A compound shaft drives three gears, as shown in Figure P6.14. Segments (1) and (2) of the shaft are hollow bronze [G = 6,500 ksi] tubes with an outside diameter of 1.75 in. and a wall thickness of 0.1875 in. Segments (3) and (4) are solid 1.00 in. diameter steel [G = 11,500 ksi] shafts. The shaft lengths are L1 = 60 in., L2 = 14 in., L3 = 20 in., and L4 = 26 in. The torques applied to the shafts have magnitudes TB = 960 lb ⋅ ft, TD = 450 lb ⋅ ft, and TE = 130 lb ⋅ ft, acting in the directions shown. The bearings shown allow the shaft to turn freely. Calculate (a) the maximum shear stress in the compound shaft. (b) the rotation angle of flange C with respect to flange A. (c) the rotation angle of gear E with respect to flange A.
A compound shaft drives three gears, as shown in Figure
P6.14. Segments (1) and (2) of the shaft are hollow bronze [G =
6,500 ksi] tubes with an outside diameter of 1.75 in. and a wall
thickness of 0.1875 in. Segments (3) and (4) are solid 1.00 in. diameter
steel [G = 11,500 ksi] shafts. The shaft lengths are L1 = 60 in.,
L2 = 14 in., L3 = 20 in., and L4 = 26 in. The torques applied to the
shafts have magnitudes TB = 960 lb ⋅ ft, TD = 450 lb ⋅ ft, and TE =
130 lb ⋅ ft, acting in the directions shown. The bearings shown allow
the shaft to turn freely. Calculate
(a) the maximum shear stress in the compound shaft.
(b) the rotation angle of flange C with respect to flange A.
(c) the rotation angle of gear E with respect to flange A.


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For Tmax=TC/J in the first part why is 16 in the numerator
![### Calculation of Maximum Torque
To determine the maximum torque, we start with the given value:
\[ \tau_{\text{max}} = 640 \, \text{lb.ft} \]
We use the formula for torque:
\[
\tau_{\text{max}} = \frac{16 \tau_{\text{max}} d_2}{\pi \left( d_2^4 - d_1^4 \right)}
\]
Substitute the known values:
\[
\tau_{\text{max}} = \frac{16 \times 640 \times 12 \times 1.75}{\pi \left( 1.75^4 - 1.375^4 \right)}
\]
Carrying out the calculations gives:
\[
\tau_{\text{max}} = 11792.58 \, \text{psi}
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