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Lecture notes on Beams, Shear force and bending moment diagram|
By Prof S S Chauhan, Mechanical Engineering, IEC College, Greater Noida, Uttar Pradesh
Bending Moment: |
Let us again consider the beam which is simply supported at the two prints, carrying loads P1, P2 and P3 and having the reactions
R1 and R2 at the supports Fig 4. Now, let us imagine that the beam is cut into two potions at the x-section AA. In a similar
manner, as done for the case of shear force, if we say that the resultant moment about the section AA of all the loads and
reactions to the left of the x-section at AA is M in C.W direction, then moment of forces to the right of x-section AA must be
?M' in C.C.W. Then ?M' is called as the Bending moment and is abbreviated as B.M. Now one can define the bending moment to be
simply as the algebraic sum of the moments about an x-section of all the forces acting on either side of the section
Sign Conventions for the Bending Moment:
For the bending moment, following sign conventions may be adopted as indicated in Fig 5 and Fig 6.
a- Uniformly varying load.
Some times, the terms ?Sagging' and Hogging are generally used for the positive and negative bending moments respectively.
Bending Moment and Shear Force Diagrams:
The diagrams which illustrate the variations in B.M and S.F values along the length of the beam for any fixed loading conditions
would be helpful to analyze the beam further.
Thus, a shear force diagram is a graphical plot, which depicts how the internal shear force? F' varies along the length of beam.
If x denotes the length of the beam, then F is function x i.e. F(x).
Similarly a bending moment diagram is a graphical plot which depicts how the internal bending moment? M' varies along the length
of the beam. Again M is a function x i.e. M(x).
Basic Relationship between the Rate of Loading, Shear Force and Bending Moment:
The construction of the shear force diagram and bending moment diagrams is greatly simplified if the relationship among load,
shear force and bending moment is established.
Let us consider a simply supported beam AB carrying a uniformly distributed load w/length. Let us imagine to cut a short slice
of length dx cut out from this loaded beam at distance? x' from the origin ?0'.
The forces acting on the free body diagram of the detached portion of this loaded beam are the following
o The shearing force F and F+ dF at the section x and x + dx respectively.
o The bending moment at the sections x and x + dx be M and M + dM respectively.
o Force due to external loading, if ?w' is the mean rate of loading per unit length then the total loading on this slice of
length dx is w. dx, which is approximately acting through the centre ?c'. If the loading is assumed to be uniformly distributed
then it would pass exactly through the centre ?c'.
This small element must be in equilibrium under the action of these forces and couples.
Now let us take the moments at the point ?c'. Such that
Next Page - Conclusion