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**Example 2: Concentrated moment**

The concentrated moment, M_{o}, is considered to be equivalent to the action of two concentrated forces, P, equal in magnitude but opposite in direction and separated by a distance L as shown in the figure 11.5. Thus,

(11.26)

We obtain the solution to this loading case by superposing the displacement field obtained in the above example for a single point load. Thus, it follows from equations (11.17) and (11.19) that the displacement due to the downward acting force at a distance, L/2 from the origin is,

(11.27)

Similarly, the displacement due to the upward acting force at a distance, âˆ’L/2 from the origin is

(11.28)

Since, the displacement is small and the material obeys Hookeâ€™s law, we can superpose the solutions as discussed in section 7.5.2. Hence, the displacement under the action of both the forces, âˆ† = âˆ†_{L/2} + âˆ†_{âˆ’L/2} evaluates to,

(11.29)

where we have used equation (11.26). When L â†’ 0 and P L â†’ M_{o} the above equation (11.29) evaluates to,

(11.30)

Having found the displacement, (11.30), the variation of the bending moment along the axis of the beam obtained from (11.2) is:

(11.31)

**Figure 11.6: Schematic of a long beam on elastic foundation subjected to uniformly distributed load of length L on either side of the mid span **

and the shear force variation computed using (11.14) is:

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