(a) Draw the shearing force and bending moment diagrams for the beam loaded as shown in the figure below. (15 marks) (b) During…

(a) Shear force and bending moment diagrams

Measured from the free end C: pin support A at 1 m, D at 3 m, E at 4 m, roller B at 6 m, F at 8 m. Loads: 2 kN/m on CD (6 kN at 1.5 m), 8 kN down at D, a clockwise couple of 6 kN·m at E, and 4 kN down at F.

Reactions: moments about A give 5R_B = 6(0.5) + 8(2) + 6 + 4(7) = 53, so R_B = 10.6 kN ↑. From ΣV = 0, R_A = 6 + 8 + 4 − 10.6 = 7.4 kN ↑.

Shear force (kN): at C it is 0 and it falls linearly to −2.0 just left of A. At A it jumps by 7.4 to +5.4 and falls linearly under the UDL to +1.4 just left of D. The 8 kN load at D takes it to −6.6, which stays constant to B (the couple does not change it). At B it jumps by 10.6 to +4.0, constant to F, where the 4 kN load brings it to zero.

Bending moment (kN·m, sagging positive): M_C = 0. M_A = −2 × 1 × 0.5 = −1.0. M_D = 7.4 × 2 − 6 × 1.5 = +5.8, the maximum sagging moment, because the shear changes sign at D. M_E = 5.8 − 6.6 × 1 = −0.8 just left of E, and the clockwise couple adds 6, so it is +5.2 just right of E. M_B = 5.2 − 6.6 × 2 = −8.0, the maximum hogging moment; from the right, −4 × 2 = −8.0 confirms it. M_F = 0.

Points of contraflexure, measured from C: on AD, 7.4(x − 1) − x² = 0 gives x = 1.19 m; on DE, 5.8 − 6.6(x − 3) = 0 gives x = 3.88 m; on EB, 5.2 − 6.6(x − 4) = 0 gives x = 4.79 m.

(b) Adequacy of ISMB 550 (span 7 m, laterally restrained)

Factored loads (γ_f = 1.5): UDL dead = 100 + 1.037 × 7 = 107.26 kN, imposed = 150 kN, so W_u = 1.5 × 257.26 = 385.9 kN. Point load P_u = 1.5 × 100 = 150 kN at mid-span. V_u = 385.9/2 + 150/2 = 267.9 kN. M_u = W_uL/8 + P_uL/4 = 337.7 + 262.5 = 600.2 kN·m.

(i) Shear: V_d = f_y h t_w/(√3 γ_m0) = 250 × 550 × 11.2/(√3 × 1.1) = 808.3 kN, above 267.9 kN, so safe. V_u is below 0.6V_d = 485 kN, so it is a low-shear case and the moment capacity needs no reduction.

(ii) Bending (plastic section, laterally supported): M_d = β_b Z_p f_y/γ_m0 = 2711.98 × 10³ × 250/1.1 = 616.4 kN·m. The limit 1.2 Z_e f_y/γ_m0 = 643.6 kN·m is not exceeded. M_d = 616.4 kN·m exceeds M_u = 600.2 kN·m, so safe, with utilisation 0.97.

(iii) Deflection under imposed load (150 kN UDL and 50 kN at mid-span), EI = 2 × 10⁵ × 64 900 × 10⁴ N·mm²: δ = 5WL³/(384EI) + PL³/(48EI) = 5.16 + 2.75 = 7.9 mm. The limit L/300 = 23.3 mm (L/360 = 19.4 mm if brittle finishes are supported) is not exceeded, so safe. Even under the total service load δ = 14.4 mm, which is within both limits.

(iv) Web buckling at the support: d = 550 − 2(19.3 + 18) = 475.4 mm and r = t_w/√12 = 3.23 mm, so KL/r = 0.7d/r = 102.9. Interpolating the table, f_cd = 107 − 0.29 × 12.4 = 103.4 MPa. Effective width = b₁ + n₁ = 100 + 550/2 = 375 mm (45° dispersion to mid-depth). F_cdw = 375 × 11.2 × 103.4 = 434.2 kN, above 267.9 kN, so safe.

The section is adequate in all four checks; bending governs at 97% utilisation.

(c) Deflection at A by the unit load method

C is fixed, BC is vertical (3.5 m, 2.5I) and BA is horizontal (2.5 m, I). Loads: 8 kN horizontal at B and 15 kN down at A. Take s from A along AB and y down from B along BC.

Real moments: AB, M = −15s; BC, M = −(37.5 + 8y). Unit load at A horizontal: AB, m = 0; BC, m = −y. Unit load at A vertical: AB, m = −s; BC, m = −2.5.

Horizontal: AB contributes nothing. δ_H = (1/2.5EI) ∫₀^3.5 (37.5 + 8y) y dy = (1/2.5EI)[37.5 × 3.5²/2 + 8 × 3.5³/3] = (229.69 + 114.33)/2.5EI = 137.6/EI (kN·m³), towards the right.

Vertical: AB gives (1/EI) ∫₀^2.5 15s² ds = 15 × 2.5³/3EI = 78.13/EI. BC gives (1/2.5EI) ∫₀^3.5 2.5(37.5 + 8y) dy = (131.25 + 49.0)/EI = 180.25/EI. So δ_V = 258.4/EI (kN·m³), downward.

The resultant is δ_A = √(137.6² + 258.4²)/EI = 292.7/EI, inclined about 62° below the horizontal. Both components are positive, so A moves along each unit load: right and down. Bending deformation only is considered.