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 144 mm apart. They must be connected by a pair of spur gears with a module of 4 mm and a speed reduction of 3.5 : 1. How many teeth N₂ must the gear have?

a = 144 mmm = 4 mmi = 3.5

N₂ =

2.Gears & gear trainsHard

An open belt connects an 80 mm pulley and a 315 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 80 mmd₂ = 315 mmC = 1000 mm

θ = °

3.Torsion of shaftsMedium

A solid brass 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 = 400 N·md = 30 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 1000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1000 rpmτ_allow = 60 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = 120 MPa and τₓᵧ = 30 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 20 MPaσ_y = 120 MPaτ_xy = 30 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 100 MPa, σᵧ = −40 MPa and τₓᵧ = −40 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 100 MPaσ_y = −40 MPaτ_xy = −40 MPaθ = 40°

τ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 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 0.25 mm in total.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaδ = 0.25 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 = 45 kNsegment 1: Ø40 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø20 mm, 1 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 = 2 mE = 200 GPa (steel)d = 30 mm

P_cr = kN

10.Column bucklingHard

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

L = 2 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 20 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 690 N·s/m. Find the damping ratio ζ.

m = 20 kgk = 150 kN/mc = 690 N·s/m

ζ =

12.Mechanical vibrationsHard

A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 15 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 = 5 kgk₁ = 60 kN/mk₂ = 15 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₂ = 56

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 = 2·144/4 = 72

Split by the ratio

N₁ = 72/(1 + 3.5) = 16

N₂ = 3.5·16 = 56

Answer

N₂ = 56

2.Gears & gear trains

Answer: θ = 166.5 °

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 angle

asin(0.1175) = 6.748°

Angle of wrap

θ = 180 − 2·6.748 = 166.5°

Answer

θ = 166.5°

3.Torsion of shafts

Answer: φ = 5.912 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·800/(39000·79522) = 0.1032 rad

φ = 0.1032·180/π = 5.912°

Answer

φ = 5.912°

4.Torsion of shafts

Answer: d = 34.35 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 50000/104.7 = 477.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·477465/(π·60)) = 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_abs = 64.15 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 = 70 MPa, R = √((−50)² + 30²) = 58.31 MPa

σ₁ = 128.3 MPa, σ₂ = 11.69 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(116.6, 128.3, 11.69)/2 = 64.15 MPa

Answer

τ_abs = 64.15 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: τ_θ = −75.88 MPa

Transformation equation

τ_θ = −70·0.9848 + −40·0.1736 = −75.88 MPa

Answer

τ_θ = −75.88 MPa

7.Axial stress & strain

Answer: P = 33.66 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 = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N

Load

P = 0.25/7.427 × 10⁻⁶ = 33660 N = 33.66 kN

Answer

P = 33.66 kN

8.Axial stress & strain

Answer: σ_max = 143.2 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·20²/4 = 314.2 mm²

σ₁ = 35.81 MPa

σ₂ = 143.2 MPa

Largest stress

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

σ_max = 143.2 MPa

Answer

σ_max = 143.2 MPa

9.Column buckling

Answer: P_cr = 40.04 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·39761/1400² = 40043 N = 40.04 kN

Answer

P_cr = 40.04 kN

10.Column buckling

Answer: σ_cr = 77.11 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 1·2000 = 2000 mm

KL/r = 2000/12.5 = 160

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/160² = 77.11 MPa

Answer

σ_cr = 77.11 MPa

11.Mechanical vibrations

Answer: ζ = 0.1992

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·√(150000·20) = 3464 N·s/m

Damping ratio

ζ = c/c_c = 690/3464 = 0.1992

Answer

ζ = 0.1992

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

12.Mechanical vibrations

Answer: f_n = 19.49 Hz

Equivalent stiffness

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

k_eq = 60 + 15 = 75 kN/m

Natural frequency

ω_n = √(75000/5) = 122.5 rad/s

f_n = 122.5/(2π) = 19.49 Hz

Answer

f_n = 19.49 Hz

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