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Thread: Determine the center of gravity

  1. #1
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    Determine the center of gravity

    Hello everyone,

    I want to calculate a center of gravity of a profile.

    I have divided the profile into different sections. See attached image below.

    Determine the center of gravity-hoeklijn.png



    The static moment relative to y I determined in the following way:
    Determine the center of gravity-uitwerking-sy.png

    Zy= Sy total / Area = 10800 + 59136 + 366,16 - (45,77)^2 / 2874,13 = 24,42 mm

    But according to the documentation of the profile lies the center of gravity of y at 24,18 mm ( See attached image below )


    Determine the center of gravity-hoeklijn-volgens-gegevens.png


    I come up with wrong values, what's wrong with my elaboration? At the center of gravity compared to x I have the same story. Can someone help me ?
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  2. #2
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    Re: Determine the center of gravity

    Your equations for S3, S4 and S5 are incorrect. I'll help you with S3, then you can use the same technique to determine S4 and S5. Also, you should be aware that the values for S3 and S4 are not equal.

    The equation of the circle that defines S3 comes from:

     (x-x_0)^2 + (y-y_0)^2 = R^2

    where (x_0,y_0) is the coordinates of the center of the circle and R is the radius. So for S3, which has a radius of 13 mm, we have:

     (x-25)^2 + (y-25)^2 = 13^2

    After some rearranging:

     y = \pm \sqrt{13^2-(x-25)^2} + 25

    Now we need to determine whether the sign in front of the square root is plus or minus. If we let x = 25, we get y = \pm 13 + 25 We're interested in only the lower value, to yield a value of y(25) = 12, so we know that the sign in front of the square root should be negative.

    Finally, the height of the fillet is equal to h = y-12:

     h = -\sqrt {13^2-(x-25)^2} + 13

    The integral for S3 is then:

    S_3 = \int_{12} ^{25}  ( -\sqrt {13^2-(x-25)^2} + 13) x dx

    Notice how the limits for the integral are not 0 to 13 as you had it - I am keeping a consistent definition of the x- and y- coordinates here.

    Try this - see what you get for S3, then use the same technique for S4 and S5.
    Last edited by ChipB; Oct 4th 2017 at 06:48 AM.
    Thanks from westerwolde
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  3. #3
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    Re: Determine the center of gravity

    Consider the following function:

    $f(x) = \begin{cases}150 & 0 \le x \le 5.5 \\ 143.5 + \sqrt{12+11x-x^2} & 5.5 \le x \le 12 \\ 25-\sqrt{-x^2+50x-456} & 12 < x \le 25 \\ 12 & 25 \le x \le 93.5 \\ 5.5+\sqrt{-x^2+187x-8700} & 93.5 \le x \le 100\end{cases}$

    Now, the area is:

    $\displaystyle \int_0^{100} f(x)dx$

    Similarly, the $Z_y$ you are trying to calculate will be:

    $\displaystyle \dfrac{ \displaystyle \int_0^{100} xf(x) dx }{ \displaystyle \int_0^{100} f(x)dx }$

    With this method, I get 24.18, which agrees with the diagram listed.
    Last edited by SlipEternal; Oct 4th 2017 at 08:56 AM.
    Thanks from westerwolde
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  4. #4
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    Re: Determine the center of gravity

    Quote Originally Posted by ChipB View Post
    Your equations for S3, S4 and S5 are incorrect. I'll help you with S3, then you can use the same technique to determine S4 and S5. Also, you should be aware that the values for S3 and S4 are not equal.

    The equation of the circle that defines S3 comes from:

     (x-x_0)^2 + (y-y_0)^2 = R^2

    where (x_0,y_0) is the coordinates of the center of the circle and R is the radius. So for S3, which has a radius of 13 mm, we have:

     (x-25)^2 + (y-25)^2 = 13^2

    After some rearranging:

     y = \pm \sqrt{13^2-(x-25)^2} + 25

    Now we need to determine whether the sign in front of the square root is plus or minus. If we let x = 25, we get y = \pm 13 + 25 We're interested in only the lower value, to yield a value of y(25) = 12, so we know that the sign in front of the square root should be negative.

    Finally, the height of the fillet is equal to h = y-12:

     h = -\sqrt {13^2-(x-25)^2} + 13

    The integral for S3 is then:

    S_3 = \int_{12} ^{25}  ( -\sqrt {13^2-(x-25)^2} + 13) x dx

    Notice how the limits for the integral are not 0 to 13 as you had it - I am keeping a consistent definition of the x- and y- coordinates here.

    Try this - see what you get for S3, then use the same technique for S4 and S5.



    Thanks for your comment
    When I calculate s3, I get up : 540,53

    I've done s5 in the following way:


    ( x- 93,5)^2 + ( y -5,5)^2 = 6,5^2

    After some rearranging:

    y= +- √ (-x^2 + 187x- 8700 ) + 5.5

    h= + √ (-x^2 + 187x- 8700 ) + 12

    S_5 = \int_{5.5} ^{12}  ( +\sqrt {6.5^2-x^2+187x-8700} + 12) x dx

    Have I done this correctly?
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  5. #5
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    Re: Determine the center of gravity

    Quote Originally Posted by SlipEternal View Post
    Consider the following function:

    $f(x) = \begin{cases}150 & 0 \le x \le 5.5 \\ 143.5 + \sqrt{12+11x-x^2} & 5.5 \le x \le 12 \\ 25-\sqrt{-x^2+50x-456} & 12 < x \le 25 \\ 12 & 25 \le x \le 93.5 \\ 5.5+\sqrt{-x^2+187x-8700} & 93.5 \le x \le 100\end{cases}$

    Now, the area is:

    $\displaystyle \int_0^{100} f(x)dx$

    Similarly, the $Z_y$ you are trying to calculate will be:

    $\displaystyle \dfrac{ \displaystyle \int_0^{100} xf(x) dx }{ \displaystyle \int_0^{100} f(x)dx }$

    With this method, I get 24.18, which agrees with the diagram listed.


    Thanks for your comment! I'm going to work on this too
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  6. #6
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    Re: Determine the center of gravity

    Quote Originally Posted by westerwolde View Post
    Thanks for your comment
    When I calculate s3, I get up : 540,53

    I've done s5 in the following way:


    ( x- 93,5)^2 + ( y -5,5)^2 = 6,5^2

    After some rearranging:

    y= +- √ (-x^2 + 187x- 8700 ) + 5.5

    h= + √ (-x^2 + 187x- 8700 ) + 12

    S_5 = \int_{5.5} ^{12}  ( +\sqrt {6.5^2-x^2+187x-8700} + 12) x dx

    Have I done this correctly?


    Hey ChipB ;

    I did not get the correct values, so I've revisited my calculation.

    For S5, I now turn to: S_5 = \int_{5.5} ^{12}  ( +\sqrt {6.5^2-(x-25)^2} + 6.5) x dx


    But I do not yet finish at 24.18. Do you see, or someone else what I'm doing wrong?
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  7. #7
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    Re: Determine the center of gravity

    Quote Originally Posted by westerwolde View Post
    Hey ChipB ;

    I did not get the correct values, so I've revisited my calculation.

    For S5, I now turn to: S_5 = \int_{5.5} ^{12}  ( +\sqrt {6.5^2-(x-25)^2} + 6.5) x dx


    But I do not yet finish at 24.18. Do you see, or someone else what I'm doing wrong?
    Where did you get those numbers from? You are saying the center of the circle is at $(25,6.5)$? That does not seem right.
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  8. #8
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    Re: Determine the center of gravity

    Quote Originally Posted by SlipEternal View Post
    Where did you get those numbers from? You are saying the center of the circle is at $(25,6.5)$? That does not seem right.
    $S_5 = \displaystyle \int_{93.5}^{100}\left(5.5+\sqrt{6.5^2-(x-93.5)^2}\right) dx = \int_{93.5}^{100} \left(5.5+\sqrt{-x^2+187x-8700}\right)dx$
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  9. #9
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    Re: Determine the center of gravity

    Quote Originally Posted by SlipEternal View Post
    $S_5 = \displaystyle \int_{93.5}^{100}\left(5.5+\sqrt{6.5^2-(x-93.5)^2}\right) dx = \int_{93.5}^{100} \left(5.5+\sqrt{-x^2+187x-8700}\right)dx$
    Oops, for $S_5$, you are looking for the amount to remove, not the amount to keep. That would be:

    $S_5 = \displaystyle \int_{93.5}^{100}\left(6.5 - \sqrt{-x^2+187x-8700}\right) dx$
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  10. #10
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    Re: Determine the center of gravity

    Quote Originally Posted by SlipEternal View Post
    Where did you get those numbers from? You are saying the center of the circle is at $(25,6.5)$? That does not seem right.


    oops stupid, I had probably the values of S3 in my head...
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  11. #11
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    Re: Determine the center of gravity

    Quote Originally Posted by SlipEternal View Post
    Oops, for $S_5$, you are looking for the amount to remove, not the amount to keep. That would be:

    $S_5 = \displaystyle \int_{93.5}^{100}\left(6.5 - \sqrt{-x^2+187x-8700}\right) dx$


    SlipEternal thanks for your help.


    S_5 = \int_{93.5} ^{100}  ( 6,5- \sqrt {-x^2+187x-8700}  x dx


    That 6.5, is that just the value of the radius? Or in what way did you determine that?
    Last edited by westerwolde; Oct 11th 2017 at 12:53 AM.
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  12. #12
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    Re: Determine the center of gravity

    Quote Originally Posted by westerwolde View Post
    SlipEternal thanks for your help.


    S_5 = \int_{93.5} ^{100}  ( 6,5- \sqrt {-x^2+187x-8700}  x dx


    That 6.5, is that just the value of the radius? Or in what way did you determine that?
    The area of the shape from x=93.5 to x=100 was given in post #8. You calculated the area of the rectangle with width 12. So, I did 12 minus the position of the y-value for the top of the shape as x moves from 93.5 to 100.

    93.5 = 100 minus the radius of the circle.
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