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 1.5 kW motor running at 960 rpm drives a pinion with 18 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.
In a planetary gear set the sun gear has 20 teeth and each planet gear has 30 teeth. The planet carrier is held fixed and the sun is driven at 1800 rpm. Find the speed nr of the ring gear (magnitude).
A hollow shaft must carry a torque of 100 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.
A solid aluminium shaft of diameter 25 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 1.5 m from A and 0.8 m from B. Find the angle of rotation φC of section C, in degrees.
A point in a loaded part has σₓ = 140 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = 80 MPa and τₓᵧ = 30 MPa (σz = 0). Find the von Mises equivalent stress σvM.
A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 70 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
A 2 kg mass is supported by a spring with a stiffness of 25 kN/m and a viscous damper with a damping coefficient of 20 N·s/m. Find the damping ratio ζ.
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 40 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
Answer: Ft = 552.6 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Answer: nr = 450 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
With nc = 0 the sun drives the ring through the planets, which act as idlers.
The ring turns opposite to the sun.
Answer: do = 22.7 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.
A solid shaft would need d = 21.68 mm; the hollow one is a little larger but much lighter.
Answer: φC = 7.495 °
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Use segment AC: it is fixed at A and carries TA.
Answer: θp = 10.9 °
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.
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
Positive angles are counter-clockwise.
Answer: σvM = 140 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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Answer: δ = 0.05563 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 = 222.7 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 = 78.48 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: σcr = 4.441 MPa
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: ζ = 0.04472
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: keq = 13.33 kN/m
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.
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
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