# Thread: Determine the center of gravity

1. ## 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.

The static moment relative to y I determined in the following way:

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 )

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 ?

2. ## 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.

3. ## 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.

4. ## Re: Determine the center of gravity

Originally Posted by ChipB
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.

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?

5. ## Re: Determine the center of gravity

Originally Posted by SlipEternal
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

6. ## Re: Determine the center of gravity

Originally Posted by westerwolde
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?

7. ## Re: Determine the center of gravity

Originally Posted by westerwolde
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.

8. ## Re: Determine the center of gravity

Originally Posted by SlipEternal
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.
$S_5 = \displaystyle \int_{93.5}^{100}\left(5.5+\sqrt{6.5^2-(x-93.5)^2}\right) dx = \int_{93.5}^{100} \left(5.5+\sqrt{-x^2+187x-8700}\right)dx$

9. ## Re: Determine the center of gravity

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

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

10. ## Re: Determine the center of gravity

Originally Posted by SlipEternal
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...

11. ## Re: Determine the center of gravity

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

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

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

That 6.5, is that just the value of the radius? Or in what way did you determine that?

12. ## Re: Determine the center of gravity

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