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​ = ?N2​

Two parallel shafts are 90 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 4 : 1. How many teeth N1 must the pinion have?

a = 90 mmm = 1.5 mmi = 4
N1 =
2.Gears & gear trainsHard
Ns​ = 24ring fixedNp​ = 30nc​ = ?

In a planetary gear set the sun gear has 24 teeth and each planet gear has 30 teeth. The ring gear is held fixed and the sun is driven at 1200 rpm. Find the speed nc of the planet carrier.

Ns = 24Np = 30ns = 1200 rpm
nc = rpm
3.Torsion of shaftsMedium
d1​ = 75 mmd2​ = 50 mmT = 2000 N·mABC2.5 m1.2 m

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 75 mm and length 2.5 m; segment BC has diameter 50 mm and length 1.2 m. A torque of 2000 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.

T = 2000 N·md1 = 75 mm, L1 = 2.5 md2 = 50 mm, L2 = 1.2 mG = 26 GPa
φC = °
4.Torsion of shaftsHard
d = 30 mmT0​ = 1000 N·mACBa = 1.5 mb = 3 m

A solid aluminium shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 1000 N·m is applied at C, 1.5 m from A and 3 m from B. Find the maximum shear stress τmax in the shaft.

d = 30 mmT0 = 1000 N·ma = 1.5 mb = 3 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −30σy​ = −10τxy​ = 10MPa

A point has σₓ = −30 MPa, σᵧ = −10 MPa and τₓᵧ = 10 MPa (tension positive). The element is rotated 45° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σx = −30 MPaσy = −10 MPaτxy = 10 MPaθ = -45°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 10σy​ = 20τxy​ = 30MPa

A point in a loaded part has σₓ = 10 MPa, σᵧ = 20 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σx = 10 MPaσy = 20 MPaτxy = 30 MPaθ = 20°
σθ = MPa
7.Axial stress & strainMedium
150 kNØ30120 GPaØ120/50600 mmδ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 50 mm, E = 120 GPa). Both are 600 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø50 mm, E = 120 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø3070 GPaØ100/70300 mm

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø30 mm, E = 200 GPa, σallow = 120 MPatube: Ø100/Ø70 mm, E = 70 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2.5 md = 40 mmcross-section

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 2.5 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 5 md = 60 mmcross-section

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

L = 5 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx10 mm4 cycles4.7 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 4 cycles later by a peak of 4.7 mm. Find the damping ratio ζ.

x0 = 10 mmx4 = 4.7 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 12k2​ = 60m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 60 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 = 12 kN/mk2 = 60 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: N1 = 24

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 = 2·90/1.5 = 120

Split by the ratio

N1 = 120/(1 + 4) = 24

Answer

N1 = 24
2.Gears & gear trains

Answer: nc = 266.7 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

Nr = Ns + 2·Np = 24 + 2·30 = 84

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(1200 − nc)/(0 − nc) = −84/24
nc = ns·Ns/(Ns + Nr) = 1200·24/108 = 266.7 rpm

Answer

nc = 266.7 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: φC = 12.17 °

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 moments

J1 = π·75⁴/32 = 3.106 × 10⁶ mm⁴
J2 = π·50⁴/32 = 613592 mm⁴

Twist of each segment

T = 2000 N·m = 2 × 10⁶ N·mm
G = 26 GPa = 26000 MPa
φAB = 2 × 10⁶·2500/(26000·3.106 × 10⁶) = 0.06191 rad
φBC = 2 × 10⁶·1200/(26000·613592) = 0.1504 rad

Total twist

φC = 0.06191 + 0.1504 = 0.2123 rad = 12.17°

Answer

φC = 12.17°
4.Torsion of shafts

Answer: τmax = 125.8 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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 1000
TA·a = TB·b
TA = 1000·3/4.5 = 666.7 N·m
TB = 1000·1.5/4.5 = 333.3 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τmax = 16·666667/(π·30³) = 125.8 MPa

Answer

τmax = 125.8 MPa
5.Mohr’s circle & stress transformation

Answer: σy′ = −10 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′ = −20 − (−10)·0 − 10·(−1) = −10 MPa

Check

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

σx′ = −30 MPa, σx′ + σy′ = −40 = σx + σy

Answer

σy′ = −10 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = 30.45 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 = 15 + −5·0.766 + 30·0.6428 = 30.45 MPa

Answer

σθ = 30.45 MPa
7.Axial stress & strain

Answer: δ = 0.07126 mm

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

Ar = π·30²/4 = 706.9 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·9346 = 1.263 × 10⁹ N

Shortening

δ = 150000·600/1.263 × 10⁹ = 0.07126 mm

Answer

δ = 0.07126 mm
8.Axial stress & strain

Answer: Pmax = 253.1 kN

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

Ar = π·30²/4 = 706.9 mm²
At = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·4006 = 4.218 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0007143) = 0.0006 (steel)

Largest load

Pmax = 0.0006·4.218 × 10⁸ = 253055 N = 253.1 kN

Answer

Pmax = 253.1 kN
9.Column buckling

Answer: Pcr = 39.69 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 1·2500 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/2500² = 39688 N = 39.69 kN

Answer

Pcr = 39.69 kN
10.Column buckling

Answer: σcr = 1.554 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·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 2·5000 = 10000 mm
KL/r = 10000/15 = 666.7

Critical stress

σcr = π²E/(KL/r)² = π²·70000/666.7² = 1.554 MPa

Answer

σcr = 1.554 MPa
11.Mechanical vibrations

Answer: ζ = 0.03003

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 = (1/4)·ln(10/4.7) = 0.1888

Damping ratio

ζ = δ/√(4π² + δ²) = 0.1888/√(39.48 + 0.03563) = 0.03003

Answer

ζ = 0.03003
12.Mechanical vibrations

Answer: fn = 13.5 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 = 12 + 60 = 72 kN/m

Natural frequency

ωn = √(72000/10) = 84.85 rad/s
fn = 84.85/(2π) = 13.5 Hz

Answer

fn = 13.5 Hz

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