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Cantilever beam point load analysis with diagram

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Require various values from cantilever beam analysis including
    Bending moment
    K stiffness
    2nd moment of inertia
    Max stress

    2. Relevant equations
    Shown below


    3. The attempt at a solution
    Here is my attempt at this question but I am unsure about units and therefore my answers?


    Bending moment

    Image attached below

    Calculated as follows:
    L=0.24M
    L2=0.02M
    P=-0.0049KN

    FX=0
    FY=0

    RAx=0 (No horizontal forces)
    +Ray-0.0049=0

    M=0
    - (0.0049) (0.25) +Ma=0
    Ma=1.176x10-3

    Second moment on inertia

    Ix=bh^3/12 = (30) (260) ^3/12 = 43.94x10^6 (439.4mm^4)


    Neutral axis (Y)

    X axis = 6mm hence Y=3mm

    Flexural formula

    To find max stress
    M/I = Sigma/Y
    M=1.176x10-3
    I=43.94x10^6
    y=3mm
    Stress= My/I (flexural formula)
    Stress = (1.176x10-3) (0.003) = 8.04x10^-14
    (43.94x10^6)

    K stiffness

    K=3EI = (3)(190x10^9)(43.94x10^6)/0.26^3 = 1.425x10^13
    l^3

    Young’s modulus formula

    E= f
    e
    Ee=f
    Strain= Stress/Young’s modulus
    Strain= 6.68/190x10^9= 3.51


    Stress

    Stress = W/Z (l-x)

    Stress = 4.9/14.64x10-3 (.26 -.24) = 6.69N

    E= stress/strain
    E=190x10^9
    Strain = Stress/E = 6.69x190x10^9 = 3.52x10^-11
    Stress = E x Strain = 190x10^9 x 3.52 x 10^-4 = 6.68

    Any help is greatly appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

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    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 21, 2013 #2

    SteamKing

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    I see a bunch of numbers which may or may not be correct.

    Without an intelligible problem statement, who knows what you have done?
     
  4. Feb 21, 2013 #3
    Im essentially trying to gain a value for the maximum stress recorded from a cantilever beam so that this can be compared to that of an FEA model. However to do this I need to obtain values for bending moment, stiffness, strain etc beforehand.

    I have no direct question for this problem other than are the values for these problems right with the information that I have provided?

    Once I know they are right I can compare the max stress value here to the values from the rig using voltage to strain conversion

    This of any help?
     
  5. Feb 21, 2013 #4

    SteamKing

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    All right, here goes:

    1. L of the beam = 0.24m Why is your moment = P * 0.25m?

    2. Neutral axis (Y): X axis = 6mm hence Y=3mm
    What does this mean?

    3. Second moment on inertia

    Ix=bh^3/12 = (30) (260) ^3/12 = 43.94x10^6 (439.4mm^4)

    It is technically the second moment of area (or moment of inertia)

    The dimensions of your beam cross section are not given in your problem statement
    It appears you are saying the beam is 260 mm deep by 30 mm wide.
    This beam has very unusual proportions, in that its width is approximately equal to its length.

    All of your other calculations depend on the accuracy of the results above.
     
  6. Feb 21, 2013 #5
    Its point load so its 0.24m from fixed end to load (P) then 0.02 after load, giving a total length of 0.26m. The load at 0.24m is 4.9N

    Y from the bending formula is distance from neutral fibres to base and so neutral axis is 3mm as total thickness is 6mm.

    Second moment of area ive now got as 4.394x10-5 m4

    dimensions of beam are as follows
    b=30mm
    h=260mm
    so its a long thin rectangular beam as usual

    thanks for your help
     
  7. Feb 21, 2013 #6

    SteamKing

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    Let's try this again:

    M = 4.9N * 0.24m = 1.176 N-m

    If the beam is 6 mm deep and 30 mm wide, then

    I = (1/12)*(30)*(6^3) = 540 mm^4

    Since 1 m = 1000 mm, then 1 m^4 = (1000 mm)^4 = 10^12 mm^4

    so I = 540 /10^12 = 5.4 * 10^-10 mm^4
     
  8. Feb 21, 2013 #7

    SteamKing

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    Check the I calc:

    I = 540 mm^4 = 5.4 * 10^-10 m^4
     
  9. Feb 21, 2013 #8
    I understood for a rectangular beam

    Ix = b*h3 / 12

    where

    b = width

    h = height

    its not the thickness but the height of the beam that is entered into the formula?
     
  10. Feb 21, 2013 #9

    SteamKing

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    The moment of inertia is calculated for the cross section of the beam.
    The cross section is taken transversely, not longitudinally.

    Your initial calculation had b = 30 mm and h = 260 mm, yet you said the NA was 3 mm from the outer fiber of the beam because the depth = 6 mm
     
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