Subject : Structural Analysis 1

Topic : Unsymmetrical Bending

Difficulty : High

Here, $\alpha=30^\circ$ $ \therefore \sin\alpha=0.5\ \ \ \ \ \ \ \cos\alpha=0.866$

The momentsof inertia of th section are

$I_{XX}=\frac{100\times150^3}{12}=\underline{28125000\ mm^4}\\ I_{YY}=\frac{150\times100^3}{12}=\underline{12500000\ mm^4} $

$\tan\beta=\frac{I_X}{I_Y}\tan\alpha=\frac{28125000}{1250000}\tan30^\circ=1.30\\ \therefore \underline{\beta=52.43^\circ}$

By resolving moment @ X & Y axes:

$\sigma=\frac{M_xY}{I_x}+\frac{M_yX}{I_y}$ ---- (1)

$\therefore M_x=M\cos\alpha=(15\times10^6\times0.866)=1299000\ Nmm,$

Here, x=50 mm; y=75 mm. for all parts A,B,C,D …