**Solving indeterminate beam by method of consistent defromation**

**Problem 7-1**

**The propped cantilever
with applied loading is shown
in figure 7-1(a). Find the support reactions and
draw Bending moment diagram. **

**Figure 7-1(a)**

**Solution:**

The free body diagram of this structure in fig
7-1(b) shows that the given propped cantilever has 3 support
reactions (A_{y}, B_{y}, M_{B}) whereas there are only 2 equations of equilibrium
(ΣF_{y} = 0 and ΣM_{z} = 0)
available for this structure. Therefore it is
statically
indeterminate of degree one.

**Figure 7-1(b)**

To solve for the unknown reaction of this
structure we need one more equation which can be obtained by
compatibility. It is evident that
deflection at A is zero.
Considering A_{y} as redundant **R **and applying
principle of superposition we get;

Total deflection at A is equal to the sum of
deflection due to applied loading and deflection due to redundant **
R **(refer to figure 7-1(c).

**Figure 7-1(c)**

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 =

The resultant bending moment diagram shown in figure 7-1(d) is drawn by superimposing the negative moment diagram over the positive moment diagram. The positive moment diagram will be parabolic (due to udl) whereas negative moment diagram will be a straight line (due to point load)

**
Fig 7-1(d|)**

The resultant BMD shows that this beam will have a point of contra-flexure (also known as point of inflection) which can be determined in a simple way by writing the equation of bending moment at a distance 'x' and equating it to zero because the bending moment is equal to zero at the point of contra-flexure.

M_{x}
= 15x -10x^{2} /2 = 0

Therefore x = 3m from the prop A,

It is clear from the resultant BMD that the maximum bending moment value is greatly reduced.

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

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