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 3 kW motor running at 2900 rpm drives a pinion with 30 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 3 kWn1 = 2900 rpmN1 = 30N2 = 60η = 96%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (30 teeth) on the input shaft turning at 960 rpm drives gear B (32 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (16 teeth) on the output shaft. Find the output speed nout.
nin = 960 rpmA = 30B = 32C = 15D = 16
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 = 100 N·md = 30 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid aluminium shaft transmits 30 kW at 300 rpm. The shear stress must not exceed 50 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.
P = 30 kWN = 300 rpmτallow = 50 MPaθallow = 0.25°/mG = 26 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 110 MPa and σᵧ = 60 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 125 MPa?
σx = 110 MPaσy = 60 MPaσ1 = 125 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = −60 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 20° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 160 MPaσy = −30 MPaτxy = −60 MPaθ = -20°
τθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 10 kNd = 20 mmL = 0.5 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 = 85 kNsegment 1: Ø35 mm, 2 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 200 GPa (steel)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 60 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 60 kNn = 2L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 10 kg mass on a spring with a stiffness of 8 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 10 kgk = 8 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 100 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 = 20 kgk1 = 100 kN/mk2 = 25 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 9.879 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 3000/303.7 = 9.879 N·m
Answer
T1 = 9.879 N·m
2.Gears & gear trains
Answer: nout = 843.8 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.
First stage
nB = 960·30/32 = 900 rpm = nC
Second stage
nout = 900·15/16 = 843.8 rpm
Answer
nout = 843.8 rpm
total ratio = 1.138 : 1
3.Torsion of shafts
Answer: φ = 0.7205 °
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) = 100·1000·800/(80000·79522) = 0.01258 rad
φ = 0.01258·180/π = 0.7205°
Answer
φ = 0.7205°
4.Torsion of shafts
Answer: d = 96.23 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·300/60 = 31.42 rad/s
T = P/ω = 30000/31.42 = 954.9 N·m
Diameter for strength
T = 954.9 N·m = 954930 N·mm
dτ = ∛(16·954930/(π·50)) = 45.99 mm
Diameter for stiffness
θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm
G = 26 GPa = 26000 MPa
dφ = (32·954930/(π·26000·4.363 × 10⁻⁶))1/4 = 96.23 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(45.99, 96.23) = 96.23 mm
Answer
d = 96.23 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τxy = 31.22 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 = (110 + 60)/2 = 85 MPa
R = σ1 − C = 125 − 85 = 40 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(40² − 25²) = 31.22 MPa
Answer
τxy = 31.22 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = 15.1 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 = −95·−0.6428 + −60·0.766 = 15.1 MPa
Answer
τθ = 15.1 MPa
7.Axial stress & strain
Answer: ε = 318.3 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·20²/4 = 314.2 mm²
σ = P/A = 10·1000/314.2 = 31.83 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 31.83/100000 = 0.0003183 = 318.3 µε
Answer
ε = 318.3 µε
8.Axial stress & strain
Answer: σmax = 422.8 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·35²/4 = 962.1 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 88.35 MPa
σ2 = 422.8 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 422.8 MPa
Answer
σmax = 422.8 MPa
9.Column buckling
Answer: Pcr = 411.2 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 = 80·50³/12 = 833333 mm⁴
Effective length
KL = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·833333/2000² = 411234 N = 411.2 kN
Answer
Pcr = 411.2 kN
10.Column buckling
Answer: d = 72.57 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 = 2·60000 = 120000 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·2800²/(π³·70000))^(1/4) = 72.57 mm
Answer
d = 72.57 mm
Check: the slenderness KL/r is 154.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 4.294 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 = √(8000/10) = 28.28 rad/s
fn = 28.28/(2π) = 4.502 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 4.502·√(1 − 0.3²) = 4.294 Hz
Answer
fd = 4.294 Hz
12.Mechanical vibrations
Answer: fn = 4.673 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 = 100 + 25 = 125 kN/m
keq = 125·20/(125 + 20) = 17.24 kN/m
Natural frequency
ωn = √(17241/20) = 29.36 rad/s
fn = 29.36/(2π) = 4.673 Hz
Answer
fn = 4.673 Hz
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