MAK205_Chapter5.ppt

2005 Pearson Education South Asia Pte Ltd
5. Torsion
1
CHAPTER OBJECTIVES
• Discuss effects of applying
torsional loading to a long straight
member
• Determine stress distribution within
the member under torsional load
• Determine angle of twist when material behaves in
a linear-elastic and inelastic manner
• Discuss statically indeterminate analysis of shafts
and tubes
• Discuss stress distributions and residual stress
caused by torsional loadings

5. Torsion
2
CHAPTER OUTLINE
1. Torsional Deformation of a Circular Shaft
2. The Torsion Formula
3. Power Transmission
4. Angle of Twist
5. Statically Indeterminate Torque-Loaded Members
6. *Solid Noncircular Shafts
7. *Thin-Walled Tubes Having Closed Cross
Sections
8. Stress Concentration
9. *Inelastic Torsion
10. *Residual Stress

5. Torsion
3
• Torsion is a moment that twists/deforms a
member about its longitudinal axis
• By observation, if angle of rotation is small, length
of shaft and its radius remain unchanged
5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT

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5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT
• By definition, shear strain is
Let x  dx and   = d
BD =  d = dx 
 = (/2)  lim ’
CA along CA
BA along BA
 = 
d
dx
• Since d / dx =  / = max /c
 = max

c
( )
Equation 5-2

5. Torsion
5
5.2 THE TORSION FORMULA
• For solid shaft, shear stress varies from zero at
shaft’s longitudinal axis to maximum value at its
outer surface.
• Due to proportionality of triangles, or using Hooke’s
law and Eqn 5-2,
 =  max

c
( )
...
 =
 max
c ∫A 2 dA

5. Torsion
6
• The integral in the equation can be represented as
the polar moment of inertia J, of shaft’s x-sectional
area computed about its longitudinal axis
 max =
Tc
J
 max = max. shear stress in shaft, at the outer surface
T = resultant internal torque acting at x-section, from
method of sections & equation of moment
equilibrium applied about longitudinal axis
J = polar moment of inertia at x-sectional area
c = outer radius pf the shaft

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• Shear stress at intermediate distance, 
 =
T
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

5. Torsion
8
Solid shaft
• J can be determined using area element in the form
of a differential ring or annulus having thickness d
and circumference 2 .
• For this ring, dA = 2 d
J = c4

2
• J is a geometric property of the circular area and
is always positive. Common units used for its
measurement are mm4 and m4.

5. Torsion
9
Tubular shaft
J = (co
4  ci
4)

2

5. Torsion
10
Absolute maximum torsional stress
• Need to find location where ratio Tc/J is maximum
• Draw a torque diagram (internal torque  vs. x along
shaft)
• Sign Convention: T is positive, by right-hand rule, is
directed outward from the shaft
• Once internal torque throughout shaft is determined,
maximum ratio of Tc/J can be identified

5. Torsion
11
Procedure for analysis
Internal loading
• Section shaft perpendicular to its axis at point
where shear stress is to be determined
• Use free-body diagram and equations of
equilibrium 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 = (co
4  ci
2)/2

5. Torsion
12
Shear stress
• Specify radial distance , measured from centre
of x-section to point where shear stress is to be
found
• Apply torsion formula,  = T /J or max = Tc/J
• Shear stress acts on x-section in direction that is
always perpendicular to 

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13
EXAMPLE 5.3
Shaft shown supported by two bearings and
subjected to three torques.
Determine shear stress developed at points A and B,
located at section a-a of the shaft.

5. Torsion
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EXAMPLE 5.3 (SOLN)
Internal torque
Bearing reactions on shaft = 0, if shaft weight
assumed to be negligible. Applied torques satisfy
moment equilibrium about shaft’s axis.
Internal torque at section a-a determined from free-
body diagram of left segment.

5. Torsion
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EXAMPLE 5.3 (SOLN)
Internal torque
 Mx = 0; 4250 kN·mm  3000 kN·mm  T = 0
T = 1250 kN·mm
Section property
J = /2(75 mm)4 = 4.97 107 mm4
Shear stress
Since point A is at  = c = 75 mm
B = Tc/J = ... = 1.89 MPa

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EXAMPLE 5.3 (SOLN)
Shear stress
Likewise for point B, at  = 15 mm
B = T /J = ... = 0.377 MPa
Directions of the stresses
on elements A and B
established from
direction of resultant
internal torque T.

5. Torsion
17
• Power is defined as work performed per unit of
time
• Instantaneous power is
• Since shaft’s angular velocity  = d/dt, we can
also express power as
5.3 POWER TRANSMISSION
P = T (d/dt)
P = T
• Frequency f of a shaft’s rotation is often reported.
It measures the number of cycles per second
and since 1 cycle = 2 radians, and  = 2f T, then
power
P = 2f T
Equation 5-11

5. Torsion
18
Shaft Design
• If power transmitted by shaft and its frequency of
rotation is known, torque is determined from Eqn
5-11
• Knowing T and allowable shear stress for
material, allow and applying torsion formula,
J
c
T
allow
=

5. Torsion
19
Shaft Design
• For solid shaft, substitute J = (/2)c4 to determine c
• For tubular shaft, substitute J = (/2)(co
2  ci
2) to
determine co and ci

5. Torsion
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EXAMPLE 5.5
Solid steel shaft shown used to transmit 3750 W from
attached motor M. Shaft rotates at  = 175 rpm and
the steel allow = 100 MPa.
Determine required diameter of shaft to nearest mm.

5. Torsion
21
EXAMPLE 5.5 (SOLN)
Torque on shaft determined from P = T,
Thus, P = 3750 N·m/s
Thus, P = T, T = 204.6 N·m
( )
 = = 18.33 rad/s
175 rev
min
2 rad
1 rev
1 min
60 s
( )
= =
J
c
 c4
2 c2
T
allow
.
.
.
c = 10.92 mm
Since 2c = 21.84 mm, select shaft with diameter of
d = 22 mm

5. Torsion
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5.4 ANGLE OF TWIST
• Angle of twist is important when analyzing reactions
on statically indeterminate shafts
 =
T(x) dx
J(x) G
∫0
L
 = angle of twist, in radians
T(x) = internal torque at arbitrary position x, found
from method of sections and equation of
moment equilibrium applied about shaft’s axis
J(x) = polar moment of inertia as a function of x
G = shear modulus of elasticity for material

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5.4 ANGLE OF TWIST
Constant torque and x-sectional area
 =
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 adding
each segment’s angle of twist:
 =
TL
JG


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5.4 ANGLE OF TWIST
Sign convention
• Use right-hand rule: torque and angle of twist are
positive when thumb is directed outward from the
shaft

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25
5.4 ANGLE OF TWIST
Internal torque
• Use method of sections and equation of moment
equilibrium applied along shaft’s axis
• If torque varies along shaft’s length, section made
at arbitrary position x along 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

5. Torsion
26
5.4 ANGLE OF TWIST
Angle of twist
• When circular x-sectional area varies along
shaft’s axis, polar moment of inertia expressed as
a function of its position x along its axis, J(x)
• If J or internal torque suddenly changes between
ends of shaft,  = ∫ (T(x)/J(x)G) dx or  = TL/JG
must be applied to each segment for which J, T
and G are continuous or constant
• Use consistent sign convention for internal torque
and also the set of units

5. Torsion
27
EXAMPLE 5.9
50-mm-diameter solid cast-iron post shown is buried
600 mm in 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 t N·mm/mm along its 600 mm
buried length. G = 40(103) GPa

5. Torsion
28
EXAMPLE 5.9 (SOLN)
Internal torque
From free-body diagram
 Mz = 0; TAB = 100 N(300 mm) = 30  103 N·mm

5. Torsion
29
EXAMPLE 5.9 (SOLN)
Internal torque
Magnitude of the uniform distribution of torque along
buried segment BC can be determined from
equilibrium of the entire post.
 Mz = 0;
100 N(300 mm)  t(600 mm) = 0
t = 50 N·mm

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EXAMPLE 5.9 (SOLN)
Internal torque
Hence, from free-body diagram of a section of the
post located at position x within region BC, we have
 Mz = 0;
TBC  50x = 0
TBC = 50x

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EXAMPLE 5.9 (SOLN)
Maximum shear stress
Largest shear stress occurs in region AB, since
torque largest there and J is constant for the post.
Applying torsion formula
max = = ... = 1.22 N/mm2
TAB c
J

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EXAMPLE 5.9 (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 segments AB and BC twist, so
A = +
TAB LAB
JG
TBC dx
JG
∫0
LBC
.
.
.
A = 0.00147 rad

5. Torsion
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5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
• A torsionally loaded shaft is statically indeterminate
if moment equation of equilibrium, applied about
axis of shaft, is not enough to determine unknown
torques acting on the shaft

5. Torsion
34
• From free-body diagram, reactive torques at
supports A and B are unknown, Thus,
 Mx = 0; T  TA  TB = 0
• Since problem is statically
indeterminate, formulate the
condition of compatibility; end
supports are fixed, thus angle of
twist of both ends should sum to
zero
A/B = 0

5. Torsion
35
• Assume linear-elastic behavior, and using load-
displacement relationship,  = TL/JG, thus
compatibility equation can be written as
TA LAC
JG
TB LBC
JG
 = 0
• Solving the equations
simultaneously, and realizing that
L = LAC + LBC, we get
TA = T
LBC
L
( ) TB = T
LAC
L
( )

5. Torsion
36
Equilibrium
• Draw a free-body diagram
• Write equations of equilibrium about axis of shaft
Compatibility
• Express compatibility conditions in terms of
rotational displacement caused by reactive
torques
• Use torque-displacement relationship, such as
 = TL/JG
• Solve equilibrium and compatibility equations for
unknown torques

5. Torsion
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EXAMPLE 5.11
Solid steel shaft shown has a diameter of 20 mm. If it
is subjected to two torques, determine reactions at
fixed supports A and B.

5. Torsion
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EXAMPLE 5.11 (SOLN)
Equilibrium
From free-body diagram, problem is
statically indeterminate.
 Mx = 0;
 TB + 800 N·m  500 N·m  TA = 0
Compatibility
Since ends of shaft are fixed, sum of angles of twist
for both ends equal to zero. Hence,
A/B = 0

5. Torsion
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EXAMPLE 5.11 (SOLN)
1.8TA  0.2TB = 750
Compatibility
The condition is expressed using the load-
displacement relationship,  = TL/JG.
.
.
.
Solving simultaneously, we get
TA = 345 N·m TB = 645 N·m

5. Torsion
40
*5.6 SOLID NONCIRCULAR SHAFTS
• Shafts with noncircular x-sections are not
axisymmetric, as such, their x-sections will bulge or
warp when it is twisted
• Torsional analysis is complicated and thus is not
considered for this text.

5. Torsion
41
*5.6 SOLID NONCIRCULAR SHAFTS
• Results of analysis for
square, triangular and
elliptical x-sections are
shown in table

5. Torsion
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EXAMPLE 5.13
6061-T6 aluminum shaft shown has x-sectional area
in the shape of equilateral triangle. Determine
largest torque T that can be applied to end of shaft if
allow = 56 MPa, allow = 0.02 rad, Gal = 26 GPa.
How much torque can be applied to a shaft of
circular x-section made from same amount of
material?

5. Torsion
43
EXAMPLE 5.13 (SOLN)
By inspection, resultant internal torque at any x-
section along shaft’s axis is also T. Using formulas
from Table 5-1,
allow = 20T/a3; ... T = 179.2 N·m
allow = 46TL/a3Gal; ... T = 24.12 N·m
By comparison, torque is limited due to angle of twist.

5. Torsion
44
EXAMPLE 5.13 (SOLN)
Circular x-section
We need to calculate radius of the x-section.
Acircle = Atriangle; ... c = 14.850 mm
Limitations of stress and angle of twist require
allow = Tc/J; ... T = 288.06 N·m
allow = TL/JGal; ... T = 33.10 N·m
Again, torque is limited by angle of twist.
Comparing both results, we can see that a shaft of
circular x-section can support 37% more torque
than a triangular one

5. Torsion
45
*5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
• Thin-walled tubes of noncircular shape are used to
construct lightweight frameworks such as those in
aircraft
• This section will analyze such shafts with a closed
x-section
• As walls are thin, we assume stress is uniformly
distributed across the thickness of the tube

5. Torsion
46
Shear flow
• Force equilibrium requires the
forces shown to be of equal
magnitude but opposite direction,
thus AtA = BtB
• This product is called shear flow
q, and can be expressed as
q = avgt
• Shear flow measures force per unit
length along tube’s x-sectional area

5. Torsion
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Average shear stress
avg = average shear stress acting over
thickness of tube
T = resultant internal torque at x-section
t = thickness of tube where avg is to be
determined
Am = mean area enclosed within
boundary of centerline of tube’s
thickness
avg =
T
2tAm

5. Torsion
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Since q = avgt, the shear flow throughout the x-
section is
q =
T
2Am
Angle of twist
Can be determined using energy methods
 = ∫
TL
4Am
2G
ds
t
O

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IMPORTANT
• Shear flow q is a product of tube’s thickness and
average shear stress. This value is constant at all
points along tube’s x-section. Thus, largest
average shear stress occurs where tube’s
thickness is smallest
• Both shear flow and average shear stress act
tangent to wall of tube at all points in a direction to
contribute to resultant torque

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EXAMPLE 5.16
Square aluminum tube as shown.
Determine average shear stress in the tube at point
A if it is subjected to a torque of 85 N·m. Also,
compute angle of twist due to this loading.
Take Gal = 26 GPa.

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51
EXAMPLE 5.16 (SOLN)
Am = (50 mm)(50 mm) = 2500 mm2
avg = = ... = 1.7 N/mm2
T
2tAm
Since t is constant except at corners,
average shear stress is same at all
points on x-section.
Note that avg acts upward on darker-
shaded face, since it contributes to
internal resultant torque T at the
section

5. Torsion
52
EXAMPLE 5.16 (SOLN)
Angle of twist
Here, integral represents length around
centerline boundary of tube, thus
 = ∫
TL
4Am
2G
ds
t
O = ... = 0.196(10-4) mm-1 ∫ ds
O
 = 0.196(10-4) mm-1[4(50 mm)] = 3.92 (10-3) rad

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5.8 STRESS CONCENTRATION
• Three common discontinuities
of the x-section are:
a) is a coupling, for connecting
2 collinear shafts together
b) is a keyway used to connect
gears or pulleys to a shaft
c) is a shoulder fillet used to
fabricate a single collinear
shaft from 2 shafts with
different diameters

5. Torsion
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• Dots on x-section indicate
where maximum shear stress
will occur
• This maximum shear stress
can be determined from
torsional stress-concentration
factor, K

5. Torsion
55
• K, can be obtained from
a graph as shown
• Find geometric ratio D/d
for appropriate curve
• Once abscissa r/d
calculated, value of K
found along ordinate
• Maximum shear stress is
then determined from
max = K(Tc/J)

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IMPORTANT
• Stress concentrations in shafts occur at points of
sudden x-sectional change. The more severe the
change, the larger the stress concentration
• For design/analysis, not necessary to know exact
shear-stress distribution on x-section. Instead,
obtain maximum shear stress using stress
concentration factor K
• If material is brittle, or subjected to fatigue
loadings, then stress concentrations need to be
considered in design/analysis.

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EXAMPLE 5.18
Stepped shaft shown is supported at bearings at A
and B. Determine maximum stress in the shaft due
to applied torques. Fillet at junction of each shaft has
radius r = 6 mm.

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58
EXAMPLE 5.18 (SOLN)
Internal torque
By inspection, moment equilibrium about axis of
shaft is satisfied. Since maximum shear stress
occurs at rooted ends of smaller diameter shafts,
internal torque (30 N·m) can be found by applying
method of sections

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59
EXAMPLE 5.18 (SOLN)
From shaft geometry, we have
D
d
r
d
2(40 mm)
2(20 mm)
6 mm)
2(20 mm)
= = 2
= = 0.15
Thus, from the graph, K = 1.3
max = K(Tc/J) = ... = 3.10 MPa

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60
EXAMPLE 5.18 (SOLN)
From experimental evidence, actual stress
distribution along radial line of x-section at critical
section looks similar to:

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61
*5.9 INELASTIC TORSION
• To perform a “plastic analysis”
for a material that has yielded,
the following conditions must
be met:
1.Shear strains in material must
vary linearly from zero at center of
shaft to its maximum at outer
boundary (geometry)
2.Resultant torque at section must
be equivalent to torque caused by
entire shear-stress distribution
over the x-section (loading)

5. Torsion
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• Expressing the loading
condition mathematically,
we get:
T = 2∫A  2 d
Equation 5-23

5. Torsion
63
A. Maximum elastic torque
• For maximum elastic shear strain Y,
at outer boundary of the shaft, shear-
strain distribution along radial line will
look like diagram (b)
• Based on Eqn 5-23,
TY = (/2) Yc3
• From Eqn 5-13,
d =  (dx/)

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64
B. Elastic-plastic torque
• Used when material starts yielding,
and the yield boundary moves inward
toward the shaft’s centre, producing
an elastic core.
• Also, outer portion of shaft forms a
plastic annulus or ring
• General formula for elastic-plastic
material behavior,
T = (Y /6) (4c3  Y
3)

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B. Elastic-plastic torque
Plastic torque
• Further increases in T will shrink the radius of
elastic core till all the material has yielded
• Thus, largest possible plastic torque is
TP = (2/3)Y c3
• Comparing with maximum elastic torque,
TP = 4TY / 3
• Angle of twist cannot be uniquely defined.

5. Torsion
66
C. Ultimate torque
• Magnitude of Tu can be determined “graphically”
by integrating Eqn 5-23

5. Torsion
67
C. Ultimate torque
• Segment shaft into finite
number of rings
• Area of ring is multiplied
by shear stress to obtain
force
• Determine torque with the
product of the force and 
• Addition of all torques for
entire x-section results in
the ultimate torque,
Tu ≈ 2 2

5. Torsion
68
IMPORTANT
• Shear-strain distribution over radial line on shaft
based on geometric considerations and is
always remain linear
• Shear-stress distribution must be determined
from material behavior or shear stress-strain
diagram
• Once shear-stress distribution established, the
torque about the axis is equivalent to resultant
torque acting on x-section
• Perfectly plastic behavior assumes shear-stress
distribution is constant and the torque is called
plastic torque

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69
EXAMPLE 5.19
Tubular shaft made of
aluminum alloy with elastic -
diagram as shown.
Determine (a) maximum torque
that can be applied without
causing material to yield,
(b) maximum torque or plastic
torque that can be applied to
the shaft.
What should the minimum
shear strain at outer radius be
in order to develop a plastic
torque?

5. Torsion
70
EXAMPLE 5.19 (SOLN)
Maximum elastic torque
Shear stress at outer fiber to be
20 MPa. Using torsion formula
Y = (TY c/J); TY = 3.42 kN·m
Values at tube’s inner wall
are obtained by proportion.

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71
EXAMPLE 5.19 (SOLN)
Plastic torque
Shear-stress distribution shown below. Applying
 = Y into Eqn 5-23:
TP = ... = 4.10 kN·m
For this tube, TP represents a 20% increase in
torque capacity compared to elastic torque TY.

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72
EXAMPLE 5.19 (SOLN)
Outer radius shear strain
Tube becomes fully plastic when shear strain at
inner wall becomes 0.286(10-3) rad. Since shear
strain remains linear over x-section, plastic strain at
outer fibers determined by proportion:
o = ... = 0.477(10-3) rad

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73
*5.10 RESIDUAL STRESS
• Residual stress distribution is calculated using
principles of superposition and elastic recovery

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74
EXAMPLE 5.21
Tube made from brass alloy with length of 1.5 m and
x-sectional area shown. Material has elastic-plastic
- diagram shown. G = 42 GPa.
Determine plastic torque
TP. What are the residual-
shear-stress distribution
and permanent twist of the
tube that remain if TP is
removed just after tube
becomes fully plastic?

5. Torsion
75
EXAMPLE 5.21 (SOLN)
Plastic torque
Applying Eqn 5-23,
When tube is fully plastic, yielding started at inner
radius, ci = 25 mm and Y = 0.002 rad, thus angle of
twist for entire tube is
TP = ... = 19.24(106) N·mm
P = Y (L/ci) = ... = 0.120 rad

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76
EXAMPLE 5.21 (SOLN)
r = (TPco)/J = ... = 104.52 MPa
Plastic torque
Then TP is removed, then
“fictitious” linear shear-stress
distribution in figure (c) must be
superimposed on figure (b). Thus,
maximum shear stress or modulus
of rupture computed from torsion
formula,
i = (104.52 MPa)(25 mm/50 mm) = 52.26 MPa

5. Torsion
77
EXAMPLE 5.21 (SOLN)
’P = (TP L)/(JG) = ... = 0.0747 rad
Plastic torque
Angle of twist ’P upon removal of TP is
+  = 0.120  0.0747 = 0.0453 rad
Residual-shear-stress distribution is shown.
Permanent rotation of tube after TP is removed,

5. Torsion
78
CHAPTER REVIEW
• Torque causes a shaft with circular x-section to
twist, such that shear strain in shaft is
proportional to its radial distance from its centre
• Provided that material is homogeneous and
Hooke’s law applies, shear stress determined
from torsion formula,  = (Tc)/J
• Design of shaft requires finding the geometric
parameter, (J/C) = (T/allow)
• Power generated by rotating shaft is reported,
from which torque is derived; P = T

5. Torsion
79
CHAPTER REVIEW
• Angle of twist of circular shaft determined from
• If torque and JG are constant, then
• For application, use a sign convention for
internal torque and be sure material does not
yield, but remains linear elastic
 =
TL
JG

 =
T(x) dx
JG
∫0
L

5. Torsion
80
• If shaft is statically indeterminate, reactive
torques determined from equilibrium,
compatibility of twist, and torque-twist
relationships, such as  = TL/JG
• Solid noncircular shafts tend to warp out of
plane when subjected to torque. Formulas are
available to determine elastic shear stress and
twist for these cases
• Shear stress in tubes determined by
considering shear flow. Assumes that shear
stress across each thickness of tube is
constant
CHAPTER REVIEW

5. Torsion
81
CHAPTER REVIEW
• Shear stress in tubes determined from
 = T/2tAm
• Stress concentrations occur in shafts when x-
section suddenly changes. Maximum shear
stress determined using stress concentration
factor, K (found by experiment and represented
in graphical form). max = K(Tc/J)
• If applied torque causes material to exceed
elastic limit, then stress distribution is not
proportional to radial distance from centerline
of shaft

5. Torsion
82
CHAPTER REVIEW
• Instead, such applied torque is related to stress
distribution using the shear-stress-shear-strain
diagram and equilibrium
• If a shaft is subjected to plastic torque, and
then released, it will cause material to respond
elastically, causing residual shear stress to be
developed in the shaft

MAK205_Chapter5.ppt

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