Question 6 of 6: Twin Acme Power Screws Raising a Sluice Gate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, 16-Mec-B1 Advanced Machine Design, May 2018. Open book, 3 hours, 100 marks. Part I (Problems 1 & 2) is compulsory; candidates answer only three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Find. (a) raising and lowering torque per screw; (b) screw rotation speed; (c) motor horsepower per screw to raise.
One of two Acme power screws: the gate load plus track friction is shared by the two screws; each also overcomes collar-bearing friction at $d_c=5$ in.
Approach. Split the 50-ton gate (with $\pm2$ tons track friction) between the two screws, apply the Acme raising and lowering torque formulas (with the thread-angle secant correction) plus the collar term, get the speed from lead and lift rate, and convert raising torque × speed to horsepower.
Load per screw. Raising, the gate weight and track friction add: $(50+2)/2=26\ \text{t}=52\,000\ \text{lb}$. Lowering, track friction subtracts: $(50-2)/2=24\ \text{t}=48\,000\ \text{lb}$.
Raising torque. For an Acme thread with half-angle $\alpha=14.5^\circ$ ($\sec\alpha=1.033$) and collar term,
$$T_R=\frac{F d_m}{2}\!\left(\frac{l+\pi\mu d_m\sec\alpha}{\pi d_m-\mu l\sec\alpha}\right)+\frac{F\mu_c d_c}{2}=\boxed{15\,493\ \text{lb}\cdot\text{in}}.$$
Lowering torque. The thread term changes sign (friction now opposes descent):
$$T_L=\frac{F d_m}{2}\!\left(\frac{\pi\mu d_m\sec\alpha-l}{\pi d_m+\mu l\sec\alpha}\right)+\frac{F\mu_c d_c}{2}=6580\ \text{lb}\cdot\text{in}.$$
Since $\pi\mu d_m\sec\alpha=0.864\gt l=0.5$, the thread term stays positive — the screw is self-locking (it will not back-drive under load).
Rotation speed. Each revolution advances the gate one lead, $l=0.5\ \text{in}$:
$$N=\frac{v}{l}=\frac{24\ \text{in/min}}{0.5\ \text{in/rev}}=\boxed{48\ \text{rpm}}.$$
Motor horsepower to raise (per screw). With $T_R$ in lb·in and $N$ in rpm,
$$\text{hp}=\frac{T_R\,N}{63\,025}=\frac{15\,493\times48}{63\,025}=\boxed{11.8\ \text{hp/screw}}.$$
Problem 6 results
Quantity
Value
Load per screw (raise / lower)
$52\,000$ / $48\,000\ \text{lb}$
Raising torque $T_R$
$15\,493\ \text{lb}\cdot\text{in}$
Lowering torque $T_L$
$6\,580\ \text{lb}\cdot\text{in}$ (self-locking)
Screw speed $N$
$48\ \text{rpm}$
Motor power to raise
$11.8\ \text{hp/screw}$
Check: a “ton” is taken as the US short ton (2000 lb), consistent with the imperial screw sizes; using a metric tonne would scale every force by 1.10. The collar friction adds $F\mu_c d_c/2$ to both directions and is a significant share of $T_R$.
