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 80 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 3 : 1. How many teeth N2 must the gear have?
a = 80 mmm = 2 mmi = 3
N2 =
2.Gears & gear trainsHard
A 4 kW motor at 1450 rpm drives gear A (32 teeth), which meshes with gear B (48 teeth). Gear C (45 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.
P = 4 kWnin = 1450 rpmA = 32B = 48C = 45D = 60ηmesh = 98%
Tout = N·m
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 = 200 N·md = 50 mmL = 1.5 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 30 kW at 1450 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
P = 30 kWN = 1450 rpmdi/do = 0.6τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 70 MPa and σᵧ = −50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 110 MPa?
σx = 70 MPaσy = −50 MPaσ1 = 110 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −30 MPa, σᵧ = 10 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = −30 MPaσy = 10 MPaτxy = −20 MPaθ = -30°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 40 kNd = 30 mmL = 2 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 = 65 kNsegment 1: Ø25 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 0.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 70 GPa (aluminium)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 90 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 90 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 40 kg mass on a spring with a stiffness of 12 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency fd of its free vibration in Hz.
m = 40 kgk = 12 kN/mζ = 0.35
fd = Hz
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 50 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 5 kgk1 = 20 kN/mk2 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: N2 = 60
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 = 2·80/2 = 80
Split by the ratio
N1 = 80/(1 + 3) = 20
N2 = 3·20 = 60
Answer
N2 = 60
2.Gears & gear trains
Answer: Tout = 50.6 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.
Output speed
nout = 1450·(32/48)·(45/60) = 725 rpm
ωout = 2π·725/60 = 75.92 rad/s
Output power and torque
Pout = 4000·0.98² = 3842 W
Tout = Pout/ωout = 3842/75.92 = 50.6 N·m
Answer
Tout = 50.6 N·m
3.Torsion of shafts
Answer: φ = 1.077 °
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) = 200·1000·1500/(26000·613592) = 0.0188 rad
φ = 0.0188·180/π = 1.077°
Answer
φ = 1.077°
4.Torsion of shafts
Answer: do = 24.89 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·1450/60 = 151.8 rad/s
T = P/ω = 30000/151.8 = 197.6 N·m
Outer diameter from the allowable stress
T = 197.6 N·m = 197572 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·197572/(π·75·0.8704)) = 24.89 mm
di = 0.6·24.89 = 14.93 mm
Answer
do = 24.89 mm
A solid shaft would need d = 23.76 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: τxy = 80 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 = (70 + (−50))/2 = 10 MPa
R = σ1 − C = 110 − 10 = 100 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(100² − 60²) = 80 MPa
Answer
τxy = 80 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −2.679 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 + −20·0.5 + −20·−0.866 = −2.679 MPa
Answer
σθ = −2.679 MPa
7.Axial stress & strain
Answer: ε = 565.9 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·30²/4 = 706.9 mm²
σ = P/A = 40·1000/706.9 = 56.59 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 56.59/100000 = 0.0005659 = 565.9 µε
Answer
ε = 565.9 µε
8.Axial stress & strain
Answer: σmax = 574.7 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 = 132.4 MPa
σ2 = 574.7 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 574.7 MPa
Answer
σmax = 574.7 MPa
9.Column buckling
Answer: Pcr = 130.6 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 = 80·50³/12 = 833333 mm⁴
Effective length
KL = 0.7·3000 = 2100 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·833333/2100² = 130550 N = 130.6 kN
Answer
Pcr = 130.6 kN
10.Column buckling
Answer: d = 95.62 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 = 2.5·90000 = 225000 N
KL = 2·3000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·225000·6000²/(π³·200000))^(1/4) = 95.62 mm
Answer
d = 95.62 mm
Check: the slenderness KL/r is 251, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 2.582 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.
Undamped natural frequency
ωn = √(12000/40) = 17.32 rad/s
fn = 17.32/(2π) = 2.757 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.757·√(1 − 0.35²) = 2.582 Hz
Answer
fd = 2.582 Hz
12.Mechanical vibrations
Answer: fn = 8.507 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
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/20 + 1/50 → keq = 20·50/(20 + 50) = 14.29 kN/m
Natural frequency
ωn = √(14286/5) = 53.45 rad/s
fn = 53.45/(2π) = 8.507 Hz
Answer
fn = 8.507 Hz
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