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 worm with 4 start(s) turns at 1450 rpm and drives a worm wheel with 48 teeth. Find the speed nw of the wheel.
z = 4Nw = 48n = 1450 rpm
nw = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (50 teeth) on the input shaft turning at 3000 rpm drives gear B (72 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (50 teeth) on the output shaft. Find the output speed nout.
nin = 3000 rpmA = 50B = 72C = 18D = 50
nout = rpm
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 100 N·m. The two ends of a gauge length of 300 mm rotate 2.81° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 100 N·md = 20 mmL = 300 mmφ = 2.81°
G = GPa
4.Torsion of shaftsHard
A solid brass shaft transmits 20 kW at 1500 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.
P = 20 kWN = 1500 rpmτallow = 40 MPaθallow = 1°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 160 MPa, σᵧ = 10 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 160 MPaσy = 10 MPaτxy = −50 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 90 MPa, σᵧ = −70 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.
σx = 90 MPaσy = −70 MPaτxy = −20 MPaθ = 60°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 50 kNd = 16 mmL = 2 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 largest normal stress σmax in the bar.
P = 105 kNsegment 1: Ø30 mm, 2 m, steel (E = 200 GPa)segment 2: Ø20 mm, 2 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 = 4 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 20 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 20 kNn = 3L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 25 kg mass on a spring with a stiffness of 150 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 = 25 kgk = 150 kN/mζ = 0.35
fd = Hz
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 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 = 100 kgk1 = 10 kN/mk2 = 5 kN/mk3 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: nw = 120.8 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.
Wheel speed
nw = n·z/Nw = 1450·4/48 = 120.8 rpm
Answer
nw = 120.8 rpm
2.Gears & gear trains
Answer: nout = 750 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 = 3000·50/72 = 2083 rpm = nC
Second stage
nout = 2083·18/50 = 750 rpm
Answer
nout = 750 rpm
total ratio = 4 : 1
3.Torsion of shafts
Answer: G = 38.94 GPa
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 moment and angle
J = π·20⁴/32 = 15708 mm⁴
φ = 2.81·π/180 = 0.04904 rad
Shear modulus
T = 100 N·m = 100000 N·mm
G = TL/(Jφ) = 100000·300/(15708·0.04904) = 38942 MPa
G = 38.94 GPa
Answer
G = 38.94 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: d = 37.15 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·1500/60 = 157.1 rad/s
T = P/ω = 20000/157.1 = 127.3 N·m
Diameter for strength
T = 127.3 N·m = 127324 N·mm
dτ = ∛(16·127324/(π·40)) = 25.31 mm
Diameter for stiffness
θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·127324/(π·39000·1.745 × 10⁻⁵))1/4 = 37.15 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(25.31, 37.15) = 37.15 mm
Answer
d = 37.15 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τmax = 90.14 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.
Radius of Mohr’s circle
τmax = R = √(((160 − 10)/2)² + −50²) = 90.14 MPa
Answer
τmax = 90.14 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −47.32 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 + 80·−0.5 + −20·0.866 = −47.32 MPa
Answer
σθ = −47.32 MPa
7.Axial stress & strain
Answer: δ = 7.105 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·16²/4 = 201.1 mm²
σ = P/A = 50·1000/201.1 = 248.7 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 248.7/70000 = 0.003553 = 3553 µε
Elongation
δ = ε·L = 0.003553·2000 mm = 7.105 mm
Answer
δ = 7.105 mm
8.Axial stress & strain
Answer: σmax = 334.2 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·30²/4 = 706.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 148.5 MPa
σ2 = 334.2 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 334.2 MPa
Answer
σmax = 334.2 MPa
9.Column buckling
Answer: Pcr = 3.504 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·39761/2800² = 3504 N = 3.504 kN
Answer
Pcr = 3.504 kN
10.Column buckling
Answer: d = 81.55 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 = 3·20000 = 60000 N
KL = 2·2500 = 5000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·60000·5000²/(π³·70000))^(1/4) = 81.55 mm
Answer
d = 81.55 mm
Check: the slenderness KL/r is 245.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 11.55 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 = √(150000/25) = 77.46 rad/s
fn = 77.46/(2π) = 12.33 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 12.33·√(1 − 0.35²) = 11.55 Hz
Answer
fd = 11.55 Hz
12.Mechanical vibrations
Answer: fn = 1.233 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 = 10 + 5 = 15 kN/m
keq = 15·10/(15 + 10) = 6 kN/m
Natural frequency
ωn = √(6000/100) = 7.746 rad/s
fn = 7.746/(2π) = 1.233 Hz
Answer
fn = 1.233 Hz
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