Uploaded by ishmaminator

Lecture 12

advertisement
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.
What is the twist prior to unloading?
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
Unloading is all elastic:



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
Unloading is all elastic:



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
Download
Random flashcards
Radiobiology

39 Cards

Radioactivity

30 Cards

Marketing

46 Cards

African nomads

18 Cards

Create flashcards