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Simply supported loaded beams

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data
    The simply supported beam shown is 10 m long with
    E = 200 × 109 Pa and I = 150 × 10–6 m4.

    Capture.PNG


    1. Determine the position and magnitude of the maximum bending
    moment.


    2. Plot a graph of the deflection along the length of the beam
    (calculate the deflection at 1 m intervals).


    3. What I value for the beam would be required to halve the maximum
    deflection of the beam?


    4. The calculation for the maximum bending moment is to be verified
    experimentally using a strain gauge bonded to the outer surface of the
    beam, at the point where the maximum bending moment occurs.
    Derive an equation which could be used to calculate the bending
    moment from the measured strain value. State the meaning of all
    symbols used in your equation.

    2. Relevant equations

    M/I = σ/y = E/R

    d2y / dx2 = M/EI


    can anyone advise on where i should start?


    Thanks.
     
  2. jcsd
  3. Oct 23, 2013 #2

    SteamKing

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    1. Set up a free body diagram for the beam.
    2. Solve for the unknown support reactions.
    3. Draw the shear diagram for the beam.
    4. Use the shear diagram to construct the bending moment diagram.
    5. Once the moment diagram is done, make the M/I diagram.
    6. Integrate the M/I diagram twice, making sure you keep the constants of integration.
    7. After doing the integrations, apply the boundary conditions for the beam to determine the constants of integration.
    8. Adjust the last integration to calculate the deflections of the beam.
     
  4. Oct 24, 2013 #3
    Thanks for your response.

    Is the free body diagram not the same as the attached diagram?
     
  5. Oct 24, 2013 #4

    SteamKing

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  6. Oct 25, 2013 #5
    Ok I think i have calculated the unknown support reactions. does this look correct?

    (3 * 20) + (7 * 30) + (5 * 10 *5) = r2 * 10

    520 = r2 * 10

    r2 = 52kN

    upward forces=downward forces

    r1 + 52 = 20 + 30 + (5 * 10)

    r1 = 48kN
     
  7. Oct 25, 2013 #6

    SteamKing

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    Looks good so far.
     
  8. Oct 26, 2013 #7
    oxon88, you didn't have to ask us if it was correct. You could check it yourself by taking moments about any axis other than the one you have already used going through R1.
     
  9. Oct 29, 2013 #8
    Thanks. I have drawn the shear force diagram and also calculated the bending moments (below).


    x=0 = 0kN-m
    x=1 = (48*1)-(5*1*0.5) = 45.5kN-m
    x=2 = (48*2)-(5*2*1) = 86kN-m
    x=3 = (48*3)-(5*3*1.5) = 121.5kN-m
    x=4 = (48*4)-(20*1)-(5*4*2) = 132kN-m
    x=5 = (48*5)-(20*2)-(5*5*2.5) = 137.5kN-m
    x=6 = (48*6)-(20*3)-(5*6*3) = 138kN-m
    x=7 = (48*7)-(20*4)-(5*7*3.5) = 133.5kN-m
    x=8 = (48*8)-(20*5)-(30*1)-(5*8*4) = 94kN-m
    x=9 = (48*9)-(20*6)-(30*2)-(5*9*4.5) = 49.5kN-m
    x=10 = (48*10)-(20*7)-(30*3)-(5*10*5) = 0kN-m
     
    Last edited: Oct 29, 2013
  10. Oct 29, 2013 #9

    SteamKing

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    I'm not sure what you are doing in Post #8.

    At x = 0, the shear force diagram has a value equal to R1, or 48 kN. The bending moment at x = 0 is equal to 0 kN-m (Note the units).

    Can you clarify?
     
  11. Oct 29, 2013 #10
    post #8 are my calculations for the bending moments along the beam. I have amended the units.
     
  12. Oct 29, 2013 #11

    SteamKing

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    The bending moments look good. Proceed with answering the questions in the OP.
     
  13. Oct 30, 2013 #12
    thanks. Please can you explain what the M/I diagram is?
     
    Last edited: Oct 30, 2013
  14. Oct 30, 2013 #13

    SteamKing

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    Just what it says. Take the ordinates of your bending moment diagram and divide by the moment of inertia of the beam (in this case I = 150 * 10^-6 m^4 according to the OP). It's not really required for this problem, since I is a constant, but when I changes along the length of the beam, an M/I diagram is useful when calculating deflection.
     
  15. Oct 30, 2013 #14
    x=0 = 0kN-m
    x=1 = 45.5kN-m / 150 * 10^-6 m^4 = 303333.3
    x=2 = 86kN-m / 150 * 10^-6 m^4 = 573333.3
    x=3 = 121.5kN-m / 150 * 10^-6 m^4 = 810000
    x=4 = 132kN-m / 150 * 10^-6 m^4 = 880000
    x=5 = 137.5kN-m / 150 * 10^-6 m^4 = 916666.7
    x=6 = 138kN-m / 150 * 10^-6 m^4 = 920000
    x=7 = 133.5kN-m / 150 * 10^-6 m^4 = 890000
    x=8 = 94kN-m / 150 * 10^-6 m^4 = 626666.7
    x=9 = 49.5kN-m / 150 * 10^-6 m^4 = 330000
    x=10 = 0kN-m


    so would the diagram be as pictured?

    Capture.PNG
     
  16. Oct 30, 2013 #15

    SteamKing

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    Looks OK. Make sure you use the proper units.
     
  17. Oct 31, 2013 #16
    ok Thankyou. What units should be used? Can you provide some guidance on how to integrate the M/I diagram twice?
     
  18. Oct 31, 2013 #17

    SteamKing

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    Obviously, the units which result from dividing bending moment M by moment of inertia I.

    The integration of the M/I diagram proceeds just like any other numerical integration: divide the x-axis into regular spacings and determine the M/I ordinates. You can use a trapezoidal rule to calculate the actual values of the integrals.

    Remember, you must include the constants of integration in your calculation. There will be two unknown constants which are determined by applying the support conditions for your beam. You can start the integration by assuming the value of the integrals is zero at x = 0.

    Since I is constant, the slope [itex]\vartheta[/itex] = (1/EI) ∫ M dx + C1
    The deflection δ = (1/EI)∫∫ M dx + C1*x + C2

    The final values for slope and deflection are determined by applying these constants of integration to your calculations of the integral values at the various x-locations.
     
  19. Nov 6, 2013 #18
    am i not able to use this formula to solve y?

    Capture.PNG
     
  20. Nov 6, 2013 #19

    SteamKing

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    Where did you get this formula? It looks like only a point load P is present. What about deflection due to the distributed load?
     
  21. Nov 7, 2013 #20
    Our learning material gives it as an example. Yes it states that the formula is used for simply supported beams at ends with concentrated load P at any point.

    Would it not be posssible to calculate the deflection for each of the loads and then sum the answers?

    For the UDL, could this formula not be used?

    Capture.PNG
     
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