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 187.5 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 1.5 : 1. How many teeth N2 must the gear have?
a = 187.5 mmm = 5 mmi = 1.5
N2 =
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turning at 1500 rpm drives gear B (60 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 18B = 60C = 48D = 60
nout = rpm
3.Torsion of shaftsMedium
A solid steel 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 = 30 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft transmits 10 kW at 1000 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.
P = 10 kWN = 1000 rpmτallow = 60 MPaθallow = 0.5°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 10 MPa and σᵧ = 0 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 20 MPa?
σx = 10 MPaσy = 0 MPaσ1 = 20 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 30 MPa, σᵧ = 30 MPa and τₓᵧ = −20 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 30 MPaσy = 30 MPaτxy = −20 MPaθ = 50°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 105 kNd = 25 mmL = 2 mE = 200 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 = 80 kNsegment 1: Ø25 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø20 mm, 2 m, copper (E = 117 GPa)
δ = mm
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)d = 25 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 80 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 2770 N·s/m. Find the damping ratio ζ.
m = 80 kgk = 150 kN/mc = 2770 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 kN/m and the lower one k₂ = 12 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 30 kN/mk2 = 12 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: N2 = 45
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 = 2·187.5/5 = 75
Split by the ratio
N1 = 75/(1 + 1.5) = 30
N2 = 1.5·30 = 45
Answer
N2 = 45
2.Gears & gear trains
Answer: nout = 360 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 = 1500·18/60 = 450 rpm = nC
Second stage
nout = 450·48/60 = 360 rpm
Answer
nout = 360 rpm
total ratio = 4.167 : 1
3.Torsion of shafts
Answer: φ = 9.006 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·2500/(80000·79522) = 0.1572 rad
φ = 0.1572·180/π = 9.006°
Answer
φ = 9.006°
4.Torsion of shafts
Answer: d = 41.12 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 = 10000/104.7 = 95.49 N·m
Diameter for strength
T = 95.49 N·m = 95493 N·mm
dτ = ∛(16·95493/(π·60)) = 20.09 mm
Diameter for stiffness
θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·95493/(π·39000·8.727 × 10⁻⁶))1/4 = 41.12 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(20.09, 41.12) = 41.12 mm
Answer
d = 41.12 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τxy = 14.14 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 = (10 + 0)/2 = 5 MPa
R = σ1 − C = 20 − 5 = 15 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(15² − 5²) = 14.14 MPa
Answer
τxy = 14.14 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = 3.473 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 = −0·0.9848 + −20·−0.1736 = 3.473 MPa
Answer
τθ = 3.473 MPa
7.Axial stress & strain
Answer: δ = 2.139 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 = 105·1000/490.9 = 213.9 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 213.9/200000 = 0.00107 = 1070 µε
Elongation
δ = ε·L = 0.00107·2000 mm = 2.139 mm
Answer
δ = 2.139 mm
8.Axial stress & strain
Answer: δ = 5.575 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·25²/4 = 490.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 163 MPa
σ2 = 254.6 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 163·1500/200000 = 1.222 mm
δ2 = σ2L2/E2 = 254.6·2000/117000 = 4.353 mm
δ = δ1 + δ2 = 5.575 mm
Answer
δ = 5.575 mm
9.Column buckling
Answer: Pcr = 3.312 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·19175/2000² = 3312 N = 3.312 kN
Answer
Pcr = 3.312 kN
10.Column buckling
Answer: σcr = 49.35 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/10 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa
Answer
σcr = 49.35 MPa
11.Mechanical vibrations
Answer: ζ = 0.3998
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·√(150000·80) = 6928 N·s/m
Damping ratio
ζ = c/cc = 2770/6928 = 0.3998
Answer
ζ = 0.3998
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.084 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 = 1/30 + 1/12 → keq = 30·12/(30 + 12) = 8.571 kN/m
Natural frequency
ωn = √(8571/50) = 13.09 rad/s
fn = 13.09/(2π) = 2.084 Hz
Answer
fn = 2.084 Hz
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