Exercise set

Power Transmission Gears, Belts, and Drives Exercises

Worked power transmission exercises for torque, gear ratios, mesh forces, belt tensions, overhung loads, reflected inertia, brakes, screw drives and torque limiters.

Branch: Mechanical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on power transmission: torque, speed, gears, belts, pulleys, drives, reflected inertia, hoists, brakes, power screws and torque limiters. The engineering task is to keep the power path traceable from motor to load and to check abnormal cases such as startup, braking, jam, overload and degraded lubrication.

Use these calculations as first-pass design and commissioning screens. Final release needs supplier ratings, guards, lockout provisions, lubrication evidence, thermal checks, vibration tests, braking tests, speed verification and maintenance access review.

Release Evidence Notes

A power-transmission calculation should state the operating case. Nominal running torque is rarely the only design point. Startup, acceleration, jam recovery, emergency stop, reversing, overload, belt slip, gearbox thermal loss and brake energy often control the release decision. Connect each calculation to a measured speed, load, current, temperature, vibration or stopping-distance test.

Engineering Boundary Notes

A gear, belt, coupling, brake, screw or torque limiter is part of a drive train. Changing ratio changes torque, speed, reflected inertia and thermal loss. A component can pass nominal torque while the drive fails during acceleration, braking or jam protection.

Common Release Mistakes

using motor nameplate power without service factor or startup torque;
confusing speed ratio with torque ratio direction;
checking gear tangential force but ignoring radial load on bearings;
sizing belt tension without overhung shaft load;
accepting brake holding torque without stopping-energy capacity;
setting a torque limiter above the downstream damage threshold;
omitting validation at worst credible load and speed.

Scenario Map

Scenario	Main calculation	Release decision
Power and speed	$T=9550P/n$	Establish design torque.
Gear reducer	ratio, output speed and torque	Select ratio and check loss.
Gear mesh	tangential and radial force	Size bearings and shafts.
Belt drive	tight/slack tension and overhung load	Check pulley and bearing loads.
Dynamic drive	reflected inertia and acceleration torque	Set motor and drive limits.
Safety function	hoist brake, screw and torque limiter	Accept, derate or block release.

Validation Package Checklist

motor speed, load torque and service factor basis;
gearbox ratio, efficiency, lubrication and thermal rating;
gear mesh, pulley and chain load paths into bearings;
acceleration, braking, reversing and jam cases;
brake stopping-energy and holding-torque evidence;
torque-limiter setting, repeatability and downstream damage threshold;
guarded commissioning test at expected load and speed.

Exercise 1: Torque From Power and Speed

A machine transmits:

P=18\ \text{kW}

at:

n=450\ \text{rpm}

Compute running torque using:

\displaystyle T=\frac{9550P}{n}

Solution

\displaystyle T=\frac{9550(18)}{450}=382\ \text{N m}

Engineering Comment

This is nominal running torque. Startup, shock and service factor should be applied before sizing downstream parts.

Plausibility Check

Lower speed at the same power gives higher torque, so hundreds of N m is plausible.

Exercise 2: Service-Factor Design Torque

Use nominal torque $382\ \text{N m}$ with service factor:

K_s=1.6

Compute design torque.

Solution

T_d=K_sT=1.6(382)=611\ \text{N m}

Engineering Comment

Use design torque for gears, belts, shafts, keys and torque limiter checks unless another abnormal case is more severe.

Plausibility Check

The service factor increases torque by $60\%$ .

Exercise 3: Gear Ratio and Output Speed

A motor runs at $1470\ \text{rpm}$ and drives a reducer with ratio $i=4.2:1$ . Compute output speed.

Solution

\displaystyle n_{out}=\frac{1470}{4.2}=350\ \text{rpm}

Engineering Comment

State the ratio convention. Confusing input/output ratio reverses speed and torque conclusions.

Plausibility Check

A reduction ratio greater than one lowers output speed.

Exercise 4: Gearbox Output Torque

Input power is $12\ \text{kW}$ , output speed is $350\ \text{rpm}$ and gearbox efficiency is $92\%$ . Compute output torque.

Solution

Output power:

P_o=0.92(12)=11.04\ \text{kW}

\displaystyle T_o=\frac{9550(11.04)}{350}=301\ \text{N m}

Engineering Comment

Efficiency loss becomes heat in the gearbox. Thermal rating should be checked, not only torque.

Plausibility Check

The output torque is higher than motor-speed torque because the reducer lowers speed.

Exercise 5: Gear Mesh Tangential Force

A spur gear transmits $T=300\ \text{N m}$ at pitch diameter:

d_p=0.160\ \text{m}

Compute tangential tooth force.

Solution

\displaystyle F_t=\frac{2T}{d_p}=\frac{2(300)}{0.160}=3750\ \text{N}

Engineering Comment

This force loads teeth, shafts and bearings. It is not only a gear-tooth stress input.

Plausibility Check

Torque divided by radius gives force; $300\ \text{N m}$ at $0.08\ \text{m}$ radius gives $3750\ \text{N}$ .

Exercise 6: Gear Radial Force

The spur gear pressure angle is $20^\circ$ and tangential force is $3750\ \text{N}$ . Estimate radial force:

F_r=F_t\tan\phi

Solution

F_r=3750\tan20^\circ=1365\ \text{N}

Engineering Comment

The radial force drives bearing load and shaft bending. It must be included in component checks.

Plausibility Check

Radial force is smaller than tangential force for a $20^\circ$ pressure angle.

Exercise 7: Belt Tension Difference

A belt pulley transmits $T=180\ \text{N m}$ with pulley diameter $0.30\ \text{m}$ . Compute tension difference.

Solution

T=(F_1-F_2)r

\displaystyle F_1-F_2=\frac{180}{0.15}=1200\ \text{N}

Engineering Comment

The difference transmits torque, but the sum of belt tensions creates overhung bearing load.

Plausibility Check

A smaller pulley would require a larger tension difference for the same torque.

Exercise 8: Tight and Slack Belt Tensions

For the belt in Exercise 7, assume tension ratio:

\displaystyle \frac{F_1}{F_2}=3.0

Find $F_1$ and $F_2$ .

Solution

F_1-F_2=1200

3F_2-F_2=1200

F_2=600\ \text{N},\quad F_1=1800\ \text{N}

Engineering Comment

High tight-side tension may control shaft bending, bearing life and belt life.

Plausibility Check

The tight side is three times the slack side and the difference is $1200\ \text{N}$ .

Exercise 9: Overhung Pulley Load

Use $F_1=1800\ \text{N}$ and $F_2=600\ \text{N}$ . Estimate radial pulley load as:

F_p=F_1+F_2

Solution

F_p=1800+600=2400\ \text{N}

Engineering Comment

The pulley radial load can govern shaft bending and bearing life even when torque capacity passes.

Plausibility Check

The bearing load is larger than the torque-transmitting tension difference.

Exercise 10: Reflected Inertia

A load inertia is:

J_L=0.80\ \text{kg m}^2

It is driven through a $4:1$ speed reduction. Reflect inertia to the motor shaft:

\displaystyle J_{ref}=\frac{J_L}{i^2}

Solution

\displaystyle J_{ref}=\frac{0.80}{4^2}=0.050\ \text{kg m}^2

Engineering Comment

Reduction makes the load easier for the motor to accelerate, but output torque and gearbox rating must still be checked.

Plausibility Check

Reflected inertia falls with the square of ratio.

Exercise 11: Acceleration Torque

Total reflected inertia at the motor is $0.065\ \text{kg m}^2$ . Required angular acceleration is $90\ \text{rad/s}^2$ . Compute acceleration torque.

Solution

T_{acc}=J\alpha=0.065(90)=5.85\ \text{N m}

Engineering Comment

Add this to load torque during acceleration. A high-inertia start may require a different drive limit than steady running.

Plausibility Check

The torque is modest because reflected inertia is small.

Exercise 12: Hoist Drum Torque

A hoist lifts $m=650\ \text{kg}$ on a drum radius $r=0.18\ \text{m}$ . Compute load torque at the drum.

Solution

F=mg=650(9.81)=6377\ \text{N}

T=Fr=6377(0.18)=1148\ \text{N m}

Engineering Comment

Holding brakes, gearbox output shaft and drum attachment must all be checked at this torque with safety factor.

Plausibility Check

Hundreds of kilograms on a sub-metre drum gives more than $1000\ \text{N m}$ .

Exercise 13: Brake Holding Torque

Use hoist drum torque $1148\ \text{N m}$ . The gearbox ratio from brake shaft to drum is $20:1$ and efficiency during holding is conservatively $0.85$ . Required safety factor is $1.8$ . Estimate brake torque.

Solution

Brake-shaft load torque:

\displaystyle T_b=\frac{1148}{20(0.85)}=67.5\ \text{N m}

With safety factor:

T_{b,req}=1.8(67.5)=121.5\ \text{N m}

Engineering Comment

Brake torque should be validated with wear, temperature, voltage loss and contaminated friction surfaces considered.

Plausibility Check

The gearbox ratio reduces brake-shaft torque, then safety factor raises the requirement.

Exercise 14: Braking Energy

A rotating load has inertia $J=1.2\ \text{kg m}^2$ and speed $900\ \text{rpm}$ . Compute kinetic energy to stop.

Solution

\displaystyle \omega=900\frac{2\pi}{60}=94.25\ \text{rad/s}

\displaystyle E=\frac{1}{2}J\omega^2=0.5(1.2)(94.25^2)=5330\ \text{J}

Engineering Comment

The brake must handle both torque and energy. Repeated stops can overheat a brake that passes holding torque.

Plausibility Check

Energy scales with speed squared, so high-speed stops are thermally important.

Exercise 15: Power Screw Raising Torque

A screw lift must raise axial load $F=12\ \text{kN}$ with lead $l=5\ \text{mm/rev}$ . Ideal torque without friction is:

\displaystyle T=\frac{Fl}{2\pi}

Compute ideal torque.

Solution

\displaystyle T=\frac{12000(0.005)}{2\pi}=9.55\ \text{N m}

Engineering Comment

Real screw torque can be much higher because friction dominates. Check efficiency and self-locking separately.

Plausibility Check

A fine lead gives large mechanical advantage and low ideal torque.

Exercise 16: Screw Efficiency Correction

Use ideal torque $9.55\ \text{N m}$ and screw efficiency $32\%$ . Estimate input torque.

Solution

\displaystyle T_{in}=\frac{9.55}{0.32}=29.8\ \text{N m}

Engineering Comment

Low screw efficiency creates heat and affects motor sizing. Backdriving and holding safety must be reviewed.

Plausibility Check

Because efficiency is below one, input torque exceeds ideal torque.

Exercise 17: Torque-Limiter Setting

A conveyor normally needs $240\ \text{N m}$ . Startup peak is $310\ \text{N m}$ . Downstream damage is expected above $470\ \text{N m}$ . Choose a tentative limiter setting with $15\%$ margin above startup but below damage threshold.

Solution

Startup margin setting:

T_s=1.15(310)=356.5\ \text{N m}

This is below damage threshold:

356.5<470\ \text{N m}

Engineering Comment

The setting transmits startup torque while protecting the downstream path. Repeatability and slip heating must be tested.

Plausibility Check

The chosen setting is above normal and startup torque, but below the damage threshold.

Exercise 18: Drive Release Gate

A drive has design torque $611\ \text{N m}$ , gearbox output rating $700\ \text{N m}$ , pulley bearing load limit $2600\ \text{N}$ against predicted $2400\ \text{N}$ , brake energy rating $6000\ \text{J}$ against predicted $5330\ \text{J}$ and torque-limiter setting $356.5\ \text{N m}$ below a $470\ \text{N m}$ damage threshold. Decide release status.

Solution

Torque margin:

700-611=89\ \text{N m}

Pulley load margin:

2600-2400=200\ \text{N}

Brake energy margin:

6000-5330=670\ \text{J}

Limiter protection margin:

470-356.5=113.5\ \text{N m}

Engineering Comment

All simplified gates pass, but pulley and brake margins are not large. Release should require load test, brake temperature check and limiter slip repeatability evidence.

Plausibility Check

Every margin is positive, so the correct decision is conditional release rather than rejection.

REF