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 60 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 4 : 1. How many teeth N₂ must the gear have?

a = 60 mmm = 1.5 mmi = 4

N₂ =

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 1440 rpm drives gear B (60 teeth). Gear C (25 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed n_out.

n_in = 1440 rpmA = 16B = 60C = 25D = 30

n_out = rpm

3.Torsion of shaftsMedium

A solid aluminium 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 = 500 N·md = 50 mmL = 2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft transmits 50 kW at 1500 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.

P = 50 kWN = 1500 rpmτ_allow = 40 MPaθ_allow = 0.25°/mG = 26 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 40 MPa, σᵧ = 30 MPa and τₓᵧ = 20 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 40 MPaσ_y = 30 MPaτ_xy = 20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 12 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 800 mmt = 12 mmp = 0.8 MPaα = 60°

σ_n = MPa

7.Axial stress & strainMedium

Find the elongation δ of the copper bar under the axial load.

P = 60 kNd = 25 mmL = 0.5 mE = 117 GPa

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 110 kNsegment 1: Ø35 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 1.5 mE = 70 GPa (aluminium)d = 50 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 130 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 130 kNn = 2.5L = 3 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

A 15 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 210 N·s/m. Find the damping ratio ζ.

m = 15 kgk = 12 kN/mc = 210 N·s/m

ζ =

12.Mechanical vibrationsHard

A 50 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 kN/m and k₂ = 20 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency f_n of vertical vibration in Hz.

m = 50 kgk₁ = 40 kN/mk₂ = 20 kN/mk₃ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₂ = 64

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 = 2·60/1.5 = 80

Split by the ratio

N₁ = 80/(1 + 4) = 16

N₂ = 4·16 = 64

Answer

N₂ = 64

2.Gears & gear trains

Answer: n_out = 320 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 = 1440·16/60 = 384 rpm = n_C

Second stage

n_out = 384·25/30 = 320 rpm

Answer

n_out = 320 rpm

total ratio = 4.5 : 1

3.Torsion of shafts

Answer: φ = 3.591 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·2000/(26000·613592) = 0.06268 rad

φ = 0.06268·180/π = 3.591°

Answer

φ = 3.591°

4.Torsion of shafts

Answer: d = 73.12 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 50000/157.1 = 318.3 N·m

Diameter for strength

T = 318.3 N·m = 318310 N·mm

d_τ = ∛(16·318310/(π·40)) = 34.35 mm

Diameter for stiffness

θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·318310/(π·26000·4.363 × 10⁻⁶))^1/4 = 73.12 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(34.35, 73.12) = 73.12 mm

Answer

d = 73.12 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_p = 37.98 °

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.

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·20, 40 − 30) = 75.96°

Angle on the element

θ_p = 37.98°

Answer

θ_p = 37.98°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_n = 16.67 MPa

Wall stresses

r = 800/2 = 400 mm

σ_h = pr/t = 0.8·400/12 = 26.67 MPa

σ_a = pr/(2t) = 13.33 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 13.33·sin²60° + 26.67·cos²60° = 16.67 MPa

Answer

σ_n = 16.67 MPa

7.Axial stress & strain

Answer: δ = 0.5224 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·25²/4 = 490.9 mm²

σ = P/A = 60·1000/490.9 = 122.2 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 122.2/117000 = 0.001045 = 1045 µε

Elongation

δ = ε·L = 0.001045·500 mm = 0.5224 mm

Answer

δ = 0.5224 mm

8.Axial stress & strain

Answer: σ_max = 972.6 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·35²/4 = 962.1 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 114.3 MPa

σ₂ = 972.6 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 972.6 MPa

Answer

σ_max = 972.6 MPa

9.Column buckling

Answer: P_cr = 94.2 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/1500² = 94203 N = 94.2 kN

Answer

P_cr = 94.2 kN

10.Column buckling

Answer: d = 80.63 mm

Required critical load

P_cr = n·P = 2.5·130000 = 325000 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·325000·2100²/(π³·70000))^(1/4) = 80.63 mm

Answer

d = 80.63 mm

Check: the slenderness KL/r is 104.2, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.2475

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·√(12000·15) = 848.5 N·s/m

Damping ratio

ζ = c/c_c = 210/848.5 = 0.2475

Answer

ζ = 0.2475

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

12.Mechanical vibrations

Answer: f_n = 3.898 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 40 + 20 = 60 kN/m

k_eq = 60·60/(60 + 60) = 30 kN/m

Natural frequency

ω_n = √(30000/50) = 24.49 rad/s

f_n = 24.49/(2π) = 3.898 Hz

Answer

f_n = 3.898 Hz

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