
Τ - d
ι
FIG. 87
2nd Moment of mass about Y Y , Iy = Mass χ k2,
ρ b2
(bd x d a ; ) ~ χ — g 12
F rom the parallel axis theorem I8 = Iy + (Mass χ χ'
(bdQ · dx^ b2^
l g ! 12
( ^
bdQ lb* dx
g
(where χ varies from zero
L
1 s t A N D 2nd M O M E N T S 91
F o r the whole solid Ig =
2f -y (j£
+ x* )
dx>
ο 2bdq (b2x a?\L/2
\ l 2+
J ) 0 ' 9 \ A Z d/o
2bdQ (b*_ L_ j _ IM
g V12 X 2
+ 3
X 8 J'
to ta l weight,
W lb* + £2
J f /62 + £
2\
Ύ V î 2 ; *
W 9 δ2 4- Ζ
2
Since / „ = — , it follows tha t k2 =
g 89 s
12
' fr2 + L
2
"Î2 '
E X A M P L E . F ind the volume of the solid generated when an equi-
lateral triangle of side a is rota ted once about the #-axis. F ind the
moment of inertia of this solid about the #-axis in terms of a,
π, g and the density ρ. g
Hence show tha t the radius of gyration about this axis, k = ~J^Q~ ·
Solution
Volume of slice = ny2 dx and y
2 = 3x
2, F ig . 88,
= 3πχ2dx.
a/2
7 = 2 / 3πχ2 dx,
υ
Xs
3 0
2π· a
3
Τ* πα
ζ
4
92 S T R E N G T H O F M A T E R I A L S
Volume of slice = ny2 dx.
Weight of slice = ny2 dxg .
Mass of slice τι y
2 dxg
FIG. 8 8
πν2 dxo ν
2
2nd Moment = — ^ χ ^ - ,
y* dx and y = y Sx, .*. y1
9πρ
2g • x
4 dx.
α/2
#4 da;, α/2
g J
9πρ α/2
0
9 πρ a5
= "5
X T " X 3 2 '
9πρα5
1st A N D 2nd M O M E N T S 93
w And / = — * » ,
9 7 2 9πρα
5 g πα?ρ
k =WÔg-X-W> W h e re * = F é ? = — :
9πρα5 4
— χ
JE; =
160 " πα3ρ
9α2
3α
"j/4Ö*
Examples I I I
1. A T-section is 1-0 in. thick throughout, 6-0 in. deep overall and has a flange width of 5-0 in. Find the position of the centroid of the section and calculate the approximate value of the 2nd Moment of area about an axis through it and parallel to the flange (3 in. above the tip of the web, 33 in
4).
2. A piece of angle iron 6 χ 6 in overall, is used as a cantilever with one side vertical. If it is 0-5 in. thick throughout, find the 2nd Moment of area of the section about a horizontal axis through the centroid (19-9 in
4).
3. Find the Radius of gyration of the section of a hollow shaft having inter-nal and external radii of 5 and 6 in. respectively. What is the percentage difference between this and the mean radius (5-52 in., 0-36 per cent).
4. Find the 2nd Moment of area of a channel section about an axis through the centroid and parallel to the web given that the section is 8 χ 4 in overall and 1-0 in. thick throughout (18*9 in
4).
5. The section of a beam is an equilateral triangle with its base horizontal. If the side of the section is 5-2 in. find the 2nd Moment of area of the section about a horizontal axis through the centroid (39-6 in
4).
6. Show that, for a triangle of height d base b the 2nd Moment of area about the base is given by bd3/12. The section of an oak beam is rectangular and 8 χ 6 in. Find the 2nd Moment of area about a diagonal (183 in
4).
4a SM
C H A P T E R IV
BEAMS II—SIMPLE BENDING
Simple Bending
A bending moment applied to a beam produces tensile and com-
pressive strains separated by an unstrained or " n e u t r a l " plane as
in Fig . 89.
In Compression
In Tension'
Load producing bendina
Neutral axis of section
FIG. 8 9
Where the beam section and this neutral plane intersect , is
known as the neutral axis of the sect ion; for the symmetrical
section shown this axis is half way down. F o r the unsymmetrical
section of F ig . 90 i ts position must be found.
I t is reasonable to assume t h a t the strain a t any point due to
bending (and hence the stress / ) is proportional to the distance y
FIG. 9 0
9 4
B E A M S I I — S I M P L E B E N D I N G 95
from the neutral plane or axis, so tha t the graph of / against y
is a straight line (shown dot ted in Fig . 90) the origin 0 being any
point on the neutral plane. ( I t will be proved later t ha t this is in
fact the case.)
L e t / be the stress in elemental area a, distant y above the N.A.
Then, compressive force on element is fa and this will increase from
zero a t 0 up to a maximum a t B , where y — h. Similarly, for ele-
ments below the neutral axis, the force will be tensile and will
increase from zero a t 0 down to a negative maximum a t A, where
y = -(d - h).
I f the section A B is to be in horizontal equilibrium, the resultant
compressive force above the neutral axis must be equal and oppo-
site to the resultant tensile force below i t ,
i .e. 27« = °-
Ε As will be shown later, / = — x y where Ε = Elas t ic i ty modulus
R and R = Radius of curvature of beam at the section.
Ey
Hence, Σ x » = 0 ,
Ε
or — Σ W = 0.
Ε .'. Σ
ay
= 0 , since — Φ 0 .
R
Now, b y definition, Σαν i
s the 1st Moment of Area of the section
about the neutral axis, and this can only be zero if this axis passes
through the centroid of the section.
Hence, to find the position of the neutral axis i t is only necessary
to locate the centroid of the section. y ar
As already shown, i ts depth is given by h = — — from the top,
where A is the to ta l sectional area (Fig. 91) .
Having determined the position of the neutral axis, the value of
the 2nd Moment of area of the section about this axis (IN.A . ) m a
y
now be found in a given case b y one of the usual methods. (As
will be seen later, this value, together with those of Ε and Μ, is
required before the stress can be evaluated.)
An expression for the stress a t any point in an elastic beam can
now be derived, the following being the chief assumptions made :
4 a*
96 S T R E N G T H O F M A T E R I A L S
FIG. 9 1
L e t the two transverse plane sections A B and CD (Fig. 92) be
a small distance dx apart when the beam is straight. The applica-
t ion of a bending moment M will cause A B and CD produced to
intersect a t some point 0 so tha t B D subtends the small angle do.
Elements such as E F above the neutral axis will be reduced in
length while similar elements below the N.A. will undergo tensile
strain.
ο
f
FIG. 9 2
1.. The material is homogeneous and isotropic.
2. The effects of shear force (which are explained la ter in the
t e x t ) are neglected so tha t a plane section perpendicular to
the axis is assumed to remain plane during bending.
3. I n neither tension nor compression is the limit of proportio-
nali ty exceeded, and in both the value of Ε is the same.
B E A M S I I — S I M P L E B E N D I N G 97
R '
Hence, Compressive stress / = Ε x Strain,
7? *
Similarly, for elements below the neutral axis,
Tensile stress / -— — . Ey
Thus, in general, / = the sign depending on
R the position.
Since R is constant for this small length of beam, it follows tha t
/ oc y as was assumed earlier. Hence / = 0 when y — 0 (i.e. zero
stress a t the neutral plane) and is a maximum when y is a maximum,
FIG. 9 3
i.e. a t the upper and lower surfaces. The stress distribution (graph
of / against y) is shown in Fig . 92. Note tha t the maxima have diffe-
rent signs and are not necessarily the same.
Now the transverse area of the element E F is b dy and this
may, for convenience, be denoted by a.
F o r the element E F in F ig . 92 :
Original length = dx = R d6, where R = radius of neutral plane.
Length after bending = (R — y) d0 = R dd — y dö.
.'. Reduct ion in length = y dO,
ydO i.e. Compressive strain = ,
98 S T R E N G T H O F M A T E R I A L S
Then, Compressive force on this element = Stress χ Section,
Ey
Ey Moment of this force about the N.A. = —— χ a X y
Ii
κ ,
I f all such similar moments are added (including those of the
tensile forces below the neutral axis which have the same sign),
the result will be what is called the Moment of resistance of the
beam.
Ε This can be written Σ "5"
α2 /
2
= 2J AV
2 since is constant,
i t R
Now, by definition, the quant i ty 2Jay2 is the 2nd Moment of the
section about the neutral axis . This is denoted b y I. Ε
Hence, Moment of resistance to bending = —I. Since a t any sec-
t ion (within the limit of proportionality) this internal moment of
resistance is equal and opposite to the applied bending moment M WFI p a n WRITE,
B u t
Ε M E M
= - RI °
r - = T
, Ey Ll L f Ε
>=ΊΓ> S O T H AT J = -R
(1)
(2)
Combining (1) and (2) gives the fundamental equation for simple bending, i.e.
L - ϋ - —
B E A M S I I — S I M P L E B E N D I N G 9 9
Since / = My/I i t can be seen that , for a given beam section,
i.e. for a given value of I, the stress varies along the beam in pro-
portion to M and across the section in proportion to y. This is
illustrated in Fig . 9 4 b y the graphs of stress distribution drawn at
equal intervals along a simply loaded cantilever.
Load W
FIG. 94
E X A M P L E . A wooden beam 1 2 χ 6 in. carries a load of 2 tonf
concentrated a t the centre of a 1 2 ft span. F ind the greatest bend-
ing stress in i t when the longer side is
(a) upright and (b) horizontal.
2 tonf 2 tonf
(a) y = 6in
FIG. 95
(b) y=3in
Solution
I n case (a) : bf_ _ 6 χ 1 2
3
~ Ϊ 2 ~ ~ Ϊ 2 8 6 4 in
4,
100 S T R E N G T H O F M A T E R I A L S
M„
Λ / »
I n case ( b ) :
WL _ ( 2 χ 1 2 )
~ 4 ~ ~ 4
Mmaxy _ 7 2 χ 6
/ 8 6 4
1 2 χ 63
1 2 = 7 2 tonf in.
0 - 5 tonf in2.
1 2
M is the same as in case (a)
7 2 χ 3
2 1 6
2 1 6 in4.
= 1 -0 tonf in2.
The stress in case (b) is twice tha t in case (a) because, although y
is halved, the value of Ib is only one quarter of tha t of Ia.
E X A M P L E . A horizontal jois t is supported over a span of 2 0 ft.
I t is 1 8 in. deep and the value of I is 1 1 5 0 in4. Determine the value
of the maximum stress if the load per foot (including the weight
of the jo is t ) is 1 - 5 tonf.
Solution
Mma, WL
( 1 - 5 x 2 0 ) ( 2 0 χ 1 2 )
8
9 0 0 tonf in ,
1-5 ton f / f t
- 2 0 f t —
FIG. 9 6
y=9 in
My _ 9 0 0 x 9
' ' m a x ^ " 7 " = 1 1 5 0
- 7 - 0 5 tonf/in2.
B E A M S I I — S I M P L E B E N D I N G 101
E X A M P L E . A jo is t has the section shown. Determine
(a) the value of I N . A. a n
d (b) the bending moment required to induce a maximum stress of 7-5 tonf in
2.
Solution
F o r the solid rectangle 6 χ 12.
6 Χ 1 23 0-875_in
1 N.A. 12
F o r each shaded piece of depth 10-25 and width 2-75,
/Λ .Α. 2-75 χ 10-25
3
12
Net 2nd Moment
6 χ 1 23
^N. A. 12
2-75 χ 10-253
12
= 864 - 4 9 3 ,
= 371 i n4.
- 6 in-
0-5 in-
12 in
FIG. 97
Bending moment M = where y = 6 in. and / = 7-5,
7-5 χ 371
- 6 ;
= 463 tonf in.
E X A M P L E . The 2nd Moment of area about the neutral axis of a beam section 20 in deep is 1670 in
4.
F ind the longest span over which a beam of this section, simply supported, could carry a uniformly distributed load of 1-5 tonf/ft with a maximum resulting bending stress of 7 tonf/in
2.
1 0 2 S T R E N G T H O F M A T E R I A L S
1*5
Load in tonf/in = — - and y = 1 0 in.
wL2 / m ax /
y
1-5 L2 7 χ 1 6 7 0
* 12 8
.'. L2
1 0 '
7 χ 1 6 7 0 χ 9 6
Ï 5
- 7 4 , 8 0 0 .
L = 2 7 3 in.
= 2 2 - 8 ft.
L inches
E X A M P L E . A flat steel strip has a thickness of 0 - 2 in. To what
radius may i t be bent if Ε = 1 3 , 0 0 0 tonf/ in2 and / > 2 0 tonf/in
2?
I f the strip is 1 - 5 in. wide and 7 2 in. long and is supported a t the
centre only, what equal loads suspended from the ends would
produce the same stress?
FIG. 9 8
B E A M S I I — S I M P L E B E N D I N G 1 0 3
Solution
f =
:. β =
Ey_
R '
Ey
t '
1 3 , 0 0 0 χ 0 - 1
y=ο·ι
2 0
6 5 in.
M = W χ 3 6 ,
bd?
1 2
1 - 5 χ 0 - 23
I = -1-5 in-
1 2
0 0 0 1 in4,
My
\ FIG. 99
or
2 0 x 2 2 4 0 =
W =
3 6 χ 0 - 1
0 0 0 1 '
2 0 χ 2 2 4 0 χ 0 0 0 1
3 6 χ 0 - 1
= 1 2 - 4 4 lbf.
FIG. 100
E X A M P L E . The flange of a T-section jois t is 5 in. wide and 0 - 5 in. th ick . T h e web is also 0 - 5 in. t h i ck and has a depth of 4 - 5 in. I f i t is used with the web vert ical and the flange uppermost, find:
(a) the position of the neutral axis (b) the value of /Ν.Λ.·
104 S T R E N G T H O F M A T E R I A L S
Solution
The section may be divided into two elements :
ax = 5 χ 0-5 with centroid a t rx = 0*25 in. from the datum,
a9 = 4-5 χ 0-5 with centroid a t r2 = 2-75 in. from the datum.
Datum
FIG. 101
The depth of the centroid G (i.e. of the Neutral Axis N.A.) is then
given b y
Zar (5 χ 0-5) 0-25 + (4-5 χ 0-5) 2-75
~~ A ~~ (5 χ 0-5) + (4-5 χ 0-5)
0-625 + 6-2
~ 2-5 - I - 2-25 '
6-825 4-75
= 1-435 inch from the top.
The 2nd Moment of area of a rectangle about an edge is given by
bd*
~ ~ 3 ~ *
.". ^N.A. 0-5 χ 3 -565
3 5 χ 1-435
3 4-5 χ 0 -935
3
^ + •
45-3 2-96
+ •
3
0-816
6 ' 0-6 0-67
= 7-55 + 4-93 - 1-23,
= 11-25 in4.
B E A M S I I — S I M P L E B E N D I N G 105
E X A M P L E . F ind a suitable spacing for the jois ts supporting a
floor on which the load is to be 336 lbf/ft2 if the max . bending
stress > 1000 lbf/in2. The span is 14 ft and the jo is ts are 12 in
deep χ 4-5 in. wide.
Solution
L e t χ = width of floor per jois t .
J
/ M 12 i
FIG. 102
2nd Moment of section 4-5 χ 1 2
3
12
648 in4.
Permissible Bending moment M = — . y
F o r a Uniform load
il y
1000 χ 648
6 '
108000 lbf/in.
WL
8 8i¥n
L
_ 8 χ 108-000
14 χ 12
= 5150 lbf.
(where L = 14 ft)
Effective area between joists = 14# f t2.
.·. To ta l load W = 14a; χ 3 3 6 .
106 S T R E N G T H O F M A T E R I A L S
Equat ing, 14α; χ 336 5 1 5 0 .
5150
14 χ 336
= 1-095 ft,
- 13-1 in.
E X A M P L E . F ind the
depth of the neutral I
axis for the T-sect ion 0-3 i n ^ f
shown. Determine the
maximum tensile and
compressive stresses
when a bending mo-
ment of 4-75 tonf inch
is applied. (Assume a
cantilever.)
Solution
Area of section
A = (3 x 0-3)
_U (4.4 χ 6 -3) ,
= 0-9 + 1-32,
= 2-22 in2.
Depth of N.A.
t 1-25 in
4 - 4 in
3-15 in
h =
- 3in-
0-3 in
FIG. 1 0 3
Ι /0-I5 in
h = 1-55 in t
2-5 in
27 ar
A 9
(3 x 0-3) 0-15 + (4-4 χ 0-3) 2-5
2-22
0 1 3 5 + 3-3
2-22
3-435 2-22 '
= 1-55 in.
3 χ 1-553 3 χ 1-25
3 0-3 χ 1-25
3
( 0-3 χ 3 1 53
r . 1 = o-l in*. 1 Χ.Λ. — ο ο ' Q
1 Q
B E A M S I I — S I M P L E B E N D I N G 107
I f the beam is used as a cantilever, the flange will be in tension.
.'. Max. tensile stress My
, where y =~ 1-55 in.
4-75 χ 1-55
5 1
: 1-45 tonf/in2.
Max. Compressive stress = My
where y = 3-15 in. ,
4-75 χ 3-15
5 1
= 2-94 tonf in2.
E X A M P L E . The shaft of a mechanical hammer is rectangular in
section and pivoted a t a point 60 in. from the head, which weighs
560 lb. I f the head is to be raised from the horizontal with an
init ial acceleration of 5 f t /s2, determine the shaft section required
to l imit the maximum bending stress to 0-75 tonf/ in2 assuming the
depth to be twice the thickness.
6 0 in •
5 f t / s2
5 6 0 lbf
Solution F I G . 1 0 4
W Accelerating force on head F = —/,
9
_ 560 χ 5
32-2
= 87 lbf.
*. Tota l force on end of arm = 87 + 5 6 0 ,
= 647 lbf.
H b
Section on XX
Bending moment M = 647 χ 6 0 ,
= 38 ,820 lbf inch about point O.
1 0 8 S T R E N G T H O F M A T E R I A L S
From the bending equation -L M
I '
0 - 7 5 χ 2 2 4 0 3 8 , 8 2 0
d/2 ~ M»/12 '
i.e., bd
3
1 6 8 0 - = 3 8 , 8 2 0 4 ·
Δ
.·. bd2
= 1 3 8 - 5 ^and b
:. d3
= 2 χ 1 3 8 - 5 ,
= 2 7 7 ,
Hence, d = 6 - 5 in ,
and b = 3 - 2 5 in .
E X A M P L E . A locomotive coupling rod is 1 0 0 in. long between the
supports, 5 in. deep and 2 in. th ick . The crankpin radius is 1 ft
and the wheel diameter 6 ft. I f the density of the material is
0 - 2 8 3 lb / in3 es t imate the maximum bending stress in the rod
when the locomotive is travelling a t 9 0 ft /s.
FIG. 1 0 5
Solution
The centrifugal force on the rod ac ts radially outwards, and,
like the weight of the rod itself, is uniformly distributed. When the
rod is in i ts lowest position, the centrifugal and gravity forces are
in the same downward direction so tha t the loading is as a t (b)
and the bending moment is a maximum.
Weight of rod W = ( 1 0 0 χ 2 - 5 ) 0 - 2 8 3 = 2 8 3 lbf.
Speed of wheel rim relative to wheel axis = ν = 9 0 ft/s.
Speed of crankpin relative to wheel axis vc = v/3 = 3 0 ft/s.
B E A M S I I — S I M P L E B E N D I N G 109
32-2 χ 1 0 '
7920 lbf.
To ta l distributed load = F + W,
= 7920 + 2 8 3 ,
= 8200 lbf. 2 χ 5 3
2nd Moment of rod section I = — — — = 20-82 in 4. iz
WL Max. bending moment M = ———, where W = 8 2 0 0 ,
8
8200 χ 100
102,500 lbf in.
Corresponding max . stress / = , where y = 2-5 in. , My
I
102,500 χ 2-5
20-82
12,300 lbf/in2.
Beam of Circular Section
A bending moment may be applied to a shaft by a force applied to a crank, pulley or toothed gear in addition to any bending effect due to the weight of the shaft itself. Since the bending stress is given b y / = Myjl i t is required to find the value of / about the neutral axis, i.e. about a diameter. The value of y is clearly the greatest radius.
F rom the theorem of three mutually perpendicular axes :
Iζ =
Ιχ + Iy a n
d clearly Iy = Ix.
/. Iz - 27„
i.e.
W V2
Centrifugal force F = (where r = 1-0 f t ) , gr
283 χ 3 02
110 S T R E N G T H O F M A T E R I A L S
Since all points on this element are distant r from the axis ZZ, by
definition :
ι Y 2nd Moment about
Γ ZZ = 2nr dr χ r2,
/ ^T/AC ^ \ \ \ \ V 2 n (
* Moment °^ w n
° le
xf ([ \—τπχ I s = / ^ r 3 d r '
γ or J 3 = . J ( Ä i - Ä i ) FIG. 1 0 7
Hence, I, ~ \ (R{ - />'!) = (D\ - D\).
F o r the circular element in F ig . 107 :
Area = Length χ Width,
= 2πτ χ dr.
I Y
FIG. 1 0 6
B E A M S I I — S I M P L E B E N D I N G 111
I n the case of a solid shaft R2 = 0 and R1 R.
and
J ;
lr
2
nR*
32
ττΧ>4
64
E X A M P L E . A telegraph pole is 9 in. dia. I f the bending stress a t
ground level is not to exceed 500 lbf/in2 find the horizontal force
which may be applied to a s tay wire 36 ft above the ground.
I f the pole were subjected to a wind load of 2 0 lbf/f t2 of projected
area, est imate the maximum bending stress which would be induc-
ed a t ground level if the pole were unstay-
ed, and 4 0 ft high.
Solution
2nd Moment of area about the neutral axis
64 '
_ π χ 94
" 64
= 322 in4.
36 ft
where
M
f
. F
: F (36 χ 12) = ÎL y
500 lbf/in2 and î / = 4 -5 in . / / / / / / / /
500 χ 322
4-5 (36 χ 12)
: 83 lbf.
γτπτττττπτ
Projected area of pole
= Height χ Diameter ,
9
4 0 f t
40 χ 12
9
= 30 f t2.
To ta l load W = 30 χ 2 0 ,
= 600 lbf.
777777777"
FIG. 1 0 8
1 1 2 S T R E N G T H O F M A T E R I A L S
Bending moment M WL
2
= 6 0 0
at ground level,
4 0 χ 1 2
( ^ )
Max. Bending stress
1 4 4 , 0 0 0 lbf in.
, My
1 4 4 , 0 0 0 χ 4 - 5
3 2 2
= 2 0 0 0 lbf/in2. (approx.)
E X A M P L E . An emergency footbridge consists of two steel pipes
6 i n . bore, 7 in. outside dia., laid parallel and 3 6 in. apart and cover-
ed with planking. The tota l load per foot is 2 2 4 lbf. Es t ima te the
maximum bending stress in the pipes when there is an additional
concentrated load of 4 4 8 0 lbf a t the midpoint of the span which is
of 1 0 ft.
2 cwt / foot 4 0 cwt
7 in 6 in
120 in •
Solution
Tota l load ( 1 0 χ 2 2 4 )
6 7 2 0 lbf ,
R 2
FIG. 1 0 9
4 4 8 0 ,
^ 1 7
^ 2 6 7 2 0
= 3 3 6 0 lbf.
At centre, J f m a x = (R1 χ 6 0 ) - ( 2 χ 1 1 2 ) 5 χ ~ ,
= ( 3 3 6 0 χ 6 0 ) - ( 1 1 2 0 χ 3 0 ) ,
= 2 0 1 , 6 5 0 - 3 3 , 6 0 0 ,
= 1 6 8 , 0 0 0 lbf in.
B E A M S I I — S I M P L E B E N D I N G 1 1 3
/ = where I 6 4
(74 _ 64) , = 5 4 - 2 5 i n
4 per pipe,
1 6 8 , 0 0 0 χ 3 - 5
5 4 - 2 5 χ 2
= 5 3 2 0 lbf / in2.
Alternatively, WL W2L
= L 8 ^ 4
(where W1
W,
Mn
Tota l distributed load,
( 2 χ 1 1 2 ) 1 0 ,
2 2 4 0 lbf ;
Concentrated load,
4 0 χ 1 1 2
: 4 4 8 0 lbf)
1 2 0 ( 2 8 0 + 1 1 2 0 ) ,
1 2 0 χ 1 4 0 0 ,
1 6 8 , 0 0 0 lbf in.
E X A M P L E . A wagon axle is carried in bearings 6 6 in. apart, the
wheel centres being 5 8 in. apart and placed symmetrically between
the bearings. I f the axle carries 1 0 tonf, find a suitable diameter for
the axle assuming a safe stress of 4 tonf/in2.
Corresponding max . bending stress
114 S T R E N G T H O F M A T E R I A L S
Solution
d
y = r
ltd*
=
5 t o n f
N - » - A
- 6 6 in -
4 in
5 tonf
58 in -
FIG. 110
Between the wheels, Constant M - 5 χ 4 ,
^ 20 tonf in.
/ = My
20 χ d
4 =
.·. 4 =
64
6 40 · λ
6 40 H I
d = 3-7 in. (say, 3 | i n . ) .
B E A M S I I — S I M P L E B E N D I N G 115
E X A M P L E . A steel pipe 12 in. inside dia., 13 in. outside dia., is
used as a beam. F ind the max . permissible central load on a span
of 8 ft if / m ax > 5 tonf/in2. Neglect the weight of the pipe.
Solution
/ n . a . = ^ ( 1 34 - 1 2
4) ,
= - £ - ( 2 8 , 5 0 0 - 2 0 , 7 0 0 ) , 64
_ π x 7800
64 '
= 383 in4.
4 8 in-
it
FIG. I l l
At centre, W
y
= 24 W tonf in , and / = 5 tonf/in2,
y = 6-5 .
5 χ 383 24Tf =
W =
6-5 '
5 χ 383
6-5 χ 2 4 '
12-28 tonf.
E X A M P L E . F ind the maximum bending stress induced b y i ts own
weight in a 9 in. dia. shaft 30 ft long when i t is simply supported
at the ends, given tha t the density of the material is 0-28 lbf/in3.
116 S T R E N G T H O F M A T E R I A L S
Solution
Weight of shaft W = (0-785 χ 92) ( 3 0 χ 12) 0-28,
= 6400 lbf.
- 3 0 f t -
FIG.112
k9in-*H
2nd Moment of section I = π χ 9
4
64
= 322 in4.
-*-"· max WL
8 where L = 360 in ,
_ 6400 χ 360
8
= 288 ,000 lbf in.
/max = MmvaLyjI, where y = 4 - 5 ,
_ 288 ,000 χ 4-5
322 = 4032 lbf/in
2.
E X A M P L E . F ind, in terms of
the diameter D, the dimen-
sions b and d of the strongest
rectangular beam which can
be cut from a cylindrical log
(Fig. 113) .
Solution
. My_ I τ
= Μ [ ^- X — ^
FIG.113
12
ΊηΡ) = M (and<Z
2 = D
2- &
2) ,
bur
6M
~~ b(D* - 62) '
B E A M S I I — S I M P L E B E N D I N G 117
F o r / to be a minimum, bD2 — b
z must be a maximum.
/ . Differentiating, D2 - 3b
2 = 0 ,
D2
Examples IV
1. Calculate the value of the horizontal load which, when applied at the top of a tubular pole, 6 in. outside dia. 0-1 in. thick and 20 ft high, will induce a maximum stress of 12,000 lbf/in
2 (136 lbf).
2. Discuss the assumptions made in the derivation of the equation for simple bending.
3. A diecast bracket is of I-section 0-2 in. thick throughout and 1-4 in. deep, the widths of upper and lower flanges being respectively 1-0 and 0-6 in. Calculate the bending moment which will induce a tensile stress of 1500 lbf/ in
2 in the upper flange and determine the corrseponding stress in the lower
flange (357 lbf in., 2350 lbf/in2).
4. Oil of density 0-0318 lbf/in3 is to be poured into a tank 5 ft square which
is supported symmetrically by the flanges of two parallel cantilevers each of T-section 5 χ 5 χ 0-5 in. If the tank is in contact with the wall and its weight is small in comparison with that of the liquid, determine at what depth of oil the maximum stress in the cantilevers will reach a figure of 10,000 lbf/in
2 (19-5 in.).
5. Estimate the uniformly distributed load which may safely be carried over a simple span of 20 ft by a timber joist 12 in. deep and 6 in. wide if the permissible stress is 800 lbf/in
2 (3850 lbf).
6. A crane is to lift a steel shaft 120 in. long and 2-0 in. dia. by means of two vertical slings. Determine the sling positions which will give rise to minimum stress and find this stress (24-8 in. from each end. 700 lbf/in
2
approx.).
5 SM
C H A P T E R V
BEAMS III—SIMPLE SHEAR
Shear Force in Beams (F)
Convention
I f the resultant force to the left of a section is upwards as in
Fig . 114 then F is taken as positive.
X X
Positive shear Positive shear
X
J
F I G . 1 1 4
Cantilever with Concentrated Load at the Free End
Considering any section X X of the cantilever in Fig . 115(a) , in
addition to causing a bending movement of — Wx the concentrated
load W also tends to force the piece of length χ downwards relative
to the remainder of the beam, i.e. to shear i t off as in F ig . 115(b ) .
The net transverse force on X X is called the Shear Force and
denoted by F.
In this case F = — W for the whole beam and i ts graph is the
horizontal straight line of Fig . 115(c ) .
Cantilever with Several Concentrated Loads
From 0 to A (Fig. 116(a)) there is no change in load and, there-
fore, F = -W1.
1 1 8
B E A M S I I I — S I M P L E S H E A R
•
119
FIG. 1 1 5
L
F I G . 1 1 6
• (α)
(b)
I F = - W
JL (c)
(α)
(b)
120 S T R E N G T H O F M A T E R I A L S
At A there is an increase in load and hence from A to B ,
F = -(W1 + W2)
Similarly from Β to C
F = -{Wx + W2 + W3).
The graph of F is therefore the series of steps shown in Fig . 116 (b ) .
Simply Supported Beam with Single Concentrated Load
(1) Load at Centre
From 0 to A (Fig. 117 (a)), the net transverse force on any sec-
tion X X W
Fa. = and is positive.
W
(a)
Β
Τ
F I G . 1 1 7
(b)
From A to B the net transverse force on any section Y Y ,
W W Fy = W = — , i.e. is negative.
2 2
The graph of F is therefore as in Fig. 117(b).
B E A M S I I I — S I M P L E S H E A R 121
(2) Load not at Centre
Moments about the L H end give :
R2L = Wa, i.e. R2 = Wa
and Wa
L
The graph of F is then as in Fig . 118 (b) and consists of two unequal
steps.
Simply Supported Beam with Several Concentrated Loads
As before Rx and R2 can be found by taking moments . F rom Ο to
A, J P 0 A = Rx (positive). At A the shear force is reduced b y an
amount equal to W1,
i.e. FX]i =B1 - W,.
Similarly,
Then,
F I G . 1 1 8
122 S T R E N G T H O F M A T E R I A L S
Similarly, FCd = R 1 - W 1 - W 2 - W ^
= R1 (W1 + W2 + Wz)
Beam with Loaded Ends Overhanging Supports
(./) Supports Symmetrically Spaced
From Ο to A the shear force ( = — W) is downwards to the left
of the section and is therefore negative, F ig . 120.
At A the shear force is reduced b y an amount W and hence be-
tween A and B , F = 0.
A t Β the shear force is reduced by a further amount W and
hence between Β and C,
F = +W.
(2) Supports not Symmetrical
I n th is case Rx and R2 must first be found by taking Moments.
The shear force between the supports will be zero only if the
bending Moments a t A and Β are equal, i .e.
W > = W2c.
Rx will then be equal to W1 and R2 t o W2. F o r all other cases the
graph of F will be as shown in Fig . 121,
F I G . 1 1 9
B E A M S I I I — S I M P L E S H E A R
W W
A Β
R=W R=W
123
-w JL
FIG. 1 2 0
W,
Ο α
W/////JU/f/ff
//ffÜ!ZZL
+w?
FIG. 121
General Case
The value of F a t any section is the ne t transverse force acting on the section. As already stated, i f this force acts downwards to the left of the section considered, i t is taken as negative and vice versa.
I n general, when there is no change in load between two sections of a beam, the shear force remains unchanged,
1 2 4
w,
S T R E N G T H O F M A T E R I A L S
W 2 W,
ν///////////77777λ
FIG. 1 2 2
E X A M P L E . F ind the shear force a t each of the points A , Β and C
in the system shown in Fig . 1 2 3 and draw the graph of F.
0-5 tonf I 2
_o_
-0-5 •
Li.
- 3 - 5 tonf
FIG. 1 2 3
B E A M S I I I — S I M P L E S H E A R
Solution
FOA = -0-5 tonf (anticlockwise shear) .
(0-5 + 1-0),
1-5 tonf.
F Λ W - (0-5 + 1-0
- 3 - 5 tonf.
2 -0 ) ,
E X A M P L E . Calculate the values of Rt and B2 in the system shown
and draw to scale the graph of F.
6 8 10 tonf
• 2 f H
— 4 f t —
— 1 2 f t -
-16 f t -
5 a S M
FIG. 1 2 4
125
126 S T R E N G T H O F M A T E R I A L S
Solution
Moments about Ο give :
1 6 i ? 2 = (6 χ 2) + (8 χ 4 ) + (10 x 1 2 ) ,
= 12 + 32 + 1 2 0 ,
- 1 6 4 .
1U - 10-25 tonf ,
and RX - 6 + 8 + 10 - 10-25,
= 13-75 tonf .
FQK = + 1 3 - 7 5 tonf.
F A N = + 1 3 - 7 5 - 6 ,
- + 7 - 7 5 tonf .
F\ic = +7*75 - 8 ,
= - 0 - 2 5 tonf .
F A ) - - 0 - 2 5 - 1 0 ,
- - 1 0 - 2 5 tonf - - R 2 . The graph of F
E X A M P L E . Draw to scale the graph of shear force for the system
shown in F ig . 125.
is then as shown above.
2 tonf 4 tonf 6 tonf 2 tonf
F
F I G . 1 2 5
B E A M S I I I — S I M P L E S H E A R 127
Solution
Moments about A give :
(R1 χ 20) + (2 χ 5) = (4 χ 5) + (6 χ 15) + (2 χ 2 5 ) .
20 R2 + 10 .= 20 + 90 + 5 0 ,
20 R2 = 1 5 0 .
E2 = 7-5 tonf .
FH = (2 + 4 + G - I - 2) - 7-5
= 6-5 tonf .
The graph of F is then as shown.
E X A M P L E . Draw to scale the graph of shear force for the system shown in F i g . 126.
Solution
Moments about A give :
(R2 χ 20) + (4 χ 5 ) ,
= (2 χ 3) + (5 χ 12) + (3 χ 1 8 ) ,
i.e. 20 R2 + 20 = 6 + 60 + 5 4 .
20 R2 = 1 0 0 ,
i.e. R2 = 5 ton f .
R1 = (4 + 2 + 5 + 3) - 5 ,
= 9 tonf .
The graph of F is then as shown.
Cantilever with Uniform Load
Considering any section X X of the cantilever in Fig . 127(a ) , χ
in addition to causing a bending moment of —wx-^-the weight
wx of the piece of length χ induces a shear force of wx on the sec-t ion. This is downwards to the left of X X and is therefore taken as negative,
i.e. Fx = — wx.
5 a*
128 S T R E N G T H O F M A T E R I A L S
4 tonf
- 8 f t - - 9 f t - - 6 f t -
· - 5 f t - -20 ft —
F
FIG. 1 2 6
FIG. 127
B E A M S I I I — S I M P L E S H E A R 129
Since w is constant F oc χ and the graph of F is a straight line as shown in F ig . 127 (b). Clearly Fmax = — wL and occurs a t the support.
Simply Supported Beam with Uniform Load
wL Load per support = ——. Considering any section X X , shear
^ wL force Fx = — wx.
At the L H support, when χ = 0, the second term is zero. _ wL
.". F = —^— and is positive.
FIG. 1 2 8
The value of .F is clearly reduced as χ increases by an amount proportional to x, so tha t the graph is a straight line. At the centre, when χ = L/2,
π wL wL F = = 0
The straight line therefore cuts the #-axis a t χ = L/2. This is also the point of maximum bending moment.
130 S T R E N G T H O F M A T E R I A L S
τι wL
Τ
χ = L, F = — wL
wL
The complete graph is therefore as shown in Fig . 128(b) .
Relation Between w, F and M
Consider an element d# of cantilever carrying a load of w per
unit length (Fig. 129(a) ) . I t s weight is wax and the graphs of F
and M are as shown in F igs 129(b) and 129(c ) .
F + dF
FIG. 1 2 9
Since the element is in equilibrium under the forces exerted on i t
by the remaining parts of the beam,
At the R H support when
(α)
Bending movement
(b)
Sheor force
(c)
B E A M S I I I — S I M P L E S H E A R 131
(2) ACW Moments = C W Moments ,
M + dM (considei
point G ) ,
dx dx i.e. M + F— + (F + dF) — = M + dM (considering
dx ΎΊ dx 1 dx ~Λ ~M M-
J^
+ j P^ ~
+ d j P~ 2 ~
= d J f'
i.e. F dx = dM (neglecting the product of small quanti t ies) ,
dM or F =
dx
Thus the slope of the bending moment graph (i.e. rate of change in M with respect to x) is equal to the shear force. F o r the cantilever shown, dM/dx (and hence F) is negative.
I t follows from the foregoing, t ha t integration of w with respect to χ will give the shear force,
i.e. F = j w dx,
= Area of load intensi ty diagram.
Similarly, integration of F with respect to χ will give the bending moment,
i.e. M = J Fdx,
= Area of shear force diagram.
I t should be noted tha t if w is not constant , i.e. the load per unit length is variable, then w must be expressed as some function of x. I f this is not possible, then the graph of w against χ must be
(1) Upward force = Downward force,
i.e. F + dF = F + wdx,
dF = w dx,
dF or —— = w.
dx
Thus the slope of the shear force graph (Le. rate of change in F
with respect to x) is constant for a uniform load. Since (from Fig . 129(c)) this slope is negative, a downward load must be taken as negative. Then
dF
dx
132 S T R E N G T H O F M A T E R I A L S
plotted and the area found b y planimeter, adding squares or mid-
ordinate rule.
Application to Simply Supported Beam with Uniform Load
Referring to Fig . 130 for any section X X . ,
wL χ Mx = ——- χ χ — wx χ — ,
2 2
wL w 2 = ~2~
X ~ YX '
which is the equation of a parabola, is always positive, and is a
continuous function.
wJL 2
ι c d (Mx) ' »"x - d x
ι c d (Mx) ' »"x - d x
0
FIG. 1 3 0
B E A M S I I I — S I M P L E S H E A R 133
As already shown,
Fx = (Mx) = slope of graph of M, which is positive
when x < L and negative when χ > —}
Δ 2
wL wx (differentiating)
w ( 4 - ) · = w x Distance of X X from centre of beam.
Hence, when
when χ = 0 ,
when χ = L,
F o r max . B .M. ,
i.e.
wL
wL (as previously obtained).
w a; = 0
i.e. — χ = 0, since w Φ 0 ,
o r
The maximum value of M is obtained b y substituting this value of χ in the equation for Μ,
i.e. i f m a x =_ x T _ Y ( _
4 8 w L
2
8 and = IT (the to ta l load) ,
at the centre, where F (i.e. \ is zero. \ ax J
134 S T R E N G T H O F M A T E R I A L S
E X A M P L E . Draw to scale the graphs of F and M for the system
shown in Pig. 131 and find the position and value of the maximum
bending moment.
Solution
Moments about L H end give :
12 R2 = (4 χ 3) + (6 χ 6-5) + (3 χ 1 0 ) ,
= 12 + 39 + 3 0 ,
= 6-75 tonf .
Rx = (4 + (3 χ 2) + 3) - 6-75,
= 13-0 - 6-75
= 6-25 tonf .
The graph of F is then as shown and cuts the horizontal a t a point
6-125 ft from the L H end. This gives the position of zero shear and
hence of max . bending moment. F o r any section X X between Β
and C, i.e. a t (x + 5) from the L H end:
Mx = 6-25 (x + 5) - 4(a + 2) - 2a;— , Δ
- 6-25* + 31-25 - 4 τ - 8 - χ2,
= 225χ - χ2 + 23 -25 .
4- (ΜΛ - 2-25 - 2χ = 0 for maximum Β.Μ. , do:
i.e. χ - 1 - 1 2 5 ft,
Hence, confirming the result obtained above, MmuLK occurs a t (x + 5 ) ,
i.e. a t 6-125 ft from 0 .
Substi tuting, l f m ax = (2-25 χ 1-125) - 1-1252 + 23-25 ,
- 2-53 - 1-27 + 23 -25 ,
= 24-5 tonf f t .
Mx = 6-25 x 3 = 18-75 tonf f t .
i f Β = (6-25 χ 5) - (4 χ 2) = 23-25 tonf f t .
Mc = (6 25 χ 8) - (4 χ 5) - (6 χ 3)
- 21-0 tonf f t .
MD = (6-75 χ 2) = 13-5 tonf f t .
The graph of M is then as shown.
B E A M S I I I — S I M P L E S H E A R 135
4 tonf
- 6 - 7 5
FIG. 131
E X A M P L E . A simply supported jois t is loaded as shown in F ig . 132 F ind the position and magnitude of the maximum bending moment and draw the graphs of F and M to suitable scales.
Solution
Moments about the L H end give :
(R2 x 40) = (25 χ 1 ) - ^ - + (3 χ 3 0 ) ,
4 0 i ? 2 = 312-5 + 9 0 ,
R2 = 10-06 tonf , whence
and R1 = 17-94 tonf .
136 S T R E N G T H O F M A T E R I A L S
From the graph of F
i t can be seen tha t zero
shear (and hence max .
B .M. ) occurs a t point
D. B y measurement
OD = 17-94 ft.
F o r any section X X between Ο and A,
MT = 17-94*; -
( * χ ΐ ) γ
= 17-94a; - — . Δ
From the shear force diagram, this is a maxi-mum when χ = 17*94
3 tonf
I tonf / f t
and
(17-94
χ 17-94)
17-942
- 161 tonf f t .
(17-94 χ 25)
25 - (25 χ 1)
= 136 ton ft.
FIG. 1 3 2
when χ = 10
MB = (10-06 χ 10) (working from R H end),
= 100-6 tonf f t ,
1 02
Mx = (17-94 χ 10) - — ,
= 179-4 - 5 0 ,
= 129-4 tonf f t .
The graph of M is then as shown.
B E A M S I I I — S I M P L E S H E A R 137
E X A M P L E . A uniform beam 3 0 ft long is simply supported a t
points 9 ft and 24 ft respectively from the L H end. I t carries a
distributed load, the intensi ty of which increases uniformly from
zero a t the L H end to 0 - 8 tonf/ft a t the R H end. Calculate the posi-
tion and magnitude of the maximum bending moment, and the
values of M a t the supports.
Solution. The loading is as shown:
X »d
c
« 9 ft -
wx
* 15 ft—
X
« 6 ft
0-8 tonfAt
F I G . 1 3 3
9 + 1 5
Weight of load A to C = ^—^ j0-8 = 9-6 tonf acting a t Ql9
Weight of load C to D = - | χ 0*8 = 2-4 tonf acting a t G2.
Moments about Β give : R2 χ 15 = (9-6 x 7) + (2-4 χ 1 7 ) ,
= 67-2 + 40 -8 ,
R0 =
1 0 8 .
7-2 tonf .
S1 = (9-6 + 2-4) - 7-2,
= 4-8 tonf .
χ χ B y proportion, loading a t any section X X , wx = — x 0-8 = — .
Art oU
Weight of shaded portion to left of X X ,
Shear force on section Fx = Rx — Wx,
i.e.
giving
= 4*8 — —- and this is zero a t the 60
point of maximum bending moment
,
x = 16-97 f t .
1 3 8 S T R E N G T H 0 E M A T E R I A L S
ΜΓ = - ( 1 0 y 0 - 5 ^ = - 2 5 - 0 tonf f t . 1 0 _
(x + 6 )
Mx = Rlx - (x + 6) 0-5 2 ,
= 8-33x - 0-25(x + 6 )2,
= 8·33.τ - 0-25(a;2 + 12x + 3 6 ) ,
- 8-33^ - 0-25a;2 - Sx - 9 ,
- 5·33.τ - 0·25.τ2 - 9 .
B .M. on section X X between Β and C,
Mx = Βλ(χ - 9) - W^j} 1 6 - 7 7
2 / 1 6 - 9 7
Hence, a t χ = 1 6 - 9 7 , Mmax - 4 - 8 ( 1 6 - 9 7 - 9 ) 6Q I — —
1 6 - 9 73
- 3 8 - 3 - 2 7 - 2 ,
= 1 1 - 1 tonf f t .
When χ = 9 , i.e. a t B , Mv> = - — - - 4 - 0 5 tonf f t . 1 8 0
9 4 3
When χ = 2 4 , i.e. a t C, Mi: ( 4 - 8 χ 1 5 ) - - — = - 5 - 0 tonf f t . 1 8 0
E X A M P L E . F ind, for the system shown, in Fig . 1 3 4
(a) the load carried b y each support, (b) the position of the two points of contraflexure,
(c) the position and value of the maximum bending moment
between the supports.
Solution
Assuming the load (wL) to be concentrated a t the centre of OC :
Moments about A give :
R2 χ 2 4 = ( 4 0 χ 0 - 5 ) ^ - β ) ,
r2 =
20 *
14
= 1 1 . 6 7 t o n f , 2 2 4
and R1 = 2 0 - 1 1 - 6 7 = 8 - 3 3 t on f .
The graph of F is as shown.
MA - - ( 6 χ 0 - 5 ) - 9 - 0 tonf f t .
B E A M S I I I — S I M P L E S H E A R
At points of contraflexure Mx = 0 ,
0-25x2 - 5·33α; + 9 = 0 (changing signs) ,
x2 - 2h32x + 36 = 0 ,
21-32 ± | / [21-322 - 4(1 χ 36)]
or
χ =
_ 21-32 + ] /3122
2
- 20 f t2 or 1-83 ft to the right of point A.
139
FIG. 1 3 4
140 S T R E N G T H O F M A T E R I A L S
Again, dMxjdx = 5-33 - 0·5χ = 0 for max . B .M. ,
0-5x = 5 -33,
χ = 10-66 f t .
I m a x = (5-33 χ 10-66) - 0 .25(10-66)2 - 9 ,
- 57 - 28-6 - 9 ,
= 19-4 tonf f t .
Note tha t this is not the greatest bending moment in the system.
E X A M P L E . Draw to scale the graphs of F and M for the system
shown and find the position and value of the max . B . M .
0-5 tonf 1-5 tonf
F I G . 1 3 5
Solution
Moments about A give :
(R2 x 6) + (0-5 χ 4) = (4 χ 1)1 + (1-5 χ 4 ) (treating the Δ
distributed load as concentrated at i ts centre) .
6 i ? 2 = 8 + 6 - 2 ,
.·. R2 = 2 tonf ,
R1 = (0-5 + (4 χ 1) + 1-5) - 2-0 ,
= 4 tonf.
The graph of F is then as shown.
MA = - ( 0 - 5 χ 4 )
= - 2 - 0 tonf f t .
MB = - ( 0 - 5 χ 8) + (4-0 χ 4) - (4 χ 1)1 , = - 4 - 0 + 1 6 0 - 8 0 ,
= + 4 - 0 tonf f t .
B E A M S I I I — S I M P L E S H E A R 141
Mx = - [ 0 - 5 ( 4 + x)] + 4-Oa; - ( a i x l ) - ,
F
= - 2 0 - C )-5x - f 4-0# — ^r- , Δ
• - 2-0 .
Point of zero shear corresponds with point of max. B.M.
0 Π 1 3-5 f t Η
M 4 - l 3 = M m a x.
0
FIG . 136
When χ = 1, i f * = 3-5 - 0-5 - 2-0 = 1-0 tonf f t .
χ = 2 , Mx = 7-0 - 2-0 - 2-0 = 3-0 tonf f t .
x = 3, Mx = 10-5 - 4-5 - 2 0 = 4-0 tonf f t .
Note: that (1) M = 0 at point D so that at this point the beam is straight and (2) M is of different sign on either side of D so that there is a change in curvature here, the deflected shape being as shown dotted.
Since, for the part A B the equation for M is a continuous func-
tion, the value of χ which makes this function a maximum, can be
found by differentiating and equating to zero.
F o r any section X X between A and Β :
142 S T R E N G T H O F M A T E R I A L S
Hence,
Thus, 4 - ( Μ τ ) = 3-5 - — = 0 , i.e. χ = 3 - 5 f t . da; 2
J l f I l l ux = (3-5 χ 3-5) - - 2
= 12-25 - 6-12 - 2 ,
- 4-13 tonf f t .
(This is not necessarily the max . B .M. in the system)
FIG. 137
E X A M P L E . The simply supported beam shown, carries a distri-
buted load which increases uniformly from zero a t the L H support
up to a maximum of 1-5 tonf/ft.
Draw the graphs of shear force and bending moment, indicating
the position of the maximum bending moment and i ts value.
Solution
Mean loading Max. loading/2 — 0-75 tonf f t .
To ta l load = (9 χ 0-75) = 6-75 tonf .
Considering this as acting a t G (the centroid of triangle OAC) which
is 6 ft from 0 , Moments about Ο give :
9f t 6 f t
F I G . 1 3 8
(R2 x 15) = 6-75 χ 6 ,
i.e. R2 = 2-70 tonf .
and R% = 4-05 tonf .
B E A M S I I I — S I M P L E S H E A R 143
X ι η
- x l - 5
jpx = R1 - weight of piece O E D ,
= R1 — area under O D ,
= * < K ~ 2 9 X l
'5
) '
4 0 5 18
F
0
* 6-95 f t— —*\ \ I \
M
-18-76 = M,*».
1 \
1 1
0
FJCI. 1 3 9
F o r any section X X between 0 and A :
Loding wx =
144 S T R E N G T H O F M A T E R I A L S
When χ = 3 , Fx = 4-05 - l
'5 * ^ = 3-3 tonf .
When x = 5, Fx = 4-05 - * '5 *
5' = 1-97tonf.
lo
F o r max . B .M. , Fx = 0 , i.e. ^-x2 = 4 -05 ,
χ2 = 48-6 and χ = 6-95 f t .
The weight of the portion O E D may be considered to ac t a t Gx
(the centroid of triangle O E D ) which is \x from X X .
1*5 / 1 Hence, Mx = Rxx Το~
χ2\Τ
χ
18 \ ο
36
dM · χ2
Note: —-r— = 4-05 = — - = 0 for max. B.M., i.e. χ — 6-95 ft as above.) dx 12
33
When χ = 3 , Mx = (4-5 χ 3) - — - 11-4 tonf f t . x 36
x = 5 , = (4-05 χ 5) - — = 16-8 tonf f t .
ou Q3 x = 9, Mx = (4-05 χ 9) - — = 16-2 tonf f t .
6-053 χ - 6 - 9 5 , i V / - (4-05 χ 6-95) 36
= 28-1 - 9-34,
= 18-76 tonf f t .
E X A M P L E . A uniform beam 20 ft long weighing 5 tonf is hinged
at the L H end and is maintained in a horizontal position, by a
vertical prop at a point 12 ft from the hinge. Between hinge and
prop there is a distributed load of 0-75 tonf/ft while a t a point
4 ft to the right of the prop is a concentrated load of 4 tonf. Draw
to scale the graphs of F and M and find (a) the value and position
of the maximum bending moment between hinge and prop and
(b) the position of the point of contraflexure.
B E A M S I I I — S I M P L E S H E A R 145
Solution
The loading diagram is as shown, the reaction a t the hinge being
vert ical .
0-75 tonf/ft
JX I 0-2 tonf/ft (beam weight) 4 tonf
Hinge
R2(Prop)
FIG. 1 4 0
Moments about the L H end give :
R2 χ 12 = (0-75 χ 1 2 ) - ^ - + (0-25 χ 2 0 ) - ^ · + (4 χ 1 6 ) , Δ Δ
= 54 + 50 + 6 4 ,
= 1 6 8 . R2 = 14 tonf and R1 = (0-75 χ 12) + (0-25 χ 20) + 4
- 1 4 ,
= 9 + 5 + 4 - 1 4
= 4 tonf.
The graph of F is thus as shown.
FIG. 141
F o r any section X X between hinge and prop : Mx = 4tx — χ
(0-75 + 0-25) j
146 S T R E N G T H O F M A T E R I A L S
t o n f f t I
FIG. 142
E X A M P L E . The variation in load intensity over an aircraft wing
is given in the following table :
Distance from wing tip in feet
0 1 3 5 10 15 20 25 30
Load intensity in lbf/ft
0 27 47 60 78 88 95 98 100
Draw the graph of load intensity over the wing using a horizontal scale of 1 in. = 5 ft and a vert ical scale of 1 in. = 50 lbf/ft.
F rom this derive the graphs of F and M.
0 | 5 ÎO 15 20 25 30
Distance from wing tip in feet
FIG. 143
42
At χ = 4 , Mmax - (4 χ 4) - — = 8 tonf f t .
When M = 0 , 4# = ^ - , i.e. a; = 8 ft, giving the position of the Δ
point of contraflexure. The graph of M with the other principal values is as shown.
B E A M S I I I — S I M P L E S H E A R 147
Section at
Total area to left of section Shear force F = Total load = (A χ 250) lbf.
Section at No. of squares
η Area in in
2
A = rc/100
Shear force F = Total load = (A χ 250) lbf.
A 77 0-77 192 Β 77 + 140 = 217 2-17 542 C 47 + 170 = 387 3-87 967 D 387 + 184 = 571 5-71 1427 Ε 571 + 195 = 766 7-66 1914 F 766 + 199 - 965 9-65 2410
The graph of F against distance from wing t ip is then as drawn
below.
Oj 5 10 15 2 0 25 3 0
Fro. 144
The bending moment M a t any section is the area under the
shear force graph to the left of the section, i.e. f F dx, so tha t having
found the areas under each section of the graph (by counting squa-
res as before) they can be added as required and converted into
bending moments as in the following table :
Solution
The shear force F a t any section is the net load to the left of the
section, i.e. — j w dx so tha t having found the areas under each
section of the graph (by counting squares) t hey can be added as
required and converted into load as in the following table :
148 S T R E N G T H O F M A T E R I A L S
Section at
Total area to left of section Bending moment M = Fax lbf ft = {A x 5000) lbf ft
Section at No. of squares
η Area in in
2
A =
Bending moment M = Fax lbf ft = {A x 5000) lbf ft
A 6 0-06 300 Β 6 + 45 = 51 0-51 2550 C 51 + 74 = 125 1-25 6250 D 125 + 120 = 245 2-45 12250 Ε 245 + 170 = 415 4-15 20750 F 415 + 217 = 632 6-32 31600
The graph of M against distance from wing t ip is then as drawn
below.
Wing t ip Wing root
2 1 0 , 0 0 0
2 0 , 0 0 0
3 0 , 0 0 0
FIG. 145
Examples V
1. A beam is simply supported over a span of 20 ft and carries a uniform wall weighing 2 tonf/ft from the left-hand support to the centre of the span. Draw the graph of shear force and determine the position of the point of zero shear. Use this figure to calculate the maximum bending moment (7-5 ft from LH end. 56-25 tonf ft.)
2. A spring is to be made from steel strip 2-0 in. wide and, when deflected, is to be in the form of a circular arc of radius 54 in. Find the strip thickness at which the maximum tensile bending stress attains a value of 17,900 lbf/in
2
and calculate the bending moment exerted by the spring at this stress value (0-064 in., 245 lbf in.).
3. A total load of 5 tonf is distributed over a simply supported span of 20 ft in such a way that the intensity of loading increases uniformly from
B E A M S I I I — S I M P L E S H E A R 149
zero at the LH end. Determine the position and magnitude of maximum bending moment (12-8 tonf ft at 11-5 ft from the LH end).
4. Over a simply supported span of 30 ft the rate of loading on a beam increases from 1-0 tonf/ft to 3 tonf/ft. Calculate the position and size of the maximum bending moment (228 tonf/ft at 16-2 ft).
5. A joist 18 ft in length carries 2 tonf/ft over the central third from which value the loading rate decreases uniformly to zero at the ends. The supports are at 3 and 6 ft from the ends. Draw to scale the graphs of M and F and state the maximum values (13 tonf ft 10 tonf).
(i S M
C H A P T E R VI
BENDING WITH DIRECT STRESS
Composite Beams
A t imber jois t may be stiffened by a steel plate (or plates) clamp-
ed rigidly to i t so tha t the section is as shown in F ig . 146 :
1 L
(a) ( b )
FIG. 1 4 6
The jois t is then said to be " f l i t ched" .
Since both components bend to the same radius when the load
is applied, then
Common radius R = —ττ^~ =
where Et and Es are the elastic moduli of t imber and steel respec-
verely and 7 t and Is are the corresponding second moments about
the common neutral axis.
Also B . M . carried by t imber M. = LL
and B .M. carried by steel = AZl.
Tota l permissible B . M . for compound beam
M = Mt + Ms.
Note. The section should be symmetrical about the neutral axis and the sections of the component parts should be symmetrical about some vertical axis.
1 5 0
B E N D I N G W I T H D I R E C T S T R E S S 151
E X A M P L E . A steel plate f in. th ick is rigidly sandwiched between
two t imber jois ts as shown. I f the modular ratio (EsjEt) is 20 and
the maximum stress in the t imber is not to exceed 1000 lbf/in2.
calculate ,
(a) the corresponding
stress in the steel and
(b) the permissible uni-
form load (per foot) .
Solution
2nd Moment for t imber
4 χ 1 23
ΝΔ-
12
= 1152 in4.
- 4 in- - 4 in-
ί s:
Permissible Β . M .
fJt
1000 χ 1152
6
= 192,000 lbf in.
FIG. 147
where / t £ 1000 lbf/in2,
. 8 in 12 in
2nd Moment for steel I, = 0-75 χ
12
32 in4.
20 χ 32
1152
= 106,700 lbf in
χ 192 ,000 ,
Stress in steel / s — Msys
106,700 χ 4
32
13,300 lbf/in2.
6*
1 5 2 S T R E N G T H O F M A T E R I A L S
W L 8 And Ms = — — where 1KS = load carried by steel = — χ Ms
8 W L 8
Mt = — - — where Wt = load carried by t imber = — χ Mt.
Tota l load W Wt I- Ws,
= ^-(Mt + Ms),
8 ( 1 9 2 , 0 0 0 + 1 0 6 , 7 0 0 ) ,
- 9 9 5 5 lbf , i.e. 4 9 7 lbf/ft .
E X A M P L E . A flitched beam 1 0 ft long and simply supported
consists of a steel channel 8 in. deep a t tached to a t imber joist
8 χ 5 in.
F ind the max. permissible central load if (E^Et) = 2 0 and the
stress in the steel > 1 0 , 0 0 0 lbf/in2.
F o r the channel I = 4 6 - 7 2 in4.
Find also the max . bending stress in the t imber a t the load
Solution
8 in
- 5 in 7 S = 4 6 - 7 2 in
4 (given).
5 χ 83
1 2 '
= 2 1 3 - 3 in4.
/ My
R KU M,
Fm. 1 4 8
EtIt
F o r steel, 1 0 0 0 = M s χ 4
4 6 - 7 2 '
4 6 - 7 2 χ 1 0 , 0 0 0
4
1 1 6 , 8 0 0 lbf in .
Mt
= J - χ χ 1 1 6 , 8 0 0 , 2 0 4 6 - 7 2
= 2 6 , 6 5 0 lbf in .
B E N D I N G W I T H D I R E C T S T R E S S 153
.'. Stress in t imber 26 ,650 χ 4
" 213-3
= 500 lbf/in2.
WL Tota l bending moment = Ms + Mt = ^
W =L(3IS + Jf t),
~ 10 χ 12
(116,800 + 2 6 , 6 5 0 ) ,
_ 4 χ 143,450
~ ~~ Ï2Ô '
= 4780 lb f .
Reinforced Concrete Beams
Concrete is a material which is relat ively weak in tension. Steel
reinforcing rods are therefore embedded in tha t part of a beam
subject to tensile strain e.g., the upper layer in a cantilever and the
lower in a simply supported beam.
General design assumptions are :
1. Perfect adhesion between reinforcement and surrounding concrete.
2. The tensile load is carried by the reinforcement alone.
3. The tensile stress in the reinforcement is uniform over i ts
section.
4 . I n the concrete strain is proportional to stress.
5. Sections which are plane before bending, remain so during bending.
Since such beams are composed of two materials having diffe-rent moduli of elast ici ty, the simple bending formula (/ == My β)
does not hold good i.e., the neutral axis does not pass through the centroid of the section.
I f , as in F ig . 149(a ) , the depth of the neutral axis is h then the shaded area ^ r e p r e s e n t s the section of concrete under compression.
154 S T R E N G T H O F M A T E R I A L S
Reinforcing rods I (section on AB) (assumed
uniform)
(a) (b) (c) (d)
FIG. 1 4 9
Assuming tha t the strain a t any point is proportional to i ts
distance from the neutral axis, then
Compressive strain in concrete a t point A _ h
Tensile strain in steel d — h
I f the depth of the reinforcement axes is d then the rods are a t
(dr-h) below the neutral axis.
B u t ec = — and e s = where Ec anidEs are the elastic moduli c Ec Es
for conrete and steel respectively, so tha t
e s ECU °r d - h Ε J,
(1)
The quan t i t y -^ - is referred to as the Steel I Concrete Modular
Ratio.
I n the steel, Tensile force Ft = fsAs shere As = S tee l section.
The average stress in the concrete above the neutral axis is / c / 2
where / c is the maximum stress and this is distributed over the
area Ac = bh. fc
. ' . i n the concrete Thrust Fc = — bh.
Δ Since there is no resultant force normal to the section, the
tota l force (compressive) in the concrete must be equal and
B E N D I N G W I T H D I R E C T S T R E S S 155
opposite to the to ta l force (tensile) in the steel
i.e. F, = Fc,
or UAs = ffbh (2)
Since, for the concrete, the stress diagram is a triangle, the resul-
t an t force Fc may be considered to ac t a t the centroid of this tri-
angle i.e. a t f h from the neutral axis . Similarly the tensile force
in the steel (P) ac t s a t a distance (d-h) from the neutral axis. Since
these forces are equal they form a couple, the arm of which is given
b y 2 a = —- h + (d — h), ο
or a — d — •— ό Hence, Magnitude of couple (i.e. Moment of resistance of beam),
M = Fta
-/•*(«-!•). M Alternat ively, Fca,
kbh
While the beam remains in equilibrium under the forces acting on
it , the Moment of resistance of the beam is equal and opposite
to the bending moment acting on the beam, i.e. to M.
Note. From Equation (2):
and from Equation ( 1 ) :
A - A . bh
Ύ K = ( Is Fs \-
d-h bh2
2m(d - h) bh2
-d bh Ύ ' where m = Er.
2md — 2mh
.·. bh2 = 2mdAs - 2mAJi
156 S T R E N G T H O E M A T E R I A L S
bh2 + (2mAs) h - (2mdAH) = 0 .
This is a quadratic equation which fixes the value of h and contains
only known quanti t ies.
E X A M P L E . The horizontal flange of a T-beam is 4 in. th ick and
60 in. wide, and the reinforcement in the rib is 15 in. below the upper
surface. I f the stress limits are 16,000 and 600 lbf/in2 8LndEJEc = 15
and the neutral axis is to coincide with the lower edge of the flange,
calculate :
(a) the required section of reinforcement,
(g) the moment of resistance of the beam,
(c) the actual max . stress in each material .
Solution h
d - h ~~ EJS
4 1 5 / c
15
f h = 4 i n Γ d = l5 in f
I
-b = 6 0 in ·
ΦΦΦΦ;
FIG. 1 5 0
/ s ^ s —
_ fc bh 60 χ 4 1
41-25 2-91 in
2.
Moment of resistance M = fsAs (d — — j ,
16,000 χ 2-91 15 -3 / '
= 16,000 χ 2-91 χ 13-67,
= 636,000 lbf in.
B E N D I N G W I T H D I R E C T S T R E S S 157
Note: If / c is taken as 6 0 0 , then / S = 6 0 0 χ 4 1 - 2 5 =
= 2 4 , 7 5 0 lbf/in2. which is outside the given limit.
E X A M P L E . A concrete slab 7-5 in. th ick has a span of 10 ft and is
reinforced by steel rods 0-5 in. dia., 6 in. apart at 1-5 in. above the
lower surface.
Es t ima te the load which may be carried per square foot, addi-
t ional to i ts own weight, given t h a t : m a x stress in steel >> 18,000 Ε
lbf/in2 and max . stress in concrete > 700 lbf/in2. —^- = 12. Densi ty
of material = 150 lbf/ft 3 (L .U. ) . E*
Solution
i K— b= 12 in—Η
t N-T— 7-5 in Τ ,
J d=6in
1-5 in
0-5 in dia. î
FIG. 1 5 1
Considering a strip of floor 1 ft wide and parallel to the reinforce-ment :
As = 2(0-785 xO-52) = 0-393 in
2.
2 ' U 2 Λ '
12h
2 χ 0-393 '
12h
and
0-785 '
h Esfc fs 12(6 - h)
d - h Ecf/ U * 72 - 12A
h
12h 72 - 12h Equat ing, Q^ ft ,
i.e. 12h2 f (12 χ 0-785) h - (72 χ 0-785) = 0 ,
h2 + 0-785/* - 4-71 = 0 ,
whence h = 1-81 in .
6 a S M
1 5 8 S T R E N G T H O F M A T E R I A L S
/ 1 2 H e n C e ' t = ¥ 7 8 5
X L-
8 1'
= 2 7 - 7 so tha t if the steel is stressed to its maximum
of 1 8 , 0 0 0 lbf/in2 then,
stress in concrete fc = — ' „ = 6 5 0 lbf/in2
/ c 27.7 ' which is within the permitted figure of 7 0 0 lbf/in
2.
M = fsAs ( d - j } = 1 8 , 0 0 0 χ 0 - 3 9 3 ^ 6 -
WL = 3 8 , 1 0 0 lbf in, and M = ——
.·. To ta l load W{ = 8
* = 2 5 4 0 lbf. 1 0 χ 1 2
B e a m weight Wh = ( 1 0 χ 1 ) - ^ - χ 1 5 0 = 9 3 7 lbf. IΔ
Useful load W = 2 5 4 0 - 9 3 7 = 1 6 0 0 lb over a span 1 - 0 ft wide,
= 1 6 0 lbf/ft2.
E X A M P L E . A concrete beam 1 2 in. wide 1 5 in. deep is reinforced
by 6 steel rods 0 - 7 5 in. dia., the centres of which are a t 2 in. from
the bot tom of the beam which is simply supported over a span of
1 8 0 in. Assume a maximum stress in the concrete of 6 0 0 lbf/in2
and determine
(a) the depth of the neutral axis, (b) the moment of resistance of the beam,
(c) the stress in the steel, (d) the to ta l permissible uniform loading.
Make the usual assumptions and take the steel/concrete modular
rat io as 1 5 .
Solution
Since / S A = l§-bh, where ^ S = 6 ( 0 - 7 8 5 χ 0 - 7 52) = 2 - 5 6 in
2.
Δ
whence / S = 1 3 6 0 / L
B E N D I N G W I T H D I R E C T S T R E S S 159
Again, — - — = -—- χ - ~ and = Modular ratio = 1 5 , a Lc fs Jtbc
_h / 600 \ _ 6-62
\ — h~ \ 1360A / ~ A 9
' ' 13 -
so tha t A2 = 6-62(13 - A)
or A2 + 6-62A - 86
= 0, giving
A = 6-55 in.
Moment of resistance,
* - τ » ( ' - ϊ ) · 600
χ 12
- « — b= 12 in •
= 13 in - N.A.
/ t \ / t \ A\ /+\ /T\ st\
î h
ι ι V|/ \l) ψ ψ \\} si/ 2 in
/ 6-55 \ x e - β β 1 8 - - J - ) ,
= 3600 χ 6-55 χ 10-8 ,
= 255 ,000 lbf in.
Stress in steel / s = 1360A,
= 1360 χ 6 -55 ,
= 8900 lbf/in2.
FIG. 1 5 2
Bending moment permissible, M = WL
8
.'. Permissible load
Or, load per foot
T Tr 8 χ 255 ,000 B M J_ £ W = —rz τη— = 5-05 tonf.
15 χ 12
5 0 5 w = 180/12
= 0-336 tonf or 750 lbf.
Bending Combined with Direct Stress
The direct compressive stress induced in a short s trut of un-
symmetrical section A b y a compressive load F act ing a t the sec-
t ion centroid G is given b y
/ - F
Αι - ~ Γ · 6 a *
160 S T R E N G T H OF M A T E R I A L S
291 ·*>Ή
B E N D I N G W I T H D I R E C T S T R E S S 161
This may be represented by a horizontal straight line 'mn'
(Fig. 153(a)) normal to the strut axis and distant an amount f(l
(to some convenient scale) from a datum MN.
I n practice, no load acts truly a t the centroid of a section, there
being inevitably some eccentr ic i ty a: (as in F ig . 153 (b)) introducing
a bending moment given b y
M - Fx.
This causes bending about the neutral axis (N.A.) of the section,
so tha t the strut deflects as in F ig . 153(c ) . The bending stress,
which is clearly compressive on the load side of the neutral axis
(i.e. is of the same sign as the direct stress) and tensile on the oppo-
site side, may be represented by the straight line pq intersecting
mn a t the neutral axis . The two maximum values are respectively,
My, ^ , . x / b = —-— at Q (compressive),
and fb = - ^ p - a t Ρ ( tensile) .
The resultant stress is obtained by superimposing these values on
the direct stress,
F My so tha t a t Q : / m ax = — H γ^- (compressive),
-, -p. £ F My2 ί F My2 \ and a t P : / m in = — — ί compressive since — > — — 1.
I f the eccentr ic i ty χ is increased for a given value of F unti l the tensile bending stress (My2\I) is equal to the direct stress, then the resultant stress a t Ρ will be zero as in Fig . 153(d) . Any further increase in χ will result in a net tensile stress a t Ρ (Fig. 153(e)) although the applied load F is compressive. The stress distribution is represented by the shaded area in Figs . 153 (a-e) and shows the effect of increasing eccentr ic i ty for a given load, i.e. of increasing the bending moment. In general, the resultant stress is the alge-braic sum of direct and bending stresses and is given by
162 S T R E N G T H O F M A T E R I A L S
I f , instead of increasing the eccentr ic i ty x, the bending moment
is increased by increasing F while keeping χ constant , the direct
stress (F/A) is increased in the same proportion as the bending
IF (increased)
FIG. 1 5 4 FIG. 1 5 5
stress (Fxyjl) so tha t the shape of the stress distribution diagram
is unaltered, although all stress values are increased as shown in
F ig . 155.
Section Modulus
I n the theory jus t given,
Max . compressive bending stress = My1/I = ~JJjj~~
The quant i ty l\yx is referred to as the Compressive Section Modulus
and denoted b y Zv Similarly I\y2 is denoted b y Z2 and called the
Tensile Section Modulus. Thus , where a component is subjected
to a combination of direct and bending stresses, the equation for the
resultant stress m a y be writ ten
, F M
where Ζ is the relevant modulus. The positive sign will give the maximum stress and this will
be of the same kind as the direct stress. The negative sign will give the least stress, the sign of which will depend on the relative magnitudes of the direct and bending stresses.
Clearly, for a section which is symmetrical about the neutral axis, the two values of Ζ are the same.
B E N D I N G W I T H D I R E C T S T R E S S 1 6 3
E X A M P L E . The axis of the load on a tensile test piece is 0 - 0 0 8 in.
from the geometrical axis. I f the tes t piece diameter is 0 - 7 5 in.
and the load a t the first sign of yielding is 9 - 1 6 tonf, find the maxi-
mum corresponding stress. Show diagrammatically how the stress
varies across the section.
FIG. 1 5 6 FIG. 157
Solution
Sect ion
Direct stress
Bending moment
2nd Moment
Distance from N.A
A = — χ 0 - 7 52 = 0 - 4 4 2 in
2.
4
F
1 9 1 6
0 - 4 4 2
M = 9 - 1 6 χ 0 0 0 8 ,
- 0 - 0 7 3 3 tonf in
π
2 0 - 7 8 tonf/in2.
7 = • _ χ 0 - 7 5 4 , 6 4
0 - 0 1 5 5 in4.
0 - 7 5 0 - 3 7 5 in.
164 S T R E N G T H O F M A T E R I A L S
.'. Sect ion modulus Ζ = — ,
Ί . . M 0 - 0 7 3 3 -, —_ , . . . 2 Bending stress — = = 1 - 7 7 tonf/m2.
Ζ 0 - 0 4 1 3
l r F M Max. stress / m ax = — + — ,
- 2 0 - 7 8 + 1 - 7 7 ,
= 2 2 - 5 5 tonf /in2.
Min. stress / I N IN = 2 0 - 7 8 - 1 - 7 7 ,
= 1 9 - 0 1 tonf/in2.
The stress distribution is then as shown in F ig . J 5 7 .
E X A M P L E . A short I-section column has a section of 1 0 - 3 in2 the
depth of which is 8 in. Draw to some convenient scale a diagram of
the stress distribution across the section when a vertical load of
2 0 tonf is applied on the web at a distance of 2 - 5 in. from the neutral
axis. Take / as 1 1 0 in4.
Solution
Bending stress *
_ 2 0 χ 2 - 5
1 1 0 / 4 '
= 1 - 8 2 tonf/in2.
F 2 0 Direct stress = — = ——- = 1 - 9 4 tonf/in
2.
A 1 0 - 3
F M Resul tant stress / = — ± —— ,
A Ζ = 1 - 9 4 ± 1 - 8 2 .
/max = 1 - 9 4 + 1 - 8 2 ,
= 3 - 7 6 tonf/in2 (at A , F ig . 1 5 8 ) .
/ M IN = 1 - 9 4 - 1 8 2 ,
- 0 - 1 2 tonf/in2 (at B ) .
B E N D I N G W I T H D I R E C T S T R E S S 165
^ 2 - 5 i n - » j2 Q
tonf
— 8 i n
FIG. 1 5 8
Datum
E X A M P L E . The maximum stress in a hollow tie 3 in. outside dia. and 0-5 in. th ick is not to exceed the mean by more than 20 per cent. Es t imate the permissible eccentr ic i ty of the load.
Solution
F o r the section, 1-5in, (Fig. 159)
π
y
' = £ - ( 34
- 24
) , 64
Ζ
A
£ ( 8 1 - 1 6 ) ,
3-19 in4
I_ _ 3-19
y ~ 1-5
0-785 ( 32 - 2
2) ,
3-93 in2.
= 2-127 i n3.
166 S T R E N G T H Ο Ε M A T E R I A L S
Max. stress = 1·2 χ Mean stress (where the Mean stress is F/A)
F M
Τ + ΊΓ M
Ζ
Fx
F
l^Z- -4- and M = Fx. A A
F_
T'
- 0 - 2 Ζ A '
0-2 χ 2 1 2 7
3-93 = 0-108 in.
\
— ι
FIG. 1 5 9
E X A M P L E . A wind pressure of 40 lbf/ft2 of projected area is to
cause no tension in the base section of a pillar built of stone having
a density of 140 lbf/ft3. I f the diameter is to be 6 ft, est imate the
height to which i t may safely be built.
Solution I [T]l r~
Weight of column
W = (0-785 x 62
) ä x 140, eftdia.-^ « -
= 3960Λ lbf.
Sect ion of column h
A = 0-785 χ 62,
= 28-25 f t2.
Direct stress
W 3960ft I
A ~ ^ 2 5 " /77mW77ffM7 = 140ft lbf/ft
2 (compressive). y I G \QQ
B E N D I N G W I T H D I R E C T S T R E S S 167
Projec ted area = 6A f t2.
To ta l wind force = 6h χ 4 0 ,
= 240JUbf .
Bending moment, M = 240h χ - ^ ,
= 1 2 0 A2l b f t .
π χ 64
2nd Moment of section J = — — — = 63-7 f t4 and y = 3 f t .
64
Bending stress My 120Ä
2 χ 3
63-7 '
5-65A2 lbf/ft
2.
W My
Resul tant stress = — ± —j— (and this is to be jus t
zero on the windward side),
W_ My
UOh = 5-65A2,
i.e.
140
24-8 f t .
E X A M P L E . Es t ima te the value of tensile load which will induce
a maximum stress of 10 tonf/ in2 in a 1·5 in. dia. steel rod. The axes
of load and rod are parallel and 0-25 in apart. Determine graphically
the distance from the rod axis a t which the resultant stress is zero.
Solution
R o d section A = 0-785 χ 1-52,
= 1-77 in2.
FIG. 161
168 S T R E N G T H O F M A T E R I A L S
2nd Moment π χ 1-5
4
6 4 ~ ~ '
0-248 i n4.
Bending moment M = 0-25F lbf in .
M 025F Bending stress Ζ 0 -248/0-75 '
= 0 - 7 6 ^ lbf/in2.
F F Direct stress -—
A 1-77 '
0 - 5 6 ^ lbf/in2.
Taking tensile stress as positive,
_ F_ M /max - - χ + ^ - '
0 - 5 6 /7 + 0-762<\
1-32^, whence F = 7-56 tonf.
0-56F - 0 - 7 6 ^ ,
- 0 - 2 χ 7-56,
—1-512 tonf/in2 (compressive).
F rom the stress distribution diagram, the resultant stress is zero
a t a distance of 0-55 in. from the axis of the bar, on the side opposite
to the load axis.
FIG. 1 6 2
/min
B E N D I N G W I T H D I R E C T S T R E S S 169
E X A M P L E . W h a t load may be applied to the cranked strut shown
if the maximum tensile stress is not to exceed 6 tonf/in2? At this
load, what is the maximum stress in the material?
Solution
Sect ion A
2nd Moment
/
χ 42, π
Τ 4π i n
2.
π
64
4π in'
χ 44,
Modulus Ζ = —
4π
ΊΓ' 2π i n
3
M 6F
Ζ ~ ~2π Μ = 6F, .·. Bending stress
and Direct stress
Resul tant stress
Taking compressive stress as positive
- 6
F F
Α
F 6F
4π 2π
F 6F
4π 2π
fd-τ)- ·
F 6π
¥ 7 5 '
- 6 in-
- 4 in d ia .
IF
FIG. 1 6 3
- 6-85 tonf .
170 S T R E N G T H O F M A T E R I A L S
• , F 6F Max. compressive stress = - j — - + ,
1
2π
6-85 + (6 χ 6-85)
= - ^ - ( 3 - 4 2 + 4 1 - 1 ) ,
44-5
7-1 tonf/in2.
E X A M P L E . A short column having the section shown supports a
concentrated load of 20 tonf applied a t point O. Calculate the stress
a t points A , B , C, and D taking the web and flanges as rectangular.
0-9 in
0 5 in -
10 in X -
- 6 i n -
|Y ID
| y
FIG. 1 6 4
2t
Solution
Sectional area A = 2(0-9 χ 6) + (8-2 χ 0 -5) ,
= 14-9 in2.
B E N D I N G W I T H D I R E C T S T R E S S 171
Direct stress 20
14-9 = 1-34 tonf/in
2.
6 χ 1 03 5-5 χ 8 ·2
3
12
500 - 2 5 3 ,
247 in4.
12
/ 0 - 9 x 63\
I i 2 r 8-2 χ 0 -5
3 / 0 - 9 χ
L 2
12 ^
= 0-855 + 32-4 ,
= 33-26 in4.
Bending moment Mx = My = 20 χ 2 ,
= 40 tonf in .
Max. bending stress about X X , fx = Myx
4 0 χ 5
where yx = 5 ,
247 '
= 0-81 tonf/ in2.
4 1 3 From the diagram
f A= - 3 - 0 7 t o n f / i n2
V - 1 - 4 5 t o n f / i n2
f c= + 5 - 7 5 t o n f / i n2
M f D= + 4 - l 3 t o n f / i n
2
FIG. 1 6 5
1 7 2 S T R E N G T H O F M A T E R I A L S
Solution
The distance of the neutral axis from the edge of the section
nearest to point A is given by :
A
( 1 5 χ 1 ) 0 - 5 + 2 ( 8 χ 1 ) 5 + 2 ( 4 - 5 χ 1 ) 9 - 5 + 2 ( 4 χ 1 ) 1 2
+ 2 ( 2 χ 1 ) 1 4 - 5
- ( 1 5 χ 1 ) + 2 ( 8 χ 1 ) + 2 ( 4 - 5 χ 1 ) + 2 ( 4 χ 1 ) + 2 ( 2 χ 1 )
_ 7 - 5 + 8 0 + 8 5 - 5 + 9 6 + 5 8
1 5 + 1 6 + 9 + 8 + 4 '
3 2 7
~" 5 2 '
= 6 - 2 9 in. ( ~ 6 - 3 in .) .
My j Max. bending stress about Y Y , fy = ——— where yy =-. 3 ,
_ 4 0 χ 4
3 3 - 2 6 '
= 3 - 6 0 tonf/ in2.
The stress distribution across faces A B , B C , CD and D A due to
the direct stress / and the bending stress fx combined is represented
by the dotted lines F G , GH, H E and E F respectively.
Superimposed on this is the bending stress fy represented on
faces D A and B C by the full lines MN and P Q drawn on E F and
GH as datum respectively. The resultant stress distribution over
faces A B and CD is then given b y the full lines N P and QM.
E X A M P L E . The main pillar of an hydraulic press has the section
shown, (Fig. 1 6 6 ) the load being applied a t point A. Es t imate the
values of the max . tensile and compressive stresses in the material
when the operating load is 5 0 tonf and draw to some scale the stress
distribution diagram.
B E N D I N G W I T H D I R E C T S T R E S S 173
Fo r a rectangle about an edge / = — - . Considering half the section : ό ^N.A. = I [(7-5 χ 6 ·3 3) - (6-5 χ 5 ·3 3) + (3-5 χ 3·7 3) - (2-5 χ 2 ·7 3)
+ (2 χ 8·7 3) - (1 χ 7·7 3) - (1 χ 2 · 7 3) ] ,
= | ( 1875 - 968 + 177 - 48 + 1310 - 456 - 2 0 ) ,
1870 - _
3 ~ '
= 623 in4,
Ι5 ίη 8 in 4 i n
-15 in -- 1 0 in-
- 6 in-
1
FIG. 1 6 6
3-5in7î
yc = 8-7 in
- 4 - 2 9 i n X ^ ^ ^ ,
F I G . 1 6 7
-£^mm^X 3-11 in Datum
174 S T R E N G T H O F M A T E R I A L S
Hence for whole section, / n . a . = 1247 in4.
Direct tensile stress = ^— = = 0-962 tonf/in2,
Bending moment , M = 50(6 + 6-3) ,
= 50 χ 12-3 ,
= 615 tonf in .
Mv 615 χ 6·3 Tensile bending stress a t Τ = —p-= — = 3-11 tonf/in
2.
/ 1247 •
Resul tant stress a t Τ = 0-962 + 3 -11 ,
= 4-072 tonf/in2 ( tensile).
. t ν x j. η ~ M y c 615 χ 8-7 Compressive bending stress a t C = — = ————
r ° / 1247 = - 4 - 2 9 tonf/in2.
Resul tant stress a t C = 0-962 - 4 -29 ,
= —3-33 tonf/in2 (compressive).
E X A M P L E . An I-section jois t 3 χ 5 in. overall has a web and
flange 0-3 in. thick. F ind the compressive stress when a short length
is axially loaded with 5 tonf. Determine the stress distribution when
one of the flanges is removed to give a T-section, if the position and
size of the load remain the same.
Solution
Area of section, A = (2 χ 3 χ 0-3) + (4-4 χ 0 -3 ) ,
= 1-80 + 1-32,
= 3 1 2 in2.
5 Compressive stress, / = - Ξ - Τ ^ Γ - ,
= 1-6 tonf/in2.
Reduced area of section A = (3 χ 0-3) + (4-4 χ 0-3)
= 0 - 9 + 1-32 = 2-22 in2.
Depth of neutral axis h = A
(3 χ 0-3) 0-15 + (4-4 χ 0-3) 2-5
1-55 in.
B E N D I N G W I T H D I R E C T S T R E S S 175
Eccent r ic i ty of load, χ = 2-50 — 1-55,
= 0-95 in .
0-3inj^
5 in 4 - 4 in
Original section
Ν —
FIG. 1 6 8
Bending moment, M = Fx,
= 5 χ 0-95,
- 4-75 t o n f i n .
J N . A . = έ (3 x 1-553) - 1 (3 χ 1-25
3) + è(0-3 χ 1·25
3)
+ J (0-3 χ 3 · 1 53) = 5-1 in
4.
/ m a x , g . + = 4 + i Z ^ i £ = 2 . 2 5 + 2 . 9 4 = g . 1 9 t o n f / i n , .
/ m in = | . 2 - 2 5 - ^ 2 - 2 5 - 1-45 = 0-80tonf/in*,
The stress distribution is therefore as shown above.
176 S T R E N G T H O F M A T E R I A L S
E X A M P L E . F ive drums each 4 ft dia. and weighing 6 cwt. are
s tacked as shown. The chassis of the wagon consists of two beams
of uniform and symmetrical section having vertical ends of similar
section rigidly at tached.
FIG. 169
The springs have a span of 2 ft, the ends being 1 ft from the
inside edges of the vert ical members. Determine the loads on the
horizontal and vert ical members, neglecting friction, and draw
the bending moment diagram. F ind also the section modulus for
each beam if the maximum bending stress is not to exceed 6 tonf/in2.
Solution
F o r upper drum A, 2F1 cos 30 = W,
W 1 , Θ
* 1
2cos 30 '
F o r lower drum B , F2 = W + 2F1 cos 3 0 ,
W
2cos 30
= 2W,
--= 12 cwt .
B E N D I N G W I T H D I R E C T S T R E S S 177
w
F o r lower drum C,
FIG.170
Fs = W + Fx cos 3 0 ,
W = W +
= 1-5JF,
= 9 cwt .
2cos 30 cos 3 0 ,
F± = 2 ^ sin 30
W
2cos 30 sin 3 0 ,
W = — t a n 3 0 ,
Δ
- 3 χ 0-577,
= 1-732 cwt .
Load per support
= 1 ( 9 + 1 2 + 9)
— 7-5 cwt .
Load in cwt
.1-73
Γ Β I D
9 12
7-5 7-5
- 3 8 7
1-73
7-5 7-5
2 3 0 0 = M m
FIG. 171
1 7 8 S T R E N G T H O F M A T E R I A L S
Hence, Ma - Mh - - ( 1 - 7 3 2 χ 1 1 2 ) 2 - - 3 8 7 lbf ft
Mc = - 3 8 7 + ( 7 - 5 χ 1 1 2 ) 1 = + 4 5 3 lbf ft
Md = - 3 8 7 + ( 7 - 5 + 1 1 2 ) 2 - ( 9 χ 1 1 2 ) 1
= + 2 8 5 lbf ft
Mc = - 3 8 7 + ( 7 - 5 χ 1 1 2 ) 5 - ( 9 χ 1 1 2 ) 4
+ ( 7 - 5 χ 1 1 2 ) 3 = + 2 3 0 0 lbf f t .
The bending moment graph is then as shown in Fig . 1 7 1 , being
symmetrical about the wagon centreline.
E X A M P L E . A short tube 0 - 9 in. outside dia. 0 - 1 in. th ick is to
carry a load the axis of which is parallel to and 0 - 5 in. from the
tube axis. Use the values of working stress given and est imate the
safe load assuming the tube to be made from (a) cast iron, (b) mild
steel. C.I. : 5 0 0 0 lbf/in2 in tension, 2 0 , 0 0 0 lbf/in
2 in compression.
M.S. : 1 5 , 0 0 0 lbf/in2 in both tension and compression.
Required modulus
2 3 0 0 χ 1 2
6 χ 2 2 4 0
= 2 - 0 5 i n3.
A
F F
- f r 1— Datum
Fia. 172
Solution
Inside diameter of tube = 0 - 7 in.,
Sect ion A = ^ ( 0 - 92 - 0 - 7
2) = 0 - 2 5 1 in
2.
B E N D I N G W I T H D I R E C T S T R E S S 179
2nd Moment of area / = (0-94 - 0-7
4) = 0-0204 in
4,
b4
0-0204 .'. Sect ion modulus Ζ = „ = 0-0454 in
3.
0-45
Bending moment M = 0-5F lbf in where F = Load.
Taking compressive stress as positive :
F M For cast iron, / m ax = — + ——-, where / m ux = 20,000 lbf/in2,
F 5F Ζ. 20 ,000 = - — - +
F
0-251 0 0 4 5 4
1 5
0-251 0 -0454 ,
= i^(3-99 + 1 1 0 )
= 1 4 - 9 9 ^ whence F = 1330 lbf .
A j , _ F M
And /min — ~^
- 5 0 0 0 = F(3-99 - 11-0) ,
= - 7 - 0 1 ^ whence F = 715 lbf .
The smaller of the two values of F is the safe load for cast iron.
Similarly, for mild steel, we have as before :
either / m ax = 1 4 - 9 9 ^ , but in this case, / m ax = + 1 5 , 0 0 0 lbf/in2,
. 15,000
·' ~ 14-99 '
= 1000 lbf approx.
or / m in = -101F, where / m in = - 1 5 , 0 0 0 lbf/in2,
15,000
7 0 1 '
2140 lbf .
The smaller of the two values of F is the safe load for mild steel.
180 S T R E N G T H O F M A T E R I A L S
Examples VI
1. Two rods 0-75 in. in diameter are used to reinforce the section of a concrete cantilever 5 in. wide and are positioned with their centres at a depth of 10 in. Calculate the bending moment required to induce in the con-crete a stress of 560 lbf/in
2 given that Ec = 1-79 χ 10
6 lbf/in
2 and that
Es = 15 Ec (59,500 lbf/in).
2. The horizontal flange of a T-section concrete beam is 18 in. wide and 4 in. thick and is reinforced by 8 rods 3/8 in. dia. with their axes 1-0 in. from the surface furthest from the web. If the web is uppermost and 4 in. thick and the total depth of the section is 18 in., calculate the moment of resistance of the beam assuming stresses of 10,000 and 500 lbf/in
2 and taking
the modular ratio as 12 (Reinforcement 128,900 lbf/in., concrete 105,300 lbf in.).
3. At a certain point the section of a casting is represented by the area enclosed by two circles 10 and 7 in. dia. with centres 1-0 in. apart. Calculate the 2nd Moment of area of the section about an axis normal to the line join-ing these centres and hence find the two values of the section modulus (298 in
4, 73-8 and 50-0 in
3).
4. A square section beam is loaded so that the neutral axis coincides with a diagonal. Show that the section modulus may be increased by removing the upper and lower apices of the section and that the maximum modulus value is obtained when the vertical diagonal has been reduced by one ninth.
5. That part of a G-clamp parallel to the screw axis is of T-section 0-125 in. thick throughout. The flange, which is nearer the screw, is 0-625 in. wide and the total depth is 0-75 in. Calculate the maximum values of tensile and com-pressive stress induced by a clamping load of 75 lbf given that the screw axis is 2-75 in. from the tip of the web (5650 and 9900 lbf/in
2).
C H A P T E R VII
TORSION
Modulus of Rigidity
The application of forces F F to the cube A B C D in Fig. 173 causes
the facte A B to turn through a small angle φ radians to the dotted
FIG. 1 7 3
position A B r This angle is defined as the shear strain.
χ Thus, Shear strain φ = radians (see Chapter I ) .
Within the elastic limit, the ratio Shear stress fs
is a constant . Shear strain φ
This constant is called the Modulus of Rigidity and denoted by G.
FIG. 1 7 4
7 S M 1 8 1
182 S T R E N G T H O F M A T E R I A L S
Thus 0 ^- or ψ - i l ψ u
Consider a solid shaft of radius R to be acted on by equal and oppo-
site couples, as shown in Fig . 174, so tha t the radius OA twists
through 0 radians with respect is the other end and adopts the
dotted position O B (Fig. 175). The state is one of pure shear.
FIG. 1 7 5
Clearly, Arc A B == R6 = Lcp,
R6 Λ '
/, _ RO . G6 χ R Λ _ _ _ , !.e. /, - L ,
= kR for a given value of Θ.
Thus for a given twist on a given shaft, the shear stress fs is propor-
tional to the radius, so tha t zero stress obtains a t point 0 .
f GO The equation is normally written — = -γ-, where r has any
value between 0 and R.
I t follows tha t if,
f's = stress a t any radius r, F ig . 176
fs = max. stress a t radius R,
T O R S I O N 183
then ' · ' '· 4 · · . R
Shear force on elemental ring
- f's2nr dr,
— — r 2ir.r j r 2nr dr, R
= -^.2nr2 dr.
Moment of this force about the polar axis
= ~^-2nr2dr x r ,
R
= -4 * 2 j w3d r .
R
Tota l moment of resistance to shear
η
Τ = A x2njr*dr,
= Ι χ 2 π ' Τ '
nR*
Hence, T =
R
nR*fK _ nD3fs
2 ~ 16
R
Radius
FIG. 1 7 6
(lbf in, if D is in inches and fs is in lbf/ in
2) .
This gives the torque which can be t ransmit ted by a given dia-
meter shaft for a given maximum shear stress.
Note: This expression does not apply to hollow shafts, for which see later.
F r o m * above, Τ = L R
nR* . The quant i ty in the brackets ,
nR* nD* i.e. —-— or _ is the polar 2nd moment of area of the shaft 32 section, J.
7*
184 S T R E N G T H O F M A T E R I A L S
f±=G6_
R L
Thus, Τ = fs R
χ J ,
i.e. fs R
Τ
T' but
Hence, fs R
Τ
Τ GO
17
This is known as the Torsion Equation and is similar to tha t for
Simple Bending.
Relation between Speed and Shaft Diameter for a Given Max.
Stress and Power
The power t ransmit ted by a shaft carrying a torque Τ lbf ft and
rotating at Ν rev/min is given by
_ 2πΝΤ
'Ρ' "~ 33,000 *
Hence, y = 33 ,000 (h.p.) 1 2π Ν
kx (-^r j if the power is constant .
F o r a solid shaft diameter d :
16 where fs = Max. shear stress,
OR d3 = χ T
7,
= k2T if fs is constant,
/. Nd* = k for a given power and max. stress.
Ï O H S I O N 185
I t follows that , for any given power and permissible shear stress,
the higher the speed, the less the required shaft diameter.
Suppose tha t 100 h.p. is t ransmit ted b y a 3 in. dia. shaft a t
50 rev/min. The diameter required to t ransmit this power a t 50 ,000
rev/min can be found from
E X A M P L E . Determine the shaft diameters required to transmit
15 h.p. a t the following speeds:
50
100 500 The stress in the shaft is not to exceed 3 tonf/in
2 and the
1000 angle of twist is not important . Assume pure torsion. 2000
5000 rev/min.
P lo t D vertically full size, against Ν horizontally and find from
the curve, the diameter of shaft necessary to transmit this power
at 300 rev/min.
Solution
r p nD*fH 33 ,000 h.p. Τ = — Γ 7 Γ - = 0 Λ7 x 12 lbl / in .
16 2πΝ
33,000 χ 15 χ 12 χ 16
2 π2 χ 3 χ 2240
7 1 6 .
Hence, D3 = , and Fig . 177 shows the graph.
i.e.
d - 50 ,000
6 '
~ 1 03 '
d = = 0-3 in.
D3N =
186 S T R E N G T H O F M A T E R I A L S
Ν
rev/min
7 1 6 = Iß
Ν D in.
5 0 1 4 - 3 0 2 - 4 3
1 0 0 7 - 1 6 1-93
5 0 0 1-43 1 1 3
1 0 0 0 0 - 7 1 6 0 - 8 9
2 0 0 0 0 - 3 5 8 0-71
5 0 0 0 0 1 4 3 0 - 5 2
2 0 0 0 3 0 0 0 4 0 0 0
Ν r e v / m i n
FIG. 177
E X A M P L E . A 4-0 in. dia. shaft 100 in. long rotates a t 1500 rev/
min.
Calculate (a) torque being transmitted, (b) power being trans-
mitted, (c) to ta l twist of shaft in degrees given tha t G = 12 χ ΙΟ6
lbf/in2 for the shaft material and tha t the measured maximum
shear stress is 10,000 lbf/in2.
Solution nd*
Polar 2nd Moment of shaft section J ----- ,
_ π x 44
= 32 '
25-1 in4.
T O R S I O N 187
Σ = 4* > Torque on shaft Τ = ^ ,
10,000 χ 25-1
2
- 125,000 lbf in. ,
- 10,400 lbf f t .
Power t ransmit ted = 2πΝΤ
33,000 '
2π χ 1500 χ 10,400
33 ,000
= 2970 h.p.
— = Angular twist θ = where R = 2-0 in. ,
10,000 χ 100
~ 12 χ ΙΟ6 χ 2 '
= 0-0416 radian,
= 0-0416 χ - ^ - , π
= 2-38°.
E X A M P L E . A solid circular shaft is to be made of material having
a modulus of rigidity of 12 x 1 0e lbf/in
2. Calculate the required
diameter i f i t is to transmit 100 h.p. a t 3000 rev/min under either
of the following conditions :
( a ) / s m a x> 6000 lbf/in2,
(b) 0 > 0ο8
χ per foot length.
Assume only the Torsion Equat ion .
Solution
Torque 33 ,000 χ h.p.
2^N lbf f t ,
= 2100 lbf in.
188 S T R E N G T H O F M A T E R I A L S
/, _ Τ 6000 _ 2100 ( a )
7~ ~ Τ ' ~dJ2~ ~ πάψ2 '
2 χ 6000 χ π # = 32 χ 2100 x d,
32 χ 2100 ri à
12,000 π 9
= 1-785,
.'. =
1-215 in. for safe stress.
— - — · 2 1 00
_ 12 x 1 06
/ 8 π
\ ( ' ~Τ~~~Ί7' - πά*β2 ' 12 \ 60 X 180 / 3
8 since 0 — deg.
32 χ 2100 8π χ ΙΟ6
i.e., d4
π # 60 χ 180
32 χ 2100 χ 60 χ 180
8π2 χ ΙΟ
6
- 9-2.
d2 = 3-03,
i.e. d = 1-74 in. for safe twist.
E X A M P L E . TWO lengths of 3 in. dia. shaft are connected by a
simple flanged coupling having 6 bolts at 8 in. pitch circle diameter.
Determine the bolt diameter so tha t the average stress in the bolts
shall be the same as the max. stress in the shaft when the la t ter
is transmitt ing 4 h.p. at 25 rev/min.
in
I
FIG. 178
T O R S I O N 189
Solution
33,000 x h.p. Τ
2πΝ
33,000 χ 4
2π χ 25
10,100 lbf/in.
12 ,
Τ
Τ 7"
Τ χ 32
where d = shaft diameter,
Τ χ 16
πί23
10,100 χ 16
π χ 33
1900 lbf/in2.
I f (i/, = bolt diameter, Resisting torque of bolts =
- 35 ,800 dl
Equat ing, 35 ,800 ig = 10 ,100 ,
dl = 0-282,
dh = 0-53 in.
6(0-785 df, 1900) χ 4 ,
E X A M P L E . The input to one end of a uniform line shaft is 30 h.p.
a t 150 rev/min. Three identical machines are belt-driven from
identical pulleys a t distances of 12 ft, 20 ft and 30 ft respectively
from the driven end. Take G = 12 χ 1 06 lbf/in
2, assume / i r a ax
= 9000 lbf/in2 and find (a) a suitable shaft diameter and (b) the
tota l angle of twist in degrees.
- 3 0 f t -
- 2 0 f t -
- 1 2 f t - -8ft- - 1 0 f t -
150 rev/min
3 0 h.p.
7 η S M
10 h.p. 10 h.p. 10 h.p.
FIG. 1 7 9
190 S T R E N G T H O F M A T E R I A L S
i.e. d3
2 χ 9000π '
7-13, whence d = 1-92 in. (say 2-0 in.).
Since the machines are identical, each pulley transmits 30 /3 , i.e.
10 h.p. The 10 ft portion thus carries ^ of the input torque and
hence the stress in i t is 9000/3 , i.e. 3000 lbf/in2.
Similarly the 8 ft portion carries f of the input torque and has a
maximum shear stress of χ 9000, i.e. 6000 lbf/in2.
0 ~-η£-, .'. Total twist - fl12 + fl8 - I - fl10,
9 θ ο ο ( 1 2 χ 1 2 ) + ^ ο ο ( 8 χ ] 2 )
Gr K
' Gr
+ i g » ( « . « . « . 12
— (10,800 + 48 ,000 + 30,000) rad, Gr
12 χ 186,000 180 , χ —^— deg, 12 χ 106 χ 1 0 π
180 0-186 χ
71
0 = 10-65 deg.
E X A M P L E . A 3*0 in. marine propeller shaft driven a t 100 rev/min is to be replaced by two identical shafts driven a t 720 rev/min and, between them, transmitt ing 80 per cent more power. I f the replacements are to carry a 20 per cent increase in maximum shear stress over the original shaft, est imate a suitable diameter.
Solution
τ m i 33 ,000 χ h.p. \ „ Λ 1 ί. Input torque Τ - — — ± — 12 lbf in ,
^ 3 , 0 0 0 x 3 0 x 12 ^ b f jn
2π χ 150
fs Τ . 9000 _ 12,600
Τ~'Τ' " ~äß~ ~ π ^ / 3 2 '
12,600 χ 32
T O R S I O N
Solution
I f Tx and N1 are the original torque and speed, T2 and N2 are the
torque and speed for replacement then, for the same power :
2πΝ1Τ1
33,000 χ 1-8
I I τ9
/ 2 π 2 \ Γ 2Τ 2
\ 33 ,000
2 Ν2
2 χ 720
1-8 χ 100 '
8-0.
I f d1 and fSl are the original diameter and maximum stress, d2 and
fS2 are corresponding values for the replacements,
nd\fS2 then
i.e.
nd\fs
1 1 6 and T9 =
16
and / e a = 1 .2 / Ä i.
or
1 2 ,
8-0 χ 1-2,
9-6.
Hence, 2-12, where dx = 3-0 in.,
3 0
2-12 '
- 1-42 in.
Hollow Shafts
I n this case the maxi-mum shear stress
fs occurs when r = . FIG. 1 8 0
7;I*
191
1 9 2 S T R E N G T H O F M A T E R I A L S
Hence, from the torsion equation :
—Γττ- = —r , where J dJ2 J '
U
nd\ nd\
3 2
π
~32~
3 2
(df-di).
Safe torque, Τ - -f^ χ i L (d\ - d\),
i.e. Τ = nfs ( df — d\
1 6 lbf in usually.
E X A M P L E . At what speed must a hollow shaft 6 0 in. long, 3 in.
inside dia. and 6 in. outside dia. rotate in order to t ransmit 5 0 0 h.p.,
assuming fs > 6 0 0 0 lbf/in2 ? Through what angle will the shaft
twist if G = 1 2 χ 1 06 lbf/in
2?
Solution
π J
( 1 2 9 6 - 8 1 ) 3 2
π x 1 2 1 5
3 2
= 1 1 9 in4,
dr τ
h.p. = 5 0 0
Required speed Ν
fsL 6 0 0 0 χ 1 1 9 2 3 8 , 0 0 0 lbf in.
Angle of twist θ =
2πΝΤ
3 3 , 0 0 0
5 0 0 χ 3 3 , 0 0 0
2π χ 2 3 8 , 0 0 0
1 3 2 rev/min.
TL
~GJ '
2 3 8 , 0 0 0 χ 6 0
1 2 χ 1 0 6 χ 1 1 9
0 - 0 1 radian,
0 - 5 7 deg.
χ 1 2 ,
T O R S I O N 193
E X A M P L E . A 4 in. dia. solid shaft is to be replaced by a hollow
shaft 6 in. outside dia. F ind the thickness required to make the
new shaft (a) of equal strength and (b) of equal weight. I n case (a)
determine the ratio of the weights of old and new shafts and in
case (b) the ratio of the strengths.
Solution
Sol id (s) Hol low (h )
FIG. 181
(a) The replacement must carry the same torque,
i.e.
16 16 \ dx
χ d*
d* d \ - d \
Weight hollow
Weight solid
d\ = d \ - dxd*,
= 64 - (6 χ 4
3) ,
= 1296 - 3 8 4 ,
= 9 1 2 .
d2 = 5-5, i.e. thickness
6 0 - 5-5 . = = 0-25 m.
2 0-785(df - d\) LQ
0-785d2 LQ
(? 62 dl - dl ö2 - Γ>·5
2
36 - 30-2 5-8 3
i 6 =i r=
8 a p p r o x -
i.e.
194 S T R E N G T H O F M A T E R I A L S
(b) The replacement must have the same cross sect ion:
i.e. 0-785(d2 - d\) = 0-785 d
2.
d \ - d \ = d2.
di = d \ - d2,
= 62 - 4
2,
- 36 - 16 ;
- 2 0 . 6 0 - 4-47
d2 = 4-47 in., i.e. thickness = ---= 0-76 in.
Torque hollow 16 \ d1 ) d\ - rff
Torque solid nfs dß*
_ 64 - 4 -47
4
6 χ 43 '
1296 - 400
6 x 6 4 '
896 7 = " 3 8 4 ' = y W °
x-
E X A M P L E . A torsion spring consists of a steel shaft anchored a t
one end and enclosed b y a steel tube to which i t is rigidly fixed a t
the other end. (Fig. 182.) The arrangement is as shown, the force being applied a t point L .
Determine, for an allowable shear stress of 10,000 lbf/in2,
(a) the rat io of the maximum shear stresses in tube and shaft,
(b) the maximum torque which may be applied,
(c) the angle through which the free end of the tube turns.
The shaft diameter is 1 in. and for the tube the diameters are 1-414 in. outside and 1-300 in. inside, the act ive lengths of each being 24 in. Assume pure torsion and t ake G = 12 χ 1 0
6 lbf/in
2.
Solution TRD
4 7T χ l
4
F o r shaft, J , - ^ L - - — = 0-0982 in4.
F o r tube, j t = JL (Z)4 _. di) = ο.0982(1·4144 - 1-3
4) ,
T O R S I O N 195
FIG. 182
= 0-0982(4 - 2 -85) ,
= 0-113 in4.
Since
hence, for the same torque
Τ
Ύ
fstJ t
fs,
k r
9
J
Ji X rs
0-0982
~ Ö 4 3 ~ X
1-23.
Maximum torque, Τ = —— f s J 10,000 χ 0-113
1-414/2
1-414
1
= 1600 lbf in.
Twist a t free end θ = ds + 6t, and each has the same length,
TL TL
+ GJS
TL
G
1600 χ 24
GJt
V 0-09 12 χ 1 0e
3200
+ 1
0982 0 113
(10-19 + 8-85), 1 0
e
_ 3200 χ 19-04
= 0-06 radian or 3-44 dec.
196 S T R E N G T H O F M A T E R I A L S
E X A M P L E . A steel shaft 2 in. dia. is closely surrounded by, and
firmly a t tached to a brass liner 3-5 in. outside diameter. I f the
permissible shear stresses for steel and brass are respectively 3
and 2 tonf/in2, est imate the power which may safely be t ransmit ted
by the compound shaft a t 400 rev/min. ( G B r a ss = 0-45 χ 6r St e ei )
Solution
fs ΟΘ — = —— and tor any one r L
J
section θ and L are constant
A r
f.
i.e., oc G
G oc r.
When, a t the common radius,
or
r = 1,
/%ax /]îmin
f Βτρίη
/ Κ max — χ /
FIG. 1 8 3
1-75
1
= 0-45 χ 3 χ 1-75 (assuming / S m a xt o be the criterion),
= 2-37 tonf/in2,
which exceeds the permitted value, i.e. / L W must be the criterion.
/ = ^ x Cr β
1
1-75
1 χ
x / i w
x 2 ,
For steel, J s = — χ 24 and T{
0-45 1-75
= 2-53 tonf/in2, which is within the limit.
/s«^s where r = 1, r
and fs = 2-53, 2-53 / π χ 2
4 '
1 \ 32
3-97 tonf in .
For brass,
and
T O R S I O N
r
2
3-5
where r = 1*75 and / B
π / 3 ·54 - 2
4 \
1-75 32 \ 3-5
= 4-28 tonf in.
2240 Tota l torque = (3-97 + 4-28) — — ,
\ Δ
= 1520 lbf f t .
2 χ 400 χ 1520 Power at 400 rev/min =
33,000
117 h.p.
Torsional Strain Energy
F rom the torsion equation for a circular shaft,
so tha t the T-θ graph has the linear form shown in Fig . 184.
FIG. 1 8 4 FIG. 1 8 5
197
π / 3 ·54 - 2
4
198 S T R E N G T H O F M A T E R I A L S
The shaded area represents the work done in twisting one end
of a shaft (Fig. 185) through an angle θ relative to the other end,
this work being (within the elastic l imit) stored in the material as
strain energy.
Torsional strain energy UT = Average torque χ Angle of twist,
= L x 0, where Τ = ^L χ θ,
Δ L GJO
2
Alternatively,
Again,
2L
Τ
UT
2
T2L
~2GJ '
Τ — χ 0 ,
χ 0 , where 0 TL
~GJ
where Τ = —
and θ
ί£ r
]jL_ rG
1 / , / w isL 2 r rG
fl LJ
~G 2 r2 '
L
where J
G X as" X
"32" a nd f2
32 '
d?_
IT
~G~ X
64
nd2L A.
4G
B u t nd
2L
— Volume of shaft, i.e. UT iL 4 0
4
x Volume
(where fs = max. shear stress).
I f G and fs are in lbf/in2, then the strain energy will be in in. lb.
Coil Springs
A spring may be compressed or extended (as is most common) or twisted axially so as either to wind or unwind it . I n each case
T O R S I O N 199
the work done in deformation (within the elastic l imit) is conserved
in the form of strain energy.
A spring m a y therefore fulfil one of the following functions:
(a) prevention of shock by absorption of kinet ic energy (e.g. vehicle
spring),
(b) restoration of a deflected mechanism to its position of rest
(e.g. indicating instrument control spring),
(c) release of energy under controlled conditions (e.g., clock spring).
Helical Spring—Effects of Axial Load
An axia l load F
(Fig. 186 (a)) can be resolved
into two components acting
a t point Ρ :
1. F cos θ normal to the
axis of the wire,
2. F sin θ parallel to the
axis of the wire.
At point Q on the wire
axis the component F cos 0
tends to produce the effect
shown in F ig . (187(a)) i.e.,
i t induces a direct shear
stress of
F cos θ . . nd%
γ. where A = ——
Λ
A 4
(d being the wire diameter) .
Since F cos Q also acts a t
an instance Ε from point Q
i t applies a torque
Τ = F cos θ x R FIG. 1 8 6
to any section X X (Fig. 186(b)) and hence over all such sections
there will be an additional shear stress increasing from zero a t
the wire axis to a maximum given by
Tr Λ d ——-, where r = —.
J Δ
(This is relatively much greater than the direct shear stress).
2 0 0 S T R E N G T H O F M A T E R I A L S
F cos0
FIG. 187
F sin Ο
Fcos0
FIG. m
(b)
(α)
(Α)
(b)
T O R S I O N 201
Considering again point Q on the wire axis, the component
i^sinö tends to produce the effect shown in F ig . 188(a) i.e., i t
F sin θ nd2
induces a direct tensile stress of , where A = ——. A 4
This component is also responsible for a bending moment M =
Fsind χ R (Fig. 188(b)) which tends to increase the curvature of
the wire and hence the coil radius R. The maximum bending stress My d nd*
is given by / = — — , where y = —- and I = —-.
As already shown T2L
Torsional strain energy UT = and Τ = FR cos θ,
(FRcose)*L
2GJ '
_ F*R2L I cos20 \
~ 2 \~GJ~I'
Bending strain energy UB = j M2 dx
ο
(see chapter on strain energy of beams). L
M2 r Since M is constant, UB = | do;
2El J ο
M2L and M = FR s inö , 2 # /
(FR sin θ) 2
£
2 7
i ^2. ß
2£ / s i n
20
^ 7
Tota l strain energy U = UT + ? 7 ß
.F272
2L / cos
2<9 \ ^
2. ß
2£ / s in
20
2 \ GJ ) 2 \ EI
F2R
2L î cos
20 s in
20
GJ EI
2 0 2 S T R E N G T H O F M A T E R I A L S
B u t Work done in deflection = Average force χ Deflection,
„ . F „ F*R*L / c o s2φ s in
20
Lquatmg, 6 = — g — +
Now cos θ = 2 π
^ç
j where ç = number of coils,
Length of wire m spring, L cos Ö
Q w w # r · , 2nFR3n / cos
2Ö sin
2Ö\
SubBtitutmg for £ gives : ί ^ _ _ ß _ + _ j
Although this and previous expressions are based upon the
assumption tha t R and θ are constant , whereas they both vary
slightly with the load F, t hey are in practice sufficiently accurate.
When θ is small ( < 10 deg say) s in2 θ is negligible and cos θ
^ 1-0. The material may thus be considered as being in a state of
pure torsion and the spring described as "c lose-coi led" .
Put t ing cosφ = 1, and ignoring sin2φ gives
2nFRH • T nd* Τ=—GJ—>
W h e r e J s = l T '
_ 2nFR*n 3 2 ~ G
X ~JVF'
64:FR3n
Whence axial deflection
T O R S I O N 203
This expression may be obtained more simply by assuming pure
torsion, as follows :
Work done by F = Torsional strain energy,
F Τ
TO
; J = y , where Τ ^ FR
(Fig. 1 8 9 ) ,
= RO and θ TL
GJ
FRL
GJ 9
= R χ FRL
GJ 9
FR2L
GJ
FR2
, and L = 2nRnm&v\y,
GJ
2nFR*n
χ 2nRn,
η co i l s
GJ ' W h e r e J = l f '
2nFR*n 3 2
G x
Fia. 189
„ , 64, F RH Hence, ο = -r-^— as before.
Gd*
E X A M P L E . Wire 6 ft long and 0 - 1 9 2 in. dia. is formed into a helical spring of 2 in. mean coil dia., a to ta l of 8 in. being allowed for the end fastenings. F ind the axial stiffness in lbf/in.
I f / S > 5 5 , 0 0 0 lbf/in2 and G = 1 2 χ 1 0
6 lbf/in
2, find the maximum
permissible axial load.
204 S T R E N G T H O F M A T E R I A L S
Solution
IIIIIIUIIIIIIIIIIIII
w
2 i n -
F I G . 1 9 0
F o r wire,
Τ
Τ
0 - 1 9 2 in
2nRn = 6 ft less 8 in
(2π χ l ) w = (6 χ 12) -
64 .'. η = —— ,
2π
= 10-2 turns.
64:FR*n
k r
or
i.e. F
, Axial stiffness,
F Gd*
~ Ύ ~ 64 RH
12 x 1 0e (0-192)
4
64 χ 13 χ 10-2
= 25 lbf in.
J =
J A =
π ί4
" 3 2 " '
π χ 0-00136
32
0-000133 i n4.
r
U_
r
rR
55,000 x 0-000133
0-192 \
2
76-5 lbf .
x 1
T O R S I O N 2 0 5
E X A M P L E . A helical spring has 1 0 close-coiled turns the outside
dia. of which is 5 t imes tha t of the wire. I f the spring is to stretch
0 - 5 in. under an axial load of 3 0 0 lbf, find (a) the necessary wire
diameter and (b) the max . shear stress.
Take G = 1 2 χ 1 06 lbf/in
2.
Solution
Mean diameter
= M - d,
= 4d,
R = 2d
Deflection
_ 64:FR3n
Gd* 9
6 4 χ 3 0 0 (2df 1 0
* * 1 2 χ ΙΟ6 χ d* '
A
F I G . 1 9 1
_ 6 4 χ 3 0 0 χ 23 χ 1 0
~a¥ ~ 1 2 χ ΙΟ6 χ 0 - 5 '
giving d = 0 - 2 5 6 in. (i.e. r = 0 - 1 2 8 i n . ) ,
d* = 0 - 0 0 4 3 ,
J = =L K
X 0 0 0 4 3'
= 0 - 0 0 0 4 2 2 in4.
Tr
Stress / , = — where Τ = FR,
_ ( 3 0 0 χ 2 χ 0 - 2 5 6 ) 0 - 1 2 8
~ 0 - 0 0 0 4 2 2 '
- 4 6 , 6 0 0 lbf/in2.
E X A M P L E . A length of wire of diameter 0 - 1 4 4 in. is formed into
a spring with a mean coil diameter of 0 - 8 7 5 in. W h a t axial load will
induce a maximum shear stress of 5 0 , 0 0 0 lbf/in2?
Es t imate the number of coils required to give a deflection of
1 - 7 5 in, a t this load. G - 1 2 χ 1 06 lbf/in
2.
206
Deflection
S T R E N G T H O F M A T E R I A L S
Τ kL } where Τ — FR, r
0-875
2 W - 0-4375i<
7.
/ , - 5 0 , 0 0 0 ,
0 1 4 4
2 = 0-072 in. ,
J „ ( 0 - 1 4 4 )4
,
— x 0-00043
0 0 0 0 0 4 2 2 in4.
0-4375 J
F
50,000 x 0 0 0 0 0 4 2 2
0 0 7 2
50,000 x 0-0000422
0 0 7 2 x 0-4375
67 lbf.
64:FR*n
Gd*ô where R
0-875
64 FR* ' 2
12 χ ΙΟ6 (0-144)
4 x 1-75
64 x 67 ' 0-875 \
3
2 J
12 x 1 0e x 0-00043 x 1-75
64 x 67 x 0-084
— 25-1 coils.
E X A M P L E . A helical compression spring is made of steel rod 0-3
in. dia, and has 10 coils of 1-7 in, outside dia, I t s free (unloaded)
Solution
giving
rL A J " r
T O R S I O N 207
Axial load in 112 224 336 448 560 672 784 1120
Compression in inches
0-26 0-51 0-77 1-04 1-29 1-55 1-81 2-59
Plo t the load vert ically on a base of compression. Selec t a con-
venient pair of values from the graph, find the slope and est imate
(a) the value of G and
(b) the twist per unit length a t maximum load.
Solution
Mean coil radius
Polar 2nd Moment
R = 1-7 - 0-3
2
0-7 in.
I F '
- ^ ( 0 - 3 ) * ,
- 0-000795 in4.
F From the graph slope, ^ = 4 2 6 .
ο
1-7 in
- 0 - 3 in
FIG. 192
4 2 6
J 1 0
δ, i n c h e s
FIG. 1 9 3
length is 10 in. A tes t gave the following figures :
2 0 8 S T R E N G T H O F M A T E R I A L S
Deflection UFRH
G =
G6
L
Gd* 9
~ d*
64(0-7)3 χ 10 χ 426
(0-3)4
11-6 x 1 06 lbf/in
2 (approx.
T_
~J9
Twist /Unit length — L
Τ
~GJ where Τ
1120 χ 0-7
FR,
11-6 χ ΙΟ6 χ 0-000795
0-085 radian/in.
a t max . load,
E X A M P L E . A helical spring is to have a mean coil diameter of
4 in. and is to deflect 6 in. under an axial load of 1250 lb, the cor-
responding shear stress being then > 25 tonf/in2.
I f wire is available in the following diameters : \, g, JJ,
and f in., find the most suitable, the number of coils required
and the actual maximum shear stress, assuming G = 12 χ ΙΟ6
lbf/in2.
Solution
r J 9
25 χ 2240 1250 χ 2
d\2
d*
π # / 3 2
1250 χ 2
. Since Τ = FR,
χ 32
Deflection
25 χ 2 χ 2240π
= 0-227.
d = 0-61 in. , say 0-625 in . (i.e. § in.)
ô = R6, where à = 6 in.
R = 2 in.
TL
2 TL
GJ
0 = GJ '
and L = 2nRn = 4πη.
Τ = 1250 x 2 = 2500 lbf in.
- 0-015 in4.
π χ 0 -6254
32
6 = 2 χ 2500 χ 4πη
12 χ ΙΟ6 χ 0 0 1 5
T O R S I O N
i.e.
Stress
6 χ 12 x 1 0e χ 0 0 1 5
500 χ 4π
17-2 coils.
Tr
2500 0-625
0 0 1 5 2
52,000 lbf/in2,
23-3 tonf/in2.
E X A M P L E . A helical spring 5 in. mean diameter is required to
absorb 6 inch tonf of energy with a maximum shear stress of
35 tonf/in2. F ind the diameter of steel rod required and the num-
ber of coils if the maximum amount of compression is to be 6 in.
Take G = 5400 tonf/ in2 ( I . Mech. E . )
Solution
Strain energy
4.Γ* 1
3 52
4G per unit volume.
6 = -377(2πΒ* x 4 r < 4(Τ \ 4
2π x 2-5n x - y - j · 4 χ 5400
6 χ 4 χ 5400 χ 4 d
2n
3 52 χ 5 π
2
i.e.,
= 8-56.
8-56
η F
Energy absorbed = — ,
F 6 = — χ 6 giving F = 2 tonf
. Δ
ι κ ζ
Ο Ο Ο Ο Ο
Ό1
Ο Ο Ι ο ο
- 5 in dia-6 in
FIG. 1 9 4
209
210 S T R E N G T H O F M A T E R I A L S
~Gd* *
64 χ 2 χ 2-53?a
5400 d*
0 0 6 1 7 ,
i.e. d* = 0-0617 η
Hence, 0-0617^
nz
Required diameter d
E X A M P L E . A petrol engine valve weighing 1 lbf is to be given a
maximum acceleration of 1000 f t / s2 occurring a t a lift of 0-205 in.,
the to ta l lift being 0-475 in. Es t ima te the maximum force in the
spring if i t has a free length of 6-5 in. and is 4-5 in. long when the
valve is shut. Determine the mean diameter of the coils and their
number, taking a wire diameter of 0-128 in. and working a t a
maximum shear stress of 50 ,000 lbf/in2. (G = 12 χ 1 0
6 lbf/ in
2) .
Solution
Ini t ia l compression = 6-5 — 4-5 = 2 in. W
At 0-205 in. lift, i.e. δ = 2-205 in. Acc . force F = /, 9
= w x 1 0 0 0'
= 31-1 lbf.
2-475 Hence a t 0-475 in lift, i.e. δ = 2-475 in. Max. force = — — χ 3 1 - 1 ,
Δ' Δ\)θ = 34-9 lbf.
Deflection
d*
η
_ 8-562
n2
8-562
" 0 0 6 1 7
= 1185 , giving η = 10-6 coils.
1 / 8 - 5 6
- j /0-806 - 0-899 in.
T O R S I O N 211
When d = 0-128 in., J = .
= 0-0000263 in4.
·— , ± —
J r r ο. Λτ >
5 0>
0 0 0 x 0-0000263
34-9 R —
giving R =
0 1 2 8 / 2
50,000 χ 0 0 0 0 0 2 6 3
34-9 χ 0 0 6 4
- 0 - 5 9 i n . , .·. D = 1-18in.
A = , Ζ/ == r
^ , where L = 2nRn,
R
.*. 2 π / ^ = - 7 — x — , fs R
i.e., w = r <
? ^ D 9 , where ό = 2-475 in . lnfsR*
0-064 χ 12 χ ΙΟ6 χ 2-475
2π χ 50 ,000 χ (0-59)2
= 17-35 coils.
E X A M P L E : A compound spring consists of two co-axial close-
coiled helical springs as follows :
Mean coil dia. Ό in. Wire dia. d in. Coils η Free length
Outer 1-75 0-192 10 4 in.
Inner 1-25 0160 8 3-25 in.
I f the combination is subjected to an axial compressive load
of 100 lbf, determine for each spring (a) change in length, (b) load
carried and (c) max . stress.
T a k e G = 12 χ 1 06 lbf/in
2.
212 S T R E N G T H O F M A T E R I A L S
Solution
64:FRhi
Od*
F .'. Stiffness λ = — =
Od*
Gd*
64:R3n '
W=IOOIbf Hence for outer spring,
12 χ 106(0 ·192)
4
0 7 5 in
3-25 in
4 in
8 χ (1-75)3 χ 10
38 lbf in .
-1-25 in
-175 in—*j
The outer must be compressed by 0-75 in before the inner begins com-pression. .'. Load to compress outer by this
amount = 0-75 χ 3 8 ,
= 28-5 lbf.
The remaining load of 71*5 lbf now fafo compresses both springs simultane-I Τ
1 ously.
I Again, for inner spring,
12 χ 106(0 -160)
4
λ2 = FIG. 1 9 5
8(1-25) 3 x 8
62-9 lbf in.
.*. Combined stiffness = λ1 + λ2 = 38 + 62-9 ,
= 100-9 lbf/in.
71-5 Common reduction
100-9 0-708 in.
F o r outer δλ = 0-75 + 0-708. F o r inner δ2 = χ = 0-708 in.
= 1-458 in . F9
Fx = Vi - 38 χ 1-458,
= 55-45 lbf .
λ2ο2, 62-9 χ 0-708,
44-55 lbf .
Τ ^ = Ί 7 Γ = 1 1 " 2 " ~7td\ = F l R l Tri*
= 55-45 χ — — χ — χ 2 π
= 34,900 lbf/in2.
— V 0 - 1 9 2 / '
T O R S I O N 2 1 3
And snndarly: = F2B2 — = 4 4 - 5 5 χ _ χ — χ ( _ )
= 3 4 , 9 0 0 lbf/in2.
E X A M P L E . An open-coiled helical spring has 2 0 coils made of 0 - 5 in. dia. steel. The mean coil radius is 5 in. and the coils make an angle of 3 0 ° with the horizontal when the unloaded spring is suspended vertically.
I f Ε = 2 · 5 £ - 3 0 χ 1 06 lbf/in
2 find the axialdeflection for a load
of 4 0 lbf (a) assuming pure torsion and (b) taking bending effects into account .
W h a t approximate percentage error results from making assump-tion (a)?
Solution
(a) Pure Torsion
à = M
g ^n , where Β = 5 i.e., B* = 1 2 5 ,
d - 0 - 5 i.e., d* = 0 - 0 6 2 5 ,
G = 3 0
* 1 06
= 1 2 χ W lbf/in2.
_ 6 4 x 4 0 x 1 2 5 χ 2 0
1 2 χ Ι Ο6 χ 0 - 0 6 2 5 '
= 8 - 5 3 in.
(b) Torsion with bending
2nFB*n / cos20 s in
2ö \ _ _ o _ _
δ= 7Γ-(—γΓΓ- + —ΈΤΓ-)> w h e re
E=25G and cos 0 \ GJ EI j
J = 21 for a circular section,
_ 2nFB*n (cos26 s in
20 \
~~ cos θ [ 2GI +
2-5GI j '
_ 2nFB*n I cos26 s in
20 \
^ GI cos Θ \ ~ ~ 2 ~ ~ +
~ ¥ 5 ~ " j
214 S T R E N G T H O F M A T E R I A L S
and cos θ = 0-866 i.e., cos20 = 0-75
sin θ = 0-5 i.e., s in20 - 0-25
64
π χ 0 -54
64
= 0-00307 in4.
2π χ 40 χ 125 χ 20 / 0-75 0-25 \
~ 12 χ ΙΟ6 χ 0-00307 χ 0-866 [ 2
+ 2-5 / '
= 19-66(0-375 + 0-1) ,
- 9-35 in.
Thus assumption (a) underestimates the deflection.
1. Find the strain energy stored in the material of a hollow shaft 48 in. long, 2-0 in. inside dia., 4-0 in. outside dia. when subjected to a torque of 800 lbf ft. Take G = 11-8 χ 10
6 lbf/in
2 (6-3 inch lbf).
2. Determine the torque which will induce in a piece of steel tube 2-0 in. outside dia., 0-125 in. thick, 30 in. long a maximum shear stress of 5 tonf/in
2. Find also the angular displacement of one end relative to the other
when this torque is applied. Take G = 5200 tonf/in2. What power could this
tube transmit under the above conditions? (7-75 tonf in., 1-65 deg, 410h.p.).
3. Design a close-coiled helical spring having a mean coil diameter of ten times the wire diameter, an axial stiffness of 120 lbf/ft and a safe deflection of 1-0 in. Assume a value for G oî 13 χ 10
6 lbf/in
2 and allow a maximum
stress of 15,000 lbf/in2 (21 turns, 1-3 in. dia.).
4. A hollow steel shaft 2-5 in. outside dia. is to be connected via a clutch, to a solid alloy shaft of the same diameter. If the torsional rigidity of the steel shaft is to be 0-8 of that of the alloy shaft, calculate the required inside dia-meter given that Oa = 0-46rs (2-1 in.).
5. The coils of an open-coiled helical spring make an angle Θ with planes normal to the spring axis. If the extension is calculated using the formula for close-coiled springs, find the value of Θ corresponding with an error of 1 per cent (8-1 deg).
Error
82
9-35 '
= 8-8 per cent (low approx).
Examples VI I
C H A P T E R VIII
COMPLEX STRESS I
Shear Stress Resulting from a Tensile Load
A tensile load F produces a direct tensile stress on planes such as
C B (Fig. 196) normal to the load axis, and this stress is given b y :
1 A '
The load m a y be re-
solved into two compo-
nents when considering
i ts effect on any plane
such as CD inclined 0 to
C B :
1. F cos 0 normal to CD
acting over the area
A /cos 0 and intro-
ducing a normal ten-
sile stress given by
i ^ c o s 0
where A = Section.
A /cos 0
F = — c o s
20 .
A FIG. 1 9 6
This is a maximum when cos θ is a maximum, i.e. when 0 = 0 .
Then F
2. F sin θ parallel to CD acting over the same area A /cos 0 and
introducing a shear stress given b y
_ .Fsine '* ~~ ~~A~i
F . = — sm I
/ A J cos θ A
costf == —^7- sin 2 0 . 2A
8* 2 1 5
216 S T R E N G T H O F M A T E R I A L S
This is a maximum when sin 26 is a maximum
i.e., when 26 = 90 deg or 6 = 45 deg.
JL = L 2A 2 '
Then,
FIG. 1 9 7
Thus on planes inclined a t 45° to the axis of
the applied stress, the induced shear stress has
half the value of the tensile stress on sections
normal to the load axis.
I t follows that , if the shear strength of a material
is less than half i ts tensile strength, the material
will fail in shear when subjected to a tensile load.
Similarly, if the shear strength is less than half
the compressive strength, failure in shear will
occur during compression. Cast iron fails in this
manner in compression (Fig. 197) .
Complementary Shear
Suppose a shear stress fSi to be induced on opposite faces of a
rectangular element by forces Fx applied as shown in Fig . 1 9 8 :
F[ inducing fS|
F2 -
1 FIG. 1 9 8
Then, clockwise shear couple = Fxx, where Fx = fSl (yz).
I f the equal anticlockwise couple required for equilibrium is
introduced by forces F2 acting (as shown dotted) on the other two
faces.
then. F2y = F,x.
C O M P L E X S T R E S S I 217
I f fSt is the shear stress induced by F2 then F2 = fSl (xz).
Hence, fs2(xz) V = f»I(Vz) x> substituting for Fx and F2.
i.e., Is, /.,· Thus if a body remains in equilibrium under the action of a
system of forces two of which induce a shear stress, an equal shear
stress is automatically in existence on planes a t 90 deg. This is
referred to as a Complementary Shear.
Applied shear
Complementary shear
(a)
From Fig .199 (a) it can be seen that , due to the two shear couples,
the diagonal face A B will be under compression. I f / , v = normal
compressive stress required for equilibrium of portion A D B , then,
resolving forces horizontally :
fx(pS χ S) cos 45 = / , χ S2, where S = length of edge,
i.e. fNpS2 x ~ fsS\
whence, / N = fx.
Clearly there will be a tensile stress (of equal magnitude to the applied shear stress) on the other diagonal section of the cube (CD) as in Fig. 199(c) .
Bulk or Volumetric Strain
As a result of an elastic linear tensile strain on a linear dimen-
sion L,
New length = L + dL9 where dL = increase,
= L + eL where e = strain,
- L(l <•)
218 S T R E N G T H O F M A T E R I A L S
Similarly, after compression,
New length — L(l e)
Suppose a rectangular element χ x y χ ζ, to be immersed in liquid
as shown in Fig . 200 . The force acting on each face will bring about a
reduction in the length of each edge
and hence in the volume, i.e., i t will
produce a volumetric strain.
I f the strains in the three mutu-
ally perpendicular directions are
ex, ey and ez then,
New volume
= x(l - ex)y(l - ey)z(l - ez),
= xyz(l + exey + eyez + ezex
FIG. 200
I f the elastic limit is not exceeded
the strains are very small and their
products may be neglected, so tha t
New volume = xyz + xyz ( — ex — ey — e.) ,
= xyz - xyz(ex + ey + β . ) .
B u t , xyz = Original volume.
.'. Volume change = New volume — Original volume
= — xyz(ex -\-ey-\-ez) (the negative sign
indicating a reduction)
-xyz(ex + ey + es) i.e., Volumetr ic strain ev
- [e.
xyz
e,, + ez)
Thus the volumetric strain is the sum of the linear strains, the
negative sign in this case indicating tha t i t is compressive.
F o r an elemental cube, χ = y = ζ.
Hence, ex = ey = ez = e, say.
Hence, ev = — 3e ,
= 3 χ Linear strain.
C O M P L E X S T R E S S I 219
E X A M P L E . Calculate the reduction in volume of a steel cylinder
2-5 in. dia., 10-0 in. long, when submerged in the sea to a depth
of 36 ,000 ft. Take Κ = 23-3 χ 1 06 lbf/ in
2 and assume tha t the
density of sea water remains constant a t 0·036 lbf/in3.
Solution
The compressive stress is equal to the pressure, i.e., i t is the
product of depth and density
f = hq where h = depth and ρ = densi ty,
= (36,000 χ 1 2 ) 0 - 0 3 6 ,
= 15,500 lbf/in2.
Volumetric strain e v = (see p. 227 ) ,
_ 15,500
23-3 χ 1 06 '
= 0-000667.
Original volume V = — χ 2 ·52 χ 1 0 ,
= 49 in3.
Reduct ion in volume = evV,
= 0-000667 χ 4 9 ,
- 0-0328 in3.
Poisson's Ratio
I n addition to the change in length in the direction of a direct load there is a simultaneous dimensional change in the two direc-tions normal to such a load. Thus a tensile load produces a tensile (longitudinal) strain along i ts own axis and a compressive (lateral) strain along the other two axes resulting in a reduction in the corresponding dimensions.
Within the elastic l imit the ratio of lateral to longitudinal strain is constant for a given material and referred to as Poisson's Ratio.
I t is denoted by σ or 1/m where, for most metals, a lies between 0-25 and 0-33, i.e. m lies between 3 and 4.
220 S T R E N G T H O F M A T E R I A L S
Thus Poisson's Ra t io
i.e. Latera l strain
Latera l strain
Longitudinal strain
σ χ Longitudinal, s train,
1
m
Ε f
mE
Since lateral strains are of opposite sign to longitudinal strains
caused by the same load, it follows tha t in the case of three mutually
perpendicular direct loads, the resultant strain in
the direction of any one of them is the algebraic sum
of the relevant longitudinal and lateral strains. fy=6'75
fv
FIG. 201
Solution
B u t ,
E X A M P L E . A tes t piece 2-0 in. wide, 0-5 in. thick
has a gauge length of 4-0 in. Take Ε = 13,500 tonf/in2
and a = 0-25 and estimate, for an axial load of
6-75 tonf :
(a) longitudinal strain,
(b) lateral strain,
(c) volumetric strain,
(d) change in volume.
TU =
6-75
2 χ 0-5
T„ _ 6-75 Ε 13,500
/ , = 0 ,
e. = - ae,j,
= - 0 - 2 5 χ 0 0 0 0 5
= - 0 0 0 0 1 2 5 .
er = eu + ec + ez,
eu + ( - 2 e , )
6-75 tonf/in2,
0 0 0 0 5 ,
0 0 0 0 5 - (2 χ 0 0 0 0 1 2 5 )
0-00025.
C O M P L E X S T R E S S I 2 2 1
Original volume V = 2 χ 0 - 5 χ 4 , =^ 4 - 0 in3,
Volume change = evV = 0 - 0 0 0 2 5 χ 0 - 4 ,
= 0 - 0 0 1 in3, (increase.)
E X A M P L E . I f the test piece in the previous question is also
subjected to an additional horizontal stress of 4 - 0 tonf/in2, the
stress in the remaining direction being zero, calculate the vertical
and horizontal strains when the second stress is (a) compressive
(b) tensile.
Solution
Vertical Strains :
(a) The horizontal compressive stress pro-
duces a reduction in width and hence
an increase in length, i.e., the lateral
strain due to i t is added to the longitu-
dinal strain due to fu.
.'. Resul tant vertical strain
*» = Ε Ε ^- , taking tensile strain as
positive,
1 3 - 5 0 0
7 - 7 5
1 3 , 5 0 0
[ 6 - 7 5 + ( 0 - 2 5 χ 4 - 0 ) ] ,
= 0 - 0 0 0 5 7 3 (tensile).
(b) The effect of fx in a vertical direction is to reduce the strain due to fy.
/// <*ίχ_ Ε '
fy =6-75
f x = 4 - 0
FIG. 2 0 2
f y = 6 -75
'•e- β , = E
1
1 3 , 5 0 0 [ 6 - 7 5 - ( 0 - 2 5 χ 4 - 0 ) ] ,
5 * 7 5
1 3 5 QQ = 0 - 0 0 0 4 2 6 (tensile) FIG. 2 0 3
* fx =4 -0
8;i SM
222 S T R E N G T H O F M A T E R I A L S
Horizontal Strains :
I n direction of fx : (a) ex = — -4- -êr (effects of fx and fy are
cumulative),
1 • [4-0 + (0-25 χ 6-75)] ,
13,500
5-69
13,500 = — 0-000421 (compressive).
1 [4-0 - (0-25 χ 6-75)] ,
13,500
2-31
13,500 0-000171 (tensile).
In direction of fz (zero applied stress) :
(a) ez=J±-.?k + ^ , Wh e r e / , = 0 ,
1 [ - ( 0 - 2 5 χ 6-75) + (0-25 χ 4 -0 ) ] ,
13,500
- 0 - 6 9
13,500 -0-000069 (compressive).
1 [ - ( 0 - 2 5 χ 6-75) - (0-25 χ 4-0)]
13,500
- 2 - 6 9 _ 0 · 0 0 0 1 9 8 (compressive).
13,500
E X A M P L E . The three mutually perpendicular stresses acting on
the faces of a rectangular block are :
fy = 5 tonf/in2, vert ically and tensile,
fx = 2 tonf/in2, horizontally and compressive,
f2 = 4 tonf/in2, horizontally and tensile.
I f Ε = 13,000 tonf/in2 and σ = 0-286 calculate the strain in
each of the three directions. I f the dimensions measured along the
axes y, χ and ζ are respectively 4-0 in., 2-0in. , and0-5in . , calculate
the change in length in each direction.
(b)
C O M P L E X S T R E S S I 223
Solution Taking tensile strain as positive :
S t ra in along y-axis ey = k + - ZJ^~ 2k
Ε 1
13,000
5 - 0-57
1300 '
- 0-00034 ( tensile) .
[5 + 0-286 (2 - 4 ) ] ,
Corresponding change in length
= 0-00034 χ 4 ,
= 0-00136 in. (increase). ^
Strain along a;-axis
FX < <IFZ
Ε Ε Ε
1 = -TTFX + *(fv + /=)L
1
13,000
4-57
[2 + 0-286
( 5 + 4 ) ] ,
13,000
~- - 0 - 0 0 0 3 5 2 (compressive).
Corresponding change in length
= - 0 0 0 0 3 5 2 χ 2 ,
= - 0 - 0 0 0 7 i n . (reduction).
S t ra in along z-axis ez = -{—— Ε Ε
FIG. 2 0 4
*FY
Ε
t =2
¥ [/, + *(/* -/,)], ι
13,000
4 - 0-86
13,000
3 1 4
[4 + 0-286 (2 - 5)]
13,000 = 0-000242 (tensile).
8 a*
2 2 4 S T R E N G T H O P M A T E R I A L S
Corresponding change in length = 0 - 0 0 0 2 4 2 χ 0 - 5 ,
== 0 - 0 0 0 1 2 in. (increase)
E X A M P L E . The shank of a punch is 1 - 0 in. dia. and during the
punching operation carries an axial compressive stress of 1 0 tonf/in2.
I f the socket restricts the lateral (i.e. radial) strain to one third
tha t of the value when unconstrained, find :
(a) the radial stress imposed by the socket ,
(b) the radial strain,
(c) the increase in diameter of the shank.
Assume Ε = 7 5 0 0 tonf/ in2 and a - 0 - 2 5 .
Solution
Clearly the lateral strain is the same in all horizontal directions
i.e. ez = er since f. = fr.
fy=IO
FIG. 2 0 5
This lateral strain is given by :
C O M P L E X S T R E S S I 225
(taking compressive strain as positive)
1 Τ Ux-<*(fu+fz)]>
= Ύ U.v -o(fy + U)], since f3 = fx.
This is equal to one third of the unconstrained value (fy acting alone),
Ε i.e.
or
ηξϋχ - <r{fy + fx)] =γ
3fx-Mfu+fx)+<rfu = °> 3/, ; -Safu - Safx + fau = 0
3/ j :( l - σ ) = 2 σ / ν ,
i.e., fx = 2 <
3 ( 1 - σ )
2 χ 0-25 χ 10
3(1 - 0-25) = 2-22 tonf/in
2
Hence, radial strain ex
1 [2-22 - 0-25(10 + 2-22) ] ,
Λ Increase in diameter
7500
2-22 - 3 0 5
75ÖÖ 5
0-83
" 7500 '
- 0 - 0 0 0 1 1 , i.e. tensile.
Radia l strain χ Diameter,
+ 0 0 0 0 1 1 χ 1-0
+ 0-00011 in.
I n the absence of the constraint, the increase in diameter would
have been 0-00011 χ 3 or 0-00033 in., i.e. Ε
Relation Between the Elastic Constants
As already shown, a shear stress fs applied to opposite faces of a
cube will, together with i ts complementary shear (shown dot ted in
Fig . 206) produce tensile and compressive stresses / v on the
diagonal faces such tha t fN = fs.
226 S T R E N G T H O F M A T E R I A L S
t s — • h H
f s (applied shear) j ι
U ί- H FIG. 2 0 6
Thus, Strain in diagonal = J ~ + ,
- £ < ι -'-»>·
Since the deformation is very small within the elastic range, the χ
increase in length in the diagonal is approximately y, i.e. —j—
nearly. So that , Strain in diagonal
Increase χ ^ ,^
Original length ]/2
1 χ χ f = — — and — = -77, where G = Rigidi ty modulus,
~ 2 Ö '
= γ - ^ - , since fs = fN.
Equat ing expressions for strain we have :
i.e., # = 267(1
Thus, for an elastic, homogeneous and isotropic material , the value
of G depends on tha t of E. Thus if, for steel, Ε is taken as 29-8 χ 1 0
6 lbf/in
2 and a as 0-287 we have ;
C O M P L E X S T R E S S I 227
Ε Modulus of rigidity G = — ,
2(1 + G)
29-8 χ ΙΟ6
~ 2(1-287) '
= 11-6 χ ΙΟ6 lbf/in
2, approx.
I f the cube is subjected to a fluid stress / on all faces, then, taking
compressive stress as positive :
Linear strain in each edge = ^— — — ,
Ε Ε Ε - i - ( l - & r ) .
B u t , Bu lk or Volumetric strain = 3 χ Linear strain,
- * ( ! - * ) .
And, if Κ = Bu lk modulus,
Volumetric strain = . Ε
f 3 / Equat ing, we have — = — (1 — 2 σ ) , Ε Ε
i.e. Ε = 3K(l - 2σ).
Thus, for steel, using the same figures for Ε and a :
Ε Bulk modulus Κ
3(1 - 2a) 9
29-8 χ 1 06
3[1 - (2 χ 0-287)] '
29-8 χ 1 06
3 χ 0-426 '
23-3 χ 1 06 lbf/in
2. (approx.)
Equa t ing the expressions for Ε :
2 0 ( 1 +σ) = 3K(l - 2 σ ) .
.·. 2G + 2Ga = 3K - 6 Ζ σ ,
or 2Ga + 6 Ζ σ = 3K - 2G}
2 2 8 S T R E N G T H O F M A T E R I A L S
whence, σ 2G + 6K
9
Ε
but, 1 + a = (from equation for E),
Ε -2G i.e. a = Ö
Equat ing the expressions for a :
Ε -2G 3K -2G
whence, Ε -
G 2G + 6K '
9GK
G + 3K
E X A M P L E . When subjected to an axial load of 4 - 0 tonf the
changes in gauge length and diameter of a standard test piece were
respectively 0 - 0 0 2 8 2 in. and 0 - 0 0 0 1 8 5 in. Es t imate the value of G.
Solution
F o r a standard test piece: Gauge length L = 2 - 0 in.,
Gauge diameter d = 0 - 5 6 4 in.,
Latera l strain = 0'*?°?3J5
= 0 - 0 0 0 3 1 ,
0 - 5 6 4
0 * 0 0 2 8 2 Longitudinal strain e = — = 0 - 0 0 1 4 1 .
.*. Poisson's Ra t io σ = ^ = 0 - 2 2 . 0 0 0 1 4 1
Area of section A = — χ 0 - 5 6 42 = 0 - 2 5 in
2.
4
4 0
.*. Axial stress / = -77^3- = 1 6 tonf/in2.
ί 1 6
Elas t ic modulus Ε = — = 41 = 1 1 , 7 0 0 tonf/in2.
Rigidi ty modulus G = n „ E
x = J1'™® = 4 8 0 0 tonf/in2. 6 J 2 ( 1 + a) 2 χ 1 - 2 2 - —
3K - 2G
C O M P L E X S T R E S S I 229
Examples V I I I
1. As a result of tests on a certain steel specimen, the values of Ε and a were found to be respectively 29-6 χ 10
6 lbf/in
2 and 0-294. Determine the
values of G and K. Find also the reduction in volume of a cylinder of this material 6 in. long, 2 in. dia. when subjected to a fluid stress of 11,200 lbf/in
2
(11-5 χ 106 and 23-95 χ 10
6 lbf/in
2, 0-009 in
3).
2. A tensile load of 2 tonf resulted in a measured extension of 0-00517 in. on a 2 in. gauge length of a test piece 0-5 in. dia. The measured twist on an 8 in. length of a similar piece of material produced by a torque of 480 lbf in. was 3-2°. Estimate from these figures the value of σ and state why this method of finding it is unsatifactory (0-3 approx.).
3. The strains registered by two gauges mounted at 90° on the surface of a brass pressure vessel 0-125 in. thick were respectively 0-00071 and 0-000167. Take σ = 0-3 and Ε = 12 χ 10
6 lbf/in
2 and estimate the reduction
in the thickness (0-000047 in.).
4. Calculate the increase in area of a piece of steel plate 12 in2 when sub-
jected to tensile stresses of 17,900 and 11,200 lbf/in2 at 90°. Take a = 0-3
and Ε = 29-6 χ 106 lbf/in
2 (0-066 in
2).
5. Describe three methods of determining the value of Poisson's Ratio and discuss their relative merits.
C H A P T E R I X
COMPLEX STRESS II
Principal Planes and Stresses
L e t two opposite faces of a block A B C D (Fig. 207) of unit
thickness be subjected to a shear stress fs and let direct stresses
fx and fy both tensile, be applied as shown, the direct stress in the
third direction being zero.
FIG. 2 0 7
F o r equilibrium there must also be a complementary shear
couple of opposite sense to the couple resulting from the applied
shear.
On any inclined section E F there must be acting a normal stress
fN together with a tangential (shear) stress fT in order tha t the
piece E F D (Fig. 208) may be in equilibrium, the direction of fT
depending on the relative magnitudes of fx and fy.
2 3 0
C O M P L E X S T R E S S I I 231
FIG. 2 0 8
Resolving forces || fN (i.e. J_ E F ) :
fN E F - fx F D sin 0 + fs F D cos 0 + / S E D sin 0 + fy E D cos θ
(all terms χ 1-0),
Λ /Ν , F D . F D _ E D . E D
FX s in2 θ + / y cos
2
FR
' 1 - cos20 + /y
1 -{- cos 20 "
L 2 + /y 2
2fs sin 0 cos 0 ,
+ / 5s i n 2 0 ,
FR cos 20 / y /// cos 20
2 ΊΓ 2 '5 '
A + 4 + (/„ /,) c q s 2 Ö + ^ g i n w (1)
Resolving J_ / N (i.e. | [ E F ) :
/ r E F = / S F D s i n 0 - fsED cos 0 - fx¥O cos 0 + / y E D s i n 0 ,
, F D . E D F D E D .
= f s sin2 θ — fs cos
20 — /ajSinO cos0 + / ys i n 0 cos0 ,
= / s( s in2 0 - cos
2 0) + (fy - fx) sin 0 cos 0 ,
= ^ ~ sin 20 - / , cos 20 (2)
232 S T R E N G T H O F M A T E R I A L S
F o r maximum or minimum normal stress -^IN = Ö, i.e.,
or
-2 sin 20) + fs(2 cos 20) = 0 differentiating equation 1,
2 / , cos 20 = (fy - / r ) sin 2 0 ,
2fs tan 20 (3)
This gives two values of 20 differing b y 180° and hence there are two values of 0 (0j and 0 2) differing by 90° .
When fT = 0 i.e., when the stress on the oblique plane is wholly direct
^ ~ f * ) sin 20 - / e cos 2 0 , 2
i.e., tan 20 2 / .
h - t.
which is the condition already
obtained for making the direct
stress / i V a maximum. Thus, \
when there is no shear on face
E F i.e., when fT = 0 the stress
diagram becomes as shown in
Fig . 209 .
Resolving \\fx:
fx¥O + / , E D = / N sin 0 E F
F D F D
i.e. fx
or
tan 6
tan 0
sin0
= /N FIG. 2 0 9
/ , = (/.v - / , ) t a n 0
Resolving J_ fx: / S F D + / ^ED - ( / A E F ) cos 0 and E D
E F
i.e.,
F D '
/.v - /// tan 0
(4)
F D
tanfl
F D
sin 0
(5)
C O M P L E X S T R E S S I I 233
U ' ι u
When fy is compressive then, changing i ts sign :
/ A W = * { / , -fy + Wx + / „ ) ' + 4 /2l } >
and is still tensile (positive) even though fy may be greater numeric-
ally than fx.
fs** = * {( /* - /*) - Wx + h? + 4 /2] } '
4 χ o g i v e s : β = (fA - fa){fN - fy),
=~ f% — ÎNÎij — ÎNÎx + fxim -FN-Mtx+TU) + /*/,„
or / , | - (fx + fu)fN + (fjy - / J ) = 0 , which is a quad-rat ic in fN.
Hence , / , = ^ + ^ * V + ^ " ^ ~ ™ , 2
= \IU + fy± M + fv)2 + 2Uu - MÂu + 4/î)]. = I {/* + fu± Wx - /i/)2 + 4/']) ·
Taking the positive sign, the maximum normal stress will be given by
/ A W = \\U + i„ + Wx - /.v)2 +
acting over a plane making an angle θχ with the applied stress fx.
This value is clearly tensile, i.e., of the same sign as fx and fy.
Taking the negative sign, the minimum normal stress will be given by
hmia = h{f, + f,-mf, - / „ ) ' + 4 / * ] } ,
acting over a plane making an angle ö 2 = θ1 + 90° with the stress fx. This is zero when
fx +fu = ]/[(fx-fv)2+*jl], i.e., when β + /» + 2/,./, = /* - 2fJy + /» + 4/?
or when /,./,, = /2.
Thus the least normal stress is tensile (positive) only if frfy > j2
s .
These two values of fN are called the Maximum and Minimum
Principal Stresses, and the planes over which they ac t (and on
which there is no shear) are called the Principal Planes.
The stresses acting on the element are then as shown in Fig . 210 .
Note: F rom E q . 5 : tanO = f* ~
t y , Λ tar i f f -
/ iW ~
fy
234 S T R E N G T H O F M A T E R I A L S
and is clearly compressive (negative) since ^[fx + fy)2 + 4/f] >
(fx — fy) always. Thus when fx and fy are of opposite sign, the
principal stresses are also.
FIG. 210
I t should be noted tha t / ^ m ax is not necessarily greater than
fN numerically, i.e. i t is possible for the greatest stress in the
material to be compressive.
Maximum Shear Stress
As already shown in E q . (2), the shear stress on any plane
inclined a t 0 to fx is given b y
fr = ^ ^ ~ *r j sin 20 - / , cos 219.
F o r maximum or minimum shear stress -L- (fr) — 0 , ad
i.e. ^ » ~ (2 cos 20) - / s ( - 2 s i n 2 0 ) - 0 ,
or (fy - fx) cos 20 + 2 / , sin 20 - 0 ,
.·. 2 / , sin 2 0 = - ( / „ - / * ) cos 20
i.e., tan 20 = - ^ ~ *r (6)
C O M P L E X S T R E S S I I 235
This is the negative reciprocal of E q . (3) so tha t the two values
of 0 (90° apart) which i t gives, are 45° in advance of θλ and 0 2
respectively. The planes of maximum shear are therefore midway
between the principal planes as shown in F ig . 2 1 1 .
FIG. 211
E.g. if 0 1 = 2 2 · 5Ο,
20 ! = 45° ,
tan 2 0 x = + 1 - 0 .
Hence, for the planes of maximum shear :
tan 20 = - 1 - 0 ,
20 = 135°,
i.e. 0 = 67-5° = 0, + 45° .
As already shown, fT = ^ L Z À j sin 20 - fs cos 20 (Eq. 2 ) ,
( — fy Han 20 cos 20 - fs cos 2 0 ,
236 S T R E N G T H O F M A T E R I A L S
and, from E q . (6), is a maximum when tan 20 ^ /,/ - F,
2/.,
/..• - /.v
' 7 m a x 2
2 / s '
Cv ~ F") cos 20 - /„ cos 20 (sub. for
V "I* I tan 20) .
= - c o s 20
= - c o s 20
4/,· + / V
4/Î
and cos 20 = + ] / - — — ^ r-, (/ t a n
2 20 + 1
±
4/Î + 1
±
4/Î
2 / .
y[(/., - / y )2 + 4 / f
Hence, /7< = ± ' ' ' max - J-
2/., (/ , - /„)» + 4/2
ç
4/«
= ± 1 - tu? + 4/Î.1, = i 2 (Difference of principal stresses)
/ A max / Λ min
Thus each value of / y m ax has the same magnitude and is the complementary shear of the other. Clearly, maximum shear stress occurs when / y m ax and FYMLN are o f opposite sign, since then,
/ y ma x = H / A W - (-/vmin)] (algebraic difference), = \ [/.V'max /-Vmin 1 '
Since all solids are three-dimensional there may be three
principal stresses acting over three mutually perpendicular
principal planes. I f the third principal stress is zero, as is usually
the case, then, if the other two principal stresses are of the same
C O M P L E X S T R E S S I I 237
sign, the maximum shear stress in the material will be given by
7'raax ~~ 2 (AVmax ^ ) - 2 either, /
/ V m a x =
2 lt\ or 0] = /Λ
2
(i f
^ m a x > / A m i nn u m e r i c a l l y ) ,
Principal Strains
I f a material is assumed isotropic, i.e. the elast ici ty is the same
in all directions, then, under complex stress, the greatest and least
direct strains will occur in the directions of the principal stresses
and will be linear functions of such stresses. Such strains are
referred to as Principal Strains.
Suppose principal stresses fx / 2 and / 3 to ac t upon the element
shown in Fig . 212. E a c h will produce the same strain as i t would
if acting independently. Thus
Principal strain in direction of ft :
fx <*h - ah 1 e Ε Ε
h Ε ~¥^2 ~'~
Similarly,
and 6 0
~Ê ~ ~E~ ^3 + / 1 ) '
FIG. 2 1 2
I f / j is the greatest principal stress and / 2 the least principal stress, then greatest principal strain difference :
, i \ fz
Hence.
h σ Ε ~Ë h - ft
Ε fl -
Ε /ι —
+ /ι>.
1 + a) and 1 + a = —-E_
2G (see p. 226)
2G
2 3 8 S T R E N G T H O F M A T E R I A L S
i.e. /i - /. = 2 / 3
/
Thus the maximum shear strain is
equal to the greatest difference in
the principal strains.
E X A M P L E . The stresses applied
a t a point in a material are as
shown. Determine the principal
stresses and the angles made with
the horizontal b y the planes over
which they ac t .
G '
= Maximum shear strain
5 t o n f / i n2
Solution
f,
fy
ÎS
AVmax =
2
- 5 ,
3 FIG. 2 1 3
\% + fv + l/[(f*- fyf + 4 / ? ) ] }
i [ 2 + ( - 5 ) + V(2- ( - 5 ) )2 + ( 4 χ 3
2) |
* [ - 3 + V(72 + 3 6 ) ]
9 - 2 2 = - 1 - 5
and
F rom E q . ( 3 ) :
- 1 - 5 + 4 - 6 1 = + 3 - 1 1 tonf/ in2 (tensile)
- 1 - 5 — 4 - 6 1 = — 6 - 1 1 tonf/in2 (compressive).
2 / , 6 tan 20 =
tu - U '
6
- 5 - ( + 2 ) '
- 0 - 8 5 7 . " - 7 '
2 0 ! = 1 8 0 - 4 0 - 6 ,
= 1 3 9 - 4 , θ1 = 6 9 - 7 deg.
Since Maximum shear stress / r m ax = — / 2 ) ,
C O M P L E X S T R E S S Π
f N i m j x= +3ΊΙ tonf/ in-
fN m i n
= ~
6' " +onf/ in
2 (Comp.)
FIG. 2 1 4
3 (Applied shear couple anticlockwise)
Complementary shear
15
F r o . 2 1 5
239
2 4 0 S T R E N G T H O F M A T E R I A L S
Alternatively, tan θ1 = /.V max
_ 3 - 1 1 - ( - 5 )
3
- 2 - 7 .
Λ 9 0 - 0 X = 2 0 - 3 deg, i.e. θχ = 6 9 - 7 deg as before.
The effect of reversing the applied shear couple i.e. of making i t
anticlockwise as in Fig . 2 1 5 is to t i l t the principal planes the other
way to get a mirror image of F ig . 2 1 4 as in F ig . 2 1 6 .
This is equivalent to measuring 0λ downward from the direction
of the horizontal stress.
E X A M P L E . A beam having the section shown in Fig . 2 1 7 sustains
a bending moment of 1 5 0 tonf in. and a shear force of 8 tonf.
Assume the shear stress to be uniform over the web section and
find the value of the maximum principal stress a t a point 5 in.
above the neutral axis.
f N = + 3 · Ι I t o n f / i n2
"max.
f N = - 6 · Ι I + o n f / i n2 (comp.)
FIG. 2 1 6
C O M P L E X S T R E S S I I 241
Solution
2nd Moment about N.A.
3-5 χ 1 23 3 125 χ 1 1
3
I = 12 12
- 503-5 - 346-5 = 157 in4.
Direct bending stress
4 M
y ν
j — — — , where y = 5 in . 150 χ 5
157 '
= 4-78 tonf/in2.
3 Web section — 11 χ —-,
8
= 4-125 in2.
ft Mean shear stress /,
I in
FIG. 2 1 7
4-125
*. Maximum principal stress :
1-95 tonf/in2.
/ A W = i [ / + l / ( /2+ 4 /
2) ] ,
= -ΐ-{4·78 + ]/[4-782 + (4 χ 1-94
2)]},
= l [4 -78 + V(22-9 χ 15-1)] ,
= £[4·78 + 6·1β] ,
10-94
12 in
- 5-47 tonf/in2.
Möhr's Circle:
From this can be determined graphically :
(a) Principal stresses / iv m ax
a n (i / ; v mi n>
(b) Max. shear stress / r m a x> (c) Normal and tangential stress on a given section.
242 S T R E N G T H O F M A T E R I A L S
Construction: Se t off 0 A X = fy (Fig. 2 1 8 ) ,
0 A 2 = fx,
A ^ j = fs = A 2 E 2 , normal to Ο Α 2Α χ .
FIG. 218
Bisec t A 2A j a t C and with C as centre draw a circle radius C E X .
J o i n Έ2ΟΕ1 and mark D 2 and D 2 . L e t D 1 C E 1 = 2Θ1 (i.e. twice the
angle between fx and plane of principal stress).
S e t off 2Θ ACW from CEX. Obtain point Ρ and project to A.
Inser t point Q. Then,
ODi = / A W = * & + / „ + UU - fy? + 4/f]} a t φί + 90) to fx, OA = fN = Normal Stress on Sect ion a t θ to fx.
O D 2 = / A W = Η/χ + /. - VtO« - fy? + 4/f]} a t (β, + 90) to /,. A P = fT = Tangent ia l stress on section a t θ to fx.
C Q = /tw = ± i V[(/x - / y )8 + 4/ s
2] a t 45° to
0 X = -| (angle DjCEj^ as measured).
When using the aforementioned construction, the following rules
must be observed :
1. S e t off tensile (i.e. positive) stresses to the right of point 0
and compressive to the left,
C O M P L E X S T R E S S I I
2 . Shear stress to be set off down
from Av and up from A 2 ,
3 . Se t off 2 0 A C W from 0ΈΧ,
4 . Measure 2Θ1 A C W from C E 2 .
E X A M P L E . The stresses applied a t
a point in a material are as shown in
Fig . 2 1 9 . F ind graphically the principal
stresses and the angles made with the
horizontal by the planes over which
they act . Ske tch the stress diagram.
15
FIG. 2 1 9
FIG. 2 2 0
243
244 S T R E N G T H O F M A T E R I A L S
Solution
f N M I N= - 6 - l l t o n f / i n2 (Comp.)
F i d . 2 2 1
(a)
fr =33
FIG. 2 2 2
The principal planes and stresses arc as shown in Fig. 221 :
E X A M P L E . F ind, using the figures of the previous example, the
effect of : • f N m j n= + 3 - 1 1 t o n f / i n
2
(a) reversing the di-
rection of fy,
(b) reversing the di-
rection of fu and fr,
(c) interchanging the
values of fy and fx in
(a) above.
Theories of Elastic Failure
I n simple tension a material is said to have failed when a per-
manent strain occurs, i.e. when the elastic l imit is exceeded. (For
steels, which obey Hooke's Law, the elastic l imit corresponds
approximately with the limit of proportionality).
C O M P L E X S T R E S S I I 245
(b)
I
FIG. 2 2 3
(c)
FIG. 2 2 4
The solution to (c) is left as an excercise.
246 S T R E N G T H O F M A T E R I A L S
The elastic limit in simple tension is usually thought of as a
definite value of tensile stress and, for simple tensile loading, the
working (operating) stress is usually limited to a value leaving a
margin of safety suitable to the conditions.
In a complex stress system there exist other quantities (see
below) one of which, according to the theory adopted, may be the
criterion of failure, i.e. lead to permanent strain. The value of this
quanti ty corresponding with the simple tensile elastic l imit is then
taken as the limiting value, with which the calculated actual value
in a given case is then compared.
The main hypotheses of failure are considered below in relation
to two-dimensional stress systems only.
1. Maximum Principal Stress Theory (Rankine)
This theory appears to hold good for brit t le materials generally
and, according to it , failure occurs when, irrespective of the values
of the other principal stress(es), the maximum principal stress
reaches the simple elastic limit stress,
i.e. when / iv m ax = ± / e (where fe = simple elastic l imit)
so that , for failure not to occur :
L{(fX-fy) + N(fX-fy)2 + ±M>fe
Note: If, numerically, / N m in > /jvm ax then the former value must be used.
If, in a given case, the principal stresses / j v m as
a n (l ÎN^
A RE
expressed as fractions of fe (i.e. as dimensionless numbers) and
plotted against one another, then the boundary within which the
plotted points must lie for safety (i.e. for failure not to occur) will
be the square E A B C D (Fig. 225) in which O E = O F = 1-0. In
other words, according to the Rankine Theory, failure will occur
when such points fall outside this perimeter, i.e. when the ratio
of either principal stress to the elastic l imit exceeds unity.
F o r combined bending and torsion (see later) i t should be noted
tha t fj\'msix is positive and fNmfn negative so tha t in this case the
fourth quadrant of Fig . 225 is the relevant one.
C O M P L E X S T R E S S I I 247
FN M I,
f.
Β 1-0 F i
Ε - 1 - 0 0 1-0
c -1-0 I
FIG. 2 2 5
Maximum Shear Stress Theory (Coulomb, Tresca and Guest)
This theory gives a good approximation where ductile materials are concerned and, according to i t , failure occurs when the maxi-mum shear stress is equal to the shear stress value ( / e/2) correspond-ing with the simple tensile elastic limit i.e. when
FR = — ·
' 1
MAX 2
Now, in general, / r m ax — ±2" (difference between greatest and least principal stresses) so that , when the two principal stresses are alike (of the same sign), the third principal stress being zero,
E i the r :
/ T W = ±i(AvMAX - ° ) IF
/ A W > /vmm numerically, 9*
248 S T R E N G T H O F M A T E R I A L S
i.e. ΊΫΨ. = ±FTM&X = ± A a t elastic limit,
•'• Aw > ±fe for ^ f e t y ( o r : - ^ - > ± 1 - 0 j .
Or: / T W = ±WN^ - 0 )
i.e. » ±FE for safety (or: > ± 1 - 0 ) .
I f the points representing the rat io of the principal stresses to
FE are plotted as before, then, for like principal stresses (1st and 3rd
quadrants), they must not be outside the boundaries E A F and
GCH (Fig. 226) .
FIG. 2 2 6
Again, when the two principal stresses are unlike, i.e. of opposite
sign (2nd and 3rd quadrants) then
FTAXNX = i 2 (Ανmax ~~ / min ) '
Since the limiting value of / r m ax is -—then, for safety: Δ
ÎNU
C O M P L E X S T R E S S I I 249
> ± 1 ,
i.e. Ji^- = > ± JL 9 thus fixing points U and V .
fe fe *
Clearly other points obtained from the above equation lie on the straight lines E H and F G so tha t the complete perimeter is E A F V G C H U E .
3. Maximum Strain Energy Theory (Haigh)
This theory also gives a good approximation where ductile materials are concerned and, according to i t , failure occurs when the strain energy per unit volume is equal to the value (β/2Ε)
corresponding with the simple elastic limit.
Suppose the three principal stresses to be fx / 2 and / 3 . Then, as already shown :
Principal strain in direction of / x ex = ~- —~ (/2 + /3) Ε Ε
Work done in lateral straining = χ La te ra l strain,
= - Γ Χ Ψ (k + h) Per unit volume,
Net work done by / 3 per unit volume = —\- — ——- (/ 2 + / 3 ) , ZE ZE
= w[/î-/iff(/. + /s)L
and since the minor principal stress is negative,
/jVmax + /iVmin > ± FE >
i.e. I ^ + h ^ u _ > ± 1
JE Je
Put t ing the first term equal to zero gives points F and H (Fig. 226) while points Ε and G result from putting the second term equal to zero.
When the two principal stresses are equal numerically then
2/Vnun 2
/ i V „
250 S T R E N G T H O F M A T E R I A L S
and work done by / 2 per unit volume = — [ / f — / 2 σ ( / 3 + / J ] , 2z^
and work done by / 3 per unit volume = —3— [/3 — /^( /χ + / 2)] . JILL/
Hence, to ta l strain energy per unit volume,
Υ = M - FMH + H) + 11- H°IH + H) + /i - HO (H + /,)],
= 2 j - [ / i2 + /! + /I - σ(/χ/2 + y, + HH + HK + HK + KM,
= [/? + / ! + / ! + 2 σ ( / 1 / 2 + HF3 + KH)L ·
Equat ing this t o — g i v e s
/? + /I + /§ - 2ff(/j/2 + / 2/ 3 + /,/,) = /J, and, if the third principal stress is zero, then
/! + /I - 2σΑ/2 = /;, or, using the previous notation,
/ Anax ~ /iVmin / mai AVmin =
/ c '
This is the equation of an ellipse the major and minor axes of
which intersect a t the origin, the major axis being inclined a t some
angle φ (Fig. 227) to the horizontal axis.
When FNMIN = 0 , then FNMIA = ±FE,
when FNMAI = 0 , then FNMIN = ±FE. Thus the ellipse passes through the points E , F , G and H (Fig. 226) .
To find the inclination φ and the magnitude of the major and
minor semi-axes, put
/ iv m ax = x c os
Ψ - Y s in
Ψ
and FNMIN = X sin φ - Y cos φ (Fig. 227)
Then, X2 cos φ + Y
2 s in
2 φ — 2 Χ Υ sin φ cos<p - f X
2 s i n
2^
+ Y2 cos
2ç> + 2 X Y sin φ cosç? — 2 σ ( Χ
2 sinç? cosç? + X Y cos
2«??
— X Y sin2ç9 — Y
2 sin φ coscp) = F
2
E,
i.e. X2 + Y
2 - 2 σ ( Χ Υ cos2<p + ( Χ
2 - Y
2) sin<p cos<p) = /
2.
C O M P L E X S T R E S S I I 251
When cos2<p = 0 then 2φ = 90 and φ = 45° . At this value of φ the
term in X Y disappears and the equation becomes
/ X2 — Y
2 \
X 2 _i_ Y2 _ 2σ I sin 2φ\ = /2 and sin 2φ = 1 0
Λ Χ 2 + Y2 _ σ{χ2 _ Υ2 ) = f2
or Χ2( 1 - α) + Υ
2( 1 + σ ) = /
2
;
Χ2 Υ
2
i.e. _ + 7 2 —= 1
·
Je Je
This is the equation of an ellipse of major semi-axis fel]'(l — σ)
passing through the origin and inclined at 45° to the horizontal axis . The minor semi-axis is / e / j / ( l + σ) at 90° to the major axis and also passing through the origin.
when a = 0-25, Major semi-axis = J * ^ ~ 1-15L, 0-75
Minor semi-axis = ^ ^ 0 - 8 9 / e .
The ellipse is shown in Fig . 228 and, for safety according to this theory, the points defined by the principal stresses must be within i t s perimeter.
252 S T R E N G T H O F M A T E R I A L S
The equation of the ellipse can be written :
I te I \ U I max / A min
fl FNM LN
*.
1-0 F
6 0-63 ! , Ε
\
i 0 8 2
/ ι / 1 /
•0
H
FIG. 2 2 8
Hence, when /A min /·>
2 σ
γ^- and is of the same sign,
te
i.e. [k^L^2(\ -σ) = 1,
whence, /ΑΓΠ 1
— ± 0 - 8 2 , when or = 0-25 . /E ^ 1/2(1 - σ
When the principal stresses are of different sign, then
7 A W \2 , »ff*
x2
whence, /Λ
= ± 1
]/2(1 +σ) ± 0 - 6 3 , when σ = 0-25 .
C O M P L E X S T R E S S I I 253
4. Shear Strain Energy Theory ( Von Mises and Hencky)
A good approximation in the case of ductile materials is also
given by this theory, which is to be preferred when the mean
principal stress (see below) is compressive. According to this theory
failure will occur when the to ta l shear strain energy is equal to the
value /?/66r a t the simple elastic limit. This energy is tha t part of
the to ta l strain energy which produces distortion, i.e. i t is the
to ta l strain energy less the strain energy due to volumetric stress.
Suppose the three principal stresses to be denoted by / j / 2
and / 3 .
Now, U = K/i + H + H) + - U) + WX - FS),
and
and
(Multiplying out will show tha t this is so.)
H = i(/i + H + H) + OFT - H) + - H) H = WX + H + H) + - H) + UH - /«)·
F I G . 2 2 9
The first term \(FX + / 2 + / 3 ) is the mean principal stress and this acting volumetrically on all three faces (Fig. 230) produces a volumetric strain unaccompanied b y distortion since the shear stress is everywhere zero.
f|+f2+f3 3
The second and third terms ac t as shown in F igs . 231 and 232 and are
3 3
FIG. 2 3 0
9 a SM
254 S T R E N G T H O F M A T E R I A L S
FIG. 2 3 2 FIG. 2 3 1
proportional to the three corresponding maximum shear stress
values since these values are themselves proportional to the
differences in principal stress. The strains due to these result in
distortion and the energy of distortion, i.e. the shear strain energy,
is therefore given by
Us = To ta l strain energy—Volumetric strain energy.
Now, Volumetric strain
ev = βι + e2 + e 3,
fi < r ( / a + h) , U o(U + fi) Ε E E Ε fi+U + h 2σ
Ε fi + /, + fs
Ε
Ε (1 - 2a)
(fi +U +fz)>
And Volumetric strain energy
Mean stress χ Strain,
Ε Ε
(1)
1 /ι + U + h χ e„
fi+f* + U v fi+U + fz 7^ X ~ Ε (1 -2a),
6E (/i + / , + hf (1 - 2σ)
1
(2)
To ta l strain energy = _ [ / « + / » + / » - 2 σ ( / 1/ 2 + / 2 / 3 + / 3 Λ ) ] (3) (from p. 250) J±
C O M P L E X S T E E S S I I
[2(/? + / I -Ι- / I ) - 2 ( / j / 2 + / 2 / 3 + y j ] ,
and 1 + σ = ,
= ~
/ a )2 + ( /a ~
/ a )a + ( /s ~
/ i ) 2 ]-
I n Simple tension a t the elastic limit f1 = fe and / 2 = f3
1
"· U s =
W( 2 / ΐ) =
~ 6 # "
Equat ing, A - [ f t - / , ) « + ( / , - / , ) · + ( / , - hf\ = A
i.e. ( / a - / 2 )2 + (/, - k)2 + (k - hf = 2/?·
I f the third principal stress is zero, then,
(A - kf + / ! + / ! = 2 / f ,
or, / Î - 2A/ 2 + / ! + / ! + /? = 2/? ,
i-e., /? + / ! - / i / , = / | .
Or, using the previous notat ion,
/ iVmax /iVmin / ^ m a x / ^ m i n =
/e
Hence, Shear strain energy
u. = u- u„
= 4 WM + {L 1 IL
~ M HK + HK + / S / L )]
- A f t + / , + / , ) « ( l - ä r ) ,
= - ^ - { 3 ( / ? + fl+ il) - 6 σ ( / χ / 2 + kh + Μι)
- (1 - 2er) [/} + / I + / f + 2 ( Λ / 2 + /2/3 + / a / x ) ] } ,
= -^W+fl+il) ( 3 - H - 2σ)
- ( / ι / 2 + / 2 / 3 + / 3 / ι ) ( 6 ( Τ + 2 - 4 σ ) ] ,
= A { ( 2 + 2σ[(/? + / f + / i ) - ( / x/ 2 + kh + kh)l},
1 + Σ
Γ Λ / / 2 ι ;2 ι 42
255
ι (6)
9 a*
256 S T R E N G T H O F M A T E R I A L S
i.e.
or
X2 - I - Y
2 -
X2 + Y
2 -
X2 ^
9
2
X2 Y
2
~~2 2"
3 Y2
whence, X
2 Y*_
9 / 2 2 /2
= /I-
= 1. (8)
The major semi-axis is thus ] /2/ e so tha t the ellipse passes through
points A and C Fig . 233 while the minor axis is | / f fe or 0 ·82/ β
approx.
f.
F
/
y/ /
/ \
g /
/
/E 0
\
Ao
H - 1 - 0
FIG. 2 3 3
Dividing through by fe, gives
/ /iVmax \2 , (jj^mmY _ /^max /A
7min _ τ (H\
\ fe ) \ fe ) fe ' ' This is the equation of an ellipse similar to the one obtained
under the previous (Haigh) theory. Put t ing / iv m ax = X cos9? — Y
sin<p and fNmm — X s m <
P + Y COSÇJ as before and substituting in
E q . (6) gives, for φ — 45 deg.
X2 — Y
2
- sin 2φ ^ /2, where sin 2ç? = 1,
C O M P L E X S T R E S S I I 257
Again, when / Λ Ί η Ιη = 0 , !ψ± = ± 1 from E q . (7) le
and when fNmBki = 0 , = ± 1 ,
le so tha t the ellipse, within which the points must lie for safety
(according to this theory) passes also through E , F , G and H which
are common to the perimeters derived from theories 1, 2 and 3.
5. Maximum Principal Strain Theory (St. Venant)
According to this theory, which should not be generally used since i t overestimates the strength of ductile materials, failure will occur when the greatest principal strain reaches the value fejE
corresponding with the simple elastic limit, i.e. when, assuming principal stresses fx / 2 and / 3 , either,
4 " - ! - ( / • + /») = ± j - . i-e. h - * ( / , + / , ) = ±U,
or A _ I L ( / , + / L ) = ±1±, i.e. / , -σ( / ,+/ 1 )= ± fe,
or A _ £ . ( / 1 + / i ) = ± A . , i.e. h-a{h + ft) = ±fc.
F o r a two-dimensional stress system, / 3 = 0. Then, either
/i - tf/2 = ±/<<> or /2 - ff/i = ±/e-Using the previous notation, the above will be written
F o r like principal stresses, when / i Y m In = 0 , fNmax = ±fe,
and when fNmax = 0 , fNlûin = ±fe.
Thus the perimeter passes through the points E F G and H as in the previous four cases.
When fNmln = fNmax then, (taking a = 0-25) ,
ι _ /A7
ma x ι ι
/Armax ^ — = t / c ;
. / iVmax _ , _ fNmln
"h 3 / „ '
FIG. 2 3 5
258 S T R E N G T H OF M A T E R I A L S
I FIG. 2 3 4
C O M P L E X S T R E S S I I 259
This fixes points J and K . F o r unlike principal stresses,when fNmin = — / A W
/ ^ m a x , l ™ / iVmax , ^ /^Vmin tnen fNm + = ± / e , whence ^ = ± • β
This fixes points R and S.
Composite Graphical Representation
B y superimposing the perimeters corresponding with each of the
five theories, a composite figure is obtained which is usually referred
to as a B e c k e r Diagram.
1. Rankine Square A B C D 2. Coulomb, Tresca, Guest Figure A F G C H E 3. Haigh Ellipse P F N G Q H M E 4 . Von Mises, Hencky Ellipse A F G C H E 5. S t . Venant Rhombus J F R G K H S E
Combined Bending and Torsion
The shear stress induced in a shaft by torsion is given by
Tr , τΦ = ——, where J =
J ' 32 '
T h e tensile and compressive bending stresses are given by
Mr _ T i d4
/ = — — , where I 64
The criteria for failure are usually :
(a) Max. principal stress, i.e. tha t on a plane where the shear stress is zero, or where the stress is entirely normal,
(b) Max. shear stress, i.e. failure is due to shear. Br i t t l e materials fail in tension—hence (a) is here used and ductile materials fail in shear—hence (b) is here used.
Consider the greatly magnified element E G F D (Fig. 236) : The shear stresses on E G and F D are accompanied by equal shears on F G and E D . The bending stress (tensile or compressive) acts normally to E G and F D .
L e t fN = normal stress, 1 required for equilibrium of
fT == tangential (shear) stress j portion E F D (Fig. 237) .
260 S T R E N G T H O F M A T E R I A L S
W
FIG. 236
f
FIG. 237
Comparing the system from which the principal stress formulae
were derived with the shaft stress system, i t will be seen tha t
/* = /
/ir = 0
Hence, substituting in \{FX + FY ± ]/L(FX ~ + 4 / ? ] } , Mr Tr
M a x . p r i n . s t r e s s , / A W = i [ / + l / ( / 2 + 4 / 1 ) ] , w h e r e / — , / , = —
Min. prin. stress, FNMIN = \[F - | / ( / 2 + 4 / 2) ] .
Note: 1. Since ^(/2 + 4/
2) is always > / it follows that fNm&K is tensile and
fNmin is compressive, i.e. principal stresses are of opposite sign. 2. For the other side of the neutral axis, the bending stress is compres-
sive. This alters the values of fNmax and fNmla but not the signs. 2fs 2fs
Again, substituting in - — ^ - 7 - , tan 2 0 = 7 ^ . Iy ~ fx /
C O M P L E X S T R E S S I I 261
This gives the two angles θ1 and 0 2 made by the principal planes
with the bending stress / .
Final ly, substituting in ± | - ftJf + 4 /2] ,
Max. shear stress /rm ax = ±j V(/2 + 4/Î)
Summary—Combined bending and torsion :
FIG. 2 3 8
Normal stress, / γ = / sin 20 - f /> sin 2 6 ,
Max./Min. prin. stresses / A W = \ ]/[f(±) V(/2 + 4 /
2) ]
( m i n )
2f on planes given by tan 26 = γ-, 90 deg apart over which
fT = 0 /
Shear stress, / r = \f sin 26 - f fs cos 2 6 ,
Max. shear stress / 7 m ax = ± \ | / ( /2 + 4 /
2) .
FIG. 2 3 9
262 S T R E N G T H O F M A T E R I A L S
On planes given by tan 2Θ = / / 2 / s , 90 deg apart and halfway between principal planes.
Equivalent Torque
As already shown,
1 Mr / a w = g" [/ + V(/2 + 4 /
2) ] , where / = — and /,
4Mr
nr* + lQM
2r
2 \6T
2r
2
(nr* ( t i t *
2M 1 4r
T r r3 2 π
4Γ
r
2 M 2 :±5- + - £ - y ( J | f i
Ufr
7 r r4/ 4
4 J f r
~πτΓ
Tr
~7~' Tr
2Tr
τη. ÎN m a x • [M + | / ( M
2 + T
2) ] ,
i.e., / λ · , M + ]j(M2 + T
2) and similarly,
Now, —— fs = —— / s = Torque for a solid shaft under shear stress
only. Since the e x p r e s s i o n — — / A m ax has the same form, it is re-Δ
ferred to as the Equivalent Twisting Moment and denoted by ΤE.
Thus, Τ Ε; = M + ]/(ilf2 + Τ
2) (Rankine Formula for brit t le
shafts.)
Again, as already shown,
1 fi - V ( / 2 + 4 / f ) , where / = _ and fs = —
(from above) ,
4 K 1 6 1 f
2r
2 1 6 T
2r
2
+ (nr* (nr*f
Λ ii
1 4r
2 Ivf-
2 nr
3 y (71/2 +y 2 ) >
C O M P L E X S T R E S S I I 263
i.e. —— / r m ax = ]/(M2 + T
2) = Equivalent torque as before.
Δ
Thus TE = ]/(M2 + T
2) (Coulomb Formula for ductile shafts.)
The above expressions apply only to solid shafts.
F o r a hollow shaft J = n(R* — r4) /2 and must be separately
calculated. I t can then be substi tuted in the equation
TE / Λ ' ΜΧ
From the foregoing an equivalent torque may be described as
tha t combined bending and twisting moment (]/(M2 + T2) or
M + ] / ( l f2 - f T
2), depending on the criterion of failure) which will
induce a given maximum principal/shear stress a t radius r when
used in the torsion equation,
V(^2 + r 3 ) /rm a. i.e., J
where ]j(M2 -f- Τ
2) — ΤE when the criterion of failure is the
maximum shear stress.
J r '
where M - f ]I(M2 + T
2) = T
7^ when the criterion of failure is the
maximum principal (tensile) stress.
Note tha t an expression for equivalent torque can be derived from each of the other theories of elastic failure.
E X A M P L E . F ind a suitable diameter for a mild steel shaft required
to t ransmit 9 h.p. a t 1 3 0 0 rev/min. given tha t the permitted
maximum shear stress is 6 0 0 0 lbf/in2. I t may be assumed tha t the
bending moment act ing is of equal magnitude to the torque.
Solution ... n m 3 3 , 0 0 0 χ h.p. „ r f I orque transmitted, Τ = —— lbf f t ,
2τιΝ
3 3 , 0 0 0 χ 9 >
2π χ 1 3 0 0
4 3 6 lbf in.
Hence this is also the value of M.
264 S T R E N G T H O F M A T E R I A L S
Since the criterion is the maximum shear stress, the Coulomb
Formula must be employed.
= ]/(4362 + 4 3 6
2) ,
= V(2 x 4 ·362 χ 1 0 0
2) ,
= 100 1/38,
= 617 and / w = 6000 lbf/in2.
3 617 χ 2
"T " π x 6000 '
= 0 0 6 5 5 . ,
whence, r = 0-403 in.,
13 d = 0-806 in , say, — - i n . dia. J
16
E X A M P L E . The maximum torque carried b y a crankshaft is
40 tonf ft and the corresponding bending moment is 28 tonf ft. I f
the safe tensile and shear stresses are respectively 6 tonf/in3. and
3-5 tonf/in2 find the required shaft diameter b y (a) Rankine
Formula , (b) Coulomb Formula .
Solution
1. Rank ine :
— / j V m ax - M + ]/(M2 + T
2) (equivalent torque)
= 28 + V (28 2 + 4 0 2 )
= 28 + 48-83 ,
= 76-83 tonf f t ,
= 922 tonf in. , and fNmax = 6 tonf/in2.
922 χ 2
π x 6
= 97-8 .
• r = 4 - 6 in,, i.e. d = 9-2 in.
C O M P L E X S T R E S S I I 265
2. Coulomb:
max |/(if
2 + T
2) ,
= 48-83 (from above) ,
= 585 tonf in. and fTmaK = 3-5 tonf/in2.
585 χ 2 * γ* —
π χ 3-5
- 106-3.
r = 4-735 in. i.e., d = 9-47 in.
A suitable diameter is therefore 9 J in.
E X A M P L E . F ind the principal stresses in a propeller shaft and the inclination to the axis of the planes over which they ac t given tha t (a) the thrust results in an axial compressive stress of 1 tonf/in
2
and (b) the engine torque results in a shear stress a t the shaft surface of 4 tonf/in
2.
f N m in = - 4 - 5 3
= 3 -53
f = -
f s= 4
FIG. 2 4 0
Solution
/ A w = i~[/ + V ( / 2 + 4 / f ) ] , = + + (4 χ 4^ ) ] ,
= * ( - ! + 1/65),
= ^ ( - 1 + 8-06) = = 3-53 tonf/ in2.
2 6 6 S T R E N G T H O F M A T E R I A L S
tan 20 = - - y - >
_ 2 x 4
- 1 ' =r. 8 .
/. 201 = 82°54 ' ,
i.e. 01 = 41°27A.
.·. 0 a = 41°27/ + 90 - 131°27'
The planes and stresses are then as shown in Fig . 240 .
E X A M P L E . I f M = 10,000 lbf ft, Τ = 12,000 lbf ft and
/ * m a X ^ 1^,000 lbf/in2 est imate a suitable diameter for the shaft
and calculate the value of the greatest tensile stress in the material .
( I . Mech. E . )
Solution
Using the Guest Formula :
TE = —— fSmax = ]/(M2 + Τ
2), where M = 10,000 lbf f t .
Δ
Τ = 12,000 lbf f t .
= ]/[(100 χ 106) + (144 x 1 0
e) ] ,
= 1 03 1/(100 + 144 ) ,
- 15,650 lbf f t . ,
= 188,000 lbf in. and fSmax = 10,000 lbf/in2.
3 _ 188,000 χ 2
' ' T
10,000 π '
= 11-94,
r = 2-29 in., i.e., d = 4-58 in.
Using the Rankine Formula to find the max . principal stress :
TE = ^fNm&x = M + y(M2 + T
2),
= 10,000 + 15 ,650 ,
= 25 ,620 lbf f t ,
= 308,000 lbf in. , (and r = 2-29 in.)
_ 308,000 χ 2 • · m a x - π χ 2. 2 93
= 16,375 lbf/in2.
C O M P L E X S T R E S S I I 267
E X A M P L E . A meta l tube 3 in. inside
dia. 0-125 in. th ick is held rigidly a t one
end in a vert ical position and a flexible
cable 0-125 in. dia. is wound on the upper-
most 2 in., the ends being brought out as
shown. Es t imate the maximum tensile
stress in the tube when the pull on each
end of the cable is 240 lbf.
Solution
Effective cable radius
FIG. 2 4 1
810 lbf in .
2 4 0
2 4 0 lbf 2 4 0
L - 2 L
FIG. 2 4 2 FIG. 2 4 3
At any point in the tube distant L from the top :
Bending moment M = 240 L - 240 (L-2),
= 240 L - 240 L + 4 8 0 ,
= 480 lbf in .
268 S T R E N G T H O F M A T E R I A L S
Equivalent torque TE = M + ]/(M2 + T
2),
= 480 + ]/(4802 + 8102),
= 480 + 1/(230,400 + 656,100),
- 480 + 1/886,500,
- 480 + 940,
- 1420 lbf in.
Outside diameter = 3 + (2 χ 0-125) = 3-25 in.
Polar moment J = (3-254 - 3 4),
= -g- (10-55 + 9) (10-55 - 9),
_ π χ 19-55 χ 1-55 32 '
= 2-98 in4.
Σ± = Ί ^ - / 1 4 20 w 3 2 5
j r > . . / A m : ix 2. 9 8 2
= 773 lbf/in2.
E X A M P L E . The shaft in the system shown is 6 in dia. and trans-
mits 250 h.p. a t 80 rev/min the maximum torque being 1-25 χ Mean.
Neglect the weight of the shaft and find the max. tensile and shear
stresses induced in i t .
12 in 9 6 i n -
2 -5 t o n f
FIG. 2 4 4
C O M P L E X S T R E S S I I 269
/ 1-25 χ 33 ,000 χ 2 5 0 \
\ 2π χ 80 j 12 = 246,000 lbf in .
Moments about the L H end give: i? 2(12 + 96) = 2-5 χ 12 ,
R2= 2
'5
* 12
= 0-278 tonf .
108
M a x . B .M. , M = 0-278 χ 2240 χ 96 ,
= 5 9 , 7 0 0 lbf. in. T T f
3
F o r failure in shear, TE = — / S m ax = ]/(M2 + T
2),
= ] /(59,7002 + 2 4 6 , 0 0 0
2) ,
= 253,000 lbf in .
.'. /.nax = 253,000 χ π χ 3
3 '
= 6000 lbf/in2 ( = 2-68 tonf / in
2) .
Fo r failure other than in shear, TE = - y Avmax/min
= M ± ]/(M2 + T
2) ,
= 59 ,700 ± 253 ,000 ,
= + 3 1 2 , 7 0 0 lbf in. ,
or = - 1 9 3 , 3 0 0 lbf in.
.·. / , V m ax = 312,700 χ = 7350 lbf/in2 (tensile),
TZ Χ ο
( = 3-28 tonf / in2) ,
2 a n d Avmin = - 1 9 3 , 3 0 0 χ — = 4550 lbf/in
2 (compressive).
71 X od
E X A M P L E . A shaft 4-0 in dia. carries a flywheel weighing 2-0 tonf
a t the centre of a simply supported span of 8-0 ft. The centroid of
the flywheel is 0 -1 in. from the axis of rotat ion. I f the shaft t ransmits
1000 h.p. a t 360 rev/min., es t imate the maximum stress a t the
centroid, (a) when this is vert ically below the axis, (b) when this is vert ically above the axis.
Solution
Max. torque, Τ = 1-25 lbf f t ,
1-25 χ 33 ,000 χ 250
270 S T R E N G T H O F M A T E R I A L S
Solution
Angular velocity
Centrifugal force
W Fe = —œ
2r,
2π χ 360 C0=—60—
= 12jz;rad/s.
ω = 1420.
1 —4-0 f t •
12 χ 2240 \ Έ, Λ Λ
( - 3 2 5 - ) 1 4 20
1640 lbf .
~Ϊ2
| 4 · 0 in
Flywheel weight
W = 2 χ 2 2 4 0 ,
= 4480 lbf . 2 ± F C
FIG. 2 4 5
The centrifugal force must be added to or subtracted from the flywheel weight when the centroid is below or above the axis respectively.
(a) Load per support = £(4480 + 1640) = 3060 lbf .
Maximum B . M . = 3060(4 χ 12) = 147,000 lbf in .
' 3 3 , 0 0 0 χ 1000 \ Torque t ransmit ted = χ 12 = 175,000 lbf in.
2π χ 360 I
Equivalent torque TE = M + ]/(M2 + T
2),
= 147,000 + ]/(147,0002 + 1 7 5 , 0 0 0
2) ,
= 147,000 + 2 2 8 , 0 0 0 ,
= 375,000 lbf in .
/ i V ma Now, ^-y-, where r = 0-1 and J = , J Δ
π2 χ 2*
C O M P L E X S T R E S S I I 2 7 1
.'. Max. principal stress / Λ o. l (
3 7 5 , 0 0 0
8π
1 4 9 0 lbf/in2 (tensile).
(b) Load per support = ± ( 4 4 8 0 - 1 6 4 0 ) - 1 4 2 0 lbf .
Maximum B . M . = 1 4 2 0 ( 4 χ 1 2 ) = 6 8 , 0 0 0 lbf in. and Τ is
unchanged.
/ . TE = 6 8 , 0 0 0 + } / 6 8 , 0 0 02 + 1 7 5 , 0 0 0
2)
= 6 8 , 0 0 0 + 1 8 8 , 0 0 0 ,
= 2 5 6 , 0 0 0 lbf in.
Hence, / 2 5 6 , 0 0 0
\ 8ττ
1 0 2 0 lbf/in2 (tensile).
E X A M P L E . A hollow shaft is to t ransmit 5 0 h.p. a t 2 5 0 0 rev/min
and must withstand a simultaneous bending moment of 2 9 0 lbf in.
I f the internal diameter is to be 0 - 6 5 of the external diameter and
the maximum principal stress is not to exceed 5 tonf/in2 calculate
suitable shaft diameters.
1 2 ,
Solution
3 3 , 0 0 0 χ h.p. Torque Τ = ^
_ f 3 3 , 0 0 0 χ 5 0 \
~ \ 2π χ 2 5 0 0 J
= 1 2 6 0 lbf in.
Equivalent twisting moment,
= 2 9 0 + ] / ( 2 9 02 + 1 2 6 0
2) ,
= 1 9 0 + 1 2 9 0 ,
= 1 5 8 0 lbf in.
272 S T R E N G T H O F M A T E R I A L S
F o r a hollow shaft, J = ( £4
- r4
) and r = 0-65Ä,
[R* - (0 -65Ä)4] ,
£4[ 1 - (0 -65 )
4] ,
(1 - 0 - 1 7 8 ) ,
nR* = l - 2 9 i ?4i n
4.
2
π
π
Τ nR*
2
0-822
Again, r£_E_
J
1580
1-29I?4 :
R3
/Λ'»
R
5 χ 2240
(where / Λ ι Μχ > 5 tonf / in2) .
R
1580
5 χ 2240 χ 1-29 '
= 0-109.
R = 0-478 in., i.e. D = 0-956 in. (say 1 in.) .
.*. Internal diameter = 0-65 in.
E X A M P L E . A 6 tonf propeller on a 9 in. dia. shaft overhangs the
bearing by 18 in. and exerts a thrust of 15 tonf when the input to
i t is 4000 h.p. a t 300 rev/min. Calculate the principal stresses a t
the point of support :
(a) a t the ends of a vert ical diameter and
(b) a t the ends of a horizontal diameter.
6 t o n f
FIG. 2 4 6
C O M P L E X S T R E S S I I 273
Solution
Bending moment,
Torque,
M = (6 χ 2 2 4 0 ) 1 8 ,
= 242 ,000 18 lbf in. ,
Τ = / 33 ,000 χ 4000 \
V 2π χ 300 / '
840,000 lbf in.
y = 4-5 in
I = 322 in4.
J χ 94,
32
644 in4.
Bending stress,
Direct stress,
My _ 242,000 χ 4-5
~~Γ ~ 322
F_
~A
± 3 3 8 0 lbf/in2.
15 x 2240 , = - 5 3 0 lbf/in2 (compres-
0-785 χ 92 sive).
ι j j . Tr 840,000 x 4-5 Shearstressduetotorsion = — - = —— = 5870 lbl/nr*.
J 644
Λ Τ Ι , . ^ . Ι ^ · 4 IF 4 6 χ 2240 Max. shear stress due to bending = — χ — - = — χ -γτ-^ζ τττ
= 282 lbf/in2. (see Chapter X I I )
F o r point A :
and fNmS]
max 2
1
F o r point C :
- 5 3 0 ) + 3380 / = 2 8 5 0 , = 5870 J
{2850 + ]/[28502 + 4(5870
2) ]} = 2 = 7475 lbf/in
2,
{2850 - ] /[28502 + 4(5870
2) ]} = - 4 6 2 5 lbf/in
2.
- 5 3 0 - 3 3 8 0 J / = - 3 9 1 0 , = 5870 1
··· fit** = γ { - 3 9 1 0 + V[3910* + 4(58702) ]} = 4245 Ibf/in
2,
274 S T R E N G T H O F M A T E R I A L S
a n d /jvmin = I T { - 3 9 1 0 - ]/[3910
2 + 4(5870
2) ]} = - 8 1 5 5 lbf/in
2.
Δ
For point Β
• f r
- 5 3 0 = / - 5870 + 282 = 6152
.'. /.v«,« = γ { - 5 3 0 + ]/[5302 + 4 ( 6 1 5 2
2) ] } , - + 5 8 8 5 lbf/in
2,
and / i V m in = - - { - 5 3 0 - }/[5302 + 4 ( 6 1 5 2
2) ] } , = - 6 4 1 5 lbf/in
2.
The maximum compressive stress is therefore 8155 lbf/in2 a t
point C and the maximum tensile stress is 7475 lbf/in2 a t point A.
E X A M P L E . Calculate for points X and Y on the shaft shown, the
principal stresses and illustrate their directions on a suitable
diagram. Assume clockwise rotation when viewed from the input
end.
100 rev/min 1«·—3ft—·+«—3 f t -
m C_ Input
•γ 150 h.p.
3 tonf
FIG. 2 4 7
- 3 f t » h 3ft—»j
I
6 in dia.
3 tonf
F o r bending only :
Compressive stress a t X = Tensile stress a t Y ,
My and / where M = 3(3 χ 12) tonf in.
= 108 tonf in.,
π
64
~2 ~ ~ 2 :
108 χ 3
y
x 64 = 62 in4,
3 in.
62
± 5 - 2 tonf/in2.
C O M P L E X S T R E S S I I 275
F o r torsion only :
where J = 21 = 124 in4,
/ 33 ,000 χ 150
~ \ 2π χ 100 12 χ 1-5,
= 142,000 lbf in.
= 63-3 tonf in.
63-3 χ 3
~ Ϊ 2 4 '
= 1-53 tonf/in2.
= I- { - 5-2 + ]/[5·22 + (4 χ 1-53
2)]}
= H - 5 - 2 + 6-0)
= - 2 - 6 + 3 0
= + 0 - 4 tonf/in2.
fantn = ~2
*6 - 3-0
= 5*6 tonf/in2.
X : h + V ( / » + 4/î)]
tan 2θα1 = r-
t Λ 2 ö n = 30°30
Γ,
2 χ 1-53
- 5 - 2 = + 0 - 5 9 .
fln = 15°15 '
At Υ = Κ + 5 - 2 + 6-0) ,
= + 5 - 6 tonf/in2.
/ Λ η , η = έ ( + 5 · 2 - 6 · 0 ) ,
= - 0 - 4 tonf/in2.
= - 0 - 5 9 .
θν1 = 74°46 ' . (see Fig . 248)
276 S T R E N G T H O F M A T E R I A L S
Examples I X
1 . Calculate the values of the principal and maximum shear stresses and the inclinations of the planes over which they act given that
fx = 8-0 tonf/in2 (horizontal),
fy = 3 0 tonf/in2 (vertical),
fz = 0, fs = 3-0 tonf/in
2 clockwise in the plane of fx and fy.
Check the values obtained by means of the Möhr Circle and insert them on a suitable sketch (9-4, 1-6 and ±3-9 tonf/in
2, θ1 = 25-1 deg).
2. State the five main theories which have been advanced with the object of setting up a criterion of elastic failure under complex stress. Discuss briefly their relative merits.
Fio. 248
C O M P L E X S T R E S S I I 277
3. The bending moment in a 2 0 in. dia. solid shaft rotating at 300 rev/min is numerically equal to 40 per cent of the torque. If the maximum shear stress in the material is 5 tonf/in
2 calculate the power being transmitted
(778 h.p.).
4. The flywheel of a single cylinder gas engine weighs 2-0 tonf, has a radius of gyration of 2-5 ft and is mounted on a 6-0 in. dia. shaft which overhangs the bearing by 1-25 ft. If, as a result of fluctuation in torque, the flywheel has a maximum instantaneous angular acceleration of 4-8 rad/s
2 determine
the maximum shear stress induced in the shaft (0-89 tonf/in2).
5. Find the ratio of twisting and bending moments in a solid circular shaft given that the minimum principal stress is numerically equal to one fifth of the maximum principal stress (TjM = ^(20)/4).
10 SM
C H A P T E R Χ
BEAMS IV—DEFLECTION
Flexure and Radius of Curvature
I n accordance with the sign convention, a clockwise bending
moment to the left of a section is taken as positive. Such a bending
moment induces a " sagg ing" curvature i.e. concave when viewed
from above as in Fig . 249 .
Positive bending moment M
Negative Q| (sagging) detlection y
Graph of slope | f
FEG. 2 4 9
P a r t of a sagging beam of varying radius of curvature, i.e.,
deflected b y a varying positive bending moment is shown in
Fig . 250 .
2 7 8
B E A M S I V D E F L E C T I O N 279
Slope of beam over element ds,
$ - = UnO, dx
·' dx* = sec 0
2
dx
differentiating,
sec20 4 - - -τ—·
ds dx
sec20 x — x sec0
R
(and sec20 = 1 + t a n
20 ) .
. 1 &y
" R dx2 (1 + tan
20)
3'
and tan 0 = dy_
dx
dx2 1 + V.
FIG. 2 5 0
Since deflection is small within the elastic limit, dy/dx is of the
order of small quantities and i ts powers may by neglected.
1 d2y
Thus, in this case — = - — - . R dx
2
Similarly, for a negative bending moment resulting in a "hogging "
curvature, i.e., convex when viewed from above:
1 d2y
— = — · Also, from the bending equation,
l M xl x . , 1 d2y M
so tha t m general : _ = ± _ = —
Hence, when / is constant,
This is known as the Differential Equation of Flexure.
ί M d2y\
Note tha t when / is not constant —— = ± Ε —— . V I dx
2 J
10*
2 8 0 S T R E N G T H O F M A T E R I A L S
I f both sides of this equation are integrated with respect to χ
then, ±EI^- = f M dx, where M is expressed in d x
J terms of x.
i.e., slope 4 - = ± —!- f]\ldx. ax EI J
Again, deflection y = ± -^j- j*JM dx.
The above general expressions may be applied to specific cases of
loading, care being taken to obtain the correct sign for d2yfdx
2.
Referring to F ig . 249 i t can be seen tha t the slope dy/dx in-
creases from negative through zero to positive so tha t d2y/dx
2 is
+ v e . in this case. Conversely, for the same axes, a hogging (nega-
t ive) bending moment makes d2yldx
2 negative.
Horizontal Cantilever with Concentrated Load at Free End
FIG. 2 5 1
F o r any section X X distant χ from the origin 0 :
MX = EI—^ = - W(L-x),
:. EI^r- = -W(LX -^L) + A, and when x = Q,^L = 0 , dx \ 2 / da;
Hence, EI^- = - W (LX -dx \ 2 /
Λ - 4 = 0 .
B E A M S I V — D E F L E C T I O N 281
dy from which the s l o p e - — a t any point can be found.
ax
f x2 1 x
z \
And, Ely = -WlL— - — — I + B, and when χ = 0 ,
y = 0, .'.B = 0.
Hence, Ely = - W ( γ χ2 -
from which the deflection y a t any point can be found.
At the free end, y = ymAX and χ = L,
Ely* -'(τ»-τ)-
(Τ ~ ~ 6 ~ ) ' = — WL*
WL3
Thus, denoting the maximum deflection ymsix by Z, we have
3 EI '
The negative sign shows tha t the deflection is downwards from the origin considered.
E X A M P L E . A horizontal cable carrying a tension of 300 lbf is
a t tached to the upper end of a wooden pole 6 in. dia. 12 ft high,
the lower end being embedded in the ground.
I f Ε = 1·0 x 1 06 lbf/in
2, find (a) the maximum bending stress
in the pole and (b) the deflection a t the top.
Solution
2nd Moment of area of section
64 ' π
144 in
X
y = ¥ = 3 i n .
64
= 63-65 in4.
and M = 300 χ 144
= 43 ,200 lbf in.
77777^
-300 Ibl
-6 in dia.
\777,
Ρίο . 252
282 S T R E N G T H O F M A T E R I A L S
.*. Ground level bending stress a t y = 3 / My_
I 9
43,200 χ 3
= 2035 lbf/in2.
Deflection a t upper end,
1
WL3
EI
300 χ 1 4 43
1-0 χ ΙΟ6 χ 63-65
= 4-7 in (approx.)
E X A M P L E . A short horizontal canti lever 12 in. long, 3 in. wide
and 5 in. deep of material having Ε = 13,500 tonf/in2 carries a
concentrated load a t the free end. I f a spirit level half-way along
registers an incline of 1 in 2000 , est imate the value of the load.
Solution
« L = 12 in - >>
1 1
1 u _
dx; 1
J Υ
i I
3 in
- 5 in
FIG. 2 5 3
2nd Moment I 12 '
_ 3 x 53
12 9
= 31-25 in4.
Mx = EIiâ== - w(L~x)> EI
ày
dx •W Lx — A,
B E A M S I V — D E F L E C T I O N 2 8 3
and when χ = 0, ^L = 0, t\ A = 0 . dx
.·. BIEL da;
dy
•W
and when χ = 6 , -τ- = — ——— , ' dx 2 0 0 0 '
1 3 , 5 0 0 χ 3 1 - 2 5
2 0 0 0 = -WÇ72 - 1 8 ) ,
whence, W = 3 - 9 tonf.
E X A M P L E . A cantilever of circular section is 4 8 in. long. Estimate the diameter necessary to prevent the maximum stress from exceeding 9 0 0 0 lbf/in
2 when there is a concentrated load of 5 0 0 0 lbf
at the free end. Calculate also the corresponding deflection at the load point, taking Ε = 2 9 x 1 0
E lbf/in
2.
Solution
FIG. 2 5 4
Bending moment, M = 5 0 0 0 χ 4 8 ,
= 2 4 0 , 0 0 0 lbf in.
/ = 9 0 0 0 lbf/in2,
y = - ~ , where d is the required diameter,
284 S T R E N G T H O F M A T E R I A L S
Since,
9000 -
My
240,000 χ 4"
nd1
64
120,000 χ 64
Hence,
9000π
= 2 7 1 .
d = 6-5 in.
r π χ 6 ·44
64
= 86 in4.
1 WL3
Deflection a t free end, Ζ = —- χ 3 EI '
1 5000 χ 4 83
χ 3 " 29 χ ΙΟ6 χ 86 '
= 0-074 in.
E X A M P L E . An I-section jois t 10 χ 6 in. overall has a web-J in.
thick and flanges ^ in. thick. I t is built into a wall with the web
vertical for use with lifting tackle , the overhang being 7 ft 6 in.
Additional support is given to the free end by a steel t ie bar 1 in.
wide and f in. thick, the upper end of which is rigidly fixed 24 ft
above. Take Ε for cantilever and tie as 13,400 tonf/in2 and est imate
the deflection produced by a concentrated load of 5 tonf at the free
end. F ind also the ratio of the tensile stress in the tie to the maxi-
mum bending stress in the jois t .
Solution F
Tensile stress in tie bar /
Extension in tie bar χ
1 χ 0-375
Ε '
F 24 χ 12 χ 0-375 Ε
—, F .
B E A M S I V — D E F L E C T I O N 285
1
"Î2
19QQ
~ Ϊ 2 ~ '
158 in4.
(6000 - 4 1 0 0 ) ,
7 f t 6 in
5 tonf
24 ft
FIG. 255
Deflection a t free end of cantilever,
Ζ = WL
3
3EI
(5 - F) 9 03
3E χ 158 '
and this is equal to the tie bar extension.
10 a SM
where W = 5 - F, L = 90 in,
F o r cantilever section,
/N.A. = - ^ - [ ( 6 x 103) - (5-625 χ 9
3) ] ,
286 S T R E N G T H O F M A T E E I A L S
Equat ing, 770 4r= ! 5 4 0 Ε
110F = 7700 - 1540.F,
7700 F =
2310 '
3 - 33 tonf.
Hence,
.'. Stress in tie
Bending stress
Z = 1 ^ 4 ^ = 0-192 in. 13,400
3-33
0-375 = 8-9 tonf/in
2.
My
.*. Stress ratio
_ (1-67 x 90) 5
158
= 4-75 tonf/in2,
8-9
4-75 = 1-87.
E x A M P L E . A cantilever is made from steel for which Ε = 13,000
tonf/in2 and has a uniform section for which / = 200 in
4. I f a
concentrated load of 5 tonf is applied a t the mid-point, calculate,
assuming only the differential equation of flexure :
(a) the slope of the beam a t the load point, (b) the deflection a t the load point, (c) the deflection a t the free end, if L = 240 in.
Solution
FIG. 2 5 6
B E A M S I V — D E F L E C T I O N 287
At any section X X :
ΜΎ = EI
dx
dx2
I20x
-5(120 - x),
r2 1 faj
+ A, and when χ = 0 , = 0 , dx
Λ ^ = 0 .
A t the load point, i.e. when χ = 1 2 0 :
EI dy_
dx - 5 (120 χ 120)
1 2 02
500 χ 144
/ . Slope *L dis-
in tegrat ing again :
- 3 6 , 0 0 0 ,
36 ,000
13,000 χ 200 '
- 0 0 1 3 8 5 .
Ely = - 5 ^
2 1 #
3
1 2 0 T - I T + B} and, when a; = 0 , y = 0 ,
.'. J5 = 0
= - 3 0 0 . T2 + —X
s
Ό
At the load point, i.e. when χ = 1 2 0 :
Ely = - ( 3 0 0 χ 120 2 ) + ( - | - x 120 3J,
- 4 , 3 3 0 , 0 0 0 + 1,450,000,
- 2 , 8 8 0 , 0 0 0
13,000 χ 200 '
— 1-107 in., i.e. downwards from the origin 0,
= Z2 (Fig. 256)
In the diagram, Z1 = Slope χ 120 ,
= - 0 - 0 1 3 8 5 χ 120 ,
- - 1 - 6 6 0 in.
10a*
288 S T R E N G T H O F M A T E R I A L S
Deflection a t free end, Z% = Zx + Z2,
= - 1 - 6 6 0 - 1-107,
= - 2 - 7 6 7 in.
E X A M P L E . I f the deflection a t the top of the 2 in. dia. pole
illustrated in Fig . 258 is to be zero and the bending stress is not
to exceed 2000 lbf/in2, find
(a) the rat io P i / P 2 and
(b) the values of Ρλ and
P 2 ( I . Mech. E . ) .
Solution
2nd Moment,
0-785 in4.
Bending moment, M = i l y
where y = 1-0 in ,
2000 χ 0-785
1 0
= 1570 lbf in.
F o r Ρ2 alone,
Mx = E I ^ = P2(l20-x)) dx*
When χ = 0 ,
7T = °> dx
0 .
When χ
dy
120 ,
EI dx
dx
1 2 02 -
7 2 0 0 P 2
1 2 02\
EI and
dy
dx
Straight part
g l
120 in
//////////{///
FIG. 2 5 7
I oft
F I G . 2 5 8
2 0 f t
'//////A
;. Ζ = 864,000 — f - · EL
120
B E A M S I V — D E F L E C T I O N
y»2 1 >
289
Integrat ing again, J J J ! , = P 1 ( 1 2 0 1 -
and y = 0 when χ = 0 , .'. Β = 0 .
When χ = 1 2 0 , „ / 1 2 0
3 1 2 0
3\
6 ) '
= 5 7 6 , 0 0 0 P 2 and y
: . z 2 = 5 7 6 , 0 0 0 A .
Tota l deflection due to P 2 , Z 3 = Z 3
= (864,000 + 576 ,000) - = f , ILL
:. Zz = 1 , 4 4 0 , 0 0 0 - ^ .
F o r Px alone, deflection a t end = Τ
P1 χ 2 4 03
X EI '
i.e., Ζ = 4 ,608,000 Pi EI '
Since this is to be equal and opposite to Z3,
4,608,000 = 1 , 4 4 0 , 0 0 0 ,
Ρτ 1-440 5
" P2~ 4-608 ~~ 16 '
Mmax = 1570 = 1 2 0 P 2 - 2 4 0 P 1 / •
[—*~p\
g where P 1 = — P 2 , /
= P* 3 / 24C ) in
= P 2( 1 2 0 - 7 5 ) , /
15 //////////////
Hence, P1 = 10-9 lbf. FIG . 2 5 9
2 9 0 S T R E N G T H O F M A T E R I A L S
- [L -x ] y =z 'mox.
Τ PIG. 2 6 0
At any section X X distant χ from the origin Ο :
L - x .τ) (-
——{L2 — 2Lx + χη,
E Id y
dx
dy
A - 0 .
EI^-= - — L2x - Lx
2 + — - ) from which
dir 2 \ 3
and when χ = 0 ,
Hence,
dy the slope ——at any point can be found.
ÇX.X
w / _ _ χ2 _ X '
3 l a :
4. _
• · ί ί » = - τ Κ - £ τ + 8 Τ ι + ί ' and when χ = 0 , y = 0,
τ τ T IT w / £2
Hence, Ai?/ w I L
2
5 = 0 .
L x
z + — a;
4 ] from which the
3 12
deflection y a t any point can be found.
Horizontal Cantilever with Uniformly Distributed Load
B E A M S I V — D E F L E C T I O N 2 9 1
A t the free end, y
:• Ely™, =
and χ = L,
2 \ 2 3 1 2 ) '
\2 3 +
1 2 / '
and wL = W (the to ta l load) ,
wL* / l
" 2 ~
wL*
8 and ym Ζ
i.e., 1 WL3
The negative sign shows tha t the deflection is downwards from the origin considered.
E X A M P L E . A uniform cantilever has a depth of 1 8 in. and a
length of 1 4 ft. A load of 4 tonf is to be concentrated a t the free end
and a second load of 6 tonf is to be uniformly distributed throughout
the length. I f the maximum bending stress is to be limited to
5 tonf/in2 find the least necessary value of / .
I f Ε = 1 3 , 0 0 0 tonf/in2, find the deflection a t the free end.
W,=4 tonf
Solution
At the fixed end, M„ ( 4 χ 1 4 ) +
= ( 5 6 + 4 2 ) 1 2 ,
= 9 8 χ 1 2 ,
= 1 1 7 6 tonf in.
( · * τ ) 1 2
292 S T R E N G T H O F M A T E R I A L S
j — tJL } where y = 9 in ,
_ 1176 χ 9
" 5 '
= 2117 in4.
The tota l deflection is the sum of the deflections due to each load
acting alone, i.e. 1 WXU 1 W2L*
3 EI 8 EI '
L3 / 4 6_\
~ " ^ T i 3~ + 8"/ '
_ (14 χ 1 2 )3 25
~ 13,000 χ 2117 X
"Î2 '
= 0-36 in.
0-25 in
E X A M P L E . A cantilever consists of a triangular steel plate of
density 0-28 lbf/in3 and having the dimensions shown. Assume only
the differential equation
of flexure and calculate
the deflection a t the free
end when loaded uniform-
ly with 4 lbf/in2 of plan
area.
The 2nd Moment of the
section varies linearly
from zero a t the free end,
so that , for any section
X X
χ Ix = —1 , where
Ε
I — value at support.
I f w = load/in2, then load -p IG
causing bending on sec-tion X X = w xbx/2 and this acts a t χβ from X X .
Hence, EL dx
2
bx ••= —w X
X x 3-
B E A M S I V — D E F L E C T I O N 293
i.e., Ε xl d
2y
L dx*
wBx2
EI
When χ
dy_
dx
6
dy wB xz
~dx =
6~~ 3~
wB χ*
wbx2
6
(1)
where b = —B, Ε
+ Ä (2)
18 4 Ax + B (3)
FIG. 2 6 3
When χ = L ]
y = 0 J " Hence, in E q . (3)
in E q . (2)
0 =
i n E q . (3)
V 18
whence A = wBL*
18
χ L* + ^ — — x L j + B, 72
whence J5 = wBIA
~24T
wB wBLz wBL*
When χ = 0 , A/ i / W
^ and = jr/ the to ta l load,
W ^3
1 2 ^ 7
To ta l load/in2 = (0-28 X 0-25) + 4-0 = 4-07 lbf.
' 6 χ 10\ W = 4-1
= 122-1 lbf.
and I = 6 χ 0 -25
3
12 128
2 9 4 S T R E N G T H O F M A T E R I A L S
Λ Deflection, 1 2 2 - 1 χ 1 0
3 1 2 8
y = 1 2 χ 3 0 χ 1 0
6 χ 1
= 0 0 4 3 5 in.
E X A M P L E . A tota l load of 2 0 tonf is distributed over a uniform
section cantilever 1 0 ft long, the intensi ty of loading varying
uniformly from zero a t the free end to a maximum a t the fixed end.
Obtain an expression for the bending moment in terms of distance
from the free end and plot this a t intervals of 2 ft over the length,
using the following scales : 1 cm = 2 ft, 1 cm = 1 0 tonf ft. Es t ima te
the deflection a t the free end, given tha t 7 = 1 6 0 0 i n4 and
Ε = 1 3 , 4 0 0 tonf/in2.
Solution
Tota l load W = Mean loading χ L,
/ M a x . rate of loading\
.'. Max. rate of loading = 2W
L
Max. rate of loading
70
tonf f t
FIG. 2 6 4
B E A M S I V — D E F L E C T I O N 2 9 5
d% R a t e of loading w
x = EI (i.e. proportional t
o x)
2W χ
EI d^r
da;4
2W
L2 χ x.
Shear force
When χ = 0
Fx = 0
. " . - 4 = 0
d3y 2W tx*
EI d
3y
dx3
W
L2 χ x*
Bending moment, Mx = EI d
2y
When χ = 0
i f , = 0
. \ 5 = 0
= - -
dx2
W
W a;3
ΊΪΧΤ + Β·
where i f = 20 tonf
and L = 1 0 f t ,
20
3 χ 1 02
= - 0·0667;ζ3.
χ χ3,
The table of bending moments is then as given in Fig . 264 .
F rom above, „Td
2y W
EI dy
dx
W fx*
and when χ = L, dx
= 0
3 L2 \ 4
L4
+ C
Hence dy
El-rr- = dx
;. Ely =
Λ + ( 7 = 0 i.e., C=—. 4 4
I 4 4 / '
(a;4 - £
4) .
3 £2
1 2 £2
IT /a ;5
1 2 L2
At any section XX:
296 S T R E N G T H O F M A T E R I A L S
when χ = L1
y = o|
.·. Ely
Hence, deflection y
and is a m a x . when χ
Simply Supported Beam ivith Central Concentrated Load
ο* * Για. 265
At any section X X distant χ from a central origin Q (Fig. 265) ,
„T&y W I L \
5 5 '
TF / a ;5 „ 4 Z
5\
= - w ( - 5 - - ^+
— ) -
1 2 Ä J £ » I 5 5 j = 0 .
12EIIß[ 5 / '
_ WL3
15EI '
2(10 χ 1 2 )3
~ _
15 χ 13,400 χ 1600 '
= — 0 4 7 0 6 in, i.e. downwards.
B E A M S I V — D E F L E C T I O N 297
and 4 " = 0 , when χ = 0 , A = 0 . ax
EI — W
<L
dx
_W(L * \
~ 2 [Τ* 2 J d?/
from which the slope - j — a t any point can be found.
W / L χ* 1 ζ3 \ _ _
Again, Ä J „ = —(— — - γ—j + 5 and y = 0
when χ = 0 , .*. -B
I f / L ^2 z
3\
= τ ( — - τ ) from which the deflection y a t any point can be found.
At a support, when L
% = γ , y = 2/max = " Ζ ,
where Ζ is the maximum deflection (at the centre) .
W/L L* 1 L*\
·'· Α / ^ = τ ( τ χ Ί Γ - β X" T J ' _ J f / Iß_ _ L*_\ ~ ~ 2 ~ \ ~ Î 6 ~ ~~ 48 ) '
if J L) V 32 96 / '
P F £3
48
1 TfX3
2/max = ~ # ] p ' upwards relative to Q.
Hence, Ζ 1 WL*
48 A /
The negative sign shows tha t the deflection is downwards relative to the origin 0 .
The fractions which precede the quant i ty WL3/EI are known as
Deflection Coefficients.
E X A M P L E . T W O self-aligning bearings 10 ft apart support a 2-5 in. dia. shaft. A t the centre of the span is a pulley to which a transverse load of 600 lbf is applied. Find, from first principles, in degrees, the inclination of the shaft a t the bearings.
298 S T R E N G T H O F M A T E R I A L S
Find also, by any method, the deflection of the shaft a t a point
3 ft from a bearing. Ε = 30 x 1 0e lbf/in
2.
Solution
F o r shaft section,
= 1-92 in4.
π / = — χ 2-5*,
L = I0 f t =120 in
FIG. 2 6 6
At any section X X ,
and
EI
d2?/ W (L \
dy M
dy_
dx
dx 2 \ 2 " 2
0 when χ = 0 , . ' . - 4 = 0 .
1 T dy W iL2 1 L
2
Ί ΊΤ dy W (L* 1 L2\
* ^ = ^ ( τ - 2 Ί γ ) ' w h e n *
. dy
L_
2
W (— —
" dx _ 2ËF\Ï 8~
WL2
WEI'
600 χ 1 2 02
16 χ 30 x 1 0e χ 1-92
= 0-00938.
B E A M S I V — D E F L E C T I O N 299
Hence, angle of inclination when χ = —
θ = t a n1 0 -00938 , = 0°36 ' .
W l L χ2 1 χ
3 \
Again, Ely = — ^— — — — j + Β and y = 0 ,
when # = 0 , Β = 0 ,
_ TT / £ a2
2 \ 4
_ IT / 120
χό
— [—— χ 2 42
2 \ 4 when χ = 24 in
Hence. 600 χ 2 4
2
2 43\
" 6 ~ ; '
i.e., a t 3ft from one end.
( 3 0 - 4 ) , 2 χ 30 x 1 0e χ 1-92
0-078 in., upwards from the lowest point, i.e.
from the centre.
Max . deflection, WL
3
EI 1
600 χ 1 2 03
48 χ 30 χ ΙΟ6 χ 1-92
= 0-375 in .
.·. Required deflection = 0-375 - 0-078,
= 0-297 in., downwards from the original
position.
E X A M P L E . Three identical joists are built into a wall a t the same level and equidistant, the distance between adjacent joists being equal t o the protruding length. A fourth jois t of the same section but twice the length is mounted across the free ends of the other three, and a vertical load concentrated a t the centre. Show tha t the central cantilever will deflect twice as much as the outers.
Solution
I f ρ = upward force exerted b y end cantilever on cross member, then force producing deflection in cross member is : W — P, when Ρ is the upward force due to centre cantilever.
300 S T R E N G T H O F M A T E R I A L S
From the diagram, Z 3 = Z2 — Zx,
1 (W - P)L* _ 1 PP 1 plz
48 # 7 3 EI 3 EI
(W -P)L* ( P - y)
48 3
2pZ,3 (W - 3p)l*
" ' 48 3
8 ( J F - 3 p ) = p ( y )3 and j = 2 ,
8 I f - 24p = 8p
3 2 ^ = SW
W i.e., ρ = — .
[sub-for (W - P) and (P - p)]
W W Hence, p=W-2p = W - -^ = - - .
Δ Δ
The central cantilever thus carries twice the load of an outer and
hence deflects twice as much.
31, = wL IL
wL (L
~ 2
(j~X)-W(-2 -X)
4
wL2
8
wxz
~ 2 ~
l/L X ,
2 2
W iL* ο r \
2{-T + x 2 - L x ) wL
2 wLx wL
2
wx* 8 + •
wLx
w/unit length
wL
2
FIG. 2 6 8
• • ^ d e »_
2 U r
when x = 0 , - ^ = 0, A = 0.
Simply Supported Beam with Uniformly Distributed Load
At any section X X distant χ from a central origin Ο :
B E A M S I V — D E F L E C T I O N 301
3 0 2 S T R E N G T H O F M A T E R I A L S
„ T dy w IL2 x
z\ , . . . t. . dv
*. EI -p— = — I -£-χ ^-1 from which the slope a t any
point can be found.
w iL2 χ
2 1 x*\
• ' • m y = Y [ - T x Y - J x T ) + B w h en *YZ0\ :. Β = ο!
_ R M> / L2x
2 X
s- \ , . , , . , .
.'. Ely = — — — I , from which y a t any point can be 2 \
8 1 21 found.
A t the support y = y ^ and χ = γ ,
w / i2 i
2 1 L
4 1
w / i2 1 £
4\
• • • Ä i y - - T l T x T - l 2 x l 6 J ' w ί L* \
~ ΊΓ\32" ~~ 102/
• ( - - — ) \ 3 2 1 9 2 /
w £4 / 1
~ 2 ~
w L4 5
2 1 9 2 and w £ = ΡΓ the to ta l load.
5 w L3
' y - = 3 8 4 X ΎΓ ( u p W a r d 8)
Λ Ζ == ~3§ïië~ ( d o w n w a r d s)-
E X A M P L E . A uniform beam of length L and depth d is to carry
a to ta l load of IT tonf uniformly distributed. I f the maximum
deflection is not to exceed L / 4 0 0 when the maximum bending stress
is 8 tonf/in2, show tha t L > 20d. The beam is simply supported
and Ε = 1 3 , 5 0 0 tonf/in2.
Solution
5 WL* „ L Maximum deflection Ζ = — - rrr— and Ζ > ——
3 8 4 EI 4 0 0
. _ L _ _ _ 5 _ W X3
' " 4ÖÖ ~ 3 8 4 ^ 7 '
B E A M S I V — D E F L E C T I O N 3 0 3
i.e., 3 8 4 EI
2Ô50" W 9 where I = My
f WL d 1
8 XT * T
L2
or
3 8 4 χ 1 3 , 5 0 0
2 0 0 0 χ 1 2 8
L = 20-25<Z.
χ Ld
E X A M P L E . A brass strip 3 in. χ 0 - 7 5 in. χ 1 2 0 in. weighing
7 6 lbf is slightly curved. The centre deflection is 1 - 6 4 in. when
supported a t the ends with the 3 in. width horizontal. This becomes
1 * 1 3 in. when the strip is turned completely over. F ind the sag due
to the original curvature and determine the value of E.
Case©
3 8 lbf
FIG. 2 6 9
Solution
I f χ — sag due to curvature,
then deflection due to weight == 1 - 6 4 -
and
Hence,
B u t deflection Ζ
where I
1 1 3 + 0 - 2 5 5
/. Ε •-
χ, in case ( 1 )
= 1 - 1 3 + x, in case ( 2 ) .
1 - 6 4 - χ = 1 - 1 3 + x,
2x = 1 - 6 4 - 1 - 1 3 ,
χ = 0 - 2 5 5 in.
5 χ
WL3
3 8 4 " EI
3 χ 0 - 7 53
1 2 0 - 1 0 5 5 in
4.
χ 7 6 χ 1 2 0
3
3 8 4 " 0-1055.Ë7 '
5 χ 7 6 χ 1 2 03
1 - 3 8 5 χ 3 8 4 χ 0 - 1 0 5 5 '
1 1 - 7 x 1 0E lbf/in
2.
304 S T R E N G T H O F M A T E R I A L S
Uniformly Loaded Cantilever, Supported at Free End
As already shown, without the support Ρ
1 WL3
8 EI
^ ι h
0 • L
k z,
/ W = total load = wL
FIG. 2 7 0
I f the free end is propped up to the same level as the wall sup-
port then
Z 2 - Z 1 = 0 9
1 PL* ~t · . . .
and is positive, i.e. upwards. where Z0
3 EI
Thus, λ ™ 3 EI
1 WL3
8 EI whence Ρ = —W.
Uniformly, Loaded Beam, Supported at Ends and Centre
As already shown, without the support Ρ
= 5
WL3
1 - 384 # J '
An upward force Ρ will produce a centre deflection
1 PL3
Z2 = + 48 ,E7
B E A M S I V — D E F L E C T I O N 305
whence,
I t follows tha t each end support carries —-^ W.
FIG. 2 7 1
Assuming the prop Ρ not to sink, i t will, if elastic, shorten by an il Ρ
amount χ = — , where / = —— and I = length of prop. Sect ion
E X A M P L E . A short vert ical elastic strut of height h and section A supports the free end of a horizontal canti lever of length L and 2nd Moment of area I.
I f the lower end of the strut does not sink when a to ta l load of W is distributed uniformly along the cantilever, show tha t the load i t carries is given b y
3 AWL* F =·
24ΖΛ + SAL*
(E has the same values for strut and cantilever.)
Hence, for the centre to be raised to the level of the ends,
Z 2 - Zx = 0 ,
i.e. 2 2 = ^ι·
1 PL* 5 WL*
48 EI "~ 384 EI
3 0 6 S T R E N G T H O F M A T E R I A L S
Solution 1 WL
3
Deflection a t free end due to W = ——=y—, downwards. 8 LI
1 FL3
Deflection at free end due to F = ————, upwards. 3 LI
FIG. 2 7 2
The difference between these deflections is the reduction (x) in the
height of the strut.
B u t , χ = , where / = direct stress in strut, L
_ . F A
" Τ x ¥ "
_ Fh 1 i f L3 1 FL
3 . t
H e n c e ' Ü F = 8" " I T ~ ¥ ·anrl £ c a n c e l s o u t -F_
• • 1 WL
3 FL
3
8Ih BIh
„ , 1 L3
01 Fl^r + m i.e.
WL3
SIh
I 3Ih + . 4 £3 \ _ 1 T £
3
\ / 8 J A '
so that ,
whence,
F(SIh + ^ L3) =
F =
3 AWL3
8 '
_ 3 £ ] f L3
_
2 4 / f c + 8 J L3
E X A M P L E . A B e a m of I section 1 4 in. deep and having I = 4 3 9 i n4
carries a load of 1 5 tonf uniformly distributed over a simple span
of 3 0 ft. I t is then propped a t the centre so tha t the deflection there
B E A M S I V — D E F L E C T I O N 3 0 7
FIG. 2 7 3
Before Rz is introduced, the centre deflection is given by
5 WL*
3 8 4 X ~ËT '
5 1 5 χ 2 2 4 0 ( 3 0 χ 1 2 )3
3 8 4 3 0 χ ΙΟ6 χ 4 3 9
1-55 in.
The upward deflection produced by R3 is given by χ
Ί KK Bs ( 3 0 χ 1 2 )3
i.e. 1-55 = - 4 - χ ν
'
4 8 ΕΙ
4 8 3 0 χ ΙΟ6 χ 4 8 9 *
1-55 χ 4 8 χ 1 0Β χ 4 8 9
Β = 3
~ 3 02 χ 1 7 2 8 '
= 2 1 , 0 0 0 lbf,
= 9 - 3 6 tonf.
D D 1 5 - 9 - 3 6 R
i = R
2 = Ö = 2 - 8 2 tonf.
is zero. F ind
(a) the load on each of the three supports,
(b) the positions of the points of contraflexure,
(c) the magnitude and position of the max . positive bending
moment,
(d) the bending moment and stress a t the centre.
Ske tch the Bending moment diagram. Ε = 3 0 x 1 0E lbf/in
2.
Solution
308 S T R E N G T H O F M A T E R I A L S
F o r any section X X , Mx = 2-S2x - (o-5x χ -jj + 9-36 (x - 1 5 ) ,
working from the L H end.
= 12Λ8χ - 0-25z2 - 140 ,
= 0 a t points of contraflexure.
12-18 ± V[12-182 - (4 χ 0-25 χ 140)]
·"' X
" 0^5 '
= 18-72 f t , or 30 ft which can be ignored.
Thus the points of contraflexure are a t (18-72 — 15) or 3-72 ft from
the centre or 11-28 ft from the ends.
Alternatively, working from the R H end,
= 2·82χ - (θ·5χ x | = 0 ,
x2 = 4 χ 2-82o:
i.e., # = 11-28 ft.
= 2-82 - 0·5.τ = 0 for max. positive B .M. , i.e. χ = 5-64 ft. da:
.·. M = (2-82 χ 5-64)
- (0-25 χ 5 -642) ,
= 15-90 - 7 - 9 5
= 7-95 tonf ft.
(posit ive).
At the centre, F l G
* 2 74
M = (2-82 χ 15) - (0-5 χ 1 5 ) 7 - 5 ,
= 13-8 tonf ft.
(13-8 χ 2240 χ 12) 7 .·. Max. stress in beam =
439
5900 lbf/in2.
T h e B . M . diagram is therefore as shown.
Beam Subjected to Uniform Bending Moment
As already shown, if the supports are symmetrically placed as
in Fig . 275 then the bending moment between the supports is
B E A M S I V — D E F L E C T I O N 309
uniform and equal to — Wl. Since M = EIjB, i t follows tha t R is
constant between the supports, i.e., this part of the beam bends into
an arc of a circle.
F o r any section X X between A and Β
MX = E I ^ = - Wl
EI dy
dx
and, when
•Wlx + A
L dy
0 = -Wl^- + A, Δ
w giving A = — x IL.
: . E I ^ - == - wix dx
WIL
from which the slope a t
any point between the
supports can be found.
4 L
»
1 x
X Z=
ymox
FIG. 2 7 5
Again, Ely = - WI χ χ
2 WILx
i.e.
Wl
wi
+ Β and, when χ = 0 , y = 0 ,
.*. 5 = 0 .
2 # /
( L * - χ2),
( 2 λ Ε — α*2),
from which the deflection a t any point between the supports can be found.
11 SM
310 S T R E N G T H O F M A T E R I A L S
The maximum deflection Ζ occurs a t the centre, when χ = —
\ L x j -Ζ wi r 2EI
WIL2
ML2
(since Wl = M) SEI ' SEI
which is positive i.e., upwards from 0 and relative to point A .
E X A M P L E . A copper rod 48 in. long 0·75 in. dia. was supported
horizontally a t points 6 in. from the ends. When loads of 25 lbf
were hung from the ends the measured deflection a t the centre was
0-1 in.
A tes t length of 6 ft was then subjected to a torque of 600 lbf in
and the twist found to be 1-078 deg. Es t imate the value of Pois-
son's Ra t io .
Solution
- 4 8 in-
25 lbf 25
— 36 in —
FIG. 2 7 6
Π oL 25 lbf
Between the supports the bending moment is uniform and is
given b y :
i f = - (25 χ 24) + (25 χ 1 8 ) ,
- - 25(24 - 18 ) ,
= - 25 χ 6
= - 150 lbf in.
Deflection a t centre, Ζ
o-i
.*. Ε
ML2 ι γ π /(Λ and 7 = _ ( 0 · 7 5 )
4,
SEI
150 χ 3 62
8 # χ 0 0 1 5 5 '
150 χ 3 62
0 1 χ 8 χ 0-0155
15-7 χ 1 0β lbf/in
2.
0-0155 in4.
B E A M S I V — D E F L E C T I O N 311
where J = -Ξ- (0-75)4 - 0-031 in
4.
Hence G 6 00 x 6
0 0 3 1 χ 0-0188 '
= 6-18 χ 1 06 lbf/in
2.
1^-7 γ 1 0e
Since Ε =2G{1 + a), l + , = _ _ = 1 4 7 .
a = 0-27.
Simply Supported Beam with Ν on-central Concentrated Load
Moments about Ο give :
R2L = Wa, i.e. R2
Wa
Similarly,
FIG. 2 7 7
Wb R, =
Wb F o r any section X X to the left of W, Mx = — - x x.
L
Τ GO τι I n torsion, — = —— where θ = 1-078 χ ——
J L 180
= 0-0188 radian.
1 1 *
312 S T R E N G T H O F M A T E R I A L S
dx L Jx dx — kW J (x - a) d(x — a)
Thus the 1st term of E q . (1) must be integrated with respect to χ
while the 2nd term must be integrated with respect to the bracket
(χ - a),
riTdy I Wb χ2 \ JTI_/ x - a \
2
1Χ>·' A / d H — ΧΎ + Λ)- kW(—2— ) • ( 2)
wrîere A is a constant .
Similarly, Ely = χ £ + A* + B) - (^J, (3)
integrating the last term with respect to (x — a) as before.
Wb F o r any section X X to the right of W, Mx = — χ χ
Ε — W (χ — a).
Thus, in general, to avoid using two equations, d2y Wb
Mx = EI = -j- x x - kW(x - α), (1)
where k = 0 for χ < a, i.e. where the bracket in the 2nd term is
negative, and k = 1 for χ > α, i.e. where the bracket is positive.
Note: If there are two loads there will be three equations, the last of which can be employed in the same manner, terms including negative brackets being omitted as required. The method applies for any number of loads and is due to Macaulay.
X
dy Wb Γ Γ Integrating, EI - j — = -j— \ xdx — kW J (x — a) dx.
ô χ α χ
B u t , kW J (x - a) dx = kW j (x - a) dx + kW j (x - a) dx, o o o
χ
= 0 + kW j (x - a) dx a
(because k = 0 , when χ < a), x - a
= kW I (x - a)d(x - a).
B E A M S I V — D E F L E C T I O N 313
At the L H end when χ = 0, y = 0 and k = 0, i.e. the term in
(x — a) is omitted. Hence, in E q . ( 3 ) :
0 = B.
At the R H end when χ = L, y = 0 and k = 1, i.e. the term in
(x — a) is included. Hence, in E q . (3)
W
i.e.,
and (L
so t ha t
0
A
a) =
Wb L*
6 (L-a)\
W Wb
~6L χ L
2
b, L = α I δ, L2 = a
2 + 2a& + 6
2.
Hence, =
6 £
6 L
Wb
W
6
6
6L
Wba2
a2 + 2a6 + &
2)
21Ta&2 J F 6
3
6 L
χ x°
foe3
W_
6
6 L
(a - a )3 +
6 L *
2Wab2
6 L 6 L
(substituting for A in E q . (3 ) ,
a2bx 2ab
2x
bx
W_
6
iL 6
- (ζ2 — a
2
- ( a2
2ab) -
(x — a )3
ix — a)
bx
2ab - x2) + (x - a)
and a2 + 2a& + b
2 = L
2,
i.e. a2 + 2a6 = L
2— b
2,
(L2 - b
2 - x
2) + (x - a)
3
Thus the downward deflection y a t any point distant χ from the L H end is given by
314 S T R E N G T H O F M A T E R I A L S
the term (x — a)z being omit ted when χ < a i.e. to the left of the
load. A t the load point, when χ = α,
y W
W
6EI
Wa2b
2
3EIL
ή - (L2 - b
2- a
2)
E and L = a + δ,
ab [a
2 I- 2ab + 62) - b
2 - a
2
The negative sign indicates tha t the deflection is downwards from
the origin considered. The deflected shape is as shown dotted in
Fig. 278 .
I 1
FIG. 278
The maximum deflection occurs when the slope dy/dx is zero and
in Fig . 278 this is clearly to the right of W, i.e. χ > a.
If , for convenience, a is made greater than b as in Fig. 279, then
zero slope occurs when χ < a.
Fie. 279
B E A M S I V — D E F L E C T I O N 315
A 6L 6 L
Wb 9 Wba2 2Wab
2
0 = — — χ χ2
or, 0 = bx2
i.e., 0 = x2
2L 6L 6L
ba2 2ab
2
3 3
a2 2ab
3 3 '
a2 + 2ab
3
L2 - b
2
3
b2
(and a2 + 2ab = L
2 - b
2),
(or 3x2 - L
2 - b
2),
P i L L
2 3L
2
Note tha t since b < —, .'. b2 < —-, i.e. 3x
2 > ——from above,
2 4 4
or χ2 >
L2
Τ L
and χ >
T o obtain the maximum deflection, the above expression for χ
may be substituted in E q . (3), the 2nd term being ignored.
The slope a t the ends may be found by calculating the value of the constant A and substituting in E q . (2), care being taken to omit terms as required. The actual inclinations of the beam (or shaft) in degrees are given by θ = t a n
- 1 dy/dx for each end value
of dy/dx, i.e. when χ = 0 and χ = L.
F o r more than one load the deflection a t any point is the sum of the deflections due to each load acting alone.
E X A M P L E . T W O self-aligning bearings 120 in. apart carry a steel countershaft 2-5 in. dia. At 4 0 in. from one end of the shaft is a pulley on which the resultant transverse load due to a bel t is
Thus, in E q . (2) when dyjdx = 0, the 2nd term is omitted and
the constant A has the value already found i.e.,
Wba2 2Wab
2
316 S T R E N G T H O F M A T E R I A L S
1 6 0 0 lbf
- 4 0 in- 8 0 in -
Fig . 2 8 0
Moments about R H end give :
1 2 0 ^ = 600 χ 8 0 , whence R1 = 400 lbf and R2 = 200 lbf.
Working from the L H end :
= (8000 χ 14,400) - (8000 χ 6 4 0 0 ) .
1 2 0 4 - - 8 0 0 02,
Λ Α = - 5 3 3 , 0 0 0 .
600 lbf. Determine in degrees, the inclinations of the shaft at the
bearings. Ε = 30 χ 1 06 lbf/in
2.
Solution
'hi X
Μτ = El^r- = 40Os - 600 (x - 40 dx
2 (1
r3 fx — 4 0 \
3
and Ely = 200— - 300 ί j +Ax+B.
For L H end, .τ < 40 , .'. 2nd term omitted. Also when
(2)
(3)
χ = 0, y = 0, ,\B = 0.
F o r R H end, χ > 40 , 2nd term included. Also when
χ = 120, y = 0 .
Hence in E q (3) above : 0 = ^2. χ 1 2 03 - 100(120 - 4 0 )
3 + 120 ,4 ,
B E A M S I V — D E F L E C T I O N 317
F o r R H end, χ = 120 and 2nd term is included since χ > 4 0 ,
dy
dx E I ^ - = (200 χ 120
2) - 300(120 - 4 0 )
2 - 533 ,000 ,
= 2 ,880,000 - 1,920,000 - 5 3 3 , 0 0 0 ,
= 4 2 7 , 0 0 0 .
dy 427 ,000
' ' dx ~ 30 χ ΙΟ6 χ 1 92
.·. ΘΗ = t a n1 0 -00742,
= 0°25 ' .
= + 0 0 0 7 4 2 ,
E X A M P L E . A beam 20 ft long and simply supported a t i ts ends
carries a load of 0-5 tonf a t a point 12 ft from the L H end. The
beam is of I-section 1\ in. χ 5 in. overall, flanges and web being
each 0-3 in. thick. Take Ε = 13,500 tonf/in2 and find
(a) the deflection under the load and (b) the maximum bending stress.
Solution 2-5 χ 5
3 2-2 χ 4 - 4
3
' N . A . - ja ja ,
= ^ ( 3 1 2 - 5 - 1 8 7 ) ,
_ 125-5
Ϊ 2 ~ '
- 10-46 in4.
I I a S M
Hence in E q . (2) above: EI^- = 200χ·2 - 300(χ· - 4 0 )
2- 5 3 3 , 0 0 0 .
dx
F o r L H end χ = 0 and 2nd term is omitted since χ < 4 0 .
E I ^ = - 5 3 3 , 0 0 0 and / = -^r(^X = 1-92 in4,
dx 64 \ 2 /
. dy 533 ,000
" d ^ - 30 χ 10* χ 1-92 ^Q
'Q Q 9 2 5
>
.·. ΘΣ = t a n1 0 -00925,
= 0°32 ' .
318 S T R E N G T H O F M A T E R T A L S
Ζ = ZEIL '
α = 12 χ 12 = 144 in.
b = 8 χ 12 = 96 in.
L - 20 χ 12 - 240 in.
-2-5 in-
0-3 in
V/////,
0-3 in 0-5 tonf
5 in
12 f t -
- 20 f t -
RL
F i g . 2 8 1
F i g . 2 8 2
.'. Ζ = 0-5 χ 1 4 4
2 χ 9 62
3 χ 13,500 χ 10-46 χ 240 '
= 0-942 in.
Moments about R H end give :
RL χ 20 = 0-5 χ 8 ,
Max. B . M . occurs a t Load point.
0-5 χ 8
20
0-2 tonf.
BL x 12 ,
0-2 χ 12 χ 12 ,
28-8 tonf in.
y 28-8 χ 2-5 / m a x
i Um a x ^ J 10*46
= 6-9 tonf/in2.
Wa*b* At the load point,
where
B E A M S I V — D E F L E C T I O N 3 1 9
E X A M P L E . A brass beam 3 6 in. long of l i n . square section is
simply supported a t i ts ends and carries a concentrated load of
2 4 lbf a t a point 6 in. from the R H end.
Take Ε = 1 4 x 1 06 lbf/in
2 and calculate
(a) the deflection a t mid span,
(b) the deflection a t the load point and
(c) the position and magnitude of the maximum deflection.
Solution
Moments about R H end give :
3 6 2 ^ = 2 4 χ 6 ,
2 4 χ 6 R1
R2 = 2 0 lbf.
0 - 5 χ 0 - 53
Ϊ 2 ~
3 6 = 4 lbf.
0 - 0 0 5 2 in4.
- 3 0 in-
R, = 4
2 4 lbf
- 6 in-
R2=20
FIG. 2 8 3
0-5 in
Έ2Ά Ό·5ίη
Mr = EI d
2y
dx2
dy E I ^ = dx
Ely =
4;r - 24(.τ - 3 0 ) .
4x2
Λ Λ (χ - 3 0 )2
2 4 - + A.
2x3
~ 3 ~
( 1 )
(2)
(3)
F o r L H end χ < 3 0 ,
y = 0, Β = 0 .
.'. 2nd term omit ted. Also when χ = 0 ,
l i a *
320 S T R E N G T H O F M A T E R I A L S
F o r R H end χ > 30 , .*. 2nd term included. Also when
x = 36 , y = 0 .
3 63 12
.·. 0 = 2 χ — — ( 3 6 - 3 0 )3 + 36-4 (sub. in E q . (3 ) ) , ό ό
= 31,104 - 864 h 36.4 , A - - 3° f * ° .
=-- - 8 4 0 .
At mid-span, χ = 18 and the 2nd term is therefore omitted in
E q . (3) above.
1 83
Hence, Ely = 2 χ — (840 χ 18 ) , ο
= 3888 - 15,120
= - 1 1 , 2 3 2 .
Π>
2 3 2 fti^r-
• •y
= - 14 χ IQ« χ 0-0052 = " « j i5
^
At the load point :
I f aa6
2 24 χ 3 0
2 χ 6
2
J 3EIL 3 χ 14 χ ΙΟ
6 χ 0-0052 χ 36 :
F o r max . deflection :
i/L*-b* Ί/ ( 3 6 + 6 ) (36 - 6) i / 4 2 x 3 0 x = ν—γ—= r — s = I/—*—
= | /420 = 20-5 in.
20· 53
Λ EIym = 2 x — (840 χ 20 -5 ) ,
= 5850 - 172,000,
= - 1 1 , 3 5 0 . 11,350
·· * ~ = - 14 χ 10« χ 0-0052 =
E X A M P L E . I f Ε = 13,600 tonf/in2, and I = 1200 in 4,
B E A M S I V — D E F L E C T I O N 321
find, for the beam shown
(a) the position and value of the maximum deflection and
(b) the deflection under each load.
4 10 t o n f
8 f t » | « » ~ 4 f f » |
R L( = 6 - 4 )
L = 2 0 f t -
FIG. 2 8 4
Solution
Moments about L H end give: RR χ 20 = (4 χ 8) + (10 χ 1 2 ) , whence RR = 77-6 tonf. and RL = 6-4 tonf.
F o r any section X X , Mx = EI = 6·4£ - i(x - 8)
- 10(χ - 12) (1)
(χ - 1 2 )2
(2)
A t L H end, when χ = Ο, y = Ο in and the 2nd and 3rd terms in E q . (1) are omit ted; .'. Β = 0 . A t R H end, when χ = 2 0 , y = 0. Hence in E q . (3),
0 = ^ χ 2 03) - ( | - χ 1 2
3) - (4- x δ 3) + 2 M >
whence,
3 " " / \ 3
A = - 3 2 6 in ft units.
The deflection is a maximum at the point of zero slope. I f this is assumed to be between the two loads, then in E q . (3),
0 = 3 ·2^2 - 2(x - 8 )
2 - 326 (neglecting 3rd term),
= 3-2x2 - 2(x
2 - I6x + 64) - 3 2 6 ,
= l-2x + 32.τ - 4 5 4 .
- 4 3 2 ± V{322 - 4 [ 1 · 2 ( - 4 5 4 ) ] } _ 32 -j- 56-6
2 χ 1-2
= 10-25 ft (ignoring negative value).
2-4
, ^ = 6 ^ - 4 i ^ - 1 0 i ^ ^ + ^ dx 2 2 2
n n a:3
ft (α - 8 )3 , (.τ - 1 2 )
3
Λ Ely = 3-2 — - 2 v ;
- 5 — — + Ax + Β Ο Ο Ο
(3)
322 S T R E N G T H O F M A T E R I A L S
Putt ing this value in E q . (3) and changing all units to inches,
1 3-2(10-25 χ 1 2 )3
2(2-25 χ 1 2 )3
(326 χ 122)(10·25 χ 12)
4!r (1980 - 1 31
- 5 8 0
° ) ' EL
ΙΟ3 χ 3830
13,600 χ 1200 = - 0 - 2 3 5 in.
Under the 4 tonf load, i.e. when χ = 96 in. the 2nd and subsequent
terms are ignored.
3-2 χ 9 63
EI
- 0-226 in.
(326 χ 122 χ 96)
Under the 10 tonf load, i.e. when χ = 144 in. the 3rd term only is
ignored.
EI
3-2 χ 1442 2 χ 4 8
3
3 (326 χ 12
2 χ 144)
= - 0-230 in.
E X A M P L E . Pa r t of the mechanism of a recording machine
consists of a steel beam 12 in. χ 0-25 in. χ 0-125 in. supported a t
A and Β as shown. Determine the load required a t point C halfway
between the supports to give a movement of 0-1 in. at the free end.
W h a t will be the ratio between this movement and tha t a t the load
point? Take Ε = 30 χ 1 06 lbf/in
2 and assume only the differential
equation of flexure.
B E A M S I V — D E F L E C T I O N 323
Solution
0-1 in
FIG. 2 8 6
EI
T7r dhj WIE \
dy _ W (Lx _ Χ2
\ Λ _ Δ
d ^ " ~ 2 " U Ύ) + Α
'
ay when x = 0, —2- == 0 , ;.A=0.
dx
L L1T dy W IL L 1 L » '
• è l - w (El Ελ -
"~dx~ JET \~4 8 /
a t B).
WL*
10ËT'
The slope a t Β is tha t of the straight part, i.e.
and
Equa t ing ,
_ 0-25 χ 0 -1253
~ Ϊ2
= 0-0000407 in4.
0-1
7
0-25 in •
t 0-125 in N-
1
FIG. 2 8 7
W χ 52
.'. W =
16 χ 30 χ ΙΟ6 χ 0 0 0 0 0 4 0 7
0-1 χ 480 χ 44-2 7 χ 25
11-1 lbf.
3 2 4 S T R E N G T H O F M A T E R I A L S
W ί Lx2 χ3 \
Integrating again, Ely = ~2~l~~g jg~J + B> w n e n
% = 0, y = 0, ,'.B=0.
At the ends where y is a max. and equal to the centre deflection,
L x = - .
W iL L2 1 L
3
2ËI \T X ~ 4 ~ " 6"
X ~ 8 ~
WL3
4SEI
1 2 1 χ 53
4 8 χ 3 0 χ Ι Ο6 χ 0 - 0 0 0 0 4 0 7
= 0 - 0 2 3 8 in.
Λ Deflection ratio = 0 - 1
0 0 2 3 8 = 4 - 2 .
E X A M P L E . A beam freely supported over a span of 2 5 ft has
concentrated loads of 2 tonf and 8 tonf a t points 5 ft and 2 1 ft
respectively from the L H end. Find, in terms of Ε and / , the
deflection a t a point 8 ft from the L H end. I f the beam is now
additional supported a t this point, so tha t all three supports are a t
same level, find the magnitude and direction of the load on each
support.
Solution
In general,
Ely =
Ely,
6
8
ΊΓ
8
1Γ
bx (L2
+ {χ - a )3
4 χ 8 ( 2 52 - 4
2 - 8
2)
2 5
3 2 ( 6 2 5 - 1 6 - 6 4 )
2 5
8 χ 2 2 χ 5 4 5
6 χ 2 5
( 8 - 2 1 )3
(neglecting negative term),
- 9 3 0 .
B E A M S I V — D E F L E C T I O N 325
W2=2 W,=8
- o 2 = 5 - - b 2= 2 0
- a , =21 Hb ,=4 -H
x=8
L = 2 5
FIG. 2 8 8
20 χ 8 ( 2 52 - 2 0
2 - 8
2)
25
160(625 - 400 - 64)
(8 - 5)*
25 + 27
_ / 160 χ 161
~~ \ 25
= 344 + 9 ,
= 3 5 3 .
. • . ^ + ^ 2 = 930 + 353 = 1 2 8 3 , giving y1 + y2
1283
Upward deflection due to R3 = ZEIL
J R O / 82 χ 17
2
, where a = 8
and b = 17 ,
R^/i EI \
2 4 7 # 3
~ ^ 7 ~
3 χ 25
^. 247 _ 1283 Equat ing, — - R3 - jß , = = 512 tonf.
247 EI * EI 7
Moments about L H end give,
(R2 χ 25) + (5-2 χ 8) = (8 χ 21) + ( 2 x 5 )
168 + 10 - 41-6 R9
25
136-4
~ 2 5 ~ '
5-46 tonf.
.·. R1 = 10 - (5-2 + 5-46)
= —0*66 tonf. i.e. downwards.
326 S T R E N G T H O F M A T E R I A L S
73 WL3
χ Z a 648
N EI,
w w
FIG. 2 8 9
i 1
W
FIG. 2 9 0
Solution
E a c h cantilever carries W and has a length of L/2.
.*. Deflection a t free end = -ί- W
EIr
WL3
2iEIv
F o r the arrangement shown,
W
6EI
bx (L
2 - b
2 - χ
2)
+ (χ - α )3
W
6EI
bx (£2 _ δ
2 _ a;2
since .τ < a.
I n the given case, χ =
κ .1 — ι
FIG. 2 9 1
.*. Centre deflection due to each load = -— W
6 t f ( / , / 2 )
X 1
Τ x Τ
Example . I f J c a n l i l c v cr = 2 / b c a m, in Fig. 289,
show tha t
B E A M S I V — D E F L E C T I O N 327
Tota l beam deflection a t
centre
2 χ W
3EIr 6 \ 9
9EIC \ 9
WL* I _1_ _
L2
Τ
L2
~4~
9EIC
WL* I 23
EIC 9 3G '
ZA = J F L
3 / 23
23 PTL3
324 X
1
EIC \ 324 ^ 2 4 / '
73 WL* χ 648
Encastré Beam with Central Concentrated Load
This type of beam has the ends built in, but without longitudinal constraint. The shape under load is as shown dotted, and, since the slope a t the ends is zero, there must be a fixing couple a t each end in addition to the reaction. This is CW a t R H end and ACW at L H end.
F rom Symmetry , R = R = W/2 and ML = MR.
F o r any section X X to the left of W, F ig . 292 ,
π τ cry vvχ
αχ 2 2
W r3 r
2
(1)
(2)
(3)
328 S T R E N G T H O F M A T E R I A L S
At the L H end, when χ = 0 , 4^ = 0 , Λ A = 0 .
dx Also when χ = 0 , ?/ = 0 , δ = 0
Λ'οίβ: All constants of integration are zero for built-in beams.
-φΞ^Ι0·75ΐη
At the centre, when
Hence in E q . (2), ο
or
W 1? L — χ Ύ - ML χ γ
WL2
16 '
WL
8
Hence E q . (2) becomes :
dx
Wx2
WLx
From this the slope a t any point to the left of W can be obtained. And E q . (3) becomes:
Ely Wx'
~I2~ - WLx2
B E A M S I V — D E F L E C T I O N 329
Prom this the deflection a t any point to the left of W can be
obtained. The deflection is a maximum a t the centre where χ = L/2.
W EIymax = - j ^ " x
L*
~ 8 ~ ~~
WL
16
* - y max EI y "96" ~ ±) 6 4 /
or Z= 192
WL3
or Z= 192 EI
4 '
i.e. downwards from the origin considered.
I n E q . ( 1 ) :
Wx
2
Wx
ML,
WL
2 8
The B . M . diagram is
thus the normal triangle
A B C less the rectangle
A D E C of height WL/S
and the ne t B . M . is re-
presented by the shaded
area.
When χ = L / 4 there
are two points of con-
traflexure, Ρ and Q.
Since
WL
8 '
= 0 .
w L
The Shear force diagram is as shown. The beam is thus twice as strong and four t imes as stiff as when simply supported.
M
y? Datunrr
t i
FIG. 2 9 3
330 S T R E N G T H O F M A T E R I A L S
Β.ΛΙ. Diagram jar Fixing Moments alone
I f a fixing moment is applied as shown a t point E , then the
reaction a t the L H end must be downwards for equilibrium. Hence
the reaction a t Ε (the moment of which is zero about E ) must be
upwards.
The directions of these reactions R,{ and R{ will be unchanged by
the introduction of a fixing moment Mr provided tha t this is less
than Mu.
Μ κ
FIG. 2 9 4
For point Ε , clearly, Clockwise Moment = Anticlockwise Moment,
i.e. MR - ML = RL χ L.
For any section X X , Fig . 295 ,
Mx = — RLx — ML ^ where RL = ——j-—-fcom abovej .
Since this is a straight line law, the B . M . diagram for unequal
fixing moments alone is as shown in Fig. 296 .
1 X
* — X
1 ^
Ι 1
FIG. 2 9 5
Encastré Beam with Non-Central Concentrated Load
The reactions will be unequal and so will the fixing moments required a t the two ends.
η . - )
FIG. 2 9 7
X X ,
M* =
ΕΙ^ΖΞ =
Ri
x -
ψ(
χ - a) - ML
EI
F o r any section X X ,
dx2
dx
(x — a)2
MLx
BL X>_ W {x - af 2 3
(1)
(2)
(3)
B E A M S I V — D E F L E C T I O N 331
When χ = L,
F I G . 2 9 6
332 S T R E N G T H O F M A T E R I A L S
And in E q . (3), 0 = ^ L3 - (L - a)
3 -
.'. —- L3 = (L — a )
3 + ( a n d £ - o = 6 ) ,
Ό Ό 2
Wa2b
Similarly, MR = ^2
Wb2 2 Wab
2
Hence from E q . (4), n L = —JJ- + — x
Wb2
• (L + 2a)
Wa2
Similarly, RR =~jjr(L +
2b)
Now, E I
^ = R
L Τ ~ Ύ{ Χ
-= ^
(Eq*
( 2 ) )'
At the R H end, when χ = L, y = 0 and —— = 0 . da;
Hence in E q . (3), 0 = - ^ " ^2 - - « )
2 ~
Λ J | l / v2 = -51 (£ _ « ) 2 + j f A£ ( a nd L - a = b)
Wb3 M L
W&3 I f , i l f f
Equat ing (5) and (4), + 3 = + 2
(which eliminates RL),
ML ML Wb* Wb3
L L I ? 1? '
Wb2
whence, ML = —jj- [L — b) and L — b = a.
Wab2
.'. Mr = —— -£ 2
(5)
• 7? W h" . 2
M' < (4)
B E A M S I V — D E F L E C T I O N 333
The maximum deflection occurs a t the point to the left of W where
dy —— = 0 and χ < a. dx
Wb2
Wab2
Hence in E q . (2), 0 = —JJ- {L + 2a) ^ „
(ignoring 2nd term and substituting for BL and ML).
Wab2 Wb
2
e. — - — χ = L 3 (L + 2a)Y
L + 2a x
or
L 2 9
2aL
L 4- 2a for max . deflection.
B y substituting the above for χ in E q . (3), the actual maximum
deflection can be found.
F rom E q . (1) , B
Mx = RL χ - ML
when χ < a.
The B . M . diagram is thus
the normal triangle A B C
less the trapezoid A D E C
due to the fixing moments.
T h e ne t B . M . is then
represented by the shaded
area, D E beingthe datum.
The Shear force dia-
gram is as shown.
E X A M P L E . The ends of
the built-in beam shown
are a t the same level, / for
the section is 387 i n4 and
Ε for the material is
13,500 tonf/in2.
Assume only the differential equation of flexure and find the
position and magnitude of the maximum deflection.
κ\//////////////////77Λ
FIG. 2 9 8
334 S T R E N G T H O F M A T E R I A L S
Solution
L e t the fixing moments and vertical reactions be as shown.
Then E I ^ = Mx = RLx - ML - \0{x - 12)
„Tdy ΒΣ 9 „ i . f - 1 2 )2
(1)
(2)
10 tonf
- 1 2 f t -
- 2 0 f t -
FIG. 2 9 9
R r r3 x
2
ο
(3)
When χ — 0, both y and dy/dx are zero so tha t both A and Β are zero.
At the R H end when χ = 2 0 , = 0 and « = 0 . dx
I n E q . (2) 0 - ~ x 2 02 - 20ML - 5(20 - 12 )
2,
Δ
:. 200RL = 2 0 i l f i + 3 2 0 ,
10 -H 1 6 .
And in E q . (3) 0 = χ 203 - χ 20
2 - - | (20 - 12)
3,
υ Δ ό 8000 π „ Λ Λ 1, 2560
.·. — - £ Λ = 200JÎZ. + — — ,
Λ n n = — M L + 0-64. (5)
(4)
B E A M S I V — D E F L E C T I O N 3 3 5
Equat ing (5) and (4), 10
+ 1-6 = ^ M L + 0 - 6 4 ,
- 20
20
1-6 - 0-64,
Hence,
ML = 19-2 tonf ft. = 230 tonf in.
10
3-52 tonf.
1-6,
When dy/dx = 0 , y is a maximum and this point clearly is between RL and W.
In E q . (2), 0 = — 19*2.1' (ignoring the term (χ — 12) ) .
Max. deflection ymilx
3 ·52
2 χ 19-2
3-52 '
10-9 ft ( = 131 in.).
1
EI 1 3 1
2
1
13,500 χ 387
- ( 2 · * . , , .
x KM2 3-52
- - 0 - 1 2 6 in .
Encastré Beam with uniformly Distributed Load
The fixing moments ML and MR are again equal while each reaction is half the to ta l load wL.
F o r any section X X to the left of the centre,
MX = M ^ = ^ - M L - W X , * - (1)
^Tày wL χ2 w x
3
ΎΊΤ wL χ3 ^r x
2 w x*
(2)
(3)
336 S T R E N G T H OF M A T E R I A L S
At the L H end, when χ = 0 ,
ax
\ A = 0 .
( - -
X w/unit length V///
WX
ο - *L
L" T
Straight
FIG. 3 0 0
and when
At the centre, when
χ = 0 ,
y = 0 ,
5 = 0 .
L
4 =o. da;
. Λ L2
Hence in E q . (2), 0 = —— χ —— 4 4
_ M r Χ —
L3
τχ—' L
M L x Y
wL3 wL
3
16
wL3
wL3
24
W J y2
12
48
— --M 16 48 / '
(= WL/12. As will be seen later from the graph, this is the max. value.)
B E A M S I V — D E F L E C T I O N 337
From this the deflection a t any point to the left of the centre can
be obtained.
The deflection is a maximum a t the centre when χ = L\2 wL L* _ wL
2 L
2 L*
% m a x - 72" x
8 - ~2Γ X Τ " 24
X 16
wL χ L* ( 1 1 1 \ . . £ = ί / , η η ν = 77T 1 — - — - ΤΓ777" I and wL = W,
1 WLZ
i.e. downwards from the origin considered. F rom E q . (1) ,
wL2
wLx w _
~ 2 2 12
The graph of M is thus the normal parabola A B C less the
rectangle of height wL2\\2 and the net B . M . is represented by the
shaded area.
The height of the shaded portion above the line D E is
w/S (L — 2xp)2, where L — 2xp = base of small parabola and
xp = distance of point of contrariexure Ρ from L H end.
w wL2 wL
2
From the graph, — (L - 2xp)2 = — — .
SL2
(L - 2xp)2 = L2 - 12
L2^
Hence E q . (2) becomes
^T dy wL 0 wL2 w 0
dx 4 12 6
F rom this the slope a t any point to the left of the centre can be
obtained. And E q . (3) becomes
^T wL _ wL2 . w .
Ely = —— χ3 - —τ—χ
2 - —x
4.
* 12 24 24
338 S T R E N G T H O F M A T E R I A L S
i.e. L - -E = 2xp, whence xp - ~ ^1 - - ^ - j ,
= 0-21L approx.
E X A M P L E . A uniform built in beam carries a distributed load of w/it over two thirds of its length, beginning a t the left hand end. F ind the values of the fixing moments given tha t for a point load the fixing moments a t left and right hand ends are respectively Wab
2jL
2 and Wa
2bjL
2 where a f b — L and a is the distance from
the load to the left hand end.
Solution
Fixing couple required a t L H end due to load element w άχ
FIG. 3 0 1
.'. To ta l fixing couple required a t L H end will be the sum of such
couples between χ = 0 and χ = 2L/3,
B E A M S I V — D E F L E C T I O N 339
ç 0 b
^
WOB2 ' / / ^ '; ; - WOB
* - L - - - - -I
— * A h L -x v
^ ^ w/unit length >^
FIG . 3 0 2
2 A/3
i.e., M l
^ l ß J ( Ι Λν
* 2 L : r2 :i
'3)
clr>
ο
w TJ2 16 4 \ 2 r_
Fixing couple required a t R H end due to load element w àx
w ax χ x2(L — x) w
= - ρ = -jj (Lx2 - -τ
3) dx.
2L/3
Λ M«=Jjf ( L a
2- *
3) dx,
0
_ W
\L — - — *LIZ
~ U \ X
~ 3 ~ ~ " T 0 '
E X A M P L E . F ind the reactions and fixing moments for the system
of loading shown in Fig . 3 0 3 , and sketch the graph of M.
340 S T R E N G T H O F M A T E R I A L S
12 tonf
FIG. 3 0 3
F o r any section X X as shown :
Mx=EI^jL= RLx - M j - \2{x - 5)-w(x - 10) { X
^ (1)
dy EI-?- = RL— - MLx - 12
dx L
2 L
2
(x - 5 )2 w (x - 1 0 )
3
+ A (2)
^ T R t XZ , r X
2 (χ - 5
3) w (a - 1 0 )
4
6 + Ax + £
(3)
When dy/dx = 0, χ = 0 and functions in brackets are ignored, .*. A = 0. Also when dy/dx — 0, y — 0 and 5 is similarly zero.
When χ = L
dx
y = ο
In E q . (2) 0 = ^ L2 - i f 7 i - 6 ( i - 5 )
2
Δ
-—(L - 1 0 )3 (and Ζ = 2 0 ) ,
2 02
0-75 - 6 ( 2 0 - 5 )
2 _ ( 2 0 - 10 )
3,
or
= 2 0 0 £ L - 2 0 i f L - 1350 - 125 ,
1475 = 200Ä,, - 2 0 J f L (4)
Solution
B E A M S I V — D E F L E C T I O N 341
·. I n E q . (3) 0 = - ^ - Z3 - ^ L* - 2(L - 5 )
3 - ^ (L - 10)*,
_ 8000 400 = -Rz, —w ML— 6750
0-75
or
6 ^ 2 24
= 1 3 3 3 J ? L - 200ML - 6750 - 3 1 3 ,
7063 = 1 3 3 3 i ? L - 2 0 0 J f L
(4) χ 10 gives 14,750 = 2 0 0 0 i ? L - 2 0 0 i f L
(6) - (5) gives 7687 = 666RL
(10 ,000) ,
(5)
(6)
i.e.
F rom (4)
RL = 11-53 tonf ,
.·. RR = (0-75 χ 10) + 12 - 11-53
- 7 -97 tonf .
20ML = (200 χ 11-53) - 1475 ,
2306 - 1475 M L = 20 '
= 41-56 tonf f t .
Moments about L H end give :
MR = ML + (12 χ 15) + (0-75 χ 1 0 )5 - (RL χ 2 0 ) ,
= 41-56 + 180 + 37-5 - 230-6 ,
= 28-46 tonf f t .
The graph of M is then as shown.
Μ I
FIG. 3 0 4
12 SM
342 S T R E N G T H O F M A T E R I A L S
E X A M P L E . A beam A B has a clear span of 3 0 ft and is built in at
A and B . I t carries a load which varies from zero a t A to 1 tonf/ft
a t Β plus a concentrated load of 1 0 tonf a t a point 1 0 ft from A . Determine the reactions and fixing moments a t A and Β and the
deflection a t the centre of the span, given tha t Ε = 1 3 , 2 0 0 tonf/in2
and I = 7 0 5 in4.
Solution
At any section X X :
Intensi ty of loading χ 1
F k j . 3 0 5
Tota l load on part of length
χ = χ χ Mean load,
Η τ χ ! » ) ·
and this ac ts a t # / 3 from X X .
1 χ
2 X 30 τ -
l 0 (x
R \X x°
EI dx
χ·
180
.2 ^4
= R x ~2
1 0 ) - MA,
10{x - 1 0 ) - Mx,
1 0 )2 - MAx - f A,
x* 10_.
EIy = Rx-3 6 0 0
J _ 0 (χ - 1 0 )
3 - Μ χ - •Ax + Β.
At the L H end when χ = 0 , y = 0 and dy/dx = 0 while the 3rd term is omitted since χ < 1 0 . Hence both A and Β are zero.
At χ = 30
= 450ÁΛ - 1125 - 2000 - 307lf A .
. '. 3 0 3 / A - 4 5 0 Α Λ - 3125
- 4500ÄA - 6750 - 13,330 - 4 5 0 i f A
4 5 0 i l f A = 4500ÄA - 20 ,080
(1)
(2) 450itf A = 4 5 0 0 ^ A - 20 ,080
F r o m ( l ) : 3 0 0 J f A - 4 5 0 0 f l A - 3 1 , 2 5 0 ,
i .e. , (2) - (1) gives 150MA = 11 ,170 ,
/ . i f A - 74-4 tonf f t .
Substi tut ing in E q . (1 ) : 4 5 0 i ? A = (30 χ 74-4) + 3125
whence RA — 11-87 tonf ,
/. R = (15 + 10) - 11-87
= 13-13 tonf .
Moments about Β give: MB =» MA + (10 χ 20) + 15
- (11-87 χ 3 0 ) ,
= 74-4 + 200 + 150 - 356 -1 ,
= 68-3 tonf f t .
( 1 5 3 \ ΊÊ5 i n
n ' 8 7 > ! T r ) - i « J τ - - r < 1 5 ^ 1 0 » '
= 6670 - 211 - 208 - 8 3 7 0 ,
= - 2 1 1 9
and Ε = 13,200 χ 1 22 tonf/ft
2,
705 I = " j 2 T
ft
4. (Note foot units.)
2119 χ 12*
' 'y " 13,200 χ 1 2
2 χ 705 '
= - 0 - 0 3 2 7 ft = - 0 - 3 9 2 in .
B E A M S I V — D E F L E C T I O N 343
344 S T R E N G T H O F M A T E R I A L S
Moment-Area Method
L e t F ig . 306 be part of a bending moment diagram and Fig . 307
the deflected shape.
FIG. 3 0 0
In tercept cut off by) tangents at a and b j |
( + v e when tangent at a ~ i ~ strikes the vertical above ' the tangent at b )
dz
'b Deflectei
shape
FIG. 3 0 7
The area under the graph of M between points a and b on the deflected shape is given by
b
A - j'Max. a
The first moment of this area about the J f - a x i s is given by b
Ax = j Mx dx, where χ = distance of centroid of A from M-axis .
S i n c e " pL^-l^M Λ [*>L = _ L fMdx dx
2 EI J dx2 EI J
i.e., dy
b _ A
~dx a
= ~ËÎ
B E A M S I V — D E F L E C T I O N 345
Thus Slope a t b — Slope a t a = I/ΕΙ χ Net Area under graph of M between a and b .
Now dz = χ άθ
dx = ΧΊΓ'
where — = M
M x dx
Αχ ; . j dz=-^jj Mxdx i.e. Ζ
a
Thus, for a given vertical, In tercept Ζ = —^-r χ 1st Moment of Area. EI
I f a p o i n t o f zero slope is taken as point b [i.e. (dy/dx)h = 0 ] , and the vert ical is drawn through point a, then the tangent from b will
FIG. 3 0 8
be horizontal as in Fig . 308 and the intercept Ζ will be the upward deflection of point a relative to point b .
Since ^ χ ( d r - a n
d ( <^τ ) b ~ ° '
Slope a t a = -φ-El
Ax and, relative to b, Upward deflection at a - Ζ — -ρ—.
El
346 S T R E N G T H O F M A T E R I A L S
I t is convenient to split the B . M . diagram into simple shapes and
find the first moment of area of each. Account must be taken of the
sign of I f , i.e. of positive and negative areas.
Simply Supported Beam with Central Concentrated Load
For the piece of beam ab :
Shaded area
WL
4
WL2
16
Λ Slope a t a
dy
L\ 1
A
EI
WL2
WEI FIG. 3 0 9
dy Intercept Ζ = Deflection a t centre where —— = 0,
dx Ax
EI '
WL2
16
WL*
4:$EI 9
WL*
where χ = ~
ο
1 L X EI
X T'
L_
Ύ L^
48EI
upwards from b,
, downwards from a.
Simply Supported Beam with Uniform Load
F o r the piece of beam ab :
2 Shaded area A = (Enclosing rectangle),
ό _2(WL L
WL2
24 '
B E A M S I V — D E F L E C T I O N 347
Uniformly Loaded Cantilever
F o r the piece of beam ab.
Shaded area A = — (Enclosing rectangle) , «5
3 \ 2
WI?
WL χ L
Υ FIG. 310
Intercept Ζ = Deflection a t centre where = 0 ,
Αχ Λ 5 L 5L = -τ—-, where χ = χ — = ——, since curve is
EI 8 2 16 u v parabolic,
WL2 1 5L
~~~24"~ x
ΎΓ x
le"' = "384"
X ~W'
uP
w a r ds f r om b-
348 S T R E N G T H O F M A T E R I A L S
Slope a t a = + - τ — , since tangent a t b strikes the vertical above a,
. (dy\ WL2
Intercept Ζ = Deflection of a relative to b,
Ax
where χ = 3L
(Since
curve is parabolic.)
WL2 3L
1 WLZ
FIG. 311
Leaf, Laminated or Plate Springs
The simplest type has one leaf and is an inverted, simply sup-
ported centrally loaded beam as shown.
Maximum stress / = My
WL t 12
A W L
2 bt*
B E A M S I V — D E F L E C T I O N 349
Central deflection
1 WL* Ζ = — - X 48 EI
9
WL* 12
4 8 # bt*
1 WL*
1~EbF'
τ
FIG. 3 1 2
Curvature
F o r a spring to be jus t flat under the operating load, i t must be given an initial curvature. The value of this may be obtained as follows :
B y geometry :
R2 = (R- Z)
2 + (AJ
R2 = R
2 — 2RZ + Z
2 + —
4 L* É
.·. 2RZ-Z* = —, — r r 4 /
Z(2R -Z) = ¥-, /
2Â=ºæ+æ> / . „ L* Ζ „_ i.
/ n L2 \ FIG. 3 1 3
\o r
~8Z A
P PR O X
- j
Thus if L is known and Ζ is specified, R may be calculated.
12 a S M
350 S T R E N G T H O F M A T E R I A L S
Multi-leaf Type
L e t a rhombus of thickness t be supported along i ts diagonal A B and loaded with W\2 a t the ends of the other diagonal of length L
as shown in Fig . 314 .
At any distance χ from the L H end :
X 2i71/l)X Width of section =
x nb =
2nd Moment /
Bending Moment M
Depth of section y
L/2 L
I 2nbx \
\ L ) X
__ nbt*x
12 ~ 6L
Wx
2 t
~2'
, My Wx t 6L 3 WL .'. Stress on section / = — — = — — χ χ —
2 2 nbt*x 2 nbt2
This is I jn of t ha t with a single rectangular plate. I t is also indepen-
dent of χ and hence is uniform throughout the span.
B E A M S I V — D E F L E C T I O N 351
I f the rhombus were cut along the dot ted lines as in F ig . 3 1 5 ( a ) .
and the pieces of width b/2 assembled as shown in F igs . 315 (b) and
315 (c ) , i t would become a mult ideaf spring having η leaves of
width b, except for the pointed ends.
To form the eye, the ends of the centre (main) leaf are made
rectangular .
( 0
xn leaves of width b
F I G . 3 1 5
Since the beam is inverted, EI à
2y
άχ2 -M.
. à2y
άχ2
M
Wx _1_ 6L
Ί Γ X Ύ + nbFx~ 9
3WL
nEbt*
Integrat ing,
12 a*
ày
dx
3WL
nEbfi χ x + A
352 S T R E N G T H O F M A T E R T A L S
and when
Hence,
Integrat ing,
and when y
dx U
' X
" 2 ' 3 W L L
• 0 = - « X T + I ' A 3
da;
2 wJ^W3
3 J F L 3 J F Z2
χ a; + • nEbt*
3WL
2 nEbt* '
nEbt* X
2 +
2
3 WL2
= 0 , x = 0, ,\B = 0.
The maximum deflection occurs when χ L / 2 .
χ .τ + B ,
W Z2
_ WL 1_ Iß_ £
- J
~ ~ m_B^s x
J X
Τ +
Ί nEbt3
3WL3
nEbt3
3 WL3
( ϊ - ϊ ) ·
8 wJEW1
Quarter Elliptic Type
I n the expressions already ob-tained, ΡΓ must be written for W/2 (Le.2Wfor i f ) a n d Z f o r X / 2 (i.e. 2 £ for L ) .
This gives :
2W χ 2L 6WL
' 4
and
nbt2
2W(2L)*
nbt2
6WL*
8 nEbt* nEbt*
E X A M P L E . A vehicle spring has plates 3 in. χ 0-375 in. and has a FIG. 3 1 6
span of 36 in. I f the stress is not to exceed 15 tonf/in
2 when the load a t mid-
span is 0-5 tonf, est imate the required number of leaves and
B E A M S I V — D E F L E C T I O N 353
calculate the deflection. Ε = 30 x 1 06 lbf/in
2. W h a t is the init ial
radius, if the spring is jus t flat under load?
Solution
0-25 font
3 WL
2 nbt2
0-25 tonf
FIG. 3 1 7
15 χ 2240 3 0-5 χ 2240 χ 36
~~2 X
3 (0 ·375)2 η '
3 χ 1120 χ 36 : 2 χ 3(0 ·375)
2 15 χ 2240
4-27.
A suitable number of plates is therefore 5 since a lesser number will
give a stress in excess of tha t specified.
0 - 3 7 5 in
3 in-
Deflection
Radius
FIG . 318
3 WL3
8 nEbt*5
3 x 1120 χ 3 63
8 χ 5 χ 30 χ ΙΟ6 χ 3 (0 -375)
3 '
0-825 in .
R = ~8Z approx
3 62
8 χ 0-825 '
: 197 in .
354 S T R E N G T H O F M A T E R I A L S
E X A M P L E . Calculate the leaf thickness for a semi-elliptic spring
if i t is to deflect 5 in. when centrally loaded with 1500 lbf and the
corresponding stress is not to exceed 75,000 lbf/in2. The effective
length is to be 40 in. F ind a suitable number of leaves and state
their width. Take Ε = 30 χ 1 06 lbf/in
2.
Solution
W= 1 5 0 0 lbf
Fro. 3 1 9
4 3 WL
nr A A A 3 1 5 00 X 40
Stress / - — x —rpr , Λ 75,000 = τΓΤΓζ 2 nbt- 2nbt
2
3 x 1500 χ 40 nbt
2 =
Deflection Ζ
75,000 χ 2 '
- 1-2.
3 WL* _ 3 χ 1500 χ 4 03
8 nEbt* ' ' * 8^(30 χ 106) bt*
3 χ 1500 χ 4 03
nbt* 5 χ 8 χ 30 χ 1 0
6
- 0-24.
nbt* _ 0-24
* ~ÜW = 1
~ 1-2
= 0-2 in.
1-2 B u t nbt
2 - 1-2, nb = — ,
tz
1-2
" (0-2)2
- 3 0 .
A suitable number of leaves would be 10 and this makes b = 3 in.
B E A M S I V — D E F L E C T I O N 3 5 5
8 x 5
4 0 in .
/ _E_ . l_-UL Ί β ~ Ύ ' " 2 " Ε
/ 7 5 , 0 0 0 χ 4 0 \
I 3 0 χ 1 06 j '
0 - 2 in.
1 / Ε T MR WL I = — — , where M =
I R' ·· Ε 5
4
Hence / 1 5 0 0 x 4 0 \ 4 0
\ 4 / 3 0 χ 1 06
B u t ,
0 - 0 2 in4.
(nb) t*
nb =
1 2
1 2 /
1 2 χ 0 0 2
( 0 - 2 )3 '
_ 1 2 χ 0 0 2
0 - 0 0 8 '
= 3 0 , as before.
E X A M P L E : A cantilever leaf spring 2 4 in. long has 6 leaves 3 in.
wide and 0 - 3 7 5 in. thick.
W h a t load will produce a max . stress of 4 0 , 0 0 0 lbf/in2 and what
will be the corresponding deflection? Ε = 3 0 χ 1 06 lbf/in
2.
Alternative Solution to Previous Example
L2
In i t ia l radius R = , where Ζ = Max. deflection. oZ
4 02
3 5 6 S T R E N G T H O F M A T E R I A L S
Solution
f = GWL
nbt2
0 - 3 7 5 i n :
3 in
4 0 , 0 0 0 =
W =
6W χ 2 4
6 χ 3 ( 0 · 3 7 5 )2 '
4 0 , 0 0 0 χ 6 χ 3 ( 0 - 3 7 5 )2
6 χ 2 4
= 7 0 0 lbf.
- 2 4 i n -
\ n = 6
FIG. 3 2 0
Deflection Ζ = 6WL*
nEbt*
6 χ 7 0 0 χ 2 43
6 χ 3 0 χ Ι Ο6 χ 3 ( 0 - 3 7 5 )
3
2 in.
E X A M P L E . A locomotive plate spring has 1 6 plates 5 in. wide and
0 - 5 in. thick, and the span is 4 0 in. Take Ε = 3 0 χ 1 06 lbf/in
2 and
calculate the stress and deflection per ton of load.
Solution
Deflection
3_ WL*
" 8 X
nEbt* '
Stress
/ 3 WL
3 χ 2 2 4 0 χ 4 P3
: 2 χ 1 6 ( 3 0 χ 1 0
6) 5 ( 0 - 5 )
3
0 - 1 7 9 in.
nbt2 '
_ 3 χ 2 2 4 0 χ 4 0
~ 2 χ 1 6 χ 5 ( 0 - 5 ) 2
= 6 7 2 0 lbf/in2 ( = 3 tonf/in
2)
16 leoves(n)
I 2 2 4 0 lbf
FIG. 3 2 1
B E A M S I V — D E F L E C T I O N 357
Summary of Conventions used in the Theory of Simple Bending
To achieve correct mathemat ical correlation between quantities
the following conventions must be observed.
The origin Ο should be taken
whenever possible a t the left of the beam as in Fig . 322 (a) which
illustrates (for the sake of argument) a simply supported beam
uniformly loaded.
Rela t ive to this natural origin the load
produces a negative deflection a t all points as in Fig . 322 (b ) .
A t all points such as Ρ to the r ight of
mid-span, àyjàx is clearly positive and the complete slope curve
(graph of àyfàx against x) is as in F ig . 322(c) .
Since the slope of the àyjàx curve (i.e. d2v/da;
2)
increases to a maximum a t mid-span and then decreases, the graph of d
2.y/dx
2 against χ is as shown in Fig . 322 (d) and is
wholly positive. F rom the differential equation of flexure: à
2y
M = EI -Tj^g- so tha t if the value of J is constant , the graph of
M (Fig. 322 (e)) will be identical in form and sign t o F ig . 322 (d),
the vert ical scale being multiplied by EI.
I f the curve of F ig . 322 (d) is differentiated
again to give à3y/àx
3 the derived graph has the form shown in
Fig . 322 (f) since the slope of the curve is initially positive and
decreases uniformly, changing sign a t mid-span.
Again, the shear force is given b y :
dM _ _r d2y
F = — — where Μ - EI —4 da; dx
2
^ da;3
The graph of F against χ (Fig. 322 (g)) is therefore identical in form and sign to tha t of F ig . 322( f ) , the vert ical scale being multiplied b y EL
358 S T R E N G T H O F M A T E R I A L S
FIG. 3 2 2
(α)
( b )
( c )
(d)
(e)
( f >
(g)
( h )
B E A M S I V — D E F L E C T I O N 359
Final ly , the load intensi ty is given by :
dF
dx
= EI
where F = EI
d'y
à3y
dx*
M x = + Rx
4- ve. m o m e n t
(»)
+ ve. shear
( b )
and since dF/dx is everywhere negative, the graph of w against
χ is the straight line of F ig . 3 2 2 ( h ) .
F rom the foregoing i t is clear that , to obtain the correct signs
for M and F the following conventions must be used :
1. To the left of any
section X X (Fig.
323(a) ) a clockwise
moment is considered
positive and vice
versa.
2. I f the resultant trans-
verse force to the left
of a section X X
(Fig. 323(b ) ) is up-
wards, then the shear
force is considered
positive and vice
versa. FIG. 323
Examples X
1. The flexural rigidity (EI) of a uniform section beam is 120,000 tonf in2.
If the beam is simply supported over a horizontal span of 96 in. and loaded vertically with 5 tonf at a point 36 in. from the left hand end, determine the deflection at mid-span (0-7 in. approx.).
2. Equal end couples of 30,000 lbf in are applied to a 72 in. length of aluminium shaft 2 0 in. dia.. If Ε =10 χ 10« lbf/in
2 find
(a) the slope at the ends (7-9 deg) (b) the deflection at the midpoint (248 in.). 3. A beam ABCD carries a uniform load from A to Β and is simply
supported at Β and C so that AB = CD. Find the ratio of AB to AD so that the slope at A and D shall be zero. (AB = 0-21 AD approx.).
4. The wind load on an unstayed vertical mast 32 ft high may be assu-med to increase from zero at a point 7 ft from the top to a maximum of 25 lbf/ft at the base. Find the value of the horizontal force which, when applied at the top, would bring about zero resultant deflection at this point. (42.5 lbf).
360 S T R E N G T H OF M A T E R I A L S
5. The span of an Encastré beam is 36 ft and the section has a 2nd Moment of Area of 400 in
4. Find the position and magnitude of the maximum
deflection due to a vertical load of 16 tonf at a point 12 ft from the left hand end. Assume Ε = 13,000 tonf/in
2. (0-87 in at 20-5 ft from LH end,).
6. A single leaf return spring is Od in. thick and has an initial radius of curvature of 60 in. The span is 30 in. while the width is 4-0 in at the centre and decreases uniformly to zero at the ends. Determine the stress in the spring when it is just flat, (25,000 lbf/in
2).
C H A P T E E X I
STRAIN E N E R G Y OF BENDING
I T HAS been shown that , in simple tension or compression the work done in straining is given by
f2
U = x Volume,
where / is the uniform tensile or compressive stress across the
section under load and is within the elastic limit.
The application of a simple bending moment to a beam induces
a varying stress across the section.
FIG. 3 2 4
F o r a small length of beam dx between two plane transverse sections, consider the transverse area dA (distant y from the neutral axis) over which the direct stress due to the application of a simple bending moment (M) is / .
Volume of element = dx χ dA.
361
362 S T R E N G T H O F M A T E R I A L S
2EI2
M2y
2
Strain energy between sections = ν ρ ρ x
^1 c
^ '
- 1 1 ± . Σ ΆΛ x y2
2EI2
(taking out the constants).
Clearly Σ^Α x y2 is the 2nd Moment of Area of the section
i.e., / .
.'. Strain energy between sections i.e. for the length da;
1 χ M
2dx.
2EI
Thus for a length of beam L :
Tota l strain energy U = JM2 dx
where M is expressed in terms of x.
Note: Tf / is not constant it must be expressed in terms of χ and included in the integral.
The foregoing is a general expression and may be applied to
part icular cases of loading, as follows :
(a) Simply Supported Beam with Central Concentrated Load
From a support to the centre
of the span : X w
X
- L -t \ 2 5
FIG. 3 2 5
w
M2 W
2x
2
f2 (My\
2
Strain energy of element - - — - χ dx χ dA, where /2 = ——
2E \ I j
χ da; χ d i .
S T R A I N E N E R G Y OF B E N D I N G 363
L/2
F o r naif the beam, £7 = / 2i£i J 4
Fo r whole beam, Î7
L / 2
\2EIJ 4 ~ ~
L W 2
U X
4
^ 2 Z3
~ 3 ~ o
12J0J 8 '
96EI '
(b) Cantilever with Uniform Load
At any section X X :
, , χ ΜΎ = wx χ —,
wx*
i f2 W W
4 ' FIG. 3 2 6
·. u 2Elj
w2x* dx
w* 2EI
w x Z , s , and i.e. w2 = —.
1 jf2
4 0 ^ 7 X
X2
W2L*
40EI '
L\
364 S T R E N G T H OF M A T E R I A L S
(c) Simply Supported Beam with Uniform Load
At any section X X :
l i r wLx χ Mx = — wx χ — ,
wLx wx2
,_0 w2L
2x
2 w
2x*
Ml = — r - + - r -
/ wLx wx2
w2L
2
X X2 + -
^ ( = ^x totol load)
FIG. 3 2 7
w2L
χ χό.
1 Γ / w2£
2
w* χ
2 # /
w2L
2 χ
3 w
2 x
5
— χΎ + —χΎ
w2L
2
w2£
X a:3 da;,
x*
2 ~ X T 1 /M )
2Z
2
T, W2
2 U l 2 " x i + ¥ x i
w2L
χ £4 ,
m W / 1
I P X3
" 2 ^ F
TF2Z
3
2 4 0 ^ 7
120
(d) Beam Subjected to Uniform Bending Moment
An example of this is tha t part of a wagon axle between the
wheels.
Since M\I = Ε/Β, Λ Β = EIjM (all of which are constant .)
Hence R is constant and the part A B bends into a circular arc .
S T R A I N E N E R G Y O F B E N D I N G 3 6 5
Thus, U = 2Elj
ο
m CA
2ËïJ dX
>
M* dx,
M2L
2EI '
E X A M P L E . A 2 in. dia. shaft is simply supported over a span of 3 0 in. and loaded a t the centre unti l the bending stress is 1 5 , 0 0 0 lbf/in
2. Calculate the value of the load and est imate the strain
energy stored. Ε = 3 0 x 1 0E lbf/in
2.
Solution
π 7
= 6 4Χ 2 4
'
y = 1 in.
/ = 1 5 , 0 0 0 lbf/in2.
|W
- 3 0 in -
0 - 7 8 5 in4.
-ψ. ^
FIG. 3 2 9
3 6 6 S T R E N G T H O F M A T E R I A L S
Max. B .M. -WL
Strain energy
and M
WL
W
U
1± y
9
i l
y '
ML Ly '
4 χ 1 5 , 0 0 0 χ 0 - 7 8 5
3 0 χ 1
1 5 7 0 lbf.
W2L
3
9QEI '
1 5 7 02 χ 3 0
3
9 6 χ 3 0 χ Ι Ο6 χ 0 - 7 8 5
2 9 - 4 in lbf
E X A M P L E . A spring steel r ibbon 0 - 1 2 5 in. wide, 0 - 0 2 in. th ick is
wound on a 2 0 in. dia. cylinder. F ind the max . stress in the steel
and the strain energy stored per foot. Take Ε = 3 0 χ 1 06 lbf/in
2.
Solution
I =
U
bd3
Ί 2 ~ '
0 1 2 5 χ 0 0 23
ΙΟ in r
12
1
/ = 0 - 1 2 5 in
FIG. 3 3 0
1 2 χ 1 0
Ey
~R~
3 0 χ Ι Ο6 χ 0 - 0 1
" ÏÔ 9
= 3 0 , 0 0 0 lbf/in2.
M2L 1 EI c%/\
, where M = —— and L = 20π, 2E1 M
EIL
2R2 9
30 χ Ι Ο6 χ 20π
0 - 0 2 ίη
S T R A I N E N E R G Y O F B E N D I N G 367
U per ft length = 0-785 χ 12
20π
0-15 in lbf.
(e) Simply Supported Beam with Non-central Concentrated Load
Moments about the L H end give
Wa,
Wa
R2L
R9
L
Λ X I » X, x 2 *2 X, x 2 *2
Χι x2
FIG. 3 3 1
η - wo
Similarly, Wb
F o r any section X X X ! in part a : il/,. = / ijA'j L
\ r ψ2^2
\ F o r part a of beam: Un = / ——— χ dx, ILL J LJ
WW
2EIL*
W2b
2a
s
QEIL2
3 6 8 S T R E N G T H O F M A T E R I A L S
Similarly, for part b of beam :
W2a
2
L2
0
0
W2a
2 I
2EIL2 I
W2a
2b*
6EIL2
F o r whole beam : U = Ua + Ub,
W2b
2a* + W
2a
2b*
.r| da:,
6EIL2
W2a
2b
2(a + 6)
£7 =
6EIL2
W2a
2b
2
and α
6 # L L
IF Since work done in straining = — y where y = deflection a t load point,
W -y
W2a
2b
2
whence y — Wa
2b
2
ZEIL 2 ü
6EIL '
(a result already obtained.)
E X A M P L E . A steel jo is t having I = 8 0 i n4 carries a uniform load
of 4 tonf/ft over a span of 1 0 ft. Es t imate the elastic strain energy,
assuming Ε = 3 0 χ 1 06 lbf/in
2.
Solution
Tota l load
Span
where
W = 4 χ 1 0 χ 2 2 4 0 ,
= 8 9 , 6 0 0 lbf.
L = 1 0 χ 1 2 ,
= 1 2 0 in.
W2L*
" 2 4 0 ^ 7 '
# = 3 0 χ 1 06 lbf/in
2,
7 - 8 0 in4.
4 t o n f / f t
- 10 f t -
FIG. 3 3 2
S T R A I N E N E R G Y O F B E N D I N G 369
He nee U = (89-6 x 1Q
3)
2 x 1 2 0
3
240 χ 30 x 1 0e χ 80 '
89 ·62 χ ΙΟ
6 χ 1728 x 1Q
3
24 χ 3 χ 8 χ ΙΟ6 χ 1 0
3
89·62 χ 1728
2 42
- 24 ,000 in lbf
- 2000 ft lbf ,
= 0-89 ft tonf.
E X A M P L E . A shaft 60 in. long 2 in. dia. carries a pulley weighing
100 lbf a t a point 20 in. from a bearing. Calculate the strain energy
and hence find the deflection a t the pulley. Ε = 30 χ 1 06 lbf/in
2.
Solution
64 I =
17 =
J T X 24
64 0-785 in
4
6EIL '
1 0 02 χ 2 0
2 χ 4P
2
6 χ 30 χ ΙΟ6 χ 0-785 χ 60
64 χ 1 08
108 χ 0-785 χ 1 08 '
0-75 in lbf.
6 0 m -
2 0 in
1 0 0 lbf
I FIG. 333
Deflection under load
2 0-75 χ 2 0 0 1 5 1 in.
E X A M P L E . A steel wire is to be of such diameter tha t , when wound on a 72 in. dia. drum, there shall jus t be no permanent strain on removal. Take Ε = 13,200 tonf/in
2, assume an elastic l imit of
14 tonf/ in2 and calculate
(a) the permissible maximum wire diameter,
(b) the 2nd Moment of area of the wire section, (c) the bending moment on the wire,
(d) the strain energy stored per turn.
370 S T R E N G T H O F M A T E R I A L S
Solution
From the bending equation :
. Ey d / = — ' where y = —,
14
. d :
R
13,200 d
~ ~ 8 6 ~ ~ X
2"'
14 χ 36 χ 2
13,200
0-076 in.
Fo r the wire section
nd*
"64~ '
π χ (0-076)4
FIG. 334
Bending moment
64
0-00000164 in4.
M -EI
13,200 χ 0-00000164
36
- 0-000603 ton in.
= 1-35 lbf in. and is uniform.
F o r uniform bending moment,
M*L Strain energy U
2ΕΙ '
1·352(72π)
2 χ 13,200 χ 2240 χ 0 0 0 0 0 0 1 6 4 '
where 72π = length per turn,
= 4-25 in lbf .
E X A M P L E . A shaft of length Lin., dia. d, carries a concentrated
load W lbf,
(a) a t the centre of a simply supported span
(b) in direct tension.
S T R A I N E N E R G Y OF B E N D I N G 371
Determine from first principles the ratio of L to d for the strain
energies in the two states to be equal.
Solution
FIG.335
(a) At any section X X
2
/ , / 2
u- = -arf M2dxx2> 0
L / 2
EI S
ψ2χ2 —: dx,
0
W2 Ι . τ
31
7/
2
4 £ 7 I 3 |o
W2L* Λ r nd*
where I = 96EI
9 " " " "
M - 64
9
2W2L*
3End* *
(b) Area of section = .
71 Volume of section = ~-d
2L.
4
372 S T R E N G T H O F M A T E R I A L S
Tensile stress / =
iL 2E
nd2
χ Volume, 0 .
Hence, if
then
Un
2W2L
3
~ ( nd2 J 2E
2W2L
End2 '
2W2L
V///J, nd
2L
Τ
3End* End2
L2 = 3d
2
or L = dp.
E X A M P L E . The 2nd Moment of area of the
section of a canti lever increases linearly from FIG. 3 3 0 zero a t the free end to I a t the support.
Deduce, from a consideration of strain energy, an expression for
the deflection a t the free end when a transverse load W is concen-
t ra ted there.
A cantilever spring consists of a triangular steel plate having
the dimensions shown:
F I G . 3 3 7
S T R A I N E N E R G Y O F B E N D I N G 373
Calculate the deflection a t the free end when a load of 60 lbf
is applied there. Ε = 30 χ 1 06 lbf/in
2.
Solution
At any section X X :
2nd Moment Ir
Bending moment
Strain energy U
W*L L* i.e. U =
W2L*
4EI
B u t U = ^WZ, where Ζ = deflection a t load point.
Equat ing,
WL3
1 W2L*
±WZ = ——, 2 4 # / '
whence,
I n the given case,
Ζ =
/ =
2JW '
bd3
w 6 χ 0 -25
3
where I = value a t support.
12
1
"Ϊ28 '
Deflection, Ζ 60 χ 1 0
3
2 χ 30 χ 1 06
128
128
1 03 '
0-128 in .
13 S M
3 7 4 S T R E N G T H O F M A T E R I A L S
Examples X I
1. A uniform beam the section of which has a 2nd Moment of area of 35 in
4 supports a dead load of 3 tonf at the centre of a simple span of 120 in.
Take Ε = 13,000 tonf/in2 and calculate
(a) the strain energy of bending (0-37 in tonf), (b) the deflection at the load point (0-24 in.).
2. Deduce an expression for the strain energy per unit length of a beam at a point where the section has a 2nd Moment of area of I and the bending
3. A steel bar of square section is subjected to a uniform bending moment (a) in a plane parallel to one of the sides and (b) in the plane of a diagonal. If the maximum stress (within the elastic limit) is / tonf/in
2 deduce in each
case (in terms of / and E) an expression for the recoverable strain energy
4. A cantilever of length L has the form of a right conical frustum, the axis being horizontal and the diameter at the free end being d. Derive an expression for the strain energy of bending due to a vertical load W at the free and given that the diameter at the built-in end is 2d. Hence deduce an expression for the deflection at the load point.
(Note: Change limits and integrate the expression for U with respect to diameter)
5. Derive, in terms of stress and Modulus of Elasticity, a formula for the strain energy of a centrally loaded leaf spring having η leaves of width b
moment is M
(a) -fijjj x Volume, (b) χ Volume.
and
and thickness t, the length of the main leaf being L
C H A P T E R X I I
SHEAR STRESS DUE TO BENDING
Shear in Beams
The shear stress produced b y a vert ical shear force is accompan-
ied by an equal complementary shear in a horizontal plane. I n a
laminated spring, this results in relative motion between the leaves.
The two shears produce tensile and compressive stresses in the
web of an I- joist which m a y cause failure by buckling.
A reo Δ
C D
M + d M
dy y
km TT A L A
E F
FIG. 3 3 9
My — b.dy
= stress x section =force due to M
( M + d M ) y j b .dy =force due t o ( M + d M )
FIG. 3 4 0
13* 3 7 5
376 S T R E N G T H O F M A T E R T A L S
a x i s : Resul tant force || N.A. =
( M +^
M ) y χ b dy — χ bdy,
dM bydy.
y max
dM .'. To ta l pull on piece of section A and length dx = j^-j-bydy.
y
This is resisted by the complementary shear stress q acting on the area b dx. The shear face produced = qb dx.
Equat ing, dM
qb dx = J — by dy ,
y
I'/max
' " • '/ 7Λ./
Λ·"
, 1· '
/
and
"UNIX
J bydy 1st Moment of area A,
where 2/ is the distance from the N.A. of the centroid of this area.
Hence q = ^ ^ , since = T7 (the shear force.) 76 dx
Clearly this also gives the value of shear stress on the vertical
section a t y from N.A.
7. Rectangular Section
It w/4y////
It -A d
FIG. 3 4 1
Considering the forces on the element distant y from the neutral
S H E A R S T R E S S D U E T O B E N D I N G 377
Shaded area A = b ^— ?/j ,
F rom the neutral axis the distance of the centroid of this area
bd3
2nd Moment / = — . I Ζ
.*. Shear stress on a vert ical section a t any distance y from the N.A.
FAy
d \( d/2 - y
2
12F (d τ - y
12
d - 2y\ y + *
yx
bd3 \ 2
12F ld-2y\( 4y + d - 2y \
bd3
6F (d - 2y) (2y -f- d)
~bd? 4 '
6F (d2 - 4y
2 ' _ 6F (d
2 - ±y
2 \
~ ~h¥\ 4 Γ bd3 \ 4
6F / d2 - y
2
I f q is plotted against y, the result is a parabola. Clearly q is a
max . when y = 0 and a min when y = d/2.
3 F 3 Put t ing j , = 0 , 2 L N A X= - — = -
E X A M P L E . A t imber jois t 3 in. wide by 6 in. deep is simply sup-ported over a span of 20 ft and carries a central concentrated load of 200 lbf. I f Ε = 1 χ 1 0
6 lbf/in
2, ca lcula te :
(a) Max. bending stress, (b) Slope a t supports in degrees,
d
378 S T R E N G T H O F M A T E R I A L S
(c) Deflection a t load point,
(d) Max. shear stress,
(e) Bending and shear stresses a t a point 2 in. above the N.A.
and 5 ft from a support.
Solution
3 in, 2 0 0 lb*
I- 5.η---•I
'31 0 m
2 0 f t 4100 lbf
FIG. 3 4 2
/ =
At the centre,
Slope a t ends,
Centre deflection,
3 χ 63
12 = 54 in
4.
M = 100 χ 10 χ 12
= 12,000 lbf in .
When y - 3 in., / m ax -12,000
54
667-0 lbf/in2.
dy WL2 200(20 χ 1 2 )
2
"cbT " 16EI ~ 16 χ 1 χ ΙΟ6 χ 54
= 0-01335 radian = 0-76 deg.
WL* Ζ = 48EI
= 0 0 1 3 3 5 χ
3
dy L
dx 3
Max. shear stress, qnVAK = — χ -r-r = - x F_
bd
20 χ 12
3 ^
3 100 3 x 6
1 065 in.
8-33 lbf/in2.
At 5 ft from support, M =- 100 χ 5 χ 12 ,
= 6000 lbf in.
When y = 2 in
A = 3 in2
y = 2-5 in.
F = 100 l b f
δ = 3in
S H E A R S T R E S S D U E TO B E N D I N G
6000 X 2
379
54
= 222-0 lbf/in2.
FAy _ 100 χ 3 χ 2-5
Λ 54 χ 3 '
= 4-63 lbf/in2.
2. / — S e c t i o n
(a) Web
The complementary shear
in the web is on longitudinal
planes parallel to the N.A.
(Fig. 343) and a t 90° to the
shear applied to the section
which is parallel t o the sur-
face of the web.
F o r any section X X dis-t an t y from the neutral axis (Fig. 344)
FIG. 343
A
- H h — b I FIG. 3 4 4
Applied shear
380 S T R E N G T H OF M A T E R I A L S
where A = shaded area above X X (including tha t of the flanges),
y = distance of centroid of this area from the N.A. (found from y = J £ a r / A ) ,
b = web thickness.
The graph of q against y is again parabolic in form, though q
is not zero where the web joins the flange, since A is not zero a t
this point. The shear stress is again a max . when y = 0, i.e. on
the neutral axis .
F I G . 3 4 5
"Jl Q m o x .
(b) Flange
Since the shear stress is parallel to the surface of the material , i t must ac t inwards from the extremities of the flange towards the centre as in Fig . 345 .
F o r any section Y Y distant χ from the centre (Fig. 346) ,
FIG. 3 4 6 9x =
FAy
lb
S H E A R S T R E S S D U E TO B E N D I N G 381
where A = shaded area to the left of Y Y , y = distance of centroid of this area from the N.A., b =r- flange thickness.
Since A is the only variable, the graph of q against a; is a straight
line, the max . shear stress being a t the centre.
In comparison with qy, qx is usually negligible.
E X A M P L E . I n I-section jois t 8 in. χ 4 in. overall may be taken
as 0-5 in. thick throughout. Es t imate the maximum shear force
which m a y be applied given t ha t the longitudinal shear stress is
not to exceed 5 tonf/in2.
Solution
4 χ 83 3-5 χ 7
3
12 12
2048 - 1200
Ϊ 2 '
- 4 in -
ΥΖΖΖΖΖ7ΖΖΖΖΖΆ
0 5 i n -
_ 848 _ H T '
= 70-67 in4.
Area of section above neutral axis
(shaded),
A = (4 χ 0-5) + (3-5 χ 0 -5 ) ,
= 2 + 1-75
= 3-75 in2.
Ν 8 in
0 - 5 in
FIG. 347
The distance of the centroid of this area from the neutral axis is
given b y :
y
(4 χ 0-5) 3-75 + (3-5 χ 0-5) 1-75
3-75
= 2 1·75
2
3-75
= 2 + 0-817,
= 2-817 in.
13 a SM
382 S T R E N G T H O F M A T E R T A L S
At the neutral axis, qn
Permissible shear force F =
F Ay
F χ 3-75 χ 2-817
70-67 χ 0-5
70-67 χ 0-5 χ 5
3-75 χ 2-817
= 16-75 tonf.
E X A M P L E . A beam has the
section shown, and carries
a shear force of 20 tonf.
Determine the maximum
shear stress and the ratio
(Ζιη;ιχ/(/ ιη<';»η·
- 5 in
Solution
5 χ 53 3 χ 3
3
12 12
= i ( 6 2 5 - 8 1 ) ,
544
- 3 in-
Fio. 348
12 '
= 45-3 in4.
Shaded area above N.A. A = | ( 52 - 3
2) ,
= 8 in2.
Λ
Γιο. 349
4 — 3 in 5 in
S H E A R S T R E S S D U E TO B E N D I N G 3 8 3
Distance of centroid of A from neutral axis,
_ 2 > » = — '
_ ( 5 χ 1 ) 2 + 2 ( 1 χ 1 - 5 ) 0 - 7 5
" 8
_ 1 0 + 2 - 2 5
" 8 '
_ 1 2 - 2 5
~ ~ ~ 8 ~ ~ '
= 1 - 5 3 1 in.
_ 2 0 x 8 x 1 - 5 3 1
4 5 - 3 χ 2 '
- 2 - 7 tonf/in2
Çmeixn
. gmax = 2 - 7
imoan 1 * 2 5
E X A M P L E . An I-section jo is t is 1 4 in. χ 8 in. overall and is
0 - 5 in. th ick throughout. Determine the rat io of max . to mean
shear stress a t a section where the upper flange thickness has been
reduced b y planing off 0 - 2 5 in. of metal . F o r the mean value, as-
sume tha t the shear is taken b y the web alone.
Solution
Ini t ia l section = 2 ( 8 χ 0 - 5 ) -f ( 1 3 χ 0 - 5 )
= 8 + 6 - 5 ,
= 1 4 - 5 in2.
Sect ion removed = 8 χ 0 - 2 5 ,
= 2 in2.
Sect ion remaining Aj — 1 4 - 5 — 2 ,
= 1 2 - 5 in2.
Web section Aw = 1 3 χ J ,
= 6 - 5 in2.
= 2 J = ^ = 1 - 2 6 tonf/in».
= 2 - 1 7 .
13a*
384 S T R E N G T H O F M A T E R I A L S
h = 7 - 8 5 in
y = 5-15 in
7-6 in
Ό 2 5 in
I 3 in
h =
« ι in
F E G . 3 5 0
Σαν (8 χ 0-25)0-125 + (13 χ 0-5)6-75 + (8 χ 0-5)13-5
12-5 0-25 - f 43-85 + 54
98-1
12-5
12-5
= 7-85 in.
Sect ion above N.A., A = (7-85 - 0-25] 0-5 + (8 χ 0-25) ,
= 3-8 + 2 0 ,
= 5-8 in2 (shaded).
F o r this area y
^ Ν . Λ . =
(7-6 χ 0-5) 7-6/2 + (8 χ 0-25) (7-6 + 0-125)
14-4 + 15-45
5-8
0-5 χ 1 33
5-8
29-85
5-8 = 5-15 in.
12 + (0-5 χ 13) 1-1
2 +
8 χ 0 -253
Ϊ 2
8 χ 0 -52
12 + (8 χ 0-25) (7-85 - 0 -125)
2 +
+ (8 χ 0-5) (5-9 - 0 -25 )2
91-5 + 7-87 + 0 0 1 + 119-4 + 0 0 8 3 3 + 1 2 8 ,
347 in4.
j-* 8 m - - H
S H E A R S T R E S S D U E TO B E N D I N G 385
= 1 1 2 .
lb I Aw Ib F 347 χ 0-5 (taking least value of b)
E X A M P L E . Show that , for a square section beam simply supported
with a diagonal horizontal, the shear stress a t a point y from the
neutral axis is given b y
F q = ( a V ^ — %y) (a)'2 + fy) where a is the length of the side.
Hence show tha t q is a max . when y = α/]/32
Solution
From similar triangles :
b ap (a/1/2) - y a/p
= 91 a
F I G , 3 5 1
FAy Ι F FAy Aw 5-8 χ 5-15 _ κ
- X " T T - = -7ΓΤΖ 7Z~Z X 6-5
386
F o r this area
S T R E N G T H OF M A T E R I A L S
b ( a = ~2\~P Ρ
1
2
^ I a χ 2 — a
ρ a
Ρ - y
1 / a
ïy , « 3 31/2 '
F o r a triangle about one edge, I =
y + τ 3]/2 1 _ 3
base χ (height)3
Λ / Ν . Λ . = 2
12
α]/2 / α ^3
12 \ p
Shear stress
F Ay F
α4/12
6F
ni
a
Ρ
12
2y
y]\ir 31/2
a
ρ
^ I a 2
( p -y
'2y a
α4 V 3 + 6
6F (ayy2 a2
a4 V 6 6
3
4y2
~ 6 ~
LL (ayp + a2 - 4y
2)
6F (ayp a2 2y
2 _ ay \
3 +
~6 3 " 3 ψ 2 / '
6i^ / ai/1/2 ( a2 2#
2 ayp \
— (2ay 1/2 + 2 a2 - 8?/*)
F2
-.—(ap-2y)(ap + ±y).
Shaded area b
S H E A R S T R E S S D U E TO B E N D I N G 387
For max . q :
= (4a V(2) - 8 t / - 2a \'(2) - 8y),
^ r < M ( 2 > - l < * ) = 0 (and 0 = ^ - 1 6 )
which is negative,
16y = 2a p a
1/32
E X A M P L E . An overhead rail is made from 2 in. square section
with the diagonal horizontal and supports a load of 0*75 tonf
Find, for the load point
(a) the max . shear stress,
(b) the mean shear stress,
(c) the shear stress a t the neutral axis.
Y o u may assume the expressions derived in the previous example.
Solution _ a
(a) q is a max. when y = 1/32 '
2
1/32 '
2
1/(2 χ 16) '
2 p ~ 4 p ~~ 4 '
0-75 = 1 Γ ( 1 ' 5 ^ 2 X 3 ] / 2 )
0-75 χ 9
32
0-212 tonf/in2.
3. Solid Circular Sectio?i
The width of the shaded area A is not constant so tha t an ex-pression for Ay in terms of constants must first be derived.
Width of strip = b1 dz and -j^j r
— cosö, i.e. bx = 2r co s0 ,
= 2r cos θ dz.
1st moment of strip = 2r cos θ dz χ ζ and ζ = r sin θ,
dz i.e. - ττΓ = r cos 0,
dd
= 2 r2 cos θ sinÖ dz and dz = r cos θ do,
- 2 r3 cos
20 sine d0 .
388 S T R E N G T H OF M A T E R I A L S
F 0*75 (b) AVERAGE = ~ ^ =
= 0 1 87 Ί Θ Ϊ ΐ Ί^ '
F c) when y = 0 , g N. A. = — (α J/2 χ a | / 2 ) ,
= — 2 a2
2 a4
_ ^
~ ä2
= (ZUVCRAGC
= 0-187 tonf/in2.
FIG. 3 5 2
S H E A R S T R E S S D U E TO B E N D I N G 389
and cos-^- = 0 .
( Ύ \ = 2 r
3 I — J u
2 du J , where u = cos0,
- 2 r3
2 r3
d^ = — sin0 d0 ,
u3
3 Jçj
= — I cosJ — — COS (fj
2r3
.'. Ay = — - c o s3 φ.
ό
FIG. 353
Since 0 varies between φ and JZ /2,
π/2
Tota l 1st Moment Ay = f 2 r3 co s
20 sin0 dö,
ψ π/2
390 S T R E N G T H O F M A T E R I A L S
Shear stress q — ———, where 1 = —— and b = 2r cos φ, lb 4
T
2 r3 „ \ 4 1
χ ( 2 r * \ 4
- — C O S30 9 — r „ _
3 7
/ π Γ4 2 r cos <P
4i^ cos2 99
3nr* and cos
2 99 = 1 — sin
2 φ,
1 - —
4 f Λ ν2
3nr2 \ r
2
which is the equation of a parabola.
4:F 4 When y = 0 , g m ux = ^ ^2 = — χ Mean shear stress.
E X A M P L E . Determine the ratio Ljd which willmake the maximum
shear stress in a short length L of solid circular shaft dia. d equal
to one quarter of the maximum bending stress when the shaft is
(a) simply supported with W a t the centre and (b) built in a t one
end and loaded with W a t the other. F o r a semicircular area, the
centroid is a t 4Λ/6π from the diameter.
FIG. 354
Solution
The shear stress is given by q = a n
d is a maximum when b —d.
S H E A R S T R E S S D U E TO B E N D I N G 391
Then
y
I
nd2
: " ΊΓ '
M
nd*
~6T
(given),
where F = Shear force.
The bending stress is given by /
d
My
when y
B u t ,
i.e.
M =
F =
9 m a x
M
/ n u
d_ 64
and is a max.
32M
nd* "
1 32M 16F
3nd2 ~JX"^cW
M 2d
F ~Ύ
WL
4
iL 2
WL 2
4 Χ
ΐ Γ
_2d
ΊΓ
whence, FIG. 355
d 3 "
M=WL\ . ΤΓ£ 2d
F=W \ " W " " 3 " '
£ 2
w
whence, FIG. 35(5
Ζ
/ y ^ - - - L •
/
392 S T R E N G T H OF M A T E R I A L S
Examples X I I
1. A T-section is used as a beam with the flange uppermost. Sketch the distribution of shear stress across the section.
2. The average shear stress over the section of a hollow beam 5 in square outside and 1-0 in. thick when supported with two sides horizontal is 2-22 tonf/in
2. Calculate the corresponding value of the maximum shear stress
and sketch the distribution graph (2-7 tonf/in2).
3. Particulars of two dissimilar beam sections are given below :
Web Flange Total Distance of N.A. from Section Thickness Thickness Depth outer edge of flange
in. in. in. in.
/ 0-20 0-35 2-5 5-0 Τ 0-50 0-50 5-0 4 0
The beams are simply supported over the same span and each is loaded at the centre. If the maximum bending stress is the same in each case, find the ratio of the maximum shear stresses. (/ = 3-4Τ approx.).
4. The vertical section of a horizontal beam is an isosceles triangle of base 4-0 in. and height 6-0 in. Draw a diagram showing the distribution of shear stress at a point where the shear load is 6 tonf. (L. U.)-
5. A beam which is initially square in section, the side of the square being 4-0 in., has a longitudinal hole 3-0 in. dia. bored symmetrically through it. Calculate the value of the maximum shear stress induced by a vertical shear load of 5 tonf, the neutral axis being horizontal and parallel to a side (1-66 tonf/in
2).
C H A P T E R X I I I
STRUTS
Struts Subject to Axial Load
When the length of a s trut is great in relation to i ts sectional
dimensions failure will not occur due to the compressive stress but
due to bending since no strut can be truly straight, no load truly
axial, and no material t ruly homogeneous.
Such bending under axial load is called buckling and the load
which produces i t is referred to as the Buckling, Crippling or
Critical Load.
When such failure occurs, the strut remains in equilibrium in the bent position, as shown in Fig . 357 :
Euler Formulae
Assuming
(a) tha t direct compressive stress is
negligible and
(b) the ends are pin-jointed (i.e. free to
change their s lope) :
d2y
M = EI dx
2 = —Py (where I = least valuefor secion).
d2y P_
da:2 + EI
0 ,
y = ο
i.e.
Rewriting,
0 , where k2 =
EI
Ρ = k2EI.
(D2 + k
2)y = 0 .
393
Γ FIG. 357
394 S T R E N G T H O F M A T E R I A L S
Hence the Auxiliary Equat ion is m2 + k
2 — 0 ,
Hence, General solution is
When χ = 0 ]
V = 0 J
m2 - — k
2,
m = 0 ± jk.
y — A cos kx + ß sin
0 - .4 cosO -I . B s i n O .
When χ = L
y = 0
- A -h 0 .
Λ = 0
.'. y = Β sin for.
0 = 5 sin kL.
The constant Β cannot be zero since this would make y zero for all
values of x.
Hence sin kL = 0 , i.e. kL — π (taking least value),
4
k2 =
whence, Ρ =
71
u — χ i£7 from above. E"
The safe load will be this value divided by a suitable safety factor.
I f one end is built in (i.e. fixed) so tha t a change in slope a t this end is prevented, then, if there is no lateral restraint a t the other end (Fig. 358) a bending moment M must be introduced to maintain equilibrium.
The strut is now equivalent to half a strut of length 2L loaded as in Fig . 357.
Hence, Critical load = •
i.e. Ρ =
π" (2L)
2
1 π2ΕΙ
1 L2
χ EI,
S T R U T S 395
I f both ends are built in as shown in Fig . 359 there will be two
points of contraflexure C, C, distant. L / 4 from each end.
The piece of strut between them, of length Lj2 is exac t ly similar
in shape to the pin-jointed strut of F ig . 357 since the bending mo-
ment a t the points C is zero. Hence for this case,
Critical load = — = — τ - χ EI,
τ ) -
4 n
2EI
I f the free end of the strut in F ig . 358 is prevented from moving
laterally by a horizontal force F (Fig. 360) then, for any point dis-
t an t χ from the fixed end :
Mx=EI$?jL= -Py + F(L-x),
I I ι
F I G . 3 5 8 F I G . 3 5 9 F I G . 3 6 0
396 S T R E N G T H OF M A T E R I A L S
R
RL Equat ing constants, k2A = RL, i.e. A = - 7 0 -
Hence, RL Rx
'Ύ2 W
R
F El — χ — [L-x).
F Hence the particular integral ν = — (L — x).
T o find the complementary function, the R H side of the equation
must be made zero.
Thus, g . + iV = 0,
or D2y + k
2y = 0 (using operator D).
d2y P F /r
d2y P F /τ
d^ +
-ÊIy = W
{ L~
x h
d2y F P
or —— -f k2y = R(L — χ), where R = ·— and i
2 = —— .
d.r2 j \ η EI EI
Assuming the particular integral ν = A -f Bx (i.e. equal in de-gree to R H side),
then, = Β dx
d2v
and - ^ = 0 .
Substi tution of these values in the equation gives
0 + k2{A + Bx) = i ? (L - x)9
i.e. 4 M + k2Bx = RL - Rx.
Equat ing coefficients of x, k2B = — R, i.e. B =
S T R U T S 397
i.e.
Also
y = u + ν
F '. y = A cos kx + Β sin 4# -f — (X — #)
d#
Ρ — 4 4 sin kx -f 5 4 cos kx —
( 1 )
(2)
when χ = 0
y = 0 0 = A cos 0 + Β sin 0
P L
P L (Sub. in (1)) ,
=4 + •
i.e.
when χ
ày
άχ
i.e.,
A = -P L
Ρ - 4 4 sin 0 + 5 4 cos 0 - — (Sub. in (2))
= 5 4 - P_
P'
Β = F
kP
0 = A cos 4 L + Β sin 4 L (Sub. in (1)) When χ = L
y = 0
since the third term becomes zero.
.'. Β sin 4 L = —A cos 4 L
or tan 4 L A_
Β *
Then the auxiliary equation is :
m 2 _|_ jfc2 = ο
m2 = — 4
2
m = 0 ± ?4
Hence the complementary function w = 4 cos kx + 5 sin 4#.
Since the general solution is the sum of particular integral ν and
complementary function u
398 S T R E N G T H ΟΓ M A T E R I A L S
kL — t a n- 1 ( —
Β
tan FL kP\
P F /
.'. kL = t an"1 kL (or tan kL = kL)
B y tr ial and error, kL = 4 4 9 radians, = 257°24 '
a n d t a n 2 5 7 ° 2 4 ' = tan 77°24 ' = 4 4 9 .
4 4 9 . k =
Ä:2 -
L
20-2
""Σ2"
Hence for this case, Critical load = 20-2
Ρ
W '
EI L
2 '
Since 2 π2 = 19-75 i t may be assumed tha t the cri t ical load is given
approximately by n
2EI
Ρ = 2 -L
2
Thus the axial crippling load is a constant χ π
2ΕΙ
IF • and this is
summarized in Fig . 361 in which the crippling loads are expressed
as a multiple of tha t for the simplest (first) case :
0 25 Ρ
" '//At///,
®
2 Ρ opprox.
FIG. 3 0 1
S T R U T S 399
In pract ice, some change in slope always occurs a t the ends, even
when 4 4
f ixed" . The constants must therefore be considered care-
fully before use.
F o r the first case, F — ——— , LP
since I = ^4&2, where A = section,
& = least radius of gyration.
I f the compressive stress P/A is plot ted vertically against the slen-
derness rat io Ljk for a given mater ial and end conditions, a curve
will be obtained from which the limiting value of L\k (correspond-
ing to the yield stress) can be obtained.
Assuming Mild Steel and pin-jointed ends, π2Ε = 9-87 χ 13 ,400 ,
= 132 ,000 .
P/Ä 132,000
Ρ/Α
ι / / 132,000 \
η Pia ) - L>k
2 66000 235 5 26500 163
10 13200 115 15 8800 94 20 6600 81 25 5300 73 30 4410 65
From the curve F ig 362 a limiting value of Ljk 81 corre-
sponds with the Mild Stee l yield stress of 20 tonf/in2.
I n other words, if L < 81 k the strut material will yield before
buckling occurs.
P_ π2Ε _
1 , e" -4 ~^ ( Z # )
2 : compressive stress a t failure,
or — - ] /π Ε -
k ~~V F/A ~ slenderness ratio.
400 S T R E N G T H OF M A T E R I A L S
100 200
L/k
FIG. 3 6 2
Consider the following sections :
R2 (a) Solid circular, about a diameter,
(b) Thin tube dia. d,
4
^ 1 6 '
.·. k = L i.e., L = 81 L = 20 d approx.
i.e., j /8 L - 81 - p - = 28-7 d approx.
Other sections may be similarly t reated.
F o r the Euler Theory, which neglects the direct stress, a slender-
ness ratio of about 80 corresponds with the Mild Steel yield point
and gives L = 20 d as the approximate l imit.
Thus in practice the Euler Formulae are used only for 4
' long
columns", i.e., having a minimum slenderness ratio of about 120.
This keeps the direct stress a t the buckling point down to about
S T R U T S 401
one third of the yield stress (see curve) and gives
L = 1204 , where h =
= 120 χ
- 30rf.
Rankine-Gordon Formula
This is an empirical formula i.e. based on experimental results, and is used where both direct and bending stresses are important, i.e. where L < 30d.
The crippling load is given by
Ρ A4
' 1 + "(ΤΊ where fc = yield stress, A section ( in
2) , L — length (in), Tc — least
radius of gyration, a — a constant which depends on material and end conditions.
F o r mild steel, the values of a for the four cases considered are
(a)
(b)
(o)
(d)
7500
1
7500
4 / 1
) Ο θ ) '
9 \ 7500
1 / 1
4 \ 7500
Note tha t case (d) which is the stiffest, has the least value of a i.e., Ρ has the greatest value.
The safe load is obtained by dividing the crippling load Ρ by a suitable safety factor .
x E X A M P L E . Compare the strengths of two steel struts 2 in 2 in. χ 100 in. and 2 in. χ 2 in. χ 30 in. given tha t Ε = 30 χ ΙΟ
6
lbf/ in2 and fc = 21 tonf/in
2.
The columns have one fixed and one free end.
402 S T R E N G T H O F M A T E R I A L S
Solution
I = 9 3
12 '
- 1-334 in4.
Euler buckling load,
Ρ = 71
fif for the long strut, 4 L
2
π2 x 30 x 1 0
s x 1-334
4 χ 1 0 02
= 9880 lbf ,
= 4-41 tonf.
FIG. 3 6 3
Rankine-Gordon
[ where k2 = -4-
\ A
Hence
1-334
22 = 0-334 , and a =
7500
21 χ 2240 χ 22
1 + / 3 0
2 \ '
7500
84 χ 2240
1 4 χ 900
7500 χ 0-334
84 χ 2240 1 + 1-44
84 χ 2240
2 4 4
= 77,150 lbf ,
34-4 tonf .
E X A M P L E . A strut having rounded ends is 12ft long and of circular section. When freely supported a t the ends a load of 9 lbf a t the centre produces a deflection of 0-375 in. Determine the Euler crit ical load. I f Ε = 30 χ 1 0
6 lbf/in
2, find the diameter of the bar.
S T R U T S 4 0 3
Solution
Ζ
EI =
Ρ =
cl*
. d
J _ WL*
4 8 El
WL*
4 8 Z "
n2EI
Ζ = 0·375 in
W=9lbf
FIG. 3 6 4
L2
2 WL* 1 71 X~4SZXT2
n2LW
4SZ 9
π2 x 1 4 4 x 9
4 8 x 0 - 3 7 5 9
> Ί2π2,
7 1 0 lbf.
P = n
2EI
L2
PL2
6 4 n2E
9
64PL2 6 4 x 7 1 0 χ 1 4 4
2
π*Ε π2 x 3 0 x 1 0
E
^ 1 - 0 2 = 1 - 0 0 5 in, say, l i n
= 1 - 0 2 .
E X A M P L E . A 1 4 s.w.g. ( 0 - 0 8 in. th ick) round tube of 1 in. out-side dia. and length 4 0 in. is to be used to t ransmit thrust to an aircraft control. I t may be assumed to be pin-jointed a t each end. Use Rankine ' s Formula to determine i ts failing load taking a - 1 / 3 0 0 0 and fc = 1 9 tonf/in
2.
Obtain the Euler buckling load for the same strut under the
same loading conditions.
404 S T R E N G T H Ο Γ M A T E R I A L S
Solution
π I = — [ I4 - (0 -84)
4] ,
^ ( 1 - 0-498) ,
0-025 in4.
Ο Ό 8 in
A = - ^ - [ l2 - (0 -84)
2] ,
4
= j - ( l - 0-706) ,
- 0-2325 in2.
* 2 = T>
0 0 2 5
0-232 '
= 0 1 0 6 .
Pn = /c4
4 0 in
-ffl 19 χ 0-232
1 1600
3000 \ 0-106 1 + •
π2ΕΙ
L2 9
_ π2 χ 13400 χ 0-025
Î6ÔO
= 2-03 tonf.
19 χ 0 2 3 2
1 + 5 0 1
I in
FIG. 3 6 7
= 0-73 tonf.
taking Ε = 13,400 tonf/in2
E X A M P L E . A s trut is composed of two T-sections riveted back to
back to form a cruciform section 6 in. χ 8 | in. overall. E a c h T-sec-
t ion is 6 in. χ gin. χ 4 | in. The effective length is 20 ft and the
ends are rigidly secured. F ind the maximum safe load using a
safety factor of 5 given tha t fc = 21 tonf / in2 and tha t a = 1/30,000
in the Rankine-Gordon Formula .
S T R U T S
Solution
0-625 χ 8-753
+ g /2·687 χ 1 ·253^
1
~ " Ϊ 2
_ 428-5
12 '
= 35-8 in4.
12
(418-0 + 10-5) ,
Z2-687 χ l - 2 53\
A 12 I
0 - 6 2 5 i n -
I 2 5 in
- 2 - 6 8 7 i n -
8 - 7 5 in
— 6 in
FIG. 366
1-25 χ
12
= 22-5 + 0-15
= 22-6 in4.
+ • 7-5 χ 0 -625
3
12
Area of section, A = (6 χ 1-25) + (7-5 χ 0 -625) ,
= 7-5 + 4 -68 ,
= 12-18 in2.
22-6 Leas t value of i,2 _ Iy\i
12-18 = 1-86.
14 S M
405
406 S T R E N G T H O F M A T E R I A L S
Rankine buckling load Ρ fcA
where L = 20 ft,
21 χ 1 2 1 8
30 ,000 \ 1
21 χ 1 2 1 8
1 + 1 033 '
2 02 χ 1 2
2
b 8 6
21 χ 12-18
^ 0 3 3 125-5 tonf.
.'. Safe load = 125-5
5
= 25-1 tonf.
E X A M P L E . S ta te the fundamental assumptions made in deriving
the Euler Formula for a strut hinged a t the ends. A straight length
of steel bar 6 ft long and 1 in. χ | in. in section is loaded axial ly
until i t buckles. Assume Euler 's formula to apply and est imate
the max . central deflection possible before the material passes
the yield point, if the la t ter is 21 tonf/ in2 and Ε = 30 χ 1 0
6 lbf/in
2.
Solution
Assumptions :
1. Uniform elast ici ty throughout,
2. No eccentr ic i ty of axial load a t ends, 3. Buckl ing load independent of degree of deformation, 4 . E las t ic limit not exceeded.
1 χ 0 -253
Leas t value of / = —τ , Λ = 0-25 in2. 12
= 0-0013 in4,
0 0 0 1 3
0 1 2 5 = 0-0104 in
3.
Buckling load, Ρ = π*ΕΙ
Ζ2
π2 χ 30 χ ΙΟ
6 χ 0-0013
7 22
= 74-5 lbf.
S T R U T S 4 0 7
Direct stress, — Ρ
A
7 4 - 5
0 - 2 5 *
2 9 8 lbf/in2.
Bending stress,
Μ __Ρδ (where Ô = ~2 -2, required deflection),
7 4 - 5 0
" 0 - 0 1 0 4 '
= 7 1 5 0 < 5 .
Permissible stress,
Ρ = 2 1 χ 2 2 4 0
Λ 4 7 , 0 4 0
δ
M
2 9 8 + 7 1 5 0 ( 5
4 7 , 0 4 0 - 2 9 8
7 Ï 5 0
4 6 , 7 4 0
7 1 5 0 *
6 - 5 3 in.
|«-K)in-»| j
0-25 in
L = 72 in
FIG. 3 6 7
E X A M P L E . A stanchion with fixed ends and 3 6 ft high, consists of an R . S . J , having a web 0 - 5 in. thick, flanges 0 - 8 8 in. th ick and measuring 1 5 in. χ 6 in. overall. E a c h flange is reinforced by a plate 1 2 in. wide and § in. th ick, b y means of rivets. Es t ima te the safe axial load, assuming a factor of safety of 5 and taking Ε = 3 0 χ 1 0
6 lbf/in
2.
Solution
The least value of / must first be found.
(12 χ 1 6 - 2 53
1 2
- 4 2 9 0 - 1 6 8 7
= 1 5 3 9 in4.
j _ ^ 6 χ 1 5 3 j _ ^ 5 - 5 χ 1 3 - 2 4 3 j
1 0 6 4
14*
408 S T R E N G T H OF M A T E R I A L S
- 0 6 2 5
"0-88 in — 15 i n —
-13-24 in-
0-5 in
6in
12 in
3 in
FIG. 3 6 8
1-25 χ 1 23 1-76 χ 6
3
12 1
12
= 180 + 31-7 + 0 1 3 8
= 211-8 in4.
+ • 13-24 χ 0 -5
3
12
This is the value required. Since both ends are fixed and L > 30 d the Euler crippling load is
given b y n
2EI
Ρ = 4 L
2
4 π2 χ 30 χ 1 0
β χ 211-8
(36 χ 1 2 )2
1,345,000 lbf,
600 tonf.
Hence safe load = 600
= 120 tonf .
S T R U T S 409
E X A M P L E . F ind the Euler critical load for a hollow C.I. column 6 in. outside dia. J in. th ick 2 0 ft long and hinged a t bo th ends. Ε = 5000 tonf/in
2.
Compare this with the Rankine-Gordon critical load taking 1
a = "Ï6ÔO"
a n d fc = 3 6 t 0 nf / i n 2'
F o r what length of column would the two formulae give the same critical load?
Solution
π
4-5*),
= — (1296 - 4 1 0 ) ,
886π
64 43-5 in
4.
Ρ = π
2ΕΙ π
2 x 5000 χ 43-5
L2 ~ 2Â& *
= 37-3 tonf. (Euler)
A = 0-785 [D2 - d
2),
= 0-785 ( 62 - 4 - 5
2) ,
= 12-4 in2.
~ A
P =
43-5
12-4
LA
3-51 ,
1 + a (L/k)2
36 x 12-4
L=240 in
4 4 6
1 + • / 2 4 0
2\ 1 + 10-25
1600 \ 3 - 5 1 /
39-8 tonf. (Rankine)
410 S T R E N G T H O F M A T E R I A L S
F o r Same Load : π
2ΕΙ
L2 1 + a (Ljk)
2 9
fcAL2 = π
2ΕΙ + π
2ΕΙα
IF9
L2
fcA-π
2ΕΙα
L2(fcA -
L2 =
k2
π2ΕαΑ)
π2ΕΙ
= π2ΕΙ
= π2ΕΙ.
and k2 1_
~Α
fcA - π2ΕαΑ
9-87 χ 5000 χ 43-5
L
(36 χ 12-4)
2 ,150,000
4 4 - 382
183 in. = 15-3 ft.
^9-87 χ 5000
= 3 3 , 6 0 0 .
1600 χ 12-4
0 - 1 2 5 m
0 · I in
y//
-1 in -
FIG. 3 7 0
" "Γ 0-5 in
Solution
Sectional area, A
E X A M P L E . A symmetrical steel
connecting rod has the section
shown and a length between
bearing centres of 10 in. Take the
crushing stress as 21 tonf/in2 and,
using appropriate values of the
constant a calculate the R a n k i n e -
Gordon crippling load (a) in the
plane of oscillation and (b) in the
plane a t 90 deg to the web.
(2 χ 0-5 χ 0-125) + (0-65 χ 0 -1) ,
0 125 + 0 0 5 7 ,
0-2 in2.
0-5 χ Ρ 0-4 χ 0 -753
12
0-5 - 0 1 6 9
12
0-331
12 9
0-0276 in4.
12
S T R U T S 4 1 1
L^Ak* Λ * -0
"0 2 76
0 - 2
= 0 1 3 8
ΙΟ2
7 2 5
ίήτγ (AND A = W ) ' ι +
2 1 χ 0 - 2 4 - 2 3 - 8 3 tonf.
lyy
7 2 5 1 - 0 9 6 6 + 7 5 0 0
/ 0 - 1 2 5 χ 0 - 53\ / 0 - 7 5 x 0 - l
3\
= 2 l 1 2 J + l 1 2 ] '
0 - 0 3 1 2 5 + 0 - 0 0 0 7 5
1 2
0 0 3 2
1 2 '
= 0 - 0 0 2 6 7 in4.
Im. = Akf. . . k1t
0 - 0 0 2 6 7
yy — ^Ltoy . . toy — ^
= 0 - 1 3 3 ,
1 02
ky 0 0 1 3 3 7 5 2 0 .
SINCE a = — x 2 1 x 0 - 2 / . 1 1
s
1 / 7520 \ + T \ 7 5 0 0 /
' * ~" 1 + 1
( 1 5 20
) \ " 4
" 7 5 00
4 - 2
1 - 2 5 1 3 - 3 6 tonf.
E X A M P L E . A tubular strut 6 0 in long 2 in. outside dia. is to carry
an axial load of 2 tonf with a factor of safety of 6 based on the
Euler crit ical load for pin-jointed ends. Es t ima te the thickness
required, given tha t Ε = 1 3 , 4 0 0 tonf/in2. Es t ima te also the factor
of safety according to the Rankine-Gordon formula, assuming
the constants to be 2 1 tonf/in2 and 1 / 7 5 0 0 .
412 S T R E N G T H O F M A T E R I A L S
Solution
Safe load = 2 tonf ,
Λ Euler crippling load Ρ
= 2 x 6
= 12 tonf.
π2ΕΙ
Ρ = — w h e r e L = 6 0 . L2
12 = π
2 χ 13,400
6 02
π
or
x EI" ( 2 * - ^ ) ,
16 - =
Em. 371
12 χ 6 02 x 64
i.e., t
13 ,400π3
= 6-654.
. & = 16 - 6-654,
= 9-345, giving d = 1
-2
-1
·7 8
- 0 - 1 2 5 in.
Area of section A = 0-785 ( 22 - 1-75)
2,
= 0-736 in2.
/ = " 6 Τ ( 2 4 _ 1 ' 7 5 4 ) ·
. — (16 - 9 -346) ,
6·65π
k2
64
I 0-362
A " 0-736
0-326 in4.
= 0-443
S T R U T S 4 1 3
Rankine crippling load = fcA
1 + a(L/k)2
2 1 χ 0 - 7 3 6
1 + 6 0
2
7 5 0 0 χ 0 4 4 3 ,
1 5 4 5
1 + 1 - 0 8 '
1 5 4 5
2 - 0 8 '
= 7 - 4 3 tonf.
.'. Corresponding safety factor Crippling load
Actual load
7 4 3
2
3 - 7
E X A M P L E . A stanchion is to be constructed from two channel sections as shown, each having Ixx — 4 - 3 i n
4 and Iyy = 5 5 i n
4 and
a section of 5 - 6 7 5 in2. The ends are to be fixed and the effective
1 length is to be 2 5 ft. Take a = — — - and L = 2 1 tonf/in2 in
° 7 5 0 0 x 4 ' the Rankine Formula and est imate the distance apart to give equal strength in both directions. F ind the safe load for a factor of 7 .
i
0 - 8 in
FIG. 3 7 2
Solution
B y the Paral le l axis theorem, from Fig . 3 7 3 ,
Ixx = 4 - 3 + 5 - 6 7 5 h2 for each section (where h = (d /2) + 0 - 8 ) . 14 a SM
4 1 4 S T R E N G T H OF M A T E R I A L S
Y
0-8 in
IY
FIG. 373
d
and
4-3 + 5 - 6 7 5 ^ + 0-8
/ d \2
= 8-6 + 1 1 - 3 5 ( — + 0 - 8 J
= Iyy = 2 x 5 5 = 1 1 0 in4.
'd \2
1 1 0 = 8-6 + 1 1 - 3 5 ^ - + 0-8 j
id \2 1 1 0 - 8 - 6
d d - + 0-8 = 2 9 9 , . ' . — = 2 1 9 ,
I = Ak2,
Buckl ing load, Ρ = -
h2 — ΊΑ
n χ —
ny
icA
8-93
d = 4 - 3 8 in.
= 9 - 6 9 .
1 + .(. 2 1 χ 1 1 - 3 5
1 1 0
1 1 - 3 5
and the ends are fixed,
1 + • 1
7 5 0 0 χ 4
2 1 χ 1 1 3 5
~~ 1 + 0 - 3 0 9 6 '
= 1 8 2 tonf.
( 2 5 χ 1 2 )2
9 - 6 9
S T R U T S 4 1 5
Hence, for a factor of safety of 7 : Safe load = 1 8 2
7
= 2 6 tonf.
E X A M P L E . A short length of mild steel tube 0 - 1 in. th ick 3 - 0 in.
outside dia. yielded a t 3 5 tonf/ in2 when tes ted in compression,
the t es t figures also giving a value of 1 3 , 5 0 0 tonf/in2 for E.
The crippling load on a length of 7 ft when tes ted as a strut
with pin-jointed ends, was 1 7 tonf. Compare this with the Euler
crippling load. F ind also the constant a in the Rankine—Gordon
formula, assuming fc in this formula to be the yield stress.
Solution
I
π χ 1 9 - 6
6 4 = 0 - 9 6 1 in
4.
6 1 - 4 ) ,
2 - 84) ,
0-1 in
L 7 χ 1 2 = 8 4 in. FIG. 374
Euler crippling load = n
2EI
L2
π2 χ 1 3 , 5 0 0 χ 0 - 9 6 1
8 Ϊ2
= 1 8 - 1 tonf.
Area of section, A = 0 - 7 8 5 ( 3 + 2 - 8 ) ( 3 - 2 - 8 ) ,
= 0 - 7 8 5 χ 5 - 8 χ 0 - 2 ,
= 0 - 9 1 in2.
14a*
/.a 1
1 Q 56
416 S T R E N G T H 0 E M A T E R I A L S
fcA
1 +
35 χ 0-91
1 + 84
2α
1 0 5 6
31-85
1 + 6685α
31-85
17 '
= 0-875,
0-875
6685 '
= 1-875,
7630
Rankine cripppling load, Ρ
Λ 17
.·. 1 + 6685 α
Λ 6685 α
α
FIG. 375
Solution
E X A M P L E . A b lock and tackle is
suspended from the apex of a tripod
as shown, and used to raise a load
W. E a c h leg makes an angle of
60 deg with the ground and consists
of a pipe 2-5 in. outside dia., 2-0in.
inside dia., 10 ft long.
Take the Rankine constants as
1/7500 and 21 tonf/ in2 and est imate
the maximum load which may be
lifted, assuming a safety factor of 5.
/ = - ^ ( 2 - 5 * - 2 - 0 * ) ,
= - £ - ( 6 - 2 5 + 4) (6-25 - 4 ) ,
A = 0-785 (2-52 - 2
2) ,
= 0-785 χ 2-25 = 1-77 in2.
/ = Ak2, Λ k
2 =
1-77
= 0-64.
S T R U T S 417
Rankine crippling load,
ι +
where £ = 120 in. ,
21 χ 1-77
1 + χ 1 2 0
2 '
7500 0-64
3 7 2
1 + 3 '
= 9-3 tonf.
.*. Safe load per strut,
9-3
5 '
= 1-86 tonf.
F FIG. 376
F cos 3 0 ,
Since there are three struts,
• iL ·' 3
Λ W = 3 χ 1-86 χ 0 -866,
= 4-82 tonf.
w Τ
τ \ /////////////////////m FlG-377
Examples X I I I
1. The force required at the end of a clutch operating lever is 56 lbf and is transmitted by means of a pin-ended steel rod 10 in. long. Take Ε = 29-8 lbf/in
2 and determine a suitable rod diameter given that the factor
of safety based on the Euler Formula is to be 10 (0-25 in.).
418 S T R E N G T H OF M A T E R I A L S
2. Draw a curve showing the theoretical relation between crippling stress and slenderness ratio up to Ljk = 150 for a material for which crushing stress and elastic modulus are respectively 10 and 4000 tonf/in
2. Show how
and explain why an experimentally obtained curve would differ from the above.
3. The section of a pin-ended strut is 0-375 χ 0-125 in. If the strut is to fail at an axial load of 356 lbf, find the required length given that Ε = 28-9 χ 10
6 lbf/in
2 (10 in.).
4. Each corner of a column 30 ft high 18 in. square consists of an angle 4 χ 4 χ 0*5 in. nominally, the angles being braced together. Calculate the Rankine-Gordon safe load using appropriate coefficients for steel and pin-jointed ends and assuming a factor of safety of 6. The centroid of an angle section is 1-17 in. from an outer side and the value of / about an axis through the centroid and parallel to a side is 5-46 in
4 (I. Mech. E.) (41 tonf).
5. Calculate, using the Rankine-Gordon Formula, the safe load for a pin-jointed strut 120 in. long composed of two steel T-sections 5 χ 2-5 χ 0-5 in.
with the 5 in. sides riveted back to back. Take a = a n
d / = 30 tonf/in2
and assume a factor of safety of 6 (10 tonf).
C H A P T E R X I V
THICK AND THIN CYLINDERS
Stresses in a Thick Tube
L e t Fig . 378 represent the section of a th ick tube under internal
and external pressure, i.e. both surfaces subject to compressive
stress. Due to the internal pressure on the tube ends there will be
an axial tensile stress fz. I n addition to this, any element as shown
in F ig . 379 will be subject to a radial stress fy and a hoop stress fx.
Since the cylindrical element in F ig . 379 is in equilibrium,
Downward force due to stress on section A B
= Upward force tending to fracture cylinder across A B
i.e., 2fx dr = fy2r - (fy + dfy) 2(r + dr)
= 2rfy-(fy + dfy){2r + 2är)
= 2rfy - (2rfy + 2r dfy + 2fy dr + 2dfy dr)
= 2rfy - 2rfy - 2rdfy - 2fydr - 2dfydr.
.*. 2fx dr = —2r dfy — 2fydr (neglecting the last t e rm) , J 1
Λ fx = - ^ " ^ 7 - fy (dividing through by d r ) ,
I f fz = axial tensile stress, then, assuming transverse sections to remain plane :
or y (1)
Axial strain, ez
fz Φ afy Ε Ε ^ Ε
i.e. 4 - - f y ) ]
4 1 9
(2)
S T R E N G T H OF M A T E R I A L S
FIG. 3 7 8
Unit length
fy + dfv
F =f, dr
FIG. 3 7 9
4 2 0
T H I C K A N D T H I N C Y L I N D E R S 421
Now the quant i ty (jx — fy) in E q . (2) must be a constant since
Ε, σ and fz are constant .
L e t fx — fy = 2a so tha t fx = ftJ + 2a . Substi tut ion of this value
in equation (1) gives
Transposing, — 2 x — dr,
i.e. J fy + a
J J r
log e (/// + α) = — 21og er -f constant and 2 log er = l og er2
\°ge(jy + ») + l o g e r2 = constant ,
log e
f 2 (/// +
a) = constant ,
^2(/// + = constant .
L e t r2( / y + a) = 6 , then — a
And = fy + 2 a , whence / r = — + a
(3)
(4)
Equat ions (3) and (4) are known as Lame's Formulai and give the
radial and hoop stresses respectively a t any radius r between R2
and Rv
At the inner surface where r = R 2 ,
fy = / 2 and .·. / a = — _ α
(and / 2 = applied internal
pressure).
At the outer surface where
r = Bl9
fy = fx and ; . f 1 = — - a
(and fx is usually zero).
Thus the constants α and b
can be found and their values substi tuted in the equation for hoop stress. FIG. 3 8 0
4 2 2 S T R E N G T H O F M A T E R I A L S
F o r zero external pressure the stress distribution is as shown
in Fig . 3 8 0 , the difference in height of the curves being 2a.
E X A M P L E . A cylinder 8 in. outside dia. and 4 in. inside dia. has
an internal fluid pressure of 3 0 0 0 lbf/in2. Calculate the longitudinal
stress (fz) assuming this to be uniform.
Solution
4 i n •—• I 3 0 0 0 = tx
FIG. 3 8 1
Area over which pressure ac ts , Ax = 0 - 7 8 5 χ 42,
Tensile force in pipe, F = 3 0 0 0 χ 0 - 7 8 5 χ 42,
= 3 7 , 7 0 0 lbf.
Area over which force acts , A2 = 0 - 7 8 5 ( 82 — 4
2) ,
= 0 - 7 8 5 ( 8 + 4 ) ( 8 - 4 ) ,
= 0 - 7 8 5 χ 1 2 χ 4 ,
= 3 7 - 7 in2.
Tensile stress =3 7
'7 0
° 3 7 - 7 '
= 1 0 0 0 lbf/in2.
E X A M P L E . The delivery pipe to a diesel engine injector is 6 - 0 mm
outside dia. and has a bore of 1 - 4 mm. Determine the maximum
hoop stress in the material if the injector orifice opens a t 2 0 0 0 lbf/ in
2.
Solution ΕΎ = 3 - 0 mm = 0 - 1 1 8 in.
R2 = 0 - 7 mm - 0 - 0 2 7 5 in.
T H I C K A N D T H I N C Y L I N D E R S
Hoop stress,
Radia l stress,
b
R=3O mm
FIG. 3 8 2
At R2:
At JV
2000 =
0 =
0 -02752
b
0 0 0 0 7 5 5
b
- a,
0-1182
b
0-0139
Substi tut ing for a in equation (1) above:
b b
— a, .*. a = 0 0 1 3 9
2000 0 0 0 0 7 5 5 0-0139 '
1 1
\ 0 0 0 0 7 5 5 0-0139
= 6(1320 - 72)
= 1248&,
.*. b 2000
1248 — = 1-6, :.a =
1-6
0 0 1 3 9 = 1 1 5 .
423
(1)
(2)
4 2 4 S T R E N G T H O F M A T E R I A L S
The hoop stress is a maximum a t R2.
I, + 1 1 5 , *
m ax ~ 0-000755
= 2120 + 1 1 5 ,
= 2235 lbf/in2.
E X A M P L E . A cylinder 8 in. outside dia. and 4 in. inside dia. has
an internal fluid pressure of 3000 lbf/in2.
F ind the max . and min. hoop stresses.
Solution
Hoop stress
Radia l stress
fx =Τ + °" r2
5 0 0 0 i
2 0 0 0
FIG. 3 8 3
At inner surface r = R2 = 2,
fy = 3 0 0 0 ,
3 0 0 0 = — - a .
T H I C K A N D T H I N C Y L I N D E R S 4 2 5
At outer surface
fy = 0,
• • • 0 = 4 - *
a = 1 6
• Κ 3 0 00
Ι Α
. . b = —— χ 1 6 ,
Min. hoop stress =
Max. hoop stress =
= 1 6 , 0 0 0 Hence a = 1 0 0 0 .
1 6 , 0 0 0
— T T " + 1 0 00
> 42
2 0 0 0 lbf/in2
1 6 , 0 0 0 + 1 0 0 0
22
5 0 0 0 lbf/in2.
E X A M P L E . A hollow forged boiler drum 7 6 - 4 in. outside dia. and
5 - 2 in. th ick was tes ted before entering service to 2 8 5 5 lbf/in2.
FIG. 3 8 4
426 S T R E N G T H O F M A T E R I A L S
Calculate the maximum and minimum hoop stresses under test .
Solution
Rt = 38-2 in.
R2 = 33 0 in.
At inner surface, r = R2 = 3 3 ,
33:
At outer surface, r = Rx — 38-2 ,
tu = 2855 = 4r - α (1)
b_
b a = 38·2
2
Substi tut ing for α in (1) above:
b 2855
3 32 38 -2
2
= b( b [ ~ W ~ ISS 2") 9
= δ(0·000915 - 0 -000685) ,
= 0-00023 &
b = 12-4 χ 1 06, a = 8 5 0 0 .
b ( 12-4 x 1 06 \
/ « Ü N - ^ + a , -( 38,22 J + 8 5 0 0 ,
= 8500 + 8 5 0 0 , = 17,000 lbf/in2.
4 b 1 1 2 4 x 1 0
6 \ rt ΛΛ
fcmax = - j j * +a>
= [ 332 j +
8 5 00 >
- 11,400 + 8 5 0 0 , = 19,900 lbf/in2.
E X A M P L E . F ind the longitudinal and hoop stresses a t the outer
surface of a hydraulic pipe 2 in. bore, 4 in. outside dia. a t the
operating pressure of 9 tonf/in2. Es t ima te also the increase in
outer dia. when under load. The pipe is steel and σ = 0-26,
Ε = 13,400 tonf/ in2.
Solution
Force on end = 9 χ 0-785 χ 22.
Area of section = 0-785(42 - 2
2) .
(2)
T H I C K AND T H I N C Y L I N D E R S
.*. Longitudinal stress,
427
U 9 χ 0-785 χ 2
2
~ 0-785 χ 12
= 3 tonf/ in2.
Radia l stress,
fy = 4- - α· A t R,
9 = - p - - a ,
a = b - 9 .
At i ^ :
0 = ^ - « ,
6 α = τ·
FIG. 3 8 5
b_
Τ 6 = 1 2 ,
Hoop stress,
At Rx:
= 9 ,
Λ a = 3 .
+ α,
~~ "22~ 3 5
= 6 tonf/in2. (and fy = 0)
Circumferential strain = 4τ — - τ τ and σ = 0-26 ,
Ε Ε 1
[6 - (0-26 χ 3)] 13,400
= 0-00039 .
'. Increase in diameter on 4 in. = 0-00039 χ 4 ,
= 0 0 0 1 5 6 in.
R. = 2 in
4 2 8 S T R E N G T H O F M A T E R I A L S
E X A M P L E . Draw the curves of radial and hoop stress on a base
of radius for a tube 6 in. outside dia. and 3 in. inside dia. when sub-
jec ted to
(a) an internal pressure of 9 0 0 0 lbf/in2 only,
(b) an external pressure of 6 0 0 0 lbf/in2 only,
(e) pressures as a t (a) and (b) together.
Solution
(a) Radia l stress fy = — a ( tensile),
Λ 9 0 0 0 ==T^~a
b b and 0= — - a , : . a = — .
Λ 9 0 0 0 = B
2 - 2 5 9 '
= 6 ( 0 - 4 4 5 - 0 1 1 1 ) ,
= 0 - 3 3 4 6 ,
.·. b = 2 6 , 9 0 0 and a = 3 0 0 0
r r2 26,900
r2
26,900 / y = — ^ - - 3 0 0 0
26,900 fx = +
3 0 00
1-5 2-25 11,900 8900 14,900 2 0 4Ό0 6720 3720 9720 2-5 6-25 4300 1300 7300 3-0 9-00 3000 0 6000
(b) Radia l stress, /„ = ^ — a (compressive),
b b
T H I C K A N D T H I N C Y L I N D E R S 429
and 6000 = — - a ,
ό
_b_ b__
~ 9 _
2-25 ' = 6(0-111 - 0-445) ,
= - 0 - 3 3 4 6 ,
- - 1 8 , 0 0 0 and a = - 8 0 0 0 .
r r2 18,000
r2
18,000 fy- r2 + 8 0 0 0
18,000 fx= ~2 8000
1-5 2-0 2-5 3 0
2-25 4-00 6-25 9-00
- 8 0 0 0 - 4 5 0 0 - 2 8 8 0 - 2 0 0 0
0 +3500 +5120 +6000
-16 ,000 -12 ,500 -10 ,880 -10 ,000
(c) F o r inner surface 9000 = 1-5
2
b F o r outer surface 6000 == — — a
ό
. • . ^ i - 9 0 0 0=A_ e 0 0 0
Ύ ( Ά - ΐ ) = 3 0 0 0' whence b = 9000
and a = - 5 0 0 0
r r2 9000 9000
fy- r2 +5< > U U
9000 fx = — - 5000
1-5 2-25 4000 +9000 - 1 0 0 0 2 0 4-00 2250 + 7250 - 2 2 7 5 2-5 6-25 1440 + 6440 - 3560 3-0 9-00 1000 + 6000 - 4 0 0 0
The diagrams of stress distribution are shown in Fig . 386 .
E X A M P L E . A hydraulic press has a ram dia. of 8 in. and operates
on an internal pressure of 1500 lbf/in2. Es t ima te the wall thickness
needed to prevent the hoop stress from exceeding 3600 lbf/in2.
F ind the increase in inside diameter under load, taking a = 0-25
and Ε = 16-8 χ 1 06 lbf/in
2.
430 S T R E N G T H OF M A T E R I A L S
Ι 5·
0 0 0Γ Hoop f, w
10,000 - / ^ S . Rodial fy 7 7 % .
5000 - ^ 7 / / 7>
2 O- (a) internal pressure only
-5° o ° - , b
9, ° / ° W I
-ιο,οοοί 7 /
10.000 ρ \ /
5000 - τ ·. ^ r ^ 7 % lf v
A Tensile ^<77/////
0 - - - ! {////// ( b) E x + e r n al Pressure only
~ - 5000 - ^ Compressive/ / / / / / / A
- 1 0 . 0 0 0 - / ' / / / / j J A * *
- 1 5 . 0 0 0 - y/>^ >Ä>
\ / ^ s " 6 0 0 0 l b f / i n2
- 2 a o o oL χ /
iQOOOr-
5 0 0 0 - ////^ZffiTTtf*
/ / / / / / / / / / / \ ^0 +h
Internal and ^ 0 *LZ / / / / / ex+ernal pressure
§ 9 0 0 0 -^7 ( /J* - 5 0 0 0 - l b f / i n
2 < ^ /
- 1 0 . 0 0 0 - \ v £ ^
\ 6 0 0 0 lb f / in2
- I 5 , 0 0 0 L \ .
FIG. 3 8 6
T H I C K AND T H I N C Y L I N D E R S
Solution
Radia l stress
Hoop stress
At R2,
b
1500 = — — a 42
R, = (4 + t ) in
f , = 0
FIG. 3 8 7
At Rl9
AtR2,
Adding (1) and ( 3 ) :
3600 = A + « 42
1βΟΟ + 3600 = Α + ^ = |
b = 5100 χ 8 ,
= 40 ,800
40 ,800 /0 Έ . _ α = — ^ 1500 (Subst . in E q .
= 2550 - 1500
- 1050
431
(1)
(2)
(3)
4 3 2 S T R E N G T H O F M A T E R I A L S
Substi tuting these values in E q . ( 2 ) :
4 0 , 8 0 0 0 = — 1 0 5 0 ,
R\
R\ 4 0 , 8 0 0
1 0 5 0
= 3 8 - 9
\ Rx = 6 - 2 5 i.e. t = 6 - 2 5 - 4 = 2 - 2 5 in.
. ι. , ± 1 5 0 0 χ 0 - 7 8 5 χ 82 , Λ , Λ 11 0
Longitudinal stress, / , = 0 - 7 8 5 ( 1 2 - 52 - 8
2)
= 1 0 42 L B F/
M 2-
.'. Circumferential strain a t inner surface
3 6 0 0
Ε ( τ x I 5 0 0 ) - (τ x i o 4 3 7 1 4
1 6 - 8 χ 1 06 9
0 - 0 0 0 2 2 1
,\ Increase in circ. = 0 - 0 0 0 2 2 1 χ 8π
·. Increase in dia. = 0 - 0 0 0 2 2 1 x 8 = 0 - 0 0 1 7 7 in .
FIG. 3 8 8
E X A M P L E . Es t imate the ratio of thickness to inside dia. required
to limit the max . hoop stress in a thick pipe to 1 -6 t imes the internal
pressure, the external pressure being negligible. I f such a tube
T H I C K AND T H I N C Y L I N D E R S 433
has an inside diameter of 4 in., est imate the increase in this diam-
eter when the internal pressure is 6 tonf/in2 given tha t Ε = 13,40
tonf/in2 and Poisson's R a t i o is 1/3*5.
Solution
Radia l stress,
b_
r2 fy
Hoop stress fx
AtRl9 0
a and
b + a.
AtR9
At Bl9
and
i.e.
giving
a =
_b_
b
Rl
_ A A Rl
D9
2-6 _ 0-6
_ Ri — R% 2R,
fx (max) : ' * 6 fp
FIG. 3 8 9
/a'max
b_
R\
\r% + R*)~~
~Rj + ~Rj
R\ + R\~ \R\ + R\J
1·6/ 2 (given).
' • e 4 ( ^ - 3 f ) '
or
1 6
~Rj
Ik R9
1-6
Rl
2 ·08 R2 — R2
2&>
1-08 0-54
I n the given case, D2 = 4 in, i.e. Dr = 8-32 in,
434 S T R E N G T H O F M A T E R I A L S
Area of section = 0·785(8·322 - 4
2)
Force on end = 6(0-785 χ 42)
6 χ 0-785 χ 42
0-785(8 · 3 22
1-81 tonf/in2.
42)
Max. hoop stress
Circumferential strain
1-6 χ Internal pressure = 1 - 6 χ 6
= 9-6 tonf /in2.
9^6
Ε
1
-α) -M (6 - 1-81)
9-6 + 3-5
Increase in diameter =
13,400
9-6 + 1-2
13,400
= 0-000806.
Increase in circumference
π 4π χ 0-000806
0-00322 in.
E X A M P L E . A steel tube 4 in. inside dia. and 6 in. outside dia. is
to have a second tube of the same material 7 in outside dia. shrunk on
i t , the shrinkage allowance being such tha t the radial pressure be-
tween the two tubes is to be 2 tonf/in2. Take Ε = 13,000 tonf/in
2
and a = 0-25 and calculate :
(a) hoop stress a t inner surface of outer tube,
(b) increase in internal diameter of outer tube,
(c) hoop stress a t outer surface of inner tube, (d) reduction in external diameter of inner tube.
Solution
F o r outer tube,
FIG. 3 9 0 giving
T H I C K A N D T H I N C Y L I N D E R S 435
And fy=2 a t r = 3 ,
' · 32 12-25
6 (0 -1111 - 0-0816) = 0 -02956 .
6 = 67-8 1 67-8 Λ at r = 3 : / r = — - + 5-53
and ο, = 5-53 32
= 7-53 + 5-53 ,
= 13-06 tonf/ in2 ( tensile) .
Increase in internal dia. = [13-06 + (2-5 χ 2 ) ] - ^ -
1 3 , 0 0 0 η
= 0-00624 in.
F o r inner tube, = 0 a t r = 2 ,
6 0 = — - a giving α = — ,
And / y = 2 a t r = 3 ,
and a = - 3 - 5 8 J 32
= - 1 - 5 9 - 3-58 ,
= - 5 - 1 7 tonf/in2
(compressive).
Reduct ion in external dia. = , ^ \„„ (δ·17 + •^-\ 13,000 \ 4 / π
= 0-00261 in.
Shrinkage allowance on diameter = 0-00624 + 0 -00261 ,
= 0-00885 in.
E X A M P L E . Determine the interference fit per in. of dia. necessary
to produce a radial pressure of 12 tonf/ in2 a t the common surface
of a compound tube. The inner and outer diameters are 4 in. and
10 in. and the inner tube has an outer diameter initially of 8 in.
W h a t must be the initial inner diameter of the outer tube? Take
Ε = 13,500 tonf/in2 and assume Lame 's Formulae.
436 S T R E N G T H OF M A T E R I A L S
FIG. 3 9 1
F o r outer tube fy = 0 a t r = 5 in.
Als
25
/„ - 12 a t r = 4 in.
12 = — - a = 6(0-0625 - 0 -04) , 4
2
= 0-02256.
6 = 533
a = 21-3 : . a t r = 4 : fx =
533 + 21 -3 ,
Hoop strain f. 54-6 12σ
Ε ^ Ε ^ 13,500 +
' 1 3 , 5 0 0
= 33-3 + 21 -3 ,
= 54-6 tonf/in2.
(1]
Solution
Hoop stress
Radia l stress
T H I C K A N D T H I N C Y L I N D E R S 437
F o r inner tube , fv = 0 a t r •-= 2 in.
:. ο b b
: α = Γ Also, fy = 12 a t r = 4 in.
.'. 12 b
δ(0·0625 -
- 0 - 1 8 7 5 6
.·. b = - 6 4 1
a = — 16 I .'. a t r — 4 :
64
= - 4 - 16
= —20 tonf/ in 2 (i.e. compressive)
Hoop strain = _ + _ / „ = - _ _ + _ _ (2)
Strain difference a t common diameter = (1) — ( 2 ) ,
54-6 12σ
+ • 13,500 13,500
20 12<7
+ 13,500 13,500
74-6 = 0 0 0 5 5 3 . 13,500
Change in inner diameter of outer tube = 0-00553 χ 8 = 0-044 in
Required initial diameter = 8 — 0-044 = 7-956 in.
Thin Cylinders
When the thickness is small in comparison with the diameter,
the hoop stress m a y be considered uniform and the radial stress
negligible. Referring to K g . 392 :
Area of element = r do χ L.
Radia l force
15 SM
= Lrad χ p.
438 S T R E N G T H OF M A T E R I A L S
FIG. 3 9 2
pd2 pd
(2)
Thus,
Hence the vessel will fail under hoop stress, i.e. split along its
fx - fz length. N . B . Max. shear stress = — .
Horizontal component = Lrάθ χ ρ χ sind
= pLr sin θ dö
Tota l force on half cylinder to the right of Y Y :
π
F = pLr / s i n O d o ,
ο
= I - c o s e i S ,
= pLr [ — cos η — ( — cos Θ)]
= pLr[-(-l) - ( - 1 ) ] ,
= pLr( + l + 1)
= 2pLr.
This is resisted by the hoop
stress acting over the section
tL on each side.
Λ 2(fjL) = 2pLr,
Tota l force on an end
nd2
= p χ — •
This is resisted by the longitu-dinal stress /- acting over the section π dt.
.. /- x ndt = ρ x —^—,
T H I C K A N D T H I N C Y L I N D E R S 439
E X A M P L E . A torpedo shell 20 in. dia. is 0-375 in. thick. Es t ima te
the hoop stress induced b y an internal pressure of 2000 lbf/in2
assuming the stress to be uniform.
Solution
Hoop stress fx = ,
2000 χ 20
~~ 2 χ 0-375 '
neglecting any external pressure,
= 53 ,300 lbf/in2.
E X A M P L E . A s team boiler 6 f t dia. is to operate a t a pressure of
300 lbf/in2 above atmospheric pressure. Es t ima te the plate thick-
ness required to limit the hoop stress to 5 tonf/in2.
Solution
Hoop stress fx
i.e. t
pd
"~2t'
pd
Ύ 300(6 χ 12)
2(5 χ 2 2 4 0 ) '
: 0-964 in. (say 1-0 in.)
Tensile Stress in a Thin Rim Due to Rotation
I f the thickness of the rim or hollow cylinder is small in relation to the diameter, then the stress m a y be assumed uniform.
Sect ion of element = r άθ χ t referring to Fig . 393 ,
Volume of element = r άθ t χ b
Weight of element = r άθ tb χ ρ, where ρ = density.
^άθώρ Mass of element =
g 15 a SM
440 S T R E N G T H OF M A T E R I A L S
.*. Centrifugal force on element = T
Ü>Q χ v_ g r
where ν = peripheral speed, ( = cor)
btqv2
άθ.
Vert ical component of this btqv
2
dö χ sin 0.
YZZA
FIG. 393
F = centrifugal F sin θ ι >*force on
[s(Q element
f ( b t ) = force per side resisting rupture
.*. Tota l vertical force tending to rupture rim across a horizontal diameter,
btqv2
btqv2
f g
btqv2
g
Μρν2
g
2btqv2
g
s inödf l ,
π
— cos θ , ο
[ — cos τι — ( —cosO)] ,
[ _ ( _ ! ) _ ( _ ι ] ) ?
T H I C K A N D T H I N C Y L I N D E R S 441
I f ν is in ft/s and ρ is in lbf /ft3, then / will be in lbf/ft
2.
E X A M P L E . Determine the maximum safe speed of a cast iron
flywheel 5 ft dia. if the density is 0-27 lbf i n3 and the safe stress
is 5000 lbf/in2.
Es t imate the bursting speed given tha t cast iron fails in tension
a t a stress of 7 tonf/in2. t ake Ε = 16 χ 1 0
6 lbf/in
2.
, ρν2 pœ
2r
2 , Λ
/ = — = - , where r = 2-5 f t . 9 9 0*27 χ 1 2 3
5000 χ 12 2 = — — χ 2 · 5 2ω 2 (working in ft units.),
i.e.,
32-2 5000 χ 32-2
w " 0-27 χ 12 χ 6-25 '
= 7 9 5 0 .
ω = 89-1 rad/s.
.". ^safe = 852 rev/min.
F o r the flywheel to " b u r s t " :
/ = 7 χ 2240 χ 1 22 lbf/ft
2.
2 7 χ 2240 χ 1 22 χ 32-2
* ' ω
0-27 χ 12 χ 625
- 2 4 , 9 0 0 .
.'. ω = 158 rad/s.
.*. Ν = 1510 rev/min.
E X A M P L E . Show tha t , for a flywheel r im made of C.I. having a
density of 0-272 lbf/ in3 the safe speed in rev/min is given by
Ν = 30 ? where / = permissible stress in lbf/in2.
outside diameter r = in ft.
and this must be equal to (stress in rim) χ (section resisting rup-ture) .
whence, / =
442 S T R E N G T H O F M A T E R I A L S
Solution
As already shown,
QV2
Stress / = - — , where υ = ων, 9
ρω2/·
2 - 2πΝ
— - and ω =-- ,„ν , 17 υΟ
" ? 3600 Χ
·
π2ρτ
2
where / is in lbf/ft2, ρ is in lbf/ft
3, r is in ft, g is in f t / s
2.
900gr(/ χ 122)
π2Γ
2(ρ χ 12
3)
putting / into lbf/in2 and ρ into lbf/in
3.
900a / χ 12π
2ρ " r
2 '
_ -ι / 900 χ 32-2 ι /_/_
|/ 12π2 x 0-272
X y r
2 '
= y 9 o o x | / l ,
= 30 j/A rev/min.
E X A M P L E . The stress in a petrol engine flywheel is not to exceed 5000 lbf/in
2 a t 4170 rev/min. Es t ima te a suitable outside dia.
taking ρ = 0-28 lbf/in3.
I f the axial width of this flywheel is to be 2 in. and the polar moment of inertia is to be 0-167 slug f t
2, determine from first prin-
ciples a suitable radial thickness.
Solution
Α ι ι 2π x 4170 . . . Angular veloci ty ω = — = 437 rad/s.
T H I C K AND T H I N C Y L I N D E R S 443
Peripheral rim speed ν = œRx, where Rx = outer radius in ft,
= 437 Rx.
Centrifugal stress, / = - — , where ρ = density in lbf/ft3
and / = stress in lbf/ft2.
.·. 5000 χ 1 22 =
R\
0-28 χ 1 23
(437ÄJ)2
32-2
5000 χ 1 22 χ 32-2
0-28 χ 1 23 χ 4 3 7
2 '
= 0 -25 .
·. R1 = 0-5 ft ( = 6 i n . ) , i.e. A = 1-0 ft.
E lementa l mass
2nd Moment
2nrdr χ bρ
9
2nrdr χ ϋρ
where b — width,
9 χ r
2,
I -
2πδρ
2π bρ
χ r3dr.
r3 dr,
2nbQ I R{ - R\
Axis
FIG. 3 9 4
444 S T R E N G T H O F M A T E R I A L S
12 32-2 \ 4 0-167 χ 12 χ 32-2
485π 2
0-0425 = 0 0 6 2 5 - Rl
Λ Rl = 0-020,
i.e. R2 = 0-375 ft,
= 4-5 in.
Hence, Radia l thickness / = Rx — i£ 2 ^ 6 — 4 -5 ,
= 1-5 in.
E X A M P L E . A flywheel is to have / = 200 slug f t2 when using
material having a density of 0-261 lbf/in3. Determine from first
principles the outside dia. diameter and radial thickness of the
rim so tha t the induced stress shall not exceed 600 lbf/in2 a t
300 rev/min. Neglect the effect of the spokes and assume an axial
width of 6 in.
Solution
Λ ι ι ·. * u ι 2π x 300 Angular velocity οι wheel, ω = 60
= 31-42 rad/s.
Peripheral speed of rim, ν = œRx
(where Rx = outer radius infeet),
= 31-42 i ^ f t / s .
QV2
Centrifugal (or hoop) stress, / = - — , where ρ is in lb/ft3,
/ is in lb/f t2.
.'. 600 χ 122 = ° '
2^
1 2 3( 3 1 - 4 2 ^ )
2
where ρ = 0-28 χ 1 23 lbf/ft
3 = 485 lbf/ft
3, δ = - ^ - f t .
_ 2 485 / 0 - 54 — Ri
Hence, 0-167 = 2π χ — - χ
i.e.
T H I C K AND T H I N C Y L I N D E R S 445
R\ 12 χ 0-261 x 3 1 - 4 2
2'
= 6-25.
Rx = 2-5 ft, i.e. outside dia. = 5 ft.
Mass of element
2nr dr bq
g /
2nd Moment
2nr dr bq
g
2nbq
g
'.I =
x r*
χ r3d r .
2nbq ί FIG. 395
r3 dr ,
n2
2nbq ( R\ - R\
g \ 4
I f / is to be in slug f t2,
6 then b = = 0-5 ft,
ρ = 0-261 χ 1 23 = 450 lb/ft
3,
R± = 2 - 5 ft (already found).
where R2 = required min. radius.
200 χ 2 χ 32-2 _
π x 0-5 χ 4 5 0
18-2 = 39 - ß4.
22f = 20 -8 .
R2 = 2-14 f t .
600 χ 32-2
446 S T R E N G T H OF M A T E R I A L S
Change in Volume Under Load
As already shown, the hoop stress fx in a thin cylinder under in-
1
ternal pressure is given by — and is twice the longitudinal stress /-. Referring to Fig. 3 9 6 :
2t
Axial strain, = ^- yy^^ I
Ε Ε /Y \ \
and fx = 2fS9 f i Λ
Hence e. = 4τ — I
\\ J
= ^ ( 1 - 2 σ ) \^===^ Circumferential or Hoop strain, a
t a
e = L - 2 k where / = Α ε ε
9 h 2 ' L^rz^ — il _ — ~ 2 $ ' FIG. 3 9 6
^ - é ( ' - ï ) -The diametral strain is the same as this, since diameter and circum-
ference increase in the same ratio. Fur ther , the diameter increases
in both directions so tha t :
.*. Rad ia l thickness, t = 2-5 — 2-14 ,
= 0-36 f t ,
= 4-32 in .
T H I C K A N D T H I N C Y L I N D E R S 447
2tE \ 2
i .e. e0=^(l-2i>-a).
UE 4tE 2tE UE '
* d ' 1 a + 2-a\
Change in Volume = Volumetric strain χ Original volume,
= evV,
Vpd
= (1*25 — σ) and is clearly an increase in
this case.
E X A M P L E . Es t imate the increase in capaci ty in a 7 ft diameter shell 16 ft in length when subjected to an internal pressure of 120 lbf/ in
2 (gauge) given tha t Ε = 30 χ 1 0
6 lbf/in
2, σ = 0-287
and tha t the skin is nominally 0-875 in. thick.
Solution
Original volume V = j (7 χ 12)2(16 χ 12) = 1-125 χ 1 0
6 in
3.
Volumetric strain, ev = - 7 ^ - (1-25 — σ ) , tE
- 1 2 0 ( 7 X 1 2 ) ( 1. 2 5 _ 0. 2 8 7 );
0-875(30 χ 106)
== 370 χ 10~6.
Change in volume = (370 χ 10"6)(1-125 χ 1 0
6) ,
= 416 in3 (increase).
E X A M P L E . An internal pressure of 750 lbf/in2 is applied to a steel
tube 6-0 in. dia. 0-125 in. thick, 24 in. long, the ends of which are plugged.
Volumetric strain, ev = ez + 2ex,
-Α<>-*>+·£Κ). pd 2apd 2pd 2apd
448 S T R E N G T H OF M A T E R I A L S
Axia l s t ra in , e. = 4 1
Circumferential strain, ev
Ε Ε 29-8 x 1 0e
χ [9000 - (0-28 χ 18,000)J ,
0 0 0 0 1 3 3 .
U «h 1 Ε Ε 29-8 χ ΙΟ
6
χ [18,000 - (0-28 χ 9000) ] ,
= 0 0 0 0 5 2 .
Volumetric strain ev = ez + 2ex,
= 0 0 0 0 1 3 3 + (2 χ 0 0 0 0 5 2 ) ,
= 0 0 0 1 1 7 .
.'. Increase in capacity of tube = 0-00117 χ 62 χ 2 4 j ,
- 0-79 in3.
Radia l strain due to hoop and axial stresses,
e - -2L^?L " Ε Ε '
29-8 x 1 0e
= - 0 - 0 0 0 2 7 .
0*28 (18,000 -f 9 0 0 0 ) ,
Volumetric strain of tube, er = ex -f e,f + e_,
= 0-00052 - 0 0 0 0 2 7 + 0 0 0 0 1 3 3 ,
= 0 0 0 0 3 8 3 .
I f Ε = 29-8 χ 1 06 lbf/in
2 and σ = 0-28, find:
(a) hoop stress,
(b) axial stress,
(c) increase in capacity,
(d) increase in volume of tube material .
Solution
Hoop stress fx = % = - 18,000 lbf/in*.
Axial stress, / . = ^- = 9000 lbf/in2.
T H I C K A N D T H I N C Y L I N D E R S 449
.·. Change in material volume = 0-000383 (π x 6 χ 24 χ 0-125)
= 0-0216 in3.
E X A M P L E . A storage t ank is made of material 1-0 in. th ick and
consists of a cylinder 8 ft long, 4 ft inside dia. the ends of which are
closed by hemispheres. I f , initially, i t is full of water a t atmospheric
pressure, determine how much additional water must be pumped
in to bring about a tes t pressure of 500 lbf/in2. Take Ε and σ for
the tank material as 29-8 χ 1 06 lbf/ in
2 and 0-287 respectively and
assume the circumferential strains a t the junct ion of cylinder and
hemisphere to be the same. F o r water assume Κ = 0-32 x 1 0e lbf/in
2.
Considering the cylindrical part first and assuming the hoop stress uniform :
Longitudinal stress, / . = = 6000 lbf/in2.
.*. Hoop strain, ev = — ^—, assuming the radial stress f%
Solution
Hoop stress, /., = — = 500 χ 48
2 χ 1 = 12,000 lbf/in
2.
pd
to be zero,
^ [12,000
10,278
(0-287 χ 6000)]
and Longitudinal strain, Ε
= — [6000 - (0-287 χ 12,000)] Ε 2556
~ Ί Γ ~
but, Volumetric strain e. = ez + 2e ;
450 S T R E N G T H OF M A T E R I A L S
Jt 7t and Original volume, V = ~^d2L = — χ 4 8 2 χ 9 6 ,
= 173,000 in3.
Hence, Increase in volume = Volumetric strain χ Original volume,
= evV,
2 3'
1 0 2 χ 173 ,000 ,
29-8 χ 1 06
- 134 in3.
Considering next the hemispherical ends :
Volumetric strain ev = 3eL., where ec has the value found above,
_ 30,834
Ε ' 4
Original volume V = -— nr3, where r = 24 in, ό
= φ π χ 2 43,
= 57 ,800 in3.
Hence, Increase in volume = evV,
30 ,834
29-8 χ 1 06
60 in3.
χ 5 7 , 8 0 0 ,
Reduct ion in volume of original water
5 0 0 (173,000 + 57,800)
0-32 χ 1 06
360 in3.
.*. Additional water necessary = 134 -\- 60 -f 360 = 554 in3.
T H I C K AND T H I N C Y L I N D E R S 451
Examples X I V
1. A cylindrical vessel 8-0 in outside dia. is made of sheet 0-2 in. thick and subjected to an internal pressure of 800 lbf/in
2. Calculate the hoop and longi-
tudinal strains when the cylinder is subjected to an axial load of 10,000 lbf
(a) in tension (ex = 0-000522., e s = 0-0000865) and (b) in compression (ex = 0-000566,. ez = -0-000221).
2. Show that the maximum shear stress occurs at the inner surface of a thick tube when inder internal pressure and that its value is \ (fx + fy) where fx and fy are respectively the hoop and radial stresses.
3. Find the ratio of thickness to inside diameter for a tube subjected to internal pressure when the ratio of internal pressure to maximum hoop stress is 0-5. Find the alteration in the thickness of such a tube when the internal pressure is 5 tonf/in
2 given that ο = 0-304 and Ε = 13,000 tonf/in
2
(L.U.) (0-73., 0-0031 in.).
4. One end of a steel tube 1-0 in. outside dia. and 0-125 in. thick is closed and the other screwed into a pressure vessel. Find what pressure will induce a maximum stress of 20,000 lbf/in
2. If the effective length is 10 in. estimate,
neglecting end effects, the increase in internal volume under this pressure. (5600 lbf/in
2, 0-0062 in
3).
5. An iron pipe 8-0 in. outside dia., 1-0 in. thick failed under test at an internal pressure of 3-1 tonf/in
2. Determine, using a factor of safety of 4,
the safe pressure for a pipe of the same material and inside diameter but 1-5 in. thick. Assume failure in both cases to be due to the maximum hoop stress. (1-07 tonf/in
2).
I N D E X
Beams composite 150 direct stress in 97 encastré 327 reinforced concrete 153 shear stress in 118
Bending moment 38 simple theory of 94, 357 with direct stress 159 with torsion 259
Flexure differential equation of 279
Lamé theory of thick pipes 421
Load critical or buckling 393 gradual 13 shared by two materials 27 shock 15 sudden 13
Columns long 393 short 27
Contraflexure 56 Cylinders
compound 434 thick 419 thin 437
Deflection coefficient 297 of beams 278 of coil springs 199 of leaf springs 348 Macaulay's method 312 moment-area method 344
Maximum principal strain theory 257 principal stress theory 246 shear strain energy theory 249 shear stress theory 246 strain energy theory 249
Modulus of Bulk 227 of Elasticity 3 of Rigidity 4, 181 of a section 162
Moment of Inertia cone 82 cylinder 84 disk 85 sphere 81
Elastic constants 225 failure, theories of 245 limit 2 modulus 3
Euler theory of struts 393
Neutral axis 94
Parallel axes theorem 67 Perpendicular axes theorem 77 Poisson's Ratio 219 Power transmitted by shafts 184
First Radius Moment of Area 59 of curvature 95, 278, 349 Moment of Mass 65 of gyration 67, 399
453
454 I N D E X
Rankine-Gordon theory of struts 401
Relation between elastic constants 225 between w, F and M 130, 358
Second Moment of Area 66 Moment of Mass 80
Shear complementary 216, 379 force 118 strain 4, 181 stress 4, 182, 234 stress due to bending 118, 375 stress due to tension 215
Simple bending assumptions 96 equation 98 theory 94
Slenderness ratio 399 Springs
close coil helical 202 leaf or laminated 348 open coil helical 199
Strain lateral 219 longitudinal 2, 219 principal 237 shear 181 volumetric 217, 227, 447
Strain energy in bending 361, 201 in tension 12 in torsion 197, 201
Stress complex 230 distribution in beams 96 due to bending 97 due to rotation 439 due to temperature change 30 due to twisting 182 in beams 94 in cylinders and pipes 419 in struts 399 Möhr circle of 241 principal 230 proof 6 shear 234 simple 1
Struts long 393 short 27
Theorem of parallel axes 67 of perpendicular axes 77
Torque equivalent 262 in hollow shaft 191 in solid shaft 182
Torsion simple theory of 181 with bending 259
Volumetric strain 217, 227, 447