Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
A belt runs on a 120 mm driving pulley at 2880 rpm. The tension is 1000 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.
In a compound gear train, gear A (18 teeth) on the input shaft turning at 3000 rpm drives gear B (36 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (40 teeth) on the output shaft. Find the output speed nout.
In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 40 N·m. The two ends of a gauge length of 250 mm rotate 7.22° relative to each other (the material is still elastic). Find the shear modulus G of the material.
A solid brass shaft transmits 50 kW at 500 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.
A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = −50 MPa and τₓᵧ = 30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −40 MPa and τₓᵧ = −50 MPa (σz = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 70 mm, E = 120 GPa). Both are 300 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 35 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
A 25 kg mass is supported by a spring with a stiffness of 15 kN/m and a viscous damper with a damping coefficient of 60 N·s/m. Find the damping ratio ζ.
A 5 kg mass hangs from spring k₃ = 5 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 40 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
Answer: P = 12.67 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Answer: nout = 450 rpm
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
total ratio = 6.667 : 1
Answer: G = 38.98 GPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
Answer: d = 73.12 mm
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Both limits must hold, so the larger diameter governs.
Here stiffness governs.
Answer: τabs = 71.59 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.
Answer: n = 2.739
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
Answer: δ = 0.04416 mm
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Answer: Pmax = 686.4 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Answer: Pcr = 313.9 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
A round section has the same I about every axis.
Answer: d = 93.8 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Check: the slenderness KL/r is 213.2, so the column is slender and Euler applies.
Answer: ζ = 0.04899
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
ζ < 1, so the system is underdamped and oscillates as it decays.
Answer: fn = 4.799 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
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