Solving indeterminate structure by method of consistent deformation
A continuous beam with simple supports at A, B and C is subjected to the loading as shown in figure 7-2(a). Find the support reactions and draw Bending moment diagram.
The free body diagram of this structure in figure 7-2(b) shows that the given beam has 4 support reactions (Cx ,Ay, By, Cy) whereas there are only 3 equations of equilibrium (ΣFx = 0; ΣFy = 0; and ΣMz = 0) available for this structure. Therefore it is statically indeterminate of degree one.
To solve for the unknown reactions of this structure we need one more equation which can be obtained by compatibility. It is evident that deflection at all the supports is zero. Considering By as redundant R and applying principle of superposition we get;
Total deflection at B is equal to the sum of deflection due to applied loading and deflection due to redundant R (refer to figure 7-2(c).
Deflection at B due to applied loading
δLoad = 5 w L4 /384 EI ↓
δLoad = 5 x 10 x 64 /384 EI ↓
Deflection at B due to Redundant R
δR = R L3 /48 EI ↑
δR = R x 63 /48 EI ↑
Taking downward deflection as negative and upward as positive and applying principle of superposition.
ΣδB = R x 63 /48 EI - 5 x 10 x 64 /384 EI
But ΣδB = 0 as the support A is a rigid support.
R x 63 /48 EI - 5 x 10 x 64 /384 EI =0
Hence R = 37.5 kN
Now we can easily determine the other reactions by applying the equations of static equilibrium;
ΣFy = 0;
Ay + 37.5 + Cy - 10 X 6 = 0;
Ay + Cy = 22.5
ΣMz = 0 (consider z-axis passing through A)
37.5 x 3 + 6 x Cy - 10 x 6 x 3 = 0;
Therefore Cy = 11.25 kN
and we get Ay = 22.5 - 11.25
Ay = 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.
Mx = 11.25 x -10x2 /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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