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Stress and Strain problems help

  1. Sep 24, 2012 #1
    Does anyone help me to solve this problem in the attached file?

    When I solved this problem, I got wrong answers.

    Attached file is the image file with the problem and answer. (Question No.3)

    I think I have to find allowable shear stress of Pin A to find the safety factor of this pin,

    since its ultimate shear strenght is already given. (vertical height is 500mm)

    However, My allowable stress for Pin A is 431.5MPa, which is far more than its ultimate

    shear stress.

    I took moment at B and C to find the shear force of Pin A. (and Pin A is in double shear)

    so I used the formula " shear stress = shear force / (area of shear plane) * 2 " to find the

    shear force, and I got the answer above(431.5Mpa) while its ultimate shear stress is 350MPa.

    I still don't get what is the problem.. (I think there are logical errors on my thought).

    Please help!

    Attached Files:

    Last edited: Sep 25, 2012
  2. jcsd
  3. Sep 24, 2012 #2
    I get safety factors similar to what is given there, but they're a little off. Could just be rounding error. The vertical height is 600 right? It's a little hard to read.

    Technically the answer will be much less, because you should use the distortion energy theory, but the problem seems just an exercise in actually finding the forces. Hopefully someone can be buggered to type out the solution for you.
  4. Sep 25, 2012 #3
    vertical height is actually 500mm.
  5. Sep 25, 2012 #4
    Could anyone can explain to me how you get the right answer? I know it's a basic question, but it's my first time to learn mechanics. It will be really helpful to me. Thx.

    Vadar2012, don't be sarcastic.
    My purpose of posting this question is to learn from people who are professional
    in this area and to develop my problem solving skills.
    Whether you get the answer or not, it's not important and its meaningless.
  6. Sep 25, 2012 #5
    Where was I sarcastic? I was just letting you know I got similar answers to what was given, but wasn't going to type it out if I had the wrong vertical height.

    To be fair I can see how I was. It was just meant to read that I couldn't be buggered. You could always type out or attach your working, and we can go through that.
  7. Sep 25, 2012 #6


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    You should sum moments about A to solve for T. Then the x component of T is the x component of A, and the y component of A is the algebraic sum of the (vert comp of the tension + the applied 15 kN force). watch signs. The resultant force of A is the sq rt of the sum of the squares. The shear force on the pin (dbl shear) is half of that. The shear stress is the shear force divided by the area.
    I see no sarcasm intended at all in Vadar2012's post. But solutions will not be typed out for you. Please show your work and we can better assist in finding your errors, or the book's errors, whatever the case might be.
  8. Sep 25, 2012 #7
    The attached files are my solution for this problem.

    I know some of words are not English, you may ignore it.

    My problem solving step for this problem is

    1) find the reaction force at A (I took moment at B and C to find out vertical and horizontal force of pin A.

    2) Since Pin A is double shear, (allowable shear stress = Force / 2* area)
    3) my allowable stress is almost 430Mpa, which is more than ultimate shear stress

    For the rope, I don't know how to solve.
    If I did any wrong, please let me know.

    Attached Files:

  9. Sep 25, 2012 #8
    I think I have misunderstanding in the meaning of "tensile strength". I just solved the problem, and got the similar answer from the problem sheet. I thought tensile strength = the allowable force of extending steel, but it was not.
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