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Stress problem- finding the diameter of a tie bar

  1. Oct 4, 2014 #1
    Hello,

    I'm having an issue trying to solve an equation- the problem is worded as such:

    "A tie bar is made of a material having an ultimate tensile strength of 231 mPa, and must carry a load of 255 kN. What is the diameter of the bar if a factor of safety of 7 is required?



    I know that the formula to find factor of safety is ultimate stress/allowable stress. So I'm thinking I need to find the allowable stress in order to find the final answer.

    So here is what I have attempted, but I feel that I'm way off:

    A = 3.1416(d^2)/4 = .25*pi*d^2


    255 kN*1000 = 255*10^3 N

    231 mPa*100 000 = 231*10^6 n/m^2


    231*10^6 N/m2= 255*10^3 N/.25*pi*d^2

    .25*pi*d^2 = 255*10^3 N/231*10*6 N/m^2

    d^2 = 255*10^3 N/.25*3.1416*231*10^6 N/m^2

    d^2 = 255 000 N/1.8143*10^8 N/m^2

    d^2 = 1.4055*10^-3m^2

    = 1.39m













     
  2. jcsd
  3. Oct 4, 2014 #2

    SteamKing

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    That's the right idea. Why didn't you do it?

    You should be using 'M' for the prefix 'mega' (x 10^6). If you use 'm', which is the prefix for 'milli' (x 10^-3), you're going to get
    confused.


    A tie rod over a meter in diameter is a pretty substantial piece of equipment. Find the allowable stress based on the safety factor, and try again.
     
  4. Oct 4, 2014 #3

    SteamKing

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    You are also showing errors in arithmetic calculations. In your calculation of the diameter, you showed

    d^2 = 1.4055*10^-3m^2

    = 1.39m

    How do you take the square root of a number less than one and wind up with a result which is greater than one? Watch the exponents on your scientific notation.
     
  5. Oct 5, 2014 #4
    Wow, yeah I messed that all up pretty good. Thanks for bringing that to my attention. Nonetheless, I was going in the wrong direction with the work I had done.

    So looking back at trying to find the diameter of this tie bar, I believe that given I have the value for ultimate tensile stress (231 MPa), the allowable load (255 kN), and a factor of safety of 7, my work leading up to the solution should be:

    231*10^6*7 = 1.617*10^9

    1.617*10^9/255*10^3 = 6 341.1765

    6 341.1765*.25 = 1 585.2941

    √(1,585.2941) = 39.8158

    =39.81 mm?
     
  6. Oct 5, 2014 #5

    SteamKing

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    There was nothing wrong with your original calculation approach, except you had omitted using the factor of safety in your calculation of allowable stress.

    In your latest calculations, you have used the factor of safety incorrectly. (A factor of safety means the allowable stress is less than the ultimate tensile stress of the material, by the factor of safety.) Please show units for all calculations.
     
  7. Oct 7, 2014 #6
    Sheesh, yeah what was I thinking? Okay, I'm quite positive that I have it figured out.

    231*10^6 Pa/7 = 33 000 000 Pa

    255 000 N/33 000 000 Pa = .007727 m^2

    .00727 m^2/3.14 = .00246 m^2

    √.00246 m^2 = .04960 m^2

    .04960 m^2*2 = .09921 m^2

    .09921 m^2*1000 m = 99.2 m

    How does that look?
     
  8. Oct 7, 2014 #7
    Oh I put the wrong units.

    09921 m^2*1000 mm = 99.2 mm
     
  9. Oct 7, 2014 #8

    SteamKing

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    You got the right answer, but you went about it in a sloppy way. Instead of writing down a bunch of numbers, you need to be explicit about what the numbers represent. After all, if someone picks up your calculation, how will they know what quantity 0.007727 m^2 represents? You also left out a factor of 4 in your calculations. Where? You can't tell very easily with your presentation.

    Look, this is a very simple calculation. If you continue to use this approach in more complex calculations or on an exam, you're probably going to lose points or make a mistake.
     
  10. Oct 8, 2014 #9
    Thanks again for your help. From hereon in, I will work on keeping my work more organized and neat.
     
    Last edited: Oct 8, 2014
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