Question (contd.) plane of loadingis inclined at $35^\circ$ with minor axis in clockwise direction. Also locate the neutral axis position.

Subject : Structural Analysis 1

Topic : Unsymmetrical Bending

Difficulty : Medium

Maximum B.M

$= \frac{WL^2}{2}=\frac{12\times3^2}{2}=\underline{54\ kNm}$

$M=54\ kNm\\ \theta=35^\circ\\ b=300\ mm\\ d=600\ mm$

Location of N.A

$I_{XX}=I_{UU}=\frac{bd^3}{12}=\frac{300\times600^3}{12}=5.4\times10^9mm^4\\ I_{YY}=I_{VV}=\frac{db^3}{12}=\frac{600\times300^3}{12}=1.35\times10^9mm^4$

The inclination of $\beta$ of the N.A

$\tan\beta=\frac{I_{uu}}{I_{vv}}\tan\theta=\frac{5.4\times10^9}{1.35\times10^9}\times\tan35\\ \beta=\tan^{-1}(2.80)=70.35^\circ$

Location of N.A is at $70.35^\circ$

Maximum bending stress

$\sigma_B=\frac{M\cos\theta}{I_{uu}}\times v+\frac{M\sin\theta}{I_{vv}}\times u$

For point C & A

$X_c=150mm\\Y_c=300mm\\X_A=-150mm\\ V_A=-300 mm\\ (\sigma_b)_c=\frac{54\times10^6\cos35}{5.4\times10^9}\times 300+\frac{54\times10^6\sin35}{1.35\times10^9}\times150\\ (\sigma_A)_c=2.45+3.44=5.89\ MPa\ (compressive)\\ (\sigma_B)_A=-2.45-3.44=-5.89\ (Tensile)$