Finding Reactions on a Supported Beam: A Differential Equation Approach

[yecko]

Homework Statement

Homework Equations

Boundary condition
EI(d^4y/dx^4)=w

The Attempt at a Solution

Boundary condition:
At A, x=0, y=0, slope=0
At C, x=L/2, y=0, shear force = -Rb
At B, x=L, y=0, moment = 0, shear force = 0

I know there are 2 distinctive formulae for A-C and C-B.

For A-C:
Load: EI(d^4y/dx^4)=w -----(1)
Shear:EI(d^3y/dx^3)=wx+A-----(2)
Moment: EI(d^2y/dx^2)=1/2*wx^2+Ax+B-----(3)
EI(dy/dx)=1/6*wx^3+1/6*Ax^2+Bx+C-----(4)
EIy=1/24*wx^4+1/6*Ax^3+1/2*Bx^2+Cx+D-----(5)

However, with this 5 equations, with 4 unknowns, and the boundary conditions, it seems Rb cannot be solved out.
May I know how should I tackle this problem?
Thank you

 

[PhanthomJay]

Eeek! Are you familiar with cantilever deflection equations that can be obtained from beam tables? One way to solve this problem is to remove the 2 roller supports and then calculate the beam deflection at those removed supports, then put them back one at a time as a unit load and calculate the deflections at those same points. Ultimately you solve for these reactions knowing that the deflection at the support points is 0.

 

[yecko]

Are u familiar with cantilever deflection equations that can be obtained from beam tables?

Actually no! I can't find any example from our lecture notes or the two recommanded textbooks...

One way to solve this problem is to remove the 2 roller supports and then calculate the beam deflection at those removed supports, then put them back one at a time as a unit load and calculate the deflections at those same points.

I know the way to deal with deflection with one support, but not two... let me guess...
first consider part A-C, and let it be a free end at C, and calculate the deflection and slope at C. Then, by y=mx, where m= slope at C, in order to calculate the deflection at B.
After finding deflection, by what way can I find the reaction force from the deflection calculated?
Thanks.

 

[haruspex]

I don't think I understand the question. If the beam is weightless, won't the deflection be downward just to the left of C, upward to the right of C, and tending to lift at B? If B is just a roller support why will contact be retained?

 

[PhanthomJay]

That is a good start. You should be able to find cantilever deflections thru a google search. And if you are familiar with the deflection method for one unknown support reaction, say R_c, then calculate the deflections at each support points in terms of R_c, then just repeat the method for the other unknown support reaction,and add up the total deflections at each support point which each add up to the deflection at that point calculated when the supports were removed. 2 equations with 3 unknowns. A bit tedious however.

 

[PhanthomJay]

I don't think I understand the question. If the beam is weightless, won't the deflection be downward just to the left of C, upward to the right of C, and tending to lift at B? If B is just a roller support why will contact be retained?

True, but the usual assumption is that a roller support can take both upward and downward loads, as if it were on a safety rail like on roller coasters.

 

[yecko]

Should I do it this way?
But as the true deflection at C is 0, should the assumed deflection at B be the difference in deflection between B and C?
Thanks

 

[haruspex]

Ok, thanks

True, but the usual assumption is that a roller support can take both upward and downward loads, as if it were on a safety rail like on roller coasters.

 

[yecko]

should the assumed deflection at B be the difference in deflection between B and C?
Thanks

 

[PhanthomJay]

View attachment 224489
Should I do it this way?
But as the true deflection at C is 0, should the assumed deflection at B be the difference in deflection between B and C?
Thanks

The only assumption you need to make is the one noted by Haruspex, assume roller can take vertical loads in both directions. Otherwise, it does not appear that you are using the tables properly. The slope of the deflected curve at the free ends is given as an an angle theta in radians, and once you know the deflection and angle at C, then the deflection at B is that deflection plus, for small angles, (theta)(L/2). Do the same for the other reaction, calculating deflection at both joints. Then add up the deflections at each point for the 3 cases and set them equal to zero. But you must watch your plus and minus signs.

 

[yecko]

Am i correct to calculate like this?
Thanks

 

[haruspex]

I held off replying to this until I was satisfied I had the solution. (It is W/28 downwards.)
I didn't use your method. I wrote out the differential equations for the two sections and solved them.

But now I am having trouble following yours, not least for readability.
You seem to have swapped B and C wrt the original diagram (which did assign A, B, C weirdly).
If you can be bothered to repost, typing in your algebra instead of posting an image, I will try to find your error(s). To make it easier, you can substitute L=2, EI=1 and abbreviate the reactions to A, B, C.