Answer:
[tex]R_A=402.018\ N[/tex]
[tex]R_B=832.782\ N[/tex]
Explanation:
Given:
Here we consider the mass of the beam as uniformly distributed load concentrated at mid-span.
Using the schematic diagram:
Balancing forces:
[tex]\sum F=0[/tex]
[tex]R_A+R_B=(40+86)\times 9.8[/tex] ..............................(1)
Balancing moment about point A:
[tex]\sum M_A=0[/tex]
[tex]4.5\times R_B=9.8\times (86\times 3.4+40\times \frac{4.5}{2})[/tex]
[tex]R_B=832.782\ N[/tex] ......................................(2)
Putting value from eq. (2) into eq. (1)
[tex]R_A+832.782=1234.8\ N[/tex]
[tex]R_A=402.018\ N[/tex]