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

Polar moments of inertia for the aluminum and bronze shaft segments are needed
for this calculation. Aluminum segment (1) is a solid 3.00 in. diameter shaft that is
60 in. long and has a shear modulus of 4,000 ksi. The polar moment of inertia for
segment (1) is
J
1
π
= (3.00 in.) = 7.952156 in.
32
4 4
Bronze segment (2) is a solid 2.00 in. diameter shaft that is 40 in. long and has a shear
modulus of 6,500 ksi. Its polar moment of inertia is
J
2
π
= (2.00 in.) = 1.570796 in.
32
4 4
The internal torque T 1 is computed by substitution of all values into Equation (e):
T
1
32 kip⋅in.
=
⎡
4
60 in. 1.570796 in.
1 + ⎛ ⎞ ⎛
⎢ ⎝
4
40 in. ⎠ ⎝
⎜
⎣
7.952156 in.
32 kip⋅in.
21.600 kip in.
⎞ 6,500 ksi 1.481481
⎠
⎟ ⎛ = = ⋅
⎝ ⎜
⎞ ⎤
4,000 ksi ⎠
⎟ ⎥
⎦
The internal torque T 2 can be found by substitution back into Equation (a):
T2 = T1
− 32 kip⋅ in. = 21.600 kip⋅in. − 32 kip⋅ in. = −10.400 kip⋅in.
Shear Stresses
Since the internal torques are now known, the maximum shear stress magnitudes can be
calculated for each segment from the elastic torsion formula [Equation (6.5)]. In calculating
the maximum shear stress magnitude, only the absolute value of the internal torque
is used. In segment (1), the maximum shear stress magnitude in the 3.00 in. diameter
aluminum shaft is
Tc 1 1
τ 1 = =
J
1
(21.600 kip⋅in.)(3.00 in./2)
7.952156 in.
4
= 4.07 ksi
Ans.
The maximum shear stress magnitude in the 2.00 in. diameter bronze shaft segment (2) is
Tc 2 2
τ 2 = =
J
2
(10.400 kip⋅in.)(2.00 in./2)
1.570796 in.
4
= 6.62 ksi
Ans.
Rotation Angle of Flange B
The angle of twist in shaft segment (1) can be expressed as the difference between the
rotation angles at the +x and −x ends of the segment:
φ = φ −φ
1
B
Since the shaft is rigidly fixed to the wall at A, φ A = 0. The rotation angle of flange B,
therefore, is simply equal to the angle of twist in shaft segment (1). Note: The proper sign
of the internal torque T 1 must be used in the angle of twist calculation. Thus,
A
φ
TL 1 1
= φ = =
JG
B 1
1 1
(21.600 kip⋅in.)(60 in.)
= 0.040744 rad = 0.0407 rad Ans.
4
(7.952156 in. )(4,000 ksi)
169