Question 2 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
National Exams — May 2016 · 07-Mec-B1 Advanced Machine Design.
Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; candidates answer any three of the four Part II problems (3–6). For study value, complete worked solutions to all six problems are provided below.
Check — assumptions stated per the exam rubric.
(1) “ton” is read as the US short ton (2000 lb), consistent with the inch/ft·min/hp unit set of Problem 2.
(2) In Problem 3 the concentrated couple Mz is taken counter-clockwise (out of the page, per the dot symbol in the figure); a clockwise reading would give RA = 0, RB = 9.5 kN.
(3) In Problem 5 the fluctuating torque is assumed transmitted over the 18 in from the drive end to the load; no stress concentration is used as the problem directs.
Question 2: Twin Acme Power Screws Raising a Sluice Gate (30 marks)
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.
