Twin Acme Power Screws Raising a Sluice Gate: 25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1), May 2016, Q2

Question 2: Twin Acme Power Screws Raising a Sluice Gate (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two 3 in single-start Acme screws sharing a 50 ton gate; track friction $\pm 2$ ton; screw friction $\mu=0.1$; collar $d_c=5$ in, $\mu_c=0.03$; lift speed $v=2$ ft/min. From Table 14-3, the 3.000 in Acme has 2 threads/in.

Given data (Problem 2)
Quantity	Symbol	Value
Threads per inch (3 in Acme)	—	2 → pitch $p=0.5$ in
Lead (single start)	$l$	$0.5$ in
Mean diameter $=d-p/2$	$d_m$	$2.75$ in
Acme half-angle	$\alpha$	$14.5^\circ$ ($\sec\alpha=1.033$)
Screw / collar friction	$\mu,\ \mu_c$	$0.10,\ 0.03$
Collar diameter	$d_c$	$5$ in
Load per screw — raise / lower	$F_R,\ F_L$	$52{,}000$ / $48{,}000$ lb

Find. Raising and lowering torque per screw, the screw rotational speed for a 2 ft/min gate, and the motor horsepower per screw to raise.

Approach. Split the 50 ton weight (plus/minus 2 ton track friction) equally between the two screws, then apply the Acme raising/lowering torque relations (thread term with the $\sec\alpha$ Acme correction, plus a collar-friction term); the lead converts lift speed to rev/min, and $\text{hp}=T\,n/63025$.

Load carried by each screw. The gate weight is shared by two screws, and track friction acts on the whole gate. Raising adds 2 ton, lowering removes 2 ton: $$F_R=\tfrac{1}{2}(50+2)(2000)=52{,}000\ \text{lb},\qquad F_L=\tfrac{1}{2}(50-2)(2000)=48{,}000\ \text{lb}.$$
Thread geometry. For a single-start 3 in Acme, $p=1/2=0.5$ in so the lead $l=p=0.5$ in, and the mean diameter $d_m=d-p/2=3.0-0.25=2.75$ in. The Acme correction uses the normal-plane half-angle $\alpha=14.5^\circ$, $\sec\alpha=1.033$.
Raising torque per screw. Summing the thread lifting torque and the collar drag, $$T_R=\frac{F_R d_m}{2}\!\left(\frac{l+\pi\mu d_m\sec\alpha}{\pi d_m-\mu l\sec\alpha}\right)+\frac{F_R\mu_c d_c}{2}.$$ Substituting $F_R=52{,}000$ lb, $$T_R=71{,}500\!\left(\frac{0.5+0.864}{8.588}\right)+3900 = 11{,}593+3900=\boxed{1.55\times10^{4}\ \text{lb}\!\cdot\!\text{in}}\ (\approx 15{,}490\ \text{lb}\!\cdot\!\text{in}).$$
Lowering torque per screw. With the thread term reversed, $$T_L=\frac{F_L d_m}{2}\!\left(\frac{\pi\mu d_m\sec\alpha-l}{\pi d_m+\mu l\sec\alpha}\right)+\frac{F_L\mu_c d_c}{2}.$$ Substituting $F_L=48{,}000$ lb gives $T_L = 2980 + 3600 = \boxed{6.58\times10^{3}\ \text{lb}\!\cdot\!\text{in}}$. Because $\pi\mu d_m\sec\alpha = 0.864 \gt l = 0.5$, the thread term stays positive: the screw is self-locking, so a driving torque (not a holding brake) is needed to lower — a desirable safety feature for a gate.
Screw rotational speed. Each revolution advances the gate one lead. For $v=2$ ft/min $=24$ in/min, $$n=\frac{v}{l}=\frac{24}{0.5}=\boxed{48\ \text{rev/min}}.$$
Motor horsepower per screw (raising). Using $\text{hp}=T_R\,n/63025$ with $T_R$ in lb·in, $$\text{hp}=\frac{15{,}490\times 48}{63025}=\boxed{11.8\ \text{hp per screw}}.$$ Equivalently $T_R\omega$ with $\omega=48(2\pi/60)=5.03$ rad/s gives $8.75$ kW.

Final results — Problem 2
Quantity	Value
Raising torque per screw, $T_R$	$\approx 1.55\times10^{4}$ lb·in (1290 lb·ft)
Lowering torque per screw, $T_L$	$\approx 6.58\times10^{3}$ lb·in (self-locking)
Screw speed, $n$	48 rev/min
Motor power per screw (raise)	11.8 hp (8.75 kW)

Topic: Acme power-screw torque with collar friction.

Key relations: $T_R=\frac{Fd_m}{2}\frac{l+\pi\mu d_m\sec\alpha}{\pi d_m-\mu l\sec\alpha}+\frac{F\mu_c d_c}{2}$; self-locking when $\pi\mu d_m\sec\alpha \gt l$; $n=v/l$; $\text{hp}=Tn/63025$.

Why this works: The thread is an inclined plane wrapped on a cylinder; the raising torque climbs the plane against friction while the lowering torque descends with friction opposing motion. The $\sec\alpha$ factor accounts for the Acme flank angle pressing the nut harder than a square thread. The collar carries the full axial thrust on a separate friction face, added as $F\mu_c d_c/2$.

Common pitfall: Forgetting to split the load between the two screws, or omitting the $\sec\alpha$ Acme correction (which under-predicts the required torque). Also, the track friction acts on the whole gate, not per screw.

Source: Shigley’s Mechanical Engineering Design, §8-2 (power screws, Eqs. 8-1 to 8-7); Juvinall & Marshek, Ch. 10.

Question 2 of 6: Twin Acme Power Screws Raising a Sluice Gate

Question 2: Twin Acme Power Screws Raising a Sluice Gate (30 marks)