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 15 kW motor running at 1440 rpm drives a pinion with 24 teeth, which meshes with a gear with 30 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 15 kWn1 = 1440 rpmN1 = 24N2 = 30η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (24 teeth) on the input shaft turns at 3000 rpm and drives gear B (50 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 480 rpm. How many teeth ND must gear D have?
nin = 3000 rpmA = 24B = 50C = 20nout = 480 rpm
ND =
3.Torsion of shaftsMedium
A solid steel 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 = 1000 N·md = 30 mmL = 1.2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 1450 rpmdo = 40 mmdi = 30 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −70 MPa, σᵧ = −10 MPa and τₓᵧ = 60 MPa (tension positive). The element is rotated 25° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −70 MPaσy = −10 MPaτxy = 60 MPaθ = 25°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = 30 MPa and τₓᵧ = −30 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −10 MPaσy = 30 MPaτxy = −30 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).
P = 85 kNd = 16 mmL = 3 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø40 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø50 mm, E = 100 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 70 GPa (aluminium)b × h = 20 × 100 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 = 3 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 3 cycles later by a peak of 6.6 mm. Find the damping ratio ζ.
x0 = 30 mmx3 = 6.6 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from spring k₃ = 80 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 60 kN/m and k₂ = 60 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 = 25 kgk1 = 60 kN/mk2 = 60 kN/mk3 = 80 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 120.6 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 15000/150.8 = 99.47 N·m
Output torque
T2 = 99.47·30/24·0.97 = 120.6 N·m
Answer
T2 = 120.6 N·m
2.Gears & gear trains
Answer: ND = 60
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 = 3000·24/50 = 1440 rpm
Teeth on gear D
ND = C·nC/nout = 20·1440/480 = 60
Answer
ND = 60
3.Torsion of shafts
Answer: φ = 10.81 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·1200/(80000·79522) = 0.1886 rad
φ = 0.1886·180/π = 10.81°
Answer
φ = 10.81°
4.Torsion of shafts
Answer: τmax = 76.66 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·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m
Polar second moment of area
J = π(40⁴ − 30⁴)/32 = 171806 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 658572·20/171806 = 76.66 MPa
Answer
τmax = 76.66 MPa
5.Mohr’s circle & stress transformation
Answer: σy′ = −66.68 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′ = −40 − (−30)·0.6428 − 60·0.766 = −66.68 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −13.32 MPa, σx′ + σy′ = −80 = σx + σy
Answer
σy′ = −66.68 MPa
6.Mohr’s circle & stress transformation
Answer: n = 6.379
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 = 10 MPa, R = 36.06 MPa
σ1 = 46.06 MPa, σ2 = −26.06 MPa, σ3 = 0
Tresca
σmax − σmin = 72.11 MPa
n = 460/72.11 = 6.379
Answer
n = 6.379
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: ε = 3613 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 = 85·1000/201.1 = 422.8 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 422.8/117000 = 0.003613 = 3613 µε
Answer
ε = 3613 µε
8.Axial stress & strain
Answer: Pmax = 672.3 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
Ar = π·40²/4 = 1257 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 100·10³·5890 = 8.404 × 10⁸ N
Governing strain
ε = min(0.0008, 0.001) = 0.0008 (steel)
Largest load
Pmax = 0.0008·8.404 × 10⁸ = 672301 N = 672.3 kN
Answer
Pmax = 672.3 kN
9.Column buckling
Answer: Pcr = 2.879 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 = 100·20³/12 = 66667 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·66667/4000² = 2879 N = 2.879 kN
Answer
Pcr = 2.879 kN
10.Column buckling
Answer: σcr = 34.27 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.5·3000 = 1500 mm
KL/r = 1500/6.25 = 240
Critical stress
σcr = π²E/(KL/r)² = π²·200000/240² = 34.27 MPa
Answer
σcr = 34.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.08007
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 = (1/3)·ln(30/6.6) = 0.5047
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5047/√(39.48 + 0.2547) = 0.08007
Answer
ζ = 0.08007
12.Mechanical vibrations
Answer: fn = 6.974 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 = 60 + 60 = 120 kN/m
keq = 120·80/(120 + 80) = 48 kN/m
Natural frequency
ωn = √(48000/25) = 43.82 rad/s
fn = 43.82/(2π) = 6.974 Hz
Answer
fn = 6.974 Hz
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