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

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
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: ε_T = α·ΔT

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

Mechanical vibrations

Natural frequency: ω_n = √(k/m), f_n = ω_n/(2π)
From static deflection: ω_n = √(g/δ_st)
Critical damping: c_c = 2√(km) = 2mω_n
Damping ratio: ζ = c/c_c
Damped frequency: ω_d = ω_n√(1 − ζ²)
Logarithmic decrement: δ = (1/n)·ln(x₀/x_n), 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

Two parallel shafts are 90 mm apart. They must be connected by a pair of spur gears with a module of 4 mm and a speed reduction of 1.5 : 1. How many teeth N₁ must the pinion have?

a = 90 mmm = 4 mmi = 1.5

N₁ =

2.Gears & gear trainsHard

An open belt connects a 125 mm driving pulley at 960 rpm to a 400 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 125 mmn₁ = 960 rpmd₂ = 400 mmC = 1000 mmμ = 0.3T₁ = 500 N

P = kW

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 50 mm and length 2.5 m; segment BC has diameter 30 mm and length 1.5 m. A torque of 1500 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 1500 N·md₁ = 50 mm, L₁ = 2.5 md₂ = 30 mm, L₂ = 1.5 m

τ_max = MPa

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 10 kWN = 500 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 100 MPa, σᵧ = −70 MPa and τₓᵧ = 60 MPa (tension positive). The element is rotated 30° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 100 MPaσ_y = −70 MPaτ_xy = 60 MPaθ = 30°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 40 MPa, σᵧ = −30 MPa and τₓᵧ = −30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 40 MPaσ_y = −30 MPaτ_xy = −30 MPaθ = 20°

τ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaδ = 0.5 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø80/Ø50 mm, E = 120 GPa, σ_allow = 70 MPa

P_max = kN

9.Column bucklingMedium

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

L = 5 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 125 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 125 kNn = 2.5L = 4 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 2 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency f_d of its free vibration in Hz.

m = 2 kgk = 15 kN/mζ = 0.7

f_d = Hz

12.Mechanical vibrationsHard

A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 5 kgk₁ = 10 kN/mk₂ = 12 kN/m

k_eq = kN/m

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: F_t = 2T/d.

Total number of teeth

N₁ + N₂ = 2a/m = 2·90/4 = 45

Split by the ratio

N₁ = 45/(1 + 1.5) = 18

Answer

N₁ = 18

2.Gears & gear trains

Answer: P = 1.812 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 = e^μθ, 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 = 2.866 rad

Slack-side tension

e^μθ = e^0.8597 = 2.362, T₂ = 500/2.362 = 211.6 N

Power

P = (500 − 211.6)·6.283 = 1.812 kW

Answer

P = 1.812 kW

3.Torsion of shafts

Answer: τ_max = 282.9 MPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Stress in each segment

T = 1500 N·m = 1.5 × 10⁶ N·mm

τ_AB = 16·1.5 × 10⁶/(π·50³) = 61.12 MPa

τ_BC = 16·1.5 × 10⁶/(π·30³) = 282.9 MPa

Answer

τ_max = 282.9 MPa

4.Torsion of shafts

Answer: d = 23.49 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 10000/52.36 = 191 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·190986/(π·75)) = 23.49 mm

Answer

d = 23.49 mm

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

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −79.46 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

σ_y′ = 15 − 85·0.5 − 60·0.866 = −79.46 MPa

Check

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

σ_x′ = 109.5 MPa, σ_x′ + σ_y′ = 30 = σ_x + σ_y

Answer

σ_y′ = −79.46 MPa

6.Mohr’s circle & stress transformation

Answer: τ_θ = −45.48 MPa

Transformation equation

τ_θ = −35·0.6428 + −30·0.766 = −45.48 MPa

Answer

τ_θ = −45.48 MPa

7.Axial stress & strain

Answer: P = 53.35 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, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·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 = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(100000·2827) = 1.415 × 10⁻⁶ mm/N

Load

P = 0.5/9.372 × 10⁻⁶ = 53348 N = 53.35 kN

Answer

P = 53.35 kN

8.Axial stress & strain

Answer: P_max = 361 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·3063 = 6.189 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0005833) = 0.0005833 (copper)

Largest load

P_max = 0.0005833·6.189 × 10⁸ = 361021 N = 361 kN

Answer

P_max = 361 kN

9.Column buckling

Answer: P_cr = 113.4 kN

Euler buckling: a slender column becomes unstable at P_cr = 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/3500² = 113394 N = 113.4 kN

Answer

P_cr = 113.4 kN

10.Column buckling

Answer: d = 70.91 mm

Required critical load

P_cr = n·P = 2.5·125000 = 312500 N

KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·312500·2800²/(π³·200000))^(1/4) = 70.91 mm

Answer

d = 70.91 mm

Check: the slenderness KL/r is 157.9, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 9.843 Hz

Free vibration of a mass–spring system: the undamped natural frequency is ω_n = √(k/m) in rad/s, and f_n = ω_n/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/c_c, with critical damping c_c = 2·√(k·m) = 2·m·ω_n. An underdamped system (ζ < 1) oscillates at ω_d = ω_n·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Undamped natural frequency

ω_n = √(15000/2) = 86.6 rad/s

f_n = 86.6/(2π) = 13.78 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 13.78·√(1 − 0.7²) = 9.843 Hz

Answer

f_d = 9.843 Hz

12.Mechanical vibrations

Answer: k_eq = 5.455 kN/m

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/k_eq = 1/10 + 1/12 → k_eq = 10·12/(10 + 12) = 5.455 kN/m

Answer

k_eq = 5.455 kN/m

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