(a) Draw the free body diagram of link ABCD and determine all reaction forces acting at A, B, C and D of the assembly shown in…

(a) Free body diagram of link ABCD and reactions

Two members meet at pin A. The upper member AE carries 480 N at E and bears on roller C. Link ABCD is pinned to the bracket at B, pressed down by roller C and propped by roller D; rollers give vertical forces only. With the lower-bar line as datum (mm): A (0, 40), B (80, 0), C (160, 0), D (200, 0), E (240, 80).

Member AE: ΣM_A = 0: C × 160 = 480 × 40, so C = 120 N. ΣFₓ and ΣF_y then give 480 N ← and 120 N ↓ on AE at A.

Link ABCD (forces at A reversed: 480 N →, 120 N ↑; C = 120 N ↓):

• ΣM about B = 0 (anticlockwise positive): 120(−80) − 480(40) − 120(80) + D(120) = 0, which gives −28 800 − 9 600 + 120D = 0, so D = 320 N (upward).
• ΣFₓ = 0: 480 + Bₓ = 0, so Bₓ = 480 N to the left.
• ΣF_y = 0: 120 − 120 + 320 + B_y = 0, so B_y = 320 N downward. The bracket pulls the link down.

Check (whole assembly, about B): 480 × 80 = 320 × 120 = 38 400 N·mm ✓

• A: 480 N →, 120 N ↑ (resultant 494.8 N)
• B: 480 N ←, 320 N ↓ (resultant 576.9 N)
• C: 120 N ↓
• D: 320 N ↑

(b) Load taken by the tie rod

Let T be the rod tension. Since the rod was initially unstrained, the net downward deflection of B must equal the rod's elongation.

Due to w on DC (deflection wL⁴/8EI and slope wL³/6EI at C; CB stays straight): δ_w = wL⁴/(8EI) + (wL³/6EI)·L = 7wL⁴/(24EI) (downward)

Deflection at B due to T. δ_T = T(2L)³/(3EI) = 8TL³/(3EI) (upward)

Rod elongation. δ_rod = TL/(AE)

Compatibility. δ_w − δ_T = δ_rod gives 7wL⁴/(24EI) − 8TL³/(3EI) = TL/(AE)

Multiply through by 24AEI/L: 7wAL³ = T(64AL² + 24I)

T = 7wAL³ / (64AL² + 24I) = 7wAL³ / [8(8AL² + 3I)] (tension)

Check. For a rigid rod (A → ∞), T = 7wL/64, the prop reaction of the equivalent propped cantilever; as A → 0, T → 0. The rod therefore acts as an elastic prop whose share of the load rises with its axial stiffness AE/L relative to the beam's end stiffness 3EI/(2L)³, and the fixing moment at D falls accordingly.

(c) Maximum bending moments at C and D (span 28 m)

W = 5 × 160 = 800 kN over 6.08 m; load 1 is the leading (leftmost) load.

ILD for M at C (a = 7 m, b = 21 m). The peak ordinate is ab/L = 7 × 21/28 = 5.25 m. It falls with slope 0.75 towards A and 0.25 towards B.

Position: average load on AC should equal the average on the span, i.e. 800 × 7/28 = 200 kN on AC. Load 1 (160 kN) is less and loads 1 + 2 (320 kN) more, so load 2 is placed over C.

Ordinates: load 1 = 0.75 × 5.48 = 4.11; load 2 = 5.25; loads 3, 4, 5 = 5.25 − 0.25 × (1.52, 3.04, 4.56) = 4.87, 4.49, 4.11. Σy = 22.83 m M_C,max = 160 × 22.83 = 3652.8 kN·m (With load 1 over C: 160 × 22.45 = 3592 kN·m, which is smaller and confirms the choice.)

ILD for M at D (mid-span): peak 14 × 14/28 = 7.0 m, slope 0.5 each side; the criterion (400 kN each side) puts load 3 over D. Ordinates: 7.0, 6.24, 6.24, 5.48, 5.48, so Σy = 30.44 m M_D,max = 160 × 30.44 = 4870.4 kN·m

(d) Weld design for ISA 65 × 65 × 6 tie (Fe 410, shop weld)

1. Full strength of the member.

• Gross-section yielding: T_dg = A f_y/γ_m0 = 744 × 250/1.1 = 169.09 kN.
• Rupture (IS 800 cl. 6.3.3, no holes): A_nc = A_go = 372 mm², β = 1.4 − 0.076(65/6)(250/410)(65/232) = 1.259 (within the 1.443 limit), so T_dn = 0.9 × 372 × 410/1.25 + 1.259 × 372 × 250/1.1 = 216.3 kN.
• Yielding governs, so the design load is P = 169.09 kN.

2. Weld strength (shop weld γ_mw = 1.25; throat = 0.7 × 4 = 2.8 mm): f_wd = f_u/(√3 γ_mw) = 410/(√3 × 1.25) = 189.37 MPa Strength per mm = 189.37 × 2.8 = 530.24 N/mm Total effective length = 169 091/530.24 = 318.9 mm

3. Balancing about the centroid (18.1 mm from the heel, 46.9 mm from the toe):

• Weld at the heel: L₁ = 318.9 × 46.9/65 = 230.1 mm
• Weld at the toe: L₂ = 318.9 × 18.1/65 = 88.8 mm

Adding 2s = 8 mm for end craters gives 240 mm along the heel and 100 mm along the toe (4 mm welds; 3 mm ≤ s ≤ ¾ × 6 = 4.5 mm ✓).

4. Block shear check (given formulae). A_vg = A_vn = (240 + 100) × 6 = 2040 mm² and A_tg = A_tn = 65 × 6 = 390 mm². T_db1 = 2040 × 250/(√3 × 1.1) + 0.9 × 390 × 410/1.25 = 267.7 + 115.1 = 382.8 kN T_db2 = 0.9 × 2040 × 410/(√3 × 1.25) + 390 × 250/1.1 = 347.7 + 88.6 = 436.3 kN Both exceed 169.09 kN, so the joint develops the member's full strength ✓.

(e) Lap splice for column bars (M25, Fe 500)

Column bars are in compression. By IS 456 cl. 26.2.5.1, the compression lap equals the compression development length (≥ 24φ), and for bars of different diameters it is based on the smaller bar, φ = 20 mm.

Bond stress: τ_bd = 1.4 × 1.6 (deformed) × 1.25 (compression) = 2.8 MPa.

Development length. L_d = φ σ_s/(4τ_bd) = 20 × 0.87 × 500/(4 × 2.8) = 776.8 mm, which exceeds 24φ = 480 mm. Provide a lap of 800 mm (40φ) above the top of the second-floor slab.

Detailing.

• The 20 mm bars sit alongside the 30 mm bars over 800 mm, after which the 30 mm bars stop.
• Keep splices away from the joint and stagger alternate bars where practicable.
• 8 φ ties at 200 c/c continue (≤ 300 mm and 16φ = 320 mm; 8 mm ≥ 30/4 = 7.5 mm). Under IS 13920, hoops are reduced to 150 mm c/c over the lap and the lap placed in the central half of the storey.

Steel check: 6–20 = 1885 mm² = 1.05% ≥ 0.8% ✓ (6–30 below = 2.36%).

An 800 mm lap with ties continued through it transfers the full force from the 30 mm bars to the 20 mm bars.