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

A 4 kW motor running at 1440 rpm drives a pinion with 16 teeth and a module of 2.5 mm. Find the tangential force F_t on the pinion teeth.

P = 4 kWn = 1440 rpmN₁ = 16m = 2.5 mm

F_t = N

2.Gears & gear trainsHard

In a compound gear train, gear A (15 teeth) on the input shaft turns at 960 rpm and drives gear B (24 teeth). Gear C (30 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 562.5 rpm. How many teeth N_D must gear D have?

n_in = 960 rpmA = 15B = 24C = 30n_out = 562.5 rpm

N_D =

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 60 mm, inner diameter 36 mm, length 1.5 m) is fixed at one end and twisted by a torque of 750 N·m at the other. Find the angle of twist φ in degrees.

T = 750 N·md_o = 60 mmd_i = 36 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 5 kWN = 500 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 150 MPa, σᵧ = 70 MPa and τₓᵧ = −30 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 = 150 MPaσ_y = 70 MPaτ_xy = −30 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 70 MPa, σᵧ = −10 MPa and τₓᵧ = −10 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 70 MPaσ_y = −10 MPaτ_xy = −10 MPaθ = 45°

τ_θ = MPa

7.Axial stress & strainMedium

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

P = 15 kNd = 10 mmL = 2.5 mE = 200 GPa

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 45 kNsegment 1: Ø40 mm, 1 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1 m, copper (E = 117 GPa)

δ = mm

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 = 4 mE = 70 GPa (aluminium)d = 60 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 2 mE = 70 GPa (aluminium)d = 60 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 5 cycles later by a peak of 1.3 mm. Find the damping ratio ζ.

x₀ = 30 mmx₅ = 1.3 mmn = 5

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 50 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 20 kgk₁ = 40 kN/mk₂ = 50 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: F_t = 1326 N

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.

Torque on the pinion

ω = 2π·1440/60 = 150.8 rad/s

T = P/ω = 4000/150.8 = 26.53 N·m

Pitch diameter

d = m·N = 2.5·16 = 40 mm

Tangential force

F_t = 2T/d = 2·26.53/0.04 = 1326 N

Answer

F_t = 1326 N

2.Gears & gear trains

Answer: N_D = 32

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.

Intermediate shaft speed

n_B = n_C = 960·15/24 = 600 rpm

Teeth on gear D

N_D = C·n_C/n_out = 30·600/562.5 = 32

Answer

N_D = 32

3.Torsion of shafts

Answer: φ = 0.7275 °

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(60⁴ − 36⁴)/32 = 1.107 × 10⁶ mm⁴

Angle of twist

T = 750 N·m = 750000 N·mm

L = 1.5 m = 1500 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 750000·1500/(80000·1.107 × 10⁶) = 0.0127 rad

φ = 0.0127·180/π = 0.7275°

Answer

φ = 0.7275°

4.Torsion of shafts

Answer: d = 18.65 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 5000/52.36 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·75)) = 18.65 mm

Answer

d = 18.65 mm

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

5.Mohr’s circle & stress transformation

Answer: θ_p = −18.43 °

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·−30, 150 − 70) = −36.87°

Angle on the element

θ_p = −18.43°

Answer

θ_p = −18.43°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: τ_θ = −40 MPa

Transformation equation

τ_θ = −40·1 + −10·0 = −40 MPa

Answer

τ_θ = −40 MPa

7.Axial stress & strain

Answer: δ = 2.387 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·10²/4 = 78.54 mm²

σ = P/A = 15·1000/78.54 = 191 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 191/200000 = 0.0009549 = 954.9 µε

Elongation

δ = ε·L = 0.0009549·2500 mm = 2.387 mm

Answer

δ = 2.387 mm

8.Axial stress & strain

Answer: δ = 3.707 mm

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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 35.81 MPa

σ₂ = 397.9 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 35.81·1000/117000 = 0.3061 mm

δ₂ = σ₂L₂/E₂ = 397.9·1000/117000 = 3.401 mm

δ = δ₁ + δ₂ = 3.707 mm

Answer

δ = 3.707 mm

9.Column buckling

Answer: P_cr = 109.9 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·636173/2000² = 109878 N = 109.9 kN

Answer

P_cr = 109.9 kN

10.Column buckling

Answer: σ_cr = 38.86 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 1·2000 = 2000 mm

KL/r = 2000/15 = 133.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/133.3² = 38.86 MPa

Answer

σ_cr = 38.86 MPa

11.Mechanical vibrations

Answer: ζ = 0.09942

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.

Logarithmic decrement

δ = (1/5)·ln(30/1.3) = 0.6278

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6278/√(39.48 + 0.3941) = 0.09942

Answer

ζ = 0.09942

12.Mechanical vibrations

Answer: k_eq = 22.22 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 = 1/40 + 1/50 → k_eq = 40·50/(40 + 50) = 22.22 kN/m

Answer

k_eq = 22.22 kN/m

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