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 = 3 mm

A 1.5 kW motor running at 960 rpm drives a pinion with 18 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.

P = 1.5 kWn = 960 rpmN1 = 18m = 3 mm
Ft = N
2.Gears & gear trainsHard
Ns​ = 20nr​ = ?Np​ = 30carrier fixed

In a planetary gear set the sun gear has 20 teeth and each planet gear has 30 teeth. The planet carrier is held fixed and the sun is driven at 1800 rpm. Find the speed nr of the ring gear (magnitude).

Ns = 20Np = 30ns = 1800 rpm
nr = rpm
3.Torsion of shaftsMedium
do​ = ?T = 100 N·m

A hollow shaft must carry a torque of 100 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.

T = 100 N·mdi/do = 0.6τallow = 50 MPa
do = mm
4.Torsion of shaftsHard
d = 25 mmT0​ = 250 N·mACBa = 1.5 mb = 0.8 m

A solid aluminium shaft of diameter 25 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 1.5 m from A and 0.8 m from B. Find the angle of rotation φC of section C, in degrees.

d = 25 mmT0 = 250 N·ma = 1.5 mb = 0.8 mG = 26 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
σx​ = 140σy​ = −10τxy​ = 30MPa

A point in a loaded part has σₓ = 140 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 140 MPaσy = −10 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = −70σy​ = 80τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = 80 MPa and τₓᵧ = 30 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = −70 MPaσy = 80 MPaτxy = 30 MPa
σvM = MPa
7.Axial stress & strainMedium
150 kNØ20100 GPaØ120/70300 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 70 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø20 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 100 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø25100 GPaØ80/60600 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 250 MPatube: Ø80/Ø60 mm, E = 100 GPa, σallow = 70 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2 md = 60 mmcross-section

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 = 2 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
PL = 5 md = 60 mmcross-section

A column of length L made of steel 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 = 200 GPa (steel)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 25 kN/mc = 20 N·s/mm = 2 kg

A 2 kg mass is supported by a spring with a stiffness of 25 kN/m and a viscous damper with a damping coefficient of 20 N·s/m. Find the damping ratio ζ.

m = 2 kgk = 25 kN/mc = 20 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 20k2​ = 40m = 40 kg

A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 40 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 40 kgk1 = 20 kN/mk2 = 40 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: Ft = 552.6 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·960/60 = 100.5 rad/s
T = P/ω = 1500/100.5 = 14.92 N·m

Pitch diameter

d = m·N = 3·18 = 54 mm

Tangential force

Ft = 2T/d = 2·14.92/0.054 = 552.6 N

Answer

Ft = 552.6 N
2.Gears & gear trains

Answer: nr = 450 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 = 20 + 2·30 = 80

Willis equation with the carrier fixed

With nc = 0 the sun drives the ring through the planets, which act as idlers.

nr = −ns·Ns/Nr = −1800·20/80 = −450 rpm
|nr| = 450 rpm

Answer

nr = 450 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: do = 22.7 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.

Outer diameter from the allowable stress

T = 100 N·m = 100000 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·100000/(π·50·0.8704)) = 22.7 mm
di = 0.6·22.7 = 13.62 mm

Answer

do = 22.7 mm

A solid shaft would need d = 21.68 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: φC = 7.495 °

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 = 250
TA·a = TB·b
TA = 250·0.8/2.3 = 86.96 N·m
TB = 250·1.5/2.3 = 163 N·m

Rotation of C

Use segment AC: it is fixed at A and carries TA.

J = π·25⁴/32 = 38350 mm⁴
G = 26 GPa = 26000 MPa
φC = TA·a/(GJ) = 86957·1500/(26000·38350) = 0.1308 rad = 7.495°

Answer

φC = 7.495°
5.Mohr’s circle & stress transformation

Answer: θp = 10.9 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·30, 140 − −10) = 21.8°

Angle on the element

θp = 10.9°

Answer

θp = 10.9°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σvM = 140 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 = 5 MPa, R = 80.78 MPa
σ1 = 85.78 MPa, σ2 = −75.78 MPa, σ3 = 0

Von Mises

σvM = √(85.78² − 85.78·(−75.78) + (−75.78)²) = 140 MPa

Answer

σvM = 140 MPa
7.Axial stress & strain

Answer: δ = 0.05563 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·20²/4 = 314.2 mm²
At = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·7461 = 8.09 × 10⁸ N

Shortening

δ = 150000·300/8.09 × 10⁸ = 0.05563 mm

Answer

δ = 0.05563 mm
8.Axial stress & strain

Answer: Pmax = 222.7 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·25²/4 = 490.9 mm²
At = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·2199 = 3.181 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0007) = 0.0007 (brass)

Largest load

Pmax = 0.0007·3.181 × 10⁸ = 222660 N = 222.7 kN

Answer

Pmax = 222.7 kN
9.Column buckling

Answer: Pcr = 78.48 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 = 2·2000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·636173/4000² = 78485 N = 78.48 kN

Answer

Pcr = 78.48 kN
10.Column buckling

Answer: σcr = 4.441 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 = π²·200000/666.7² = 4.441 MPa

Answer

σcr = 4.441 MPa
11.Mechanical vibrations

Answer: ζ = 0.04472

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.

Critical damping

cc = 2·√(k·m) = 2·√(25000·2) = 447.2 N·s/m

Damping ratio

ζ = c/cc = 20/447.2 = 0.04472

Answer

ζ = 0.04472

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: keq = 13.33 kN/m

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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/20 + 1/40 → keq = 20·40/(20 + 40) = 13.33 kN/m

Answer

keq = 13.33 kN/m

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