If the span of beam is 6m, locate the neutral axis and hence find the stresses at each corners of the beam.

Maximum B.M = $\frac{WL^2}{8}=\frac{10\times6^2}{8}={45\ kN}$

$\underline{M=45\times10^6\ Nmm}\\ \theta=30^\circ$

Location of N.A

$I_{XX}=I_{UU}=\frac{bd^3}{12}=\frac{80\times120^3}{12}=11.52\times10^6\ mm^4\\ I_{YY}=I_{VV}=\frac{db^3}{12}=\frac{120\times80^3}{12}=5.12\times10^6\ mm^4$

The inclination of the neutral axis

$\tan\beta=\frac{I_{UU}}{I_{VV}}\tan\theta=\frac{11.52\times10^6}{5.12\times10^6}\tan30=1.30\\ \beta=\tan^{-1}1.30=52.43^\circ \\ \underline{\beta=52.43^\circ}$

Maximum bending stress

$\sigma_b=\frac{M\cos\theta}{I_{XX}}v+\frac{M\sin\theta}{I_{YY}}u\\ At point B, u_B=-40V_B=-60\\ (\sigma_b)_B=\frac{45\times10^6\cos30}{11.52\times10^6}\times-60+\frac{45\times\sin30}{5.12\times10^6}\times(-40)\\ (\sigma_b)_B=-386.73\ N/mm^2(T)$

At D, $u_B=40mm \ \ \ \ \ v_B=60\ mm$

Bending stress at point D

$(\sigma_b)_D=\frac{M\cos\theta}{I_{XX}}\times v_D+\frac{M\sin\theta}{I_{YY}}\times u_D\\ =\frac{45\times10^6\cos30}{11.52\times10^6}\times60+\frac{45\times10^6\sin30}{5.12\times10^6}\times40\\ (\sigma_b)_D=202.97+175.78\\ \underline{(\sigma_b)_D=378.75\ N/mm^2}(C)$