Deriving Bending Moment Diagram from Shear Force Diagram. Interesting Observation

bugatti79

Folks,

I was checking to see was it possible to determine the bending moment diagram from the shear force diagram.
Seemingly one can do it for simply supported beams but if we consider a rigidly supported beam on either end its not possible to derive the bending moment diagram from the shear force diagram.

Can anyone confirm this is true?

See attached schematic of a simply and rigidly supported beams.

Travis_King

Well...I mean you can if you what type of beam you are studying.

The positive shear diagram part indicates increasing bending moment, not necessarily positive, just as the negative shear indicates decreasing moment, even though the moment may still be positive.

If you know it is rigidly supported, then you know what type of bending you will see. You should therefore know that you will have changes in the direction of concavity (that isn't a legitimate engineering concept, but you know what I mean). If you know that, you can derive the bending diagram pretty easily.

bugatti79

Well, if you use attachment 0087 as an example and let f=2 and l=2, then the M1=M2=1/2....
I dont see hows its possible to obtain this moment at the built in ends solely from the shear force diagram........?

Thanks

AlephZero

The BM diagram is the integral of the SF diagram.

But when you integrate a function, you have an arbitrary constant whcih is is determined by the bending moment applied at the ends of beam (including bending moments created by the restraints on the beam).

For a simply supported beam, the bending moment is zero at both ends, but in general the bending moment at a built-in end is not zero. If you ignore that fact, you get the "wrong" answer when you try to derive the BM diagram from the SF diagram.

bugatti79

Ok, good point. So you have basically confirmed my observation. Ie, one cannot use the SF diagram alone to determine the BM for situations where BM not 0 at ends. The SF force equations needs to be integrated and using BC's such as dy/dx=0 at built in ends to determine the constant of integration...

Thanks

bugatti79

I have one other query.
How does one sum the moments to be 0 for the rigidly supported case. Summing all moments should be 0. In the free body digram we have M1 and M2 and the (force*dist). Ie, M1 and M2 have to react the applied (force*dist). How do we show this balance? its easily done for the simply supported case.

Thanks

SteamKing

The summation of moments is the same procedure regardless of boundary conditions. However, the reaction moments M1 and M2 for the fixed-end beam cannot be calculated by using the equations of statics alone. This type of beam is statically indeterminate, and additional equations must be developed from strength of materials to allow for the calculation of the reaction moments (and forces as well).

bugatti79

Thank you SteamKing,

I actually know how to derive the 'fixed end beam' reaction moment and forces expressions by using the beam curvature relation and BC's etc.

Lets assume we know all the reaction moments and force applied to the fixed end beam which I currently have via a piece of software, ie it can calculate indeterminate problems. The software calculates moments about all 3 axis.

About any one axis I was expecting all moments to sum to 0 just like the forces but they dont...why? This really mystifies me and casts doubt on my understanding of mechanics....

SteamKing

I don't know why your software doesn't calculate sum of moments = 0 for all three axes.
The fact that they don't could indicate an error in either the set up of the problem or the calculation of the forces and moments.

If you would care to post the problem and your data, more helpful comments could be developed.

bugatti79

Well...just before I do that...if we could focus on the rigid beam attachment in which equestions are given for the fixing moments M1 and M2 as a function of the the applied load anywhere along the length of the beam.

Even with this simple set up...I have difficult getting summation of all moments =0. I tried it for F=2 and L=2 giving M1=M2=1/2. These 2 together should balance the applied moment which is at x=l/2=2/2 giving an applied moment of 1/2 at midspan.

Its not making sense....Im obviously misunderstanding the problem....

thanks

SteamKing

In doing your calculations for M1 and M2, you must take the sign of the moments into account. The table pictured in the second figure gives M1 = M2 = Fl/8, but it is important to realize that M1 and M2 are acting in opposite directions.

For the figure shown, summing moments about A, M1 and R2*l act CCW. F*(l/2) and M2 act CW. If the magnitudes of the moments are calculated properly, the sum of the moments should equal 0 after applying the proper sign convention. If they don't, there is a mistake in the calculation.

bugatti79

Thanks SteamKing,

I should have spotted this. I completely overlooked the reaction forces that also generate moments....tut tut! :-)

Now I will folow up on the software and keep an update on what happens.

bugatti79

I confirm that the software predicts moments correctly. Thanks