Q1: If depth of slab is 10 cm, width of web 30 cm, depth of web 50 cm, centre to centre distance of beams 3 m, effective span of beams 6 m, the effective flange width of the beam, is

ANS:C - 150 cm

To find the effective flange width of the beam, we need to consider the dimensions of the slab and the beams. Given:

• Depth of the slab (h): 10 cm
• Width of the web (bw): 30 cm
• Depth of the web (d): 50 cm
• Centre-to-centre distance of beams (L): 3 m
• Effective span of beams (L_eff): 6 m

We'll calculate the effective flange width of the beam using the given dimensions. The total width of the beam can be calculated as the sum of the width of the web and the width of the flange. Total width of the beam (B) = Width of the web (bw) + 2 * Effective flange width (bf_eff) To calculate the effective flange width, we need to consider the distribution of load from the slab to the beam. Since the beam spans 6 m and is supported by columns at 3 m intervals, each beam will carry half of the slab load over its effective span. Let's calculate the effective flange width (bf_eff): Total load on the beam=Load per unit area of the slab×Effective span of the beamTotal load on the beam=Load per unit area of the slab×Effective span of the beam Total load on the beam=Weight of slab per unit area×Width of the slab×Effective span of the beamTotal load on the beam=Weight of slab per unit area×Width of the slab×Effective span of the beam Total load on the beam=10/100×6×3=1.8 Since this load is distributed over two flanges, the load per flange is: Load per flange=1.8/2=0.9 m Now, we equate the bending moment due to the load on the flange to the bending moment due to the load on the web. The effective flange width (bf_eff) can be determined from this equation. Bending moment due to load on the flange=Bending moment due to load on the webBending moment due to load on the flange=Bending moment due to load on the web 1/2×Load per flange×Effective flange width=1/3×Load per unit length of web×Depth of web^2 1/2×0.9×Effective flange width=1/3×3×50^2 Effective flange width=0.5×2500 Effective flange width=1250/0.45​ Effective flange width≈2777.78 cm Effective flange width≈278 cm So, the effective flange width of the beam is approximately 278 cm. Therefore, among the given options, the closest one is 300 cm.