A simple supported beam shown in fig is subject to a set of 4 concentrated load which moves from left to right, determine a.Absolute max shear force (S.F) b.Absolute max bending moment (B.M)

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written 6.2 years ago by

teamques10 ★ 64k

• modified 6.0 years ago

1)Absolute Max +ve Shear

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$ SF= 80 \times 1+60 \times \frac{16}{18}+80 \times \frac{14}{18}+100 \times \frac{11}{18}$

$SF= 256.67KN$

2) Absolute Max -ve Shear

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$SF= -[80 \times \frac{11}{18}+60 \times \frac{13}{18}+80 \times \frac{15}{18}+100 \times 1$

$SF=-[258.88]KN$

3) Absolute Max B.M.

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$By \hspace{1mm} Varignon's \hspace{1mm} theorem$

$80 \times 3+60 \times 5+80 \times 7=320 \times a$

$a=3.43m$

[from O (origin)]

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$d=(a-3)=3.43-3=0.43$

$i.e \hspace{1mm} distance \hspace{1mm} between \hspace{1mm} resultant \hspace{1mm} load \hspace{1mm} and \hspace{1mm} nearest \hspace{1mm} load \hspace{1mm} 80KN=0.43$

$Place \hspace{1mm} the \hspace{1mm} vichels \hspace{1mm} in \hspace{1mm} such \hspace{1mm} a \hspace{1mm} way \hspace{1mm} that \hspace{1mm} 80KN \hspace{1mm}load \hspace{1mm} must \hspace{1mm} be \hspace{1mm} at (\frac{l}{2}+\frac{d}{2})...\hspace{1mm} from \hspace{1mm} left \hspace{1mm} end\hspace{1mm} A.$

$=\frac{18}{2}+\frac{0.43}{2}$

$=9.215 from \hspace{1mm} A$

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$\frac{0.43}{2}=0.215$ from center

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$\frac{ab}{l}=\frac{9.215*8.785}{18}=4.49m$

$\frac{4.49}{8.785}=\frac{y_1}{5.785}=2.97$

$\frac{4.49}{9.215}=\frac{y_2}{7.215}=3.51$

$\frac{4.49}{9.215}=\frac{y_3}{5.215}=2.54$

$Max \hspace{1mm} BM= 80 \times 2.54+60 \times 3.51+80 \times 4.49+100 \times 2.97$

$Max \hspace{1mm} BM= 1070KNm$

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