A worm and worm-wheel drive is chosen for a farm machine mainly because it:

gives a large speed reduction in a single stage and can be self-locking

has the highest mechanical efficiency of all gear drives

connects parallel shafts without axial thrust

allows the two shafts to run at the same speed

To measure the torque on a rotating shaft with strain gauges, the gauges are bonded:

at ±45° to the shaft axis, because pure torsion gives principal strains in those directions

parallel to the shaft axis, where the torsional strain is largest

around the circumference, to sense the change in diameter

on the bearing housing, since the shaft itself is rotating

Which of the following are overload safety devices used in farm machinery?

Spring-trip standard on a cultivator tine

Flywheel on a thresher cylinder

Machine Design for Farm Machinery: Gears, Belts, Chains, Overload Safety Devices and Measurement on Machine Elements

The first chapter of Section 2, Farm Machinery, of the GATE Agricultural Engineering (AG) paper. The section has two sub-headings that are really two subjects — the design and selection of the machine elements every implement is built from, and the implements themselves — so it is written as two chapters. This one covers the first: how power is carried from a tractor’s power take-off to a working part through gears, belts and pulleys, and chains and sprockets; the overload safety devices that stop a stone or a stump from wrecking a machine; the shafts, keys, couplings, bearings and knuckle joints that carry the loads; and how force, stress, torque, speed, displacement and acceleration are measured on those elements. Every formula is followed by the numerical GATE sets on it.

1. Power, torque and gear trains

Every drive calculation starts from P = 2πNT/60, with P in watts, N in rev/min and T in N·m. A tractor PTO delivering 30 kW at the standard 540 rev/min therefore carries T = 30 000 × 60/(2π × 540) = 530.5 N·m. For the same power, torque rises as speed falls — which is why a reduction gearbox needs a thicker output shaft than input shaft.

Two meshing gears must have the same module m = D/z (pitch-circle diameter over number of teeth) and hence the same circular pitch p = πm. Their pitch-line speeds are equal, so N₁D₁ = N₂D₂ and the velocity ratio is N₁/N₂ = z₂/z₁. In a simple train the idler gears change only the direction; the ratio is set by the first and last gears. In a compound train, two gears on one shaft turn together, and the overall ratio is the product of the stage ratios: N_out/N_in = (product of driver teeth)/(product of driven teeth).

Gear types used on farm machines
Type	Shafts	Typical use and property
Spur	parallel	gearboxes, metering drives; simple, noisy at high speed
Helical	parallel (or crossed)	smoother and quieter than spur, but produce axial thrust
Bevel	intersecting, usually at 90°	tractor differential and right-angle drives on rotavators
Worm and worm wheel	non-intersecting, at 90°	large reduction in one stage; can be self-locking when the lead angle is small
Rack and pinion	rotation to translation	steering, adjustment mechanisms

⚠️ Teeth and speed go the opposite way

A 20-tooth pinion at 1500 rev/min driving a 60-tooth gear gives 1500 × 20/60 = 500 rev/min, not 4500. The larger gear always turns slower, and the torque goes up by the same factor (less friction losses).

2. Belts and pulleys

Ignoring slip and belt thickness, a belt has one speed along its whole length, so N₁D₁ = N₂D₂. With belt thickness t the effective diameters are D + t; with a total slip of s per cent, N₂/N₁ = (D₁/D₂)(1 − s/100). Flat belts transmit power by friction on the rim; V-belts wedge into a grooved sheave, and the wedge action multiplies the normal force, so a V-belt transmits more power for the same tension and suits the short centre distances of threshers, chaff cutters and pumps.

At the point of impending slip the ratio of tight-side to slack-side tension is T₁/T₂ = e^(μθ) for a flat belt, where μ is the coefficient of friction and θ the angle of lap in radians on the smaller pulley. For a V-belt with groove half-angle β the ratio becomes e^(μθ/sin β), larger because sin β < 1. The power transmitted is P = (T₁ − T₂)v. At high speed the belt’s own mass adds a centrifugal tension T_c = mv² (m in kg per metre), which reduces the tension available for driving; the power is greatest when T_c equals one-third of the maximum allowable tension.

Worked: a flat belt with μ = 0.3 and θ = 180° (π rad) has T₁/T₂ = e^(0.3π) = 2.566. With T₁ = 1000 N the slack side carries 389.7 N, and at a belt speed of 10 m/s the power is (1000 − 389.7) × 10 = 6.10 kW.

⚠️ θ in radians, on the smaller pulley

Putting θ = 180 into e^(μθ) gives an absurd ratio of about e⁵⁴. And the lap that limits the drive is the one on the smaller pulley of an open belt, since that is where the belt slips first; a crossed belt has equal laps on both.

3. Chains and sprockets

A roller chain gives a positive (slip-free) drive, so the velocity ratio is exactly N₁/N₂ = z₂/z₁ in sprocket teeth. The chain speed is v = zpN/60 m/s, with p the chain pitch in metres, and the pitch-circle diameter of a sprocket is D = p/sin(180°/z). Because the chain wraps the sprocket as a polygon rather than a circle, its speed pulsates once per tooth — the polygonal (chordal) effect — and the pulsation shrinks as the number of teeth grows, which is why small sprockets with very few teeth are avoided on fast drives.

Chains dominate the ground-wheel drives of seed drills and planters, where the metering rate must stay in fixed proportion to forward travel, and the slow, high-torque drives of harvesters. Belts win where shock must be absorbed or a drive must slip safely under overload.

4. Overload safety devices

A field machine meets loads its designer cannot bound — a buried stone under a plough, a lump in a baler, a jammed thresher cylinder — so its drive line carries a deliberate weak link. The syllabus names these devices as a group, and each protects in a different way:

Shear pin (shear bolt): a pin sized to fail in shear at a set torque; cheap and certain, but the machine stops until the pin is replaced.
Friction slip clutch (torque limiter): spring-loaded friction plates that slip above a set torque and re-engage by themselves when the overload passes.
Ratchet (cam) type overload clutch: spring-loaded jaws that ride out of engagement and click, giving an audible warning.
Spring-trip and auto-reset standards on tillage tines and ploughs: the tine swings back over an obstruction against a spring and returns to depth afterwards.
Overrunning (free-wheel) clutch: lets a heavy rotor such as a rotary mower or a baler flywheel keep turning when the PTO stops, so its inertia does not drive the tractor.

Designing a shear pin is a single line of strength of materials. The pin sits at radius r from the shaft axis, so the shearing force at the design torque is F = T/r, and in single shear F = τ_u·(πd²/4). For T = 250 N·m at r = 50 mm, F = 5000 N; with an ultimate shear strength of 300 MPa, d = √(4F/(πτ_u)) = 4.61 mm. In double shear the area doubles, so the diameter falls by √2.

🎯 Why the shear pin uses the ultimate strength

Every other element is sized with a factor of safety so that it does not fail. The shear pin is the one element sized to fail — at the chosen torque, and before anything else — so it is designed on its ultimate shear strength with no safety factor.

5. Shafts, keys, couplings, bearings and knuckle joints; measuring what they carry

A solid shaft in pure torsion has maximum shear stress τ = 16T/(πd³). Most farm shafts also carry bending from pulley and gear loads, and the combined case is handled with the equivalent twisting moment Tₑ = √(M² + T²) (maximum-shear-stress theory, used with τ) and the equivalent bending moment Mₑ = ½[M + √(M² + T²)] (maximum-normal-stress theory, used with σ = 32Mₑ/(πd³)). With M = 3 kN·m and T = 4 kN·m, Tₑ = 5 kN·m and Mₑ = 4 kN·m.

A key transmits the shaft torque to the hub and is checked in shear (on the area width × length) and in crushing (on half its height × length). Couplings join shafts — rigid (flange, sleeve) where alignment is exact, flexible (bushed-pin, jaw) where it is not, and the universal (Hooke’s) joint of the PTO drive line where the shafts meet at an angle; a single universal joint makes the output speed fluctuate, and two joints phased correctly cancel it. A knuckle joint connects two rods in tension, as in tie rods and linkage, and allows angular movement in one plane; its pin is checked in double shear. Rolling bearings are selected by their dynamic load rating C: the rated life is L₁₀ = (C/P)³ million revolutions for ball bearings and (C/P)^(10/3) for roller bearings, where P is the equivalent load.

Measuring loads and motion on machine elements
Quantity	Transducer	Principle
Force, stress	strain gauges on a load cell or dynamometer ring	resistance change ΔR/R = GF·ε, read on a Wheatstone bridge
Torque	strain gauges on the shaft at ±45° to its axis; slip rings or telemetry	pure torsion produces principal strains along the 45° directions
Speed	tachometer, magnetic or optical pick-up, encoder	pulses per revolution counted against time
Displacement	LVDT, potentiometer	induced voltage or resistance proportional to position
Acceleration	piezoelectric accelerometer	charge proportional to the force on a seismic mass

🧠 Strain-gauge arithmetic

A gauge of resistance 120 Ω and gauge factor 2 under a strain of 500 microstrain changes by ΔR = GF·ε·R = 2 × 500 × 10⁻⁶ × 120 = 0.12 Ω. Changes that small are why the gauge is always read in a bridge, never with an ohmmeter.

Key takeaways

P = 2πNT/60 converts every drive: 30 kW at the 540 rev/min PTO speed is 530.5 N·m.
Gear speed ratio is the inverse of the teeth ratio, and a compound train multiplies its stage ratios.
T₁/T₂ = e^(μθ) for a flat belt, e^(μθ/sin β) for a V-belt, θ in radians; power is (T₁ − T₂)v.
A shear pin is sized to fail at a set torque on its ultimate strength: F = T/r = τ_u·πd²/4 in single shear.
Combined bending and torsion use Tₑ = √(M² + T²); a ball bearing lives (C/P)³ million revolutions.
Torque on a shaft is measured with strain gauges at 45° to the axis, read through a bridge.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A 20-tooth pinion running at 1500 rev/min drives a 60-tooth gear. What is the speed of the gear, in rev/min?
Numerical answer — type the value.
Show answer
Answer: 500
Pitch-line speeds are equal, so N₂ = N₁z₁/z₂ = 1500 × 20/60 = 500 rev/min. Multiplying by the teeth ratio instead of dividing gives 4500, which would make the larger gear spin faster — impossible for meshing gears.
In a compound gear train, gear A (20 teeth) drives gear B (50 teeth); gear C (15 teeth), on the same shaft as B, drives gear D (45 teeth). If A runs at 1800 rev/min, what is the speed of D, in rev/min?
Numerical answer — type the value.
Show answer
Answer: 240
N_D = N_A × (z_A/z_B) × (z_C/z_D) = 1800 × (20/50) × (15/45) = 1800 × 0.4 × 0.3333 = 240 rev/min. B and C share a shaft, so they turn at the same 720 rev/min; treating B as driving D directly (1800 × 20/45 = 800) ignores the second stage.
A flat belt has a coefficient of friction of 0.3 and an angle of lap of 180° on the smaller pulley. The tight-side tension is 1000 N and the belt speed is 10 m/s. Neglecting centrifugal tension, what power does it transmit, in kW (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 6.1
θ = π rad, so T₁/T₂ = e^(0.3π) = e^0.9425 = 2.566 and T₂ = 1000/2.566 = 389.7 N. P = (T₁ − T₂)v = 610.3 × 10 = 6103 W = 6.10 kW. Using θ = 180 in degrees makes T₂ vanish and gives 10 kW, and using T₁ alone (T₁v) gives the same wrong 10 kW.
A sprocket of 18 teeth on a roller chain of pitch 12.7 mm runs at 300 rev/min. What is the chain speed, in m/s (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 1.14
Each revolution passes z pitches of chain, so v = zpN/60 = 18 × 0.0127 × 300/60 = 1.143 m/s. Using πDN/60 with D taken as z × p (the polygon perimeter mistaken for a diameter) overstates it by a factor of π.
A tractor PTO delivers 30 kW at its standard speed of 540 rev/min. What torque does the PTO shaft carry, in N·m (to one decimal place)?
Numerical answer — type the value.
Show answer
Answer: 530.5
T = 60P/(2πN) = 60 × 30 000/(2π × 540) = 1 800 000/3392.9 = 530.5 N·m. Dropping the 2π gives 3333 N·m, and taking the 1000 rev/min PTO standard by mistake gives 286.5 N·m.
A shear pin placed at a radius of 50 mm from the shaft axis must fail in single shear when the torque reaches 250 N·m. The ultimate shear strength of the pin material is 300 MPa. What is the pin diameter, in mm (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 4.61
Shear force F = T/r = 250/0.05 = 5000 N. Area A = F/τ_u = 5000/(300 × 10⁶) = 1.667 × 10⁻⁵ m², so d = √(4A/π) = 4.607 × 10⁻³ m = 4.61 mm. Assuming double shear halves the area and gives 3.26 mm; applying a safety factor defeats the purpose of a pin meant to fail at 250 N·m.
A worm and worm-wheel drive is chosen for a farm machine mainly because it:
1. gives a large speed reduction in a single stage and can be self-locking
2. has the highest mechanical efficiency of all gear drives
3. connects parallel shafts without axial thrust
4. allows the two shafts to run at the same speed
Show answer
Answer: A — gives a large speed reduction in a single stage and can be self-locking
A single-start worm advances the wheel by one tooth per revolution, so ratios of 20:1 to 60:1 come in one stage, and with a small lead angle the wheel cannot drive the worm back. The price is sliding friction, which makes worm drives among the least efficient — the tempting "highest efficiency" option is the opposite of the truth.
To measure the torque on a rotating shaft with strain gauges, the gauges are bonded:
1. at ±45° to the shaft axis, because pure torsion gives principal strains in those directions
2. parallel to the shaft axis, where the torsional strain is largest
3. around the circumference, to sense the change in diameter
4. on the bearing housing, since the shaft itself is rotating
Show answer
Answer: A — at ±45° to the shaft axis, because pure torsion gives principal strains in those directions
Pure shear on the shaft surface has principal stresses of +τ and −τ at 45° to the axis, so gauges at ±45° in a full bridge read torsion while cancelling bending and axial strain. A gauge along the axis sees no normal strain from torsion at all; the rotating signal is brought out through slip rings or telemetry.
Which of the following are overload safety devices used in farm machinery?
1. Shear pin on a PTO drive line
2. Friction slip clutch
3. Spring-trip standard on a cultivator tine
4. Flywheel on a thresher cylinder
Show answer
Answer: A — Shear pin on a PTO drive line; B — Friction slip clutch; C — Spring-trip standard on a cultivator tine
The shear pin breaks, the slip clutch slips and the spring-trip standard swings back — each limits the load an obstruction can apply. A flywheel does the opposite: it stores energy to carry the cylinder through load peaks, and in a jam that stored energy adds to the damage unless an overload device is fitted.
A ball bearing has a dynamic load rating of 30 kN and carries an equivalent load of 3 kN at 1000 rev/min. What is its rated (L₁₀) life, in hours (to the nearest hour)?
Numerical answer — type the value.
Show answer
Answer: 16667
L₁₀ = (C/P)³ = (30/3)³ = 1000 million revolutions. At 1000 rev/min that is 10⁹/(1000 × 60) = 16667 h. Using the roller-bearing exponent 10/3 gives 2154 million revolutions, and forgetting the cube gives only 10 million.
A strain gauge of resistance 120 Ω and gauge factor 2.0 is bonded to a shaft that is strained by 500 microstrain. What is the change in its resistance, in ohms?
Numerical answer — type the value.
Show answer
Answer: 0.12
GF = (ΔR/R)/ε, so ΔR = GF·ε·R = 2.0 × 500 × 10⁻⁶ × 120 = 0.12 Ω. Treating 500 microstrain as 500 × 10⁻³ gives 120 Ω — a hundredfold-plus error that would mean the gauge doubled its resistance.
For a V-belt with groove half-angle β, the ratio of tight-side to slack-side tension at the point of slipping is:
1. e^(μθ)
2. e^(μθ sin β)
3. e^(μθ/sin β)
4. e^(μθ/cos β)
Show answer
Answer: C — e^(μθ/sin β)
The wedge in the groove raises the normal reaction by 1/sin β, so the effective coefficient becomes μ/sin β and T₁/T₂ = e^(μθ/sin β), larger than the flat-belt e^(μθ). Multiplying by sin β would make a V-belt grip less than a flat belt, the reverse of why V-belts are used.
A shaft carries a bending moment of 3 kN·m and a torque of 4 kN·m. What is the equivalent twisting moment, in kN·m?
Numerical answer — type the value.
Show answer
Answer: 5
Tₑ = √(M² + T²) = √(9 + 16) = 5 kN·m. The equivalent bending moment is a different quantity, ½(M + Tₑ) = 4 kN·m, and simply adding M + T gives 7 kN·m, which overstates the combined effect.