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Compressive and Tensile Stresses of an 'I' Beam

  1. Dec 26, 2011 #1
    1. The problem statement, all variables and given/known data

    A symmetric I section beam is 60 mm deep with a second moment of area of 663x10-9 m^4 and a cross sectional area of 1600 mm^2. It is subject to a bending moment of 1.2 kN.m and an axial force of 25 kN (tension). Find the maximum tensile and compressive stresses.


    2. Relevant equations

    σ=My/I

    3. The attempt at a solution

    So everything is given except the 'y'. I am unsure how to work this out as I am only given the depth.

    Also could someone explain the relevance of the given axial force?
     
  2. jcsd
  3. Dec 26, 2011 #2

    PhanthomJay

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    The 'y' distance is measured from the centroid of the cross section. For what value of y is the bending stress a maximum in tension? In compression? What is the formula for axial stresses? The stresses from each are algebraically additive.
     
  4. Dec 26, 2011 #3

    nvn

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    123321: y is one half of the I-beam depth. The tensile axial force is used to compute the uniform axial stress, which is part of the stress on the I-beam.
     
  5. Dec 26, 2011 #4
    Thanks for the quick replies guys!

    So from you's have said, am I right to say that the y for compression is 30 mm?

    How about the y for tension? Will it also be 30 mm because it is symmetrical?

    I've never heard of axial stress, is it the same as torsion?

    I'm so confused...
     
    Last edited: Dec 26, 2011
  6. Dec 26, 2011 #5

    AlephZero

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    Didn't your course deal with the axial stress in a rod? A beam can also act like a rod.

    You need to find the stress from the axial load (uniform tension) and the bending stress (tension and compression) separately, then combine the two.
     
  7. Dec 26, 2011 #6
    Nope, never heard of it.

    How do I find the two bending stresses?
     
  8. Dec 26, 2011 #7
    Can someone help please? I've been working on this question all day and been getting nowhere...
     
  9. Dec 26, 2011 #8

    nvn

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    123321: Hint 1: Try the equation listed in post 1. Hint 2: Look for a uniform axial stress equation in your text book.
     
  10. Dec 26, 2011 #9
    Yeah, I'm still confused about the the two y's, for compression and tension. Can anyone explain?
     
  11. Dec 26, 2011 #10
    I don't have a book and nothing's coming up on Google. :(
     
  12. Dec 26, 2011 #11

    nvn

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    Hint 3: Regarding post 9, use plus and minus y, to obtain the two bending stresses.
     
  13. Dec 26, 2011 #12
    Are they both 30 mm because it is a symmetrical beam? But ones -ve and the other +ve?
     
  14. Dec 26, 2011 #13

    nvn

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    Yes.
     
  15. Dec 26, 2011 #14
    @123321,

    A few tips:

    1. Draw the section out, identify its Neutral Axis & Centroid. A neutral axis is the plane or axis through a section at which zero strain occurs. (At least in Euler Beam Theory, which you are dealing with at the moment.)

    2. Grab a ruler or something and bend it. Note the behaviour. What happens at the top and what happens at the bottom? To make it easier for you to see it, place two dots about 2 cm apart from each other on the top and bottom. Apply the bending. What happens to the dots? This should tell you what stresses are in that portion of the "beam."

    3. The symbol y in that equation (which is known as the bending equation) is a 'distance' variable. It starts at the neutral axis of the beam, where it is zero and 'travels' to where you want to investigate the stresses above or below the neutral axis.
     
  16. Dec 27, 2011 #15
    That rings a bell.

    Thanks
     
  17. Dec 27, 2011 #16
    So I've found the max tensile and compressive stresses to be ±54.3 MPa

    Any clues as to which one is +ve and which one is -ve?
     
  18. Dec 27, 2011 #17
    That is purely your choice, just remember which one you made positive and which one you made negative. However, by convention tensile stresses are usually taken as +ve.

    Just remember stress is taken as a vector (though this is not 100% correct) and therefore it has a magnitude and direction. (σ = F/A, the bold symbols are vectors.)
     
  19. Dec 27, 2011 #18
    OK, thanks

    Do you know how I should deal with the axial force? Someone previously mentioned that I should deal with the compressive/tensile stress and axial force separately and then combine at the end. I'm confused by this suggestion.
     
  20. Dec 27, 2011 #19
    Oh I know how to deal with it, but you need to figure that one out. ;-)

    Here are a couple of questions that you should ask yourself:

    Q: What types of forces do you get? (Hint: There are 3, but 2 have the word "force" attached to them where the other doesn't.)

    Q: What type of stresses do these forces produce? (Hint: There are only 2 types of stresses)

    Q: What are the units of stress? (Hint: What is MPa [Megapascals] in SI units? This one should answer your question on how to deal with the axial force.)

    This is correct. Stresses are vectors and therefore vector algebra applies.
     
    Last edited: Dec 27, 2011
  21. Dec 27, 2011 #20
    1. Normal force, shear force and torsion

    2. Normal and shear stress

    3. MPa = MN/m^2 so do I divide axial force by the cross-sectional? Then how do I combine it with the tensile/compressive stresses?
     
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