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 150 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 3 : 1. How many teeth N2 must the gear have?

a = 150 mmm = 5 mmi = 3
N2 =
2.Gears & gear trainsHard
d1​ = 160 mmd2​ = 315 mm2880 rpmC = 600 mmT1​ = 1500 NT2​P = ?

An open belt connects a 160 mm driving pulley at 2880 rpm to a 315 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 160 mmn1 = 2880 rpmd2 = 315 mmC = 600 mmμ = 0.3T1 = 1500 N
P = kW
3.Torsion of shaftsMedium
d = 12 mmT = 50 N·m200 mmφ = 3.52°

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 200 mm rotate 3.52° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 12 mmL = 200 mmφ = 3.52°
G = GPa
4.Torsion of shaftsHard
d = 30 mmT0​ = 400 N·mACBa = 0.8 mb = 0.5 m

A solid brass shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 0.8 m from A and 0.5 m from B. Find the reaction torque TA at wall A.

d = 30 mmT0 = 400 N·ma = 0.8 mb = 0.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 140σy​ = 70τxy​ = 60MPa

A point in a loaded part has σₓ = 140 MPa, σᵧ = 70 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 140 MPaσy = 70 MPaτxy = 60 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​τxy​surface of the shaft

A solid shaft of diameter 60 mm carries an axial tensile force of 80 kN together with a torque of 1000 N·m. At the surface of the shaft, find the largest principal stress σ1.

d = 60 mmF = 80 kNT = 1000 N·m
σ1 = MPa
7.Axial stress & strainMedium
150 kNØ20100 GPaØ120/60300 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 60 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/Ø60 mm, E = 100 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
40 kN12300 mm400 mmØ20Ø60σ2​ = ?

A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of copper (diameter 60 mm, length 400 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: copper, Ø60 mm, 400 mm, E = 120 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
PL = 6 md = 50 mmcross-section

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

L = 6 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 20 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)d = 20 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 150 kN/mc = 180 N·s/mm = 8 kg

An 8 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 180 N·s/m. Find the damping ratio ζ.

m = 8 kgk = 150 kN/mc = 180 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 80k2​ = 100m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 10 kgk1 = 80 kN/mk2 = 100 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: N2 = 45

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 = 2·150/5 = 60

Split by the ratio

N1 = 60/(1 + 3) = 15
N2 = 3·15 = 45

Answer

N2 = 45
2.Gears & gear trains

Answer: P = 20.95 kW

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.

Angle of wrap

θ = 2.883 rad

Slack-side tension

eμθ = e0.8648 = 2.374, T₂ = 1500/2.374 = 631.7 N

Power

P = (1500 − 631.7)·24.13 = 20.95 kW

Answer

P = 20.95 kW
3.Torsion of shafts

Answer: G = 79.96 GPa

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 moment and angle

J = π·12⁴/32 = 2036 mm⁴
φ = 3.52·π/180 = 0.06144 rad

Shear modulus

T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·200/(2036·0.06144) = 79957 MPa
G = 79.96 GPa

Answer

G = 79.96 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: TA = 153.8 N·m

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 = 400
TA·a = TB·b
TA = 400·0.5/1.3 = 153.8 N·m
TB = 400·0.8/1.3 = 246.2 N·m

Answer

TA = 153.8 N·m
5.Mohr’s circle & stress transformation

Answer: τmax = 69.46 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((140 − 70)/2)² + 60²) = 69.46 MPa

Answer

τmax = 69.46 MPa
6.Mohr’s circle & stress transformation

Answer: σ1 = 41.64 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.

Stresses at the surface

σ = F/A = 80000/(π·60²/4) = 80000/2827 = 28.29 MPa
τ = 16T/(πd³) = 16·1000000/(π·60³) = 23.58 MPa
στxyCσ1​σ2​

Mohr’s circle

C = σ/2 = 14.15 MPa
R = √(14.15² + 23.58²) = 27.5 MPa
σ1 = C + R = 41.64 MPa
σ2 = C − R = −13.35 MPa

Answer

σ1 = 41.64 MPa
7.Axial stress & strain

Answer: δ = 0.04939 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 − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·8482 = 9.111 × 10⁸ N

Shortening

δ = 150000·300/9.111 × 10⁸ = 0.04939 mm

Answer

δ = 0.04939 mm
8.Axial stress & strain

Answer: σ2 = 11.35 MPa

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.

Stiffnesses

k₁ = 200000·314.2/300 = 209440 N/mm
k₂ = 120000·2827/400 = 848230 N/mm

Share of segment 2

F2 = 40000·848230/1.058 × 10⁶ = 32079 N

Stress

σ2 = 32079/2827 = 11.35 MPa (compression)

Answer

σ2 = 11.35 MPa
9.Column buckling

Answer: Pcr = 67.29 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·306796/3000² = 67288 N = 67.29 kN

Answer

Pcr = 67.29 kN
10.Column buckling

Answer: σcr = 8.812 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·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²
r = √(I/A) = √(7854/314.2) = 5 mm
KL = 0.7·2000 = 1400 mm
KL/r = 1400/5 = 280

Critical stress

σcr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa

Answer

σcr = 8.812 MPa
11.Mechanical vibrations

Answer: ζ = 0.08216

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·√(150000·8) = 2191 N·s/m

Damping ratio

ζ = c/cc = 180/2191 = 0.08216

Answer

ζ = 0.08216

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

12.Mechanical vibrations

Answer: keq = 180 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 80 + 100 = 180 kN/m

Answer

keq = 180 kN/m

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