(a) A cantilever beam ABCD, as shown in the above figure, is carrying a uniformly distributed load of 10 kN/m between B & C and a…

(a) The beam is fixed at A with AB = BC = CD = 2 m; AC has 2I and CD has I. The stiffness affects only deflection, not the statics.

Support reactions (whole beam as the free body, anticlockwise moments positive):

• ΣFy = 0: R_A − 10 × 2 = 0, so R_A = 20 kN upward.
• ΣM_A = 0: M_A − 20 × 3 − 50 = 0, so M_A = 110 kN·m anticlockwise (hogging at the wall).

FBD of support A: the wall pushes the beam up with 20 kN and applies a 110 kN·m anticlockwise couple to it; equal and opposite actions (20 kN down, 110 kN·m clockwise) act on the wall.

FBD of end D: the free end carries only the applied 50 kN·m clockwise couple. Taking segment CD as the free body, the action at C is a 50 kN·m anticlockwise couple with zero shear, so CD is in pure bending.

FBD of member BC (cut just right of B and just left of C):

• At B (action of AB on BC): shear 20 kN upward and moment 70 kN·m anticlockwise (hogging), since the loads to the right give M_B = 50 + 20 × 1 = 70 kN·m.
• On BC: UDL 10 kN/m, resultant 20 kN downward at 1 m from B.
• At C (action of CD on BC): shear 0 and moment 50 kN·m clockwise.
• Checks: ΣFy = 20 − 20 + 0 = 0; ΣM_B = 70 − 20 × 1 − 50 = 0.

Bending moment (hogging throughout): 110 kN·m at A, falling linearly to 70 kN·m at B, parabolically to 50 kN·m at C, and constant at 50 kN·m from C to D.

(b) The section is a 600 × 600 mm square with corners A (top-left), B (top-right), C (bottom-right) and D (bottom-left). The 500 kN load acts at P, 100 mm right of and 100 mm above the centroid O, i.e. towards B.

• Area = 600 × 600 = 3.6 × 10⁵ mm²; Z_x = Z_y = 600 × 600²/6 = 3.6 × 10⁷ mm³.
• Direct stress: P/A = 500 × 10³ / 3.6 × 10⁵ = 1.389 N/mm² (compression).
• Moments: M_x = M_y = 500 × 0.1 = 50 kN·m, so the extreme-fibre bending stress about each axis is M/Z = 50 × 10⁶ / 3.6 × 10⁷ = 1.389 N/mm².
• σ = P/A ± M_x/Z_x ± M_y/Z_y (compression positive; fibres on the side of P are compressed more):
• B (towards P in both directions): 1.389 + 1.389 + 1.389 = 4.167 N/mm² compression.
• A (towards P vertically, away horizontally): 1.389 + 1.389 − 1.389 = 1.389 N/mm² compression.
• C (towards P horizontally, away vertically): 1.389 − 1.389 + 1.389 = 1.389 N/mm² compression.
• D (away from P in both directions): 1.389 − 1.389 − 1.389 = −1.389 N/mm², i.e. 1.389 N/mm² tension.

Why D goes into tension: for no tension the load must lie within the kern, which for a rectangle requires 6e_x/b + 6e_y/d ≤ 1 (a rhombus with vertices at b/6 = 100 mm on the axes). Here 6(100)/600 + 6(100)/600 = 2 > 1, so P lies outside the kern. The neutral axis (σ = 0) is the line x + y = −300 mm, joining the mid-points of sides AD and DC; the triangle beyond it, at D, is in tension. Loaded at O as intended, the whole section would carry a uniform 1.389 N/mm².

(c) The influence line for bending moment at C (a = 6 m, b = 24 m, L = 30 m) is a triangle with peak ordinate ab/L = 6 × 24/30 = 4.8 m at C; y = 0.8x left of C and y = 0.2(30 − x) right of C.

For a UDL shorter than the span, M_C is greatest when the section divides the load in the same ratio as it divides the span: 1.2 m of load left of C and 4.8 m right (1.2 : 4.8 = 6 : 24). The load then occupies 4.8 m to 10.8 m from A, and the ordinates under its two ends are equal: 0.8 × 4.8 = 3.84 and 0.2 × (30 − 10.8) = 3.84. Moving the load either way would drop a higher ordinate than it picks up, so this position gives the maximum.

Area of ILD under the load = ½(3.84 + 4.8)(1.2) + ½(4.8 + 3.84)(4.8) = 5.184 + 20.736 = 25.92 m².

M_max at C = w × area = 20 × 25.92 = 518.4 kN·m.

(d) The bracket is bolted by four bolts in a 2 × 2 group: gauge 120 mm (x = ±60 mm about the column centre line) and pitch 160 mm (y = ±80 mm). P acts 300 mm from the centroid of the group, and A is the upper bolt on the load side. The plate is treated as rigid (elastic method).

• r = √(60² + 80²) = 100 mm for every bolt; Σr² = 4 × 100² = 40 000 mm².
• Direct shear: P/4 = 0.25P on each bolt, downward.
• Twisting moment: M = 300P N·mm (clockwise).
• Torsional shear on A acts perpendicular to its radius: F_t = M·r/Σr² = 300P × 100/40 000 = 0.75P. Its horizontal component is M·y/Σr² = 300P × 80/40 000 = 0.60P, and its vertical component is M·x/Σr² = 300P × 60/40 000 = 0.45P, downward (the same sense as the direct shear).
• Resultant on A: R = √[(0.60P)² + (0.25P + 0.45P)²] = √(0.36 + 0.49) P = 0.922P.
• Set R equal to the bolt value: 0.922P = 45.3 kN, so P = 49.1 kN.

The two bolts on the load side (A and the one below it) are equally critical; each bolt on the far side carries only √(0.60² + 0.20²) P = 0.63P.

(e) Transverse reinforcement (lateral ties or helical spirals) in an RC column does the following:

• Prevents buckling of the longitudinal bars. Ties shorten the unsupported length of each compression bar, so the bars reach their design stress without bowing outward and spalling the cover.
• Confines the concrete core. Closely spaced ties, and especially helical spirals, restrain lateral expansion under axial compression, which raises the strength and the ultimate strain of the core. IS 456 (cl. 39.4) allows a helically reinforced column 1.05 times the strength of a similar tied column when the helix meets the volume requirement.
• Gives ductility. Confinement lets the column undergo large deformations without sudden failure; IS 13920 therefore requires special confining links near beam–column joints in seismic zones.
• Resists shear. Under wind or earthquake loads the ties act as shear reinforcement, like stirrups in a beam.
• Holds the cage together. Ties keep the longitudinal bars in position during concreting and restrain lap splices against splitting.

Detailing rules (IS 456, cl. 26.5.3.2):

• Tie diameter: not less than one-quarter of the largest longitudinal bar diameter, and not less than 6 mm.
• Pitch: not more than the least lateral dimension of the column, 16 times the smallest longitudinal bar diameter, or 300 mm.
• Restraint: every corner bar and alternate bars must be held by a tie corner; bars within 75 mm of a restrained bar need not be tied separately.