A simply supported beam has a span of 15m UDL of 40KN/m, 5m long crosses the girder from left to right. Draw ILD for S.F and B.M at a section 6m from left hand.

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written 7.8 years ago by

teamques10 ★ 70k

modified 3.8 years ago by

sagarkolekar ★ 11k

Use this diagram to calculate to maximum S.F and B.M at this section.

structural analysis

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written 7.8 years ago by

teamques10 ★ 70k

• modified 7.6 years ago

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$ILD \hspace{1mm} for \hspace{1mm} S.F_c$

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$1. Max \hspace{1mm} S.F_c:$

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$(i) Maximum \hspace{1mm} +ve \hspace{1mm} SF \hspace{1mm} at \hspace{1mm} C:$

$\frac{0.6}{9}=\frac{y_1}{4}$

$y_1=0.267$

$Max \hspace{1mm} +ve S.F_c= 40 \times (\frac{5}{2}(0.6+0.267))$

$Max \hspace{1mm}+ve S.F_c= 86.7KN$

$(ii) Maximum \hspace{1mm} -ve \hspace{1mm} SF \hspace{1mm} at \hspace{1mm} C:$

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$\frac{y_2}{1}=\frac{0.4}{6}$

$y_2=0.067$

$Max \hspace{1mm} -ve S.F_c= 40 \times …

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