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 5.5 kW motor running at 1500 rpm drives a pinion with 30 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.
P = 5.5 kWn1 = 1500 rpmN1 = 30N2 = 40η = 96%
T2 = N·m
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 12 teeth and each planet gear has 15 teeth. The ring gear is held fixed and the sun is driven at 1440 rpm. Find the speed nc of the planet carrier.
Ns = 12Np = 15ns = 1440 rpm
nc = rpm
3.Torsion of shaftsMedium
A solid brass 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 = 250 N·md = 50 mmL = 0.8 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 60 mm is fixed at A. A torque of 500 N·m acts at B, 1.5 m from A, and a torque of 750 N·m in the opposite direction acts at the free end C, a further 2.5 m along. Find the maximum shear stress τmax in the shaft.
d = 60 mmTB = 500 N·mTC = 750 N·mAB = 1.5 mBC = 2.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 20 MPa, σᵧ = 10 MPa and τₓᵧ = −30 MPa (tension positive). The element is rotated 20° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = 20 MPaσy = 10 MPaτxy = −30 MPaθ = -20°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 10 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 1500 mmt = 10 mmp = 3 MPaα = 20°
σn = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 115 kNd = 25 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 = 50 kNsegment 1: Ø25 mm, 2 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1 m, steel (E = 200 GPa)
σmax = MPa
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 = 2.5 mE = 70 GPa (aluminium)b × h = 40 × 60 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 5 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 5 kNn = 3L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 50 kg mass on a spring with a stiffness of 8 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
m = 50 kgk = 8 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
A 50 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 5 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 = 50 kgk1 = 12 kN/mk2 = 5 kN/mk3 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 44.82 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 5500/157.1 = 35.01 N·m
Output torque
T2 = 35.01·40/30·0.96 = 44.82 N·m
Answer
T2 = 44.82 N·m
2.Gears & gear trains
Answer: nc = 320 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 = 12 + 2·15 = 42
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(1440 − nc)/(0 − nc) = −42/12
nc = ns·Ns/(Ns + Nr) = 1440·12/54 = 320 rpm
Answer
nc = 320 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: φ = 0.4789 °
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) = 250·1000·800/(39000·613592) = 0.008358 rad
φ = 0.008358·180/π = 0.4789°
Answer
φ = 0.4789°
4.Torsion of shafts
Answer: τmax = 17.68 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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 750 N·m
TAB = TC − TB = 750 − 500 = 250 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 750 N·m
τmax = 16·750000/(π·60³) = 17.68 MPa
Answer
τmax = 17.68 MPa
5.Mohr’s circle & stress transformation
Answer: σy′ = −8.114 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 − 5·0.766 − (−30)·(−0.6428) = −8.114 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = 38.11 MPa, σx′ + σy′ = 30 = σx + σy
Answer
σy′ = −8.114 MPa
6.Mohr’s circle & stress transformation
Answer: σn = 211.8 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 1500/2 = 750 mm
σh = pr/t = 3·750/10 = 225 MPa
σa = pr/(2t) = 112.5 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
σn = 112.5·sin²20° + 225·cos²20° = 211.8 MPa
Answer
σn = 211.8 MPa
7.Axial stress & strain
Answer: ε = 2343 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 = 115·1000/490.9 = 234.3 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 234.3/100000 = 0.002343 = 2343 µε
Answer
ε = 2343 µε
8.Axial stress & strain
Answer: σmax = 159.2 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 = 101.9 MPa
σ2 = 159.2 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 159.2 MPa
Answer
σmax = 159.2 MPa
9.Column buckling
Answer: Pcr = 35.37 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 = 60·40³/12 = 320000 mm⁴
Effective length
KL = 1·2500 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·320000/2500² = 35373 N = 35.37 kN
Answer
Pcr = 35.37 kN
10.Column buckling
Answer: d = 48.59 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 = 3·5000 = 15000 N
KL = 2·3000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·15000·6000²/(π³·200000))^(1/4) = 48.59 mm
Answer
d = 48.59 mm
Check: the slenderness KL/r is 494, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 1.845 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 = √(8000/50) = 12.65 rad/s
fn = 12.65/(2π) = 2.013 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.013·√(1 − 0.4²) = 1.845 Hz
Answer
fd = 1.845 Hz
12.Mechanical vibrations
Answer: fn = 2.458 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 = 12 + 5 = 17 kN/m
keq = 17·40/(17 + 40) = 11.93 kN/m
Natural frequency
ωn = √(11930/50) = 15.45 rad/s
fn = 15.45/(2π) = 2.458 Hz
Answer
fn = 2.458 Hz
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