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Please Can anyone help me on this beam Problem.

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to know the correct equations for a pinned-pinned beam and a fixed-fixed beam with 3 point loads(450N, 0.3 from the left,450 0.7 from left and 450 1.1 from the left, the beam is 1.4m long, with maximum deflection and bending stresses
    But I have to use the second moment of area for a hollow shaft(100mm diameter, 6mm thickness) which i have as 107.6.(I=pi(100^4-94^4/64) and a youngs modulus of 206e9

    2. Relevant equations
    I am sure you have to work out each point load seperately and add them together.
    But at the moment i have, 3pa^3b^2/3EI(3a+b)^2 for a cantilever fixed at both ends.

    [b3. The attempt at a solution[/b]
    The answers I have for a cantilver fixed at both endsare 1.66x10^-13, 1.534x10-12 and 1.607x10-12, so adding them together i have 3.307x0-12 for deflection, not sure if this is right.
     
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  3. Oct 4, 2013 #2

    mfb

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    Please post the full problem statement, including a sketch if one is given. It is hard to figure out what you have/want to do just based on your description.

    Your answers should have some units.
     
  4. Oct 4, 2013 #3
    Sorry, I have to model a thin walled shaft with 2 bearings, one at each end, as a simply supported beam, but have to work out the maximum deflection for pinned and fixed at both ends.
    The shaft is 100mm outside diameter, with a 6mm thickness of the wall.
    The bearings I have to model as both pinned and both fixed( cantilever), i have worked out the cantilever, but not sure if i have the right equation for maximum deflection.
    I have 3 point loads in the drawing, so i have worked out 3 pinned at both ends and 3 fixed at both ends, as there is 3 different point loads.
    I have the second moment of area of a hollow shaft as I=pi(D^4-d^4)/64 which I have 1.76X10^6mm^4,E is 206GPa and like in the drawing below the forces are 450N.
    My 3 answers for the deflection for a beam fixed at both ends are:
    1.66x10^-13mm, 1.534x10-12mm and 1.607x10-12mm,added together this gives me a deflection of 3.307x0-12mm.
     

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  5. Oct 4, 2013 #4

    SteamKing

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    I would suggest that you check your units. You have a toxic stew of newtons, meters, GPa, and mm^4 running around. It's easy to get them mixed up. Your values of deflection look really low if they are in mm.
     
  6. Oct 4, 2013 #5

    nvn

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    stevo1978: (1) Use all consistent units (N, mm, MPa). Therefore, first convert all of your dimensions to mm, and convert your tensile modulus of elasticity to E = 206 000 MPa.

    (2) Always leave a space between a numeric value and its following unit symbol. E.g., 100 mm, not 100mm. See the international standard for writing units (ISO 31-0).

    (3) You said your wall thickness is t = 6 mm. Therefore, your shaft inside diameter would be 100 - 2*t = 88 mm, not 94 mm. Therefore, your second moment of area is currently wrong. Try again.

    (4) You do not have a cantilever fixed at both ends. Therefore, your deflection formula currently looks wrong. You need deflection formulas for a fixed-end beam. Try again.

    (5) Hint 1: For the fixed-end beam, if you do everything right, I currently get a total midspan deflection of y = 6.253 + 15.888 + 6.253 = 28.39 micrometers. See if you can match this answer.
     
  7. Oct 5, 2013 #6
    Thanks for your help nvn, I really appreciate it, and will have a good go at the deflections, but just want to mention GPa is a SI unit and length the standard SI unit is the metre, just like Density is the kg/m^3 and Force is the Newton.
    You have helped me lots in the deflections, and I thank you.
     
  8. Oct 5, 2013 #7

    mfb

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    Pa (Pascal) is the unit. GPa is the unit multiplied by 10^9.
    SI-units are designed to have conversion factors of 1 where possible, so Pa=1N/m^2 = 1kg/(ms^2).
     
  9. Oct 5, 2013 #8
    Thanks for that mfb, when working out a deflection of a beam then, what is the x, is it the span of the load?
    If you have a formula of F*b^2*X^2/6EI*L^3(3*a+b)-(L*a) for a fixed beam, I have everything except x. and what is the deflection formula for a pinned-pinned beam?
    Also if nvb said its 88 mm not 94 mm(thickness of 6mm on a hollow cylinder).
    The second moment of area of a holow cylinder is I=pi(100^4-88^4)/64 is this the right formula?
    i much appreciate everybodies help on this.
     
  10. Oct 5, 2013 #9

    SteamKing

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    Formulas for beam deflections can be found in most structural handbooks, and their may be a table in your textbook. The formulas can also be googled.
     
  11. Oct 5, 2013 #10

    nvn

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    (6) Uppercase Pascal is a man, whereas pascal (Pa) is a unit of pressure or stress. Always use correct capitalization and spelling of units. Please see the links I provided in item 2 of post 5.

    (7) Uppercase Newton is a man, whereas newton (N) is a unit of force.

    (8) ms means millisecond, whereas m·s means metre second. Two unit symbols multiplied together must be separated by an asterisk, a raised dot, or a space. Please see the links in item 2 of post 5.

    (9) Always leave a space between a numeric value and its following unit symbol. E.g., 1 kg/(m·s^2), not 1kg/(m·s^2). Please see the links I provided in item 2 of post 5.

    stevo1978: x is the horizontal coordinate (position) along the beam, usually measured from the left-hand end. E.g., your first 450 N applied load is located at x = 300 mm. Your second moment of area formula in post 8 looks correct.

    You can use either of the following sets of consistent units.

    (1) newton (N), meter (m), second (s), pascal (Pa), kilogram (kg), joule (J).
    (2) newton (N), millimeter (mm), second (s), megapascal (MPa), tonne (t), millijoule (mJ).

    Item 1 is standard SI. Item 2 is often my favorite. If in doubt, use item 1.

    Therefore, convert everything to either item 1 (N, m, Pa), or else convert everything to item 2 (N, mm, MPa). But do not combine item 1 with item 2.

    Do not combine any inconsistent units.
     
    Last edited: Oct 5, 2013
  12. Oct 5, 2013 #11

    nvn

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    stevo1978: Very good. Your formula is almost correct. It only has two (or more) typographic mistakes. It instead should be [F*(b^2)(x^2)/(6*E*I*L^3)]*{[(3*a + b)*x] - (3*L*a)}. Check your source again, to see if it matches this formula.
     
    Last edited: Oct 5, 2013
  13. Oct 6, 2013 #12
    Cheers nvn for the help and everybody else on the forum really appreciate it.
     
  14. Oct 6, 2013 #13
    About the First Force of 450 N being at x = 300mm, does this mean the 2nd Force is 700mm and the third 1.1 m at x?
     
  15. Oct 6, 2013 #14

    nvn

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    stevo1978: Yes, that is correct. Your applied loads are located at x = 300 mm, x = 700 mm, and x = 1100 mm.

    But keep in mind, x in the deflection formulas is the location where you want to compute deflection, such as at x = 700 mm; x in the deflection formulas is not necessarily where an applied load is applied.
     
    Last edited: Oct 6, 2013
  16. Oct 6, 2013 #15
    Thanks nvn, looking through the beam deflection calculations all i see is a pinned one end and roller the other end, i cannot see a pinned pinned deflection formula anywhere.
     
  17. Oct 6, 2013 #16

    nvn

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    stevo1978: Pinned at one end, and a roller at the other end, is essentially the same as pinned-pinned. Use pinned at one end, and a roller at the other end.
     
    Last edited: Oct 6, 2013
  18. Oct 6, 2013 #17
    Ok will do, thanks for your advice.
     
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