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 162 mm apart. They must be connected by a pair of spur gears with a module of 3 mm and a speed reduction of 5 : 1. How many teeth N1 must the pinion have?
a = 162 mmm = 3 mmi = 5
N1 =
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 30 teeth and each planet gear has 20 teeth. The ring gear is held fixed and the sun is driven at 600 rpm. Find the speed nc of the planet carrier.
Ns = 30Np = 20ns = 600 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 = 400 N·md = 40 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft transmits 20 kW at 1000 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.
P = 20 kWN = 1000 rpmτallow = 60 MPaθallow = 1°/mG = 80 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −10 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −10 MPaσy = 0 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 5 mm holds gas at a gauge pressure of 1.2 MPa. It is made from plate welded along a helical seam at 25° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 800 mmt = 5 mmp = 1.2 MPaα = 25°
σn = MPa
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 20 kNd = 20 mmL = 1 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 70 kNsegment 1: Ø30 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, copper (E = 117 GPa)
δ = mm
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 = 4 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 100 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 100 kNn = 3L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 4 cycles later by a peak of 1.9 mm. Find the damping ratio ζ.
x0 = 40 mmx4 = 1.9 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from spring k₃ = 30 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 25 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 = 5 kgk1 = 50 kN/mk2 = 25 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: N1 = 18
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 = 2·162/3 = 108
Split by the ratio
N1 = 108/(1 + 5) = 18
Answer
N1 = 18
2.Gears & gear trains
Answer: nc = 180 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 = 30 + 2·20 = 70
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(600 − nc)/(0 − nc) = −70/30
nc = ns·Ns/(Ns + Nr) = 600·30/100 = 180 rpm
Answer
nc = 180 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: φ = 2.806 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·1200/(39000·251327) = 0.04897 rad
φ = 0.04897·180/π = 2.806°
Answer
φ = 2.806°
4.Torsion of shafts
Answer: d = 34.36 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·1000/60 = 104.7 rad/s
T = P/ω = 20000/104.7 = 191 N·m
Diameter for strength
T = 191 N·m = 190986 N·mm
dτ = ∛(16·190986/(π·60)) = 25.31 mm
Diameter for stiffness
θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·190986/(π·80000·1.745 × 10⁻⁵))1/4 = 34.36 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(25.31, 34.36) = 34.36 mm
Answer
d = 34.36 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τmax = 11.18 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.
Radius of Mohr’s circle
τmax = R = √(((−10 − 0)/2)² + 10²) = 11.18 MPa
Answer
τmax = 11.18 MPa
6.Mohr’s circle & stress transformation
Answer: σn = 87.43 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 = 800/2 = 400 mm
σh = pr/t = 1.2·400/5 = 96 MPa
σa = pr/(2t) = 48 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 = 48·sin²25° + 96·cos²25° = 87.43 MPa
Answer
σn = 87.43 MPa
7.Axial stress & strain
Answer: δ = 0.9095 mm
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·20²/4 = 314.2 mm²
σ = P/A = 20·1000/314.2 = 63.66 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 63.66/70000 = 0.0009095 = 909.5 µε
Elongation
δ = ε·L = 0.0009095·1000 mm = 0.9095 mm
Answer
δ = 0.9095 mm
8.Axial stress & strain
Answer: δ = 6.28 mm
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·30²/4 = 706.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 99.03 MPa
σ2 = 618.9 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 99.03·2000/200000 = 0.9903 mm
δ2 = σ2L2/E2 = 618.9·1000/117000 = 5.29 mm
δ = δ1 + δ2 = 6.28 mm
Answer
δ = 6.28 mm
9.Column buckling
Answer: Pcr = 56.06 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 round section has the same I about every axis.
I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/2800² = 56060 N = 56.06 kN
Answer
Pcr = 56.06 kN
10.Column buckling
Answer: d = 49.63 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 = 3·100000 = 300000 N
KL = 0.7·2000 = 1400 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·1400²/(π³·200000))^(1/4) = 49.63 mm
Answer
d = 49.63 mm
Check: the slenderness KL/r is 112.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1204
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 = (1/4)·ln(40/1.9) = 0.7618
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7618/√(39.48 + 0.5803) = 0.1204
Answer
ζ = 0.1204
12.Mechanical vibrations
Answer: fn = 10.42 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 = 50 + 25 = 75 kN/m
keq = 75·30/(75 + 30) = 21.43 kN/m
Natural frequency
ωn = √(21429/5) = 65.47 rad/s
fn = 65.47/(2π) = 10.42 Hz
Answer
fn = 10.42 Hz
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