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 4 kW motor running at 1440 rpm drives a pinion with 20 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 4 kWn1 = 1440 rpmN1 = 20N2 = 60η = 95%
T2 = N·m
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 1200 rpm. Find the speed nc of the planet carrier.
Ns = 30Np = 20ns = 1200 rpm
nc = rpm
3.Torsion of shaftsMedium
A solid aluminium shaft 1 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 3°. Find the smallest diameter d that meets this limit.
T = 1500 N·mL = 1 mG = 26 GPaφallow = 3°
d = mm
4.Torsion of shaftsHard
A hollow shaft transmits 20 kW at 1000 rpm. Find the maximum shear stress τmax in the shaft.
P = 20 kWN = 1000 rpmdo = 40 mmdi = 28 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −70 MPa, σᵧ = 60 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −70 MPaσy = 60 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 90 MPa, σᵧ = −80 MPa and τₓᵧ = −40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 90 MPaσy = −80 MPaτxy = −40 MPaθ = 25°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 70 kNd = 25 mmL = 0.5 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 300 mm, E = 70 GPaP = 30 kN
σ2 = 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 mE = 70 GPa (aluminium)b × h = 40 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed one cycle later by a peak of 18.7 mm. Find the damping ratio ζ.
x0 = 40 mmx1 = 18.7 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 25 kgk1 = 12 kN/mk2 = 80 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 75.6 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 = 4000/150.8 = 26.53 N·m
Output torque
T2 = 26.53·60/20·0.95 = 75.6 N·m
Answer
T2 = 75.6 N·m
2.Gears & gear trains
Answer: nc = 360 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.
(1200 − nc)/(0 − nc) = −70/30
nc = ns·Ns/(Ns + Nr) = 1200·30/100 = 360 rpm
Answer
nc = 360 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: d = 57.88 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 = 1500 N·m = 1.5 × 10⁶ N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = 3·π/180 = 0.05236 rad
J = TL/(Gφ) = 1.5 × 10⁶·1000/(26000·0.05236) = 1.102 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 57.88 mm
Answer
d = 57.88 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: τmax = 20 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·1000/60 = 104.7 rad/s
T = P/ω = 20000/104.7 = 191 N·m
Polar second moment of area
J = π(40⁴ − 28⁴)/32 = 190984 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 190986·20/190984 = 20 MPa
Answer
τmax = 20 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 71.59 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 = √(((−70 − 60)/2)² + 30²) = 71.59 MPa
Answer
τmax = 71.59 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −90.83 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.
Transformation equation
τθ = −85·0.766 + −40·0.6428 = −90.83 MPa
Answer
τθ = −90.83 MPa
7.Axial stress & strain
Answer: δ = 0.6094 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·25²/4 = 490.9 mm²
σ = P/A = 70·1000/490.9 = 142.6 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 142.6/117000 = 0.001219 = 1219 µε
Elongation
δ = ε·L = 0.001219·500 mm = 0.6094 mm
Answer
δ = 0.6094 mm
8.Axial stress & strain
Answer: σ2 = 17.59 MPa
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.
Stiffnesses
k₁ = 200000·314.2/600 = 104720 N/mm
k₂ = 70000·1257/300 = 293215 N/mm
Share of segment 2
F2 = 30000·293215/397935 = 22105 N
Stress
σ2 = 22105/1257 = 17.59 MPa (compression)
Answer
σ2 = 17.59 MPa
9.Column buckling
Answer: Pcr = 46.06 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 rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 50·40³/12 = 266667 mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·266667/2000² = 46058 N = 46.06 kN
Answer
Pcr = 46.06 kN
10.Column buckling
Answer: σcr = 39.34 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.7·4000 = 2800 mm
KL/r = 2800/12.5 = 224
Critical stress
σcr = π²E/(KL/r)² = π²·200000/224² = 39.34 MPa
Answer
σcr = 39.34 MPa
11.Mechanical vibrations
Answer: ζ = 0.1201
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 = (1/1)·ln(40/18.7) = 0.7604
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7604/√(39.48 + 0.5781) = 0.1201
Answer
ζ = 0.1201
12.Mechanical vibrations
Answer: keq = 10.43 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/12 + 1/80 → keq = 12·80/(12 + 80) = 10.43 kN/m
Answer
keq = 10.43 kN/m
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