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Mechanical Principles -- Assignment question

  1. Dec 6, 2016 #1
    • Thread moved from the technical forums, so no Homework Help Template is shown.
    Can anyone help me? I have a question with 3 part answers on my Mechanical Principles assignment. I believe I have answered question part 'a' and 'b' but I have been stuck on question part 'c' for days!!
    So the question is:

    Q5. The simply supported beam shown in FIG 4 is 5meters long with a Young’s Modulus of 210GN m-2 The cross section of the beam is as shown in FIG 5 Assignment Fig4 and Fig5.png .


    (a) Draw the shear force diagram for the beam
    Solution:

    Shear force diagram MP1.png

    (b) Determine the position and magnitude of the maximum bending moment.
    Solution:
    Giving the moments at 1 meter intervals..
    Moments.png

    (c) Plot a graph of deflection along the length of the beam (calculate the deflection at 1m intervals).
    Solution so far..

    I understand as it has a U.D.L over a 4 meter spread we can focus the weight at its half way point,
    4 * 10 kN = 40kN at 2m(half way) from R1.
    As there already is a load at 2meters from R1 of 10 meters, This should increase the total load to 50 kN?

    The Equations I believe are relevant to this are M/I = E/R where:
    M= bending moment
    I = The Second Moment of Inetria
    E = The Youngs Modulus (210GN m-2)
    R = Radius of curvature of the beam

    M/EI = d2y/dx2

    I = WL3/3EYmax

    I think I need to work this out using Macauley's method, and it looks difficult. Can anyone advise on nice simple formula to use for each meter interval of deflection?
     
  2. jcsd
  3. Dec 7, 2016 #2

    haruspex

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    Is that adequate? The peak moment might be somewhere else.
    For part c, if you have an equation for d2y/dx2, valid over some range, how do you find the change in dy/dx over that range?
     
  4. Dec 7, 2016 #3
    Thats the question I have been set. adequate or not I am required to give the answer in 1 meter intervals. it looks like a long method of working out is required. I will spend some time on it this afternoon and see if I can get anywhere
     
  5. Dec 7, 2016 #4
    Ahh I see, You are suggesting my answer in part 'b' is not adequate? not the question set in part 'c' I apologise, Maybe I should go back and work out the bending moment at a smaller interval between 3 meters and 3.5 meters. That way I would have a more accurate answer.
     
  6. Dec 7, 2016 #5
    New Moments.png
    With that bit sorted, I am still stuck on question part C
     
  7. Dec 7, 2016 #6

    haruspex

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    You have the right answer to b now, but that is not the best way. For each interval (i.e. the span between two points where something in the loading changes suddenly) you should write an agebraic expression for how the moment varies across it. You can then differentiate them to find the peak moment in each interval. That way you do not have to guess how finely to cut it up.

    That method becomes vital in answering part c. You did not try to answer my question. If you know the slope of a curve (dy/dx) at all points over some interval, how do you calculate the change in y across the interval?
     
  8. Dec 8, 2016 #7
    Thanks for you help so far, I was working on part 'c' yesterday but I didnt get round to posting my work so far. So here it is: Macaulays method.png
    So I believe the equations above would provide me with the correct answers if I input all the known measurements into them?

    I can only use the information I am provided in my current and previous lessons to figure out the answers to the set questions. Unless it is an oversite on my part, I am not aware of an equation given in these lessons, which would allow me to obtain the answer to question part 'b' in a different way. My process of elimination method was used in a previous assignment and I have just transfered the method to suit this situation. Could you advise of a better way of solving part 'b' as I have already demonstrated an understanding of the subject and provided a correct answer?

    Thanks
     

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  9. Dec 8, 2016 #8

    haruspex

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    That's the right approach, but there are several things I do not understand in your working.
    Where exactly is x measured from? Is your first equation only valid for a point in a specific interval?
    Having x-a and x-b looks odd. I would have expected a + sign in one of them.
    What about the distributed load? I don't see that in the equation.
     
  10. Dec 8, 2016 #9
    Ok, Today I have spend alot of time going over and over Macaulays method and I believe I have come up with the answer which I can represent in the 1 meter intervals as required. Hopefully you will be able to see all the information in my working out for this to make sense?
    Macaulays method working out pt 1.png Macaulays method working out pt2.png
    So to continue this method I would simply substitute the value of x for 4,3,2 and 1 which should give me the results in 1 meter intervals as required?

    Thanks
     
  11. Dec 8, 2016 #10

    haruspex

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    That's a lot better, but you have not handled the distributed load correctly. See https://www.codecogs.com/library/engineering/materials/beams/macaulay-method.php.
    Also, I trust you realise that when evaluating these expressions for some value of x, any term for which the bracket (x-ai) is negative should be ignored.
     
  12. Dec 12, 2016 #11
    Thanks for all your help on this one. I will get round to re-calculating my answer at some point today. Its hard juggling a full time job and looking after my 1year old whilst taking on a HNC!

    I will post my workings later on today hopefully.
     
  13. Dec 12, 2016 #12
    Dealing with UDL.png
    Does this look like the correct method to follow, pre- integration of course?

    or would it be the better to deal with each as a seperate equation and subtract the results at the end?
     
  14. Dec 12, 2016 #13

    haruspex

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    Yes, that all looks right. I would not have put the R1 term in the distributed component, but you avoided writing 2R1 when you combined them, so that's ok.
     
  15. Dec 13, 2016 #14
    Can you expand on your comment, "I would not have put the R_1 term in the distributed component"?
    Is it the way I have expressed the value in an algebraic form 'R_1' as simply 'R' would surfice or do I need to change my calculation method in a way which does not include the 33 kN which the R_1 term is representing?
     
  16. Dec 13, 2016 #15

    haruspex

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    Perhaps I misunderstood your approach. I guess you treated it as two different problems, one with point loads and one with a distributed load, but each having an end support. So when you merged them you still had just the one instance of R.
    Previously I read it as one equation that included all point forces (including end support) and one for all distributed (which would not include R), then directly adding them.

    Same result.
     
  17. Dec 13, 2016 #16
    So the equation solution above would be the correct method to follow?

    When I collect all the results I will post them and plot a graph as required
     
  18. Dec 13, 2016 #17

    haruspex

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    Yes, it should work. Just bear in mind what I wrote at the end of post #10
     
  19. Dec 14, 2016 #18
    I have come up against another problem. In post 13 I gave an equation to deal with the UDL load and algibraically expressed the value of (x-a) and (x-b).

    Then combining the two equations with the method for dealing with a point load, which also contains an algebraic expression of (x-a) and (x-b). However, as x = distance between 0 and 5 suits both equations, the relevancy of a and b in the U.D.L equation become unclear.

    Are these values (a and b) in the UDL equation the start and finish point of the UDL, so 0-4. As they represent the distance from R to the first and second point load in the second equation therefore should be expressed as a different algebraic term. Maybe c and d for example?
     
  20. Dec 14, 2016 #19

    haruspex

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    Yes, sorry, I omitted to check you had the bounds right for the UDL. The two constants in the UDL terms should be the offsets from 0 (left hand end) to the start and end points of the UDL. So, 0 and 4, not 2 and 4.
     
  21. Dec 16, 2016 #20
    ok for my full revised workings including integration of the two combined formula, Whats do you think to my solution?
    Final attempt 1.png final attempt 2.png
     
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