MECHANICS OFFourth Edition
MECHANICS OF MATERIALS
CHAPTER
MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T. DeWolf
Torsion
Lecture Notes:J. Walt OlerTexas Tech Universityy
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Contents
Introduction
Torsional Loads on Circular Shafts
Statically Indeterminate Shafts
Sample Problem 3.4
Net Torque Due to Internal Stresses
Axial Shear Components
Design of Transmission Shafts
Stress Concentrations
Shaft Deformations
Shearing Strain
Stresses in Elastic Range
Plastic Deformations
Elastoplastic Materials
Residual StressesStresses in Elastic Range
Normal Stresses
Torsional Failure Modes
Residual Stresses
Examples 3.08, 3.09
Torsion of Noncircular Members
Sample Problem 3.1
Angle of Twist in Elastic Range
Thin-Walled Hollow Shafts
Example 3.10
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Torsional Loads on Circular Shafts
• Interested in stresses and strains of circular shafts subjected to twisting
lcouples or torques.
• Turbine exerts torque T on the shaft.
• Generator creates an equal and
• Shaft transmits the torque to the generator.
• Generator creates an equal and opposite torque T’.
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Net Torque Due to Internal Stresses
• Net of the internal shearing stresses is an internal torque, equal and opposite to the
dAdFT
applied torque,
• Although the net torque due to the shearing stresses is known, the distribution of the stresses is not.
• Distribution of shearing stresses is statically indeterminate – must consider shaft deformations.
• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional l d t b d if
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loads cannot be assumed uniform.
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Axial Shear Components
• Torque applied to shaft produces shearing stresses on the faces perpendicular to the
iaxis.
• Conditions of equilibrium require the existence of equal stresses on the faces of the
• The existence of the axial shear components
existence of equal stresses on the faces of the two planes containing the axis of the shaft.
The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.
• The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.
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Shaft Deformations
• From observation, the angle of twist, of the shaft is proportional to the applied torque and to the shaft lengthto the shaft length.
L
T
• When subjected to torsion, every cross section of a circular shaft remains plane and undistorted.
• Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric.
• Cross-sections of noncircular (non-axisymmetric) shafts are distorted when subjected to torsion.
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Shearing Strain
• Consider an interior section of the shaft. As a torsional load is applied, an element on the i t i li d d f i t h b
tan in radianxy xy
interior cylinder deforms into a rhombus.
• Since the ends of the element remain planar, the shear strain is equal to angle of twist
• For small value of , It follows that
the shear strain is equal to angle of twist.
AA L
AA or AA L
L (and are in radians)
AA
AA
• Shear strain is proportional to twist and radius
maxmax and L
c
L
(Shown next page)
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cLtan AA
L
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Shearing Strain
c
At shaft surface, Max shearing strain
maxcL
Shearing strain at a distance from the shaft axisShearing strain at a distance from the shaft axis
Shear strain inside the shaft, LL
maxmaxShear strain at the shaft's surface, Lc
L cLL
maxmax LL
c c
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maxc
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Stresses in Elastic Range• Multiplying previous eq. by the shear modulus,
max Gc
G c
From Hooke’s Law, G , so
The shearing stress varies linearly421 cJ max
c
The shearing stress varies linearly with the radial position in the section.
F h ll li d
max max2
From or c c
• For hollow cylinder
41
422
1 ccJ
2
11 min max
2
Substitute ccc
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J = Polar moment of inertiamin 1
max 2
cc
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Stresses in Elastic Range
• Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section
Jc
dAc
dAT max2max
the torque on the shaft at the section,
Th lt k th l ti t i
dAJ 2maxc
• The results are known as the elastic torsion formulas,
andmaxTTc
dAdFT
and max JJ
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Normal Stresses• Elements with faces parallel and perpendicular
to the shaft axis are subjected to shear stresses only (Element a). Normal stresses, shearing stresses or a combination of both may be found for other orientations (Element b).
• Consider an element at 45o to the shaft axis,
max0max
45
0max0max
22
245cos2
o
AA
AF
AAF
0 2AA
• Element a is in pure shear. • Element c is subjected to a tensile stress on
• Note that all stresses for elements a and c have
jtwo faces and compressive stress on the other two.
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the same magnitude.
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Torsional Failure Modes
• Ductile materials generally fail in shear. Brittle materials are weaker inshear. Brittle materials are weaker in tension than shear.
• When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis.
• When subjected to torsion, a brittle specimen breaks along planesspecimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft
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gaxis.
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Sample Problem 3.1
SOLUTION:
• Cut sections through shafts ABd BC d f iand BC and perform static
equilibrium analyses to find torque loadings.
Shaft BC is hollow with inner and outer
• Apply elastic torsion formulas to find minimum and maximum stress on shaft BC.
Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown,
• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the g ,
determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD
required diameter.
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if the allowable shearing stress in these shafts is 65 MPa.
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SOLUTIONSample Problem 3.1SOLUTION:• Cut sections through shafts AB and BC
and perform static equilibrium analysis to find torque loadingsto find torque loadings.
CDAB
ABx
TT
TM
mkN6
mkN60 mkN20
mkN14mkN60
BC
BCx
T
TM
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A l l ti t i f l t Gi ll bl h i t dSample Problem 3.1• Apply elastic torsion formulas to
find minimum and maximum stress on shaft BC.
• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter.
4441
42 045.0060.0
22
ccJ mkN665 34max
MPaTcJ
Tc
46 m1092.13
22
109213
m060.0mkN2046
22max
J
cTBCm109.38 3
32
42
c
ccJ
mm8772 cd
MPa2.86m1092.13 46
J
mm45min1min c
mm8.772 cd
dAB = dBC
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MPa7.64
mm60MPa2.86
min
2max
c
MPa7.64
MPa2.86
min
max
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290 N-m
Fixed end
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2.8 kN-m
Free end(T = 0)(T = 0)
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Why ? Show on next page !!!CCD AB
B
rT Tr
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Gear CGear C Gear BGear B
rc=240 mm FFTCD
rB=80 mm
BCT=TAB= 900 N-m
FCD
TCD = (F)(rc) TAB = 900 = (F)(rB)
CD ABT TF C B
CCD AB
r rrT Tr
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Br
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Angle of Twist in Elastic Range• Recall that the angle of twist and maximum
shearing strain are related,
Lc max and max J
TJ
Tc L
max
• In the elastic range, the shearing strain and shear are related by Hooke’s Law,
T
max JJ
maxmax
Tc cG JG L
• Equating the expressions for shearing strain and solving for the angle of twistsolving for the angle of twist,
JGTL
The relation shows that, within the elastic range, the angle of twist is proportional to the torque T applied to the shaft. This is accordance with the experimental evidence. The
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previous eqs also provides us with a convenient method for determining the modulus of rigidity of a given materials.
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Angle of Twist in Elastic Range
• If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotationsfound as the sum of segment rotations
i ii
iiGJLT
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(A to B)
(A to C)
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See from A to C
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A/BB/C
A to B B to C
B/C
A/B
A to C
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A to C
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Sample Problem 3.4
SOLUTION:
• Apply a static equilibrium analysis on th t h ft t fi d l ti hithe two shafts to find a relationship between TCD and T0 .
• Apply a kinematic analysis to relate
Two solid steel shafts are connected• Find the maximum allowable torque
on each shaft – choose the smallest.
the angular rotations of the gears.
Two solid steel shafts are connected by gears. Knowing that for each shaft G = 77 GPa and that the allowable shearing stress is 55 MPa, determine
• Find the corresponding angle of twist for each shaft and the net angular rotation of end A
o eac s a t c oose t e s a est.
g(a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which
rotation of end A.
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end A of shaft AB rotates.
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Sample Problem 3.4SOLUTION:
• Apply a static equilibrium analysis on the two shafts to find a relationship
• Apply a kinematic analysis to relate the angular rotations of the gears.the two shafts to find a relationship
between TCD and T0 .the angular rotations of the gears.
60
CCD AB
B
rT Tr
T T
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222.73
CD O
CD O
T T
T T
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Fi d th T f th i Fi d th di l f t i t f h
Sample Problem 3.4• Find the T0 for the maximum
allowable torque on each shaft –choose the smallest.
• Find the corresponding angle of twist for each shaft and the net angular rotation of end A.
D = 0 (Fix)
8.05°
10.2°
61.8
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inlb5610 T
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D = 0 (Fix)
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Design of Transmission Shafts
• Determine torque applied to shaft at specified power and speed,
fTTP 2
fPPT
fTTP
2
2
• Designer must select shaft• Find shaft cross-section which will not
exceed the maximum allowable shearing stress,
• Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable g
h ftlid3
max
TJJ
Tc
without exceeding allowable shearing stress.
Angular velocity 2 f
1 Hz(rpm) Hz60 rpm
f
shafts hollow2
shaftssolid2
max
41
42
22
max
3
Tcccc
J
cc
Angular velocity, 2 f
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p 2 max22 cc
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Watts = N-m/s
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Statically Indeterminate Shafts
• Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B.at A and B.
• From a free-body analysis of the shaft,
which is not sufficient to find the end torques. The problem is statically indeterminate.
• Divide the shaft into two components which• Divide the shaft into two components which must have compatible deformations,
• Substitute into the original equilibrium equation,
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Stress Concentrations
• The derivation of the torsion formula,
JTc
max
assumed a circular shaft with uniform cross-section loaded through rigid end plates.
• The use of flange couplings, gears, and pulleys attached to shafts by keys in keyways and cross-section discontinuities
• Experimental or numerically determined t ti f t li d
keyways, and cross section discontinuities can cause stress concentrations.
JTcKmax
concentration factors are applied as
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Plastic Deformations• With the assumption of a linearly elastic material,
JTc
max
• If the yield strength is exceeded or the material has a nonlinear shearing-stress-strain curve, this expression does not hold.
• Shearing strain varies linearly regardless of material properties. Application of shearing-stress-strain curve allows determination of stress distribution
• The integral of the moments from the internal stress distribution is equal to the torque on the shaft at the
ti
curve allows determination of stress distribution.
cc
ddT0
2
022
section,
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Elastoplastic Materials• At the maximum elastic torque,
YYY ccJT 3
21
cL Y
Y
• As the torque is increased, a plastic region ( ) develops around an elastic core ( )Y Y
Y
YL
3
3
41
34
3
3
413
32 11
cT
ccT Y
YY
Y
YL
Y
cc
3
3
41
34 1
Y
YTT
• As , the torque approaches a limiting value,0Y
torque plastic TT YP 34
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Residual Stresses
• Plastic region develops in a shaft when subjected to a large enough torque.
Wh th t i d th d ti f t• When the torque is removed, the reduction of stress and strain at each point takes place along a straight line to a generally non-zero residual stress.
• On a T- curve, the shaft unloads along a straight line to an angle greater than zero.
• Residual stress found from principle of superposition.p p p p
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0 dA
JTc
m
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Example 3.08/3.09
SOLUTION:
• Solve Eq. (3.32) for Y/c and evaluate the elastic core radius.
• Solve Eq. (3.36) for the angle of twist.
A solid circular shaft is subjected to a torque at each end. Assuming that the shaft is made of an l l i i l i h
mkN6.4 T• Evaluate Eq. (3.16) for the angle
which the shaft untwists when the torque is removed. The permanent t i t i th diff b t thelastoplastic material with
and determine (a) the radius of the elastic core, (b) the angle of twist of the shaft When the
MPa150YGPa77G
• Find the residual stress distribution by
twist is the difference between the angles of twist and untwist.
angle of twist of the shaft. When the torque is removed, determine (c) the permanent twist, (d) the distribution of residual stresses
a superposition of the stress due to twisting and untwisting the shaft.
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of residual stresses.
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SOLUTIONExample 3.08/3.09SOLUTION:• Solve Eq. (3.32) for Y/c and
evaluate the elastic core radius13
• Solve Eq. (3.36) for the angle of twist
YY
31
341 3
3
41
34
Y
YYY T
Tcc
TT
m1025 3214
21 cJ
49-
3
Pa1077m10614m2.1mN1068.3
YY
YY
JGLT
cc
m10614
m1025
49
22
YY
YY
JTcT
cJ
o33
3
8.50rad103.148rad104.93
rad104.93
Pa1077m10614
Y
m1025
m10614Pa101503
496
Y
YY
T
cT
J 630.0
o50.8
mkN68.3
630.06836.434
31
Y
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68.3 c
mm8.15Y
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Example 3.08/3.09
• Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed The
• Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaftthe torque is removed. The
permanent twist is the difference between the angles of twist and untwist.
twisting and untwisting the shaft
m10614
m1025mN106.449-
33max
JTc
3 21N1064
JGTL
MPa3.187
3
949
3
6.69rad108.116
Pa1077m1014.6m2.1mN106.4
o1 81
69.650.8
pφ
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1.81o81.1p
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Torsion of Noncircular Members• Previous torsion formulas are valid for
axisymmetric or circular shafts.
Pl ti f i l• Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly
TLT
• For uniform rectangular cross-sections,
linearly.
• At large values of a/b, the maximum
Gabcabc 32
21
max
shear stress and angle of twist for other open sections are the same as a rectangular bar.
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Thin-Walled Hollow Shafts• Summing forces in the x-direction on AB,
flowshear
0
qttt
xtxtF
BBAA
BBAAx
shear stress varies inversely with thickness
qBBAA
dAqpdsqdstpdFpdM 20
• Compute the shaft torque from the integral of the moments due to shear stress
tAT
qAdAqdMT
2
220
tA2
t
dsGA
TL24
• Angle of twist (from Chapter 11)
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tGA4
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Example 3.10Extruded aluminum tubing with a rectangular cross-section has a torque loading of 2.7 kN .m. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 4 mm and wall thicknesses of (b) 3 mm on AB and CD and 5 mm on CD and BD.
SOLUTION:
D t i th h fl th h th• Determine the shear flow through the tubing walls.
• Find the corresponding shearing stress p g gwith each wall thickness .
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Example 3.10SOLUTION:
• Determine the shear flow through the tubing walls
• Find the corresponding shearing stress with each wall thickness.
tubing walls.
With a uniform wall thickness,
With a variable wall thicknessWith a variable wall thickness
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