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Homework Help: Stress & Strain (Finding diameter, extention & thickness)

  1. Jun 13, 2012 #1


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    Hi guys i have a exam coming up soon and stress is starting to kick in. Im good withy stress and strain, but just confused on this two questions and don't understand what im doing wrong. I've tried every formula and i just cant seem to get question 6 out. And question 5 same problem i understand they are similar queations but this one has tensile and compressive force given to you and i dont understand what to do with them. I can upload question 2 help with Q5 but i dont it will help mush its just a truss. The two questions are in the attachment. The answears for this questions are underneath the question.

    Please have a look at them and tell me what you think, it will also be much appriciated if can write down a solution for them.

    Thanks, Dom

    Attached Files:

  2. jcsd
  3. Jun 13, 2012 #2
    Hello Dom, welcome to Physics Forums.

    What did you make the force at B?
    Are you working out your levers correctly?
    Do you think this force on the bar at B is less than, greater than or equal to P?
  4. Jun 13, 2012 #3


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    Im not sure mate, ive tried working it out by taking the moment of it like you would on a truss but i cant apply it correctly. Is this the right approach or am i way of track? (Or do i use stress/strain formulas and solve for P)
  5. Jun 13, 2012 #4
    What do you mean, solve for P?

    You are given P in the question P=250
  6. Jun 13, 2012 #5


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    Yeah thats exerted at D, but wouldnt there be a different force B or (is it something to do with equillibrium)
  7. Jun 13, 2012 #6
    You can either use the lever formula or calculate the force at B (call it B) by taking moments about C.

    Either way you should find that B is not equal to P

    So try to answer my question or explain what your difficulty is in doing this because I think this is where your problem lies.

    Hint I think B = 6P

    Can you confirm this?
  8. Jun 13, 2012 #7


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    RB=833.33 ??? or try again
  9. Jun 13, 2012 #8

    Where did this come from for moments about C?

    What do you know about levers?
  10. Jun 13, 2012 #9


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    The input force is higher than the output force
  11. Jun 13, 2012 #10
    So which is larger and why do you think the pivot is between the input and output??
  12. Jun 13, 2012 #11


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    Oooops the pivot is definately not between them. Do i need to have a different value for where i have 3600 (-3600xRB)+(12000x250)=0
  13. Jun 13, 2012 #12
    Perhaps this sketch will help.

    The end of the bar not attached to the wall is cranked out by the lever so that

    b moves to b' and B moves to B'. Thus B moves a distance ΔL equal to the extension.

    What is the moment of P about C (you have this correct)

    What is the moment of B about C ? You should not think in terms of the wall reactions but about the force, B that is acting (in which direction) on the bar.

    Attached Files:

    Last edited: Jun 13, 2012
  14. Jun 13, 2012 #13


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    Thanks so much for taking your time to do that but im now confused on how you take the moment. Will you have to use sin and cos
  15. Jun 13, 2012 #14
    OK lets start at the beginning.

    The lever is pulled to the right at b with horizontal force P.

    The vertical (perpendicuar distance) from the line of action of P to a pivot c is 1200.
    So the moment of P about c is 1200 * P clockwise.

    The lever is attached to the bar at B so pulls the bar to the right with force B.
    Thus the bar pulls the lever to the left with force B (by Newtons Third Law - Action and Reaction are equal and opposite).

    B is a horizontal force (like P)

    OOPs! So acting on the edit: lever [STRIKE]bar[/STRIKE] is a force B to the left.
    The vertical or perpendicular distance from B to the pivot is 200
    Since B on the lever is to the left, the moment about c is anticlockwise and equal to 200 * B

    Thus 200B = 1200P :

    If you insert the given value for P can you now find B?
    Last edited: Jun 13, 2012
  16. Jun 13, 2012 #15
    How are we doing?
  17. Jun 13, 2012 #16


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    Hi yep thats good thankyou, i had that before but i thought B=1500 but i thought it was to big. Do you have a suggestion on how to procede from here because this is the spot where im actually stuck given that i did the same as you before.
  18. Jun 13, 2012 #17
    OK so now considering the bar.

    The bar is extended ΔL by a force B = 1500N

    Now that you have the force you can calculate the stress. Make sure you know the difference between the diameter and radius (a common error)

    [tex]Stress = \frac{{Force}}{{X\sec tionalArea}} = \frac{{Force}}{{\left( {\frac{{\pi {d^2}}}{4}} \right)}}[/tex]

    Once you have the stress you can use it to find the extension using the definition of E and the length of the bar that was given.

    [tex]E = \frac{{Stress}}{{Strain}} = \frac{{Stress}}{{\frac{{Extension}}{{OriginalLength}}}}[/tex]

    If you make the substitutions you will recover the answers you pencilled in at the bottom of the question.
  19. Jun 13, 2012 #18


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    Hi i got both of the answears out and they are both right thankyou. Do you think you have time to help me with the 2nd question?
  20. Jun 13, 2012 #19
    I have to go now but I will be about later today.

    Perhaps someone else might help or you can look later.

    Post it as a new thread.

    go well in your exam
  21. Jun 13, 2012 #20
    This question has further lessons to give, beyond just the asked for answer.

    Think about the mechanism and what is actually happening.

    If you consider moment equilibrium about the B, where the rod meets the lever you have a balance where the applied force, P opposes the moment generated where the bottom of the lever pushes on the hinge point c and the hinge pushes back

    200Hc = 1000P, where Hc is the horizontal hinge reaction.

    This leads to Hc = 5P = 1250N.

    Obviously this is consistent with horizontal equilibrium for the lever since

    1500 = 1250 + 250.

    But what it is saying is that most of the necessay force to pull out the bar comes from the use of the lever, not from your own 250N pull.

    This just shows the power of levers.
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