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 81 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 3.5 : 1. How many teeth N₂ must the gear have?

a = 81 mmm = 1.5 mmi = 3.5

N₂ =

2.Gears & gear trainsHard

An open belt connects an 80 mm driving pulley at 720 rpm to a 200 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 80 mmn₁ = 720 rpmd₂ = 200 mmC = 1000 mmμ = 0.3T₁ = 1000 N

P = kW

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 300 mm rotate 10.82° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 12 mmL = 300 mmφ = 10.82°

G = GPa

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 300 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 10 kWN = 300 rpmτ_allow = 40 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 160 MPaσ_y = −30 MPaτ_xy = −40 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = −70 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.

σ_x = 160 MPaσ_y = −70 MPaτ_xy = 40 MPaθ = 60°

σ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 70 mm, E = 120 GPa). Both are 500 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 120 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø80/Ø60 mm, E = 70 GPa, σ_allow = 50 MPa

P_max = kN

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 = 1 mE = 200 GPa (steel)b × h = 25 × 60 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 15 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 15 kNn = 3L = 2 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed one cycle later by a peak of 13.3 mm. Find the damping ratio ζ.

x₀ = 25 mmx₁ = 13.3 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 80 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 10 kgk₁ = 25 kN/mk₂ = 80 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₂ = 84

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 = 2·81/1.5 = 108

Split by the ratio

N₁ = 108/(1 + 3.5) = 24

N₂ = 3.5·24 = 84

Answer

N₂ = 84

2.Gears & gear trains

Answer: P = 1.798 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 = 3.022 rad

Slack-side tension

e^μθ = e^0.9065 = 2.476, T₂ = 1000/2.476 = 404 N

Power

P = (1000 − 404)·3.016 = 1.798 kW

Answer

P = 1.798 kW

3.Torsion of shafts

Answer: G = 39.02 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 = 10.82·π/180 = 0.1888 rad

Shear modulus

T = 50 N·m = 50000 N·mm

G = TL/(Jφ) = 50000·300/(2036·0.1888) = 39018 MPa

G = 39.02 GPa

Answer

G = 39.02 GPa

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

4.Torsion of shafts

Answer: d = 34.35 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 10000/31.42 = 318.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·318310/(π·40)) = 34.35 mm

Answer

d = 34.35 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_max = 103.1 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.

Radius of Mohr’s circle

τ_max = R = √(((160 − −30)/2)² + −40²) = 103.1 MPa

Answer

τ_max = 103.1 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = 22.14 MPa

Transformation equation

σ_θ = 45 + 115·−0.5 + 40·0.866 = 22.14 MPa

Answer

σ_θ = 22.14 MPa

7.Axial stress & strain

Answer: δ = 0.06431 mm

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

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·4006 = 6.22 × 10⁸ N

Shortening

δ = 80000·500/6.22 × 10⁸ = 0.06431 mm

Answer

δ = 0.06431 mm

8.Axial stress & strain

Answer: P_max = 177.2 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·2199 = 2.953 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0007143) = 0.0006 (steel)

Largest load

P_max = 0.0006·2.953 × 10⁸ = 177186 N = 177.2 kN

Answer

P_max = 177.2 kN

9.Column buckling

Answer: P_cr = 314.7 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 = 60·25³/12 = 78125 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·78125/700² = 314720 N = 314.7 kN

Answer

P_cr = 314.7 kN

10.Column buckling

Answer: d = 36.92 mm

Required critical load

P_cr = n·P = 3·15000 = 45000 N

KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·45000·2000²/(π³·200000))^(1/4) = 36.92 mm

Answer

d = 36.92 mm

Check: the slenderness KL/r is 216.7, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.09994

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 = (1/1)·ln(25/13.3) = 0.6311

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6311/√(39.48 + 0.3983) = 0.09994

Answer

ζ = 0.09994

12.Mechanical vibrations

Answer: f_n = 16.31 Hz

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k₁ + k₂.

k_eq = 25 + 80 = 105 kN/m

Natural frequency

ω_n = √(105000/10) = 102.5 rad/s

f_n = 102.5/(2π) = 16.31 Hz

Answer

f_n = 16.31 Hz

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