• For the given beam, the effective force of the distributed load$AC = 50 \times 5= 250\ kN.$
• $It\ acts\ at\ a\ distance\ of \dfrac {5}2 = 2.5\ m from\ A.$
• effective force of the distributed load,$FE = \dfrac {1} 2 \times 120 \times4 = 240kN$
• $It\ acts\ at\ a\ distance\ of \dfrac {2 \times 4}3 = \dfrac {8} 3\ m\ from\ F.$
• $As\ the\ beam\ is\ in\ equilibrium,\ \sum M_B=0 $
• $Hence, 250 \times 1.5-150sin30 \times (1+1)-150-240\times (3+1.33)+R_F\cos30 \times (1+1+1+4)=0$

$7 R_F\cos 30=-375 + 300\sin30 + 150 + 240 \times 4.33$

• $Hence, R_F= 159.1837\ kN$
• $As\ the\ beam\ is\ in\ equilibrium,\ \sum F_Y=0 $
• $Hence, -250 +R_B\sin \alpha-150sin30 -240+R_F\cos30 =0$
• $Hence, R_B\sin \alpha= 250 +150sin30 +240-159.1837cos30 $
• $R_B\sin \alpha= 427.1429 \cdots(1)$
• $As\ the\ beam\ is\ in\ equilibrium,\ \sum F_X=0 $
• $Hence, R_B\cos \alpha-150cos30 -R_F\sin30 =0$
• $Hence, R_B\cos \alpha=150cos30 +159.1837\sin30$

$R_B\cos \alpha= 209.4957 \cdots (2)$

• Now square and add (1) and (2)...

${R_B}^2.\cos^2\alpha + {R_B}^2.\sin^2\alpha = (209.4957)^2 + (427.1429)^2$

${R_B}^2(\cos^2\alpha + \sin^2\alpha)= 226339.4558$

$As,\ (\cos^2\alpha + \sin^2\alpha = 1);$

$Hence, R_B=475.7515\ kN$

• $To\ find\ the\ angle,\ Divide (1)\ by\ (2); \dfrac {R_B\sin \alpha} {R_B\ cos\alpha} = \dfrac {427.1429} {209.4957}$
• $Hence, \alpha=tan ^{-1}(2.0389) = 63.8740\ ^\circ$
• $Hence, Support\ Reaction\ at\ B\ = 475.7515\ kN\ and\ \alpha= 63.8740\ ^\circ\ ; Support\ Reaction\ at\ F\ = 159.1837\ kN\ and\ \angle= 30\ ^\circ$