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Solid Mechanics Question - Help !

  1. Nov 7, 2009 #1
    hi all :biggrin:

    i apologize in advance for the poor englsih, the question is translated from another language.

    A round beam (radius R) is harnessed on the left side to a wall.
    The beam is made up of two parts glued together by a thin layer of glue - the attachment plane is described in the picture as AB.
    A 'glue fail' happens when the SIZE of the shear strain on plane AB is bigger the Sigma0 (some unkown value).

    wjdqzozzcnyl.jpg

    On the right side of the beam theres an equaly spred strain q=q0*e3 (note that q0 might be negetive as well).

    What is the allowed value range of q0 so that there wont be a glue failure? (express your answer by using R and Sigma0)

    this question was given as a challenge, and is a part of a course called solid mechanics 2 that deals more with shear strain and strain matrices. I realy dont even have a clue of how to begin, i cant see how the radius R plays a part in the answer..

    help, please?
     
  2. jcsd
  3. Nov 8, 2009 #2

    Mapes

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    Hi dislect, welcome to PF. Do you know the relationship between force, area, and stress, and can you draw a free-body diagram and write the static equilibrium equations? These would go a long way in helping you solve the problem.

    The translation looks pretty good, but "strain" should probably be "load" (or "force"), "distributed load," or "stress." Is is possible to find out which, since they imply different loading configurations?
     
  4. Nov 8, 2009 #3
    Hi Mapes, thanks!

    yhtthznmdhnj.jpg

    It's weird but I think they actually ment strain and not force. Well, thats the word they chose to describe it as. In the first course (Solid mechanics 1) we dealt a lot with forces and free-body diagrams. Now its more about strain tensors on different planes, and thats the part that bugs me because I know that the solution as something to do with it..

    * I copied the parameter they signed for the maximum sheer strain that AB plane can tolerate (AKA Sigma0)
     
  5. Nov 8, 2009 #4

    Mapes

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    OK, great. Now using the tangential force on AB, calculate the shear stress at the joint and compare this to [itex]\sigma_0[/itex] ([itex]\sigma[/itex] is almost universally used to denote stress, not strain).
     
  6. Nov 8, 2009 #5
    Well thats the part I'm stuck in :)
    How do I do that? how does the momentum take place if i dont know any distances on the beam?
    Whem I'm summing all the forces do i count q0 as is or as q0*pi*R^2, and how do I get a range of values of q0 in such way?
     
  7. Nov 8, 2009 #6

    Mapes

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    Moments? Don't worry about the moments; you just want to turn the tangential force v into a shear stress by dividing it by the cross-sectional area (which is not a circle!).

    There's no way to tell without knowing what q0 is (e.g., a force, a stress...). I'm quite sure it's not a strain, despite the translation. I would assume it's a force.
     
  8. Nov 8, 2009 #7
    If I translate it "as is" then q0 is said to be an equaly devided 'load' (that might be negetive as well)

    Hmm.. about the sheer stress thats steping back to trigonometrics :)
    How do I find that cross-sectional area? its sort of eliptic.
     
  9. Nov 8, 2009 #8

    Mapes

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    You got it!
     
  10. Nov 8, 2009 #9
    So the cross-sectional area is 2*Pi*(R^2)
    I still dont realy get the question. It's like I'm solving it with you in baby steps and cant see the all picture :-(
     
  11. Nov 8, 2009 #10

    Mapes

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    In the problem, you're pulling or pushing on a beam with circular cross section. There's an angled, glued joint within the beam that can only sustain a certain amount of shear stress. Your job is to calculate how hard you can pull or push on the beam without making the joint fail. So you need to connect the axial load [itex]q_0[/itex] to the rupture shear stress [itex]\sigma_0[/itex]. Does this help?
     
  12. Nov 8, 2009 #11
    So its something like V*2*pi*(R^2)*sin(30) = q0*pi*(R^2) ?
     
  13. Nov 8, 2009 #12

    Mapes

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    I'm not sure where this is coming from. Are you summing the forces in the [itex]e_3[/itex] direction? There's no 30° angle in this problem.
     
  14. Nov 8, 2009 #13
    That what I tried.. casting V on e3
     
  15. Nov 8, 2009 #14
    Mapes, could you help me with the static equilibrium part?
    I fell like I'm getting lost in this question, and I have no other examples to rely on (its the first 'challenge question' that the lecturer gave). Plus, It's just the first part of the question :) in the second part they change q0 to pure moment.
     
  16. Nov 8, 2009 #15

    Mapes

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    I think you need to find the relationship between V and q0. I'm not sure where sin(30) comes in. A couple hints I can give are that (1) V is smaller in magnitude than q0 (assuming q0 is the axial force applied to the beam end) and that (2) V equals q0 when the cut is horizontal but is zero when the cut is vertical. Does this help figure out the trigonometry part?
     
  17. Nov 10, 2009 #16
    Good morning Mapes!
    Me and a friend are going to sit and do our best with the question, and later on today i'll upload our solution trial for your confirmation :-)

    thanks for all the patience, I realy appreciate it!

    btw, just today we realised that q0 is an already spread force, meaning we dont have to devide it by A and the same goes for the V(sheer)!
     
  18. Nov 10, 2009 #17
    Ok, so as I said q0 is an already spread force and therefor -< the strain ( no need to devide by A).

    This is what we did:

    lltocjtywyzr.jpg

    * Is our assumption correct about AC? if so, how can we justify it?
    * Is our solution correct? :-)

    Thanks Mapes!
     
  19. Nov 10, 2009 #18

    Mapes

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    Right away, let me reiterate that a 3-4-5 right triangle does not contain any 30° angles.

    It looks like you're getting closer.

    EDIT: Try solving it for an arbitrary angle [itex]\theta[/itex]. Think about how the answer must change as the angle approaches 0° or 90°.
     
  20. Nov 10, 2009 #19
    Hmm I'm not sure I understand why its not a 30 degree angle..
    and if its not, what does the 3-4-5 triangle give me and how do i express the n vector of plane AB?
     
  21. Nov 10, 2009 #20
    Ok I'll organize my questions:

    *Why does this triangle dont have a 30 degree angle? i thought that any 3-4-5 triangle has 30-60-90 degrees in it.
    *The overall solution for the question is right? I mean besides the angle part
    *In the second part of the question that change the strain q0 into pure moment T. Out of curiosity, where would the glue fail first on plane AB (while applying moment T)? My intuition says that it should fail all over the edges of plane AB simultaneously but I cant realy explain it. Is it true? and if so, how can I justify my saying?

    Once again, I'ts important for me to emphasize how much I appreciate your help Mapes :-)
     
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