Due to symmetry, each of the two supports will carry half the load, giving,
A_{y} =
B_{y} = 3(2)/2 = 3 kN
The moment equation for the first span, AB, is found by cutting the span at distance x from the left, and summing moments. This gives,
M_{1} + 3x(x/2)  3x = 0
M_{1} = 3x  1.5x^{2} kNm
Now that the moment equation is known for the span, it can be integrated once to find the beam rotation, and a second time for beam deflection,
The deflection, v1, for the span AB is know at x = 0 and x = 2 m. Using these two boundary conditions, gives
v_{1}(x=0)
= 0.5 (0)^{3}
 0.125 (0)^{4} + C_{1} (0) + C_{2}
==>
C_{2}
= 0
v_{1}(x=2)
= 0.5 (2)^{3}  0.125 (2)^{4} + C_{1} (2) + 0 _{}
==>
C_{1} = 1
This give the beam rotation as
EIv' = (1.5x^{2}  0.5x^{3 }1) kNm^{2}
