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
Two parallel shafts are 90 mm apart. They must be connected by a pair of spur gears with a module of 2.5 mm and a speed reduction of 3.5 : 1. How many teeth N2 must the gear have?
a = 90 mmm = 2.5 mmi = 3.5
N2 =
2.Gears & gear trainsHard
In a compound gear train, gear A (36 teeth) on the input shaft turning at 960 rpm drives gear B (60 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.
nin = 960 rpmA = 36B = 60C = 18D = 72
nout = rpm
3.Torsion of shaftsMedium
A solid steel shaft 1 m long carries a torque of 400 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.
T = 400 N·mL = 1 mG = 80 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
A hollow shaft transmits 20 kW at 3000 rpm. Find the maximum shear stress τmax in the shaft.
P = 20 kWN = 3000 rpmdo = 40 mmdi = 24 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 140 MPa and σᵧ = 40 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 180 MPa?
σx = 140 MPaσy = 40 MPaσ1 = 180 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −80 MPa, σᵧ = 60 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = −80 MPaσy = 60 MPaτxy = −60 MPaθ = 30°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 75 kNd = 25 mmL = 1.5 mE = 100 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 = 30 kNsegment 1: Ø25 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 200 GPa (steel)b × h = 20 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 15 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 80 N·s/m. Find the damping ratio ζ.
m = 15 kgk = 10 kN/mc = 80 N·s/m
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 12 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 12 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: N2 = 56
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.
Total number of teeth
N1 + N2 = 2a/m = 2·90/2.5 = 72
Split by the ratio
N1 = 72/(1 + 3.5) = 16
N2 = 3.5·16 = 56
Answer
N2 = 56
2.Gears & gear trains
Answer: nout = 144 rpm
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.
First stage
nB = 960·36/60 = 576 rpm = nC
Second stage
nout = 576·18/72 = 144 rpm
Answer
nout = 144 rpm
total ratio = 6.667 : 1
3.Torsion of shafts
Answer: d = 37.35 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.
Required polar moment
T = 400 N·m = 400000 N·mm
L = 1 m = 1000 mm
G = 80 GPa = 80000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 400000·1000/(80000·0.02618) = 190986 mm⁴
Diameter
d = (32J/π)1/4 = 37.35 mm
Answer
d = 37.35 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: τmax = 5.82 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·3000/60 = 314.2 rad/s
T = P/ω = 20000/314.2 = 63.66 N·m
Polar second moment of area
J = π(40⁴ − 24⁴)/32 = 218755 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 63662·20/218755 = 5.82 MPa
Answer
τmax = 5.82 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 74.83 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 = (140 + 40)/2 = 90 MPa
R = σ1 − C = 180 − 90 = 90 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(90² − 50²) = 74.83 MPa
Answer
τxy = 74.83 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −96.96 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
σθ = −10 + −70·0.5 + −60·0.866 = −96.96 MPa
Answer
σθ = −96.96 MPa
7.Axial stress & strain
Answer: ε = 1528 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 = 75·1000/490.9 = 152.8 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 152.8/100000 = 0.001528 = 1528 µε
Answer
ε = 1528 µε
8.Axial stress & strain
Answer: σmax = 265.3 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 = 61.12 MPa
σ2 = 265.3 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 265.3 MPa
Answer
σmax = 265.3 MPa
9.Column buckling
Answer: Pcr = 5.264 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 = 2·2500 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·66667/5000² = 5264 N = 5.264 kN
Answer
Pcr = 5.264 kN
10.Column buckling
Answer: σcr = 3.084 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/7.5 = 800
Critical stress
σcr = π²E/(KL/r)² = π²·200000/800² = 3.084 MPa
Answer
σcr = 3.084 MPa
11.Mechanical vibrations
Answer: ζ = 0.1033
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·√(10000·15) = 774.6 N·s/m
Damping ratio
ζ = c/cc = 80/774.6 = 0.1033
Answer
ζ = 0.1033
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 24 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 = 12 + 12 = 24 kN/m
Answer
keq = 24 kN/m
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