# Lecture 12

```Torsion – Recap
Linear Elastic Behavior
Equilibrium: T 

A
  dA
 ()
T
=
Kinematics:  
Material Behavior: Linear elastic  G
y

x
O
Shear stress distribution:

T
J

L
T

Torque Twist relation : T 
GJ

L
R.V. Kukta and K. S. Kirane, 2018
Torsion – Recap
Plastic Behavior
Equilibrium: T 

A
  dA
 ()
=
T
Material Behavior: elastic perfectly plastic

y
Y

O
G
x
1

Kinematics:  

Shear stress distribution depends on T
L
T

TY  12 c 3 Y
Torque Twist relation
TY < T < TP
TP  23 c 3 Y
4 
Y3 
T  TY 1  3 
3  4 
R.V. Kukta and K. S. Kirane, 2018
Numerical Example
An elastoplastic shaft of radius c=5 cm and length
L=1 m is loaded to a torque of T=35 kN.m and
released. Given G=200 GPa and Y=150 MPa,
determine the yield torque, plastic torque, size of
the elastic core, permanent twist and residual
stress.
T
What amount of torque causes yielding?
TY     dA  
A
c
 Y dA
2 Y
c 0 0
c
 12  Y c 3  29.45 kN  m

Y
A

c
2
 2  d d 

c
0
 3 d
The applied torque is enough to cause yielding.
R.V. Kukta & K. S. Kirane, 2018
Numerical Example
An elastoplastic shaft of radius c=5 cm and length
L=1 m is loaded to a torque of T=35 kN.m and
released. Given G=200 GPa and Y=150 MPa,
determine the yield torque, plastic torque, size of
the elastic core, permanent twist and residual
stress.
T
Does it yield completely?
TP     dA   Y dA
A
A
 Y 
c
0

2
0
c
d d  2 Y   2 d
0
 23  Y c 3  39.27 kN  m
It has not fully yielded.
R.V. Kukta, 2017
Numerical Example
An elastoplastic shaft of radius c=5 cm and length
L=1 m is loaded to a torque of T=35 kN.m and
released. Given G=200 GPa and Y=150 MPa,
determine the yield torque, plastic torque, size of
the elastic core, permanent twist and residual
stress.
T
What is the size of the elastic core?
T     dA  
A

2 Y
Y
AY

Y
0

 Y dA   Y dA
A
Y
P
c
 3 d  2 Y   2 d
Y
 12  Y Y3  23  Y (c 3  Y3 )
 23  Y c 3  16  Y Y3  35 kN  m
Y  3.788 cm
Note: this is the torque supported by the elastic core.
R.V. Kukta, 2017
Numerical Example
T
An elastoplastic shaft of radius c=5 cm and length
L=1 m is loaded to a torque of T=35 kN.m and
released. Given G=200 GPa and Y=150 MPa,
determine the yield torque, plastic torque, size of
the elastic core, permanent twist and residual
stress.
Apply elastic twist formula to the
elastic core.
Y  3.788 cm
Tcore  12  Y Y3  12.81 kN  m

Tcore L 2Tcore L

 0.0198  1.13
4
GJ
GY
R.V. Kukta, 2017
Numerical Example



Y
Y
c

+
c

=
c

*
max
Residual twist:
  1.13
Y  3.788 cm
 Y  150 MPa
TL  2TL

 14 .02
GJ Gc
Tm)
Tc
178
.)(
321MPa
2(max
35kN
m



max
3



1
.
02
J c 4
(200
 max (5 cm*)  max Y
 * GPa)
 2(35 kN
  m)
Y  c
c
 (5 cm) 3

  1.13 1.02  0.11
(178.3 MPa )(3.788 cm)
 *   178.3 MPa
5 cm
 135.1 MPa
R.V. Kukta, 2017
Numerical Example



Y
Y
c

+
c

=
1
2
c

*
max
Residual twist:
  1.13
  1.02
Y  3.788 cm
 Y  150 MPa
 max  178.3 MPa
 *  135.1 MPa
  1.13 1.02  0.11
 1  150 MPa  135.1 MPa
 14.9 MPa
 2  150 MPa  178.3 MPa
 28.3 MPa
R.V. Kukta, 2017
```

39 Cards