Transcript Tutorial 6 MECH 101

Tutorial 6
MECH 101
Liang Tengfei
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1
Outline
 Common problems in HW #3
 Stress in an inclined plane
 Torsi0nal shaft
Stress in an inclined plane
Determine the stress acting on inclined
section. Where the area of cross section is
1mm2, the angle is 30o. P=500N
  375MPa
  217MPa
1
cos   (1  cos 2 )
2
2
sin  cos  
 max   x , at   0;
1
2
 max   x , at   450 ;
1
(sin 2 )
2
practice
  30.960
Pmax  1.53KN
Basic concepts about torsional loading of shafts
One important assumption: when a circular shaft, whether solid or tubular,
is subjected to torsion, each cross section remains plane and simply
rotates about the axis of the member.
Torsional Shaft
d ( x )
s  r
dx
d
 ( x)  : the rate of change of the angle of twist
dx
 s  r ( x)

  


d ( x )
  
  ( x)   s r
dx

T      dA 
s
A
for uniform shaft,
  const 

L
r
  dA 
2
A
s
r
J
Example 1
A shaft shown below is attached to a wall at its left end and is in equilibrium
when subjected to the two torques. The shear modulus is 12,000 ksi.
Determine
(a) The maximum shearing stress in the shaft.
(b) The rotation of end B of the 6-in segment with respect to end A
(c) The rotation of end C of the 4-in segment with respect to end B
(d) The rotation of end C with respect to end A.
Solution
FBD at A
TAB  20  5  0
FBD at B
 TAB  15kip  ft
TBC  5  0  TBC  5kip  ft
From the above analysis, we can obtain a torque diagram which
represents the torque distribution in the shaft.
Solution
(a) Maximum shearing stress
We have known
What is JAB and JBC ?
Solution
(b) The rotation of end B of the 6-in segment with respect to end A
Apply the equation for rotation
B / A
T
 L 
L
GJ
TAB LAB  15(12)(9)(12)


 0.012733rad
GAB J AB 12000(127.23)
(c) The rotation of end C of the 4-in segment with respect to end B
Similarly
C / B 
TCB LCB 5(12)(5)(12)

 0.011938rad
GCB J CB 12000(25.13)
(d) The rotation of end C with respect to end A
C / A  C / B  B / A  (0.011938)  (0.012733)
 0.000795rad.
Example 2