Question 6 of 6: Twin Acme power screws raising a sluice gate — torque, speed and power (20 marks)

Question 6: Twin Acme power screws raising a sluice gate — torque, speed and power (20 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 screws share the gate load; the thread geometry comes from the Acme table on page 4 of the paper, whose 3.000 in row reads 2 threads per inch, pitch 0.500 in, pitch diameter 2.750 in:

Find. (a) The raising and lowering torque per screw; (b) the screw rotational speed; (c) the motor power per screw for raising.

Approach. Work out the axial load carried by one screw in each direction of travel, take the lead from the thread table, apply the Acme power-screw torque relations with the thread half-angle correction, add the collar friction torque separately, then get the speed from the lead and the travel rate and the power from raising torque times angular velocity.

1. Axial load on one screw, each direction. Track friction opposes motion, so it adds to the load being lifted and subtracts from the load being lowered. Sharing equally between the two screws: $$F_{\text{raise}}=\frac{100\,000+4000}{2}=52\,000\ \text{lb}\ \left(26\ \text{ton}\right),\qquad F_{\text{lower}}=\frac{100\,000-4000}{2}=48\,000\ \text{lb}\ \left(24\ \text{ton}\right)$$ Equal sharing assumes the two screws are driven in synchronism and the gate stays square in its guides; a jammed or mistimed gate would load one screw far more heavily, which is why such installations use a mechanical tie shaft or electrically synchronised drives.
2. Thread geometry from the table. Single-threaded means the lead equals the pitch, and the 3.000 in Acme row gives 2 threads per inch: $$p=\frac{1}{2}=0.500\ \text{in},\qquad l=p=0.500\ \text{in},\qquad d_m=2.750\ \text{in}$$ The tabulated pitch diameter is consistent with $d_m=d-p/2=3.000-0.250=2.750$ in, a useful check that the right table row has been read. The lead angle is $$\lambda=\arctan\frac{l}{\pi d_m}=\arctan\frac{0.500}{\pi\left(2.750\right)}=3.31^\circ$$ so this is a shallow, high-mechanical-advantage screw.
3. The Acme half-angle correction. On a square thread the normal force is purely axial, but an Acme thread is flanked at $\alpha=14.5^\circ$, so the normal force is inclined and the friction force it generates is larger by $1/\cos\alpha$: $$\sec\alpha=\frac{1}{\cos14.5^\circ}=1.0326$$ The recurring friction group in both torque expressions is therefore $$\pi\mu d_m\sec\alpha=\pi\left(0.1\right)\left(2.750\right)\left(1.0326\right)=0.8921\ \text{in}$$
4. Raising torque per screw. The thread torque plus the collar torque: $$T_R=\frac{F\,d_m}{2}\!\left[\frac{l+\pi\mu d_m\sec\alpha}{\pi d_m-\mu l\sec\alpha}\right]+\frac{F\mu_cd_c}{2}$$ $$T_{R,\text{thread}}=\frac{52\,000\left(2.750\right)}{2}\left[\frac{0.500+0.8921}{\pi\left(2.750\right)-0.1\left(0.500\right)\left(1.0326\right)}\right]=71\,500\left[\frac{1.3921}{8.5878}\right]=11\,591\ \text{lb}\cdot\text{in}$$ $$T_{R,\text{collar}}=\frac{52\,000\left(0.03\right)\left(5.0\right)}{2}=3900\ \text{lb}\cdot\text{in}$$ $$\boxed{\ T_R=11\,591+3900=15\,491\ \text{lb}\cdot\text{in}=1291\ \text{lb}\cdot\text{ft}\ \left(1750\ \text{N}\cdot\text{m}\right)\ }$$ The collar accounts for a quarter of the total even at $\mu_c=0.03$, purely because its 5 in effective diameter is nearly twice the thread’s mean diameter — a plain bronze collar at $\mu_c=0.15$ would add 19 500 lb$\cdot$in and more than double the torque, which is exactly why roller thrust bearings are specified.
5. Lowering torque per screw. Lowering reverses the direction of the thread-friction term while the collar term, which depends only on the magnitude of the thrust, keeps its sign: $$T_L=\frac{F\,d_m}{2}\!\left[\frac{\pi\mu d_m\sec\alpha-l}{\pi d_m+\mu l\sec\alpha}\right]+\frac{F\mu_cd_c}{2}$$ $$T_{L,\text{thread}}=\frac{48\,000\left(2.750\right)}{2}\left[\frac{0.8921-0.500}{8.6394+0.0516}\right]=66\,000\left[\frac{0.3921}{8.6910}\right]=2978\ \text{lb}\cdot\text{in}$$ $$T_{L,\text{collar}}=\frac{48\,000\left(0.03\right)\left(5.0\right)}{2}=3600\ \text{lb}\cdot\text{in}$$ $$\boxed{\ T_L=2978+3600=6578\ \text{lb}\cdot\text{in}=548\ \text{lb}\cdot\text{ft}\ \left(743\ \text{N}\cdot\text{m}\right)\ }$$
6. Confirm the screws are self-locking. The bracketed thread term for lowering came out positive, which is the algebraic signature of self-locking: torque must be supplied to bring the gate down, so the gate cannot run away under its own weight. The direct test is $$\pi\mu d_m\sec\alpha=0.8921\ \text{in}\ \gt\ l=0.500\ \text{in}\qquad\text{(equivalently }\mu\sec\alpha=0.1033\gt\tan\lambda=0.0579\text{)}$$ so the screws hold the gate at any position with the motors de-energised and no separate holding brake is needed — an important safety property for a dam gate. The thread efficiency when raising is correspondingly low: $$\eta=\frac{Fl}{2\pi T_R}=\frac{52\,000\left(0.500\right)}{2\pi\left(15\,491\right)}=26.7\ \%$$ which is normal and indeed necessary for a self-locking screw: any screw with $\eta$ above 50 % will overhaul.
7. Rotational speed. One turn advances the nut by one lead, so $$n=\frac{v}{l}=\frac{24\ \text{in/min}}{0.500\ \text{in/rev}}=\boxed{\ 48\ \text{rev/min}\ }$$ This is slow enough to be reached with a worm or helical gearbox from a standard 1750 rev/min motor at a ratio of about 36:1.
8. Motor power per screw for raising. Using the customary US horsepower relation with torque in lb$\cdot$in and speed in rev/min: $$\text{hp}=\frac{T_Rn}{63\,025}=\frac{15\,491\times48}{63\,025}=\boxed{\ 11.8\ \text{hp per screw}\ \left(8.8\ \text{kW}\right)\ }$$ Cross-checking from first principles, $\omega=2\pi\left(48\right)/60=5.027$ rad/s and $T_R\omega=15\,491\times5.027/12=6489\ \text{ft}\cdot\text{lb/s}$, which divided by 550 gives 11.8 hp — the same figure. Total installed shaft power is therefore about 23.6 hp for the pair, before gearbox losses and motor service factor; specifying two 15 hp motors would leave sensible margin for starting the gate from rest and for a partially silted track.