NAME Find the Reaction forces at support A and B 300 lbsft 1

NAME: Find the Reaction forces at support A and B. -300 lbs/ft 1500 lbs 5 10\' 6\'

Solution

F_x = 0:    H_A = 0
M_A = 0:   
- P₁*5 - q₁*8*(15 + 8/2) - (U₁^left *6/2) * (29 - (2/3)*6) + R_B*29 = 0
M_B = 0:   
- R_A*29 + P₁*24 + q₁*8*(14 - 8/2) + (U₁^left *6/2) * (29 - 29 + (2/3)*6) = 0
Solve this system of equations:
H_A = 0 (lb)
Calculate reaction of roller support about point B:
R_B = ( P₁*5 + q₁*8*(15 + 8/2) + (U₁^left *6/2) * (29 - (2/3)*6)) / 29 = ( 1500*5 + 300*8*(15 + 8/2) + 900 * 25) / 29 = 2606.90 (lb)
Calculate reaction of pin support about point A:
R_A = ( P₁*24 + q₁*8*(14 - 8/2) + (U₁^left *6/2) * (29 - 29 + (2/3)*6)) / 29 = ( 1500*24 + 300*8*(14 - 8/2) + 900 * 4) / 29 = 2193.10 (lb)
The sum of the forces is zero: F_y = 0:    R_A - P₁ - q₁*8 - (U₁^left *6)/2 + R_B = 2193.10 - 1500 - 300*8 - (300*6)/2 + 2606.90 = 0