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 93.75 mm apart. They must be connected by a pair of spur gears with a module of 2.5 mm and a speed reduction of 4 : 1. How many teeth N₁ must the pinion have?

a = 93.75 mmm = 2.5 mmi = 4

N₁ =

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 15 teeth and each planet gear has 15 teeth. The planet carrier is held fixed and the sun is driven at 1800 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 15N_p = 15n_s = 1800 rpm

n_r = rpm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 1000 N·m. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

T = 1000 N·md_i/d_o = 0.8τ_allow = 50 MPa

d_o = mm

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 1800 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 10 kWN = 1800 rpmτ_allow = 40 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 10 MPa, σᵧ = 80 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 = 10 MPaσ_y = 80 MPaτ_xy = −30 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 10 MPa, σᵧ = 110 MPa and τₓᵧ = 20 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 10 MPaσ_y = 110 MPaτ_xy = 20 MPaσ_Y = 460 MPa

n =

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 30 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø120/Ø70 mm, E = 120 GPa, σ_allow = 100 MPa

P_max = kN

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 = 6 mE = 70 GPa (aluminium)b × h = 20 × 50 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 125 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 125 kNn = 2.5L = 6 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

An 80 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 980 N·s/m. Find the damping ratio ζ.

m = 80 kgk = 12 kN/mc = 980 N·s/m

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 50 kN/m. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 20 kgk₁ = 5 kN/mk₂ = 50 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₁ = 15

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 = 2·93.75/2.5 = 75

Split by the ratio

N₁ = 75/(1 + 4) = 15

Answer

N₁ = 15

2.Gears & gear trains

Answer: n_r = 600 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 15 + 2·15 = 45

Willis equation with the carrier fixed

With n_c = 0 the sun drives the ring through the planets, which act as idlers.

n_r = −n_s·N_s/N_r = −1800·15/45 = −600 rpm

|n_r| = 600 rpm

Answer

n_r = 600 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: d_o = 55.67 mm

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.

Outer diameter from the allowable stress

T = 1000 N·m = 1 × 10⁶ N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1 × 10⁶/(π·50·0.5904)) = 55.67 mm

d_i = 0.8·55.67 = 44.54 mm

Answer

d_o = 55.67 mm

A solid shaft would need d = 46.7 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 18.9 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 10000/188.5 = 53.05 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·53052/(π·40)) = 18.9 mm

Answer

d = 18.9 mm

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

5.Mohr’s circle & stress transformation

Answer: θ_p = −69.7 °

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, 10 − 80) = −139.4°

Angle on the element

θ_p = −69.7°

Answer

θ_p = −69.7°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 4.04

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 60 MPa, R = 53.85 MPa

σ₁ = 113.9 MPa, σ₂ = 6.148 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 113.9 MPa

n = 460/113.9 = 4.04

Answer

n = 4.04

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: P = 33.26 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 = 706.9 mm²

Flexibilities

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

L₂/(E₂A₂) = 400/(100000·706.9) = 5.659 × 10⁻⁶ mm/N

Load

P = 0.4/1.203 × 10⁻⁵ = 33264 N = 33.26 kN

Answer

P = 33.26 kN

8.Axial stress & strain

Answer: P_max = 827.9 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 120·10³·7461 = 9.935 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0008333) = 0.0008333 (copper)

Largest load

P_max = 0.0008333·9.935 × 10⁸ = 827941 N = 827.9 kN

Answer

P_max = 827.9 kN

9.Column buckling

Answer: P_cr = 2.559 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 = 50·20³/12 = 33333 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·33333/3000² = 2559 N = 2.559 kN

Answer

P_cr = 2.559 kN

10.Column buckling

Answer: d = 73.4 mm

Required critical load

P_cr = n·P = 2.5·125000 = 312500 N

KL = 0.5·6000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·312500·3000²/(π³·200000))^(1/4) = 73.4 mm

Answer

d = 73.4 mm

Check: the slenderness KL/r is 163.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.5001

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·√(12000·80) = 1960 N·s/m

Damping ratio

ζ = c/c_c = 980/1960 = 0.5001

Answer

ζ = 0.5001

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

12.Mechanical vibrations

Answer: f_n = 2.399 Hz

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 = 1/5 + 1/50 → k_eq = 5·50/(5 + 50) = 4.545 kN/m

Natural frequency

ω_n = √(4545/20) = 15.08 rad/s

f_n = 15.08/(2π) = 2.399 Hz

Answer

f_n = 2.399 Hz

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