Question 9.CS.1: Motor-Belt-Drive Shaft Design for Steady Loading A motor tra......

Reactions at bearings. From Equation (1.15), the torque applied by the pulley to the motor shaft equals

kW =\frac{F 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 force 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 \text { and } F_2=572  N

Applying the equilibrium equations to the free-body diagram of the shaft (Figure 9.4(b)), we have

\Sigma M_A=3432(0.3)-R_B(0.23)=0, \quad R_B=4476.5  N

\Sigma F_y=R_A+R_B-3432=0, \quad R_A=1044.5  N

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 9.4(c)) 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

\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}

and \sigma _y = 0. For the case under consideration, Equation (3.33) reduces 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}       (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.