(a) Determine the forces in members AC, DE and GH of the truss loaded and supported as shown in the figure below : (10…

(a)

• The truss is planar, statically determinate, and symmetric about midspan. The loading is also symmetric.
• Total vertical load = 5 + 10 + 10 + 10 + 5 = 40 kN.
• Vertical reactions: R_A = R_H = 40/2 = 20 kN upward. No horizontal load, so the horizontal reaction at A is zero.
• Use the method of joints. Take tension as positive. Each diagonal has horizontal and vertical projections of 2 m, so it is at 45° and its vertical and horizontal components are equal to F/√2.
• At joint A, the net external vertical force is 20 - 5 = 15 kN upward.
• Member AC is horizontal; member AB is inclined 45° downward to the right.
• Vertical equilibrium at A: ΣF_y = 15 - F_AB/√2 = 0 F_AB/√2 = 15 kN F_AB = 15√2 kN tension.
• Horizontal equilibrium at A: ΣF_x = F_AC + F_AB/√2 = 0 F_AC + 15 = 0 F_AC = -15 kN.
• The negative sign means compression.
• By symmetry, at joint H the diagonal HF has the same force, F_HF = 15√2 kN tension, and its horizontal component is 15 kN. Horizontal equilibrium gives F_GH = -15 kN, i.e. compression.
• At joint D, the only vertical member is DE. The applied load is 10 kN downward. Taking tension in DE as pulling D downward, ΣF_y = -10 - F_DE = 0 F_DE = -10 kN.
• The negative sign means compression.
• A quick check at joint B shows that the 15 kN horizontal component of AB is balanced by BE in tension, and the 15 kN vertical component is balanced by CB in compression, so the assumed member forces are consistent.

F_AC = 15 kN compression, F_DE = 10 kN compression, F_GH = 15 kN compression.

(b)

• Assume the beam is simply supported, the section is the stated rectangular section, there are no prestress losses, and the behaviour is linear elastic.
• Width b = 160 mm, depth D = 400 mm.
• Area A = bD = 160 × 400 = 64,000 mm².
• Moment of inertia I = bD³/12 = 160 × 400³/12 = 8.5333 × 10⁸ mm⁴.
• Section modulus S = I/(D/2) = bD²/6 = 160 × 400²/6 = 4.2667 × 10⁶ mm³.
• Prestress force P = 200 kN = 2.00 × 10⁵ N.
• The centroid is at 200 mm from the bottom. The cable is at 100 mm from the bottom, so the eccentricity is e = 100 mm below the centroid.
• Use superposition of the axial prestress force and the bending moment caused by its eccentricity. The cable is straight, so this prestress moment is constant along the span.
• Uniform axial stress: P/A = 200,000/64,000 = 25/8 = 3.125 N/mm² compression.
• Bending stress due to eccentricity: Pe/S = 200,000 × 100 / 4.2667 × 10⁶ = 75/16 = 4.6875 N/mm².
• Because the prestressing force is below the centroid, it increases compression at the bottom fibre and reduces compression, or causes tension, at the top fibre.
• Prestress stresses: top: 3.125 - 4.6875 = -1.5625 N/mm², i.e. 1.5625 N/mm² tension. bottom: 3.125 + 4.6875 = 7.8125 N/mm² compression.
• The uniformly distributed load is w = 5 kN/m over an effective span L = 8 m.
• Midspan moment for a simply supported beam: M = wL²/8 = 5 × 8²/8 = 40 kN·m = 4.0 × 10⁷ N·mm.
• Flexural stress: M/S = 4.0 × 10⁷ / 4.2667 × 10⁶ = 75/8 = 9.375 N/mm².
• Sagging moment gives compression at the top fibre and tension at the bottom fibre.
• Superpose the prestress and load stresses: top: -75/16 + 25/8 + 75/8 = 125/16 = 7.8125 N/mm² compression. bottom: 75/16 + 25/8 - 75/8 = -25/16 = -1.5625 N/mm², i.e. 1.5625 N/mm² tension.
• The units N/mm² are equivalent to MPa.

Top fibre stress = 7.8125 MPa compression; bottom fibre stress = 1.5625 MPa tension.

(c)

[(c)] The figure printed on the original question paper for this part could not be recovered from the scan, so this part is not answered here.

(d)

• Use Newton’s law of viscosity: τ = μ du/dy.
• The plate is very thin, so its thickness is neglected. The gaps are small and the velocity profile in each glycerine layer is linear. The outer surfaces are stationary and the plate moves at V = 0.5 m/s.
• For a gap h, du/dy = V/h, so τ = μV/h. The shear force on one face is τA. The two faces act in the same direction to resist the drag, so the total force is F = A μ V (1/h_1 + 1/h_2), where A = 0.75 m² and μ = 0.785 N-s/m².
• No edge effects are considered because the surfaces are large and the plate is thin.

(i)

• The plate is equidistant from the two surfaces, so each gap is h_1 = h_2 = 25/2 = 12.5 mm = 0.0125 m.
• Shear stress on each face: τ = 0.785 × 0.5 / 0.0125 = 31.4 N/m².
• Force on both faces: F = 2 × 31.4 × 0.75 = 47.1 N.

F = 47.1 N.

(ii)

• The plate is 10 mm from one surface, so h_1 = 10 mm = 0.010 m, h_2 = 25 - 10 = 15 mm = 0.015 m.
• Shear stress on the 10 mm gap side: τ_1 = 0.785 × 0.5 / 0.010 = 39.25 N/m².
• Shear stress on the 15 mm gap side: τ_2 = 0.785 × 0.5 / 0.015 = 26.1667 N/m².
• Total force: F = (39.25 + 26.1667) × 0.75 = 49.0625 N.

F = 49.0625 N, approximately 49.1 N.

(e)

• The moisture content is not required for relative density; only the dry densities are needed.
• Field dry density: ρ_d,field = 779/426 = 1.82864 g/cm³.
• Since 1 g/cm³ corresponds to 9.81 kN/m³, γ_d,field = 1.82864 × 9.81 = 17.9389 kN/m³.
• Loose state: 400 g occupies 276 cm³. ρ_d,loose = 400/276 = 1.44928 g/cm³, γ_d,loose = 1.44928 × 9.81 = 14.2174 kN/m³.
• This is the minimum dry density and corresponds to the maximum void ratio e_max.
• Tamped state: 400 g occupies 212 cm³. ρ_d,dense = 400/212 = 1.88679 g/cm³, γ_d,dense = 1.88679 × 9.81 = 18.5094 kN/m³.
• This is the maximum dry density and corresponds to the minimum void ratio e_min.
• Use the void ratio relation for a soil of specific gravity G_S: e = G_S γ_w / γ_d - 1, with G_S = 2.70 and γ_w = 9.81 kN/m³.
• Field void ratio: e_field = 2.70 × 9.81 / 17.9389 - 1 = 0.47651.
• Maximum void ratio: e_max = 2.70 × 9.81 / 14.2174 - 1 = 0.86300.
• Minimum void ratio: e_min = 2.70 × 9.81 / 18.5094 - 1 = 0.43100.
• Relative density: D_r = (e_max - e_field)/(e_max - e_min) × 100 = (0.86300 - 0.47651)/(0.86300 - 0.43100) × 100 = 89.47%.
• The field dry density lies between the loose and dense dry densities, so a relative density between 0 and 100% is expected; the result is consistent.

Relative density of the field soil = 89.5%.