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Determine the shear centre as shown in the figure
written 6.2 years ago by | • modified 2.3 years ago |
Subject : Structural Analysis 1
Topic : Shear Centre
Difficulty : High
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written 6.2 years ago by | • modified 2.3 years ago |
Subject : Structural Analysis 1
Topic : Shear Centre
Difficulty : High
written 6.2 years ago by | • modified 6.0 years ago |
$ I=I_{XX_1}-I_{XX_2}=\left[ \frac{100\times400^3}{12}\right]-\left[ \frac{(100\times20)(400-2\times20)}{12}\right]\\ I=(533.33\times10^6)-(311.04\times10^6)\\ \underline{I=222.3\times10^6mm^4}$
Distance of shear centre (e)
$e=\frac{h^2b^2t}{4I}=\frac{400^2\times100^2\times20}{4\times22.3\times10^6}=35.98mm\\ \boxed{e=35.98mm}\underline{Ans} $