Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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

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

  1. Find the polar moment of area: J = πd⁴/32 for a solid shaft.
  2. Shear stress at the surface: τ = T·r/J, with r the outer radius.
  3. Angle of twist: φ = TL/(GJ) in radians.
  4. 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

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. 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
N1​ = 18, m = 2.5 mm

A 3 kW motor running at 1450 rpm drives a pinion with 18 teeth and a module of 2.5 mm. Find the tangential force Ft on the pinion teeth.

P = 3 kWn = 1450 rpmN1 = 18m = 2.5 mm
Ft = N
2.Gears & gear trainsHard
A = 15B = 80C = 36D = 80

A 1.1 kW motor at 1450 rpm drives gear A (15 teeth), which meshes with gear B (80 teeth). Gear C (36 teeth) on the same shaft as B drives gear D (80 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.

P = 1.1 kWnin = 1450 rpmA = 15B = 80C = 36D = 80ηmesh = 97%
Tout = N·m
3.Torsion of shaftsMedium
T = 100 N·md = 30 mmL = 0.8 mφ = ?

A solid aluminium 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 = 26 GPa
φ = °
4.Torsion of shaftsHard
d = 20 mmTB​ = 250 N·mTC​ = 200 N·mABC2.5 m3 m

A solid aluminium shaft of diameter 20 mm is fixed at A. A torque of 250 N·m acts at B, 2.5 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 3 m along. Find the maximum shear stress τmax in the shaft.

d = 20 mmTB = 250 N·mTC = 200 N·mAB = 2.5 mBC = 3 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 10σy​ = −20τxy​ = −10MPa

A point has σₓ = 10 MPa, σᵧ = −20 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 30° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σx = 10 MPaσy = −20 MPaτxy = −10 MPaθ = 30°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 130σy​ = 10τxy​ = 50MPa

A point in a thin plate is in plane stress: σₓ = 130 MPa, σᵧ = 10 MPa and τₓᵧ = 50 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 130 MPaσy = 10 MPaτxy = 50 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Ø 20 mmBrass0.5 mP = 55 kN

Find the elongation δ of the brass bar under the axial load.

P = 55 kNd = 20 mmL = 0.5 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
Ø 35 mmBrass1 mØ 20 mmAluminium1.5 mP = 115 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 115 kNsegment 1: Ø35 mm, 1 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
PL = 1.5 m30 × 40 mmcross-section

A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 1.5 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 40 mmcross-section

A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 2 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 80 kN/mζ = 0.4m = 10 kg

A 10 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.

m = 10 kgk = 80 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 40k2​ = 40m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 40 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 10 kgk1 = 40 kN/mk2 = 40 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 878.1 N

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.

Torque on the pinion

ω = 2π·1450/60 = 151.8 rad/s
T = P/ω = 3000/151.8 = 19.76 N·m

Pitch diameter

d = m·N = 2.5·18 = 45 mm

Tangential force

Ft = 2T/d = 2·19.76/0.045 = 878.1 N

Answer

Ft = 878.1 N
2.Gears & gear trains

Answer: Tout = 80.78 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.

Output speed

nout = 1450·(15/80)·(36/80) = 122.3 rpm
ωout = 2π·122.3/60 = 12.81 rad/s

Output power and torque

Pout = 1100·0.97² = 1035 W
Tout = Pout/ωout = 1035/12.81 = 80.78 N·m

Answer

Tout = 80.78 N·m
3.Torsion of shafts

Answer: φ = 2.217 °

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/(26000·79522) = 0.03869 rad
φ = 0.03869·180/π = 2.217°

Answer

φ = 2.217°
4.Torsion of shafts

Answer: τmax = 286.5 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.

Internal torques

Sense of TC taken as positive.

TBC = TC = 200 N·m
TAB = TC + TB = 200 + 250 = 450 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|max = 450 N·m
τmax = 16·450000/(π·20³) = 286.5 MPa

Answer

τmax = 286.5 MPa
5.Mohr’s circle & stress transformation

Answer: σy′ = −3.84 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 for the y′ face

σy′ = −5 − 15·0.5 − (−10)·0.866 = −3.84 MPa

Check

The sum of the normal stresses does not change with rotation.

σx′ = −6.16 MPa, σx′ + σy′ = −10 = σx + σy

Answer

σy′ = −3.84 MPa
6.Mohr’s circle & stress transformation

Answer: n = 1.6

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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.

στxyCσ1​σ2​

Principal stresses

C = 70 MPa, R = 78.1 MPa
σ1 = 148.1 MPa, σ2 = −8.102 MPa, σ3 = 0

Tresca

σmax − σmin = 156.2 MPa
n = 250/156.2 = 1.6

Answer

n = 1.6

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: δ = 0.8754 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 = 55·1000/314.2 = 175.1 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 175.1/100000 = 0.001751 = 1751 µε

Elongation

δ = ε·L = 0.001751·500 mm = 0.8754 mm

Answer

δ = 0.8754 mm
8.Axial stress & strain

Answer: δ = 9.039 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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 119.5 MPa
σ2 = 366.1 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 119.5·1000/100000 = 1.195 mm
δ2 = σ2L2/E2 = 366.1·1500/70000 = 7.844 mm
δ = δ1 + δ2 = 9.039 mm

Answer

δ = 9.039 mm
9.Column buckling

Answer: Pcr = 161.1 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 = 40·30³/12 = 90000 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·90000/1050² = 161136 N = 161.1 kN

Answer

Pcr = 161.1 kN
10.Column buckling

Answer: σcr = 69.09 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 = 0.5·2000 = 1000 mm
KL/r = 1000/10 = 100

Critical stress

σcr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σcr = 69.09 MPa
11.Mechanical vibrations

Answer: fd = 13.05 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 = √(80000/10) = 89.44 rad/s
fn = 89.44/(2π) = 14.24 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 14.24·√(1 − 0.4²) = 13.05 Hz

Answer

fd = 13.05 Hz
12.Mechanical vibrations

Answer: fn = 14.24 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 40 + 40 = 80 kN/m

Natural frequency

ωn = √(80000/10) = 89.44 rad/s
fn = 89.44/(2π) = 14.24 Hz

Answer

fn = 14.24 Hz

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