**Solving indeterminate structure by method of consistent deformation**

**Problem 7-2**

**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. **

**Figure 7-2(a)**

**Solution:**

The free body diagram of this structure in figure
7-2(b) shows that the given beam has 4 support
reactions (C_{x },A_{y}, B_{y}, C_{y}) whereas there are only
3 equations of equilibrium
(ΣF_{x} = 0; ΣF_{y} = 0; and ΣM_{z} = 0)
available for this structure. Therefore it is
statically
indeterminate of degree one.

**Figure 7-2(b)**

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 B_{y} 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).

**Figure 7-2(c)**

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,

**Fig 7-2(d)**

You can also use our deflection calculator for finding the deflection to solve this problem

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