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​ = 25, m = 5 mm

A 5.5 kW motor running at 1450 rpm drives a pinion with 25 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.

P = 5.5 kWn = 1450 rpmN1 = 25m = 5 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 500 mm1450 rpmC = 1200 mmθ = ?

An open belt connects a 100 mm pulley and a 500 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 100 mmd2 = 500 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
T = 250 N·md = 25 mmL = 1.2 mφ = ?

A solid brass 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 = 250 N·md = 25 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 75 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 75 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 140σy​ = −20MPaτxy​ = ?

A point has σₓ = 140 MPa and σᵧ = −20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 190 MPa?

σx = 140 MPaσy = −20 MPaσ1 = 190 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = −40τxy​ = −20MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −40 MPa and τₓᵧ = −20 MPa (σz = 0). The material has a yield strength of 420 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = −10 MPaσy = −40 MPaτxy = −20 MPaσY = 420 MPa
n =
7.Axial stress & strainMedium
Ø 12 mmBrass3 mP = 25 kN

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

P = 25 kNd = 12 mmL = 3 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
Ø 30 mmSteel2 mØ 20 mmBrass1 mP = 70 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 = 70 kNsegment 1: Ø30 mm, 2 m, steel (E = 200 GPa)segment 2: Ø20 mm, 1 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 2.5 md = 25 mmcross-section

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

L = 2.5 mE = 70 GPa (aluminium)d = 25 mm
Pcr = kN
10.Column bucklingHard
P = 10 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 10 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 10 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 50 kN/mc = 600 N·s/mm = 80 kg

An 80 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 600 N·s/m. Find the damping ratio ζ.

m = 80 kgk = 50 kN/mc = 600 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 60k2​ = 20k3​ = 20m = 10 kg

A 10 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 60 kN/m and k₂ = 20 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 = 10 kgk1 = 60 kN/mk2 = 20 kN/mk3 = 20 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 579.5 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·1450/60 = 151.8 rad/s
T = P/ω = 5500/151.8 = 36.22 N·m

Pitch diameter

d = m·N = 5·25 = 125 mm

Tangential force

Ft = 2T/d = 2·36.22/0.125 = 579.5 N

Answer

Ft = 579.5 N
2.Gears & gear trains

Answer: θ = 160.8 °

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(0.1667) = 9.594°

Angle of wrap

θ = 180 − 2·9.594 = 160.8°

Answer

θ = 160.8°
3.Torsion of shafts

Answer: φ = 11.49 °

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) = 250·1000·1200/(39000·38350) = 0.2006 rad
φ = 0.2006·180/π = 11.49°

Answer

φ = 11.49°
4.Torsion of shafts

Answer: d = 45.99 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·500/60 = 52.36 rad/s
T = P/ω = 75000/52.36 = 1432 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·1.432 × 10⁶/(π·75)) = 45.99 mm

Answer

d = 45.99 mm

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

5.Mohr’s circle & stress transformation

Answer: τxy = 102.5 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.

Required radius

C = (140 + (−20))/2 = 60 MPa
R = σ1 − C = 190 − 60 = 130 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(130² − 80²) = 102.5 MPa

Answer

τxy = 102.5 MPa
6.Mohr’s circle & stress transformation

Answer: n = 8.4

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 = −25 MPa, R = 25 MPa
σ1 = 0 MPa, σ2 = −50 MPa, σ3 = 0

Tresca

σmax − σmin = 50 MPa
n = 420/50 = 8.4

Answer

n = 8.4

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: ε = 2210 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·12²/4 = 113.1 mm²
σ = P/A = 25·1000/113.1 = 221 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 221/100000 = 0.00221 = 2210 µε

Answer

ε = 2210 µε
8.Axial stress & strain

Answer: σmax = 222.8 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·30²/4 = 706.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 99.03 MPa
σ2 = 222.8 MPa

Largest stress

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

σmax = 222.8 MPa

Answer

σmax = 222.8 MPa
9.Column buckling

Answer: Pcr = 2.12 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 1·2500 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·19175/2500² = 2120 N = 2.12 kN

Answer

Pcr = 2.12 kN
10.Column buckling

Answer: d = 29.81 mm

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.

Required critical load

Pcr = n·P = 2.5·10000 = 25000 N
KL = 0.7·2500 = 1750 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·25000·1750²/(π³·200000))^(1/4) = 29.81 mm

Answer

d = 29.81 mm

Check: the slenderness KL/r is 234.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.15

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·√(50000·80) = 4000 N·s/m

Damping ratio

ζ = c/cc = 600/4000 = 0.15

Answer

ζ = 0.15

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

12.Mechanical vibrations

Answer: fn = 6.366 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 = 60 + 20 = 80 kN/m
keq = 80·20/(80 + 20) = 16 kN/m

Natural frequency

ωn = √(16000/10) = 40 rad/s
fn = 40/(2π) = 6.366 Hz

Answer

fn = 6.366 Hz

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