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Question (Mechanics of Materials)

  1. Oct 10, 2008 #1
    Hi guys, I am attaching a pdf file with the respective question it might be silly for many of you, but I would rather this question now, and move on with the problem.
    The pdf file is from a class from mechanics of materials, I understand the procedure and how formulas are plugged and how the problem is solved, but I don't understand how the distributed load (150 kN/m changed to 90 kN) the instructor at class didn't want to answer and evade the question. (I circled the slide with the doubt about the load)
    Your answer will be very helpful to clarify my basic knowledge of this class.
    Thanks in advance
    keep the great work

    Attached Files:

  2. jcsd
  3. Oct 10, 2008 #2


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    The distributed load of 150kN/meter over the 0.600m length of the rigid bar yields a total resultant load of (150kN/m)*(0.600m) = 90kN. Thus, the distributed load of 150 kN/m over the 0.6m length is represented as a concentrated load of 90kN at the centroid of the rigid bar.
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