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 4 kW motor running at 1450 rpm drives a pinion with 16 teeth, which meshes with a gear with 50 teeth. Ignore losses. Find the torque T1 on the motor shaft.
An open belt connects a 100 mm driving pulley at 720 rpm to a 400 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 50 mm and length 1.2 m; segment BC has diameter 40 mm and length 0.5 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
A solid steel shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 500 N·m is applied at C, 3 m from A and 2 m from B. Find the reaction torque TA at wall A.
A point in a loaded part has σₓ = 0 MPa, σᵧ = −10 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 6 mm holds gas at a gauge pressure of 0.5 MPa. It is made from plate welded along a helical seam at 25° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.
A bar is made of a steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
A column of length L made of aluminium is pinned 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 130 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 4 cycles later by a peak of 0.5 mm. Find the damping ratio ζ.
A 50 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 kN/m and k₂ = 30 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: T1 = 26.34 N·m
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.
Answer: P = 3.407 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: τmax = 19.89 MPa
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.
Answer: TA = 200 N·m
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).
Answer: θp = −42.14 °
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: |τn| = 7.98 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
Answer: P = 37.7 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: σmax = 373 MPa
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
The thinner segment has the smaller area, so it carries the larger stress.
Answer: Pcr = 3.598 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 rectangle buckles about its weak axis, using the smaller dimension cubed.
Answer: d = 166.4 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 240.4, so the column is slender and Euler applies.
Answer: ζ = 0.1184
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.
Answer: fn = 2.502 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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