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 L⁴ /384 EI   ↓

        δ_Load = 5 x 10 x 6⁴ /384 EI   ↓

Deflection at B due to Redundant R

       δ_R  = R L³ /48 EI   ↑

      δ_R  = R x 6³ /48 EI   ↑

Taking downward deflection as negative and upward as positive and applying principle of superposition.

       Σδ_B = R x 6³ /48 EI  - 5 x 10 x 6⁴ /384 EI

      But Σδ_B = 0 as the support A is a rigid support.

Therefore

      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;

  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 -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,

Fig 7-2(d)

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