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 63 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 2.5 : 1. How many teeth N1 must the pinion have?

a = 63 mmm = 2 mmi = 2.5
N1 =
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 200 mm1500 rpmC = 1200 mmT1​ = 800 NT2​P = ?

An open belt connects an 80 mm driving pulley at 1500 rpm to a 200 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 80 mmn1 = 1500 rpmd2 = 200 mmC = 1200 mmμ = 0.3T1 = 800 N
P = kW
3.Torsion of shaftsMedium
T = 250 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 = 250 N·md = 25 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 60 mmT0​ = 250 N·mACBa = 2 mb = 0.5 m

A solid brass shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 2 m from A and 0.5 m from B. Find the angle of rotation φC of section C, in degrees.

d = 60 mmT0 = 250 N·ma = 2 mb = 0.5 mG = 39 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
σx​ = 0σy​ = 90τxy​ = −30MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = 90 MPa and τₓᵧ = −30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 0 MPaσy = 90 MPaτxy = −30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 0σy​ = 110τxy​ = 30MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σx = 0 MPaσy = 110 MPaτxy = 30 MPaθ = 40°
σθ = MPa
7.Axial stress & strainMedium
P = ?12500 mm400 mmØ40Ø50δ = 0.25 mm

A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 400 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 400 mm, E = 110 GPaδ = 0.25 mm
P = kN
8.Axial stress & strainHard
20 kN12300 mm300 mmØ20Ø30σ2​ = ?

A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 300 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 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: brass, Ø30 mm, 300 mm, E = 100 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 4 md = 40 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 = 4 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 20 mmcross-section

A column of length L made of steel 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 = 200 GPa (steel)d = 20 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx10 mm4 cycles2.8 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 4 cycles later by a peak of 2.8 mm. Find the damping ratio ζ.

x0 = 10 mmx4 = 2.8 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 5k2​ = 100m = 40 kg

A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 100 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 40 kgk1 = 5 kN/mk2 = 100 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: N1 = 18

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 = 2·63/2 = 63

Split by the ratio

N1 = 63/(1 + 2.5) = 18

Answer

N1 = 18
2.Gears & gear trains

Answer: P = 3.008 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 = 3.042 rad

Slack-side tension

eμθ = e0.9125 = 2.49, T₂ = 800/2.49 = 321.2 N

Power

P = (800 − 321.2)·6.283 = 3.008 kW

Answer

P = 3.008 kW
3.Torsion of shafts

Answer: φ = 2.334 °

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) = 250·1000·500/(80000·38350) = 0.04074 rad
φ = 0.04074·180/π = 2.334°

Answer

φ = 2.334°
4.Torsion of shafts

Answer: φC = 0.1155 °

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 = 250
TA·a = TB·b
TA = 250·0.5/2.5 = 50 N·m
TB = 250·2/2.5 = 200 N·m

Rotation of C

Use segment AC: it is fixed at A and carries TA.

J = π·60⁴/32 = 1.272 × 10⁶ mm⁴
G = 39 GPa = 39000 MPa
φC = TA·a/(GJ) = 50000·2000/(39000·1.272 × 10⁶) = 0.002015 rad = 0.1155°

Answer

φC = 0.1155°
5.Mohr’s circle & stress transformation

Answer: τmax = 54.08 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 = √(((0 − 90)/2)² + −30²) = 54.08 MPa

Answer

τmax = 54.08 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = 74.99 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

σθ = 55 + −55·0.1736 + 30·0.9848 = 74.99 MPa

Answer

σθ = 74.99 MPa
7.Axial stress & strain

Answer: P = 65.08 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

A₁ = 1257 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(110000·1963) = 1.852 × 10⁻⁶ mm/N

Load

P = 0.25/3.841 × 10⁻⁶ = 65080 N = 65.08 kN

Answer

P = 65.08 kN
8.Axial stress & strain

Answer: σ2 = 14.98 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 = 100000·706.9/300 = 235619 N/mm

Share of segment 2

F2 = 20000·235619/445059 = 10588 N

Stress

σ2 = 10588/706.9 = 14.98 MPa (compression)

Answer

σ2 = 14.98 MPa
9.Column buckling

Answer: Pcr = 62.01 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/2000² = 62013 N = 62.01 kN

Answer

Pcr = 62.01 kN
10.Column buckling

Answer: σcr = 25.18 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 = π²·200000/280² = 25.18 MPa

Answer

σcr = 25.18 MPa
11.Mechanical vibrations

Answer: ζ = 0.05058

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(10/2.8) = 0.3182

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3182/√(39.48 + 0.1013) = 0.05058

Answer

ζ = 0.05058
12.Mechanical vibrations

Answer: fn = 1.737 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/5 + 1/100 → keq = 5·100/(5 + 100) = 4.762 kN/m

Natural frequency

ωn = √(4762/40) = 10.91 rad/s
fn = 10.91/(2π) = 1.737 Hz

Answer

fn = 1.737 Hz

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