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 1440 rpm drives a pinion with 20 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.
P = 2.2 kWn = 1440 rpmN1 = 20m = 5 mm
Ft = N
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turns at 1440 rpm and drives gear B (48 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 150 rpm. How many teeth ND must gear D have?
nin = 1440 rpmA = 15B = 48C = 24nout = 150 rpm
ND =
3.Torsion of shaftsMedium
A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 75 mm and length 1.5 m; segment BC has diameter 40 mm and length 0.5 m. A torque of 2000 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 2000 N·md1 = 75 mm, L1 = 1.5 md2 = 40 mm, L2 = 0.5 m
τmax = MPa
4.Torsion of shaftsHard
A solid shaft must transmit 50 kW at 750 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 50 kWN = 750 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −30 MPa, σᵧ = 60 MPa and τₓᵧ = −60 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −30 MPaσy = 60 MPaτxy = −60 MPaθ = -30°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = −60 MPa and τₓᵧ = −50 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = −80 MPaσy = −60 MPaτxy = −50 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa). Both are 400 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.
P = 50 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 70 GPaL = 400 mm
δ = mm
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 = 55 kNsegment 1: Ø25 mm, 2 m, steel (E = 200 GPa)segment 2: Ø20 mm, 1.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 85 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 85 kNn = 2L = 5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed one cycle later by a peak of 7.3 mm. Find the damping ratio ζ.
x0 = 10 mmx1 = 7.3 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 25 kN/m and the lower one k₂ = 100 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 25 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Ft = 291.8 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·1440/60 = 150.8 rad/s
T = P/ω = 2200/150.8 = 14.59 N·m
Pitch diameter
d = m·N = 5·20 = 100 mm
Tangential force
Ft = 2T/d = 2·14.59/0.1 = 291.8 N
Answer
Ft = 291.8 N
2.Gears & gear trains
Answer: ND = 72
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 = 1440·15/48 = 450 rpm
Teeth on gear D
ND = C·nC/nout = 24·450/150 = 72
Answer
ND = 72
3.Torsion of shafts
Answer: τmax = 159.2 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.
Stress in each segment
T = 2000 N·m = 2 × 10⁶ N·mm
τAB = 16·2 × 10⁶/(π·75³) = 24.14 MPa
τBC = 16·2 × 10⁶/(π·40³) = 159.2 MPa
Answer
τmax = 159.2 MPa
4.Torsion of shafts
Answer: d = 34.35 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·750/60 = 78.54 rad/s
T = P/ω = 50000/78.54 = 636.6 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·636620/(π·80)) = 34.35 mm
Answer
d = 34.35 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: σy′ = −14.46 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.
Transformation equation for the y′ face
σy′ = 15 − (−45)·0.5 − (−60)·(−0.866) = −14.46 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = 44.46 MPa, σx′ + σy′ = 30 = σx + σy
Answer
σy′ = −14.46 MPa
6.Mohr’s circle & stress transformation
Answer: n = 4.082
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 = −70 MPa, R = 50.99 MPa
σ1 = −19.01 MPa, σ2 = −121 MPa, σ3 = 0
Von Mises
σvM = √((−19.01)² − (−19.01)·(−121) + (−121)²) = 112.7 MPa
n = 460/112.7 = 4.082
Answer
n = 4.082
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.03014 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.
Areas
Ar = π·30²/4 = 706.9 mm²
At = π(120² − 70²)/4 = 7461 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 70·10³·7461 = 6.637 × 10⁸ N
Shortening
δ = 50000·400/6.637 × 10⁸ = 0.03014 mm
Answer
δ = 0.03014 mm
8.Axial stress & strain
Answer: σmax = 175.1 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·20²/4 = 314.2 mm²
σ1 = 112 MPa
σ2 = 175.1 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 175.1 MPa
Answer
σmax = 175.1 MPa
9.Column buckling
Answer: Pcr = 124.4 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 = 80·30³/12 = 180000 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·180000/1000² = 124357 N = 124.4 kN
Answer
Pcr = 124.4 kN
10.Column buckling
Answer: d = 74.82 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 = 2·85000 = 170000 N
KL = 0.5·5000 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·170000·2500²/(π³·70000))^(1/4) = 74.82 mm
Answer
d = 74.82 mm
Check: the slenderness KL/r is 133.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.05003
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 = (1/1)·ln(10/7.3) = 0.3147
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3147/√(39.48 + 0.09904) = 0.05003
Answer
ζ = 0.05003
12.Mechanical vibrations
Answer: keq = 20 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 = 1/25 + 1/100 → keq = 25·100/(25 + 100) = 20 kN/m
Answer
keq = 20 kN/m
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