Please see attached sheet for the help required.
I have an answer of 15KN and the length as 2.667. Please can anyone help and show workings out so I can use it as a check,
Many Thanks.
For the beam to be in equilibrium the sum of all forces acting in the vertical direction (or the y-direction in an x-y coordinate system) must equal 0. So:
$\displaystyle \sum F_y = 0 = R_A + R_B - 10KN - 15KN - 30KN$
You know that R_B = 40KN, so you can solve for R_A.
Also to be in equilibrium the sum of moments (torques) about any point must equal 0. If you take sum of moments about support A you get:
$\displaystyle \sum T_A = 0 = 10KN(0.5m) -15KN(1m) + R_B(1.5m) -30KN(d-1)$
Again, given that R_B = 40KN, you can now solve for d.