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 60 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 4 : 1. How many teeth N2 must the gear have?

a = 60 mmm = 1.5 mmi = 4
N2 =
2.Gears & gear trainsHard
d1​ = 125 mmd2​ = 500 mm960 rpmC = 1200 mmT1​ = 1200 NT2​P = ?

An open belt connects a 125 mm driving pulley at 960 rpm to a 500 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1200 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 125 mmn1 = 960 rpmd2 = 500 mmC = 1200 mmμ = 0.35T1 = 1200 N
P = kW
3.Torsion of shaftsMedium
do​ = ?T = 750 N·m

A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.

T = 750 N·mdi/do = 0.8τallow = 50 MPa
do = mm
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 75 kW at 500 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 75 kWN = 500 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 130τxy​ = −10MPa

A point has σₓ = 130 MPa, σᵧ = 0 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 30° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σx = 130 MPaσy = 0 MPaτxy = −10 MPaθ = 30°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 100σy​ = 50τxy​ = −20MPa

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = 50 MPa and τₓᵧ = −20 MPa (σz = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 100 MPaσy = 50 MPaτxy = −20 MPaσY = 275 MPa
n =
7.Axial stress & strainMedium
Ø 30 mmCopper0.5 mP = 110 kN

Find the elongation δ of the copper bar under the axial load.

P = 110 kNd = 30 mmL = 0.5 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
80 kN12500 mm500 mmØ40Ø50σ2​ = ?

A steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 50 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: brass, Ø50 mm, 500 mm, E = 100 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2.5 md = 30 mmcross-section

A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 2.5 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 2.5 md = 20 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 = 2.5 mE = 200 GPa (steel)d = 20 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 10 kN/mζ = 0.35m = 50 kg

A 50 kg mass on a spring with a stiffness of 10 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency fd of its free vibration in Hz.

m = 50 kgk = 10 kN/mζ = 0.35
fd = Hz
12.Mechanical vibrationsHard
k1​ = 12k2​ = 10m = 10 kg

A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 10 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 10 kgk1 = 12 kN/mk2 = 10 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: N2 = 64

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 = 2·60/1.5 = 80

Split by the ratio

N1 = 80/(1 + 4) = 16
N2 = 4·16 = 64

Answer

N2 = 64
2.Gears & gear trains

Answer: P = 4.737 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 = 2.828 rad

Slack-side tension

eμθ = e0.9897 = 2.691, T₂ = 1200/2.691 = 446 N

Power

P = (1200 − 446)·6.283 = 4.737 kW

Answer

P = 4.737 kW
3.Torsion of shafts

Answer: do = 50.58 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 = 750 N·m = 750000 N·mm
1 − k⁴ = 1 − 0.8⁴ = 0.5904
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·750000/(π·50·0.5904)) = 50.58 mm
di = 0.8·50.58 = 40.46 mm

Answer

do = 50.58 mm

A solid shaft would need d = 42.43 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 56.71 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·40)) = 56.71 mm

Answer

d = 56.71 mm

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

5.Mohr’s circle & stress transformation

Answer: σy′ = 41.16 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′ = 65 − 65·0.5 − (−10)·0.866 = 41.16 MPa

Check

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

σx′ = 88.84 MPa, σx′ + σy′ = 130 = σx + σy

Answer

σy′ = 41.16 MPa
6.Mohr’s circle & stress transformation

Answer: n = 2.948

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 = 75 MPa, R = 32.02 MPa
σ1 = 107 MPa, σ2 = 42.98 MPa, σ3 = 0

Von Mises

σvM = √(107² − 107·42.98 + 42.98²) = 93.27 MPa
n = 275/93.27 = 2.948

Answer

n = 2.948

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

7.Axial stress & strain

Answer: δ = 0.665 mm

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 = 110·1000/706.9 = 155.6 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 155.6/117000 = 0.00133 = 1330 µε

Elongation

δ = ε·L = 0.00133·500 mm = 0.665 mm

Answer

δ = 0.665 mm
8.Axial stress & strain

Answer: σ2 = 17.87 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 = 200000·1257/500 = 502655 N/mm
k₂ = 100000·1963/500 = 392699 N/mm

Share of segment 2

F2 = 80000·392699/895354 = 35088 N

Stress

σ2 = 35088/1963 = 17.87 MPa (compression)

Answer

σ2 = 17.87 MPa
9.Column buckling

Answer: Pcr = 25.63 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/1750² = 25628 N = 25.63 kN

Answer

Pcr = 25.63 kN
10.Column buckling

Answer: σcr = 31.58 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 = 0.5·2500 = 1250 mm
KL/r = 1250/5 = 250

Critical stress

σcr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa

Answer

σcr = 31.58 MPa
11.Mechanical vibrations

Answer: fd = 2.108 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 = √(10000/50) = 14.14 rad/s
fn = 14.14/(2π) = 2.251 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 2.251·√(1 − 0.35²) = 2.108 Hz

Answer

fd = 2.108 Hz
12.Mechanical vibrations

Answer: keq = 5.455 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 = 1/12 + 1/10 → keq = 12·10/(12 + 10) = 5.455 kN/m

Answer

keq = 5.455 kN/m

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