Concentrated Moment Civil Engineering (CE) Notes | EduRev

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Example 2: Concentrated moment

 The concentrated moment, Mo, 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,

Concentrated Moment Civil Engineering (CE) Notes | EduRev                 (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,

Concentrated Moment Civil Engineering (CE) Notes | EduRev          (11.27)

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

Concentrated Moment Civil Engineering (CE) Notes | EduRev                 (11.28)

Since, the displacement is small and the material obeys Hooke’s law, we can superpose the solutions as discussed in section 7.5.2. Hence, the displacement under the action of both the forces, ∆ = ∆L/2 + ∆−L/2 evaluates to,

 

Concentrated Moment Civil Engineering (CE) Notes | EduRev                  (11.29)

 where we have used equation (11.26). When L → 0 and P L → Mo the above equation (11.29) evaluates to,

 Concentrated Moment Civil Engineering (CE) Notes | EduRev             (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:
 

Concentrated Moment Civil Engineering (CE) Notes | EduRev               (11.31)

 

Concentrated Moment Civil Engineering (CE) Notes | EduRev

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:

Concentrated Moment Civil Engineering (CE) Notes | EduRev

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