Giáo trình Strength of Materials - Chapter 9: Torsion

CHAPTER 9: TORSION 
9.0 Introduction 
9.1 Torsional loads on circular shafts 
9.2 Torsion of noncircular members 
9.3 Helical spring under axial load 
9.1 INTRODUCTION 
A bar subjected to loadings has internal torsion moment only on cross sections: Mz 
x 
y 
z 
Mz > 0 (counter clockwise) 
Eye’s direction 
9.1 INTRODUCTION 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
• Interested in stresses and strains of 
circular shafts subjected to twisting 
couples or torques 
• Generator creates an equal and 
opposite torque T’ 
• Shaft transmits the torque to the 
generator 
• Turbine exerts torque T on the shaft 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
NET TORQUE DUE TO INTERNAL STRESSES 
   dAdFT 
• Net of the internal shearing stresses is an 
internal torque, equal and opposite to the 
applied torque, 
• Although the net torque due to the shearing 
stresses is known, the distribution of the stresses 
is not 
• Unlike the normal stress due to axial loads, the 
distribution of shearing stresses due to torsional 
loads can not be assumed uniform. 
• Distribution of shearing stresses is statically 
indeterminate – must consider shaft 
deformations 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
AXIAL SHEAR COMPONENTS 
• Torque applied to shaft produces shearing 
stresses on the faces perpendicular to the 
axis. 
• 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. 
• Conditions of equilibrium require the 
existence of equal stresses on the faces of the 
two planes containing the axis of the shaft 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
SHAFT DEFORMATIONS 
• From observation, the angle of twist of the 
shaft is proportional to the applied torque and 
to the shaft length. 
L
T




• When subjected to torsion, every cross-section 
of a circular shaft remains plane and 
undistorted. 
• Cross-sections of noncircular (non-
axisymmetric) shafts are distorted when 
subjected to torsion. 
• Cross-sections for hollow and solid circular 
shafts remain plain and undistorted because a 
circular shaft is axisymmetric. 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
SHEARING STRAIN 
• Consider an interior section of the shaft. As a 
torsional load is applied, an element on the 
interior cylinder deforms into a rhombus. 
• Shear strain is proportional to twist and radius 
maxmax and 




cL
c

L
L

  or 
• It follows that 
• Since the ends of the element remain planar, 
the shear strain is equal to angle of twist. 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
STRESSES IN ELASTIC RANGE 
J
c
dA
c
dAT max2max



   
• Recall that the sum of the moments from 
the internal stress distribution is equal to 
the torque on the shaft at the section, 
4
2
1 cJ 
 414221 ccJ  
 and max
J
T
J
Tc 
 
• The results are known as the elastic torsion 
formulas, 
• Multiplying the previous equation by the 
shear modulus, 
max

 G
c
G 
max


c

From Hooke’s Law,  G , so 
The shearing stress varies linearly with the 
radial position in the section. 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
NORMAL STRESSES 
• Elements with faces parallel and perpendicular 
to the shaft axis are subjected to shear stresses 
only. Normal stresses, shearing stresses or a 
combination of both may be found for other 
orientations. 
 
max
0
0max
45
0max0max
2
2
245cos2
o 





A
A
A
F
AAF
• Consider an element at 45o to the shaft axis, 
• Element a is in pure shear. 
• Note that all stresses for elements a and c have 
the same magnitude 
• Element c is subjected to a tensile stress on 
two faces and compressive stress on the other 
two. 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
TORSIONAL FAILURE MODES 
• Ductile materials generally fail in 
shear. 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 planes 
perpendicular to the direction in 
which tension is a maximum, i.e., 
along surfaces at 45o to the shaft 
axis. 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
EXAMPLES 9.01 
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, 
determine (a) the minimum and maximum 
shearing stress in shaft BC, (b) the 
required diameter d of shafts AB and CD 
if the allowable shearing stress in these 
shafts is 65 MPa. 
SOLUTION: 
• Cut sections through shafts AB 
and BC and perform static 
equilibrium analysis to find 
torque loadings 
• Given allowable shearing stress 
and applied torque, invert the 
elastic torsion formula to find the 
required diameter 
• Apply elastic torsion formulas to 
find minimum and maximum 
stress on shaft BC 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
EXAMPLES 9.01 
SOLUTION: 
• Cut sections through shafts AB and BC 
and perform static equilibrium analysis 
to find torque loadings 
 
CDAB
ABx
TT
TM


mkN6
mkN60    
mkN20
mkN14mkN60


BC
BCx
T
TM
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
EXAMPLES 9.01 
• Apply elastic torsion formulas to 
find minimum and maximum 
stress on shaft BC 
      
46
444
1
4
2
m1092.13
045.0060.0
22



ccJ
  
MPa2.86
m1092.13
m060.0mkN20
46
2
2max




J
cTBC
MPa7.64
mm60
mm45
MPa2.86
min
min
2
1
max
min






c
c
MPa7.64
MPa2.86
min
max




• Given allowable shearing stress and 
applied torque, invert the elastic torsion 
formula to find the required diameter 
m109.38
mkN6
65
3
3
2
4
2
max



c
c
MPa
c
Tc
J
Tc


mm8.772  cd
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
ANGLE OF TWIST IN ELASTIC RANGE 
• Recall that the angle of twist and maximum 
shearing strain are related, 
L
c
 max
• In the elastic range, the shearing strain and shear 
are related by Hooke’s Law, 
JG
Tc
G
 maxmax


• Equating the expressions for shearing strain and 
solving for the angle of twist, 
JG
TL

• If the torsional loading or shaft cross-section 
changes along the length, the angle of rotation is 
found as the sum of segment rotations 

i ii
ii
GJ
LT

9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
SUMMARY 
4
2
1 cJ 
 414221 ccJ  
P 
M0 
z 
x 
Mz = M0 
c
J
M zmax

J
M z
max exists on the 
perimeter of the 
circular section 
The relative angle of twist (góc xoắn tỷ 
đối), 
GJ
M z
The angle of twist (góc xoắn), 
GJ
LM z
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
STATICALLY INDETERMINATE SHAFTS 
• Given the shaft dimensions and the applied 
torque, we would like to find the torque reactions 
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. 
ftlb90  BA TT
ftlb90
12
21  AA T
JL
JL
T
• Substitute into the original equilibrium equation, 
AB
BA T
JL
JL
T
GJ
LT
GJ
LT
12
21
2
2
1
1
21 0  
• Divide the shaft into two components which 
must have compatible deformations, 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
EXAMPLE 9.02 
Two solid steel shafts are connected 
by gears. Knowing that for each shaft 
G = 11.2 x 106 psi and that the 
allowable shearing stress is 8 ksi, 
determine (a) the largest torque T0 
that may be applied to the end of shaft 
AB, (b) the corresponding angle 
through which end A of shaft AB 
rotates. 
SOLUTION: 
• Apply a static equilibrium analysis on 
the two shafts to find a relationship 
between TCD and T0 
• Find the corresponding angle of twist 
for each shaft and the net angular 
rotation of end A 
• Find the maximum allowable torque 
on each shaft – choose the smallest 
• Apply a kinematic analysis to relate 
the angular rotations of the gears 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
EXAMPLE 9.02 
SOLUTION: 
• Apply a static equilibrium analysis on 
the two shafts to find a relationship 
between TCD and T0 
 
 
0
0
8.2
in.45.20
in.875.00
TT
TFM
TFM
CD
CDC
B





• Apply a kinematic analysis to relate 
the angular rotations of the gears 
CB
CC
B
C
B
CCBB
r
r
rr



8.2
in.875.0
in.45.2



9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
EXAMPLE 9.02 
• Find the T0 for the maximum 
allowable torque on each shaft – 
choose the smallest 
 
 
 
 
in.lb561
in.5.0
in.5.08.2
8000
in.lb663
in.375.0
in.375.0
8000
0
4
2
0
max
0
4
2
0
max




T
T
psi
J
cT
T
T
psi
J
cT
CD
CD
AB
AB




inlb5610 T
• Find the corresponding angle of twist for each 
shaft and the net angular rotation of end A 
  
   
  
   
 
oo
/
oo
o
64
2
/
o
64
2
/
2.2226.8
26.895.28.28.2
95.2rad514.0
psi102.11in.5.0
.24in.lb5618.2
2.22rad387.0
psi102.11in.375.0
.24in.lb561










BABA
CB
CD
CD
DC
AB
AB
BA
in
GJ
LT
in
GJ
LT






o48.10A
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
DESIGN OF TRANSMISSION SHAFTS 
• Principal transmission shaft 
performance specifications are: 
- power 
- speed 
• Determine torque applied to shaft at 
specified power and speed, 
f
PP
T
fTTP


2
2


• Find shaft cross-section which will not 
exceed the maximum allowable 
shearing stress, 
 
   shafts hollow
2
shafts solid
2
max
4
1
4
2
22
max
3
max





T
cc
cc
J
T
c
c
J
J
Tc



• Designer must select shaft 
material and cross-section to 
meet performance specifications 
without exceeding allowable 
shearing stress. 
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS 
STRESS CONCENTRATIONS 
• The derivation of the torsion formula, 
assumed a circular shaft with uniform 
cross-section loaded through rigid end 
plates. 
J
Tc
max
J
Tc
Kmax
• Experimental or numerically determined 
concentration factors are applied as 
• The use of flange couplings, gears and 
pulleys attached to shafts by keys in 
keyways, and cross-section discontinuities 
can cause stress concentrations 
9.2 TORSION OF NONCIRCULAR MEMEBERS 
TORSION OF NONCIRCULAR MEMBERS 
• At large values of a/b, the maximum 
shear stress and angle of twist for other 
open sections are the same as a 
rectangular bar. 
Gabc
TL
abc
T
3
2
2
1
max  
• For uniform rectangular cross-sections, 
• Previous torsion formulas are valid for 
axisymmetric or circular shafts 
• Planar cross-sections of noncircular 
shafts do not remain planar and stress 
and strain distribution do not vary 
linearly 
9.2 TORSION OF NONCIRCULAR MEMEBERS 
SUMMARY FOR TORSION OF RECTANGULAR MEMBERS 
b 
h 
Mz 
z 
1 
max 
1 
1 
2max hb
M z

 
max1  
3hb
M z

 
, ,  : Tabulated factors depend on the ratio h/b 
9.2 TORSION OF NONCIRCULAR MEMEBERS 
THIN-WALLED HOLLOW SHAFTS 
• Summing forces in the x-direction on AB, 
 shear stress varies inversely with thickness 
   
flowshear 
0


qttt
xtxtF
BBAA
BBAAx


   
tA
T
qAdAqdMT
dAqpdsqdstpdFpdM
2
22
2
0
0






• Compute the shaft torque from the integral 
of the moments due to shear stress 

t
ds
GA
TL
24

• Angle of twist (accepted, from energy 
method) 
9.2 TORSION OF NONCIRCULAR MEMEBERS 
EXAMPLE 9.03 
Extruded aluminum tubing with a rectangular 
cross-section has a torque loading of 24 kip-
in. Determine the shearing stress in each of 
the four walls with (a) uniform wall thickness 
of 0.160 in. and wall thicknesses of (b) 0.120 
in. on AB and CD and 0.200 in. on CD and 
BD. 
SOLUTION: 
• Determine the shear flow through the 
tubing walls 
• Find the corresponding shearing stress 
with each wall thickness 
9.2 TORSION OF NONCIRCULAR MEMEBERS 
EXAMPLE 9.03 
SOLUTION: 
• Determine the shear flow through the 
tubing walls 
  
  in.
kip
335.1
in.986.82
in.-kip24
2
in.986.8in.34.2in.84.3
2
2


A
T
q
A
• Find the corresponding shearing 
stress with each wall thickness 
 with a uniform wall thickness, 
in.160.0
in.kip335.1

t
q

ksi34.8
with a variable wall thickness 
in.120.0
in.kip335.1
 ACAB 
in.200.0
in.kip335.1
 CDBD 
ksi13.11 BCAB 
ksi68.6 CDBC 
9.3 HELICAL SPRING UNDER AXIAL LOAD 
INTRODUCTION 
9.3 HELICAL SPRING UNDER AXIAL LOAD 
INTRODUCTION 
9.3 HELICAL SPRING UNDER AXIAL LOAD 
INTRODUCTION 
9.3 HELICAL SPRING UNDER AXIAL LOAD 
INTRODUCTION 
9.3 HELICAL SPRING UNDER AXIAL LOAD 
INTERNAL FORCES 
F 
F 
D 
F 
T 
F 
9.3 HELICAL SPRING UNDER AXIAL LOAD 
d : Wire diameter 
D : Mean coil diameter 
n : Active coils 
L : Free length 
𝝀 : Deflection 
P : Load 
k : Spring constant (stiffness) 
G : Shear modulus 
τ : Torsional corrected stress 
K : Application correction factor 
BASIC FORMULAS FOR SPRING 
3
4
8nD
Gd
k 
3
8
d
PD
K

 
1
25.0













d
D
d
D
K
4
8
Gd
PDn

OR