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

$\therefore M_x=M\cos\alpha=(15\times10^6\times0.866)=12990000\ Nmm$,

including compression at C & D & tension at A & B.

$M_y=M\sin\alpha=15\times10^6\times0.5=7500000$ Nmm,

including tension at C,B & compression at A,D

$\sigma_x=\frac{M_xy}{I_x}=\frac{12990000\times75}{28125000}=34.64\ N/mm^2\\ \sigma_y=\frac{M_yx}{I_y}=\frac{7500000\times50}{12500000}=30\ N/mm^2 $