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Deflection in framework

  1. Sep 28, 2011 #1

    Jus

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    Hi there,

    The first thing I wanted to say (as I said the last time I created a new thread) is that this is NOT A HOMEWORK QUESTION!

    From the rules laid out, it says that homework for undergraduates and college students should go in the homework section. I have a degree in Mechanical Engineering and have been working as an professional engineer for 12 years. However, I have recently been offered a new job and I am just brushing up on the work I did at university.

    I have attached 2 files. One (2.45.doc) is a scanned copy of a question from the book that I am currently going through and the other is my attempt at the solution to the problem. The book is 'Mechanics Of Materials' by F.Beer. I have obviously removed any unnecessary information from the scanned page of the book as I do not want to infringe on any copyright issues. Please also be aware that I would not normally do this and have resorted to this because I have nobody else to ask and the solution to this particular question is not in the back of the book.

    The second attachment (2.45 sol.pdf) is my attempt at the solution.

    View attachment 2.45 sol.pdf

    View attachment 2.45.doc

    Moving on.... I'd be very grateful if somebody could have a look at this question and verify if my answer is correct or not. The question is concerning deflections / deformations in members that form part of a structure. As I have said, I have tried to answer the question but I would really like someone to verify this for me - afterall, there's no point in continuing to calculate the answers to other similar questions if the method is incorrect.

    Again, please forgive me if this question is not meant to be in this section. I genuinely thought that this would be ok to do. If not then please let me know and I will move it to the homework section.

    Thanks in advance to anyone / everyone who can help me on this.

    Kind Regards,

    Justin
     
    Last edited: Sep 28, 2011
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  3. Sep 28, 2011 #2

    nvn

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    Jus: Your answer currently appears incorrect (I think). You have bar AF rotating clockwise, and you show bar DE pulling on bar AF; but then you obtain a positive value for Fde. Something seems amiss. Keep trying.
     
    Last edited: Sep 29, 2011
  4. Sep 29, 2011 #3

    Jus

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    Hi nvn,

    It's an interesting point that you've made. I did notice this myself and I think that you are right, Fde should be pointing the other way in the free body diagram. But this is where I'm getting confused - surely it wouldn't matter which way the forces point in the first place as when you apply the equilibrium equations an incorrectly directioned force would just result in a negative value. For example, in my calculation I have assumed all forces in members to be tensile (arrows moving AWAY from member AF), in which case Fde (if it was compressive, it would just yield a negative result....wouldn't it??). Sorry if I've asked a stupid question.
     
  5. Sep 29, 2011 #4

    nvn

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    Jus: I have not pinpointed the exact reason why your Fde did not turn out negative, due to limited time. Your statements in post 3 are correct. What do you get for Ff? After that, check summation of horizontal forces, and summation of moment, to ensure both summations equal zero. And check the slope of member AF, between three points, to ensure the segments are collinear. I wonder if you have a sign error somewhere? Fde should come out negative, it seems. Myself, I currently got Fde = -1600 N. Keep trying.
     
  6. Sep 29, 2011 #5

    Jus

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    Hi nvn,

    Please accept my apologies if I have been over-persistant with the question. It's not a quick thing to calculate and I appreciate that you have a lot of other work to do...sorry for mithering.

    I get the exact same answer as you when I change the direction of the force on Fde. 1600N (remember in my calculation this means that the 1600N is the reaction force which in turn naturally acts against the opposing force, hence mine is a positive value as it acts against the member AF and in the same direction that I have my force acting in on the FBD).

    I'll keep looking in to it. Thank you so much for your help and guidance nvn, it is very much appreciated.

    Kind Regards,

    Jus
     
  7. Sep 29, 2011 #6

    nvn

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    Jus: In reality, my Fde points to the left in my FBD of member AF. Hence, I actually got Fde = 1600 N. I said Fde = -1600 N, only here in post 4, to match your FBD in post 1.

    Nonetheless, if you changed the direction of Fde in your post 1 FBD, then yes, Fde = 1600 N. When you said "FBD" in post 5, I assume you meant a new FBD, not your exact FBD in post 1.

    I will be interested if you pinpoint what caused the answer to change when you switched the direction of Fde in your FBD. As mentioned in post 3, it seems it should have worked out correctly either way. First, try the checks mentioned in post 4, to see how your post 1 solution checks out.
     
  8. Sep 29, 2011 #7
    Just to confirm 1600 to the left is good for the reaction on the lever since the lower link is in compression.

    To avoid sign convention issues I have used an energy method rather than your moments.
    It leads to the same equation, but you might like to compare the methods.
    Note that since we are applying P slowly and elastically energy = 0.5 force times distance.

    Your compatibility calculation can be tidied up as can your substitution - there is some easy cancelling available.

    Note also that if we consider a rotation angle alpha and some trigonometry this readily drops out.

    I am worried about nvn statement of horiz equilibrium. This frame is statically indeterminate so the invocation in post 4 cannot be carried out.

    go well
     

    Attached Files:

  9. Sep 29, 2011 #8

    nvn

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    Studiot: Regarding your last paragraph in post 7, statically indeterminate structures are in static equilibrium. The checks in post 4 are correct, and should be used by Jus.

    Jus: I think I now pinpointed your mistake in post 1. It is a sign error, which I suspected in post 4. In your post 1 attached file, change delta_de = xd to delta_de = -xd. Implement this correction, and let's see what answers you then get throughout your post 1 attached file. (I have not tried this yet. Therefore, I would like to know what answers it gives you throughout your solution.)
     
  10. Sep 30, 2011 #9
    We do not know the reaction forces at the hinge, F.

    Jus has eliminated these from the equations by taking moments about F.

    My energy approach also eliminates these since the hinge undergoes zero displacement and therefore the hinge forces do no work.

    However these force do appear in the equation for horizontal equilibrium.

    Thus, having calculated the other horizontal forces by other means, we can calculate the horizontal hinge reaction but cannot use horizontal equilibrium as a check.
     
  11. Sep 30, 2011 #10

    Jus

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    Guys, thanks so much for your help on this.

    Studiot, I was actually practising the method laid out in the chapter I am going through at the moment which isn't actually using the energy method - I haven't revised that chapter yet. But thanks for conforming the answer of 1600N.

    What was really bothering me was that when I use my FBD (putting all the forces in the tensile sense) the answer does not come out at 1600N.... From memory, it comes out at 2400N. But the way that I thought of tackling the problem (and please forgive me if this is the wrong way of thinking) was to assume that everything is in tension and then if any values come out negative this simply means the member is in compression.....but this does not seem to be the case here and I can't for the life of me spot any sign errors in the equilibrium moment equation.

    Should I be changing the direction of the arrow Fde so that this points towards the main member AF? I always used to assume that everything is in tension so that I did not get confused with my sign convention. Is this the wrong way of thinking? I understand that Fde will be a compressive force just from looking at the drawing but I would prefer not to balance the forces mentally and stick to a method...but then why does the calculation come out differently than if the arrow was pointing the other way??

    Sorry to burden you with this guys, thanks again for your help.

    Regards,

    Jus
     
    Last edited: Sep 30, 2011
  12. Sep 30, 2011 #11

    Jus

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    nvn....this is an interesting point. I did think of this when I was going through the calculation again. Should the deformation be '-'ve instead of '+'ve because it is a contraction rather than an expansion? I agree, this is a definite possibilty. But what I don't understand is why the calculation does not automatically give me the right numbers ('+'ve or '-'ve) if I have stuck to a rigorous sign convention? Is it because, taking horizontal right as positive, I should have included the necessary signs in the deformations (deltas) as well?
     
  13. Sep 30, 2011 #12
    Good morning, jus.

    Since you are worrying about assuming a force direction here is the full SP.
    You should know that assuming all tension will work if you do it properly.

    For a statically determine structure the three equations of equilibrium are sufficient to solve for the forces.
    These equations only involve the forces and numerical coefficients (determined) from the geometry) so we can assume directions and these will come out as + or - in the analysis to tell us if our initial assumption was correct or not.

    When we have a statically indeterminate structure solve it by introducing additional unknowns and compensating additional equations. These unknowns may also be signed quantities (they are in this case) and must be entered into the equations as such.

    In this problem we start with two force unknowns and one equation (derived from moments or energy considerations) I have called them P1 and P2. They are signed quantities.

    To this we add three deflections [itex]\delta[/itex]0; [itex]\delta[/itex]1; [itex]\delta[/itex]2. These are also signed quantities.

    making 5 unknowns in all.

    Compatibility yields two equations from the geometry of the situation.

    Elasticity yields two further equations, making up the 5.

    Now you could place all these equations into a matrix and solve algebraically.
    If you do this, paying proper regard to the signs, you will have a solution set that will take care of directions for you, in the resulting signs.

    nvn was correct, it is the signs of the deflections that make the difference - the product of two negatives is a positive.

    I fuly admit that I have short-cutted this process in my analysis.
     
  14. Oct 1, 2011 #13

    nvn

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    No, your deformations (deltas) are OK. Elongation (tension) is generally positive.

    Your only mistake is a missing negative sign in one of your two compatibility equations. Notice on p. 1 of your attached file, you wrote two compatibility equations: delta_cb = xc, and delta_de = xd. Your first compatibility equation, delta_cb = xc, is already correct. Your second compatibility equation, delta_de = xd, is missing a negative sign, and should instead be, delta_de = -xd.

    There is nothing wrong with your method. Your solution in post 1 requires almost no change, except for one sign that you inadvertently omitted, mentioned above, and mentioned in post 8. You cannot say delta_de is positive (elongation) and claim it is in the direction of positive xd. Do you not see that elongation (tension) of member DE is -xd? Add your missing negative sign, as described in post 8. After that, see and perform post 4. What do you get for Ff?
     
  15. Jul 15, 2012 #14

    Jus

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    Hi nvn,

    Sorry for the late reply to your post, I was made redundant and only recently found a new job!

    I agree entirely with what you said and understand exactly where I've gone wrong and how you and Studiot were correct. Thanks again for all your help. Jus.
     
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