How Do You Calculate Reaction Forces in Multi-Support Structures?

[Bill92]

structural question please HELP!

Hi all,
new to this physics forum business.
I am wondering if anyone can tell me how to find reaction forces in a structure where there is more than two supports. I am familiar with the conventional method of taking the moment about a point to find one reaction, and then summing forces in the y direction to find the other. But I am unsure where to start if there is more than two unknowns.

Please see the example I have attached.

 

[Simon Bridge]

You can extend the same method over each span.
http://www.worldscibooks.com/etextbook/p187/p187_chap06.pdf
... p257. :)

 

[Bill92]

Thanks simon,

I checked out the link and read what it had to say, something along the lines of three moment theorem, beam deflection and mohr's circe, heaps of equations, it looks like a nightmare though!

was this the method you were referring to, or is there an easier way?

 

[Simon Bridge]

That was the one - the book makes a meal of it.
Look at how it handles 2-support beams and compare with how you are used to doing it. It'll be basically the same method handled more... well... it makes it look harder than it is. But hey, it looks great in papers!

 

[Studiot]

I checked out the link and read what it had to say, something along the lines of three moment theorem, beam deflection and mohr's circe, heaps of equations, it looks like a nightmare though!

was this the method you were referring to, or is there an easier way?

What is your interest in this question, which is clearly coursework ?

Simon has offered a link to one of the best texts around on this subject (Rees) and you should know there is no easy way.

Further for your information, Simon's book refers to Mohr's theorems, not Mohr's circle which is a totally different thing. These theorems are more commonly known as the Area Moment method.

Do you understand the terms statically determinate or statically indeterminate?

 

[Bill92]

Ah okay, thanks Simon.

Studiot, we were given the question to solve on a computer program, and we have to check the reactions/bending moments using hand calcs. Thanks for your clarification between mohrs circle and his theorem's, damn geniuses coming up with 30 theorems each haha. I now understand that the problem is statically indeterminate but still can't nut out how to solve the question, I'll have to thoroughly read through the link and see how I get on.

Thanks for your help.

 

[Studiot]

Calculating the unknown reactions analytically is statically indeterminate yes.

But checking stated values, what's the problem?

This can be handled by simple rigid body mechanics.

The vertical force equilibrium condition

Ʃsupport reactions = Ʃloads.

And moment equilibrium about each support in turn.

Ʃsupport reaction moments = Ʃload moments

Here is a simpler example than yours but it shows the principle.

You do not need to learn the three moment equation or moment distribution to check something - just use your existing knowledge.
I'm sure if you needed more theory your tutors would have given it to you - you should trust them.

Final comment there are further results attributed to Mohr in structures viz the Williot - Mohr diagram.

go well

 

[afreiden]

OP, whenever you see textbook problem with a bunch of internal pins, it usually means that your beam is statically determinate, NOT statically indeterminate.

Split the 14' beam up at each pin - such that you have three segments - a 6' segment, another 6' segment, and a 2' segment.

How many total unknowns? 4 support reactions + 2 shears at the locations of the pins (these shears act equal an opposite on the segment to the left and to the right of each pin) = 6 unknowns.

How many relevant equations? ƩM,ƩFy for each segment. That totals to 6 equations.

You can easily solve for everything.
Hint: You should find that the support reaction to the far right must be zero.


I'm sure the previous posters understand all of this, but they mis-read your original post.

 

[Studiot]

I'm sure the previous posters understand all of this, but they mis-read your original post.

Yes I did see the 'pins' but decided that the object was to reinforce the need for checking if a computer's output is believable.

You might be right, the circles may represent intentional pins in which case good catch.

 

[Bill92]

Thanks guys,

studiot, as much as I would like to use that method to "check" the computer generated reactions, unfortunately I have to obtain the same reactions using hand calculations, sorry I should have further clarified this.

Afreiden, I tried cutting the beam into the sections as you suggested, when considering the far left 6' beam, I made the four meters of distributed load into a point load of 40kN at 2m from point B. I summed the moments about point B and found that Ay = -36kN (pulling down). This answer is different to the one generated on the program which states that Ay is -66kN.
*please see my paint drawing attached

Also, I am confused with what to do with the load at the second hinge where I am cutting (this load is 10kN).

Thanks for all your help!

 

[afreiden]

There's no moment at the cut, but there is a shear - that you've neglected.
I recommend you draw all three segments with all forces (known and unknown) before proceeding.

 

[Bill92]

Okay, I see what you mean about the advantage of drawing all three sections.

After doing this, I considered the 2' span at the end of the beam (including shear force) and as you predicted, found Dy to be zero, taking my obtained shear force as 10kN up, and using it as shear for the next section, equal and opposite (shear at hinge = 10kN down), I again summed the moments and the forces in y direction but could not obtain the correct reactions.

Hear are the correct reactions from left to right: Ay = -66kN, By = 129kN, Cy = 35kN, Dy= 0. (obtained from computer program)

 

[AlephZero]

After doing this, I considered the 2' span at the end of the beam (including shear force) and as you predicted, found Dy to be zero...

That's a good way to start. (You can forget about the right hand section of the beam, because it isn't carrying any loads at all.)

Then you can take the center section (between the two pins) and take moments about the left hand end to find the reaction at the support. Then balancing the vertical forces will give you the shear at the left hand pin.

Now you know all the forces on the left hand section so you can take moments about each support to find the reactions at the other support.

 

[Bill92]

Thanks AlephZero,

I did as you said, taking the center section between the two pins, and summing the momments abouty the left end and found Cy to equal 40kN (not 35 which is the correct answer, I've attached a paint drawing of the section, do I need to include a shear force at the right hand end, if so how do I obtain this force?

 

[Studiot]

In the absence of further help from the Science Advisor, I can confirm that your set of reactions is correct.

I did as you said, taking the center section between the two pins, and summing the momments abouty the left end and found Cy to equal 40kN (not 35 which is the correct answer, I've attached a paint drawing of the section, do I need to include a shear force at the right hand end, if so how do I obtain this force?

However your paint sketch is incorrect. You have not followed the advice from afreiden or AZ.

How did you calculate the end reaction R_D as zero?

Draw the beam and since you have labelled the supports A, B, C, D, label the left hinge E and the right hinge F.

Call the reactions R_A, R_B, R_C, R_D

and the hinge shears V_E, and V_F

That makes 6 unknowns as afreiden says. However you can easily establish that R_D =0 How?

So that leaves 5.

By careful choice of equations on the three free bodies suggested by afreiden you do not have to calculate the hinge shears.

Now what is the standard method for excluding a force from an equation?

Hint take a moment about its point or line of action.

So what equations can you come up with?

Further hint, if you have a copy of Marshal & Nelson have a look at pages 41 to 43.