1
MECHANICS OF MATERIALS REVIEW
Notation:
a = normal stress (psi or Pa)= shearing stress (psi or Pa)
€ = strain (in/in or rn/rn)
o = deformation (in. or m)= moment of inertia
of area (in4 or m4)
J = polar moment of
inertia of area (in4 or m4)N = revolutions per mm.
E = modulus of elasticity (psi or Pa)G = shearing modulus (psi or Pa)p = Poisson’s ratioa = coefficient of thermal
expansion (1°F or 1°C)M = bending moment in beamsT = torque
= temperature change (°F or °C)H.P. = horsepower (1 H.P. = 550 ft-lb/sec)
General Information:factor of safety = (failure-producing load)1(actual load or allowed load)thermal strain = = att
Section 1: Centric Loads
a. Axial Loaded MembersIf the load, P, passes through the centroid of the resisting
section:
axial stress = a =A
axial strain = =L
P
-f
_-j
lithe material is also elastic, then,
stress-strain relation: c = 2.E
alL PLaxial deformation: 8 = — = —
- E AE
2
b. Average shear stress in pins (or rivets)
double shear:
single shear: r =
—* P/2
P
P / 2
P
; P
Section 2: Torsion of Circular SectionsIi a shaft has a circular cross section and thematerial is elastic, the shearing stress in theshaft varies linearly with the distance from thecenter of the shaft (p) and is given by
shearing stress: t =
J
Also the angle of twist () of one end relative to theother is given by
. tL TLangle of twist: 0 = — = —
pG JG
The maximum shearing stress in the shaft is at theoutside surface and equals:
Tr3tmax
=
The oolar moment of inertia is
icr4solid section: J =
hollv section: J = (r -r14)
Also the shaft has maximum and minimumnormal stresses which are tensile andcompressive and act at 450 to the shaft axis.Their values are
Tr3lens 0camp max
—
\\
1_•
0c0mp Tm0,
/•fl TmQ
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Section 3: Beam Behavior
a. Bending Stresses (elastic)
3
If the loads on a beam act in its plane of a
symmetry and the beam is elastic, thebending stresses acting normal to the L._._Licross section vary linearly with the
S..rfoc,
distance from the neutral axis (N.A.) andhave the value
)
bending stresses: a =
In the absence of axial loads
centrodial axis = neutral axJJ
In the sketch the cross section is shown rectangular. However, the crosssection, in general, can be circular, triangular, etc. Many sections such asT-, I-, H-sections can be found in handbooks. lithe section is notstandard, you must be prepared to derive the centroidal location as wellas the value for 1NA• The maximum bending stress occurs at the fiberwhere (My) is maximum.
maximum bending stress: a = Z = section modulus =max z
b. Shearing Stresses in Beams
The transverse and longitudinalshearing stress in a beam is
iven by
V
4JA1
where Q is the first moment of the shaded area with respect th thecentroidal axis if the shearing stress is evaluated at line 1-2
Now Q = [(c-a)t][(c+a)12] {for rectangular section}
The maximum shearing stress will occur where Q/t is maximum. 0 isalways maximum at the centroidal surface. However, Q/t may or may notbe maximum at the centroidal surface. You check all possibilities.
Shear flaw = q = .2 (lb/in.)
4
You must be able to draw shear
__________
and moment diagrams for beamsin order to determine themaximum values of M and V.
b. Beam Deflectioris.
__________
The deflections of beams isdetermined from the equation
Ely”
where primes denote derivatives with respect to x and M is the bendingmoment equation. Integration of this equation determines the slopeequation.
E[Y’=fMx+Cij
Additional integration gives the deflection equation or the equation of theelastic curve
Ely = f[fMdE)dx + C1x + C2
The constants C1 and C2 are found from the slope and deflectionboundary conditions.
The solution to these equations for many different types of supports andloadings can be found in handbooks. Using the known handbooksolutions and the principle of superposition many other problems can besolved.
For slender columns L’Ir> 100 (steel) the buckling loads andstresses are given by
2EJ it2Ep = 0 =Cr
(L’)2 (L’Ir)2
where L’ is the effective length of a pinned-pinned columnand r is the radius of gyration r = /(IIA)
fixed-fixed L’ = L/2fixed-free L’ = 2L
fixed-pinned L’ = 0. 7L
L
Section 4: Bucklino
PC,
(
P_
See handbooks for codes for columns of intermediate length.
m26
Section 5a: Stresses at a Point: Mohr’s Circle for Stresses
STRESS
5
0+00
=
a 2
2I 1 —
oYJ%J I, 2 ry0 +0x y +
2
0 +0cos 20 - t sin 20
2 = 0a + R
= ax - at,,1 sin 26 + ti,, cos 20
2
tan 20 =
(a — a,,)
2
= 0a - R
Rotation on Mohr1scircle is twice the rotation on the element and in the same direction
The top half of the plane corresponds to shear stresses trying to rotate the elementclockwise; the lower half to shear stresses which try to rotate the elementcounterclockwise
For plane stress the third principal stress is zero= !4(o - an); where o, a, and a are labled right to left along the horizontalaxis.
Radius = R = = R
2226
Section 5b: Strains at a Point: Mohr’s Circle for Strains
+ € ya 2
V STAIN
2
2
6
€ +Ex y +
2
€ +€ycos
220 - sin 20
2€ =€ +Rmax a
YxY — €x — Ey
2 2sin 20 + cos 20
2- R
Yy
tan26= 2
-
2
Rotation on Mohr’s circle is twice the rotation on the element and in the same direction
The top half of the plane corresponds to shear strains trying to rotate the directionclockwise; the lower half to shear strains which try to rotate the directioncounterclockwise
For plane strain the third principal stress is zero
Ymax = (e - c); where , c, and c are tabled right to left along the horizontalaxis
Radius=R=2
I -
N 2y=2R
22Z6
C(1)
a,C)
0,
a,04.-
4-a
Va,E(I,(I)(1
a,
a,
VC(U.c
VCD
(I)CD
U)
U)
(0000.2
(I)0
(U)
;;a,
CDCD
CDQ
E.9
.9.9
a)U
)(0
(0a,
(U>
-
++
•0t
>‘-
CDG
)CD
—4-a
U)
c’ac
U)
(E.9
..9
0C
U)
U)
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QU
)a,
++
4-
‘-Q
Ua)L
Cl)
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CDE
-Cw
L_
fi
c,ac’i
qU
)U
)U
)C
)..—
00
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aC
))(
0w
c)(a)
‘-6
rII
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-oC
)
iø
(a)(a)
Ø
U)
C0(Ua,.E
>—
f
Q.
EU)
-cC)
I)>
—-—
-—
——
-—
U)
0a.
0a.-a,EE0C.)
>‘
C(U
(a)+0
Ci)+(a)
oC
%J
0(i)’
ftft
Itx
(a)(a)
>-
(I)0(a)
RC
a)(a)
Ca)C
’J
IIII
Ca)Ci)
>-
x
0CDtJ
>N
(a)
(a)
II
0(a)
0(a)(a)
><
0CDL
)
8
Section 6: Stress-Strain Relations
If the material is linear-elastic (Le, stress proportional to strain) and isotropic, then
ç = —
+= (1 +v)(j _2v)[(1 — v)E + +
=— v(a÷a)j
= (l+v)(1_2v)[(1 )€ +
=—
= (1÷v)(1 _2v)[(1 —v)€ +
y=t,JG
y=t/G t=Gy
y=tG t,1=Gy
If we have plane stress where: o = = = 0, then
ç = vci)= (i V2)(€x
+ v€)
=— var)
= (i _v2)’+ v€)
= —_(a+o)
V oz=o
txy = GYXY
where p is the value of Poisson’s ratio determined from a uniaxial tensile or
compression test= €lateral
G is the shearing modulus of elasticity G = tly, and related to E and p by E = 2(1 + IL)G.
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9
Pressure Vessels
p = pressure r = radius t = wall thicknessRelationships are valid for thin-walled vessels (hr < 0.1)
For any coordinate system on the outside surface
= (1 )(€X ÷
pr—÷p
t
€ = (€ ÷c)r(1—!.t)
Spherical:
o = pr/2t; (In any direction)
= o = pr/2t
= -p; (on the inside surface)
pr—+p2t
= 2(on the inside surface)
I fpr€ = = —l—(I
-
E[21
1
/
A ,
Hoop
Cylindrical: (A=axial; H=hoop)
= pr/2t ; 0H = pr/t
GPIOH °P20A
= -p (on the inside surface)
tmx2
iIpr(1 —
=
rpr€ = _l_(
A E121- 2k)]
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10
Statically Determinant 2-force Members
P4 f -
Equilibrium — - F1 + F2 - F3 + F4 = 0P = F1 (Tension) = F4 - F3 + F2P = F1 - F2 (Tension) = F4 - F3
= F1 - F2 + F3 (Tension) = F4PLL’ A1-JAB “AB
Thermal— r
UAB- 1AB 1AB
Statically Indeterm inant 2-Force Members
p p
.AU
gid i....[
L = LengthE= Modulus of ElasticityA = Areaa = Coefficient of Thermal expansion
= Temperature change
•19
-
Stiffness:
=
____
LAaAM)
Stiffness:AE
(8- LA UA 7)
A8E8(8 - Cl ÷ LB a8 7)
Equilibrium:A + P8 = P
- Cl + La8tT)
Equilibrium:PA(a) + PB(b) = P(b+c)
Member A was assumed in tension,therefore subtract LaSTMember B was assumed incompression, therefore add LatT
11
Torsion
Stress= Tr/J
Itt 4 4 Itt 4 4I = —d - d, = —ir —
32’ 0 2’°T = Torquer = radiusL = Length
Rotation
JGG = Shear Modulus (Modulus of Rigidity)
Power Transmission
Power = (Torque)(Angular Velocity); (Angular Velocity must be in radians/seconds)1 hp = 550 ft-lb/sec1 RPM = (2rT/60) rad/sec
Transmitting 100 hp at 3000 RPM, what is the torque?
= lOOhp (550fl_ThIsec3O0ORPM hp
Indtrminant Torsion
RPM= 175fl-lb
2it—md/sec60
TA+TB=T
TB =
___
T1 + T11 = TJGr =
= JI!!e
Brass sleeveAluminum Care
z:zzzz®
b + °
I
U
0
I—
T,e
12
Beams
- bh3
12- hb3
12= bh3
+ Ad2a12
I-Beam
= (1.5x3)3÷
____
+ (4X1X2)2J 39375j,4
Qcentro,d (1 5X1 .5)( 1) + (4)(1X2) = 9.6875 jpi3
QA = (4)(1)(2) = 8 in
The width (t) at the centroid or A is 1.5’. At the bottom or
______________________
top the width is 4’.
Failure is most likely where V or M is a maximum asobtained from the shear and moment diagrams.
a) Top or Bottom, Pure Bending, a = My/Ib) Centroid, Pure Shear, t = VQ/Itc) Wide-Flange Intersection (Point A), combination of bending and shear.
Find principal stresses and maximum shear stress.
o = My/It = VQ/ItM = Bending Momenty = distance from centroidal axis of bending
= 2’ Area Moment Of InertiaQ = 1 Area Moment = A9t = width
r -/ A
fl
//
D
0—- - -
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13
Beam Deflections
1. Find the support reactions usingStatics.
2. Find the moment as a piecewise 325 k
continuous function of x.
3. Integrate twice to get the deflection. ‘V.
_______
325kip D-xH -10---
Region 1: 0 x 5’ Region 2: 5’ x 10’
M1 = 3.25x - M2 = 7.5 - 0.75x2 ,,3
E101 = 3.25 .i - .i+ El02 = 7.5x - 0.75.i_ + C2
2 6 23 4 2 3
Ely1 = 3.25 - .-. + C1 x + D, Ely2 = 7.5 .-. - 0.75 + C2x + D26 24 2 6
4. Solve for the constants of integration using the boundary and continuity conditions.Boundary: y(0) = 0; y2(10) = 0Continuity: ø(5) = 02(5); y(5) = Y2(5)
5. The above equations can then be used to solve for the slope or deflection at anypoint along the beam.
Superposition - useful for getting the deflection at a point.
1 kp/ft 1 /ftIII H HH H 11111 H 11TH 11111 HIll HI H H Ill
5- 5- 5. 5, 5 5,1 <p 1 kip. 6 1 kip, 62
Center Deflection: = ö1 + Y2ö2 L Same cemter deflection I
1 kip/ft I kip/ft I kip/ftHHIIIIIIHHIII!IInhIIITII wit +
.
14
Indeterminant Beams
1. Remove enough supports to make the beam statically determinant, and replacethem with unknown loads.
2. Require that the unknown loads beand slopes.
or
such that they give the appropriate deflections
P must be such that the deflection of theright end is zero.
M must be such that the slope atthe left end is zero.
orM2
M1 and M2 must make the slopesat both ends zero.
P and M must make both the slope anddeflection at the right end zero.
p1
8 = wL48 El
- PL3-
3E1
/ //
P = 18.75 kipsWall Reactions
= 0 = 18.75 + R - 50R=31.25kips I
= 0 = -(50)(5) + (18.25)(10) + MM = 62.5 kip-ft (CCW)
5 Hft
U
2i22€
15
REVIEW PROBLEMS
1. An aluminum bar having a constant cross sectional area of ¼ in2 carries the axialloads applied at the positions shown. Find the total deformation of the bar.
a. _0.0192 in.
b. 0.2880 in. 80001b 6ft 60001b— 2000th
c. _0.3264 in.E 10x108d. _0.3840 in.
e. _None of these.
2. A steel rod with a cross sectional area of ½ in2 is stretched between two rigid walls.The temperature coefficient of strain is 6.5 x lO4in.Iin./°F and E is 30 x lO6psi. Ifthe tensile load is 2000 lb at 80° F, find the tensile load at 0°F.
a. _5800 lb.
b. _7800 lb.
c. _8800 lb.
d. ..9800 lb.
e. _19,600 lb.
3. The composite bar shown is firmly attached to unyielding supports at the ends andis subjected to the axial load P shown. lithe aluminum is stresses to 10,000 psi,find the stress in the steel.
a. _1000 lb.1511.
k ‘)ññf’i IA
______________
Id. _.Id’.JW JId.
frJ. St. -
c. _5000 lb.-i A1.5 11 A—20 112
lnrwri1.. 6 —
U. — I V,VVV Ii). E lOx 10 P& E = 30 x lo6p&
e. _20,000 lb.
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16
4. Find the length L to make the relative total angle of twist of both ends of the shaftequal zero.
a._3ft.
b._4ft.
6. A hollow aluminum shaft and a solidsteel shaft are rigidly connected ateach end. This compound shaft is thenloaded as shown. Determine themaximum shearing stress in eachmaterial and the angle of twist of thefree end. Assume elastic behavior.Gaijm = 4 X 106 psiG =l2xlO6psi
HL
fl
/.
e
I /I I
. I IA’.d‘ /
t. 62800 lb
b
1000 lb
E•
c._6ft.
d. 9 ft.
e. _12 ft.
5. Determine the shearing stress atpoints A and B which are at theinside and outside surface of thehollow shaft. Assume elasticbehavior.
__6+T_
1. 62800 lb
2/2Z6
17
7. Determine the maximum bending moment in the beam.
a._3600 ft-lb.
b. — 5400 ft-lb.
c. — 7200 ft-lb.
d. _8100 ft-lb.
4Ø lb/Fl
uwwwI— 6Fi 6-ri
e._4050 ft-lb.
8. Find the maximum transverse shearing force in the beam shown.
a. — 450 lb.
b. — 1800 lb.
C. — 2250 lb.
d._3600lb.
e. — 4050 lb.
gee b zr1
5-ri
9. By means of strain gages, the flexural stresses are found to be -12,000 psi at Aand +4000 psi at B. Assuming the elastic limit of the material has not beenexceeded, find the flexural stress at the bottom of the beam.
a. — 6000 psi.
b._8000 psi.
c. — 9000 psi.
d. — 10,000 psi.
e. — 12,000 psi.
10. For the cast iron beam shown, the maximum permissible compressive stress is12,000 psi and the maximum permissible tensile stress is 5000 psi. Find themaximum safe load P applied to the beam as shown.
A $e-ci r 4-ft.
a._220lb.
b. — 333 lb.
c....12501b. [d._3000lb.
S.c,o.’ A
e. _7500 lb.
18
11. A 12-inch, 35-lb I-beam 30 ft. long is supported at 5 ft. from each end and carries auniformily distributed load of 1600 lbs per ft including its own weight. Determinethe maximum flexural stress in the beam.
12. Find the maximum vertical shearing force which may be applied to a box beamhaving the cross section shown without exceeding a horizontal shearing stress of500 psi.
a._3065lb.—
b._4000lb.
c. 6000 lb.
d._61301b.
e._63001b.
13. Find the reaction at the rightend of the beam shown.
a._wLJS
b.wU4
c. 3wLI8
_______________________
14. Two beams, simply supported at their ends, jointly support a load P = 3500 lb.applied to the upper 6-ft. beam at its midpoint. The beams are identical except forlength and cross at their midpoints. Find the load carried by the lower 9-ft. beam.
a. 700 lb
b._8001b
c. — 1000 lb
d. 1750 lb
e. _2700 lb
NA
‘—4- IN—
6-IN
o orino +L
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19
15. Determine the deflection at the end of this L 1 0 ftbeam. 200x0 b—in
— I ft’fTh
i
____ ____
P=iQ,000 b
16. The critical Euler load for the pin-endedslender column restrained at the midpoint asshown in Fig. A is 1000 lb. What is the criticalEuler load for the same column with themidpoint restraint removed as shown in Fig.B.
a.
________
250 lb.
b.
______
5001b.
C.
______
7501b. Fig. A Fg.d.
______
l000lb.
e.
_______
4000 lb.
17. A rectangular bar is loaded as shown. Find the maximum tensile stress developedover section A-A.
4ft 6ft —I12,000 b A r’4 12,000 b
2Z6
20
18. For stress conditions on the element shown, findthe principal stresses and the plane on which themaximum principal stress acts.
19. A circular shaft of brittle material subjected to torsion fractures along a 45° angle.Failure is due to what kind of stress?
a.
________
Shearing stress.
b.
_________
Compressive stress.
c.
_________
Tensile stress.
d. Combined stress.
e.
________
None of these.
20. Which has the higher shear stress for a given elastic torque?
a. a one-inch diameter rod, or b. a two-inch diameter rod.
21. Identical rods of aluminum and steel are each subjected to the same elastic torque.Which rod will have the higher shear stress?
a. steel b. aluminum c. both the same stress
22. If G represents the modulus of rigidity (or shear modulus of elasticity), E is modulusof elasticity, and p is Poisson’s ratio, which of the following statements is true forany homogeneous material?
a.
_______
G is independent of E.
b.
_______
G is 0.4E.
C.
_______
G is 0.5E.
d.
________
G depends upon both E and p.
e.
________
None of these.
3DOD DS
DS
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21
ANSWERS EIT REVIEW
MECHANICS OF MATERIALS
1. C
2. d
3. e
4. d
5. T8=5,333.3 psi TA=2,6€6PS
6. Ts = 81,528 TA = 54,352 psi
8 = 1.086 rad.
7. e
8. c
9. b
10. d
11. ,=19,100psi
12. d
13. c
14. b
15. 19.8 in.
16. a
17. 8,000 psi
18. umax = 605 psi 0mãn = 6606 psi
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