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 2.2 kW motor running at 960 rpm drives a pinion with 24 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.
P = 2.2 kWn = 960 rpmN1 = 24m = 5 mm
Ft = N
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 18 teeth and each planet gear has 18 teeth. The ring gear is held fixed and the sun is driven at 1800 rpm. Find the speed nc of the planet carrier.
Ns = 18Np = 18ns = 1800 rpm
nc = rpm
3.Torsion of shaftsMedium
A solid aluminium shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 500 N·md = 50 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 15 kW at 1800 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 15 kWN = 1800 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −80 MPa, σᵧ = −60 MPa and τₓᵧ = 50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −80 MPaσy = −60 MPaτxy = 50 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = −10 MPa and τₓᵧ = 40 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = 80 MPaσy = −10 MPaτxy = 40 MPa
σvM = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 75 kNd = 16 mmL = 2.5 mE = 200 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 = 115 kNsegment 1: Ø35 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 200 GPa (steel)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 4 cycles later by a peak of 18.3 mm. Find the damping ratio ζ.
x0 = 50 mmx4 = 18.3 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 20 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 100 kN/mk2 = 20 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Ft = 364.7 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.
Torque on the pinion
ω = 2π·960/60 = 100.5 rad/s
T = P/ω = 2200/100.5 = 21.88 N·m
Pitch diameter
d = m·N = 5·24 = 120 mm
Tangential force
Ft = 2T/d = 2·21.88/0.12 = 364.7 N
Answer
Ft = 364.7 N
2.Gears & gear trains
Answer: nc = 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.
Ring gear teeth
Nr = Ns + 2·Np = 18 + 2·18 = 54
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(1800 − nc)/(0 − nc) = −54/18
nc = ns·Ns/(Ns + Nr) = 1800·18/72 = 450 rpm
Answer
nc = 450 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: φ = 1.437 °
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 second moment of area
J = π·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·800/(26000·613592) = 0.02507 rad
φ = 0.02507·180/π = 1.437°
Answer
φ = 1.437°
4.Torsion of shafts
Answer: d = 18.9 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.
Torque
ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s
T = P/ω = 15000/188.5 = 79.58 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·79577/(π·60)) = 18.9 mm
Answer
d = 18.9 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 50.99 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.
Radius of Mohr’s circle
τmax = R = √(((−80 − −60)/2)² + 50²) = 50.99 MPa
Answer
τmax = 50.99 MPa
6.Mohr’s circle & stress transformation
Answer: σvM = 110 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 = 35 MPa, R = 60.21 MPa
σ1 = 95.21 MPa, σ2 = −25.21 MPa, σ3 = 0
Von Mises
σvM = √(95.21² − 95.21·(−25.21) + (−25.21)²) = 110 MPa
Answer
σvM = 110 MPa
7.Axial stress & strain
Answer: ε = 1865 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·16²/4 = 201.1 mm²
σ = P/A = 75·1000/201.1 = 373 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 373/200000 = 0.001865 = 1865 µε
Answer
ε = 1865 µε
8.Axial stress & strain
Answer: σmax = 1017 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·35²/4 = 962.1 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 119.5 MPa
σ2 = 1017 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 1017 MPa
Answer
σmax = 1017 MPa
9.Column buckling
Answer: Pcr = 55.52 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 = 50·30³/12 = 112500 mm⁴
Effective length
KL = 0.5·4000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·112500/2000² = 55517 N = 55.52 kN
Answer
Pcr = 55.52 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 1·5000 = 5000 mm
KL/r = 5000/20 = 250
Critical stress
σcr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa
Answer
σcr = 31.58 MPa
11.Mechanical vibrations
Answer: ζ = 0.03996
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.
Logarithmic decrement
δ = (1/4)·ln(50/18.3) = 0.2513
Damping ratio
ζ = δ/√(4π² + δ²) = 0.2513/√(39.48 + 0.06314) = 0.03996
Answer
ζ = 0.03996
12.Mechanical vibrations
Answer: keq = 16.67 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/100 + 1/20 → keq = 100·20/(100 + 20) = 16.67 kN/m
Answer
keq = 16.67 kN/m
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