Method to calculate beam deflection

[ErikaPanda]

How do you get deflections of a fully restrained beam? :)
I already solved for the propped reactions and end moments.. I'm not sure how to do that part. :/
ありがとうございます。 :)

 

[Dr. Courtney]

Usually beam constraints are considered as points where they are fixed: one end only, both ends, middle, etc.

If all the points are fixed, of course it won't deflect.

 

[ErikaPanda]

Usually beam constraints are considered as points where they are fixed: one end only, both ends, middle, etc.

If all the points are fixed, of course it won't deflect.

Thank you so much for the reply :)
We have two loads on the beam and our professor wants us to get the defections at those points.. but he hasn't taught us anything so...

 

[SteamKing]

Thank you so much for the reply :)
We have two loads on the beam and our professor wants us to get the defections at those points.. but he hasn't taught us anything so...

The beam is not statically determinate with both ends fixed. Before you can find the deflections between the fixed ends, you must solve for the reaction forces and moments which put the beam in equilibrium.

Since there are two unknown reaction forces and two unknown reaction moments total, you have only two equations of static equilibrium which you can write for this beam. You need to develop two additional equations to allow you to calculate these unknown reactions. These additional equations can be formulated knowing what the slope and deflection of the beam must be at each end.

You can use the integration method to calculate the slope and the deflection of the beam in terms of the unknown reactions, and then apply the boundary conditions for the slope and deflection at the ends to solve for these unknowns.

PS: Please don't use all caps in your thread title. That is considered shouting at PF and is against the Rules.

 

[ErikaPanda]

The beam is not statically determinate with both ends fixed. Before you can find the deflections between the fixed ends, you must solve for the reaction forces and moments which put the beam in equilibrium.

Since there are two unknown reaction forces and two unknown reaction moments total, you have only two equations of static equilibrium which you can write for this beam. You need to develop two additional equations to allow you to calculate these unknown reactions. These additional equations can be formulated knowing what the slope and deflection of the beam must be at each end.

You can use the integration method to calculate the slope and the deflection of the beam in terms of the unknown reactions, and then apply the boundary conditions for the slope and deflection at the ends to solve for these unknowns.

PS: Please don't use all caps in your thread title. That is considered shouting at PF and is against the Rules.

Thank you, Sir.
Hmm, I think I've already obtained those equations.. but if I solve for the deflections by area-moment method, would the procedure be the same as that in cantilever beams? If it isn't much of a bother, it can't be read but would you mind looking at how the problem looks like, Sir?

About the title, I'm really sorry.. I had no idea. Thank you. :)

 

[SteamKing]

Thank you, Sir.
Hmm, I think I've already obtained those equations.. but if I solve for the deflections by area-moment method, would the procedure be the same as that in cantilever beams? If it isn't much of a bother, it can't be read but would you mind looking at how the problem looks like, Sir?

About the title, I'm really sorry.. I had no idea. Thank you. :)

I'm sorry, but your attached image is too small and blurry for me to read.

After I re-read your OP, I realized that you had solved for the unknown reactions for this beam. Once you do that, it is a simple matter to construct the bending moment diagram for the beam and then find the slope and deflection by integration. I'm more familiar with the integration method than the area-moment method.

Since the slope and deflection for this beam are both zero at each end, you can assume that you are dealing with a cantilever beam initially, as long as you make sure the conditions at the opposite end of the beam match the initial conditions for the slope and deflection. In this integration method, this is usually accomplished by selecting the proper constants of integration for the slope and deflection integrals.

 

[ErikaPanda]

I'm sorry, but your attached image is too small and blurry for me to read.

After I re-read your OP, I realized that you had solved for the unknown reactions for this beam. Once you do that, it is a simple matter to construct the bending moment diagram for the beam and then find the slope and deflection by integration. I'm more familiar with the integration method than the area-moment method.

Since the slope and deflection for this beam are both zero at each end, you can assume that you are dealing with a cantilever beam initially, as long as you make sure the conditions at the opposite end of the beam match the initial conditions for the slope and deflection. In this integration method, this is usually accomplished by selecting the proper constants of integration for the slope and deflection integrals.

Well then, I guess I have to try it both ways.. lol
Thank you so much for all the help, Sir! Thank youuu.