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 160 mm apart. They must be connected by a pair of spur gears with a module of 4 mm and a speed reduction of 3 : 1. How many teeth N1 must the pinion have?

a = 160 mmm = 4 mmi = 3
N1 =
2.Gears & gear trainsHard
d1​ = 125 mmd2​ = 450 mm2880 rpmC = 500 mmθ = ?

An open belt connects a 125 mm pulley and a 450 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 125 mmd2 = 450 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
d1​ = 50 mmd2​ = 25 mmT = 100 N·mABC1.5 m1 m

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 50 mm and length 1.5 m; segment BC has diameter 25 mm and length 1 m. A torque of 100 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.

T = 100 N·md1 = 50 mm, L1 = 1.5 md2 = 25 mm, L2 = 1 mG = 26 GPa
φC = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 100 kW at 300 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 100 kWN = 300 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −50σy​ = −20τxy​ = 10MPa

A point in a loaded part has σₓ = −50 MPa, σᵧ = −20 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = −50 MPaσy = −20 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
60°D = 500 mm, t = 10 mm, p = 1.5 MPap

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 10 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.

D = 500 mmt = 10 mmp = 1.5 MPaα = 60°
σn = MPa
7.Axial stress & strainMedium
Ø 16 mmCopper3 mP = 25 kN

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

P = 25 kNd = 16 mmL = 3 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
P = ?Ø2570 GPaØ80/50500 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 200 MPatube: Ø80/Ø50 mm, E = 70 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2 m25 × 100 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 = 2 mE = 200 GPa (steel)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 50 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 = 3 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 4 kN/mζ = 0.3m = 8 kg

An 8 kg mass on a spring with a stiffness of 4 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.

m = 8 kgk = 4 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
k1​ = 15k2​ = 5k3​ = 12m = 20 kg

A 20 kg mass hangs from spring k₃ = 12 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 kN/m and k₂ = 5 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 = 20 kgk1 = 15 kN/mk2 = 5 kN/mk3 = 12 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: N1 = 20

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 = 2·160/4 = 80

Split by the ratio

N1 = 80/(1 + 3) = 20

Answer

N1 = 20
2.Gears & gear trains

Answer: θ = 142.1 °

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.

Belt angle

asin(0.325) = 18.97°

Angle of wrap

θ = 180 − 2·18.97 = 142.1°

Answer

θ = 142.1°
3.Torsion of shafts

Answer: φC = 6.285 °

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·50⁴/32 = 613592 mm⁴
J2 = π·25⁴/32 = 38350 mm⁴

Twist of each segment

T = 100 N·m = 100000 N·mm
G = 26 GPa = 26000 MPa
φAB = 100000·1500/(26000·613592) = 0.009402 rad
φBC = 100000·1000/(26000·38350) = 0.1003 rad

Total twist

φC = 0.009402 + 0.1003 = 0.1097 rad = 6.285°

Answer

φC = 6.285°
4.Torsion of shafts

Answer: d = 74 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.

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s
T = P/ω = 100000/31.42 = 3183 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·3.183 × 10⁶/(π·40)) = 74 mm

Answer

d = 74 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: τmax = 18.03 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 = √(((−50 − −20)/2)² + 10²) = 18.03 MPa

Answer

τmax = 18.03 MPa
6.Mohr’s circle & stress transformation

Answer: σn = 23.44 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 500/2 = 250 mm
σh = pr/t = 1.5·250/10 = 37.5 MPa
σa = pr/(2t) = 18.75 MPa

Transform to the seam

Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.

σn = 18.75·sin²60° + 37.5·cos²60° = 23.44 MPa

Answer

σn = 23.44 MPa
7.Axial stress & strain

Answer: ε = 1063 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·16²/4 = 201.1 mm²
σ = P/A = 25·1000/201.1 = 124.3 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 124.3/117000 = 0.001063 = 1063 µε

Answer

ε = 1063 µε
8.Axial stress & strain

Answer: Pmax = 267.9 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 − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·3063 = 3.126 × 10⁸ N

Governing strain

ε = min(0.001, 0.0008571) = 0.0008571 (aluminium)

Largest load

Pmax = 0.0008571·3.126 × 10⁸ = 267933 N = 267.9 kN

Answer

Pmax = 267.9 kN
9.Column buckling

Answer: Pcr = 257 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 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 0.5·2000 = 1000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·130208/1000² = 257021 N = 257 kN

Answer

Pcr = 257 kN
10.Column buckling

Answer: σcr = 8.567 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/12.5 = 480

Critical stress

σcr = π²E/(KL/r)² = π²·200000/480² = 8.567 MPa

Answer

σcr = 8.567 MPa
11.Mechanical vibrations

Answer: fd = 3.395 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 = √(4000/8) = 22.36 rad/s
fn = 22.36/(2π) = 3.559 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 3.559·√(1 − 0.3²) = 3.395 Hz

Answer

fd = 3.395 Hz
12.Mechanical vibrations

Answer: fn = 3.082 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 = 15 + 5 = 20 kN/m
keq = 20·12/(20 + 12) = 7.5 kN/m

Natural frequency

ωn = √(7500/20) = 19.36 rad/s
fn = 19.36/(2π) = 3.082 Hz

Answer

fn = 3.082 Hz

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