Computing Lateral Deflection of a Shaft

[Ballena Joseph]

How do I compute the horizontal and vertical lateral deflection of the shaft?

 

[Asymptotic]

Google "lateral deflection of shaft". This PDF on shaft design from the University of Northern Illinois looks like a good starting point.

 

[Dr.D]

You don't compute the deflections until you know the loading, as a general rule. What loads the shaft in your case of interest?

 

[Ballena Joseph]

You don't compute the deflections until you know the loading, as a general rule. What loads the shaft in your case of interest?

I have computed loads from shear and moment diagram.

 

[Dr.D]

If you have the moment function, then you are ready to integrate E*I*y'' =M(x) to get the deflection y(x).

If the shaft is uniform, this is quite possibly something you can do in closed form. If it is not uniform, it may be necessary to resort to numerical integration to solve the DE. C.R. Mischke developed a neat method for dealing with many point loads on a non-uniform shaft; you may want to look for it in the literature.

 

[Ballena Joseph]

If you have the moment function, then you are ready to integrate E*I*y'' =M(x) to get the deflection y(x).

If the shaft is uniform, this is quite possibly something you can do in closed form. If it is not uniform, it may be necessary to resort to numerical integration to solve the DE. C.R. Mischke developed a neat method for dealing with many point loads on a non-uniform shaft; you may want to look for it in the literature.

I can't understand. I want a sample problem in order for me to understand all that you said. And also I need a reference or textbook with proper explanation in lateral deflection. One of my friend suggests that the book of P. H. Black and O. E. Adams, Jr. will help help me regarding with my problem about lateral deflection.

 

[Dr.D]

You said that you have the moment diagram, M(x). Then, at that point you have a second order ODE that governs deflection,
E*I*y'' = M(x)
In principle, all that is required is to integrate twice. Assuming that E and I are constants, this comes down to
E*I*y' = int(M(x), dx) + c1
E*I*y = int(int(M(x), dx), dx) + c1*x + c2
where c1 and c2 are determined by the boundary conditions. That is really all there is to it.

 

[Ballena Joseph]

You said that you have the moment diagram, M(x). Then, at that point you have a second order ODE that governs deflection,
E*I*y'' = M(x)
In principle, all that is required is to integrate twice. Assuming that E and I are constants, this comes down to
E*I*y' = int(M(x), dx) + c1
E*I*y = int(int(M(x), dx), dx) + c1*x + c2
where c1 and c2 are determined by the boundary conditions. That is really all there is to it.

Do you have a sample problem, Sir? Because I really don't understand just by the given formula.

 

[Dr.D]

I've showed you how to work the problem in a fairly general way. Now, I think it is time for you to put forth some effort and work the specific case of interest on your own. I know how to work this problem, but I cannot learn it for you.

 

[Ballena Joseph]

I've showed you how to work the problem in a fairly general way. Now, I think it is time for you to put forth some effort and work the specific case of interest on your own. I know how to work this problem, but I cannot learn it for you.

Okay Sir, thanks for all of your concern and help. I will give my best just to try to understand my own problem.