Just wondered if someone was able to give the homework the once over as im not feeling confident i have this right. 1. I have a simply supported beam 10metres in length, supported at 2 metres in from both ends. there is a 5Kn U.D.L along the entire length of the beam and downward forces of 10Kn 3 metres in from the LHS and 20Kn at the far RHS of the beam Determine the vertical reactions at the supports. Ok, first of all i worked out the U.D.L and its position. Total U.D.L= 10(length of beam) x 5 (Load)= 50Kn. This force is then added to the beam as a downward force of 50Kn in the middle of the distance between the 2 supports. Moments about A: (force x distance from support Ra) = (Support Rb x distance between the 2 supports) (10x1)+(50x3)+(20x8) = (Rbx6) 10+150+160 = 6Rb 320 = 6Rb Rb = 320/6 Total upward force at support Rb = 53.3Kn Moments about B: (force x distance from support Rb) = (support Ra x distance between the 2 supports) (20x2)+(Rax6)=(10x5)+(50x3) 40+6Ra=50+150 6Ra=50+150-40 6Ra=160 Ra=160/6 Total upward force at support Ra = 26.7Kn To check the answer, upward forces = 53.3Kn +26.7Kn=80Kn downward forces = 10Kn+50Kn+20Kn=80Kn I'm pretty happy i have this right as my upward forces are the same as my downward forces putting the beam in equilibrium. I have other Q's but ill do 1 at a time.