Deflection at B due
to applied loading
δ_{Load } = 5 w L^{4} /384 EI
**↓**
δ_{Load } = 5 x 10 x 6^{4}
/384 EI
**↓**
Deflection at B due to Redundant **R **
**
**
δ_{R}** **
= R L^{3} /48 EI
**↑**
**
**
δ_{R}** **
= R x 6^{3} /48 EI
**↑**
Taking downward deflection as negative and upward as positive and
applying principle of superposition.
Σδ_{B} = R x 6^{3} /48 EI - 5 x 10 x
6^{4} /384 EI
But Σδ_{B} = 0 as the support A is a rigid support.
Therefore
R x 6^{3} /48 EI - 5 x 10 x 6^{4} /384 EI
=0
Hence **R = 37.5 kN**
Now we can easily determine the other reactions by applying the
equations of static equilibrium;
ΣF_{y}
= 0;
A_{y} + 37.5 + C_{y} - 10 X 6 = 0;
A_{y} + C_{y} = 22.5
ΣM_{z}
= 0 (consider z-axis passing through A)
37.5 x 3 + 6 x C_{y} - 10 x 6 x 3 = 0;
Therefore **C**_{y} = 11.25 kN
and we get A_{y} = 22.5 - 11.25
**A**_{y} = 11.25 kN
** **The
bending moment diagram is shown in figure 7-2(d)
The
resultant BMD is drawn by superposing the diagrams of UDL and point load. It also shows the maximum Bending moment is greatly reduced which is = -56.25 +45 = -11.25 kNm. The BMD also shows that this beam will have two __points of
contra-flexure__ which can be determined in a simple way by writing
the equation of bending moment at a distance x from A.
M_{x}
= 11.25 x -10x^{2} /2 = 0
(from A to B)
Therefore x = 2.25m from the support A, and
similarly another point of contra-flexure is at a distance of 2.25m
from C,
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