Question

The box frame is subjected to a uniform distributed loading w along each of its sides. Determine the moment developed in each corner. Neglect the deflection due to axial load. EI is constant. 

The image is in this link https://ibb.co/74pwHpL

Answer 1

The fixed end moment

$FEM_{AB}= -FEM_{BA}= \frac{l^2}{12}; FEM_{AD}= -FEM_{DA}= \frac{l^2}{12}$

For symmetric loading about x-axis and y- axis

$\theta_A=-\theta_B=-\theta_D$

Using the slope deflcetion method at AB

$M_{AB}=\frac{2EI}{l}(2 \theta_A+\theta_B)+FEM_{AB}$

$M_{AB}=\frac{2EI}{l}(2 \theta_A-\theta_A)+\frac{wl^2}{12}=\frac{2EI}{l}(2 \theta_A)+\frac{wl^2}{12}$

Using the slope deflcetion method at AD

$M_{AD}=\frac{2EI}{l}(2 \theta_A+\theta_D)+FEM_{AD}$

$M_{AD}=\frac{2EI}{l}(2 \theta_A-\theta_A)-\frac{wl^2}{12}=\frac{2EI}{l}(2 \theta_A)-\frac{wl^2}{12}$

$\sum M_A=0 \implies M_{AB}+M_{AD}=0$

$\frac{2EI}{l}(2 \theta_A)+\frac{wl^2}{12}+\frac{2EI}{l}(2 \theta_A)-\frac{wl^2}{12}=0$

$\theta_A=0$

$M_{AB}=\frac{wl^2}{12}$

$M_{AD}=-\frac{wl^2}{12}$

Moment developed at each corner is $\frac{wl^2}{12}$