Question 3.8: A solid steel shaft ABC of 50 mm diameter (Fig. 3-31a) is dr...
A solid steel shaft ABC of 50 mm diameter (Fig. 3-31a) is driven at A by a motor that transmits 50 kW to the shaft at 10 Hz. The gears at B and C drive machinery requiring power equal to 35 kW and 15 kW, respectively.
Compute the maximum shear stress τ_{max} in the shaft and the angle of twist Φ_{AC} between the motor at A and the gear at C. (Use G = 80 GPa.)
Torques acting on the shaft. We begin the analysis by determining the torques applied to the shaft by the motor and the two gears. Since the motor supplies 50 kW at 10 Hz, it creates a torque T_{A} at end A of the shaft (Fig. 3-31b) that we can calculate from Eq. (3-40):
P=2\pi fT ( f = Hz= s^{-1}) (3-40)
T_{A}=\frac{P}{ 2\pi fT} =\frac{50 KW}{2\pi \left(10 Hz\right) }=796 N⋅mIn a similar manner, we can calculate the torques T_{B} and T_{C} applied by the gears to the shaft:
T_{B}=\frac{P}{ 2\pi f} =\frac{35 KW}{2\pi \left(10 Hz\right) }=557 N⋅mT_{C}=\frac{P}{ 2\pi f} =\frac{15 KW}{2\pi \left(10 Hz\right) }=239 N⋅m
These torques are shown in the free-body diagram of the shaft (Fig. 3-31b). Note that the torques applied by the gears are opposite in direction to the torque applied by the motor. (If we think of T_{A} as the “load” applied to the shaft by the motor, then the torques T_{B} and T_{C} are the “reactions” of the gears.)
The internal torques in the two segments of the shaft are now found (by inspection) from the free-body diagram of Fig. 3-31b:
T_{AB}=796 N⋅m T_{BC}=239 N⋅m
Both internal torques act in the same direction, and therefore the angles of twist in segments AB and BC are additive when finding the total angle of twist. (To be specific, both torques are positive according to the sign convention adopted in Section 3.4.)
Shear stresses and angles of twist. The shear stress and angle of twist in segment AB of the shaft are found in the usual manner from Eqs. (3-12) and (3-15):
\tau _{max}=\frac{16T}{\pi d^{3} } (3-12)
\phi =\frac{TL}{GI_{P} } (3-15)
\tau _{AB}=\frac{16T_{AB} }{\pi d^{3} } =\frac{16\left(796 N⋅m\right) }{\pi \left(50 mm\right)^{3} }=32.4 MPa\phi _{AB}=\frac{T_{AB}L_{AB} }{GI_{P} } =\frac{\left(796 N⋅m\right)\left(1.0 m\right) }{\left(80 GPa\right)\left(\frac{\pi }{32} \right) \left(50 mm\right)^{4} } =0.0162 rad
The corresponding quantities for segment BC are
\tau _{BC}=\frac{16T_{BC} }{\pi d^{3} } =\frac{16\left(239 N⋅m\right) }{\pi \left(50 mm\right)^{3} }=9.7 MPa\phi _{BC}=\frac{T_{BC}L_{BC} }{GI_{P} } =\frac{\left(239 N⋅m\right)\left(1.2 m\right) }{\left(80 GPa\right)\left(\frac{\pi }{32} \right) \left(50 mm\right)^{4} } =0.0058 rad
Thus, the maximum shear stress in the shaft occurs in segment AB and is
\tau _{max} =32.4 MPaAlso, the total angle of twist between the motor at A and the gear at C is
\phi _{AC}=\phi _{AB} + \phi _{BC}= 0.0162 rad + 0.0058 rad = 0.0220 rad = 1.26°
As explained previously, both parts of the shaft twist in the same direction, and therefore the angles of twist are added.
