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HOMEWORK 37.B - SP 24
31 thoughts on “HOMEWORK 37.B - SP 24”
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Prof. Jones did say this is due Sunday night (4/21), not Monday because of quiet week.
On Gradescope it says that the due date is Monday midnight, though, I don't mind the extra day 🙂
Unfortunately for the timing, I don't think we will be getting an extra day to turn this in. Gradescope now has it due Sunday Night.
For this problem, are we going to have a y centroid below the z axis? The only component is the negative triangle area multiplied by distance from circle centroid, since the distance of the circle centroid to the z axis is 0.
Also, is the total inertia going to be the inertia of the triangle subtracted from the inertia of the circle?
Yes the centroid will be slightly below the Z axis since material is missing above the Z axis. Thus the I value will be subtracting the (I+A(d^2)) for the triangle.
Yes I think the centroid will be below the axis because there is more of the shape below the axis than above.
I think that's right. Since we have a hole inside a circle that is above the z axis, eventually we should get a negative value for y centroid.
Since the first part of your question is well answered already, for the second part yes, the total inertia will subtract that of the triangle from the circle as the triangle's area does not exist and its hypothetical second area MOI calculated must be subtracted to obtain the accurate value
To find the Y-position of the centroid, I know we have to find the area of the shaded region first, but will we have to calculate the centroid of the 4 shapes and use those values to find the centroid?
It can be done as a single circle and a single equalateral triangle, no need to do 4 shapes
Correct, you only need to use two shapes, but the triangle is not an equilateral triangle (the base and height are the same length, not the base and the other two sides),
Right, but it is an isosceles, so we can use that equations for our Iy.
you can just use the centroid of the circle and triangle
Thanks everyone, that made my equations a lot more simpler.
Yeah you just need to subtract that of the triangle from that of the circle
To find the y position of the centroid, you can sum the area of the circle multiplied by its centroid's y-position and the area of the triangle (negative, as it's cut out) multiplied by its centroid's y-position and divide that sum by the sum of the absolute values of the area of the circle and the area of the triangle. You can also imagine finding the centroid of the figure by concieving of the equivalent case where there is a circle of the same size and position with a triangle added to it as opposed to cut out of it, just with the triangle mirrored about the z-axis.
Does about the z-direction mean we do it about the z-axis (i.e. using y I values)?
I think that's correct, yeah. Since we're finding the Y-position of the centroid, we can just pretend the Z is "X" when looking at the document, and use all of our Iy equations. That seemed to work for me.
I found it helpful to use the method of composite sections, break down the greyed area into simpler shapes, such as rectangles and semicircles. Then, find the centroids and second moments of area for each shape individually. Finally, use the composite section method to find the centroid and second moment of area for the composite shape.
This is also the approach I used and I found it helpful to be able to do each step individually to help make sure I didn't make any mistakes.
Were you guys able to simplify your final expression down much? I had a hard time simplifying the polynomial expression and ended up converting my final answer to a decimal.
Yes, this is what I did. Since second area moment has a difficult algebra, I wrote my final expression for it along with the decimal value. I got the decimal value by putting my algebraic expression to my calculator.
Does anyone know what the second area moment is for the section about the centroid? Is it the same as the 2nd moment of inertia in the equation of sigma = My/I?
the second area moment is the same as the area moment of inertia, which is otherwise known as the area moment of inertia (not the second moment of inertia), which is indeed the I found in the equation sigma = My/I.
For this problem, I used method of composite functions to break down the parts into easier shapes. Then I found the y centroids then found the second moment of area for composite shape and wrote it as a decimal.
At the first place, I was confused as to whether adding or substracting between two shapes to get second area moment. After all, I decided them to substract the circle by triangle. Am I on the right track?
After finding the location of the centroid, you can subtract the second moment of area of the triangle from the second moment of area of the circle about the centroid along the z axis to get the overall second area moment. (2nd Area Moment of Circle - 2nd Area Moment of Triangle = Overall 2nd Area Moment)
I was checking my gradescope and I am a little confused about what the area moment is for the section at the centroid?
Is it possible to use the integration method for this question? and would that be easier to do rather than composite sections?
Yes, you should be able to use the integration method for this question but usually composite sections I have found to be easier fo most shapes.
For these problems, it helps to check if your answer for the centroid makes logical sense. Because above the z axis and below are identical except for the hole, it makes sense for the centroid to be slightly below the axis