PROBLEMS ON TORSION
SIGN CONVENTION (REVIEW)
FREE-BODY DIAGRAMS (REVIEW)
TORQUE DIAGRAM (REVIEW)
Example1: The gears attached to the fixed-end steel shaft are subjected to the
torques shown in figure. If the shear modulus of elasticity is 80 GPa and the
shaft has a diameter of 14 mm, determine the displacement of the tooth P on
gear A. The shaft turns freely within the bearing at B.
Soln.
Internal Torque: By inspection, the torques in segments, AC, CD, and
DE are different yet constant throughout each segment. Free-body
diagrams of appropriate segments of the shaft are shown in Figure (b):
 M  0 gives
 150  280  40  TDE  0
or , TDE  170 N .m
But, from the right hand rule and the
established sign convention, we get
TDE = - 170 N.m
Similarly,
-150 + 280 – TCD = 0; OR, TCD = 130 N.m
But, from the right hand rule and the
established sign convention, we get
TCD = - 130 N.m
And, -150 + TAC = 0, OR, TAC = 150 N.m and is POSITIVE
The polar moment of inertia for the shaft is
J = π/2(0.007 m)4 = 3.77 (10-9) m4. Applying the formula for the angle of twist for
each segment and adding the results algebraically, we have:
 AE
TL
( 150)( 0.4)


GJ
3.77(10 9 )  80(10 9 )
( 130)( 0.3)

3.77(10 9 )  80(10 9 )
( 170)( 0.5)

 0.212rad
9
9
3.77(10 )  80(10 )
Since the answer is negative, by the right-hand rule the thumb is directed
toward the end E of the shaft, and therefore, gear A will rotate as shown
in figure (d)
The displacement of tooth P on gear A is
SP  AE r  (0.212rad )(100mm)  21.2mm
Remember that this analysis is valid only if the shear stress does not
exceed the proportional limit of the material.
Problem 2: A hollow member AC has the outer diameter of 60 mm and the
inner diameter of 40 mm. If the member is loaded as shown, determine:
(a) the maximum shear stress
(b) the angle of twist of end C relative to the fixed end, and
(c) the angle of twist of end C relative to point B
Use G = 80 GPa.
Problem 3: Determine the power that can be transmitted by a turbine driven
solid circular shaft of 35 mm in diameter if it is to rotate at 500 rpm and the
maximum allowable shear stress on the shaft is limited to 60 MPa.
Problem 4: Determine the diameter of a solid
steel shaft to transmit 30 MW at a speed of
1500 rev/min if the angle of twist is limited to 1
degree for every twenty diameters of length.
Take G = 80 GN/m2.
Problem 5: A shaft 50 mm in diameter and 0.75 m long has a concentric hole
drilled for a portion of its length. Find the maximum length and diameter of
the hole so that when the shaft is subjected to a torque of 1.67 kN m, the
maximum shearing stress will not exceed 75 kN/m2 and the total angle of
twist will not exceed 1.50 deg.
Take G = 80 GPa.