Transcript Introduction

Torsional Shaft
Function: transmit torques from one plane to the other.
Design of torsional shafts: stress and deformation
T
T
T
Deformation of a Torsional Shaft with
Circular Cross Section
Questions: After applied a torque T at section C,
1. What happens to the longitudinal member ab?
2. Does the cross section C change its shape?
3. Is there any out-of-plane deformation at C?
4. What about sections A and B,
where B is halfway away from A and C?
5. What happens to the angle between ab and cd?
Deformation of a Torsional Shaft with
Circular Cross Section
Answers:
Shear Strain - on the Surface of the Shaft
bb '  r
 ,  is the angle of twist, r is the radius of the cross section
ab
L
ee '  2  r  r
 


ae
 L 2 L
 
ee '
d d
r
,
is the rate of change of the angle of twist
ae
dx dx
d

   r
dx
 
Shear Strain - inside the Shaft
dd ' 
 

 
cd
L
 is the radial distance of element cd from the center.
Shear strain is linearly proportional to the radial distance.
It reaches its maximum value at the surface of the shaft.
Torsional Shaft - Shear Stress
  
Shear strain:
Shear Stress:    G  G
 0  0
  r  = s  Gr
 

r
s
Elastic torsion formula:

s
s
2
T    s ds      ds   J
s
e.g.
r
r
r
r1
r2
s
r
J    2ds    2 2 d  2   3d 
s
s
s
J is the polar second moments of inertial
Torsional Shaft - Torque Diagram
Example 1: Draw the torque diagram for the following shaft. Find the maximum
stresses inside the shaft.
Torsional Displacement
Angle of twist: 
  

L
 
   L   L


G
 
T
J

With respect to a reference cross section located
L away from the targeted cross section.
Assumptions:
• uniform cross section (J is constant)
• material properties are constant
• carrying an internal torque of T
Comparison with axially loaded member

PL
EA
T L
GJ
Torsional Displacement
Multiple shafts:
   i  
i
i
Ti  Li
Gi J i
Continuously changing shafts:
x
L
Tdx
GJ
0
  x  
Torsional Displacement
Example 2: Find the twist angle at cross section D if the cross section A
is fixed.
Torsional Shaft - Statically
Indeterminate Problems
Problem solving procedure:
1. Static equilibrium conditions:
2. Deformation relations:
3. Load and deformation relations:
Equation of equilibrium
Equation of compatibility
Torque - displacement equation
Torsional Shaft - Statically
Indeterminate Problems
Example 3: A shaft shown below is fixed between two walls. The outer diameter
of the shaft is 20 mm and the diameter of the cavity of 16 mm. Find the reactions.
Torsional Shaft - Statically
Indeterminate Problems
Example 4: A hollow circular aluminum alloy (G = 4000 ksi) cylinder has a steel
(G = 12,000 ksi) core as shown in the figure. The steel and aluminum parts are securely
connected at the ends.
(1) If T = 80 kip  in, what are the torques applied at aluminum and steel respectively?
(2) If the allowable stresses in the steel and aluminum must be limited to 12 ksi and 10ksi,
determine the maximum torque that can be applied to the right end of the composite shaft.
(3) Find the rotation of the right end of the composite shaft when the torque of (2) is
applied.
Torsional Shaft - Statically
Indeterminate Problems
Example 5: A circular bar AB with ends fixed against rotation has a hole extending
for half of its length (see figure). The outer diameter of the bar is d2 = 100 mm and the
diameter of the hole is d1 = 80 mm. The total length of the bar is L = 1250 mm.
At what distance x from the left-hand end of the bar should a torque T0 be applied so
that the reactive torques at the supports will be equal?
Torsional Shaft - Stresses on Inclined
Planes
Take an element of dx, dy and dz out from the shaft:
yx
M
xy
xy
o
 0   xy dydz  dx   yx dxdz  dy
  xy   yx
yx
Conclusion: If a shear stress exists at a point on any plane, a shear stress of the
same magnitude must also exist at this point on an orthogonal plane.
xy
nt
yx
n
Ft
Vy
Fn  Fn  0
F  0
t
Vx
  n dA   xy  dA cos   sin    yx  dA sin   cos   0
  nt dA   xy  dA cos   cos    yx  dA sin   sin   0
 n   xy sin 2
 nt   xy cos 2