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

Question

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.

Accepted Answer

Single start thread, so the lead, L, is equal to the pitch, P.

N = 5 teeth per inch,

[latex]P=\frac{1}{N}=\frac{1}{5}=0.2 ext { in. }[/latex]

L = 0.2 in. = 5.08 mm,
[latex]d_p = 1.15 in. = 29.21 mm[/latex],

[latex]d_c = 1.75 in. = 44.45 mm[/latex].

Assume sliding friction,µ = 0.15.

The torque to lift the load is given by Eqn (16.27).

[latex]T_{u}=\frac{P d_{p}\left(\mu \pi d_{p}+L \cos \alpha ight)}{2\left(\pi d_{p} \cos \alpha-\mu L ight)}+\mu_{c} P \frac{d_{c}}{2}[/latex] (16.27)

[latex]T=\frac{4000 imes 0.02921}{2}\left(\frac{0.15 \pi 0.02921+0.00508 \cos 14.5}{\pi 0.02921 \cos 14.5-0.15 imes 0.00508} ight)+0.15 imes 4000 \frac{0.04445}{2}[/latex]

[latex]=58.42 \frac{0.01376+4.9182 imes 10^{-3}}{0.08884-7.62 imes 10^{-4}}+13.34=12.39+13.34=25.73 N m[/latex]

The torque to lower the load is given by Eqn (16.28),

[latex]T_{d}=\frac{P d_{p}\left(\mu \pi d_{p}-L \cos \alpha ight)}{2\left(\pi d_{p} \cos \alpha+\mu L ight)}+\mu_{c} P \frac{d_{c}}{2}[/latex] (16.28)

[latex]T=58.42 \frac{0.01376-4.9182 imes 10^{-3}}{0.08884+7.62 imes 10^{-4}}+13.34=5.765+13.34=19.11 N m[/latex]

[latex]\eta=\frac{P L}{2 \pi T}[/latex]

[latex]\eta_{ ext {screw }}=\frac{4000 imes 0.00508}{2 \pi 12.39}=0.261[/latex]

[latex]\eta_{ ext {both }}=\frac{4000 imes 0.00508}{2 \pi 25.73}=0.126[/latex]

The design will be self-locking if

[latex]\mu \geq \frac{L}{\pi d_{p}} \cos \alpha=\frac{0.00508}{\pi 0.02921} \cos 14.5=0.05359[/latex]

m = 0.15 so the design is self-locking.

Question 16.3: A self-locking power screw is required for a screw jack. An ...

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