 2001, W. E. Haisler
1
Chapter 12: Torsion of Circular Bars
Torsion of Circular Bars (Chapter 12)
Consider a long, slender bar of length L with a circular
cross-section that is subjected to torsion by applied
concentrated and distributed torques:
y
x
Mt
z
L
The x-axis is placed at the centroid of the cross-section. We
consider two circular cross-sections:
 2001, W. E. Haisler
2
Chapter 12: Torsion of Circular Bars
1) Solid
2) Tube
The cross-section has an important geometrically property;
the polar moment of inertia, J:
J   r 2dA   r 2rdrd   ( y 2  z 2 )dydz  I zz  I yy
A
 ( D04  Di4 )
For a solid bar with diameter D, J   D . Also, J 
32
32
Before developing a theory for how a bar twists and
deforms under a torque loading, it is very instructive to
experimentally observe the deformation pattern of a twisted
4
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
3
bar as shown in the following photograph. The undeformed
bar has straight lines that run the length of bar as well as
circular lines around the circumference.
Circular bar twisted by end torques
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
4
Note that these lines form a pattern of squares on the surface
of the bar. After twisting (lower photograph), the straight lines
spiral around the bar and the circular lines remain circles. For a
small area on the curved surface (say one of the squares), the
spiraling lines would be straight if the curved surface were laid
out flat. Each square is now a parallelogram, which suggest that
a shear (shear stress) has been applied to the square.
Have to Apply 4 Equations: Equilibrium, Constitutive,
Kinematics and B.C.s
Kinematics
So to begin the development of the theory, we scribe a line on
the surface of the bar that runs the length of the bar from a to b.
For convenience, we assume the left end of the bar is fixed from
rotation. As the bar is twisted by a torque of Mt on the right, the
line 0-a will rotate to position 0-a’. Experimentally, it is
 2001, W. E. Haisler
5
Chapter 12: Torsion of Circular Bars
observed for a circular cross-section that line 0-a remains a
straight line as it is rotated to position 0-a’. Looking at the
end cross-section where Mt is applied, we see the line b-a
has rotated CCW to b-a’ by an angle .
line 0a moves
to 0a’
a
y
0
Mt
a’
Mt
uy
a’
x
z
z
uz
y
a
u

a’
b
z
y
a

r
x
x
Cartesian
Polar (r- in y-z plane)
 represents the angle of twist for the cross-section located
at x (assuming the cross-section at x=0 was fixed).
 2001, W. E. Haisler
6
Chapter 12: Torsion of Circular Bars
Based on the physical observation discussed earlier, we
postulate the following displacements (in Cartesian coords.).
 Since we saw no motion in the x direction, then u x  0 .
 For small angular rotations, we can relate the motion in the
y-z plane, i.e., u y and u z , through geometry to the angle of
z
twist  and write u y   z and u z   y .
uy 
These are obtained from the approximations:
u y
uz
tan    
and tan    
. The negative in
y
z
u y is because u y is negative (down) for positive  .
 These assumptions are reasonable if  is not too
large. We note that the angle of twist is a function
of x so that    ( x) .
a’
a’
uz

y
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
In polar coordinates, we assume ur  0
y
a
and u x  0 [no displacement in radial
u
and axial (x) direction]. These last two

a’ r
assumptions are equivalent to saying
z
that the diameter of the bar does not
x
increase during twisting and the bar
does not change length, which is
consistent with experimental
observation. Finally, from geometry the circumferential
displacement of a point is proportional to the angle of twist
 and it's radial position, r: u  r . Polar coordinates
should be much easier to work with.
In order to determine the strain for these displacements,
consider the experiment referred to above. If one rolls out
7
 2001, W. E. Haisler
8
Chapter 12: Torsion of Circular Bars
the curved surface of the bar into a flat surface, we have the
following (for a length of the bar between x and x+x):
x
x  x
x
a

a’
u
y
u
x

a’
z
a
r


x

In the x-y plane, the angle  represents the change in right
angle for one of the original squares and will define the
strain  x (or engineering shear strain,  x ).
 2001, W. E. Haisler
9
Chapter 12: Torsion of Circular Bars
We can write the following for the engineering shear strain:
 x   
1  u x x
tan


 u
x
x
 u
 

x x
 u
x
x
u

x
But the displacement is given by : u  r so that

 x  r
x
and
 x 
1
2 x

1 r 
2 x
.
Note that the above assumes that the bar is prismatic (r is a
constant).
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
10
If we want to do this more precisely, we can use the tensor
strain definitions in polar coordinates to obtain (recall that
ur  0, u x  0 , u  r ):
 rr
ur

 0,
r
 x 
1  u
2  x
u x

 0,
x
ur 1 u
  
0
r r 
1 u x  1 

 2r ,
r  
x

  x  2 x  r
x
1  1 ur
2  r 
 xx
u u
 r 


r
r
1  ur u x 
 rx  

0

2  x
r 

  0,

 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
Constitutive Relation
For an elastic, isotropic material, we can write the stressstrain relation as
 x
E
E
 x 
 x 
 E  x  G x
(1   )
(1   ) 2
2(1   )
where
G = shear modulus =
E
2(1 )
11
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
y
Equilibrium (Cons. Of Ang. Momentum) dA=r drd
The torque Mt(x) on the cross-section at z
location (x) must be in equilibrium with
the internal moment produced by the
z
shear stress  x :
 x
r
x
12
Mt
M t   r x dA
A
cross-section at x
Now substitute the constitutive and
kinematic relations into the equilibrium equation:
d
d 2
M t   r x dA   r (G x )dA   rG (r )dA   G r dA
A
A
A
A dx
dx
Both G and the angle of twist  are a constant for a given
cross-section located at x so that we have:
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
13
d
2
Mt  G
r
dA. The integral is a geometrical property

dx A
of the cross-section A called J, the polar moment of inertia,
so that J   r 2dA and thus we have
A
d
M t  JG
dx
The above equilibrium equation can be integrated between
any two points on the bar, say x0 and xL to obtain:
 ( xL )   ( x0 )  
xL
x0
Mt
dx
JG
Note that M t  M t ( x ) . You must know  ( x0 ) [as a
boundary condition] to determine  ( xL ) !!!
 2001, W. E. Haisler
14
Chapter 12: Torsion of Circular Bars
Example: Now consider a circular elastic bar with diameter
D, length L, shear modulus G, and with applied torques M t
at each end. The left end is held fixed from rotation.
Determine the rotation at the free end (x=L).
y
elastic bar, diameter = D
=0 at left end (fixed)
Mt
x
Mt
z
L
Mt
internal torque
diagram
x
x=L
d
We start with equilibrium: M t  JG . Integrate the ODE
dx
from x=0 to x=L to obtain:
 2001, W. E. Haisler
15
Chapter 12: Torsion of Circular Bars
 ( L)   (0)  
L Mt
dx .
JG
For this problem, the torque M t is a constant along the
length of the bar (and equal to the applied torque at each
end). For a prismatic bar, J is also a constant. Hence we can
write:
0 Mt L
Mt L
 ( L)   (0) 
dx   (0) 

JG 0
JG
0
Since the boundary condition is that the bar is fixed at x=0,
 (0)  0 . Letting  L   ( L)
Mt L
L 
JG
 L is angle of twist for a bar of length L with a constant
torque, M t , with one end fixed.
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
16
To determine stress, recall that the constitutive equation is
 x  G x
and the stress-strain equation is

 x  r
x
Hence the shear stress is
d
 x  Gr
dx
But from the equilibrium equation we have
d
d M t
M t  JG

or
dx
dx JG
Hence the shear stress becomes:
Mtr
 x 
J
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
17
Summary of important equations:
J   r 2dA
A
d M t

dx JG



 General equations when M t  M t ( x)
xL M t
 ( xL )   ( x0 )  
dx 

x0 JG
Note that  ( xL ) above is a relative twist; i.e., the angle of
twist of one end of a bar relative to its other end.
Mt L 
L 
 Angle of twist for constant M t at ends
JG 
Mtr
 x 
J
 2001, W. E. Haisler
18
Chapter 12: Torsion of Circular Bars
Example: A circular bar with diameter of 0.5 in, length of
50 in, and made of steel (G=11.5x106 psi) carries a torque of
30 ft-lb.
D=0.5 in
30 ft-lb
steel
30 ft-lb
50 in
Determine the angle of twist of end relative to the other and
the maximum shear stress.
4
4

D

(0.5
in
)
J   r 2dA 

 0.00614in 4
A
32
32
Mt L
(30 x12 in  lb)(50in)
L 

 0.255rad  14.6deg
4
6
JG (.00614in )(11.5 x10 psi)
M t r (30 x12in  lb)(0.25in)
 x 


14,658
psi
J
0.00614in 4
 2001, W. E. Haisler
19
Chapter 12: Torsion of Circular Bars
Distributed Torque.
mt ( x)
Consider the torsion of a
circular bar with distributed
M t ( x)
mt ( x)
torque of mt ( x) applied
along it's length [units of
dx
torque/length]. For moment
x
equilibrium, we have:
M t ( x  dx)  M t ( x)  mt dx  0
Divide by dx and take the limit to obtain
 Mt
 mt  0
x
M t ( x  dx )
x  dx
 2001, W. E. Haisler
20
Chapter 12: Torsion of Circular Bars
The above can be integrated to obtain M t ( x) . We can then
substitute this M t ( x) into the differential equation for  and
integrate from x1 to x2 to obtain  ( x2 ) :
x2
 ( x2 )   ( x1)   ( M t ( x) / JG )dx
x1
You must know  ( x1) as a boundary condition.
d
Alternately, we could substitute M t  JG
into
dx
 Mt

d
 mt  0 to obtain
( JG )  mt  0 . This last
x
x
dx
equation can be integrated twice to obtain . You obtain the
same result either way.
 2001, W. E. Haisler
21
Chapter 12: Torsion of Circular Bars
Torsion of Circular Bars – Example Solutions
Example 1. The aluminum circular bar below has a
constant diameter of 0.5 in. and a shear modulus of 4 million
psi. Determine the 1) angle of twist at points B and C and 2)
max shear stress in section AB and BC.
75 in-lb
40 in-lb
A
x
20 in
B
C
35 in
a) First, determine the internal torque (Mt) as a function of
x.
 2001, W. E. Haisler
22
Chapter 12: Torsion of Circular Bars
Since torque is applied only at point B and C, the internal
torque will be constant between A and B and between B and
C. Assume the internal torque in section A-B is M t1 and in
B-C is M t 2 (note: assume Mt is positive).
M
t1
A
M
1
B
t1
M
t2
B
M
2
t2
C
Make cuts between A and B and between B and C, and draw
free-body diagrams as below:
 2001, W. E. Haisler
23
Chapter 12: Torsion of Circular Bars
75 in-lb
40 in-lb
A
Mt
1
B
free-body 1
Mt
2
C
free-body 2
Starting with the free-body 2, we write moment (torque)
equilibrium equations:
free  body # 2 :  M  0  75  M t 2  M t 2  75 in-lb
free  body #1:  M  0  M t 2  40  M t1  M t1  M t 2  40
 75  40  35 in-lb
This structure is STATICALLY DETERMINANT since we
could find all internal torques by equilibrium alone. The
internal torque diagram can now be drawn:
 2001, W. E. Haisler
24
Chapter 12: Torsion of Circular Bars
Mt
75 in-lb
35
x
0
20
55 in.
b) Determine the twist of each section.
35
35
75
75
1
A
2
B
J1  J 2   D 
32
4
 (0.5)4
B
 0.00613in 4
C
32
M t1L1
35(20)

 0.0285rad
twist of bar 1: 1 
J1G1 0.00613(4 x106 )
 2001, W. E. Haisler
Chapter 12: Torsion of Circular Bars
25
M t 2 L2
75(35)
twist of bar 2: 2 

 0.107rad
6
J 2G2 0.00613(4 x10 )
A = rotation at A = 0 (boundary condition)
B = rotation at A + twist of bar 1 = A + 1 = 0 + 0.0285
= 0.0285 rad = 1.63 deg
C = rotation at B + twist of bar 2 = B + 2
= 0.0285 + 0.107
= 0.136 rad = 7.77 deg
An alternate method for part b is to integrate the torque
diagrams:
 2001, W. E. Haisler
B   A  
Chapter 12: Torsion of Circular Bars
B Mt
A
JG
dx  0  
20"
0"
20
35
dx

(0.00143)
x
0
0.00613(4 x106 )
 0.0285rad  1.64deg
C Mt
55"
75
C   B  
dx  0.0285rad  
dx
B JG
20" 0.00613(4 x106 )
55
 0.0285  0.0031x 20
26
 0.0285  0.107
 0.136rad  7.77 deg
c) Determine the maximum stress in each section
M t1r 1 35in  lb(0.5 / 2in)
 x 1 

 1, 427 psi
4
J1
0.00613in
M t 2 r 2 75(0.5 / 2)
 x 2 

 3,059 psi
J2
0.00613
 2001, W. E. Haisler
27
Chapter 12: Torsion of Circular Bars
A reminder of what the torsion equation tells you.
For a bar with a torque M t ( x) ,
d M t

.
dx JG
Integrate from A to B location:
A d  A JGt dx .
B
B
J, G
Mt
M
A
x
Mt
B
L
Thus, we obtain the expression:
B Mt
B   A  
dx . If the integrand is constant,
A JG
B Mt
 M L
 M L
dx or  t  is the
then B   A   t  . This last expression says that the term 
A JG
 JG  AB
 JG  AB
difference in the angle of twist from point A to point B (or it is the twist of B relative to A). If point A is
 M L
fixed, we have B   t  .
 JG  AB
----------------------------------------------------------------------------------------------------------------------------Consider the following problem has constant
internal torques in each segment as shown
M t1
Mt2
M t3
( M t1 in section AB, etc):
A
B
C
D
 M L
Angle of twist at B: B   A   t    A  B / A (  B / A is relative twist of bar AB; end B relative to A)
 JG  AB
 M L
Angle of twist at C: C  B   t   B  B / C ( B / C is relative twist of bar BC)
 JG  BC
 M L
Angle of twist at D: D  C   t   C  D / C ( D / C is relative twist of bar CD)
 JG CD
What is the angle of twist at point D?
M L
M L
M L
M L
M L
M L
D  C   t   B   t    t    A   t    t    t 
 JG CD
 JG  BC  JG CD
 JG  AB  JG  BC  JG CD
  A  B / A  C / B  D / C
From Boundary Conditions:  A  D  0 . Thus, for this problem: B / A  C / B  D / C  0 (i.e., twist
of D relative to A is zero since A and D are fixed).
----------------------------------------------------------------------------------------------------------------------------Very similar to axial bars in tension.
PL
1  elongation of bar 1  1 1 ,
P1
P2
A1E1
A
B
C
PL
 2  elongation of bar 2  2 2 .
A2 E2
Displacement of point B = displacement of point A + 1
Displacement of point C = displacement of point B +  2 = displacement of point A + 1 +  2
From Boundary Conditions, displacement of point A is zero.