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Shear and moment diagram for the beam

  1. Jan 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Draw the shear and moment diagram for the beam

    2. Relevant equations

    ∑Fy=0 ∑M=0; ΔV=∫w(x)dx ΔM=∫V dx

    3. The attempt at a solution I am firstly finding the external forces as any statics problem(Ay=13.5kN upwards, By=13.5kN upwards).Then I found the equations of the lines for the two loading forces.The first one is w(x)=-3x
    and the second one w(x)=3x-18.To find the V, I am integrating these equations and from what I get I drew the graph of V,of course jumping 13.5kN upwards in both sides.For V(x) I get 13.5-(3x^2)/2 for the first loading and (3x^2)/2-18x for the second, but when drawing the moment diagram for the second part at x=6 I do not get moment zero as it should be, my equation there is x^3/2-9x^2.What have I done wrong?
     

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  3. Jan 5, 2014 #2

    nvn

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    bigu01: Your V1(x) equation, for the first half of the beam, is correct. But your V2(x), for the second half of the beam, currently appears to be incorrect. Try plugging in some x values into your V2(x) equation, such as x = 6 m, and I think you might see it is currently incorrect. Try again.

    (1) By the way, always leave a space between a numeric value and its following unit symbol. E.g., 13.5 kN, not 13.5kN. See the international standard for writing units (ISO 31-0).

    (2) Always leave one space after the period (.), question mark (?), or exclamation mark (!) at the end of each sentence.
     
    Last edited: Jan 5, 2014
  4. Jan 5, 2014 #3

    SteamKing

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    Even better, leave two spaces between sentences. Also, when using commas (,) to separate items or clauses within a sentence, leave a space after the comma; ditto for semicolons.
     
  5. Jan 5, 2014 #4

    nvn

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    SteamKing: Thanks for your comment. Using a single space character between sentences is correct. Using two space characters between sentences is actually incorrect, except perhaps in a few cases using monospaced font (but even that has pretty much been phased out). And monospaced font is rarely used these days, for paragraph text.

    From wikipedia:

    "[A single space character between sentences] is the current convention in most countries that use the ISO basic Latin alphabet for published and final written work, as well as digital media. [Two space characters between sentences] ... was gradually replaced by the single space convention in published print ... and ... digital media."​

    From wikipedia:

    "From around 1950, [a single space character between sentences] became standard in books, magazines, and newspapers ... Most style guides indicate that [a single space character between sentences] is proper for final or published work today, and most publishers require manuscripts to be submitted as they will appear in publication -- single sentence spaced. Writing sources typically recommend that prospective authors remove extra spaces before submitting manuscripts ..."​

    From Manual for Writers and Editors, Merriam-Webster, 1998, p. 274:

    "A single space [character] should follow [end-of-sentence] periods and other end punctuation."​

    From Chicago Manual of Style:

    "One space [character], not two, should be used between two sentences."​

    Virtually all authoritative language style guides state there should be only one space character between sentences. E.g., from USA GPO Style Manual, 2008:

    "A single ... space [character] will be used between sentences. This applies to all types of composition."​


    Agreed. Good point.
     
    Last edited: Jan 6, 2014
  6. Jan 6, 2014 #5
    I am plugging for the second value, and I can see that it is wrong since the momentum is not going to zero at the end. I had this problem with other questions also, if there is discontinuity or any shifting I am not getting the momentum to zero by my formulas, however calculating the area I am getting correct results. Hopefully I used your suggestions correctly when writing this reply.
     
  7. Jan 6, 2014 #6

    nvn

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    bigu01: If you want to show, step by step, how you derived your V2(x), then we might be able to check your math and see where the current mistake occurred. If you plug x = 3 m into your V1(x) and V2(x), do they give the same value?
     
  8. Jan 6, 2014 #7
    Okay. First of all, I am finding the line equation for the second load, which is y = 3x - 18, knowin that V(X) is the integral of it, I am integrating and geting V2(x)= 1.5x^2 - 18x ( from 3 to 6 ). At this point for x= 6 m , I should get a -13.5 kN, since I have a 13.5 kN upwards at x=6m, but I am not getting that, knowing that my V2(X) is wrong I am not getting a right result for M2(x) too.
     
  9. Jan 6, 2014 #8

    nvn

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    bigu01: Hint 1: Integrating w2(x) = 3*x - 18 gives, V2(x) = 1.5*x^2 - 18*x + C2, not 1.5*x^2 - 18*x. Remember? Can you solve for C2, using a boundary condition? Try again.
     
  10. Jan 6, 2014 #9
    Okay. I got as C2= 40.5 , then I integrated again to get the M2(x) to zero at x = 6 m, but still was not getting the zero, re-reading your reply I remembered that I should add another constant and solve for zero. So my last equation became , M2(x) = 0.5*x^3 - 9*x^2 + 40.5*x - 27. I believe not adding the constants was causing me problems with other questions too. Thank you very much.
     
  11. Jan 6, 2014 #10

    nvn

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    Excellent work, bigu01. Your answer is correct.
     
    Last edited: Jan 6, 2014
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