Deflection at A due
to applied loading = wL^{4} /8EI = (10 x 4^{4})/8EI
↓
Deflection at A due to redundant R = (R x 4^{3})/3EI
↑
Taking downward deflection as negative and upward as positive and
applying principle of superposition.
Σδ_{A} = (R x 4^{3})/3EI
 (10 x 4^{4})/8EI
But Σδ_{A} = 0 as the support A is a rigid support.
Therefore (R x 4^{3})/3EI 
(10 x 4^{4})/8EI = 0
R
= {(10 x 4^{4})/8EI} / { (
4^{3})/3EI}
Hence R = 15 kN
Now we can easily determine the other reactions
B_{y
} = 10 x 4  15 = 25 kN
M_{B}=
15x4  10x4x2 =  20 kNm
(clockwise)
The bending moment diagram is shown in figure 71(d)
Fig 71(d)
The
resultant BMD shows that this beam will have a point of
contraflexure which can be determined in a simple way by writing
the equation of bending moment at a distance 'x';
M_{x}
= 15x 10x^{2} /2 = 0
Therefore x = 3m from the prop A,
You can also use our
Slope
deflection calculator for different combinations of load on
cantilever.
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