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 150 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 5 : 1. How many teeth N₁ must the pinion have?

a = 150 mmm = 2 mmi = 5

N₁ =

2.Gears & gear trainsHard

In a compound gear train, gear A (12 teeth) on the input shaft turning at 3000 rpm drives gear B (24 teeth). Gear C (25 teeth) is fixed to the same shaft as B and drives gear D (50 teeth) on the output shaft. Find the output speed n_out.

n_in = 3000 rpmA = 12B = 24C = 25D = 50

n_out = rpm

3.Torsion of shaftsMedium

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 = 25 mmL = 0.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1450 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 80 MPa, σᵧ = −40 MPa and τₓᵧ = −20 MPa (tension positive). The element is rotated 50° 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 = 80 MPaσ_y = −40 MPaτ_xy = −20 MPaθ = 50°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa (σ_z = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 20 MPaσ_y = −70 MPaτ_xy = −30 MPaσ_Y = 355 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 300 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, Ø30 mm, 300 mm, E = 200 GPasegment 2: copper, Ø60 mm, 300 mm, E = 120 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 50 MPa

P_max = kN

9.Column bucklingMedium

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 6 mE = 70 GPa (aluminium)b × h = 40 × 60 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium 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 = 3 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

A 50 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 150 N·s/m. Find the damping ratio ζ.

m = 50 kgk = 12 kN/mc = 150 N·s/m

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 20 kgk₁ = 100 kN/mk₂ = 80 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: N₁ = 25

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 = 2·150/2 = 150

Split by the ratio

N₁ = 150/(1 + 5) = 25

Answer

N₁ = 25

2.Gears & gear trains

Answer: n_out = 750 rpm

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.

First stage

n_B = 3000·12/24 = 1500 rpm = n_C

Second stage

n_out = 1500·25/50 = 750 rpm

Answer

n_out = 750 rpm

total ratio = 4 : 1

3.Torsion of shafts

Answer: φ = 3.735 °

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.

Polar second moment of area

J = π·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·500/(80000·38350) = 0.06519 rad

φ = 0.06519·180/π = 3.735°

Answer

φ = 3.735°

4.Torsion of shafts

Answer: d = 18.86 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 15000/151.8 = 98.79 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·98786/(π·75)) = 18.86 mm

Answer

d = 18.86 mm

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

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 50.12 MPa

Transformation equation for the y′ face

σ_y′ = 20 − 60·(−0.1736) − (−20)·0.9848 = 50.12 MPa

Check

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

σ_x′ = −10.12 MPa, σ_x′ + σ_y′ = 40 = σ_x + σ_y

Answer

σ_y′ = 50.12 MPa

6.Mohr’s circle & stress transformation

Answer: n = 3.282

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 σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = −25 MPa, R = 54.08 MPa

σ₁ = 29.08 MPa, σ₂ = −79.08 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 108.2 MPa

n = 355/108.2 = 3.282

Answer

n = 3.282

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

7.Axial stress & strain

Answer: P = 66.53 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 = 706.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(120000·2827) = 8.842 × 10⁻⁷ mm/N

Load

P = 0.2/3.006 × 10⁻⁶ = 66528 N = 66.53 kN

Answer

P = 66.53 kN

8.Axial stress & strain

Answer: P_max = 379.8 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·4006 = 5.317 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0007143) = 0.0007143 (aluminium)

Largest load

P_max = 0.0007143·5.317 × 10⁸ = 379796 N = 379.8 kN

Answer

P_max = 379.8 kN

9.Column buckling

Answer: P_cr = 24.56 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 60·40³/12 = 320000 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·320000/3000² = 24564 N = 24.56 kN

Answer

P_cr = 24.56 kN

10.Column buckling

Answer: d = 76.44 mm

Required critical load

P_cr = n·P = 2.5·105000 = 262500 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·262500·2100²/(π³·70000))^(1/4) = 76.44 mm

Answer

d = 76.44 mm

Check: the slenderness KL/r is 109.9, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.09682

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.

Critical damping

c_c = 2·√(k·m) = 2·√(12000·50) = 1549 N·s/m

Damping ratio

ζ = c/c_c = 150/1549 = 0.09682

Answer

ζ = 0.09682

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

12.Mechanical vibrations

Answer: k_eq = 44.44 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 = 1/100 + 1/80 → k_eq = 100·80/(100 + 80) = 44.44 kN/m

Answer

k_eq = 44.44 kN/m

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