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- Thread starter fonseh
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Mark44

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Do you mean "line of equal area"?## Homework Statement

For this shape , it's clear that the centroid and the horizontal line of equal axis lies on the same horizontal line , am i right ????

## Homework Equations

## The Attempt at a Solution

I'm not sure . correct me if i am wrong . [/B]

Anyway, yes, symmetry should convince you that you're right if there's no other information given.

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Are you familiar with plastic analysis ? Zp here is the plastic modulus .Do you mean "line of equal area"?

Anyway, yes, symmetry should convince you that you're right if there's no other information given.

I use another method to do , but i get different answer , why ?

Here's my working : , Zp = Sum of area x ( difference between centroid of particluar area and the equal area axis )

Zp = [ (130)(20)(150-10) + (150)(20)(150/2) ] x 2 = 1178000 , but the ans provided is only 929040

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Mark44

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No, I'm not, but I am familiar with the concept of centroids.Are you familiar with plastic analysis ?

My answer is based on the fact that the shape in the drawing is symmetric about its horizontal midline.

fonseh said:Zp here is the plastic modulus .

I use another method to do , but i get different answer , why ?

Here's my working : , Zp = Sum of area x ( difference between centroid of particluar area and the equal area axis )

Zp = [ (130)(20)(150-10) + (150)(20)(150/2) ] x 2 = 1178000 , but the ans provided is only 929040

If you had a rectangular piece of some uniform, rigid material 150 mm by 300 mm, its centroid would be at the center of the piece, at a point 75 mm to the right of the left edge, and 150 mm above the lower edge. If you cut out a rectangle 130 mm by 260 mm to form a "C" shape as in your drawing, the centroid of the piece you remove would be at its center, and the centroid of the remaining C-shaped piece would be along the horizontal midline, but a bit right of where it was for the original uncut piece of material.

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PhanthomJay

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