Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Plan the Solution
The internal torques in the three shaft segments will be determined
from free-body diagrams and equilibrium equations. The
elastic torsion formula [Equation (6.5)] will be used to compute
the maximum shear stress in each segment once the internal
torques are known. The angle-of-twist equations [Equations
(6.12) and (6.14)] will be used to determine the twisting in individual
shafts as well as the rotation angles at gears B, C, and D.
y
A
(1)
900 N.m
B
(2)
600 N.m
T 3 (3)
0.85 m 1.00 m 0.70 m
C
250 N.m
D
x
SOLUTION
Equilibrium
Consider a free-body diagram that cuts through shaft segment (3)
and includes the free end of the shaft. A positive internal torque
T 3 is assumed to act in segment (3). The equilibrium equation
obtained from this free-body diagram gives the internal torque in
segment (3) of the shaft:
∑ M = 250 Nm ⋅ − T = 0
x 3
∴ T = 250 Nm ⋅
3
Similarly, the internal torque in segment (2) is found from an
equilibrium equation obtained from a free-body diagram that
cuts through segment (2) of the shaft. A positive internal torque
T 2 is assumed to act in segment (2). Thus,
And for segment (1),
∑ M
∑ M = 250 Nm ⋅ − 600 Nm ⋅ − T = 0
x 2
∴ T = −350 Nm ⋅
2
= 250 Nm ⋅ − 600 Nm ⋅ + 900 Nm ⋅ − T = 0
x 1
∴ T = 550 Nm ⋅
1
A torque diagram is produced by plotting these three results.
Polar Moments of Inertia
The elastic torsion formula will be used to compute the maximum
shear stress in each shaft segment. For this calculation, the
polar moments of inertia must be computed for each segment.
Segment (1) is a solid 36 mm diameter shaft. The polar moment
of inertia for this shaft segment is
J
1
π
= (36 mm) = 164,895.9 mm
32
4 4
Shaft segment (2), which is a solid 30 mm diameter shaft, has a
polar moment of inertia of
J
2
π
= (30 mm) = 79,521.6 mm
32
4 4
The polar moment of inertia for shaft segment (3), which is a
solid 25 mm diameter shaft, has a value of
J
3
π
= (25 mm) = 38,349.5 mm
32
4 4
y
y
y
A
A
A
(1)
900 N.m
B
T 2 (2)
600 N.m
C
(3)
0.85 m 1.00 m 0.70 m
T 1 (1)
B
(2) (3)
C
0.85 m 1.00 m 0.70 m
(1)
900 N.m
600 N.m
0.85 m 1.00 m 0.70 m
550 N.m
900 N.m
B
−350 N.m
600 N.m
(2) (3)
C
250 N.m
250 N.m
D
250 N.m
D
250 N.m
Internal torque diagram for compound shaft.
D
x
x
x
149