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Question: Calculate the max B.M & S.F. at 8m from left support
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4 point loads 8,15,15,&10 KN have center to center spacing of 2m between conjugative loads and they travels a girder of 30m span from left to right with 10KN loading.


Subject : Structural Analysis 1

Topic : Influence line diagram.

Difficulty : High

sa1(74) • 281 views
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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 220 written 7 months ago by gravatar for awari.swati831 awari.swati831140
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1)Maximum (+ve)S.F.:-

$1^{st}$trial:-

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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

enter image description here

$S.F_C=10\times\frac{18}{30}+15\times \frac{20}{30}+15\times \frac{22}{30}-8\times\frac{6}{30}$

$S._C=25.4KN$

In two trial $1^{st}$one is maximum so $S.F_C$=30.2KN

Max+ve S.F=30.2KN


2.Max(-ve)$S.F_C$

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$S.F_C=-[10\times\frac{8}{30}+15\times\frac{4}{30}+15\times\frac{6}{30}+8\times\frac{2}{30}]$

$S.F_C=-[8.2KN]$


3.Max $B.M_C$:-

enter image description here

$\frac{5.87}{22}=\frac{y_1}{20}=5.33\\ \frac{5.87}{22}=\frac{y_2}{18}=4.80\\ \frac{5.87}{8}-\frac{y_3}{6}=4.40$

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 $\frac{8}{8}$ $\frac{40}{22}$ $w_2 avg \gt w_1 avg$

$Max BM_C=8\times4.40+15\times5.87+15\times5.33+10\times4.80$

$BM_C=251.2KNm$

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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 220 written 7 months ago by gravatar for awari.swati831 awari.swati831140
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