written 6.7 years ago by
teamques10
★ 68k
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•
modified 6.7 years ago
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1)Maximum (+ve)S.F.:-
$1^{st}$trial:-
Max+ve S.F.at C:-
$S.F_C=8\times\frac{22}{30}+15\times\frac{20}{30}+15\times\frac{18}{30}+10\times\frac{16}{30}$
$S.F_C=30.2KN$
$II^{nd}$ trial:-
In this first trial above max ordinate $\frac{22}{30}$ was not getting multiply max load 15KN So let us have another trial by keeping 15KN load
$S.F_C=10\times\frac{18}{30}+15\times \frac{20}{30}+15\times \frac{22}{30}-8\times\frac{6}{30}$
$S._C=25.4KN$
In two trial $1^{st}$one is maximum so $S.F_C$=30.2KN
Max+ve S.F=30.2KN
2.Max(-ve)$S.F_C$
$S.F_C=-[10\times\frac{8}{30}+15\times\frac{4}{30}+15\times\frac{6}{30}+8\times\frac{2}{30}]$
$S.F_C=-[8.2KN]$
3.Max $B.M_C$:-
$\frac{5.87}{22}=\frac{y_1}{20}=5.33\\
\frac{5.87}{22}=\frac{y_2}{18}=4.80\\
\frac{5.87}{8}-\frac{y_3}{6}=4.40$
Loading Crossing |
Avg.Load |
|
Remark |
|
AC |
BC |
|
|
$\frac{w_1}{Avg}$ |
$\frac{w_2}{Avg}$ |
|
10 |
$\frac{38}{8}$ |
$\frac{10}{22}$ |
$w_1 avg \gt w_2 avg$ |
15 |
$\frac{23}{8}$ |
$\frac{25}{22}$ |
$w_1 avg \gt w_2 avg$ |
15 |
$\frac{8}{8}$ |
$\frac{40}{22}$ |
$w_2 avg \gt w_1 avg$ |
$Max BM_C=8\times4.40+15\times5.87+15\times5.33+10\times4.80$
$BM_C=251.2KNm$