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 1.5 kW motor running at 1500 rpm drives a pinion with 30 teeth, which meshes with a gear with 40 teeth. Ignore losses. Find the torque T1 on the motor shaft.
P = 1.5 kWn1 = 1500 rpmN1 = 30N2 = 40η = 100%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (48 teeth) on the input shaft turns at 1450 rpm and drives gear B (60 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 348 rpm. How many teeth ND must gear D have?
nin = 1450 rpmA = 48B = 60C = 24nout = 348 rpm
ND =
3.Torsion of shaftsMedium
A solid steel shaft 1.5 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.
T = 1000 N·mL = 1.5 mG = 80 GPaφallow = 1°
d = mm
4.Torsion of shaftsHard
A hollow shaft transmits 10 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.
P = 10 kWN = 500 rpmdo = 40 mmdi = 32 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 60 MPa and σᵧ = 10 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 120 MPa?
σx = 60 MPaσy = 10 MPaσ1 = 120 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = 90 MPa and τₓᵧ = 10 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = 20 MPaσy = 90 MPaτxy = 10 MPa
σvM = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 120 kNd = 25 mmL = 2.5 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 15 kNsegment 1: Ø25 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 80 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 80 kNn = 2L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 2 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 230 N·s/m. Find the damping ratio ζ.
m = 2 kgk = 40 kN/mc = 230 N·s/m
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 20 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 30 kN/mk2 = 20 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 9.549 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 1500/157.1 = 9.549 N·m
Answer
T1 = 9.549 N·m
2.Gears & gear trains
Answer: ND = 80
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.
Intermediate shaft speed
nB = nC = 1450·48/60 = 1160 rpm
Teeth on gear D
ND = C·nC/nout = 24·1160/348 = 80
Answer
ND = 80
3.Torsion of shafts
Answer: d = 57.51 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.
Required polar moment
T = 1000 N·m = 1 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 1 × 10⁶·1500/(80000·0.01745) = 1.074 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 57.51 mm
Answer
d = 57.51 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: τmax = 25.74 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.
Torque
ω = 2πN/60 = 2π·500/60 = 52.36 rad/s
T = P/ω = 10000/52.36 = 191 N·m
Polar second moment of area
J = π(40⁴ − 32⁴)/32 = 148384 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 190986·20/148384 = 25.74 MPa
Answer
τmax = 25.74 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 81.24 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 = (60 + 10)/2 = 35 MPa
R = σ1 − C = 120 − 35 = 85 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(85² − 25²) = 81.24 MPa
Answer
τxy = 81.24 MPa
6.Mohr’s circle & stress transformation
Answer: σvM = 83.67 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.
Principal stresses
C = 55 MPa, R = 36.4 MPa
σ1 = 91.4 MPa, σ2 = 18.6 MPa, σ3 = 0
Von Mises
σvM = √(91.4² − 91.4·18.6 + 18.6²) = 83.67 MPa
Answer
σvM = 83.67 MPa
7.Axial stress & strain
Answer: ε = 3492 µε
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·25²/4 = 490.9 mm²
σ = P/A = 120·1000/490.9 = 244.5 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 244.5/70000 = 0.003492 = 3492 µε
Answer
ε = 3492 µε
8.Axial stress & strain
Answer: σmax = 132.6 MPa
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
Same force in both segments
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
A1 = π·25²/4 = 490.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 30.56 MPa
σ2 = 132.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 132.6 MPa
Answer
σmax = 132.6 MPa
9.Column buckling
Answer: Pcr = 47.98 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 rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 60·50³/12 = 625000 mm⁴
Effective length
KL = 2·1500 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·625000/3000² = 47977 N = 47.98 kN
Answer
Pcr = 47.98 kN
10.Column buckling
Answer: d = 47.42 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.
Required critical load
Pcr = n·P = 2·80000 = 160000 N
KL = 0.7·2500 = 1750 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·160000·1750²/(π³·200000))^(1/4) = 47.42 mm
Answer
d = 47.42 mm
Check: the slenderness KL/r is 147.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.4066
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·√(40000·2) = 565.7 N·s/m
Damping ratio
ζ = c/cc = 230/565.7 = 0.4066
Answer
ζ = 0.4066
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 50 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.
Equivalent stiffness
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 30 + 20 = 50 kN/m
Answer
keq = 50 kN/m
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