Question 16.3: A self-locking power screw is required for a screw jack. An ...
A self-locking power screw is required for a screw jack. An initial proposal is to use a single start 1.25-5 Acme power screw. The axial load is 4000 N and collar mean diameter is 1.75 in. Determine the lifting and lowering torques, the efficiency of the power screw, and whether the design proposal is self-locking.
Single start thread, so the lead, L, is equal to the pitch, P.
N = 5 teeth per inch,
P=\frac{1}{N}=\frac{1}{5}=0.2 \text { in. }L = 0.2 in. = 5.08 mm,
d_p = 1.15 in. = 29.21 mm,
d_c = 1.75 in. = 44.45 mm.
Assume sliding friction,µ = 0.15.
The torque to lift the load is given by Eqn (16.27).
T_{u}=\frac{P d_{p}\left(\mu \pi d_{p}+L \cos \alpha\right)}{2\left(\pi d_{p} \cos \alpha-\mu L\right)}+\mu_{c} P \frac{d_{c}}{2} (16.27)
T=\frac{4000 \times 0.02921}{2}\left(\frac{0.15 \pi 0.02921+0.00508 \cos 14.5}{\pi 0.02921 \cos 14.5-0.15 \times 0.00508}\right)+0.15 \times 4000 \frac{0.04445}{2}
=58.42 \frac{0.01376+4.9182 \times 10^{-3}}{0.08884-7.62 \times 10^{-4}}+13.34=12.39+13.34=25.73 N m
The torque to lower the load is given by Eqn (16.28),
T_{d}=\frac{P d_{p}\left(\mu \pi d_{p}-L \cos \alpha\right)}{2\left(\pi d_{p} \cos \alpha+\mu L\right)}+\mu_{c} P \frac{d_{c}}{2} (16.28)
T=58.42 \frac{0.01376-4.9182 \times 10^{-3}}{0.08884+7.62 \times 10^{-4}}+13.34=5.765+13.34=19.11 N m
\eta=\frac{P L}{2 \pi T}
\eta_{\text {screw }}=\frac{4000 \times 0.00508}{2 \pi 12.39}=0.261
\eta_{\text {both }}=\frac{4000 \times 0.00508}{2 \pi 25.73}=0.126
The design will be self-locking if
\mu \geq \frac{L}{\pi d_{p}} \cos \alpha=\frac{0.00508}{\pi 0.02921} \cos 14.5=0.05359m = 0.15 so the design is self-locking.