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 2.2 kW motor at 1450 rpm drives a worm with 1 start(s). The worm wheel has 36 teeth and the drive has an efficiency of 60%. Find the output torque T_w on the wheel shaft.

z = 1N_w = 36n = 1450 rpmP = 2.2 kWη = 60%

T_w = N·m

2.Gears & gear trainsHard

An open belt connects a 200 mm driving pulley at 1440 rpm to a 250 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.3 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 = 1440 rpmd₂ = 250 mmC = 1200 mmμ = 0.3T₁ = 500 N

P = kW

3.Torsion of shaftsMedium

A solid steel 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 = 200 N·md = 50 mmL = 2.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 20 kW at 750 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 20 kWN = 750 rpmτ_allow = 40 MPaθ_allow = 1°/mG = 39 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 20 MPa, σᵧ = −30 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 20 MPaσ_y = −30 MPaτ_xy = 30 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −70 MPa, σᵧ = −80 MPa and τₓᵧ = −60 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −70 MPaσ_y = −80 MPaτ_xy = −60 MPaθ = 20°

τ_θ = MPa

7.Axial stress & strainMedium

Find the elongation δ of the brass bar under the axial load.

P = 25 kNd = 30 mmL = 2.5 mE = 100 GPa

δ = mm

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 = 35 kNsegment 1: Ø35 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1.5 m, brass (E = 100 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 3 mE = 70 GPa (aluminium)b × h = 20 × 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 110 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 110 kNn = 3L = 4 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 50 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 570 N·s/m. Find the damping ratio ζ.

m = 50 kgk = 40 kN/mc = 570 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 100 kN/m and k₂ = 10 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 = 25 kgk₁ = 100 kN/mk₂ = 10 kN/mk₃ = 20 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T_w = 313 N·m

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Wheel speed

n_w = n·z/N_w = 1450·1/36 = 40.28 rpm

Output torque

ω_w = 2π·40.28/60 = 4.218 rad/s

T_w = η·P/ω_w = 0.6·2200/4.218 = 313 N·m

Answer

T_w = 313 N·m

2.Gears & gear trains

Answer: P = 4.565 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 = 3.1 rad

Slack-side tension

e^μθ = e^0.93 = 2.534, T₂ = 500/2.534 = 197.3 N

Power

P = (500 − 197.3)·15.08 = 4.565 kW

Answer

P = 4.565 kW

3.Torsion of shafts

Answer: φ = 0.5836 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2500/(80000·613592) = 0.01019 rad

φ = 0.01019·180/π = 0.5836°

Answer

φ = 0.5836°

4.Torsion of shafts

Answer: d = 44.18 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 20000/78.54 = 254.6 N·m

Diameter for strength

T = 254.6 N·m = 254648 N·mm

d_τ = ∛(16·254648/(π·40)) = 31.89 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·254648/(π·39000·1.745 × 10⁻⁵))^1/4 = 44.18 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(31.89, 44.18) = 44.18 mm

Answer

d = 44.18 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_max = 39.05 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 = √(((20 − −30)/2)² + 30²) = 39.05 MPa

Answer

τ_max = 39.05 MPa

6.Mohr’s circle & stress transformation

Answer: τ_θ = −49.18 MPa

Transformation equation

τ_θ = −5·0.6428 + −60·0.766 = −49.18 MPa

Answer

τ_θ = −49.18 MPa

7.Axial stress & strain

Answer: δ = 0.8842 mm

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·30²/4 = 706.9 mm²

σ = P/A = 25·1000/706.9 = 35.37 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 35.37/100000 = 0.0003537 = 353.7 µε

Elongation

δ = ε·L = 0.0003537·2500 mm = 0.8842 mm

Answer

δ = 0.8842 mm

8.Axial stress & strain

Answer: σ_max = 111.4 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·35²/4 = 962.1 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 36.38 MPa

σ₂ = 111.4 MPa

Largest stress

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

σ_max = 111.4 MPa

Answer

σ_max = 111.4 MPa

9.Column buckling

Answer: P_cr = 6.266 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·40000/2100² = 6266 N = 6.266 kN

Answer

P_cr = 6.266 kN

10.Column buckling

Answer: d = 85.92 mm

Required critical load

P_cr = n·P = 3·110000 = 330000 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·330000·4000²/(π³·200000))^(1/4) = 85.92 mm

Answer

d = 85.92 mm

Check: the slenderness KL/r is 186.2, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.2015

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·√(40000·50) = 2828 N·s/m

Damping ratio

ζ = c/c_c = 570/2828 = 0.2015

Answer

ζ = 0.2015

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

12.Mechanical vibrations

Answer: f_n = 4.141 Hz

Equivalent stiffness

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

k₁₂ = 100 + 10 = 110 kN/m

k_eq = 110·20/(110 + 20) = 16.92 kN/m

Natural frequency

ω_n = √(16923/25) = 26.02 rad/s

f_n = 26.02/(2π) = 4.141 Hz

Answer

f_n = 4.141 Hz

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