(a) A prestressed concrete T-beam having the cross-section of flange 1500 mm wide and 200 mm thick, rib of 300 mm wide and 1200…

(a) Section: flange 1500 × 200 mm and rib 300 mm wide. The 1200 mm is taken as the overall depth (rib 1000 mm below the flange), which gives the round section properties the data are set up for.

• A = 1500 × 200 + 300 × 1000 = 6.0 × 10⁵ mm²
• Centroid from the top: y_t = (3 × 10⁵ × 100 + 3 × 10⁵ × 700)/6 × 10⁵ = 400 mm, so y_b = 800 mm
• I = 1500 × 200³/12 + 3 × 10⁵ × 300² + 300 × 1000³/12 + 3 × 10⁵ × 300² = 8.0 × 10¹⁰ mm⁴
• Z_t = I/y_t = 2.0 × 10⁸ mm³; Z_b = I/y_b = 1.0 × 10⁸ mm³

Mid-span moments (span 18 m):

• Self-weight g = 0.6 m² × 25 = 15 kN/m, so M_g = 15 × 18²/8 = 607.5 kN·m.
• Live load: M_q = 20 × 18²/8 = 810 kN·m.
• Total: M = 1417.5 kN·m.

Which stages the limits apply to. If both limits are applied at service (top = 0, bottom = 5 MPa with P = 0.84P_i), the two equations give e = 1751 mm, far outside the section (y_b = 800 mm), so that reading is impossible. The limits are therefore applied at the two critical stages at mid-span: no tension at the top fibre at transfer (P_i acting with M_g only), and 5 MPa compression at the bottom fibre at service (0.84P_i with the full M).

Top fibre at transfer:

P_i/A − P_i e/Z_t + M_g/Z_t = 0, which gives P_i(e − Z_t/A) = M_g, i.e. P_i(e − 333.3) = 607.5 × 10⁶ … (1)

Bottom fibre at service:

0.84P_i(1/A + e/Z_b) − M/Z_b = 5, which gives 0.84P_i(Z_b/A + e) = M + 5Z_b, i.e. P_i(e + 166.7) = 2282.7 × 10⁶ … (2)

Subtracting (1) from (2): 500 P_i = 1675.2 × 10⁶, so

• P_i = 3350 kN
• e = 333.3 + 607.5 × 10⁶/3.350 × 10⁶ = 515 mm below the centroid (285 mm above the soffit)
• Effective force P_e = 0.84 × 3350 = 2814 kN

Stress check at mid-span (compression positive):

• Transfer, top: 5.58 − 8.62 + 3.04 = 0.00 MPa; bottom: 5.58 + 17.24 − 6.08 = 16.75 MPa.
• Service, top: 0.84 × (5.58 − 8.62) + 7.09 = 4.54 MPa; bottom: 0.84 × (5.58 + 17.24) − 14.18 = 5.00 MPa.

Both conditions are met and there is no tension at mid-span. At the supports (M = 0) a straight tendon would cause about 3.04 MPa tension at the top at transfer; this is normally controlled by debonding or sheathing the tendons near the ends.

(b) Geometry: triangles ABC, BCD and CDE are equilateral with AC = CE = 3 m, so every member is 3.0 m long. AE = 500 mm² × 200 kN/mm² = 1.0 × 10⁵ kN. Reactions: R_A = R_E = 50 kN.

Real forces N (tension +):

• Joint C (given): AC = CE = +28.87 kN and BC = CD = +57.74 kN. Check: 2 × 57.74 × sin 60° = 100 kN.
• Joint A: N_AB sin 60° + 50 = 0, so AB = −57.74 kN (compression); horizontally 28.87 − 57.74 cos 60° = 0.
• Joint B: AB (compression) pushes B with (28.87 →, 50 ↑) and BC (tension) pulls it with (28.87 →, 50 ↓), so BD must supply 57.74 kN ←: BD = −57.74 kN (compression).
• By symmetry DE = −57.74 kN.

Virtual forces n (1 kN down at C): the loading pattern is the same, so n = N/100.

• AB, DE, BD: N = −57.74, n = −0.5774, L = 3 m, NnL = 100.0 each (300.0)
• BC, CD: N = +57.74, n = +0.5774, L = 3 m, NnL = 100.0 each (200.0)
• AC, CE: N = +28.87, n = +0.2887, L = 3 m, NnL = 25.0 each (50.0)
• ΣNnL = 550.0 kN·m

δ_C = ΣNnL/(AE) = 550.0/(1.0 × 10⁵) m = 5.5 mm downward.

Energy check: ½ × 100 × δ = ΣN²L/(2AE) gives the same 5.5 mm. The top chord BD carries 57.74 kN compression and contributes 1.0 mm; it is not a zero-force member.

(c) A tension member can fail in the member itself or at its end connection. IS 800:2007 (Section 6) takes the design strength as the least of the following:

• Gross-section yielding (cl. 6.2). Away from the holes the whole section reaches f_y and the member stretches excessively. T_dg = A_g f_y/γ_m0 (γ_m0 = 1.10). This failure is ductile and gives warning.
• Net-section rupture (cl. 6.3). The member fractures across the weakest section through the bolt holes, which may be a zig-zag path through staggered holes (A_n = [b − n d_h + Σp²/4g] t). T_dn = 0.9 A_n f_u/γ_m1 (γ_m1 = 1.25). This failure is sudden and less ductile.
• Shear lag in angles and tees. When only one leg is connected, stress in the outstanding leg lags near the connection, so it is not fully effective: T_dn = 0.9 A_nc f_u/γ_m1 + β A_go f_y/γ_m0.
• Block shear (cl. 6.4). A block of the connected plate or leg tears out: shear along the bolt line plus tension across to the edge. T_db is the lesser of A_vg f_y/(√3 γ_m0) + 0.9 A_tn f_u/γ_m1 and 0.9 A_vn f_u/(√3 γ_m1) + A_tg f_y/γ_m0.
• Connection failure. Bolts can fail in shear or bearing and welds can fracture; the connection is designed so that this does not govern.

Gross yielding is the preferred mode because it is ductile, so connections are detailed to make rupture and block shear stronger than yielding.