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Shear force diagrams simple stuff hey?

  1. Jan 6, 2007 #1
    shear force diagrams simple stuff hey!?

    hey guys this is probably very straight forward stuff for u but i have an example question on a loaded beam. there is a UDL, 2 pure couples and 2 fixed points.

    now how do i do the shear force and bendin moment diagram for this

    really lookin forward to your ideas thanks
     
  2. jcsd
  3. Jan 6, 2007 #2

    radou

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    A sketch of the idea would do some good.

    In general, the relation dM(x)/dx = T(x) can be very useful when constructing shear force diagram. (T represents the shear force, and M the bending moment.)

    Set up the equations of equilibrium for the beam at some point x to get the bending moment function M(x), after calculating the reactions.
     
  4. Jan 6, 2007 #3
    heres the pic of everything

    how do i include the couples in the diagrams i havent encountered these b4
     

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  5. Jan 6, 2007 #4

    Pyrrhus

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    First, what is the method you were taught to draw the diagram?

    Graphing the piecewise function and/or the area method?
     
  6. Jan 6, 2007 #5
    i think ud describe it as the graphing piecewise function although i havent heard it described as tht b4 ne im from the uk so not my fault :smile:

    basically this is what i use

    http://www2.umist.ac.uk/construction/intranet/teaching/ul222/exp/sfbmdex.htm

    is there a "better" way im sorta teachin myself this stuff really ne help would b hugely appreciated :smile: iv got another problem im stuck on but i wona give it a proper go b4 askin learn better tht way if u can solve something yourself
     
  7. Jan 6, 2007 #6

    Gokul43201

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    So, if you're looking at piecewise section from the left end, you would, as usual have M(x) = sum of reaction to moments from -4kN, R(Ay) and the distributed load, for 0<=x<8, and beyond that point, you also include the reactions to the pure couples, so that M(x) = all these contributions + couple at C, for 8<=x<18, and M(x) = all of previous contributions + couple at E, for x>=18
     
  8. Jan 6, 2007 #7
    thanks for such quick responses!!

    i think i need to practice these diagrams alot!! are there any good websites out there or books, i have a mechanics book but doesnt really touch on the diagram side of things i havent really managed to find alot on the web.

    so basically i jus add the couples to the other loads at C and E, how would this look visually?

    this is a pretty awesome forum i read tons of threads already learned alot!
    if you dont mind me asking what do all you guys do for a living??
     
  9. Jan 6, 2007 #8

    radou

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    Here are qualitative sketches of what your bending moment and shear force diagrams should look like. (The pictures look terrible, but I guess you'll live on it. :wink:)
     
  10. Jan 7, 2007 #9
    wow those 2 images are great, the shear force was exactly how i pictured it and the bending moment explained the couples

    thanks for all ur help
     
  11. Jan 7, 2007 #10
    Moments about a Clockwise:

    (10 x 5) + [(0.5 x 10) x 5] = (4x3) + (3x15) x (R/b x20)

    75 = 57+(R/b x 20)

    R/b= 18 / 20

    R/b = -0.9

    This doesnt seem right, what have I done wrong??

    to find reaction R/a using all Upward=Downward

    R/a + 4 + 3 = 5 + 10 + 0.9

    R/a = 8.9

    help
    :grumpy:
     
  12. Jan 7, 2007 #11

    radou

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    I'm too lazy to check your calculation, but I can demonstrate how it should look like. Let's set the sum of moments around point B to equal zero, to get the reaction Ra at point A (which is assumed to point 'upwards'). Assume clockwise moments are positive. We have:

    -3 + 10 - 0.5*10*15 + Ra*20 - 4*23 = 0.
     
  13. Jan 7, 2007 #12
    so ur asuming that Rb will equal 0? or is there another way of findin Rb

    this is real interesting really making me think logically about things, i love learning this stuff!
     
  14. Jan 7, 2007 #13

    radou

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    I'm not assuming Rb will equal 0. We only set the sum of moments about the point B to equal 0, right? So, the reaction Rb does not produce a moment at this point. Thus, we have one equation with one unknown, Ra, which we can solve easily.

    Further on, you can easily google-up some basic information on the construction of internal force diagrams. If there's something specific you wish to know about, I'll be glad to help.
     
  15. Jan 7, 2007 #14
    You wouldn't happen to be doing Dr. Badi's Mechanics courswork at the University of Hertfordshire would you, studentlife?

    :p
     
  16. Jan 7, 2007 #15
    this stuff is probably a little too late for me, but its all very useful, I'm done all those classes for now where I have to draw those damn shear force diagrams and bending moments
     
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