Calculating a centroid: parabolic pre-stressed concrete beams

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KamiMunchkins

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Hello, i'm trying to do this practice question but i can't seem to calculate the centroid. The answers are on the bottom right of the picture
For 1a I integrated the equation from 0 to 250, multiplied by 2 and subtracted that value from the are of a square (500x800). I think this is inefficient and that a faster method should be possible. I tried rearranging it to x = and then integrating it from 0 to 800, but when multiplying by 2 i get a wrong answer.
For 1b i have been given formulas on the second picture but no matter which one i use i get wrong answers. I'm really stuck and could appreciate the help, thank-you!

Picture 1
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Picture 2
1679842014740.png
 
KamiMunchkins said:
Hello, i'm trying to do this practice question but i can't seem to calculate the centroid. The answers are on the bottom right of the picture
For 1a I integrated the equation from 0 to 250, multiplied by 2 and subtracted that value from the are of a square (500x800). I think this is inefficient and that a faster method should be possible. I tried rearranging it to x = and then integrating it from 0 to 800, but when multiplying by 2 i get a wrong answer.
For 1b i have been given formulas on the second picture but no matter which one i use i get wrong answers. I'm really stuck and could appreciate the help, thank-you!
Click to expand...
Please show your actual work (for both parts), so we can be sure we understand what you did, and find an error.
 
Dr.Peterson said:
Please show your actual work (for both parts), so we can be sure we understand what you did, and find an error.
Click to expand...
so i don't have a camera otherwise i'd have attached a picture of my working. If possible could you explain to me how to find the centroid as when i use the formula provided i end up with wrong solutions. Thats the part thats really throwing me off...
 
KamiMunchkins said:
i don't have a camera otherwise i'd have attached a picture of my working
Click to expand...
Hello. An image is not required. You may type your steps. Perhaps, the issue is that you've made a simple mistake. We would like to check your work. ?
[imath]\;[/imath]
 
I get a different answer for b. Key says (0, 580) but I get (0, 480), in centimeters. Maybe someone here can check.
 
The y-axis is mislabeled. They are short by a factor of 10 because [imath]0.128(250)^2 = 8000[/imath], so I get a different answer for a).
 
BigBeachBanana said:
The y-axis is mislabeled. They are short by a factor of 10 because [imath]0.128(250)^2 = 8000[/imath], so I get a different answer for a).
Click to expand...
I noticed that too about 800 vs. 8,000, in millimeters. For a, I got 2.667x10^6 in cubic mm, which is the same as 2.667x10^3 in cubic cm.
 
limiTS said:
Sorry, I meant square mm and square cm.
Click to expand...
I think the question is erroneous in a few ways so I'm going to show my solution instead of talking in the dark.
[math]\int_{-250}^{250}8000-0.128x^2 ~ dx = 2.6667 \times 10^{6} ~ \text{mm}^2[/math][math]2.6667 \times 10^{6} ~ \text{mm}^2 \times \dfrac{1 \text{cm}^2}{100 ~ \text{mm}^2} = 2.6667 \times 10^{4}\text{ or } 26,667 \text{cm}^2[/math]
 
BigBeachBanana said:
I think the question is erroneous in a few ways so I'm going to show my solution instead of talking in the dark.
[math]\int_{-250}^{250}8000-0.128x^2 ~ dx = 2.6667 \times 10^{6} ~ \text{mm}^2[/math][math]2.6667 \times 10^{6} ~ \text{mm}^2 \times \dfrac{1 \text{cm}^2}{100 ~ \text{mm}^2} = 2.6667 \times 10^{4}\text{ or } 26,667 \text{cm}^2[/math]
Click to expand...
should i just move to a different question then? i'm trying to do practice for an upcoming exam in April
 
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