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 5 : 1. How many teeth N₁ must the pinion have?

a = 81 mmm = 1.5 mmi = 5

N₁ =

2.Gears & gear trainsHard

In a compound gear train, gear A (25 teeth) on the input shaft turning at 3000 rpm drives gear B (60 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed n_out.

n_in = 3000 rpmA = 25B = 60C = 12D = 72

n_out = rpm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 100 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter d_o.

T = 100 N·md_i/d_o = 0.7τ_allow = 60 MPa

d_o = mm

4.Torsion of shaftsHard

A solid brass shaft transmits 20 kW at 1500 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 20 kWN = 1500 rpmτ_allow = 80 MPaθ_allow = 0.75°/mG = 39 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 130 MPa and σᵧ = 20 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 190 MPa?

σ_x = 130 MPaσ_y = 20 MPaσ₁ = 190 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 80 MPa, σᵧ = 20 MPa and τₓᵧ = 50 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 80 MPaσ_y = 20 MPaτ_xy = 50 MPaθ = 15°

τ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 70 mm, E = 100 GPa). Both are 400 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 100 GPaL = 400 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 80 MPa

P_max = kN

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 = 6 mE = 70 GPa (aluminium)b × h = 40 × 60 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 25 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 25 kNn = 2L = 1.5 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 200 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 1900 N·s/m. Find the damping ratio ζ.

m = 200 kgk = 50 kN/mc = 1900 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 30 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 25 kgk₁ = 20 kN/mk₂ = 30 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: N₁ = 18

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 + 5) = 18

Answer

N₁ = 18

2.Gears & gear trains

Answer: n_out = 208.3 rpm

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.

First stage

n_B = 3000·25/60 = 1250 rpm = n_C

Second stage

n_out = 1250·12/72 = 208.3 rpm

Answer

n_out = 208.3 rpm

total ratio = 14.4 : 1

3.Torsion of shafts

Answer: d_o = 22.35 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 = 100 N·m = 100000 N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·100000/(π·60·0.7599)) = 22.35 mm

d_i = 0.7·22.35 = 15.65 mm

Answer

d_o = 22.35 mm

A solid shaft would need d = 20.4 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 39.92 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 20000/157.1 = 127.3 N·m

Diameter for strength

T = 127.3 N·m = 127324 N·mm

d_τ = ∛(16·127324/(π·80)) = 20.09 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·127324/(π·39000·1.309 × 10⁻⁵))^1/4 = 39.92 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(20.09, 39.92) = 39.92 mm

Answer

d = 39.92 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_xy = 101 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.

Required radius

C = (130 + 20)/2 = 75 MPa

R = σ₁ − C = 190 − 75 = 115 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(115² − 55²) = 101 MPa

Answer

τ_xy = 101 MPa

6.Mohr’s circle & stress transformation

Answer: τ_θ = 28.3 MPa

Transformation equation

τ_θ = −30·0.5 + 50·0.866 = 28.3 MPa

Answer

τ_θ = 28.3 MPa

7.Axial stress & strain

Answer: δ = 0.0401 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·25²/4 = 490.9 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·4006 = 4.987 × 10⁸ N

Shortening

δ = 50000·400/4.987 × 10⁸ = 0.0401 mm

Answer

δ = 0.0401 mm

8.Axial stress & strain

Answer: P_max = 337.4 kN

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 + 70·10³·4006 = 4.218 × 10⁸ N

Governing strain

ε = min(0.0008, 0.001143) = 0.0008 (steel)

Largest load

P_max = 0.0008·4.218 × 10⁸ = 337407 N = 337.4 kN

Answer

P_max = 337.4 kN

9.Column buckling

Answer: P_cr = 12.53 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 = 60·40³/12 = 320000 mm⁴

Effective length

KL = 0.7·6000 = 4200 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·320000/4200² = 12533 N = 12.53 kN

Answer

P_cr = 12.53 kN

10.Column buckling

Answer: d = 27.46 mm

Required critical load

P_cr = n·P = 2·25000 = 50000 N

KL = 0.7·1500 = 1050 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·1050²/(π³·200000))^(1/4) = 27.46 mm

Answer

d = 27.46 mm

Check: the slenderness KL/r is 152.9, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.3004

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·√(50000·200) = 6325 N·s/m

Damping ratio

ζ = c/c_c = 1900/6325 = 0.3004

Answer

ζ = 0.3004

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

12.Mechanical vibrations

Answer: k_eq = 50 kN/m

Equivalent stiffness

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

k_eq = 20 + 30 = 50 kN/m

Answer

k_eq = 50 kN/m

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