Calculating the support reactions of a 2 span continuous pipe.

[elbulldogio]

Dear all,

I have been given a problem that involves a 3m length of pipe filled with water, that has a single support(support B on attached diagram) 2.5 m from the left leaving a span of 0.5m from support B to the right side of the pipe. The pipe is to be conneted to the other pipes in the line using slip on sleeves with integral gaskets.

I need to calculate how much shear occurs at the pipe ends that the slip on sleeves need to resist. So I have assumed the following:-
1. The sleeves act as simple supports to each end of the pipe.
2. Pipe stresses have been assumed as acceptable
3. Environmental and seismic effects are neglected.

Thus the problem now becomes a continuous beam with 2 unequal spans that is subjected to a udl of 397.65N/m (weight of pipe + weight of water).

The theory I have used is on the attachment, but the figures I achieve seem incorrect. I cannot see how the reaction at support B is larger than the total weight of the pipe.

I have been looking at this so long now that no matter how hard I try I cannot see the problem.

Could somebody please look at my theory and help me locate the error. Also please inform me if I have used the right theory in the firstplace.

Sorry sounds stupid, but I am not a piping engineer!!!

Many thanks in advance.

 

[Frank1344]

You need to have at least one fixed point in your system.
The system you are working on is an indeterminate system. We need to have more assumption to make it determinate.

 

[msquared48]

1. Free up the support at "C" allowing it to deflect up, then calculate the reations at "A" and "B".

2. Knowing that deflection at "C", Put a downward force "P" at "C" and calculate how much force "P" will need to be to deflect the pipe the same amount as in "1" above.

3. Knowing "P", caculate the associated reactions at "A" and "B".

4. Sum the "A" and "B" reactions from "1" and "2".

5. Sum the composite "A" and "B" values and compare this to the total weight of the pipe. The difference will be the reaction at "C".



Mike McCann
MMC Engineering

 

[BigInch]

As long as the sum of vertical forces = 0, reactions can be whatever they need to be to balance the forces... ALL the forces.

The load at B is not only the result of the uniform distributed load. You have a long span from A to B that is introducing moments into the pipe. The internal moment at B from span A-B puts a prying action at B, if you try to hold the pipe down at C. In effect, the reaction at B becomes the sum of the distributed load forces going there (approximately 1/2 of the load from A to B, plus all of load on B to C), PLUS the prying force at C. Is it not so?

We are more connected to everyone in the world than we've ever been before, except to the person sitting next to us. Lisa Gansky

 

[prex]

Can't follow you fully, as I don't understand all of your symbols.
However, as you could check at Xcalcs: Continuous beam: Unif.load:
-M_B is correct
-Bending Moment at mid point of AB is OK
-Bending Moment at mid point of BC is wrong
-the formulae for M_B after 'Then to find support reactions' are wrong (you need to include the shear for them to be correct)
-all of the support reactions are wrong
-there's an obvious mistake in the last formula

prex

http://www.xcalcs.com

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[desertfox]

Hi elbulldogio

I've calculated using the same theory that you've used, ie the "Three Moment Equation" and here I my results below.
The value of Mb in your calculations is correct and for Ra and Rc my value's are 392.6785N and 422.5075N respectively and these agree with the link that prex posted however the reaction at Rb I calculate to be 377.764N and the calculator from prex gives a value of -0.601kN or -601N, that said I can't get the vertical forces to sum to zero on prex's link, maybe I've made an error which someone will be kind enough to point out.

desertfox

 

[desertfox]

Hi again

I think I found my error, on page 2 of my calcs I wrote:-

-Rc*l2+ wl2^2/2 = Mb It should have been: Rc*l2- wl2^2/2 = Mb so the plus and minus were in the wrong place, however I actually worked the maths correctly just wrote the equation down wrong.
So Ra= 392.678N, Rb= 377.764N and Rc= 422.5075N.

desertfox

 

[prex]

desertfox, you have a mistake also for R_C, that equals -422.5 , so that R_B=1222 N.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[desertfox]

hi prex

Good spot, I see my error now thanks!

Regards

desertfox

 

[elbulldogio]

Frank 1344; msquared48; BigInch; prex and desertfox, I thank you all for your input to my query. Sorry I did not thank you all much sooner than this but have been away for a while.

Many thanks to you all.