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 an 80 mm driving pulley at 1450 rpm. The tension is 1000 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.

d₁ = 80 mmn₁ = 1450 rpmT₁ = 1000 NT₂ = 300 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 200 mm driving pulley at 1500 rpm to a 500 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 200 mmn₁ = 1500 rpmd₂ = 500 mmC = 1200 mmμ = 0.35T₁ = 500 N

P = kW

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 400 mm rotate 8.66° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 12 mmL = 400 mmφ = 8.66°

G = GPa

4.Torsion of shaftsHard

A solid aluminium shaft transmits 50 kW at 750 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 50 kWN = 750 rpmτ_allow = 40 MPaθ_allow = 0.5°/mG = 26 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 80 MPa, σᵧ = −40 MPa and τₓᵧ = 10 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 = 80 MPaσ_y = −40 MPaτ_xy = 10 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 60 mm carries an axial compressive force of 40 kN together with a torque of 2000 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 60 mmF = −40 kNT = 2000 N·m

σ₁ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 50 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: brass, Ø50 mm, 300 mm, E = 100 GPaδ = 0.5 mm

P = kN

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 = 65 kNsegment 1: Ø25 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø12 mm, 2 m, brass (E = 100 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 3 mE = 200 GPa (steel)b × h = 30 × 50 mm

P_cr = kN

10.Column bucklingHard

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

L = 1 mE = 70 GPa (aluminium)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

An 8 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency f_d of its free vibration in Hz.

m = 8 kgk = 50 kN/mζ = 0.4

f_d = Hz

12.Mechanical vibrationsHard

A 50 kg mass hangs from spring k₃ = 30 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 = 100 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 = 100 kN/mk₃ = 30 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 4.252 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 = 6.074 m/s

Power

P = (1000 − 300)·6.074 = 4252 W = 4.252 kW

Answer

P = 4.252 kW

2.Gears & gear trains

Answer: P = 4.999 kW

Angle of wrap

θ = 2.891 rad

Slack-side tension

e^μθ = e^1.012 = 2.751, T₂ = 500/2.751 = 181.8 N

Power

P = (500 − 181.8)·15.71 = 4.999 kW

Answer

P = 4.999 kW

3.Torsion of shafts

Answer: G = 39 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 = 8.66·π/180 = 0.1511 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·400/(2036·0.1511) = 39000 MPa

G = 39 GPa

Answer

G = 39 GPa

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

4.Torsion of shafts

Answer: d = 73.12 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 50000/78.54 = 636.6 N·m

Diameter for strength

T = 636.6 N·m = 636620 N·mm

d_τ = ∛(16·636620/(π·40)) = 43.28 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·636620/(π·26000·8.727 × 10⁻⁶))^1/4 = 73.12 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(43.28, 73.12) = 73.12 mm

Answer

d = 73.12 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_p = 4.731 °

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(2·10, 80 − −40) = 9.462°

Angle on the element

θ_p = 4.731°

Answer

θ_p = 4.731°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ₁ = 40.61 MPa

Stresses at the surface

σ = F/A = −40000/(π·60²/4) = −40000/2827 = −14.15 MPa

τ = 16T/(πd³) = 16·2000000/(π·60³) = 47.16 MPa

Mohr’s circle

C = σ/2 = −7.074 MPa

R = √((−7.074)² + 47.16²) = 47.68 MPa

σ₁ = C + R = 40.61 MPa

σ₂ = C − R = −54.76 MPa

Answer

σ₁ = 40.61 MPa

7.Axial stress & strain

Answer: P = 114.7 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 = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·1963) = 1.528 × 10⁻⁶ mm/N

Load

P = 0.5/4.357 × 10⁻⁶ = 114750 N = 114.7 kN

Answer

P = 114.7 kN

8.Axial stress & strain

Answer: σ_max = 574.7 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·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 132.4 MPa

σ₂ = 574.7 MPa

Largest stress

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

σ_max = 574.7 MPa

Answer

σ_max = 574.7 MPa

9.Column buckling

Answer: P_cr = 24.67 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 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·112500/3000² = 24674 N = 24.67 kN

Answer

P_cr = 24.67 kN

10.Column buckling

Answer: σ_cr = 69.09 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·1000 = 500 mm

KL/r = 500/5 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σ_cr = 69.09 MPa

11.Mechanical vibrations

Answer: f_d = 11.53 Hz

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.

Undamped natural frequency

ω_n = √(50000/8) = 79.06 rad/s

f_n = 79.06/(2π) = 12.58 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 12.58·√(1 − 0.4²) = 11.53 Hz

Answer

f_d = 11.53 Hz

12.Mechanical vibrations

Answer: f_n = 3.538 Hz

Equivalent stiffness

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

k₁₂ = 40 + 100 = 140 kN/m

k_eq = 140·30/(140 + 30) = 24.71 kN/m

Natural frequency

ω_n = √(24706/50) = 22.23 rad/s

f_n = 22.23/(2π) = 3.538 Hz

Answer

f_n = 3.538 Hz

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