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
d1​ = 120 mmd2​ = 300 mm960 rpmT1​ = 1500 NT2​ = 300 NP = ?

A belt runs on a 120 mm driving pulley at 960 rpm. The tension is 1500 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.

d1 = 120 mmn1 = 960 rpmT1 = 1500 NT2 = 300 N
P = kW
2.Gears & gear trainsHard
A = 18B = 48C = 40D = 45

In a compound gear train, gear A (18 teeth) on the input shaft turning at 3000 rpm drives gear B (48 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D (45 teeth) on the output shaft. Find the output speed nout.

nin = 3000 rpmA = 18B = 48C = 40D = 45
nout = rpm
3.Torsion of shaftsMedium
T = 500 N·md = 25 mmL = 0.5 mφ = ?

A solid steel 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 = 500 N·md = 25 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 20 kW at 500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 20 kWN = 500 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 30σy​ = 110τxy​ = 20MPa

A point has σₓ = 30 MPa, σᵧ = 110 MPa and τₓᵧ = 20 MPa (tension positive). The element is rotated 25° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σx = 30 MPaσy = 110 MPaτxy = 20 MPaθ = 25°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 80σy​ = −80τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = −80 MPa and τₓᵧ = −50 MPa (σz = 0). The material has a yield strength of 300 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 80 MPaσy = −80 MPaτxy = −50 MPaσY = 300 MPa
n =
7.Axial stress & strainMedium
Ø 30 mmAluminium0.5 mP = 90 kN

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

P = 90 kNd = 30 mmL = 0.5 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
Ø 30 mmSteel1 mØ 12 mmCopper2 mP = 100 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 100 kNsegment 1: Ø30 mm, 1 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
PL = 1 m50 × 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 = 1 mE = 200 GPa (steel)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 40 mmcross-section

A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 4 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx40 mm4 cycles0.9 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 4 cycles later by a peak of 0.9 mm. Find the damping ratio ζ.

x0 = 40 mmx4 = 0.9 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 30k2​ = 20m = 100 kg

A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 20 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 = 100 kgk1 = 30 kN/mk2 = 20 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 7.238 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.

Belt speed

v = 6.032 m/s

Power

P = (1500 − 300)·6.032 = 7238 W = 7.238 kW

Answer

P = 7.238 kW
2.Gears & gear trains

Answer: nout = 1000 rpm

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

First stage

nB = 3000·18/48 = 1125 rpm = nC

Second stage

nout = 1125·40/45 = 1000 rpm

Answer

nout = 1000 rpm

total ratio = 3 : 1

3.Torsion of shafts

Answer: φ = 4.669 °

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) = 500·1000·500/(80000·38350) = 0.08149 rad
φ = 0.08149·180/π = 4.669°

Answer

φ = 4.669°
4.Torsion of shafts

Answer: d = 31.89 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 = 20000/52.36 = 382 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·381972/(π·60)) = 31.89 mm

Answer

d = 31.89 mm

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

5.Mohr’s circle & stress transformation

Answer: σy′ = 80.39 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′ = 70 − (−40)·0.6428 − 20·0.766 = 80.39 MPa

Check

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

σx′ = 59.61 MPa, σx′ + σy′ = 140 = σx + σy

Answer

σy′ = 80.39 MPa
6.Mohr’s circle & stress transformation

Answer: n = 1.59

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 = 0 MPa, R = 94.34 MPa
σ1 = 94.34 MPa, σ2 = −94.34 MPa, σ3 = 0

Tresca

σmax − σmin = 188.7 MPa
n = 300/188.7 = 1.59

Answer

n = 1.59

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

7.Axial stress & strain

Answer: ε = 1819 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·30²/4 = 706.9 mm²
σ = P/A = 90·1000/706.9 = 127.3 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 127.3/70000 = 0.001819 = 1819 µε

Answer

ε = 1819 µε
8.Axial stress & strain

Answer: δ = 15.82 mm

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·12²/4 = 113.1 mm²
σ1 = 141.5 MPa
σ2 = 884.2 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 141.5·1000/200000 = 0.7074 mm
δ2 = σ2L2/E2 = 884.2·2000/117000 = 15.11 mm
δ = δ1 + δ2 = 15.82 mm

Answer

δ = 15.82 mm
9.Column buckling

Answer: Pcr = 308.4 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 = 60·50³/12 = 625000 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·625000/2000² = 308425 N = 308.4 kN

Answer

Pcr = 308.4 kN
10.Column buckling

Answer: σcr = 49.35 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/10 = 200

Critical stress

σcr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa

Answer

σcr = 49.35 MPa
11.Mechanical vibrations

Answer: ζ = 0.1493

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(40/0.9) = 0.9486

Damping ratio

ζ = δ/√(4π² + δ²) = 0.9486/√(39.48 + 0.8998) = 0.1493

Answer

ζ = 0.1493
12.Mechanical vibrations

Answer: fn = 3.559 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 = 30 + 20 = 50 kN/m

Natural frequency

ωn = √(50000/100) = 22.36 rad/s
fn = 22.36/(2π) = 3.559 Hz

Answer

fn = 3.559 Hz

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