(a) A vertical member in a truss is subjected to the following member force (factored) : (i) Compressive force = 30 kN (dead and…

(a) Number of bolts for the truss member joint (10 marks)

The member carries a factored compression of 30 kN and a factored tension of 150 kN, so the tension governs. Two ISA 100 × 100 × 8 are connected on either side of a 10 mm gusset plate, so each bolt is in double shear. Data: d = 16 mm (hole d₀ = 18 mm), grade 4.6 (f_ub = 400 MPa), γ_mb = 1.25, k_b = 0.6, and a plate of fu = 410 MPa (Fe 410 steel, assumed). The threaded area is A_nb = 0.78 × (π/4) × 16² = 0.78 × 201.1 = 156.8 mm², and the shank area is A_sb = 201.1 mm².

Shear capacity (IS 800:2007 cl. 10.3.3), with one shear plane through the threaded part and one through the shank:

V_dsb = (f_ub/√3)(n_n A_nb + n_s A_sb)/γ_mb = (400/√3)(156.8 + 201.1)/1.25 = 66.1 kN.

Bearing capacity (cl. 10.3.4): the bearing thickness is the smaller of the gusset (10 mm) and the two angle legs (2 × 8 = 16 mm), so t = 10 mm.

V_dpb = 2.5 k_b d t f_u/γ_mb = 2.5 × 0.6 × 16 × 10 × 410/1.25 = 78.7 kN.

Bolt value = the smaller = 66.1 kN.

Number of bolts = 150/66.1 = 2.27, so provide 3 bolts of 16 mm, grade 4.6, in a line. Check the spacing: pitch not less than 2.5d = 40 mm (provide 50 mm) and edge distance not less than 1.5d₀ = 27 mm (provide 30 mm).

(b) Loading and bending moment diagram from the shear force diagram (20 marks)

The shear force diagram is read from the figure as follows: at A the ordinate is 10 kN, plotted below the axis; it varies linearly, crossing zero at B, to 2 kN above the axis at C (4 m from A); it is constant at 2 kN above the axis from C to D (3 m); it steps up to 7 kN above the axis at D and stays constant to E (2 m); it drops to 2 kN below the axis at E and is constant to F (2 m), closing to zero at F. The 10 kN plotted below the axis at A is the upward reaction of the hinge, so ordinates below the axis are taken as positive in this reading and the loads are gravity loads (reversing the sign convention would reverse every load).

Reconstruction of the loads:

• At A: a jump of 10 kN, an upward reaction R_A = 10 kN at the hinge support.
• A to C: the shear changes linearly by 12 kN over 4 m, so a uniformly distributed load of 12/4 = 3 kN/m acts over AC (total 12 kN).
• C to D: constant shear, so no load.
• At D: a jump of 5 kN (from 2 to 7), a downward point load of 5 kN at D.
• D to E: constant, no load.
• At E: a jump of 9 kN (from 7 kN above the axis to 2 kN below it), the upward reaction R_E = 9 kN. This locates the second hinge support at E, 9 m from A.
• E to F: constant 2 kN, no load; at F the shear closes to zero through a downward point load of 2 kN at the free end.

Check: total downward load = 3 × 4 + 5 + 2 = 19 kN = 10 + 9. Moments about A: 12 × 2 + 5 × 7 + 2 × 11 = 81 = 9 × 9. ✓.

Bending moment (sagging positive, M = area of the shear diagram):

• A: M = 0 (hinge).
• B: shear is zero at x = 10/3 = 3.33 m from A, where the moment is maximum: M_B = 10 × 3.33 - 3 × 3.33²/2 = 16.67 kN·m.
• C: M_C = 10 × 4 - 3 × 4²/2 = 40 - 24 = 16.0 kN·m.
• D: M_D = 16.0 - 2 × 3 = 10.0 kN·m.
• E: M_E = 10.0 - 7 × 2 = -4.0 kN·m (hogging, at the support).
• F: M_F = -4.0 + 2 × 2 = 0.

Shape of the curve: from A to C it is a parabola (the shear varies linearly under the UDL) with a maximum at B; C to D is a straight line falling slowly (shear 2 kN); D to E is a steeper straight line (shear 7 kN); E to F is a straight line rising (shear 2 kN).

Point of contraflexure: between D and E, where M changes sign: 10 - 7x = 0, so x = 1.43 m from D, i.e. 8.43 m from A.

(c) Cantilever roof slab of the shelter (20 marks)

Data: slab 150 mm thick, projection 1500 mm on each side of a 250 mm central beam, M20, Fe500, ceiling plaster and roof finish 1.5 kN/m², inaccessible roof.

Effective depth. Exposure is moderate (a roof exposed to rain), so the nominal cover is 30 mm; assume 10 mm bars: d = 150 - 30 - 5 = 115 mm. Effective length of the cantilever (IS 456 cl. 22.2): the length to the face of the support plus half the effective depth = 1500 + 115/2 = 1557 mm.

Loads on a 1 m strip:

• Self-weight: 0.15 × 25 = 3.75 kN/m
• Ceiling plaster and finish: 1.5 kN/m
• Live load on an inaccessible roof (IS 875 Part 2): 0.75 kN/m
• Total w = 6.0 kN/m; factored w_u = 1.5 × 6.0 = 9.0 kN/m.

Design moment: M_u = w_u L²/2 = 9.0 × 1.557²/2 = 10.9 kN·m per metre width.

Steel. M_u/bd² = 10.9 × 10⁶/(1000 × 115²) = 0.826. From the table, p_t = 0.193 at 0.8 and 0.219 at 0.9, so by interpolation p_t = 0.200%.

A_st = 0.200 × 1000 × 115/100 = 230 mm² per metre. The minimum steel is 0.12% of bD = 180 mm², so 230 mm² governs.

Deflection check. The basic ratio for a cantilever is L/d = 7. Try 10 mm at 250 mm (314 mm²): f_s = 0.58 × 500 × 230/314 = 212 N/mm² and p_t = 0.27%, for which the chart gives a modification factor of about 1.85; the permitted L/d = 7 × 1.85 = 12.9, while the actual L/d = 1557/115 = 13.5, so it fails. Provide 10 mm at 200 mm c/c: A_st = 393 mm² (p_t = 0.34%), f_s = 0.58 × 500 × 230/393 = 170 N/mm², modification factor = 2.0 (the upper limit of the chart), permitted L/d = 7 × 2.0 = 14.0 > 13.5, so the deflection is acceptable.

Shear check. At d from the face: V_u = 9.0 × (1.5 - 0.115) = 12.5 kN, so τ_v = 12.5 × 10³/(1000 × 115) = 0.11 N/mm². For p_t = 0.34% and M20, τ_c = 0.40 N/mm² (IS 456 Table 19), and for a slab of depth 150 mm the factor k = 1.30, so kτ_c = 0.52 N/mm² > τ_v, and no shear reinforcement is needed.

Other checks. Maximum spacing of main bars: 3d = 345 mm and 300 mm, so 200 mm is within the limit. Development length L_d = 0.87 f_y φ/(4τ_bd) = 0.87 × 500 × 10/(4 × 1.92) = 566 mm; the top bars run continuously over the beam from one cantilever to the other, so anchorage is assured. Distribution steel: 0.12% of bD = 180 mm² per metre, so provide 8 mm at 250 mm c/c (201 mm²).

Detail: 150 mm slab; top main bars 10 mm at 200 mm c/c continuous over the beam and extending to the free edges with 30 mm cover; 8 mm at 250 mm c/c distribution bars below and at right angles to the main bars.