Centroids Method of Composite Areas
A small boy swallowed some coins and was taken to a hospital. When his grandmother telephoned to ask how he was a nurse said 'No change yet'.
Centroids
¢ Previously, we developed a general formulation for finding the centroid for a series of n areas
n xA ∑ ii i=1 x = n A ∑ i i=1 2 Centroids by Composite Areas Monday, November 12, 2012
1 Centroids
¢ xi was the distance from the y-axis to the local centroid of the area Ai
n xA ∑ ii i=1 x = n A ∑ i i=1 3 Centroids by Composite Areas Monday, November 12, 2012
Centroids
¢ If we can break up a shape into a series of smaller shapes that have predefined local centroid locations, we can use this formula to locate the centroid of the composite shape n xA ∑ ii i=1 x = n A ∑ i 4 Centroids by Composite Areas i=1 Monday, November 12, 2012
2 Centroid by Composite Bodies
¢ There is a table in the back cover of your book that gives you the location of local centroids for a select group of shapes ¢ The point labeled C is the location of the centroid of that shape.
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Centroid by Composite Bodies
¢ Please note that these are local centroids, they are given in reference to the x and y axes as shown in the table.
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3 Centroid by Composite Bodies
¢ For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ¢ The x-centroid would be located at 0 and the y-centroid would be located at 4r 3π 7 Centroids by Composite Areas Monday, November 12, 2012
Centroid by Composite Bodies
¢ If we wanted the centroid with respect to another axis, say along the top of the semicircle and along the left edge, the values in the table couldn’t be used exactly y x
C
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4 Centroid by Composite Bodies
¢ The table would give you the distance of C above the base of the semicircle, but that isn’t the distance from the centroid to the x-axis
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Centroid by Composite Bodies
y x
C 4r/3π
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5 Centroid by Composite Bodies
¢ Since the radius of the semicircle, in this case the distance to the y-centroid would be
⎛⎞4r yr=−⎜⎟ − ⎝⎠3π
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Centroid by Composite Bodies
y x
⎛⎞4r yr=−⎜⎟ − ⎝⎠3π
C 4r/3π
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6 Centroid by Composite Bodies
¢ By the same logic, the distance to the x- centroid would be
xr=
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Centroid by Composite Bodies
y
r x
r-(4r/3π)
C 4r/3π
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7 An Example
¢ Lets start with an example problem and see how this develops n xA ∑ ii i=1 x = n 1 in A ∑ i i=1
1 in
1 in 1in
3 in
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An Example
¢ We want to locate both the x and y centroids n xA ∑ ii i=1 x = n 1 in A ∑ i i=1
1 in
1 in 1in
3 in
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8 An Example
¢ There isn’t much of a chance of developing a function that is easy to integrate in this case n xA ∑ ii i=1 x = n 1 in A ∑ i i=1
1 in
1 in 1in
17 Centroids by Composite 3Areas in Monday, November 12, 2012
An Example
¢ We can break this figure up into a series of shapes and find the location of the local n centroid of each xA ∑ ii i=1 x = n 1 in A ∑ i i=1
1 in
1 in 1in
18 Centroids by Composite 3Areas in Monday, November 12, 2012
9 An Example
¢ There are multiple ways to do this as long
as you are consistent n xA ∑ ii i=1 x = n 1 in A ∑ i i=1
1 in
1 in 1in
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An Example
¢ First, we can develop a rectangle on the left side of the diagram, we will label that as area 1, A n 1 xA ∑ ii i=1 x = n 1 in A ∑ i i=1 A 1 in 1
1 in 1in
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10 An Example
¢ A second rectangle will be placed in the
bottom of the figure, we will label it A2
n xA ∑ ii 1 in i=1 x = n A A ∑ i 1 in 1 i=1
A 2 1 in 1in
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An Example
¢ A right triangle will complete the upper
right side of the figure, label it A3
n xA ∑ ii 1 in i=1 x = n A A ∑ i 1 in 1 i=1 A3 A 2 1 in 1in
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11 An Example
¢ Finally, we will develop a negative area to remove the quarter circle in the lower left
hand corner, label it A4 n xA ∑ ii 1 in i=1 x = n A A ∑ i 1 in 1 i=1 A3 A A 4 2 1 in 1in
23 Centroids by Composite3 Areas in Monday, November 12, 2012
An Example
¢ We will begin to build a table so that keeping up with things will be easier ¢ The first column will be the areas n xA 1 in ∑ ii i=1 ID Area x = n 2 (in ) A A 2 1 in ∑ i 1 A i 1 1 A3 = A2 3
A3 1.5 A A 1 in A4 -0.7854 4 2 1in
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12 An Example
¢ Now we will calculate the distance to the local centroids from the y-axis (we are
calculating an x-centroid) n
xAii ID Area xi 1 in ∑ 2 i 1 (in ) (in) = x = n A1 2 0.5 Ai A2 3 2.5 1 in ∑ A1 A i=1 A3 1.5 2 3
A4 -0.7854 0.42441 A 1 in A4 2 1in
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An Example
¢ To calculate the top term in the expression we need to multiply the entries in the last
two columns by one another n xA ∑ ii i 1 ID Area xi xi*Area 1 in = x = n (in2) (in) (in3) A A1 2 0.5 1 ∑ i A2 3 2.5 7.5 1 in i=1 A1 A A3 1.5 2 3 3
A4 -0.7854 0.42441 -0.33333 A 1 in A4 2 1in
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13 An Example
¢ If we sum the second column, we have the
bottom term in the division, the total area
n
ID Area xi xi*Area xA ∑ ii 2 3 (in ) (in) (in ) x = i=1 A1 2 0.5 1 1 in n
A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 A -0.7854 0.42441 -0.33333 1 in A 4 1 A3 5.714602 A 1 in A4 2 1in
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An Example
¢ And if we sum the fourth column, we have
the top term, the area moment
ID Area xi xi*Area n (in2) (in) (in3) xAii A1 2 0.5 1 ∑ i=1 A2 3 2.5 7.5 x = 1 in n A3 1.5 2 3 A A -0.7854 0.42441 -0.33333 ∑ i 4 i=1 5.714602 11.16667 1 in A 1 A3
A 1 in A4 2 1in
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14 An Example
¢ Dividing the sum of the area moments by
the total area we calculate the x-centroid
n ID Area x x*Area i i xAii (in2) (in) (in3) ∑ x = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3
x bar 1.9541 A 1 in A4 2 1in
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An Example
¢ You can always remember which to divide by if you look at the final units, remember
that a centroid is a distance n ID Area x x*Area i i xAii (in2) (in) (in3) ∑ x = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3
x bar 1.9541 A 1 in A4 2 1in
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15 An Example
¢ We can do the same process with the y
centroid
n ID Area x x*Area i i yAii (in2) (in) (in3) ∑ y = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3
x bar 1.9541 A 1 in A4 2 1in
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An Example
¢ Notice that the bottom term doesn’t change, the area of the figure hasn’t
changed n ID Area x x*Area i i yAii (in2) (in) (in3) ∑ y = i=1 A1 2 0.5 1 1 in n A2 3 2.5 7.5 A ∑ i A3 1.5 2 3 i=1 1 in A4 -0.7854 0.42441 -0.33333 A1 A 5.714602 11.16667 3
x bar 1.9541 A 1 in A4 2 1in
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16 An Example
¢ We only need to add a column of yi’s
ID Area xi xi*Area yi (in2) (in) (in3) (in) n A1 2 0.5 1 1 yA A2 3 2.5 7.5 0.5 ∑ ii i=1 A3 1.5 2 3 1.333333 y = n A4 -0.7854 0.42441 -0.33333 0.42441 A ∑ i 5.714602 11.16667 i=1
x bar 1.9541 1 in
1 in A 1 A3
A 1 in A4 2 1in
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An Example
¢ Calculate the area moments about the x-
axis
ID Area xi xi*Area yi yi*Area n (in2) (in) (in3) (in) (in3) yAii A1 2 0.5 1 1 2 ∑ A 3 2.5 7.5 0.5 1.5 i=1 2 y = n A3 1.5 2 3 1.333333 2 Ai A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333 ∑ 5.714602 11.16667 i=1
x bar 1.9541 1 in
1 in A 1 A3
A 1 in A4 2 1in
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17 An Example
¢ Sum the area moments
ID Area xi xi*Area yi yi*Area (in2) (in) (in3) (in) (in3) n A1 2 0.5 1 1 2 yA A2 3 2.5 7.5 0.5 1.5 ∑ ii i 1 A3 1.5 2 3 1.333333 2 = y = n A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333 A 5.714602 11.16667 5.166667 ∑ i i=1 x bar 1.9541 1 in
1 in A 1 A3
A 1 in A4 2 1in
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An Example
¢ And make the division of the area
moments by the total area
ID Area xi xi*Area yi yi*Area (in2) (in) (in3) (in) (in3) n
A1 2 0.5 1 1 2 yA ∑ ii A2 3 2.5 7.5 0.5 1.5 y = i=1 A 1.5 2 3 1.333333 2 n 3 A A4 -0.7854 0.42441 -0.33333 0.42441 -0.33333 ∑ i 5.714602 11.16667 5.166667 i=1
x bar 1.9541 y bar 0.904117 1 in
1 in A 1 A3
A 1 in A4 2 1in
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18 Example Problem 9-60
• Problem 9-60
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Example Problem 9-61
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19 Homework
¢ Problem 9-64 ¢ Problem 9-65 ¢ Problem 9-71
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