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 worm with 3 start(s) turns at 1440 rpm and drives a worm wheel with 72 teeth. Find the speed n_w of the wheel.

z = 3N_w = 72n = 1440 rpm

n_w = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (30 teeth) on the input shaft turns at 1440 rpm and drives gear B (60 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 192 rpm. How many teeth N_D must gear D have?

n_in = 1440 rpmA = 30B = 60C = 16n_out = 192 rpm

N_D =

3.Torsion of shaftsMedium

A solid aluminium 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 = 100 N·md = 30 mmL = 1.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 1000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1000 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 80 MPa, σᵧ = 90 MPa and τₓᵧ = 50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 80 MPaσ_y = 90 MPaτ_xy = 50 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 8 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1000 mmt = 8 mmp = 0.8 MPaα = 55°

σ_n = MPa

7.Axial stress & strainMedium

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

P = 90 kNd = 12 mmL = 0.5 mE = 200 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 = 120 kNsegment 1: Ø30 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, aluminium (E = 70 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)d = 80 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 45 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 45 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed one cycle later by a peak of 35.3 mm. Find the damping ratio ζ.

x₀ = 40 mmx₁ = 35.3 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 40 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 10 kgk₁ = 12 kN/mk₂ = 40 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n_w = 60 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.

Wheel speed

n_w = n·z/N_w = 1440·3/72 = 60 rpm

Answer

n_w = 60 rpm

2.Gears & gear trains

Answer: N_D = 60

Intermediate shaft speed

n_B = n_C = 1440·30/60 = 720 rpm

Teeth on gear D

N_D = C·n_C/n_out = 16·720/192 = 60

Answer

N_D = 60

3.Torsion of shafts

Answer: φ = 4.157 °

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) = 100·1000·1500/(26000·79522) = 0.07255 rad

φ = 0.07255·180/π = 4.157°

Answer

φ = 4.157°

4.Torsion of shafts

Answer: d = 31.89 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·75)) = 31.89 mm

Answer

d = 31.89 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_max = 50.25 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 = √(((80 − 90)/2)² + 50²) = 50.25 MPa

Answer

τ_max = 50.25 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 33.22 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 0.8·500/8 = 50 MPa

σ_a = pr/(2t) = 25 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 25·sin²55° + 50·cos²55° = 33.22 MPa

Answer

σ_n = 33.22 MPa

7.Axial stress & strain

Answer: δ = 1.989 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·12²/4 = 113.1 mm²

σ = P/A = 90·1000/113.1 = 795.8 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 795.8/200000 = 0.003979 = 3979 µε

Elongation

δ = ε·L = 0.003979·500 mm = 1.989 mm

Answer

δ = 1.989 mm

8.Axial stress & strain

Answer: σ_max = 596.8 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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 169.8 MPa

σ₂ = 596.8 MPa

Largest stress

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

σ_max = 596.8 MPa

Answer

σ_max = 596.8 MPa

9.Column buckling

Answer: P_cr = 441 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/3000² = 440978 N = 441 kN

Answer

P_cr = 441 kN

10.Column buckling

Answer: d = 56.86 mm

Required critical load

P_cr = n·P = 2.5·45000 = 112500 N

KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·112500·3000²/(π³·200000))^(1/4) = 56.86 mm

Answer

d = 56.86 mm

Check: the slenderness KL/r is 211.1, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.01989

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.

Logarithmic decrement

δ = (1/1)·ln(40/35.3) = 0.125

Damping ratio

ζ = δ/√(4π² + δ²) = 0.125/√(39.48 + 0.01562) = 0.01989

Answer

ζ = 0.01989

12.Mechanical vibrations

Answer: k_eq = 9.231 kN/m

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 = 1/12 + 1/40 → k_eq = 12·40/(12 + 40) = 9.231 kN/m

Answer

k_eq = 9.231 kN/m

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