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 belt runs on a 100 mm driving pulley at 720 rpm. The tension is 1000 N on the tight side and 250 N on the slack side. Find the power P transmitted, in kW.

d₁ = 100 mmn₁ = 720 rpmT₁ = 1000 NT₂ = 250 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 160 mm pulley and a 200 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 160 mmd₂ = 200 mmC = 800 mm

θ = °

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 250 mm rotate 8.12° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 12 mmL = 250 mmφ = 8.12°

G = GPa

4.Torsion of shaftsHard

A solid aluminium shaft transmits 30 kW at 500 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 30 kWN = 500 rpmτ_allow = 60 MPaθ_allow = 0.75°/mG = 26 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 150 MPa, σᵧ = 20 MPa and τₓᵧ = 10 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 150 MPaσ_y = 20 MPaτ_xy = 10 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 130 MPa, σᵧ = 80 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σ_x = 130 MPaσ_y = 80 MPaτ_xy = 30 MPaθ = 15°

σ_θ = MPa

7.Axial stress & strainMedium

Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).

P = 10 kNd = 16 mmL = 1.5 mE = 100 GPa

ε = µε

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 = 105 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1.5 m, aluminium (E = 70 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 6 mE = 200 GPa (steel)d = 25 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 20 mm is followed 3 cycles later by a peak of 9.4 mm. Find the damping ratio ζ.

x₀ = 20 mmx₃ = 9.4 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 100 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 = 12 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 100 kgk₁ = 100 kN/mk₂ = 12 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 2.827 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.

Belt speed

v = 3.77 m/s

Power

P = (1000 − 250)·3.77 = 2827 W = 2.827 kW

Answer

P = 2.827 kW

2.Gears & gear trains

Answer: θ = 177.1 °

Belt angle

asin(0.025) = 1.433°

Angle of wrap

θ = 180 − 2·1.433 = 177.1°

Answer

θ = 177.1°

3.Torsion of shafts

Answer: G = 26 GPa

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 moment and angle

J = π·12⁴/32 = 2036 mm⁴

φ = 8.12·π/180 = 0.1417 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·250/(2036·0.1417) = 25996 MPa

G = 26 GPa

Answer

G = 26 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d = 64.35 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 30000/52.36 = 573 N·m

Diameter for strength

T = 573 N·m = 572958 N·mm

d_τ = ∛(16·572958/(π·60)) = 36.5 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·572958/(π·26000·1.309 × 10⁻⁵))^1/4 = 64.35 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(36.5, 64.35) = 64.35 mm

Answer

d = 64.35 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_abs = 75.38 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 85 MPa, R = √(65² + 10²) = 65.76 MPa

σ₁ = 150.8 MPa, σ₂ = 19.24 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(131.5, 150.8, 19.24)/2 = 75.38 MPa

Answer

τ_abs = 75.38 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 141.7 MPa

Transformation equation

σ_θ = 105 + 25·0.866 + 30·0.5 = 141.7 MPa

Answer

σ_θ = 141.7 MPa

7.Axial stress & strain

Answer: ε = 497.4 µε

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·16²/4 = 201.1 mm²

σ = P/A = 10·1000/201.1 = 49.74 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 49.74/100000 = 0.0004974 = 497.4 µε

Answer

ε = 497.4 µε

8.Axial stress & strain

Answer: σ_max = 928.4 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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 83.56 MPa

σ₂ = 928.4 MPa

Largest stress

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

σ_max = 928.4 MPa

Answer

σ_max = 928.4 MPa

9.Column buckling

Answer: P_cr = 102.5 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 = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/3500² = 102511 N = 102.5 kN

Answer

P_cr = 102.5 kN

10.Column buckling

Answer: σ_cr = 4.371 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 0.7·6000 = 4200 mm

KL/r = 4200/6.25 = 672

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/672² = 4.371 MPa

Answer

σ_cr = 4.371 MPa

11.Mechanical vibrations

Answer: ζ = 0.04002

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 = (1/3)·ln(20/9.4) = 0.2517

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2517/√(39.48 + 0.06334) = 0.04002

Answer

ζ = 0.04002

12.Mechanical vibrations

Answer: k_eq = 10.71 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 + 1/12 → k_eq = 100·12/(100 + 12) = 10.71 kN/m

Answer

k_eq = 10.71 kN/m

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