(a) Determine the maximum tensile force in member DI of the truss shown below due to the series of three moving loads shown in…

(a) Maximum tension in member DI (15 marks)

Take A at x = 0 and G at x = 24 m, with top joints at 4 m spacing and depth 3 m. DI joins I(8, 0) to D(12, 3), so its length is 5 m and its vertical component is 3/5. Cut the panel through CD, DI and IJ and take the left part.

For a unit load at x, R_A = (24 − x)/24. Vertical equilibrium of the left part gives R_A − P_left + (3/5)F_DI = 0, so F_DI = (5/3)(P_left − R_A), tension positive.

Influence line:

• x ≤ 8 m: F = 5x/72, which is 5/9 at x = 8.
• x ≥ 12 m: F = −5(24 − x)/72, which is −5/6 at x = 12.
• Between 8 and 12 m the line is straight, and it crosses zero at x = 9.6 m.

The loads move from G towards A, so the 125 kN load leads. The maximum occurs with a load at the peak of the influence line (x = 8 m):

• 50 kN at 8 m (loads at 3, 5, 8 m): (125 × 15 + 100 × 25 + 50 × 40)/72 = 6,375/72 = 88.5 kN.
• 100 kN at 8 m (loads at 6, 8, 11 m): (125 × 30 + 100 × 40 − 50 × 35)/72 = 83.3 kN.
• 125 kN at 8 m: 17.4 kN.
• 50 kN at 9.6 m: 85.8 kN.

Maximum tensile force in DI = 88.5 kN, with the 50 kN load over joint I and the other two loads on the left of it.

(b) Laminar flow between parallel plates (15 marks)

The profile is parabolic: u = u_max(1 − 4y²/h²), with y from the centre line, h = 0.10 m, μ = 2.453 N·s/m², u_max = 1.5 m/s.

• Discharge per metre width: q = (2/3) u_max h = (2/3)(1.5)(0.10) = 0.10 m³/s per m.
• Velocity gradient: du/dy = −8u_max y/h². At the plates (y = ±0.05 m) it is ±60 s⁻¹.
• Shear stress at the plates: τ = μ|du/dy| = 2.453 × 60 = 147.2 N/m².
• Pressure drop over L = 20 m: u_max = Δp h²/(8μL), so Δp = 8 × 2.453 × 20 × 1.5/0.10² = 58,872 N/m² ≈ 58.9 kPa. (Check: Δp = 2τL/h = 58,872 N/m².)
• Velocity at 0.02 m from a plate: y = 0.05 − 0.02 = 0.03 m, so u = 1.5[1 − 4(0.03)²/0.10²] = 0.96 m/s.

(c) Stability of the retaining wall (20 marks)

Work per metre length. The wall has height 6 m, top width 1 m and base 3 m, with a vertical back. Use Rankine's theory with a horizontal surface and neglect passive pressure.

Active thrust. K_a = (1 − sin 35°)/(1 + sin 35°) = 0.271. P_a = ½γH²K_a = 0.5 × 17.5 × 36 × 0.271 = 85.4 kN/m, acting 2 m above the base.

Weights (distances from the toe):

• Rectangle 1 × 6 m: W₁ = 144 kN at 2.5 m.
• Triangle 2 × 6 m: W₂ = 144 kN at 1.33 m.
• ΣW = 288 kN, and x̄ = (144 × 2.5 + 144 × 1.33)/288 = 1.92 m.

Overturning. M_o = 85.4 × 2 = 170.7 kN·m. M_r = 288 × 1.92 = 552 kN·m. FS = 552/170.7 = 3.23 (> 2, safe).

Sliding. Resistance = 0.5 × 288 = 144 kN. FS = 144/85.4 = 1.69 (> 1.5, safe).

Soil pressure. The resultant acts at x = (552 − 170.7)/288 = 1.32 m from the toe, so e = 1.5 − 1.32 = 0.18 m < B/6 = 0.5 m. The base is in full compression. q_max = (288/3)(1 + 6 × 0.18/3) = 129.8 kPa at the toe. q_min = 96(1 − 0.36) = 62.2 kPa at the heel. Both are below 150 kPa, so the foundation is safe.