(a) (i) The acceleration of a particle along a straight line is defined by a = (2t – 9) m/s², where t is in seconds. At t = 0, s…

(a)(i) Rectilinear motion with a = 2t − 9 (10 marks)

Velocity. v = ∫ a dt = t² − 9t + C₁. At t = 0, v = 10, so C₁ = 10 and v = t² − 9t + 10.

Position. s = ∫ v dt = t³/3 − 4.5t² + 10t + C₂. At t = 0, s = 1, so C₂ = 1 and

s = t³/3 − 4.5t² + 10t + 1.

(I) Position at t = 9 s. s = 243 − 364.5 + 90 + 1 = −30.5 m (that is, 30.5 m on the negative side of the origin).

(III) Velocity at t = 9 s. v = 81 − 81 + 10 = +10 m/s.

(II) Total distance. The particle reverses direction where v = 0: t² − 9t + 10 = 0 gives t = (9 ± √41)/2 = 1.298 s and 7.702 s. v is positive for t < 1.298 s, negative between 1.298 s and 7.702 s, and positive again after that. The positions at the turning points are

• s(0) = 1 m
• s(1.298) = 0.730 − 7.587 + 12.984 + 1 = 7.127 m
• s(7.702) = 152.28 − 266.92 + 77.02 + 1 = −36.627 m
• s(9) = −30.5 m

Distance = |7.127 − 1| + |−36.627 − 7.127| + |−30.5 + 36.627| = 6.127 + 43.754 + 6.127 = 56.0 m.

The displacement is only −31.5 m (from 1 m to −30.5 m), so the total distance differs from it.

(a)(ii) Thin cylindrical pressure vessel (10 marks)

Data: r = 0.6 m, t = 18 mm, p = 2.8 MPa, E = 200 GPa, ν = 0.30. Since r/t = 33 is large, the thin-wall formulae apply. The weld angle of 55° affects only the stress on the weld seam and is not needed for the quantities asked.

(I) Stresses.

• Circumferential (hoop): σ_c = pr/t = 2.8 × 600/18 = 93.3 MPa.
• Longitudinal: σ_L = pr/2t = 46.7 MPa.

(II) Maximum shear stresses. The radial stress is negligible (0 at the outer surface, −p at the inner), so the principal stresses are σ₁ = 93.3, σ₂ = 46.7 and σ₃ ≈ 0 MPa.

• Maximum in-plane shear stress: (σ_c − σ_L)/2 = 23.3 MPa.
• Maximum out-of-plane shear stress (the largest, in the plane of σ_c and the radial direction): σ_c/2 = 46.7 MPa.

(III) Strains.

• ε_c = (σ_c − νσ_L)/E = (93.33 − 0.3 × 46.67)/200 000 = 3.97 × 10⁻⁴.
• ε_L = (σ_L − νσ_c)/E = (46.67 − 0.3 × 93.33)/200 000 = 9.33 × 10⁻⁵.

(b)(i) Plane stress with unknown σ_y and τ_xy (10 marks)

The normal stress on a plane at angle θ counter-clockwise from the x-axis is

σ_θ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 cos 2θ + τ_xy sin 2θ

Write S = (σ_x + σ_y)/2 and D = (σ_x − σ_y)/2, so S = 110 − D. The two conditions give:

• θ = 32°, 2θ = 64° (cos 0.43837, sin 0.89879): S + 0.43837 D + 0.89879 τ_xy = 37.
• θ = 48°, 2θ = 96° (cos −0.10453, sin 0.99452): S − 0.10453 D + 0.99452 τ_xy = −12.

Substituting S = 110 − D:

−0.56163 D + 0.89879 τ_xy = −73 … (1)

−1.10453 D + 0.99452 τ_xy = −122 … (2)

From (1), τ_xy = −81.22 + 0.62486 D. Putting this in (2): −1.10453 D − 80.78 + 0.62144 D = −122, so −0.48309 D = −41.22 and D = 85.34 MPa.

σ_y = σ_x − 2D = 110 − 170.67 = −60.7 MPa (compressive)

τ_xy = −81.22 + 0.62486 × 85.34 = −27.9 MPa

Check. S = 24.66. At 32°: 24.66 + 37.41 − 25.07 = 37.0. At 48°: 24.66 − 8.92 − 27.74 = −12.0. Both conditions are satisfied. The negative τ_xy means it acts opposite to the sense drawn in the figure.

(b)(ii) Damped natural frequency of the rotational system (10 marks)

Model. The massless rigid bar rotates about O by a small angle θ. The spring k at distance a, the damper c at distance b and the mass m at distance l all move vertically, by aθ, bθ and lθ. Gravity only fixes the static equilibrium position and drops out of the equation of motion.

Equation of motion. Taking moments about O:

• Inertia torque: (m l²) θ̈, since the mass velocity is l θ̇ and its acceleration l θ̈.
• Damper torque: (c b θ̇) b = c b² θ̇.
• Spring torque: (k a θ) a = k a² θ.

m l² θ̈ + c b² θ̇ + k a² θ = 0

Frequencies. Comparing with I θ̈ + c_t θ̇ + k_t θ = 0, with I = m l², c_t = c b² and k_t = k a²:

• Undamped natural frequency: ω_n = √(k_t/I) = (a/l) √(k/m).
• Damping ratio: ζ = c_t/(2√(k_t I)) = c b²/(2 a l √(k m)).
• Damped natural frequency:

ω_d = ω_n √(1 − ζ²) = √( k a²/(m l²) − c² b⁴/(4 m² l⁴) )

This is valid for an underdamped system, that is ζ < 1, or c < 2 a l √(k m)/b². For ζ ≥ 1 the motion is not oscillatory.

(c) Glass-epoxy composite (10 marks)

Data: E_f = 72.5 GPa, E_m = 3.5 GPa, V_f = 0.35 and V_m = 0.65.

(i) Moduli. Loading along the fibres gives equal strain in fibre and matrix (rule of mixtures, Voigt):

E_L = E_f V_f + E_m V_m = 72.5 × 0.35 + 3.5 × 0.65 = 25.375 + 2.275 = 27.65 GPa

Loading across the fibres gives equal stress in both (Reuss):

1/E_T = V_f/E_f + V_m/E_m = 0.35/72.5 + 0.65/3.5 = 0.004828 + 0.185714 = 0.190542 GPa⁻¹

E_T = 5.25 GPa

(ii) Strains for 50 MPa on 250 mm². The load is 50 × 250 = 12.5 kN. Since the fibres are continuous, the load is taken along the fibres, so fibre and matrix have the same strain:

ε = σ/E_L = 50/27 650 = 1.81 × 10⁻³ in both fibre and matrix.

The stresses are σ_f = 72 500 × 1.808 × 10⁻³ = 131.1 MPa and σ_m = 3500 × 1.808 × 10⁻³ = 6.33 MPa. The loads carried are fibres 131.1 × 0.35 × 250 = 11.47 kN (92%) and matrix 6.33 × 0.65 × 250 = 1.03 kN, which add up to 12.5 kN.

Transverse loading (for comparison). With the stress transverse to the fibres, both carry 50 MPa, so ε_f = 50/72 500 = 0.69 × 10⁻³ and ε_m = 50/3500 = 14.3 × 10⁻³, and the composite strain is 50/5248 = 9.5 × 10⁻³. The matrix strains about 21 times as much as the fibre.