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Strain in different direction

  1. Oct 20, 2016 #1
    1. The problem statement, all variables and given/known data
    In this question , I am not convinced that δAD = δAC cos theta

    2. Relevant equations


    3. The attempt at a solution
    , i think it should be
    δAC = δAD cos theta , am i right ?
    I think so because we normally get the non-vertical and non -horizontal line and then cos the angle or sin the angle to get the length in x and y direction , right ?
     

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  2. jcsd
  3. Oct 20, 2016 #2
    No. You're not right. Call L the distance between C and D. Does this distance change?
     
  4. Oct 20, 2016 #3
    no
    no , the distance doesnt chnage , so ?
     
  5. Oct 20, 2016 #4
    Call x the distance from A to C. Using the Pythagorean theorem, in terms of L and x, what is the distance between A and D?
     
  6. Oct 20, 2016 #5
    AD = sqrt ( L^2 + x^2 )
     
  7. Oct 20, 2016 #6
    Good. Now, if the distance between A and C changes by dx, in terms of L, x, and dx, by how much does the distance between A and D change?
     
  8. Oct 20, 2016 #7
    can you
    can you give some hint , i have no idea
     
  9. Oct 20, 2016 #8
    Have you had calculus?
     
  10. Oct 20, 2016 #9
    yes , and ?
     
  11. Oct 20, 2016 #10
    Apparently, that wasn't a big enough hint. What is the derivative with respect to x of ##\sqrt{L^2+x^2}##?
     
  12. Oct 20, 2016 #11
    0.5(2x) (1/ ##\ sqrt{L^2+x^2}## )
     
  13. Oct 20, 2016 #12
    What is 0.5(2x) equal to?
     
  14. Oct 20, 2016 #13
    X
     
  15. Oct 20, 2016 #14
    You really need to learn to use LaTex to display your equations. The current way you do it is totally unacceptable. Here is the link: https://www.physicsforums.com/help/latexhelp/
    Use LaTex from now on.

    As far as the derivative of ##\sqrt{L^2+x^2}## is concerned, you obtained:$$\frac{x}{\sqrt{L^2+x^2}}$$Is that what you meant?
     
  16. Oct 20, 2016 #15
    Yes, so , what are you trying to say?
     
  17. Oct 20, 2016 #16
    You still don't see it, huh?

    Well, if dx is the change in the length of AC, then the change in length of AD is ##\frac{x}{\sqrt{L^2+x^2}}dx##.
    Or equivalently, if ##\delta_{AC}## is the change in length of AC, then the change in length of AD is ##\frac{AC}{AD}\delta_{AC}##:
    $$\delta_{AD}=\frac{AC}{AD}\delta_{AC}$$
    But,$$\frac{AC}{AD}=\cos{\theta}$$
    Therefore,$$\delta_{AD}=\delta_{AC}\cos{\theta}$$
     
  18. Oct 20, 2016 #17
    Do you mean AC (del AC ) = xdx ?
     
  19. Oct 20, 2016 #18
    ok , but , i still dun know why Ac = Ad cos theta , where theta is between AC and AD ..... because normally , we will make the 'slanted line ' to get the horizontal x -direction and vertical (y-direction ) , am i right ?
     
    Last edited: Oct 20, 2016
  20. Oct 20, 2016 #19
    one more thing, how could AD = AC cos theta , where cos theta = 4/5 ,
    it's given that AD = 2 , AC = 1.6


    AC cos theta = 1.6(4/5) = 1.28 which is not = AD (2m)
     
  21. Oct 20, 2016 #20
    i dont understand this part , we know that δ = PL / AE , we also know that δAD = (4/5)δAC , so , shouldnt (4/5)δAC = (4/5)(1.6) / AE ?
     

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