The effective width of the flange of the simply supported indepen

The effective width of the flange of the simply supported indepen
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The effective width of the flange of the simply supported independent T-beam shown below having a span of 7.5 m is

A. 616 mm

B. 1033 mm

C. 2650 mm

D. 1500 mm

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

Concept:

The width of the flange with constant flexural stress equal to the actual flexural compressive stress which leads to the same longitudinal compressive force as due to the original stress distribution is called effective width of flange.

For Independent L-beam or T-beam:-

For L-Beam:-

\({b_f} = {b_w} + \frac{{0.5\;{\ell _0}}}{{\left( {\frac{{{\ell _0}}}{b} + 4} \right)}}\; \not > b\)

For T-Beam:-

\({b_f} = {b_w} + \frac{{{\ell _0}}}{{\left( {\frac{{{\ell _0}}}{b}} \right) + 4}}\; \not > b\)

Calculation:

b = 1500 mm, l0 = 7500 mm, bw = 200 mm

\({b_f} = {b_w} + \frac{{{\ell _0}}}{{\left( {\frac{{{\ell _0}}}{b}} \right) + 4}} \not > b\)

\(\therefore {b_f} = 200 + \left( {\frac{{7.5 \times 1000}}{{\frac{{7500}}{{1500}} + 4}}} \right)\) = 1033.33 mm

Important Point:

For continuous L beam or T beam:

For L – Beam:

\({b_f} = {b_w} + 3{D_f} + \frac{{{\ell _0}}}{{12}}\)

\( \not > {b_w} + \frac{{{\ell _1}}}{2}\)

For T Beam:

\({b_f} = {b_w} + 6{D_f} + \frac{{{\ell _0}}}{6}\)

\(\not > \;{b_w} + \frac{{{l_1} + {\ell _2}}}{2}\)