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

An 11 kW motor running at 1440 rpm drives a pinion with 25 teeth and a module of 2 mm. Find the tangential force F_t on the pinion teeth.

P = 11 kWn = 1440 rpmN₁ = 25m = 2 mm

F_t = N

2.Gears & gear trainsHard

In a compound gear train, gear A (25 teeth) on the input shaft turning at 1500 rpm drives gear B (30 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (50 teeth) on the output shaft. Find the output speed n_out.

n_in = 1500 rpmA = 25B = 30C = 18D = 50

n_out = rpm

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 100 N·m. The two ends of a gauge length of 400 mm rotate 3.74° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 100 N·md = 20 mmL = 400 mmφ = 3.74°

G = GPa

4.Torsion of shaftsHard

A hollow shaft must transmit 75 kW at 1500 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

P = 75 kWN = 1500 rpmd_i/d_o = 0.5τ_allow = 75 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −30 MPa, σᵧ = 0 MPa and τₓᵧ = 10 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 = −30 MPaσ_y = 0 MPaτ_xy = 10 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries an axial compressive force of 80 kN together with a torque of 1500 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 40 mmF = −80 kNT = 1500 N·m

σ₁ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 50 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 70 GPaL = 600 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 60 MPa

P_max = kN

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 = 2.5 mE = 200 GPa (steel)b × h = 40 × 60 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 105 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 105 kNn = 2.5L = 3 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 3 cycles later by a peak of 18.8 mm. Find the damping ratio ζ.

x₀ = 40 mmx₃ = 18.8 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 100 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 = 100 kgk₁ = 12 kN/mk₂ = 100 kN/mk₃ = 40 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 2918 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 = 11000/150.8 = 72.95 N·m

Pitch diameter

d = m·N = 2·25 = 50 mm

Tangential force

F_t = 2T/d = 2·72.95/0.05 = 2918 N

Answer

F_t = 2918 N

2.Gears & gear trains

Answer: n_out = 450 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 = 1500·25/30 = 1250 rpm = n_C

Second stage

n_out = 1250·18/50 = 450 rpm

Answer

n_out = 450 rpm

total ratio = 3.333 : 1

3.Torsion of shafts

Answer: G = 39.01 GPa

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 moment and angle

J = π·20⁴/32 = 15708 mm⁴

φ = 3.74·π/180 = 0.06528 rad

Shear modulus

T = 100 N·m = 100000 N·mm

G = TL/(Jφ) = 100000·400/(15708·0.06528) = 39011 MPa

G = 39.01 GPa

Answer

G = 39.01 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d_o = 32.58 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 75000/157.1 = 477.5 N·m

Outer diameter from the allowable stress

T = 477.5 N·m = 477465 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·477465/(π·75·0.9375)) = 32.58 mm

d_i = 0.5·32.58 = 16.29 mm

Answer

d_o = 32.58 mm

A solid shaft would need d = 31.89 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: θ_p = 73.15 °

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.

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(2·10, −30 − 0) = 146.3°

Angle on the element

θ_p = 73.15°

Answer

θ_p = 73.15°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ₁ = 91.71 MPa

Stresses at the surface

σ = F/A = −80000/(π·40²/4) = −80000/1257 = −63.66 MPa

τ = 16T/(πd³) = 16·1500000/(π·40³) = 119.4 MPa

Mohr’s circle

C = σ/2 = −31.83 MPa

R = √((−31.83)² + 119.4²) = 123.5 MPa

σ₁ = C + R = 91.71 MPa

σ₂ = C − R = −155.4 MPa

Answer

σ₁ = 91.71 MPa

7.Axial stress & strain

Answer: δ = 0.06314 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·20²/4 = 314.2 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·5890 = 4.752 × 10⁸ N

Shortening

δ = 50000·600/4.752 × 10⁸ = 0.06314 mm

Answer

δ = 0.06314 mm

8.Axial stress & strain

Answer: P_max = 257.4 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·4006 = 3.432 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0008571) = 0.00075 (steel)

Largest load

P_max = 0.00075·3.432 × 10⁸ = 257414 N = 257.4 kN

Answer

P_max = 257.4 kN

9.Column buckling

Answer: P_cr = 101.1 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 = 1·2500 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·320000/2500² = 101065 N = 101.1 kN

Answer

P_cr = 101.1 kN

10.Column buckling

Answer: d = 64.6 mm

Required critical load

P_cr = n·P = 2.5·105000 = 262500 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·262500·1500²/(π³·70000))^(1/4) = 64.6 mm

Answer

d = 64.6 mm

Check: the slenderness KL/r is 92.88, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.04002

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 = (1/3)·ln(40/18.8) = 0.2517

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2517/√(39.48 + 0.06334) = 0.04002

Answer

ζ = 0.04002

12.Mechanical vibrations

Answer: f_n = 2.732 Hz

Equivalent stiffness

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

k₁₂ = 12 + 100 = 112 kN/m

k_eq = 112·40/(112 + 40) = 29.47 kN/m

Natural frequency

ω_n = √(29474/100) = 17.17 rad/s

f_n = 17.17/(2π) = 2.732 Hz

Answer

f_n = 2.732 Hz

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