Classes #3 & #4
Civil Engineering Materials – CIVE 2110
Torsion
Fall 2010
Dr. Gupta
Dr. Pickett
1
Torque and
Torsional Deformation
Torque is a MOMENT
applied about the LONG axis
of a circular shaft.
Shaft twists about the LONG axis.
Circles remain circles.
Each Longitudinal grid line
deforms into a HELIX,
that intersects the circles
at EQUAL angles.
Torque and
Torsional Deformation
Torque is a MOMENT
applied about the LONG axis
of a circular shaft.
Shaft length & radius
do NOT change.
Ends of shaft remain flat,
perpendicular to long axis.
Radial lines remain STRAIGHT.
Torque and
Torsional Deformation
If shaft is FIXED at one end
and
Torque applied at other end,
Radial lines
- remain straight
- rotate thru an angle of twist
 (x)
- angle of twist varies along
length of shaft
Torque and
Torsional Deformation
An element located x
from the back end of the shaft,
will have different
angles of twist (rotations)
on its front and back faces.
 ( x)
and
 ( x)  
The difference in angles of twist
causes
Torsional SHEAR STRAINS.
Angles change from 90˚.
lim '


 
   C  A _ along_ CA
2
 B  A _ along _ BA
in the horizontal plane
BD
Tan  
x
thus
BD  xTan   x
Angles change from 90˚.
lim '


 
   C  A _ along_ CA
2
 B  A _ along _ BA
in the vertical plane
BD
Tan  

thus
BD  Tan   
setting
BD  BD
gives x  

thus   
x
For all elements on the
cross section at x,
dx
and d
are
the sam e
thus
d
is constant over the cross  sec tion
dx
consequently
shear strain varies linearily with 
from zero at center
to maxim um at outside edge

    max
c
 c
For all elements on the
cross section at x,
shear strain varies linearily with 
from zero at center
to maxim um at outside edge   c

    max
c
 max
Torsional Shear
Stress
For circular shafts,
hollow or solid,
If material is
- Homogeneous
- Linear-Elastic
Hooke’s Law gives:
  G
Thus, like Shear Strain,
Shear Stress,
varies linearly with
Radial distance from center of shaft.


    max
c
Torsional Shear
Stress
Each elemental area, dA
located at ρ from the center,
will have an internal Force, dF,
dF   (dA)
that will produce an internal
Resisting Torque, T.
  

T    dF   dA      max (dA)
c

A
A 
T
 max
c
2


A

    max
c
Torsional Shear
Stress
Polar Moment of Inertia, J:
for a Solid Shaft;
sin ce circum ference  2
the area of a
is
dA  (2 )d
ring
thickness  d
of
2
 4
4
J    dA    (2 )d 2   d 
(   c)  c
4
2
A
0
0
c
c
2
2
3
for a hollow shaft
J


c
2
4
outside
c
4
inside


    max
c
Axial Shear Stress
(solid shaft)
In order for any element
to be in equilibrium
the internal torque, T,
must develop an
Axial Shear Stress,
equal to the
Radial Shear Stress,
also
varying Linearly with radial position.
 axial   radial
T   

   max
J c
Axial Shear Stress
(hollow shaft)
In order for any element
to be in equilibrium
the internal torque, T,
must develop an
Axial Shear Stress,
equal to the
Radial Shear Stress,
also
varying Linearly with radial position.
 axial   radial
T   

   max
J c
Axial Shear Stress
(wood shaft )
In order for any element
to be in equilibrium
the internal torque, T,
must develop an
Axial Shear Stress,
equal to the
Radial Shear Stress,
also
varying Linearly with radial position.
 axial   radial
T   

   max
J c
Torque and Torsional Deformation
If shaft is FIXED at one end
and
Torque applied at other end,
Radial lines
- remain straight
- rotate thru an angle of twist
 (x)
- angle of twist varies along
length of shaft
Torque and
Torsional Deformation
An element located x
from the back end of the shaft,
will have different
angles of twist (rotations)
on its front and back faces.
 ( x)
and
 ( x)  
The difference in angles of twist
causes
Torsional SHEAR STRAINS.
Angle of TWIST
Angles change from 90˚.
lim '


 
   C  A _ along_ CA
2
 B  A _ along _ BA
in the horizontal plane
BD
Tan  
x
thus
BD  xTan   x
Angle of TWIST - Angles change from 90˚.
lim  '


 
   C  A _ along _ CA
2
 B  A _ along _ BA
in the vertical plane
BD
Tan  

thus
BD  Tan   
setting
BD  BD
gives x  
thus 
x

 
Angle of TWIST
thus 
recalling  
x


G
 
and  
T
J
T
if shaft has _ constant radius and constant _ applied_ Torque
GJ
or _ if Torque or radius or m aterial var y along shaft length
T ( x)  ( x)
then  
G ( x) J ( x)

then
and  
T ( x)  ( x)  x 
T ( x)
x



G ( x) J ( x)    G ( x) J ( x)
L
T ( x)
x
G ( x) J ( x)
0
int egrating over the length gives   
usually the m aterial
is homogeneous so G  constant
L
1 T ( x)
thus   
dx
G 0 J ( x)
and
if
radius and Torque are constant  
TL
GJ
Angle of TWIST
if
there are changes
in Torque or radius or m aterial
TL
then   
GJ
Material must be:
-
Homogeneous
Isotropic
Linear Elastic
Stress < (Yield Stress = Proportional Limit)
Angle of TWIST
Sign Convention: Positive direction is:
- Point right hand thumb outward from shaft
- Right hand fingers curl in positive rotation