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
d1​ = 120 mmd2​ = 200 mm1450 rpmn2​ = ?

A 120 mm pulley at 1450 rpm drives a 200 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.

n1 = 1450 rpmd1 = 120 mmd2 = 200 mms = 3 %
n2 = rpm
2.Gears & gear trainsHard
A = 12B = 24C = 16D = 60

A 2.2 kW motor at 960 rpm drives gear A (12 teeth), which meshes with gear B (24 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.

P = 2.2 kWnin = 960 rpmA = 12B = 24C = 16D = 60ηmesh = 95%
Tout = N·m
3.Torsion of shaftsMedium
T = 400 N·md = 25 mmL = 0.5 mφ = ?

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 = 25 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 40 mmTB​ = 250 N·mTC​ = 500 N·mABC3 m1.2 m

A solid steel shaft of diameter 40 mm is fixed at A. A torque of 250 N·m acts at B, 3 m from A, and a torque of 500 N·m in the same direction acts at the free end C, a further 1.2 m along. Find the maximum shear stress τmax in the shaft.

d = 40 mmTB = 250 N·mTC = 500 N·mAB = 3 mBC = 1.2 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 130σy​ = 40τxy​ = 30MPa

A point has σₓ = 130 MPa, σᵧ = 40 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 20° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σx = 130 MPaσy = 40 MPaτxy = 30 MPaθ = -20°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = −70τxy​ = −40MPa

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −70 MPa and τₓᵧ = −40 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 70 MPaσy = −70 MPaτxy = −40 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 30 mmCopper3 mP = 95 kN

Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 95 kNd = 30 mmL = 3 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
Ø 35 mmBrass1 mØ 12 mmBrass2 mP = 35 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 35 kNsegment 1: Ø35 mm, 1 m, brass (E = 100 GPa)segment 2: Ø12 mm, 2 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 5 md = 60 mmcross-section

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 = 5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
PL = 6 md = 30 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 = 6 mE = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx50 mm1 cycle34.3 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed one cycle later by a peak of 34.3 mm. Find the damping ratio ζ.

x0 = 50 mmx1 = 34.3 mmn = 1
ζ =
12.Mechanical vibrationsHard
k1​ = 5k2​ = 80k3​ = 40m = 50 kg

A 50 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 80 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 = 50 kgk1 = 5 kN/mk2 = 80 kN/mk3 = 40 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 843.9 rpm

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Speed without slip

1450·120/200 = 870 rpm

With slip

n2 = 870·0.97 = 843.9 rpm

Answer

n2 = 843.9 rpm
2.Gears & gear trains

Answer: Tout = 148.1 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 = 960·(12/24)·(16/60) = 128 rpm
ωout = 2π·128/60 = 13.4 rad/s

Output power and torque

Pout = 2200·0.95² = 1986 W
Tout = Pout/ωout = 1986/13.4 = 148.1 N·m

Answer

Tout = 148.1 N·m
3.Torsion of shafts

Answer: φ = 3.735 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·500/(80000·38350) = 0.06519 rad
φ = 0.06519·180/π = 3.735°

Answer

φ = 3.735°
4.Torsion of shafts

Answer: τmax = 59.68 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 = 500 N·m
TAB = TC + TB = 500 + 250 = 750 N·m

Maximum shear stress

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

|T|max = 750 N·m
τmax = 16·750000/(π·40³) = 59.68 MPa

Answer

τmax = 59.68 MPa
5.Mohr’s circle & stress transformation

Answer: σy′ = 69.81 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′ = 85 − 45·0.766 − 30·(−0.6428) = 69.81 MPa

Check

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

σx′ = 100.2 MPa, σx′ + σy′ = 170 = σx + σy

Answer

σy′ = 69.81 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 139.6 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. 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 = 0 MPa, R = 80.62 MPa
σ1 = 80.62 MPa, σ2 = −80.62 MPa, σ3 = 0

Von Mises

σvM = √(80.62² − 80.62·(−80.62) + (−80.62)²) = 139.6 MPa

Answer

σvM = 139.6 MPa
7.Axial stress & strain

Answer: ε = 1149 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·30²/4 = 706.9 mm²
σ = P/A = 95·1000/706.9 = 134.4 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 134.4/117000 = 0.001149 = 1149 µε

Answer

ε = 1149 µε
8.Axial stress & strain

Answer: σmax = 309.5 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·12²/4 = 113.1 mm²
σ1 = 36.38 MPa
σ2 = 309.5 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 309.5 MPa

Answer

σmax = 309.5 MPa
9.Column buckling

Answer: Pcr = 35.88 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/3500² = 35879 N = 35.88 kN

Answer

Pcr = 35.88 kN
10.Column buckling

Answer: σcr = 4.318 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.5·6000 = 3000 mm
KL/r = 3000/7.5 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa

Answer

σcr = 4.318 MPa
11.Mechanical vibrations

Answer: ζ = 0.05987

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(50/34.3) = 0.3769

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3769/√(39.48 + 0.142) = 0.05987

Answer

ζ = 0.05987
12.Mechanical vibrations

Answer: fn = 3.712 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 = 5 + 80 = 85 kN/m
keq = 85·40/(85 + 40) = 27.2 kN/m

Natural frequency

ωn = √(27200/50) = 23.32 rad/s
fn = 23.32/(2π) = 3.712 Hz

Answer

fn = 3.712 Hz

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