Due to symmetry, each of the two supports will carry half the load, giving,
Ay =
By = 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,
M1 + 3x(x/2) - 3x = 0
M1 = 3x - 1.5x2 kN-m
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
v1(x=0)
= 0.5 (0)3
- 0.125 (0)4 + C1 (0) + C2
==>
C2
= 0
v1(x=2)
= 0.5 (2)3 - 0.125 (2)4 + C1 (2) + 0
==>
C1 = -1
This give the beam rotation as
EIv' = (1.5x2 - 0.5x3 -1) kN-m2
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