Question 9.CS.1: Motor-Belt-Drive Shaft Design for Steady Loading A motor tra...
Motor-Belt-Drive Shaft Design for Steady Loading
A motor transmits the power P at the speed of n by a belt drive to a machine (Figure 9.4a). The maximum tensions in the belt are designated by F_{1} and F_{2} with F_{1} \gt F_{2}. The shaft will be made of cold-drawn AISI 1020 steel of yield strength S_{y}. Note that design of main and drive shafts of a gear box will be considered in Case Study 18.5. Belt drives are discussed in details in Chapter 13.
Find: Determine the diameter D of the motor shaft according to the energy of distortion theory of failure, based on a factor of safety n with respect to yielding.
Given: Prescribed numerical values are
L=230 mm \quad a=70 mm \quad r=51 mm \quad P=55 kW \\ n_0=4500 rpm\quad S_y=390 MPa ( from Table B.3) \quad n=3.5
Assumptions: Friction at the bearings is omitted; bearings act as simple supports. At maximum load F_{1} = 5F_{2}.
Reactions at bearings. From Equation 1.15, the torque applied by the pulley to the motor shaft equals
kW =\frac{F_t V}{1000}=\frac{T n}{9549} (1.15)
T_{A C}=\frac{9549 P}{n_0}=\frac{9549(55)}{4500}=116.7 N \cdot m
The forces transmitted through the belt is therefore
F_2-\frac{F_1}{5}=\frac{T_{A C}}{r}=\frac{116.7}{0.051}=2288 N
or
F_1=2860 N \quad \text { and } \quad F_2=572 N
Applying the equilibrium equations to the free-body diagram of the shaft (Figure 9.4b), we have
\begin{array}{c} \sum M_A=3432(0.3)-R_B(0.23)=0, \quad R_B=4476.5 N \\ \sum F_y=-R_A+R_B-3432=0, \quad R_A=1044.5 N \end{array}
The results indicate that R_{A} and R_{B} act in the directions shown in the figure.
Principal stresses. The largest moment takes place at support B (Figure 9.4c) and has a value of
M_B=3432(0.07)=240.2 N \cdot m
Inasmuch as the torque is constant along the shaft, the critical sections are at B. It follows that
\begin{array}{c} \tau=\frac{16 T}{\pi D^3}=\frac{16(116.7)}{\pi D^3}=\frac{1867.2}{\pi D^3} \\ \sigma_x=\frac{32 M}{\pi D^3}=\frac{32(240.2)}{\pi D^3}=\frac{7686.4}{\pi D^3} \end{array}
and \sigma _{y} = 0. For the case under consideration, Equation 3.33 reduce to
\sigma_{\max , \min }=\sigma_{1.2}=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{x y}^2} (3.33)
\begin{aligned} \sigma_{1,2} &=\frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2+\tau^2} \\ &=\frac{3843.2}{\pi D^3} \pm \frac{1}{\pi D^3} \sqrt{\frac{(7686.4)^2}{4}+(1867.2)^2} \\ &=\frac{1}{\pi D^3}(3843.2 \pm 4272.8) \end{aligned}
from which
\sigma_1=\frac{8116}{\pi D^3} \quad \sigma_2=-\frac{429.6}{\pi D^3} (b)
Energy of distortion theory of failure. Through the use of Equation 6.14,
\left[\sigma_1^2-\sigma_1 \sigma_2+\sigma_2^2\right]^{1 / 2}=\frac{S_y}{n}
This, after introducing Equation (b), leads to
\frac{1}{\pi D^3}\left[(8116)^2-(8116)(-429.6)+(-429.6)^2\right]^{1 / 2}=\frac{390\left(10^6\right)}{3.5}
Solving,
D = 0.0288 m = 28.8 mm
Comment: A commercially available shaft diameter of 30 mm should be selected.