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 4 kW motor running at 1440 rpm drives a pinion with 24 teeth and a module of 2 mm. Find the tangential force F_t on the pinion teeth.

P = 4 kWn = 1440 rpmN₁ = 24m = 2 mm

F_t = N

2.Gears & gear trainsHard

An open belt connects an 80 mm driving pulley at 960 rpm to a 400 mm pulley; the centres are 1000 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 = 80 mmn₁ = 960 rpmd₂ = 400 mmC = 1000 mmμ = 0.3T₁ = 500 N

P = kW

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 = 500 N·md = 50 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft transmits 10 kW at 750 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 10 kWN = 750 rpmτ_allow = 60 MPaθ_allow = 0.5°/mG = 80 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 130 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 130 MPaσ_y = −70 MPaτ_xy = −30 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 12 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 1200 mmt = 12 mmp = 3 MPaα = 55°

|τ_n| = MPa

7.Axial stress & strainMedium

Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 75 kNd = 30 mmL = 3 mE = 117 GPa

ε = µε

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 30 kNsegment 1: Ø25 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, aluminium (E = 70 GPa)

δ = mm

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)d = 30 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 60 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 60 kNn = 2L = 1 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 5 cycles later by a peak of 6.4 mm. Find the damping ratio ζ.

x₀ = 12 mmx₅ = 6.4 mmn = 5

ζ =

12.Mechanical vibrationsHard

A 50 kg mass hangs from spring k₃ = 100 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 60 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 = 50 kgk₁ = 25 kN/mk₂ = 60 kN/mk₃ = 100 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 1105 N

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.

Torque on the pinion

ω = 2π·1440/60 = 150.8 rad/s

T = P/ω = 4000/150.8 = 26.53 N·m

Pitch diameter

d = m·N = 2·24 = 48 mm

Tangential force

F_t = 2T/d = 2·26.53/0.048 = 1105 N

Answer

F_t = 1105 N

2.Gears & gear trains

Answer: P = 1.148 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 = 2.82 rad

Slack-side tension

e^μθ = e^0.8461 = 2.33, T₂ = 500/2.33 = 214.6 N

Power

P = (500 − 214.6)·4.021 = 1.148 kW

Answer

P = 1.148 kW

3.Torsion of shafts

Answer: φ = 1.796 °

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) = 500·1000·1500/(39000·613592) = 0.03134 rad

φ = 0.03134·180/π = 1.796°

Answer

φ = 1.796°

4.Torsion of shafts

Answer: d = 36.92 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 10000/78.54 = 127.3 N·m

Diameter for strength

T = 127.3 N·m = 127324 N·mm

d_τ = ∛(16·127324/(π·60)) = 22.11 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 80 GPa = 80000 MPa

d_φ = (32·127324/(π·80000·8.727 × 10⁻⁶))^1/4 = 36.92 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(22.11, 36.92) = 36.92 mm

Answer

d = 36.92 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_max = 104.4 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 = √(((130 − −70)/2)² + −30²) = 104.4 MPa

Answer

τ_max = 104.4 MPa

6.Mohr’s circle & stress transformation

Answer: |τ_n| = 35.24 MPa

Wall stresses

r = 1200/2 = 600 mm

σ_h = pr/t = 3·600/12 = 150 MPa

σ_a = pr/(2t) = 75 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.

|τ_n| = (150 − 75)·sin 55°·cos 55° = 35.24 MPa

Answer

|τ_n| = 35.24 MPa

7.Axial stress & strain

Answer: ε = 906.9 µε

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 = 75·1000/706.9 = 106.1 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 106.1/117000 = 0.0009069 = 906.9 µε

Answer

ε = 906.9 µε

8.Axial stress & strain

Answer: δ = 3.048 mm

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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 61.12 MPa

σ₂ = 149.2 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 61.12·1500/100000 = 0.9167 mm

δ₂ = σ₂L₂/E₂ = 149.2·1000/70000 = 2.132 mm

δ = δ₁ + δ₂ = 3.048 mm

Answer

δ = 3.048 mm

9.Column buckling

Answer: P_cr = 3.052 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 = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·39761/3000² = 3052 N = 3.052 kN

Answer

P_cr = 3.052 kN

10.Column buckling

Answer: d = 33.36 mm

Required critical load

P_cr = n·P = 2·60000 = 120000 N

KL = 1·1000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·1000²/(π³·200000))^(1/4) = 33.36 mm

Answer

d = 33.36 mm

Check: the slenderness KL/r is 119.9, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.02001

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 = (1/5)·ln(12/6.4) = 0.1257

Damping ratio

ζ = δ/√(4π² + δ²) = 0.1257/√(39.48 + 0.01581) = 0.02001

Answer

ζ = 0.02001

12.Mechanical vibrations

Answer: f_n = 4.825 Hz

Equivalent stiffness

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

k₁₂ = 25 + 60 = 85 kN/m

k_eq = 85·100/(85 + 100) = 45.95 kN/m

Natural frequency

ω_n = √(45946/50) = 30.31 rad/s

f_n = 30.31/(2π) = 4.825 Hz

Answer

f_n = 4.825 Hz

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