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
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A 5.5 kW motor running at 960 rpm drives a pinion with 12 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.
P = 5.5 kWn1 = 960 rpmN1 = 12N2 = 40η = 96%
T2 = N·m
2.Gears & gear trainsHard
A 2.2 kW motor at 2900 rpm drives gear A (12 teeth), which meshes with gear B (32 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (20 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
P = 2.2 kWnin = 2900 rpmA = 12B = 32C = 16D = 20ηmesh = 97%
Tout = N·m
3.Torsion of shaftsMedium
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 300 mm rotate 8.66° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 40 N·md = 12 mmL = 300 mmφ = 8.66°
G = GPa
4.Torsion of shaftsHard
A solid steel shaft of diameter 40 mm is fixed at A. A torque of 400 N·m acts at B, 0.5 m from A, and a torque of 250 N·m in the opposite direction acts at the free end C, a further 1 m along. Find the maximum shear stress τmax in the shaft.
d = 40 mmTB = 400 N·mTC = 250 N·mAB = 0.5 mBC = 1 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −20 MPa and σᵧ = −10 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 5 MPa?
σx = −20 MPaσy = −10 MPaσ1 = 5 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries an axial compressive force of 180 kN together with a torque of 750 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 50 mmF = −180 kNT = 750 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 50 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: brass, Ø50 mm, 300 mm, E = 100 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 50 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø50 mm, 500 mm, E = 100 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 15 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 150 N·s/m. Find the damping ratio ζ.
m = 15 kgk = 40 kN/mc = 150 N·s/m
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 100 kN/m and k₂ = 20 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.
m = 10 kgk1 = 100 kN/mk2 = 20 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 175.1 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.
Motor torque
ω₁ = 2π·960/60 = 100.5 rad/s
T1 = P/ω₁ = 5500/100.5 = 54.71 N·m
Output torque
T2 = 54.71·40/12·0.96 = 175.1 N·m
Answer
T2 = 175.1 N·m
2.Gears & gear trains
Answer: Tout = 22.72 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.
Output speed
nout = 2900·(12/32)·(16/20) = 870 rpm
ωout = 2π·870/60 = 91.11 rad/s
Output power and torque
Pout = 2200·0.97² = 2070 W
Tout = Pout/ωout = 2070/91.11 = 22.72 N·m
Answer
Tout = 22.72 N·m
3.Torsion of shafts
Answer: G = 39 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.
Polar moment and angle
J = π·12⁴/32 = 2036 mm⁴
φ = 8.66·π/180 = 0.1511 rad
Shear modulus
T = 40 N·m = 40000 N·mm
G = TL/(Jφ) = 40000·300/(2036·0.1511) = 39000 MPa
G = 39 GPa
Answer
G = 39 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 250 N·m
TAB = TC − TB = 250 − 400 = −150 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 250 N·m
τmax = 16·250000/(π·40³) = 19.89 MPa
Answer
τmax = 19.89 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 19.36 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.
Required radius
C = (−20 + (−10))/2 = −15 MPa
R = σ1 − C = 5 − (−15) = 20 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(20² − 5²) = 19.36 MPa
Answer
τxy = 19.36 MPa
6.Mohr’s circle & stress transformation
Answer: σ1 = 9.252 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.
Stresses at the surface
σ = F/A = −180000/(π·50²/4) = −180000/1963 = −91.67 MPa
τ = 16T/(πd³) = 16·750000/(π·50³) = 30.56 MPa
Mohr’s circle
C = σ/2 = −45.84 MPa
R = √((−45.84)² + 30.56²) = 55.09 MPa
σ1 = C + R = 9.252 MPa
σ2 = C − R = −100.9 MPa
Answer
σ1 = 9.252 MPa
7.Axial stress & strain
Answer: P = 114.7 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.
Areas
A₁ = 706.9 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·1963) = 1.528 × 10⁻⁶ mm/N
Load
P = 0.5/4.357 × 10⁻⁶ = 114750 N = 114.7 kN
Answer
P = 114.7 kN
8.Axial stress & strain
Answer: σ2 = 5.876 MPa
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.
Stiffnesses
k₁ = 200000·1257/400 = 628319 N/mm
k₂ = 100000·1963/500 = 392699 N/mm
Share of segment 2
F2 = 30000·392699/1.021 × 10⁶ = 11538 N
Stress
σ2 = 11538/1963 = 5.876 MPa (compression)
Answer
σ2 = 5.876 MPa
9.Column buckling
Answer: Pcr = 441 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.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 1·3000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/3000² = 440978 N = 441 kN
Answer
Pcr = 441 kN
10.Column buckling
Answer: σcr = 31.58 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.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.5·5000 = 2500 mm
KL/r = 2500/10 = 250
Critical stress
σcr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa
Answer
σcr = 31.58 MPa
11.Mechanical vibrations
Answer: ζ = 0.09682
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.
Critical damping
cc = 2·√(k·m) = 2·√(40000·15) = 1549 N·s/m
Damping ratio
ζ = c/cc = 150/1549 = 0.09682
Answer
ζ = 0.09682
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 6.59 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.
Equivalent stiffness
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 100 + 20 = 120 kN/m
keq = 120·20/(120 + 20) = 17.14 kN/m
Natural frequency
ωn = √(17143/10) = 41.4 rad/s
fn = 41.4/(2π) = 6.59 Hz
Answer
fn = 6.59 Hz
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