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 6.8 years ago by

teamques10 ★ 68k

• modified 6.6 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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$\frac{y_2}{1}=\frac{0.4}{6}$

$y_2=0.067$

$Max \hspace{1mm} -ve S.F_c= 40 \times [\frac{5}{2}(0.4+0.067)]$

$\hspace{10mm}= 46.7KN$

$2. Max \hspace{1mm} BM \hspace{1mm} at \hspace{1mm} C:$

$ I.L.D for BM_c$

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$\frac{3.6}{9}=\frac{y_1}{6}=2.4$

$\frac{3.6}{6}=\frac{y_1}{4}=2.4$

$To \hspace{1mm} find \hspace{1mm} x \hspace{1mm} by \hspace{1mm} similar \hspace{1mm} triangle$

$\frac{x}{5-x}=\frac{6}{9}$

x=2m

$Max BM_c=40 \times [\frac{2}{2}(2.4+3.6)]+40 \times [\frac{3}{2}(3.6+2.4)]$

$Max BM_c= 600KN.m$

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