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

Polar moments of inertia for the shafts are needed for this calculation. Shaft (1) is a
solid 2.00 in. diameter shaft, and shaft (3) is a solid 2.50 in. diameter shaft. The polar
moments of inertia for these shafts are as follows:
J
J
1
3
π
= (2.00 in.) = 1.570796 in.
32
4 4
π
= (2.50 in.) = 3.834952 in.
32
4 4
Both shafts have the same length, and both have the same shear modulus. Therefore,
Equation (h) reduces to
⎡ 40 teeth
T1
1 + ⎛ ⎝ ⎜ ⎞
⎢
60 teeth ⎠
⎟
⎣
2 4
⎛ 3.834952 in.
() 1
⎝
⎜
1.570796 in.
⎞ ⎤
(1) T (2.085070) 18kip in.
⎠
⎟ ⎥ = = ⋅
⎦
4 1
From this equation, the internal torque in shaft (1) is computed as T 1 = 8.6328 kip · in.
Then, substitute this result back into Equation (d) to find that the internal torque in
shaft (3) is T 3 = −14.0508 kip · in.
Shear Stresses
The maximum shear stress magnitudes in the three shafts can now be calculated from the
elastic torsion formula:
Tc 1 1
(8.6328 kip⋅in.)(2.00 in./2)
τ 1 = =
= 5.50 ksi
Ans.
4
J 1.570796 in.
1
Tc 2 2
τ 2 = =
J
2
(18kip⋅in.)(2.00 in./2)
1.570796 in.
4
= 11.46 ksi
Ans.
Tc 3 3
(14.0508 kip⋅in.)(2.50 in./2)
τ 3 = =
= 4.58 ksi Ans.
4
J
3.834952 in.
3
Since only the shear stress magnitudes are required here, the absolute value of T 3 is
used.
Rotation Angle of Gear E
The rotation angle of gear E is equal to the angle of twist in shaft (3):
φ
TL 3 3
( −14.0508 kip⋅in.)(24 in.)
= φ = =
=− 0.007328 rad =−0.00733 rad
4
JG 3 3 (3.834952 in. )(12,000 ksi)
Ans.
E 3
Rotation Angle of Gear C
The rotation angle of gear C is equal to the rotation angle of gear B plus the additional
twist that occurs in shaft (2):
φC = φB + φ2
The rotation angle of gear B is equal to the angle of twist in shaft (1):
φ
TL 1 1
(8.6328 kip⋅in.)(24 in.)
= φ = =
= 0.010992 rad
4
JG (1.570796 in. )(12,000 ksi)
B 1
1 1
Note: The rotation angle of gear B can also be found from the rotation angle of gear E:
NE
60
φB
=− φ E =− ( − 0.007328 rad) = 0.010992 rad
N 40
B
179