(a) Determine the horizontal component of deflection of joint D of the truss loaded as shown in the figure. The cross-sectional…

(a) Horizontal deflection of joint D (Castigliano)

Apply a fictitious horizontal load H at D (to the right). Then Δ_Dx = Σ F (∂F/∂H) L/(AE) at H = 0, where F is the real member force (tension positive).

Real forces for the 50 kN load at C: taking moments about A, R_B × 4 = 50 × 8, so R_B = 100 kN ↑ and A_y = 50 kN ↓. Joint C: F_CD × 3/5 = 50, so F_CD = +83.33 kN, and F_BC = −66.67 kN. Joint D: F_AD = +83.33 kN and F_DB = −100 kN. Joint A: F_AB = −66.67 kN.

Unit horizontal load at D: joints C and A give u_BC = u_CD = u_AB = 0. At D, u_AD = 5/4 and u_DB = −3/4 (with R_B = 0.75 ↑, A_y = 0.75 ↓, A_x = 1 ←).

Only AD and DB contribute. With E = 200 kN/mm²:

AD: (83.33 × 1.25 × 5000)/(390 × 200) = 6.68 mm.

DB: (−100 × −0.75 × 3000)/(575 × 200) = 1.96 mm.

Δ_Dx = 6.68 + 1.96 = 8.63 mm, to the right (the direction of H).

(b) Collapse load of the propped cantilever

The beam is propped at A and fixed at B. The mechanism has hinges at B and at a section C, distance x from A. Let C deflect by δ. Then the rotation of AC is δ/x, the rotation of CB is δ/(l − x), the hinge rotation at C is δ/x + δ/(l − x), and at B it is δ/(l − x).

Internal work = M_P [δ/x + 2δ/(l − x)] = M_P δ (l + x)/[x(l − x)].

External work = load × area of the deflected shape = P × (δ l/2).

Equating, P = 2M_P (l + x)/[l x (l − x)].

Minimising with respect to x gives x² + 2lx − l² = 0, so x = (√2 − 1)l = 0.414 l. Substituting, (l + x) = 1.414 l and x(l − x) = 0.2426 l², so

P_c = (6 + 4√2) M_P/l² ≈ 11.66 M_P/l² per metre length.

(c) Circular tank of 450 kL, flexible base (wall design)

Water depth h = 4.0 − 0.25 = 3.75 m. From πD²h/4 = 450, D² = 450 × 4/(π × 3.75) = 152.8, so D = 12.36 m internal diameter. Total wall height = 4.0 m.

With a flexible (sliding) base the wall is free at the bottom, so it acts as a ring with no bending; the hoop tension is greatest at the base. With γ = 9.8 kN/m³: T = γhD/2 = 9.8 × 3.75 × 12.36/2 = 227 kN/m.

Hoop steel (σ_s = 115 MPa): A_st = 227 000/115 = 1975 mm²/m, both faces together. Provide 12 mm bars at 110 mm c/c on each face, A_st = 2056 mm²/m. Steel stress = 227 100/2056 = 110.5 MPa, safe.

Wall thickness from the concrete tension limit (σ_ct = 1.2 MPa, m = 13.33): t = 200 mm gives equivalent area = 1000 × 200 + 12.33 × 2056 = 225 350 mm²/m, so σ_ct = 227 100/225 350 = 1.01 MPa, below 1.2 MPa; safe.

Upper part: tension falls with depth. For the top 1.75 m, T = 9.8 × 1.75 × 6.18 = 106 kN/m, needing 922 mm²/m; 12 mm bars at 200 mm c/c on each face (1131 mm²/m) suffice.

Vertical steel: minimum 0.3% of gross area = 600 mm²/m; provide 8 mm bars at 160 mm c/c on each face (628 mm²/m).

Detailing: clear cover 45 mm on the water face; hoop bars lapped in a staggered way; vertical bars tied to the hoops; a sliding joint with a water stop at the wall–base junction.

Final: D = 12.36 m, total height 4.0 m, t = 200 mm; hoops 12 mm at 110 mm c/c each face in the lower 2 m and at 200 mm c/c above; vertical 8 mm at 160 mm c/c each face.