1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What did I do wrong? Bending moments & deflection question

  1. Jan 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem shown as image
    jan2010 Q5.jpg
    Q5. a)

    2. Relevant equations
    Also shown in image

    3. The attempt at a solution
    Made a cut between A and the z-line, so ∑M = 0 = M - wx(x/2) = M - (wx2)/2 ⇒ M = (wx2)/2

    and d2y/dx2 = -M/EI

    ∴ d2y/dx2 = - (wx2)/2EI

    So dy/dx = ∫ d2y/dx2 dx = -w/2EI ∫ x2 dx = -wx3/6EI + c1

    I was fine up to this point. My answer here agrees with the answer in the question ( equation 5.2) as x = L.

    So what I got next:

    y = ∫ dy/dx dx = -w/6EI ∫ x3 dx + ∫ c1 dx = -wx4/24EI + c1x + c2

    Letting x = L

    dy/dx = -wL3/6EI + c1 (as in the question)

    and

    y = -wL4/24EI + c1L + c2 (which is different to the answer given in the question)

    How/why is the answer to equation 5.2 wL4/8EI ??

    I probably made a silly mistake somewhere or I'm just too tired :P

    Thanks for any help.
     

    Attached Files:

  2. jcsd
  3. Jan 8, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Your integrations are correct from what I can see. The only thing you haven't done is to use the boundary conditions for the cantilever to calculate what c1 and c2 must be. I think once you do this, you should be able to get the answers you seek.
     
  4. Jan 8, 2015 #3
    So boundary conditions between L and 0?
     
  5. Jan 8, 2015 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You've got only two boundary conditions with this beam: both are located at the fixed end. Do you know what they are?
     
  6. Jan 8, 2015 #5
    Ohh. The reaction forces?
     
  7. Jan 8, 2015 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, the boundary conditions are not the reaction forces. These can be determined by statics.

    Think about what you are trying to find here. You want to calculate expressions which give the slope and deflection of the beam. What can you say about the slope and deflection of the cantilever beam at the fixed end?
     
  8. Jan 8, 2015 #7
    At the fixed end, there's no slope and max deflection?
     
  9. Jan 8, 2015 #8

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Come again? Max. deflection at the fixed end? Is that your final answer?
     
  10. Jan 8, 2015 #9
    Yeah I have no idea :P Sorry
     
  11. Jan 8, 2015 #10

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If the right end of the cantilever in the problem is fixed, it can't move. What must the slope and deflection both be at the fixed end?
     
  12. Jan 8, 2015 #11
    Would it be 0?
     
  13. Jan 8, 2015 #12

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

  14. Jan 8, 2015 #13
    Then surely c1 = wL3/6EI for the first one? Which will all cancel out to 0? I'm confused where to go next :(
     
  15. Jan 8, 2015 #14

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It only cancels out when x = L. What happens when x = 0, at the free end?
     
  16. Jan 8, 2015 #15
    Then c1 = 0
     
  17. Jan 8, 2015 #16

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    That's the trivial solution to the value of c1.

    I mean, what happens to the equations for the slope and deflection when you determine c1 and c2 based on the boundary conditions of slope = deflection = 0 at x = L? Since c1 and c2 are constants of integration, they have the same values for 0 ≤ x ≤ L.
     
  18. Jan 9, 2015 #17
    So -wx3/6EI + c1 = -wx4/24EI + c1x + c2 = 0
    I'm still struggling to know where to go next.
     
  19. Jan 9, 2015 #18

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You don't set the slope and deflection equations equal to one another. You use the fact that the slope = 0 when x = L to find c1, then you substitute this value of c1 into the deflection equation, set x = L, and knowing the deflection = 0 there, then you solve for the value of c2.

    Once you have solved for the values of c1 and c2, then you substitute these into the equations for slope and deflection, such that the value of the slope or deflection of the cantilever at any point depends only on the length L, the loading w, and the position x along the cantilever from the free end.
     
  20. Jan 9, 2015 #19
    So -wx3/6EI + c1 = 0 ⇒ c1 = wx3/6EI

    ∴ -wx4/24EI + c1x + c2 = -wx4/24EI + wx4/6EI + c2 = 0 ⇒ c2 = -wx4/8EI

    Am I right so far?
     
  21. Jan 9, 2015 #20

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, the slope = 0 only when x = L. The deflection = 0 only when x = L
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: What did I do wrong? Bending moments & deflection question
Loading...