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​ = 25N2​ = 32

A 2.2 kW motor running at 960 rpm drives a pinion with 25 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 97%. Find the torque T1 on the motor shaft.

P = 2.2 kWn1 = 960 rpmN1 = 25N2 = 32η = 97%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 250 mm720 rpmC = 500 mmT1​ = 600 NT2​P = ?

An open belt connects a 100 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 100 mmn1 = 720 rpmd2 = 250 mmC = 500 mmμ = 0.3T1 = 600 N
P = kW
3.Torsion of shaftsMedium
T = 400 N·md = 20 mmL = 1.2 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 = 400 N·md = 20 mmL = 1.2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 20 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 20 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 40σy​ = 50MPaτxy​ = ?

A point has σₓ = 40 MPa and σᵧ = 50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 75 MPa?

σx = 40 MPaσy = 50 MPaσ1 = 75 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 60σy​ = −20τxy​ = −20MPa

A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = −20 MPa and τₓᵧ = −20 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 60 MPaσy = −20 MPaτxy = −20 MPa
σvM = MPa
7.Axial stress & strainMedium
150 kNØ20110 GPaØ100/70600 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 70 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 110 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø3070 GPaØ120/50300 mm

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø30 mm, E = 200 GPa, σallow = 160 MPatube: Ø120/Ø50 mm, E = 70 GPa, σallow = 70 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2 m30 × 40 mmcross-section

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

L = 2 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
P = 10 kNL = 4 md = ?cross-section

A solid round column of length L made of aluminium is fixed at both ends. It must carry 10 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 10 kNn = 3L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
k = 150 kN/mc = 690 N·s/mm = 5 kg

A 5 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 690 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 150 kN/mc = 690 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 60k2​ = 100m = 100 kg

A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 100 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 = 60 kN/mk2 = 100 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T1 = 21.88 N·m

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.

Motor torque

ω₁ = 2π·960/60 = 100.5 rad/s
T1 = P/ω₁ = 2200/100.5 = 21.88 N·m

Answer

T1 = 21.88 N·m
2.Gears & gear trains

Answer: P = 1.297 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 = 2.84 rad

Slack-side tension

eμθ = e0.8521 = 2.345, T₂ = 600/2.345 = 255.9 N

Power

P = (600 − 255.9)·3.77 = 1.297 kW

Answer

P = 1.297 kW
3.Torsion of shafts

Answer: φ = 21.89 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·1200/(80000·15708) = 0.382 rad
φ = 0.382·180/π = 21.89°

Answer

φ = 21.89°
4.Torsion of shafts

Answer: d = 29.6 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·75)) = 29.6 mm

Answer

d = 29.6 mm

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

5.Mohr’s circle & stress transformation

Answer: τxy = 29.58 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 = (40 + 50)/2 = 45 MPa
R = σ1 − C = 75 − 45 = 30 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(30² − 5²) = 29.58 MPa

Answer

τxy = 29.58 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 80 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. 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 = 20 MPa, R = 44.72 MPa
σ1 = 64.72 MPa, σ2 = −24.72 MPa, σ3 = 0

Von Mises

σvM = √(64.72² − 64.72·(−24.72) + (−24.72)²) = 80 MPa

Answer

σvM = 80 MPa
7.Axial stress & strain

Answer: δ = 0.1788 mm

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·20²/4 = 314.2 mm²
At = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·4006 = 5.034 × 10⁸ N

Shortening

δ = 150000·600/5.034 × 10⁸ = 0.1788 mm

Answer

δ = 0.1788 mm
8.Axial stress & strain

Answer: Pmax = 636.5 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·30²/4 = 706.9 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·9346 = 7.956 × 10⁸ N

Governing strain

ε = min(0.0008, 0.001) = 0.0008 (steel)

Largest load

Pmax = 0.0008·7.956 × 10⁸ = 636487 N = 636.5 kN

Answer

Pmax = 636.5 kN
9.Column buckling

Answer: Pcr = 44.41 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 = 40·30³/12 = 90000 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·90000/2000² = 44413 N = 44.41 kN

Answer

Pcr = 44.41 kN
10.Column buckling

Answer: d = 43.37 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 = 3·10000 = 30000 N
KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·30000·2000²/(π³·70000))^(1/4) = 43.37 mm

Answer

d = 43.37 mm

Check: the slenderness KL/r is 184.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.3984

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·√(150000·5) = 1732 N·s/m

Damping ratio

ζ = c/cc = 690/1732 = 0.3984

Answer

ζ = 0.3984

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

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 60 + 100 = 160 kN/m

Natural frequency

ωn = √(160000/100) = 40 rad/s
fn = 40/(2π) = 6.366 Hz

Answer

fn = 6.366 Hz

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