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Mechanics of material people Stresses

  1. Dec 29, 2012 #1
    In this question below in the 2 files, i have a problem with the ( Q ) in the shearing stress.. can someone help me and explain for me how they get [Q=(32)(30)(15)] for point (a) and how about point (b) also !!?! what i know that Q=A.y , how did they get (32)(30) for the area and (15) for the height !!


    Last edited: Dec 29, 2012
  2. jcsd
  3. Dec 30, 2012 #2
    I can see why it's confusing start with point b first and work backwards. Keep in mind we're splitting the beam up into sections and then using the centroid of those new sections as our points.

    The distance from point b to the neutral axis of the entire shape is 22.5. 22.5 is the centroid of a rectangle that starts at 15 and ends at 30. The 32 is a cross sectional width common to both points a and b.

    In the case of point a, the distance to the neutral axis of the entire shape is 0 using the above method which obviously does not work. So the distance to edge (30) serves the same purpose as the 22.5 above.

    Since we know point a is where the max shear is, you can always use (for rectangles) TAUmax= 3V/2A to confirm.

    That's derived from (for rectangles):




    Check out :http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec=04.3&page=theory


  4. Dec 31, 2012 #3
    oh thank you this was really helpful.. but now i'm facing a big problem which is how to take the section for a point in a beam for calculating the shearing stress [TAU=V.Q/Ib] or shear flow [q=V.Q/I] .. what is the points that i have to follow to know the section that i have to take for calculating the Area.. the Q.. etc.

    check those files to see some problems that i have a problem with taking the section for every point:
  5. Dec 31, 2012 #4
    Hi wmsaqqa,

    I may be able to clear up some of your confusion. You posted a lot of problems, so let me just focus on one of them: Problem 6.35, part b.

    It is hard to see in your figure, but I believe that the question (6.35, part b) is similar to my example here: http://utsv.net/mm_shear_flow.html

    The moment of inertia is always the total moment of inertia about the neutral axis (perpendicular to the load), no matter where you are calculating the shear. This is why the moment of inertia, I, is the same for #6.35a and #6.35b.

    When calculating the first moment, Q, you always multiply the area of interest by the distance to the neutral axis, even if you are not interested in the shear at the neutral axis. Look at the area of interest in my problem in the above link. Now, use the dimensions given in your problem #6.35, and you should be able to see that Q=[area of interest]*distance to neutral axis, which is Q=[56mm*6mm]*37mm.

    The shear flow would be q=VQ/I, exactly as I show in my link (http://utsv.net/mm_shear_flow.html). They ask for the shear stress in #6.35b, which is the shear flow divided by the total distance within the cross section that the shear flow is acting over. I'm sorry, but that is the best way I can put it in words. Hopefully you can conceptualize what I mean.

    This thickness, [tex]t_b[/tex], is 2 times the thickness of the top member in your "box beam," as it would be in my example in my link if I were to calculate shear stress. So, [tex]t_b[/tex]=2*6mm=12mm. Note that this result (12mm) has nothing to do with the thickness of the side members of the "box beam," which is also 12mm, by unfortunate coincidence.

    You should be able to arrive at the desired solution now.

    Does that make sense?
  6. Jan 1, 2013 #5
    Ahh yes.. Thanks alot this was really helpful and enough and thanks again for giving me some of your time.
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