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5. Torsion
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CHAPTER OBJECTIVES
Discuss effectsof applyingtorsional loading to a long straight
member
Determine stress distribution
within the member under torsionalload
Determine angle of twist when material behaves
in a linear-elastic and inelastic manner
Discuss statically indeterminate analysis of shafts
and tubes
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CHAPTER OUTLINE
1. Torsional Deformation of a Circular Shaft
2. The Torsion Formula
3. Absolute Maximum Torsional Stress
4. Power Transmission
5. Angle of Twist
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Torsion is a moment that twists/deforms a
member about its longitudinal axis
Effect??Its effect is of primary concern in the
design of axles or drive shafts used in vehicles
and machinery
For example: When torque is applied to a circular
shaft (eg: rubber material), the circles and
longitudinal grid lines originally marked on theshaft tend to distort with circles remain circles
- grid lines deforms a helix
(Fig5.1a-b)
5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT
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Torsion is a moment that twists/deforms a
member about its longitudinal axis
By observation, if angle of rotation is small, length
of shaftand its radius remain unchanged
5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT
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5.1 Torsional Deformation
If the shaft is fixed at one and a torque is applied to its
other end, it will distorted with the cross-section at a
distance x from the fixed end of the shaft will rotate
through an angle (angle of twist).
x
y
zf(x)
x
f(x)
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5.1 Torsional Deformation
HOW THIS
DISTORTIONSTRAINS THE
MATERIAL??
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Geometry of Deformation
g
r
qo
r
T
Q
R
P
L
Assumption: Cross-sections which are plane
before twisting remain plane
and undistorted after twisting.
PQ Longitudinal
generator
gshear strain at radial
distance r(rad.)
qangle of twist (radians)
grq LQR arcCircular (1)
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DEFORMATION OF A CIRCULAR SHAFT
By definition, shear strain is
Let x dx andf= df
BD=rdf= dx g
g= (/2) lim qCA along CA
BA along BA
g =rdf
dx
Sincedf/ dx = g/r= gmax/c
g = gmaxr
c( )Equation 5-2
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Stress Strain Relation
G
P.L.t
g
Assumption: Shear stress does not
exceed the proportional
limit.
Gshear modulus (N/m)
Torsion Formulae
gt G
Substituting Eq. (3) into Eq. (2) results in
(2)
L
G
LG
q
r
trqt
or (3)
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5.2 THE TORSION FORMULA
For solid shaft, shear
stress varies from zero at
shafts longitudinal axis to
maximum value at its outer
surface. Due to proportionality of
triangles, or using Hookes
law and Eqn 5-2,
t = tmaxrc( ).
.
.
T=tmax
cAr2dA
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dAc
T
cwhere
dAdMT
dMT
M
A
A A
A
z
2max
max
0
0
rt
tr
t
rt
5.2 Torsion Formulas
TZ
r
J: Polar Moment of Inertia
Solid Shaft:
Tubular Shaft: ( )44
4
2
2
io ccJ
cJ
c
dM
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5.2 THE TORSION FORMULA
The integral in the equation can be represented asthe polar moment of inertiaJ, of shafts x-sectionalarea computed about its longitudinal axis
tmax
=Tc
J
tmax= max. shear stress in shaft, at the outer surface
T= resultant internal torque acting at x-section, from
method of sections & equation of momentequilibrium applied about longitudinal axis
J= polar moment of inertia at x-sectional area
c= outer radius of the shaft
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5.2 THE TORSION FORMULA
Shear stress at intermediate distance,r
t=Tr
J
The above two equations are referred to as the
torsion formula
Used only if shaft is circular, its material
homogenous, and it behaves in an linear-elastic
manner
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5.2 THE TORSION FORMULA
Solid shaft
Jcan be determined using area element in the formof a differential ring or annulus having thickness drand circumference 2r.
For this ring,dA = 2rdr
J= c4
2
Jis a geometric property of the circular area andis always positive. Common units used for itsmeasurement are mm4and m4.
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5.2 THE TORSION FORMULA
Tubular shaft
J= (co4ci
4)2
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5.2 THE TORSION FORMULA
Absolute maximum torsional stress
Need to find location where ratio Tc/Jis maximum
Draw a torque diagram (internal torque tvs.xalongshaft)
Sign Convention: Tis positive, by right-hand rule, isdirected outward from the shaft
Once internal torque throughout shaft is determined,maximum ratio of Tc/Jcan be identified
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5.2 THE TORSION FORMULA
Procedure for analysis
Internal loading
Section shaft perpendicular to its axis at pointwhere shear stress is to be determined
Use free-body diagram and equations ofequilibrium to obtain internal torque at section
Section property
Compute polar moment of inertia and x-sectional
area For solid section,J= c4/2
For tube,J= (co4 ci
4)/2
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5.2 THE TORSION FORMULA
Procedure for analysis
Shear stress
Specify radial distancer, measured from centre
of x-section to point where shear stress is to be
found Apply torsion formula, t= Tr/Jor tmax= Tc/J
Shear stress acts on x-section in direction that is
always perpendicular tor
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EXAMPLE 5.3
Shaft shown supported by two bearings and
subjected to three torques.
Determine shear stress developed at pointsAandB,
located at section a-aof the shaft.
5 T i
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EXAMPLE 5.3 (SOLN)
Internal torque
Bearing reactions on shaft = 0, if shaft weightassumed to be negligible. Applied torques satisfy
moment equilibrium about shafts axis.
Internal torque at section a-adetermined from free-body diagram of left segment.
5 T i
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EXAMPLE 5.3 (SOLN)
Internal torque
Mx= 0; 4250 kNmm 3000 kNmm T= 0
T= 1250 kNmm
Section property
J= /2(75 mm)4= 4.97107mm4
Shear stress
Since pointAis atr= c= 75 mmtA= Tc/J= ... = 1.89 MPa
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EXAMPLE 5.3 (SOLN)
Shear stress
Likewise for pointB,atr= 15 mm
tB= Tr/J= ... = 0.377 MPa
Directions of the stresses
on elementsAandB
established from
direction of resultantinternal torque T.
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5.2 Torsion Formulas
EXERCISE F5-1 & F5-2
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5.2 THE TORSION FORMULA
EXERCISE F5-5 & F5-6
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Problem 1 (5-2)
750Nm
c=100mm
a=75mm
25mm
Torque carried by the shaded section ?
( )MPa4793.0
025.01.02
)1.0(75044
max
t
JTc
( )
Nm515
075.01.02
)10(4793.0 446
max
cT
J
aT
c
a
a
a
a
a
tt
Nm5152
2
1.0
075.0
3max
1.0
075.0 max
rrt
rrr
rtrt
dc
d
c
dATa
(b)
(a)
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Problem 2 (5-56)
300Nm500Nm
200Nm
400Nm
300Nm
500Nm200Nm
400Nm
( )
( )MPa7.75
015.02
015.0400
4
maxmax
tJ
cT
68Nm
20Nm
49Nm35Nm
68Nm
20Nm
49Nm
35Nm
132Nm
( )
( )MPa38.5
025.02
025.0132
4
maxmax
tJ
cT
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Poweris defined as work performed per unit of
time
Instantaneous power is
Since shafts angular velocity = dq/dt, we can
also express power as
5.3 POWER TRANSMISSION
P= T (dq/dt)
P= T
Frequencyfof a shafts rotation is often reported.
It measures the number of cycles per second
and since 1 cycle = 2 radians, and = 2f T, thenpower
P= 2fTEquation 5-11
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5.3 Power Transmission
fTP
TP
dt
dTP
q
2
Power:
1watts(W)=1Nm/s1hp=550ftlb/s
where = angular velocity(rad/s)
f = frequency (cycle/s)
fgmax
JG
TL
G
Lc
J
Tc
cL
GJ
Tc
f
ffg
gt
/
wheremax
max
max
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Shaft Design If power transmitted by shaft and its frequency of
rotation is known, torque is determined from Eqn
5-11
Knowing Tand allowable shear stress formaterial, tallowand applying torsion formula,
5.3 POWER TRANSMISSION: Shaft Design
J
c
T
tallow= 5-12
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Shaft Design
For solid shaft, substituteJ= (/2)c4to determine c
For tubular shaft, substituteJ= (/2)(co2 ci
2)to
determine coand ci
5.3 POWER TRANSMISSION
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EXAMPLE 5.5
Solid steel shaft shown used to transmit 3750 Wfrom
attached motor M. Shaft rotates at = 175 rpmandthe steel tallow= 100 MPa.
Determine required diameter of shaft to nearest mm.
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EXAMPLE 5.5 (SOLN)
Torque on shaft determined fromP= T,
Thus,P= 3750 Nm/s
Thus,P= T, T= 204.6 Nm
( ) = = 18.33 rad/s175 revmin
2rad
1 rev
1 min
60 s( )
= =J
c
c4
2 c
T
tallow..
.
c= 10.92 mm
Since 2c= 21.84 mm, select shaft with diameter of
d= 22 mm
5-12
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EXERCISE
A steel tube having an outer diameter of 62.5mmis used to transmit 3kW when turning at 27
rev/min.
Determine the inner diameter d of the tube the
nearest multiples of 5mm if the allowable shear
stress is tallow= 70MPa
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5.4 ANGLE OF TWIST
Angle of twistis important when analyzing reactions
on statically indeterminate shafts
f =T(x) dx
J(x) G0
L
f = angle of twist, in radians
T(x)= internal torque at arbitrary positionx, found
from method of sections and equation of
moment equilibrium applied about shafts axis
J(x)= polar moment of inertia as a function ofx
G= shear modulus of elasticity for material
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5.4 ANGLE OF TWIST
Constant torque and x-sectional area
f =TL
JG
If shaft is subjected to several different torques, or x-
sectional area or shear modulus changes suddenly
from one region of the shaft to the next, then apply
Eqn 5-15 to each segment before vectorially addingeach segments angle of twist:
f =TL
JG
5-15
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5.4 ANGLE OF TWIST
Sign convention
Use right-hand rule: torque and angle of twist arepositive when thumb is directed outward from the
shaft
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5.4 ANGLE OF TWIST
Procedure for analysis
Internal torque
Use method of sections and equation of moment
equilibrium applied along shafts axis
If torque varies along shafts length, section madeat arbitrary positionxalong shaft is represented
as T(x)
If several constant external torques act on shaft
between its ends, internal torque in each segment
must be determined and shown as a torque
diagram
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5.4 ANGLE OF TWIST
Procedure for analysis
Angle of twist
When circular x-sectional area varies along
shafts axis, polar moment of inertia expressed as
a function of its positionxalong its axis,J(x) IfJor internal torque suddenly changes between
ends of shaft, f= (T(x)/J(x)G) dxor f= TL/JGmust be applied to each segment for whichJ, T
and Gare continuous or constant
Use consistent sign convention for internal torque
and also the set of units
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EXAMPLE 5.7
50-mm-diameter solid cast-
iron post shown is buried 600mmin soil. Determine
maximum shear stress in the
post and angle of twist at its
top. Assume torque about to
turn the post, and soil exerts
uniform torsional resistance of
tNmm/mm along its 600 mmburied length. G= 40(103) MPa
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EXAMPLE 5.7 (SOLN)
Internal torque
From free-body diagram
Mz= 0; TAB= 100 N(300 mm) = 30 103Nmm
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EXAMPLE 5.7 (SOLN)
Internal torque: Magnitude of the uniform
distribution of torque along buried segmentBCcanbe determined from equilibrium of the entire post.
Mz= 0;
100 N(300 mm) t(600 mm) = 0
t= 50 Nmm
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EXAMPLE 5.7 (SOLN)
Internal torque
Hence, from free-body diagram of a section of the
post located at positionxwithin regionBC, we have
Mz= 0;
TBC50x= 0
TBC= 50x
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EXAMPLE 5.7 (SOLN)
Maximum shear stress
Largest shear stress occurs in regionAB, since
torque largest there andJis constant for the post.
Applying torsion formula
tmax = = 1.22 x106N/m2= 1.22 MPaT
AB cJ
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EXAMPLE 5.7 (SOLN)
Angle of twist
Angle of twist at the top can be determined relative
to the bottom of the post, since it is fixed and yet is
about to turn. Both segmentsABandBCtwist, so
fA = +
TABLAB
JG
TBC dx
JG0LBC
.
.
.
fA= 0.00147 rad
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EXERCISE
45
The assembly is made of A-36 steel and consists of aSolid rod 15 mm in diameter connected to the inside
of a tube using rigid disk at B. Determine the angle of
twist A. The tube has an outer diameter of 30 mmwall thickness of 3mm
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Problem 5-61 and 62
A
D
B
C
50Nm
30Nm
15mm
30mm
3mm
D
B
C
A
rad04706.0
)10(75
)3.0(50
)10(75
)3.0(8099
///
ABBC
AB
ABAB
BC
BCBC
BACBACA
JJ
GJ
LT
GJ
LT
ffff
rad00681.0)10(75
)3.0(809
//
BC
BC
BCBCBCDCB
J
GJ
LTfff
( )( )4
44
0075.02
012.0015.0
2
AB
BC
J
J
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EXERCISE F5-7/F5-8/F5-10/F5-11
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CHAPTER REVIEW
Torque causes a shaft with circular x-section to
twist, such that shear strain in shaft isproportional to its radial distance from its centre
Provided that material is homogeneous and
Hookes law applies, shear stress determinedfrom torsion formula, t= (Tc)/J
Design of shaft requires finding the geometric
parameter, (J/C) = (T/tallow)
Power generated by rotating shaft is reported,from which torque is derived; P = T
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CHAPTER REVIEW
Angle of twist of circular shaft determined from
If torque andJGare constant, then
For application, use a sign convention for
internal torque and be sure material does notyield, but remains linear elastic
f =TL
JG
f =T(x) dx
JG0L
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If shaft is statically indeterminate, reactive
torques determined from equilibrium,compatibility of twist, and torque-twist
relationships, such as f= TL/JG
Solid noncircular shafts tend to warp out ofplane when subjected to torque. Formulas are
available to determine elastic shear stress and
twist for these cases
Shear stress in tubes determined byconsidering shear flow. Assumes that shear
stress across each thickness of tube is
constant
CHAPTER REVIEW
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CHAPTER REVIEW
Shear stress in tubes determined from
t= T/2tAm
Stress concentrations occur in shafts when x-
section suddenly changes. Maximum shear
stress determined using stress concentrationfactor,K(found by experiment and represented
in graphical form). tmax= K(Tc/J)
If applied torque causes material to exceed
elastic limit, then stress distribution is notproportional to radial distance from centerline
of shaft
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CHAPTER REVIEW
Instead, such applied torque is related to stress
distribution using the shear-stress-shear-straindiagram and equilibrium
If a shaft is subjected to plastic torque, and
then released, it will cause material to respondelastically, causing residual shear stress to be
developed in the shaft