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​ = 60

An 11 kW motor running at 1450 rpm drives a pinion with 25 teeth, which meshes with a gear with 60 teeth. Ignore losses. Find the output torque T2 on the gear shaft.

P = 11 kWn1 = 1450 rpmN1 = 25N2 = 60η = 100%
T2 = N·m
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 500 mm1500 rpmC = 800 mmθ = ?

An open belt connects an 80 mm pulley and a 500 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 80 mmd2 = 500 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
d = ?T = 750 N·mL = 0.5 m

A solid aluminium shaft 0.5 m long carries a torque of 750 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 750 N·mL = 0.5 mG = 26 GPaφallow = 0.5°
d = mm
4.Torsion of shaftsHard
d = ?75 kW, 1450 rpm

A solid brass shaft transmits 75 kW at 1450 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 75 kWN = 1450 rpmτallow = 80 MPaθallow = 1°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 20σy​ = 50MPaτxy​ = ?

A point has σₓ = 20 MPa and σᵧ = 50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 105 MPa?

σx = 20 MPaσy = 50 MPaσ1 = 105 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 160σy​ = 50τxy​ = 50MPa

A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = 50 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 = 160 MPaσy = 50 MPaτxy = 50 MPaσY = 300 MPa
n =
7.Axial stress & strainMedium
P = ?12600 mm500 mmØ40Ø40δ = 0.2 mm

A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 40 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø40 mm, 500 mm, E = 120 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
P = ?Ø4070 GPaØ120/50300 mm

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø40 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø50 mm, E = 70 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
PL = 4 m20 × 60 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)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
P = 105 kNL = 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 105 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 105 kNn = 2.5L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 10 kN/mc = 110 N·s/mm = 2 kg

A 2 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 110 N·s/m. Find the damping ratio ζ.

m = 2 kgk = 10 kN/mc = 110 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 30k2​ = 100m = 10 kg

A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 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 = 10 kgk1 = 30 kN/mk2 = 100 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T2 = 173.9 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 11000/151.8 = 72.44 N·m

Output torque

T2 = 72.44·60/25·1 = 173.9 N·m

Answer

T2 = 173.9 N·m
2.Gears & gear trains

Answer: θ = 149.6 °

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(0.2625) = 15.22°

Angle of wrap

θ = 180 − 2·15.22 = 149.6°

Answer

θ = 149.6°
3.Torsion of shafts

Answer: d = 64.05 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.

Required polar moment

T = 750 N·m = 750000 N·mm
L = 0.5 m = 500 mm
G = 26 GPa = 26000 MPa
φ = 0.5·π/180 = 0.008727 rad
J = TL/(Gφ) = 750000·500/(26000·0.008727) = 1.653 × 10⁶ mm⁴

Diameter

d = (32J/π)1/4 = 64.05 mm

Answer

d = 64.05 mm

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

4.Torsion of shafts

Answer: d = 52.14 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·1450/60 = 151.8 rad/s
T = P/ω = 75000/151.8 = 493.9 N·m

Diameter for strength

T = 493.9 N·m = 493929 N·mm
dτ = ∛(16·493929/(π·80)) = 31.56 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·493929/(π·39000·1.745 × 10⁻⁵))1/4 = 52.14 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(31.56, 52.14) = 52.14 mm

Answer

d = 52.14 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τxy = 68.37 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 = (20 + 50)/2 = 35 MPa
R = σ1 − C = 105 − 35 = 70 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(70² − 15²) = 68.37 MPa

Answer

τxy = 68.37 MPa
6.Mohr’s circle & stress transformation

Answer: n = 1.673

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 = 105 MPa, R = 74.33 MPa
σ1 = 179.3 MPa, σ2 = 30.67 MPa, σ3 = 0

Tresca

σmax − σmin = 179.3 MPa
n = 300/179.3 = 1.673

Answer

n = 1.673

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

7.Axial stress & strain

Answer: P = 35.07 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 = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(120000·1257) = 3.316 × 10⁻⁶ mm/N

Load

P = 0.2/5.703 × 10⁻⁶ = 35069 N = 35.07 kN

Answer

P = 35.07 kN
8.Axial stress & strain

Answer: Pmax = 543.3 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·40²/4 = 1257 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·9346 = 9.056 × 10⁸ N

Governing strain

ε = min(0.0006, 0.001429) = 0.0006 (steel)

Largest load

Pmax = 0.0006·9.056 × 10⁸ = 543338 N = 543.3 kN

Answer

Pmax = 543.3 kN
9.Column buckling

Answer: Pcr = 19.74 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·40000/2000² = 19739 N = 19.74 kN

Answer

Pcr = 19.74 kN
10.Column buckling

Answer: d = 75.9 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 = 2.5·105000 = 262500 N
KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·262500·3500²/(π³·200000))^(1/4) = 75.9 mm

Answer

d = 75.9 mm

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

11.Mechanical vibrations

Answer: ζ = 0.3889

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·√(10000·2) = 282.8 N·s/m

Damping ratio

ζ = c/cc = 110/282.8 = 0.3889

Answer

ζ = 0.3889

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

12.Mechanical vibrations

Answer: fn = 7.646 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 = 1/30 + 1/100 → keq = 30·100/(30 + 100) = 23.08 kN/m

Natural frequency

ωn = √(23077/10) = 48.04 rad/s
fn = 48.04/(2π) = 7.646 Hz

Answer

fn = 7.646 Hz

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