(a) (i) A simple girder of 20 m span is traversed by a moving uniformly distributed load of 6 m long with an intensity of 2 kN/m…

(a)(i) Moving UDL on a 20 m girder

Data: span L = 20 m; UDL w = 2 kN/m over a length a = 6 m; section X at x = 4 m from A.

Influence lines for section X:

• Bending moment: a triangle with peak ordinate x(L − x)/L = 4 × 16/20 = 3.2 m at X.
• Shear force:
• From A to just left of X, the line falls from 0 to −4/20 = −0.2.
• Just right of X it is +16/20 = +0.8, falling to 0 at B.

Maximum BM at X. The load is placed so that X divides it in the same ratio as it divides the span (4 : 16). This puts 1.2 m of the load left of X and 4.8 m right of it, i.e. the load occupies x = 2.8 m to 8.8 m. The IL ordinate is 2.24 at both ends of the load.

• Area under the IL covered by the load = ½(2.24 + 3.2)(1.2) + ½(3.2 + 2.24)(4.8) = 3.264 + 13.056 = 16.32 m²
• M_max = w × area = 2 × 16.32 = 32.64 kN·m

Maximum shear at X.

• Positive: the head of the load is placed just right of X, so the load covers x = 4 to 10 m, where the IL ordinates are 0.8 and 0.5.
• V⁺_max = 2 × ½(0.8 + 0.5) × 6 = +7.8 kN
• Negative: the tail of the load is placed at X, but only 4 m of the load can lie on AX.
• V⁻_max = −2 × ½ × 4 × 0.2 = −0.8 kN

The maximum shear force at X is therefore 7.8 kN, positive.

Absolute maximum BM. For a single UDL shorter than the span, the greatest moment occurs at mid-span, with the load placed symmetrically about the centre.

• R_A = 12/2 = 6 kN
• M_abs = 6 × 10 − 2 × 3 × 1.5 = 51 kN·m

Equivalently, M = w a (L/4)(1 − a/2L) = 2 × 6 × 5 × 0.85 = 51 kN·m. A numerical scan over every section and every load position confirms this value.

(a)(ii) Maximum force in member BC

Section. Cut through members IH, BH and BC. Take moments about H (x = 16 m), where the other two cut members meet. The loads are applied at the top-chord panel points and H is itself a panel point, so

F_BC = M_H/h, where M_H is the beam bending moment at x = 16 m and h = 6 m.

Influence line for F_BC. It is a triangle over the 32 m span with peak ordinate (16 × 16/32)/6 = 1.333 under H. Solving the complete truss (all joint equations) with a unit load at J, I, H, G and F gives ordinates 0, 0.667, 1.333, 0.667 and 0, all tensile, which confirms this.

Maximum force.

• Point load placed at H: 80 × 1.333 = 106.67 kN
• UDL covering the whole span: 8.5 × (½ × 32 × 1.333) = 8.5 × 21.333 = 181.33 kN

F_BC,max = 106.67 + 181.33 = 288 kN (tension). Since the influence line has no negative portion, BC never goes into compression.

(b) Laced column of 2 ISMC 350, back to back

Effective length. The column is held in position but free to rotate at both ends, so K = 1.0 and KL = 8 m in both directions.

Properties about z–z (unaffected by spacing):

• I_zz = 2 × 10,000 = 20,000 cm⁴
• A = 2 × 5440 = 10,880 mm²
• r_zz = √(20,000 × 10⁴/10,880) = 135.6 mm

Spacing for maximum load. The load is greatest when the column is equally strong about both axes, i.e. I_yy = I_zz. Let S be the clear back-to-back distance between the webs, so each channel's centroid is S/2 + 24.4 mm from the axis:

2[434 × 10⁴ + 5440 (S/2 + 24.4)²] = 20,000 × 10⁴

(S/2 + 24.4)² = 17,585, so S/2 + 24.4 = 132.6 mm and S = 216.4 mm

Adopt S = 220 mm back to back. This gives I_yy = 20,521 cm⁴ > I_zz, so z–z governs.

Slenderness.

• KL/r = 8000/135.6 = 59.0
• For laced columns, IS 800 (cl. 7.6.1.5) raises this by 5 %: λ_e = 1.05 × 59.0 = 61.96

Design stress (interpolating the table):

f_cd = 168 − (168 − 152)(1.96/10) = 164.9 N/mm²

Capacity. The tabulated f_cd already includes γ_m0 = 1.10, so it is not divided again:

P_d = f_cd × A = 164.9 × 10,880 = 1794 kN (≈ 1.79 MN)

Lacing requirement. A single channel between lacing points must satisfy KL₁/r_min ≤ 50 and ≤ 0.7λ_e = 43.4, where r_min = √(434 × 10⁴/5440) = 28.2 mm. The lacing-point spacing must therefore not exceed about 1.22 m, so that no channel buckles locally before the column as a whole.

(c) Stresses at mid-span of the post-tensioned beam

Section properties:

• A = 300 × 600 = 1.8 × 10⁵ mm²
• Z = bD²/6 = 300 × 600²/6 = 18 × 10⁶ mm³
• Eccentricity e = 300 − 100 = 200 mm

Prestress:

• Initial: P_i = 500 × 1400 = 700 kN
• Final, after 15 % loss: P_e = 0.85 × 700 = 595 kN

Moments (unit weight of concrete taken as 25 kN/m³):

• Self-weight g = 0.3 × 0.6 × 25 = 4.5 kN/m, so M_g = 4.5 × 10²/8 = 56.25 kN·m
• Live load: M_q = 7 × 10²/8 = 87.5 kN·m

Initial stage (P_i + self-weight):

• P/A = 3.889 N/mm²
• Pe/Z = 7.778 N/mm²
• M_g/Z = 3.125 N/mm²
• Top: 3.889 − 7.778 + 3.125 = −0.76 N/mm² (tension)
• Bottom: 3.889 + 7.778 − 3.125 = +8.54 N/mm² (compression)

Final stage (P_e + self-weight + live load):

• P/A = 3.306 N/mm²
• Pe/Z = 6.611 N/mm²
• M/Z = 143.75 × 10⁶/18 × 10⁶ = 7.986 N/mm²
• Top: 3.306 − 6.611 + 7.986 = +4.68 N/mm² (compression)
• Bottom: 3.306 + 6.611 − 7.986 = +1.93 N/mm² (compression)

Interpretation: At transfer there is a small tensile stress of 0.76 N/mm² at the top. This is acceptable in a Type 2 member; in a Type 1 member it would have to be removed by slightly reducing the eccentricity. Under service loads the whole section stays in compression, so the beam behaves as a fully prestressed (Type 1) member at mid-span.