Strange Phenomenon in 3-Beam Structure Analysis

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Femme_physics
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Something strange happening here. I'm isolating the beams (which I'm allowed to do) and I get a different result for Dx with each different beam! Check out what happens when I section it at CE, and check out what happens when I section it at AD


Homework Statement



This structure is made out of 3 beams connected by joint. Tip A of the horizontal beam is supported by an immobile support, and tip E of beam CE is supported on a completely smooth horizontal surface. There are two forces acting on the structure, F and Q. The measurements are in meters.

http://img251.imageshack.us/img251/3385/sketch6.jpg

F = 2000 [N]
Q = 3000 [N]

The Attempt at a Solution



Attached.



Also, the "official answers" the book gives are:

Ax = 2000[N]
Ay = 1000 [N]
Bx = 200 [N]
By = 400 [N]
Ex = 0
Ey = 2000 [N]
Dx = 1800 [N]
Dy = 1600 [N]
Cx = 200 [N]
Cy = 400 [N]
 
Attachments
  • Handwritten work on graph paper: a truss diagram with supports at A and E, load F = 2000 [N] and load Q = 3000 [N], followed by equilibrium equations and solved reactions Ey = 2000 [N], Ay = 1000 [N], Ax = 2000 [N], Dx = 1900 [N], and By = 400 [N], Dy = 1600 [N].
    error1.jpg
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  • Handwritten free-body diagrams on grid paper with equilibrium equations: section CE with a moment sum and Dx = 2200 [N]; section BC with force and moment sums giving Cy = 400 [N], Bx = 200 [N] and Cx = 200 [N].
    error2.jpg
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Discussion



In section CE your plus and minus signs are a bit mixed up.

For starters this is the only section where you turn with the clock instead of against the clock.
I would suggest always turning against the clock (the "right hand" rule).

More importantly, in your equation your F turns the wrong way.
 


Yes, sorry, I feel stupid now-- I forgot to refresh the equation from my absent-minded errors, let me rewrite here the corret equation.

Section CE
Sum of all moments on C = 0 ; 2Ey+1.5Dy-2Dx+2000 = 0

Ey counterclockwise
Dy counterclockwise
Dx clockwise
Force counterwise

That according to all the other calculations

Dx here is even worst then I thought, it's 4200 [N]! Way off the actual result.
 


Femme_physics said:
Yes, sorry, I feel stupid now-- I forgot to refresh the equation from my absent-minded errors, let me rewrite here the corret equation.

Section CE
Sum of all moments on C = 0 ; 2Ey+1.5Dy-2Dx+2000 = 0

Ey counterclockwise
Dy counterclockwise
Dx clockwise
Force counterwise

That according to all the other calculations

Dx here is even worst then I thought, it's 4200 [N]! Way off the actual result.

Oops, my bad, Dy is the wrong way too.
According to Newton's 3rd law, action=-reaction.
You did draw Dx the other way, but not Dy.
 


I knew it would come back to this, and I'm ready with my answer. My equation at the AD section tells me that my Dy is the correct direction, pointing up. I reviewed the equations there again, I see no errors.
 


Femme_physics said:
I knew it would come back to this, and I'm ready with my answer. My equation at the AD section tells me that my Dy is the correct direction, pointing up. I reviewed the equations there again, I see no errors.

Well, Q is pulling section AD down, meaning Dy on section CE will be down as well.
As a reaction, the floor at E will be pushing section CE up, and effectively section AD as well.

So Dy will be down on section CE, and will be up on section AD.
 


Well, Q is pulling section AD down, meaning Dy on section CE will be down as well.

Wait, but the result I got at AD strictly tells me that I was right to make Dy pointing up. How can I just ignore it? Or do you mean, that on different beams that joint forces can be pointing at different directions? Doesn't seem to make intuitive sense to me...
 


Femme_physics said:
Wait, but the result I got at AD strictly tells me that I was right to make Dy pointing up. How can I just ignore it? Or do you mean, that on different beams that joint forces can be pointing at different directions? Doesn't seem to make intuitive sense to me...

The forces on a joint always point in opposite directions.
It is similar to when gravity pulls at you, the floor has to push up, otherwise you would fall through the floor :P.
 


Ooooooooooooooh. I think I get it. Sorry - my intuitive mechanical sense of normal force is escaping me from all these new calculations tactics we're learning. Lovely to see how this connects to pass material, and makes physical sense. So, just to verify, isolating beams, the forces are actually OPPOSITE at the joint when we look at each different beam individually, right?
 


Femme_physics said:
Ooooooooooooooh. I think I get it. Sorry - my intuitive mechanical sense of normal force is escaping me from all these new calculations tactics we're learning. Lovely to see how this connects to pass material, and makes physical sense. So, just to verify, isolating beams, the forces are actually OPPOSITE at the joint when we look at each different beam individually, right?

Yep :cool:!
 


Ahhh thank you very much Serena...very helpful as usual. :) Haven't gotten around to fathoming the solving tactics of isolation, joints, sections. I guess I mixed them up a bit! But I got it organized in my head now. Anyway, thanks again Serena.
 


I like Serena said:
Yep :cool:!

I have to ask, is it always the case? Are there exceptions? For instance, support reactions don't seem to change even while isolating the beam.
 
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Femme_physics said:
I have to ask, is it always the case? Are there exceptions? For instance, support reactions don't seem to change even while isolating the beam.

What do you mean with a support reaction?

Note that in the joint of 2 sections there are 2 forces at work: the first section that exerts a force on the second section, and the second section that exerts a force on the first section. If the forces do not cancel each other out, the joint will break apart.
Of course this is very interesting, because the possibility of the joint breaking apart is one of the reasons to do all these calculations: is the joint big enough and the material strong enough?
 


:) Yes indeed. I actually noticed that the supports (A and E in this sketch) don't change when you look at the entire structure as a rigid body, and when you isolate them. Their reaction is always the same, but thinking about it it makes sense actually. Thanks^^
 


Femme_physics said:
:) Yes indeed. I actually noticed that the supports (A and E in this sketch) don't change when you look at the entire structure as a rigid body, and when you isolate them. Their reaction is always the same, but thinking about it it makes sense actually. Thanks^^

Did you already make a free body diagram of the entire earth? :wink:
 


Sorry to harp on the issue, but I've noticed a sharp discrepancy in one exercise with respect to a joint that seems to point at the same direction. IMO, there's a mistake in the exercise.

I uploaded both my solution, and the manual's solution for comparison. The joint I'm referring to is joint B. At beam AB, and at beam CD, it points the same direction.


The engine weighs 250 kg, we need to basically calculate Ay, Ax, Cx. Cy. Dx, Dy, and Md. (in the solution manual, F = Ra). D is fixed, A, B and C are typical pin joints. The measurements are in mm.
http://img832.imageshack.us/img832/1076/jointsq.jpg
 
Attachments
  • Hebrew-labelled free-body diagram of a bent frame with points E, A, B, C and D, dimensions 1200, 400, 800 and 400, a force F at 63.43 degrees at A, and a mg load at E, followed by worked moment equations and a separate free-body diagram of part CE with forces Cx and Cy.
    jointsa1.jpg
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  • Free-body diagram of a rigid frame with pin joints C, B and D, showing forces Cx, Cy, F at 63.43 degrees, Dx, Dy and moment MD, with dimensions 800 mm and 400 mm, followed by equilibrium equations and a results box listing F = 10968.39 N, Cx = 4906.06 N, Cy = -7357.50 N, Dx = 4903.65 N, Dy = 9811.87 N, MD = -3924.85 N·m.
    jointsb1.jpg
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  • Handwritten solution to a structural statics problem on graph paper, showing a beam diagram with load W = 24 517.5 N, moment equations, and force balance equations solving for Ay = 9810 N, Cx = 4905 N, Cy = 7357.5 N, RA = 10 967.9134 N, and Py = 17 167.5 N, with the word 'wrong' written near the bottom.
    jointsmine.jpg
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Femme_physics said:
Sorry to harp on the issue, but I've noticed a sharp discrepancy in one exercise with respect to a joint that seems to point at the same direction. IMO, there's a mistake in the exercise.

I uploaded both my solution, and the manual's solution for comparison. The joint I'm referring to is joint B. At beam AB, and at beam CD, it points the same direction.


The engine weighs 250 kg, we need to basically calculate Ay, Ax, Cx. Cy. Dx, Dy, and Md. (in the solution manual, F = Ra). D is fixed, A, B and C are typical pin joints. The measurements are in mm.

Hi golden girl, :smile:


In your section EC all forces and their directions are in order.

In your section AB the forces at A as well as the forces in B should be reversed (because the force in A is a reactive force). The calculation however will come out the same.

Then in your section CD, the forces at B should again be reversed.
Also the force Dy should be reversed.

Note that since Dx, Dy, and W are the only external forces, you can already say that Dx=0 and Dy=-W.
 


Hi wonderwoman :)

Ah...typical, I made a booby somewhere. I think I got very confused because I didn't see any obvious supports when looking at the structure as a whole. Then I just peeked at the solution and everything went downhill from there. I'm used to the drawing of those half-circle thingies that are attached to the ground. (We call them smachims in Hebrew). Anywho, thank you very much for sorting this out for me! I'll rework the exercise tomorrow, quite tired from solving 10 others, heh.

Thank you dear... ^^
 


Serena, when I look at the entire first as a rigid body, I understand that I should ignore the reactions are the joints. In that case, where do I draw the force that resists the weight of the engine? Was I correct to pick A? In fact, I didn't quite pick A, I just took it from the solution paper. I'm entirely confused from the picture at which point do I draw the vectors that resist the engine when looking at the construct as a whole. Can you help me see it?

You also say that Dx equals 0, but the answer manual conflicts it.
 


Femme_physics said:
You also say that Dx equals 0, but the answer manual conflicts it.

Yes, I'm afraid the answer manual is wrong.
The first page is ok, but the second page contains a series of errors.

Let me list the errors on the second answer page:
1. Everywhere where it says 90°-63.43°, it should be 63.43°.
2. This will result in: Dx=0 N and Dy=2452.5 N (which we already expected).
3. The half-circle drawn near B represent the moment MD exerted by the bottom frame. Note that it is chosen in the negative direction. However in the ∑MD formula there is a plus sign in front of MD, which means the half circle should have pointed the other way.

Femme_physics said:
Serena, when I look at the entire first as a rigid body, I understand that I should ignore the reactions are the joints. In that case, where do I draw the force that resists the weight of the engine? Was I correct to pick A? In fact, I didn't quite pick A, I just took it from the solution paper. I'm entirely confused from the picture at which point do I draw the vectors that resist the engine when looking at the construct as a whole. Can you help me see it?

The external force that resists the engine is the force exerted by the floor on the bottom of the frame. In the second answer sheet this is modeled by the forces Dx, Dy, and the moment MD in point D.
There should be a chapter about support fixtures (smachims?) that explains about this.

It could for instance also have been modeled by an upward force by the floor on the bottom of the frame straight under the engine.

In your own answer you came out on the wrong Dy, because you inverted By.
 


See, this is what happens when you just copy things from the solution...you get bogged down by mistakes other ppl make. This seems like a fairly simple problem once I understand that D is a fixed support to the ground. Somehow I completely missed that. Thank you again Serena. Let me redo that if you will, I think I can pull it off now.
 


I found a 4th mistake, the distance between BD in the solution is 400. In the question it's 600.

Never blindly trust a solution book!
 


Femme_physics said:
I found a 4th mistake, the distance between BD in the solution is 400. In the question it's 600.

Never blindly trust a solution book!

Very good! :smile:
I have to admit that I missed that!

It also means there must be another wrong answer in the second answer sheet... :confused:
 


It could, I reworked it all according to the premise that D is a fixed support, and I think I got it all correct now. Can you give me your Serena seal of mighty all-physics-like-stuff approval?
 
Attachments
  • Handwritten solution on graph paper titled 6 ON FP 2007, showing a frame with a horizontal beam EC, diagonal member AC, vertical member CB and support D, with equilibrium equations for forces and moments, the value Dy = 2452.5 N, Cy = 9810 N, Bx = 6131.25 N, and a separate beam AB diagram.
    correction1.jpg
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  • Handwritten structural analysis on grid paper: EC beam with a load W and support A, equation sum Fx = 0 giving Cx = 6131.25 N, then CD beam with equation sum Fx = 0 giving Dx = 0, moment equation sum MD = 0 giving M = 4905 N·m, and RA = 13709.623 N, with Hebrew annotations.
    correction2.jpg
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Femme_physics said:
It could, I reworked it all according to the premise that D is a fixed support, and I think I got it all correct now. Can you give me your Serena seal of mighty all-physics-like-stuff approval?

Sorry to disappoint you. :shy:

In your first page you write: ∑MC=W·1600 - Cy·400.

I think you meant: ∑MA=W·1200 - Cy·400.

This will yield a different result for Cy and then all the subsequent calculations change...
 


Ugh, I'll tell you why this happened. I started by doing the sum of all moments on A, then I suddenly decided that I don't really like C, that it's kinda ugly and it the corner, and that I like A better (even though it doesn't matter). So, changed it and forgot to change the figures. Ugh, that's why being a girl in mechanics is bad (lol).


I'll rework that.
 


Femme_physics said:
Ugh, I'll tell you why this happened. I started by doing the sum of all moments on A, then I suddenly decided that I don't really like C, that it's kinda ugly and it the corner, and that I like A better (even though it doesn't matter). So, changed it and forgot to change the figures. Ugh, that's why being a girl in mechanics is bad (lol). I'll rework that.

Btw, I'm having difficulty reading these Arabic sentences. And since these are photos I can't copy+paste them into a translator website! :wink:
[EDIT]I did do that with you blogspot, and it's rather tricky with this right-to-left reading stuff![/EDIT]
 
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I like Serena said:
Btw, I'm having difficulty reading these Arabic sentences. And since these are photos I can't copy+paste them into a translator website! :wink:
[EDIT]I did do that with you blogspot, and it's rather tricky with this right-to-left reading stuff![/EDIT]

hehe :)

Well, I just wrote notes to my classmates so that's why it's in Hebrew (ahem, yes, Hebrew, not Arabic). Since I'm uploading the solutions (and corrections) to my blog I make sure to write detailed explanation of what I'm doing (your help doesn't only help me, it helps others!). For instance, in the first page I just wrote that "I personally believe there's a mistake in the answer book". In the second page I wrote "since there's a moment in the equation, it's critical to use meters and not milimeters, or translate the result of the moment force to meters later."

I will definitely add your name with gratitude to these papers when I re-scan them. I like Serena from physicsforums.com definitely deserves a whole lot of the credit here! :)

Oh, and here's my last fix, attached, before I go to bed. All seems to be in order.
 
Attachments
  • Handwritten solution sheet titled 6 ON ITP 2007 with a three-beam frame diagram, equilibrium equations for beams EC, CD and AB, and boxed results such as Dy = 2452.5 N, Cy = 7357.5 N, Dy = 9810 N and Bx = 4905 N.
    fix1.jpg
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  • Handwritten solution notes in Hebrew and mathematics: sum of Fx = 0 gives Cx = 4905 N for beam EC; sum of Fx for CD gives Dx = 0; sum of moments about D gives M_D = 3924 N·m; and Ay² + Ax² = RA² gives RA = 10967.9 N.
    fix2.jpg
    25.9 KB · Views: 453


Femme_physics said:
hehe :)

Well, I just wrote notes to my classmates so that's why it's in Hebrew (ahem, yes, Hebrew, not Arabic). Since I'm uploading the solutions (and corrections) to my blog I make sure to write detailed explanation of what I'm doing (your help doesn't only help me, it helps others!). For instance, in the first page I just wrote that "I personally believe there's a mistake in the answer book". In the second page I wrote "since there's a moment in the equation, it's critical to use meters and not milimeters, or translate the result of the moment force to meters later."

I will definitely add your name with gratitude to these papers when I re-scan them. I like Serena from physicsforums.com definitely deserves a whole lot of the credit here! :)

Oh, and here's my last fix, attached, before I go to bed. All seems to be in order.

Let me give you a few minor parting comments then, and sleep well! :)

On your first page it would look better if you put the MD half-circle in as well, because otherwise the frame would fall over. With the half-circle added, you would show all external forces and moments.

Since you drew Bx on section CD on your first page, you should also draw Cx, to be consistent and to be in equilibrium.

On your second page you drew the moment at D. However as you drew it, it is inconsistent with your equation. You should draw the arrowhead in the opposite direction.

Sleep well! :smile: