14.1 Centroids Problem Set
CENTROIDS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The location of the π₯ βcoordinate for the centroid from the origin of the composite section illustrated below is closest to:
A. 45 B. 55 C. 65 D. 75
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PROBLEM 2: The location of the centroid relative to the origin of the composite section illustrated below is closest to:
A. (2.7, 0) B. (0, 1.2) C. (0, 2.7) D. (2.7, 3.5)
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CENTROIDS | SOLUTIONS Complete the following to reinforce your understanding of the concept covered in this module.
SOLUTION 1: The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM. In this problem, we are presented that BODY as:
The ORIGIN is located at the location where the X and Y AXES INTERSECT. It isnβt IDENTIFIED using any NOMENCLATURE, so if we hadnβt known the basic Made with by Prepineer | Prepineer.com
DEFINITION of where an ORIGIN resides in your typical CARTESIAN COORDINATE SYSTEM, then this problem would be impossible. We are analyzing a COMPOSITE BODY, which refers generally to any BODY that can be BROKEN DOWN in to smaller SUB REGIONS composed of more SIMPLE SHAPES, such as such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. These GEOMETRIC RELATIONSHIPS are present in the TABLES provided to us under the subject of STATICS on page 69 through 71 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Although not directly delineated as presented to us, this COMPOSITE BODY CLEARLY has FOUR locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. When we are working with these types of BODIES, itβs often times more convenient to OVERESTIMATE an AREA with a SIMPLE, DEFINABLE SHAPE and then SUBTRACT another SIMPLE, DEFINABLE SHAPE to adjust for the OVERESTIMATION to achieve the CUMALTIVE CHARACTERISTICS. In other words, as we will see with this COMPOSITE BODY, we are able to ADD or SUBTRACT REGIONS from any ESTIMATED SHAPE as long as we APPLY the CORRECT SIGN to the AREA of the REGION when we are bringing everything back together at the end.
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AREAS being SUBTRACTED are always a NEGATIVE VALUE, otherwise, we would be directly impacting the real location of the CENTROID, CENTER OF GRAVITY, or the CENTER OF MASSβ¦whichever one we are directly dealing with at the time. Once we SECTION this BODY out, and with all the appropriate data defined, we will determine the LOCATIONS of the CENTROID for each SHAPE individually, and as it relates to the COORDINATE AXIS in which the COMPOSITE BODY resides. We will then bring everything back together by ADDING AND SUBTRACTING them to establish the PROPERTIES of the COMPOSITE BODY as a whole. Generally speaking, the process of working through this COMPOSITE BODY problem will lay out like this: STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS STEP 2: CALCULATE the AREA for each REGION STEP 3: CALCULATE the DISTANCE from the ORIGIN to the CENTROID of each SHAPE STEP 4: MULTIPLY the value of the AREA by its DISTANCE from the ORIGIN for each REGION STEP 5: SUM UP the VALUES of all the AREAS and NOTE that value at the BOTTOM of the AREA column
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STEP 6: SUM UP the VALUES in the LAST COLUMN and NOTE the value at the BOTTOM of that column STEP 7: COMPUTE the CENTROID COORDINATE Letβs run through this problem STEP BY STEP:
STEP 1: LIST EACH OF THE AREAS THAT MAKE UP THE DISCRETE REGION, INCLUDING ANY HOLES OR SUBTRACTED REGIONS Our first step is to LIST ALL of the AREAS that make up the DISCRETE REGION of the COMPOSITE OBJECT. For illustration purposes, we will be color coding each shape that we will evaluate in the problem. Each SUB-REGION will have its OWN INDIVIDUAL AREA and CENTROID, as they have unique GEOMETRY and DIMENSIONS as they relate back to the ORIGIN of our COORDINATE SYSTEM.
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Our FOUR REGIONS will be:
We will analyze each REGION independently, then when combined, we will use the GEOMETRY and CENTROID of each area, to calculate the CENTROID of the COMPOSITE STRUCTURE, object, or body of interest, as provided in the problem statement.
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The DIMENSIONS, as defined, lay out like this:
We have also included the VARIOUS NOMENCLATURE for the AREAS and CENTROIDS that exist in each of the DISCRETE REGIONS. REGION 1, our PURPLE REGION, is a HALF CIRCLE:
REGION 2, our YELLOW REGION, is a FULL CIRCLE:
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REGION 3, our GREEN REGION, is a RECTANGLE:
REGION 4, our ORANGE REGION, is a TRIANGLE:
We will be OVERESTIMATING both REGION 1 and REGION 3 and adjusting those REGIONS by SUBTRACTING the entire REGION 2, which is a cut out in our COMPOSITE BODY. To calculate a CENTROID COORDINATE, we will need a separate table for each X and Y CENTROID DIMENSION.
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However, since we are only asked for the X-COORDINATE in this problem, we will only be creating a single TABLE, identifying each REGION along with the COLUMN HEADINGS as shown below: X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle) 2 (circle) 3 (rectangle) 4 (triangle) Come exam day, this table doesnβt have to be fancy, it can be scratched out in its most rudimentary form, the point is to keep everything organized and in the forefront for use as we progress through each of the steps.
STEP 2: CALCULATE THE AREA FOR EACH REGION The second step of our process requires us to CALCULATE the AREA for EACH REGION, and populate our coordinate table that we set up in step 1. The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, provides a series of TABLES on pages 69 through 71 that include GENERAL FORMULAS for defining the CENTROIDS, and various other CHARACTERISTICS of the most COMMON SHAPES we will encounter come exam day.
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Specifically, each SHAPE will be defined for us in the FIRST COLUMN of the TABLE with the CENTROID and AREA FORMULAS defined for us in the SECOND COLUMN. We will be heavily relying on the use of these TABLES as we move forward. Revisiting the COMPOSITE BODY as we have it broken up at this point, we have:
It is important to recognize the ORIENTATION of each SHAPE as well as the AXIS OF SYMMETRY, as the GENERAL FORMULAS are based on the GEOMETRY and DIMENSIONS of a given SHAPE.
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Flipping back to the TABLES outlining the various SHAPES we are dealing with here, the FIRST COLUMN defines the shape, whether itβs a TRIANGLE, RECTANGLE, etc., and the SECOND COLUMN will give us the GENERAL FORMULAS to determine the AREA and CENTROID of that shape:
It is important to recognize the ORIENTATION of the SHAPE and AXIS OF SYMMETRY, as the formulas are based on this geometry and dimensions of a given shape. It is also IMPORTANT to realize that a GEOMETRIC SHAPE may not be defined in the
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Reference Handbook. In this case, REGION 1, shown in purple, we are given a HALF CIRCLE, which is not provided directly in the Reference Handbook. However, we know the area of a circle is ππ 1 , which means that a half circle will have half the calculate area of a full circle: This GENERAL FORMULA would then be: ππ 1 π 60 π΄3 = = 2 2
1
= 5,645
From our schematic, we know that: π = 40 And plugging this in, we get: π 60 π΄3 = 2
1
= 5,645 ππ1
We can now repeat this same process of referencing the TABLE, getting the GENERAL FORMULA, and quantifying the AREA of each SUB-REGION. Doing so for our REGION 2, which is a CIRCLE, we get: π΄1 = ππ 1 = π 40
1
= β5,026
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For our REGION 3, which is a SQUARE, we get: π΄7 = πβ = 120 80 = 9,600 For our REGION 4, which is a TRIANGLE, we get:
π΄< =
πβ 120 60 = = 3,600 2 2
Plugging in the calculated values for each SUB-REGION area into our X-CENTROID COORDINATE TABLE, we find get: X-CENTROID COORDINATE TABLE Region
Area (an)
1 (half circle)
5,645
2 (circle)
-5,026
3 (rectangle)
9,600
4 (triangle)
3,600
xn
xn an
STEP 3: CALCULATE THE DISTANCE FROM THE ORIGIN TO THE CENTROID OF EACH SHAPE We can now move on to Step 3 where we will CALCULATE the DISTANCE from the ORIGIN TO the CENTROID of EACH SHAPE we have defined.
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Illustrating each of the CENTROID locations, we have:
We will reference the same exact TABLE we just used as well as the same ROW and COLUMN for each of our SUB-REGIONS. For repetition purposes and to become familiar with the NCEES Reference Handbook, we will not illustrate this table, but rather challenge you to crosscheck as we move through each measurement. Starting with REGION 1, we know that the X-CENTROID will fall at the CENTER of the HALF CIRCLE as it lies on the X-AXIS, this point is located at 60mm from the Y-AXIS, giving us: π₯=> = 60 ππ
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For REGION 2, same as with the HALF CIRCLE, the X-CENTROID will fall at the CENTER of the CIRCLE as it lies on the X-AXIS, this point is located at 60ππ from the Y-AXIS as well: π₯=? = 60 ππ For REGION 3 is our RECTANGLE, again, the X-CENTROID will fall at the HALF WAY point of the BASE WIDTH, this point is located at 60ππ from the Y-AXIS: π₯=@ = 60 ππ Lastly, for REGION 4, our TRIANGLE, this one is different from the others, as itβs XCENTROID coordinate is located at a THIRD of its BASE WIDTH, or:
π₯=A =
π 120 = = 40 ππ 3 3
Plugging all of these values for each SUB-REGION in to our COORDINATE TABLE, we find the table is written as: X-CENTROID COORDINATE TABLE
Region
Area (an)
xn
1 (half circle)
5,645
60
2 (circle)
-5,026
60
3 (rectangle)
9,600
60
4 (triangle)
3,600
40
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xn an
STEP 4: MULTIPLY THE VALUE OF THE AREA BY ITS DISTANCE FROM THE ORIGIN FOR EACH REGION Moving on to STEP 4, for each individual region, we now need to take our SECOND COLUMN, the AREA, and MULTIPLY it by the THIRD COLUMN, the DISTANCE FROM THE ORIGINβ¦and plug this product in to our final column in the coordinate table. Highlighting the TWO COLUMNS in which we are MULTIPLYING as GREEN, we carry out this calculation and plug our RESULTS in to the last column, giving us:
X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle)
5,645
60
338,700
2 (circle)
-5,026
60
-301,560
3 (rectangle)
9,600
60
576,000
4 (triangle)
3,600
40
144,000
It is important to note that the NEGATIVE VALUE for the CIRCLE carries through in the calculation to represent the holeβ¦this is our ADJUSTMENT to the OVERESTIMATION of the FIRST and THIRD REGIONS.
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STEP 5: SUM UP THE VALUES OF ALL THE AREAS AND NOTE THAT VALUE AT THE BOTTOM OF THE AREA COLUMN Now for step 5, we SUM all the AREAS that we have defined up to this point. This CALCULATION takes in to account the AREA of ALL REGIONS that reside in the COLUMN we have highlighted GREEN: X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle)
5,645
60
338,700
2 (circle)
-5,026
60
-301,560
3 (rectangle)
9,600
60
576,000
4 (triangle)
3,600
40
144,000
= 13,819
STEP 6: SUM UP THE VALUES IN THE LAST COLUMN AND NOTE THE VALUE AT THE BOTTOM OF THAT COLUMN For step 6, we do the same, SUMMING all the products that we calculated in step 4 and provide the SUM at the bottom of our table.
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This CALCULATION takes in to account the COLUMN we have highlighted GREEN: X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle)
5,645
60
338,700
2 (circle)
-5,026
60
-301,560
3 (rectangle)
9,600
60
576,000
4 (triangle)
3,600
40
144,000
= 13,819
= 757,140
STEP 7: COMPUTE THE CENTROID COORDINATE In our final step, we now have all the data we need to define the X-COORDINATE for the CENTROID of this COMPOSITE BODY. The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Remember, that the CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM.
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The LOCATION of the CENTROID of an AREA is wholly a function of itβs GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
π₯B= =
πBD = π΄
π₯E πE π΄
Where π₯B= is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
π¦B= =
πBH = π΄
π¦E πE π΄
Where π¦B= which is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN Y-AXIS. We are only concerned with the X-COORDINATE in this problem, so letβs focus in on that GENERAL FORMULA:
π₯B= =
πBD = π΄
π₯E πE π΄
We much donβt care for the FIRST HALF of this FORMULA, but rather the SECOND HALF, such that:
π₯B= =
π₯E πE π΄
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From our COORDINATE TABLE we have developed in this problem, we know that: β’
π¦E πE = 13,819
β’ π΄ = 757,140 We can just plug and chug with our calculators to find that the X-COORDINATE of the CENTROID for this COMPOSITE BODY is calculated as:
π¦B= =
757,140 13,819
Or: π¦B= = 54.8 ππ Putting this value in to our SCHEMATIC, we have:
The correct answer choice is B. ππ
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SOLUTION 2: The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM. In this problem, we are presented that BODY as:
The ORIGIN is located at the location where the X and Y AXES INTERSECT. It isnβt IDENTIFIED using any NOMENCLATURE, so if we hadnβt known the basic DEFINITION of where an ORIGIN resides in your typical CARTESIAN COORDINATE SYSTEM, then this problem would be impossible. We are analyzing a COMPOSITE BODY, which refers generally to any BODY that can be BROKEN DOWN in to smaller SUB REGIONS composed of more SIMPLE SHAPES,
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such as such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. These GEOMETRIC RELATIONSHIPS are present in the TABLES provided to us under the subject of STATICS on page 69 through 71 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Although not directly delineated as presented to us, this COMPOSITE BODY CLEARLY has FIVE locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. The COMPOSITE SHAPES here will be a number of RECTANGLES with HORIZONTAL and VERTICAL ORIENTATIONS. Once we SECTION this BODY out, and with all the appropriate data defined, we will determine the LOCATIONS of the CENTROID for each SHAPE individually, and as it relates to the COORDINATE AXIS in which the COMPOSITE BODY resides. We will then bring everything back together by ADDING them to establish the PROPERTIES of the COMPOSITE BODY as a whole. Generally speaking, the process of working through this COMPOSITE BODY problem will lay out like this: STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS
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STEP 2: CALCULATE the AREA for each REGION STEP 3: CALCULATE the DISTANCE from the ORIGIN to the CENTROID of each SHAPE STEP 4: MULTIPLY the value of the AREA by its DISTANCE from the ORIGIN for each REGION STEP 5: SUM UP the VALUES of all the AREAS and NOTE that value at the BOTTOM of the AREA column STEP 6: SUM UP the VALUES in the LAST COLUMN and NOTE the value at the BOTTOM of that column STEP 7: COMPUTE the CENTROID COORDINATE Lets run through this problem STEP BY STEP:
STEP 1: LIST EACH OF THE AREAS THAT MAKE UP THE DISCRETE REGION, INCLUDING ANY HOLES OR SUBTRACTED REGIONS Our first step is to list all of the areas that make up the DISCRETE REGION of the COMPOSITE BODY. For illustration purposes, we will be color coding each shape that we will evaluate in the problem. Each SUB-REGION will have its OWN INDIVIDUAL AREA and CENTROID, as they have DIFFERENT GEOMETRY and DIMENSIONS.
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When combined, we will use the GEOMETRY and CENTROID of each area, relative to the ORIGIN as established, to calculate the CENTROID of the COMPOSITE STRUCTURE, object, or body of interest, as provided in the problem statement. Breaking up our T-SECTION COMPOSITE BODY, we have:
Each SUBREGION will have its own AREA and its own CENTROID such that:
To ORGANIZE our calculations for defining the COMPOSITE CENTROIDAL COORDINATES, we will need to develop a SEPARATE TABLE for each X and Y CENTROIDAL DIMENSION.
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Through OBSERVATION, we can conclude that SYMMETRY exists along the Y-AXIS, or in other terms, the X-COORDINATE of our CENTROID will fall in line with the ORIGIN or rather: π₯B= = 0 Recall that an OBJECT is said to be SYMMETRICAL about an AXIS if we are able to establish that AXIS and confirm that the GEOMETRY that resides on one side is a REFLECTION, or MIRRORED IMAGE, of what resides on the opposite side. Most shapes that will be encountered on the FE EXAM will be of SYMMETRICAL nature. Letβs move on to defining our Y-COORDINATE. To determine the Y-CENTROID LOCATION, we will start with creating a table with the column headings as shown: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
1 (rectangle) 2 (rectangle) 3 (rectangle) 4 (rectangle) 5 (rectangle)
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yn an
Come exam day, this table doesnβt have to be fancy, it can be scratched out in its most rudimentary form, the point is to keep everything organized and in the forefront for use as we progress through each of the steps.
STEP 2: CALCULATE THE AREA FOR EACH REGION The second step of our process requires us to CALCULATE the AREA for EACH REGION, and populate our coordinate table that we set up in step 1. The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, provides a series of TABLES on pages 69 through 71 that include GENERAL FORMULAS for defining the CENTROIDS, and various other CHARACTERISTICS of the most COMMON SHAPES we will encounter come exam day. Specifically, each SHAPE will be defined for us in the FIRST COLUMN of the TABLE with the CENTROID and AREA FORMULAS defined for us in the SECOND COLUMN. We will be heavily relying on the use of these TABLES as we move forward.
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Revisiting the COMPOSITE BODY as we have it broken up at this point, we have:
It is important to recognize the ORIENTATION of each SHAPE as well as the AXIS OF SYMMETRY, as the GENERAL FORMULAS are based on the GEOMETRY and DIMENSIONS of a given SHAPE.
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Flipping back to page 96 of the The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we will want to specifically hone in on the SECOND COLUMN titled AREA & CENTROID, which is highlighted:
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The first SUBREGION we will consider is the PURPLE RECTANGLE, focusing in on the information and formulas highlighted here in green:
Noting that the GENERAL FORMULA for the AREA of this particular region is: A = bh
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And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: b = 3 in h = 1 in Taking these DIMENSIONS and plugging them in to this GENERAL FORMULA, we get: A = bh = 3 in 1 in = 3 in1 With a base of 3 ππ and a height of 1 ππ, we plug these numbers in to the formula and get an area of 3 in1 β¦which we place in to our Y-COORDINATE table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
3
yn
yn an
2 (rectangle) 3 (rectangle) 4 (rectangle) 5 (rectangle) We can now repeat this same process of referencing the TABLE, getting the GENERAL FORMULA, and quantifying the AREA of each SUB-REGION. Flipping back to page 96 of the The NCEES Supplied Reference Handbook, Version 9.4 Made with by Prepineer | Prepineer.com
for Computer Based Testing, we will want to again hone in on the SECOND COLUMN titled AREA & CENTROID, which we highlighted previously, for each of the REGIONS. Doing so for our REGION 2, we get: π΄1 = π β = 1 3 = 3 For our REGION 3, we get: π΄7 = π β = 8 1 = 8 For our REGION 4, we get: π΄< = π β = 1 3 = 3 For our REGION 5, we get: π΄Q = π β = 3 1 = 3
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Plugging in the calculated values for each SUB-REGION area into our Y-CENTROID COORDINATE TABLE, we find get: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
3
2 (rectangle)
3
3 (rectangle)
8
4 (rectangle)
3
5 (rectangle)
3
yn
yn an
STEP 3: CALCULATE THE DISTANCE FROM THE ORIGIN TO THE CENTROID OF EACH SHAPE We can now move on to Step 3 where we will CALCULATE the DISTANCE from the ORIGIN TO the CENTROID of EACH SHAPE we have defined.
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We know that the CENTROID will some DISTANCE π¦B= from the X-AXIS. Starting with our GREEN REGION and flipping back to our tables, we see that the formula for the Y-CENTROID of a RECTANGLE is defined:
Taking this GENERAL FORMULA for the Y-CENTROID of this particular region is: π¦= = β/2
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And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: h = 1 in We calculate the CENTROID of region 1 using the formula for a horizontal rectangle:
yT =
h 1 in = = .5 in 2 2
Plugging this value in to our table we now have: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
1 (rectangle)
3
0.5
2 (rectangle)
3
2.5
3 (rectangle)
8
4.5
4 (rectangle)
3
2.5
5 (rectangle)
3
0.5
yn an
We can now repeat this same process of referencing the TABLE, getting the GENERAL FORMULA, and quantifying the AREA of each SUB-REGION. Flipping back to page 96 of the The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we will want to again hone in on the SECOND COLUMN titled AREA & CENTROID, which we highlighted previously, for each of the REGIONS.
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Moving to REGION 2, we know that the CENTROID will be some distance yT from the bottom of its shape PLUS the HEIGHT of REGION 1 since it is between it and the π₯ βaxis. This gives us a CENTROID COORDINATE of:
yT ? = 1 +
h 3 = 1 + = 2.5 2 2
Moving to REGION 3, the centroid will still remain at half the height, as all rectangles, but itβs also elevated 4 inches from the π₯ βaxis, so this must be accounted for. Plugging in our dimensions, we find that this centroid is located at 4.5 inches from the π₯ βaxis:
yT @ = 1 + 3 +
h 1 = 1 + 3 + = 4.5 2 2
Now we could continue on to REGION 4 and 5 just as we have done up to this point, that will work, but you can HACK time off the solution by recognizing the SYMMETRY of the COMPOSITE OBJECT in relation to the Y-AXIS. The Y-COORDINATE for the CENTROIDS of REGION 4 and 5 must be identical to those of REGION 1 and 2.
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For REGION 4:
yT A = 1 +
h 3 = 1 + = 2. 5 2 2
For REGION 5:
yT V =
h 1 = = 0.5 2 2
Plugging in the derived values for each SUB-REGION area into our Y-CENTROID COORDINATE TABLE, we find get: Y-CENTROID COORDINATE TABLE
Region
Area (an)
yn
1 (rectangle)
3
0.5
2 (rectangle)
3
2.5
3 (rectangle)
8
4.5
4 (rectangle)
3
2.5
5 (rectangle)
3
0.5
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yn an
STEP 4: MULTIPLY THE VALUE OF THE AREA BY ITS DISTANCE FROM THE ORIGIN FOR EACH REGION Moving on to STEP 4, for each individual region, we now need to take our SECOND COLUMN, the AREA, and MULTIPLY it by the THIRD COLUMN, the DISTANCE FROM THE ORIGINβ¦and plug this product in to our final column in the coordinate table. Highlighting the TWO COLUMNS in which we are MULTIPLYING as GREEN, we carry out this calculation and plug our RESULTS in to the last column, giving us: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
3
0.5
1.5
2 (rectangle)
3
2.5
7.5
3 (rectangle)
8
4.5
36
4 (rectangle)
3
2.5
7.5
5 (rectangle)
3
0.5
1.5
STEP 5: SUM UP THE VALUES OF ALL THE AREAS AND NOTE THAT VALUE AT THE BOTTOM OF THE AREA COLUMN Now for step 5, we SUM all the AREAS that we have defined up to this point.
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This CALCULATION takes in to account the AREA of ALL REGIONS that reside in the COLUMN we have highlighted GREEN: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
3
0.5
1.5
2 (rectangle)
3
2.5
7.5
3 (rectangle)
8
4.5
36
4 (rectangle)
3
2.5
7.5
5 (rectangle)
3
0.5
1.5
= 20
STEP 6: SUM UP THE VALUES IN THE LAST COLUMN AND NOTE THE VALUE AT THE BOTTOM OF THAT COLUMN For step 6, we do the same, SUMMING all the products that we calculated in step 4 and provide the SUM at the bottom of our table.
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This CALCULATION takes in to account the COLUMN we have highlighted GREEN: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
3
0.5
1.5
2 (rectangle)
3
2.5
7.5
3 (rectangle)
8
4.5
36
4 (rectangle)
3
2.5
7.5
5 (rectangle)
3
0.5
1.5
= 20
= 54
STEP 7: COMPUTE THE CENTROID COORDINATE In our final step, we now have all the data we need to define the Y-COORDINATE for the CENTROID of this COMPOSITE BODY. The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Remember, that the CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM.
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The LOCATION of the CENTROID of an AREA is wholly a function of itβs GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
π₯B= =
πBD = π΄
π₯E πE π΄
Where π₯B= is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
π¦B= =
πBH = π΄
π¦E πE π΄
Where π¦B= which is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN Y-AXIS. We know the X-COORDINATE already, due to SYMMETRY, so letβs focus in on the GENERAL FORMULA for the Y-COORDINATE, which is:
π¦B= =
πBH = π΄
π¦E πE π΄
We much donβt care for the FIRST HALF of this FORMULA, but rather the SECOND HALF, such that:
π¦B= =
π¦E πE π΄
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From our COORDINATE TABLE we have developed in this problem, we know that: β’
π¦E πE = 54
β’ π΄ = 20 We can just plug and chug with our calculators to find that the Y-COORDINATE of the CENTROID for the T-SECTION is calculated as:
π¦B= =
54 20
Or: π¦B= = 2.7 ππ Identifying this on our schematic, we have:
The correct answer choice is C.(π, π. π)
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PROBLEM 1: The location of the π₯ βcoordinate for the centroid from the origin of the composite section illustrated below is closest to:
A. 45 B. 55 C. 65 D. 75
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PROBLEM 2: The location of the centroid relative to the origin of the composite section illustrated below is closest to:
A. (2.7, 0) B. (0, 1.2) C. (0, 2.7) D. (2.7, 3.5)
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CENTROIDS | SOLUTIONS Complete the following to reinforce your understanding of the concept covered in this module.
SOLUTION 1: The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM. In this problem, we are presented that BODY as:
The ORIGIN is located at the location where the X and Y AXES INTERSECT. It isnβt IDENTIFIED using any NOMENCLATURE, so if we hadnβt known the basic Made with by Prepineer | Prepineer.com
DEFINITION of where an ORIGIN resides in your typical CARTESIAN COORDINATE SYSTEM, then this problem would be impossible. We are analyzing a COMPOSITE BODY, which refers generally to any BODY that can be BROKEN DOWN in to smaller SUB REGIONS composed of more SIMPLE SHAPES, such as such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. These GEOMETRIC RELATIONSHIPS are present in the TABLES provided to us under the subject of STATICS on page 69 through 71 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Although not directly delineated as presented to us, this COMPOSITE BODY CLEARLY has FOUR locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. When we are working with these types of BODIES, itβs often times more convenient to OVERESTIMATE an AREA with a SIMPLE, DEFINABLE SHAPE and then SUBTRACT another SIMPLE, DEFINABLE SHAPE to adjust for the OVERESTIMATION to achieve the CUMALTIVE CHARACTERISTICS. In other words, as we will see with this COMPOSITE BODY, we are able to ADD or SUBTRACT REGIONS from any ESTIMATED SHAPE as long as we APPLY the CORRECT SIGN to the AREA of the REGION when we are bringing everything back together at the end.
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AREAS being SUBTRACTED are always a NEGATIVE VALUE, otherwise, we would be directly impacting the real location of the CENTROID, CENTER OF GRAVITY, or the CENTER OF MASSβ¦whichever one we are directly dealing with at the time. Once we SECTION this BODY out, and with all the appropriate data defined, we will determine the LOCATIONS of the CENTROID for each SHAPE individually, and as it relates to the COORDINATE AXIS in which the COMPOSITE BODY resides. We will then bring everything back together by ADDING AND SUBTRACTING them to establish the PROPERTIES of the COMPOSITE BODY as a whole. Generally speaking, the process of working through this COMPOSITE BODY problem will lay out like this: STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS STEP 2: CALCULATE the AREA for each REGION STEP 3: CALCULATE the DISTANCE from the ORIGIN to the CENTROID of each SHAPE STEP 4: MULTIPLY the value of the AREA by its DISTANCE from the ORIGIN for each REGION STEP 5: SUM UP the VALUES of all the AREAS and NOTE that value at the BOTTOM of the AREA column
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STEP 6: SUM UP the VALUES in the LAST COLUMN and NOTE the value at the BOTTOM of that column STEP 7: COMPUTE the CENTROID COORDINATE Letβs run through this problem STEP BY STEP:
STEP 1: LIST EACH OF THE AREAS THAT MAKE UP THE DISCRETE REGION, INCLUDING ANY HOLES OR SUBTRACTED REGIONS Our first step is to LIST ALL of the AREAS that make up the DISCRETE REGION of the COMPOSITE OBJECT. For illustration purposes, we will be color coding each shape that we will evaluate in the problem. Each SUB-REGION will have its OWN INDIVIDUAL AREA and CENTROID, as they have unique GEOMETRY and DIMENSIONS as they relate back to the ORIGIN of our COORDINATE SYSTEM.
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Our FOUR REGIONS will be:
We will analyze each REGION independently, then when combined, we will use the GEOMETRY and CENTROID of each area, to calculate the CENTROID of the COMPOSITE STRUCTURE, object, or body of interest, as provided in the problem statement.
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The DIMENSIONS, as defined, lay out like this:
We have also included the VARIOUS NOMENCLATURE for the AREAS and CENTROIDS that exist in each of the DISCRETE REGIONS. REGION 1, our PURPLE REGION, is a HALF CIRCLE:
REGION 2, our YELLOW REGION, is a FULL CIRCLE:
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REGION 3, our GREEN REGION, is a RECTANGLE:
REGION 4, our ORANGE REGION, is a TRIANGLE:
We will be OVERESTIMATING both REGION 1 and REGION 3 and adjusting those REGIONS by SUBTRACTING the entire REGION 2, which is a cut out in our COMPOSITE BODY. To calculate a CENTROID COORDINATE, we will need a separate table for each X and Y CENTROID DIMENSION.
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However, since we are only asked for the X-COORDINATE in this problem, we will only be creating a single TABLE, identifying each REGION along with the COLUMN HEADINGS as shown below: X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle) 2 (circle) 3 (rectangle) 4 (triangle) Come exam day, this table doesnβt have to be fancy, it can be scratched out in its most rudimentary form, the point is to keep everything organized and in the forefront for use as we progress through each of the steps.
STEP 2: CALCULATE THE AREA FOR EACH REGION The second step of our process requires us to CALCULATE the AREA for EACH REGION, and populate our coordinate table that we set up in step 1. The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, provides a series of TABLES on pages 69 through 71 that include GENERAL FORMULAS for defining the CENTROIDS, and various other CHARACTERISTICS of the most COMMON SHAPES we will encounter come exam day.
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Specifically, each SHAPE will be defined for us in the FIRST COLUMN of the TABLE with the CENTROID and AREA FORMULAS defined for us in the SECOND COLUMN. We will be heavily relying on the use of these TABLES as we move forward. Revisiting the COMPOSITE BODY as we have it broken up at this point, we have:
It is important to recognize the ORIENTATION of each SHAPE as well as the AXIS OF SYMMETRY, as the GENERAL FORMULAS are based on the GEOMETRY and DIMENSIONS of a given SHAPE.
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Flipping back to the TABLES outlining the various SHAPES we are dealing with here, the FIRST COLUMN defines the shape, whether itβs a TRIANGLE, RECTANGLE, etc., and the SECOND COLUMN will give us the GENERAL FORMULAS to determine the AREA and CENTROID of that shape:
It is important to recognize the ORIENTATION of the SHAPE and AXIS OF SYMMETRY, as the formulas are based on this geometry and dimensions of a given shape. It is also IMPORTANT to realize that a GEOMETRIC SHAPE may not be defined in the
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Reference Handbook. In this case, REGION 1, shown in purple, we are given a HALF CIRCLE, which is not provided directly in the Reference Handbook. However, we know the area of a circle is ππ 1 , which means that a half circle will have half the calculate area of a full circle: This GENERAL FORMULA would then be: ππ 1 π 60 π΄3 = = 2 2
1
= 5,645
From our schematic, we know that: π = 40 And plugging this in, we get: π 60 π΄3 = 2
1
= 5,645 ππ1
We can now repeat this same process of referencing the TABLE, getting the GENERAL FORMULA, and quantifying the AREA of each SUB-REGION. Doing so for our REGION 2, which is a CIRCLE, we get: π΄1 = ππ 1 = π 40
1
= β5,026
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For our REGION 3, which is a SQUARE, we get: π΄7 = πβ = 120 80 = 9,600 For our REGION 4, which is a TRIANGLE, we get:
π΄< =
πβ 120 60 = = 3,600 2 2
Plugging in the calculated values for each SUB-REGION area into our X-CENTROID COORDINATE TABLE, we find get: X-CENTROID COORDINATE TABLE Region
Area (an)
1 (half circle)
5,645
2 (circle)
-5,026
3 (rectangle)
9,600
4 (triangle)
3,600
xn
xn an
STEP 3: CALCULATE THE DISTANCE FROM THE ORIGIN TO THE CENTROID OF EACH SHAPE We can now move on to Step 3 where we will CALCULATE the DISTANCE from the ORIGIN TO the CENTROID of EACH SHAPE we have defined.
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Illustrating each of the CENTROID locations, we have:
We will reference the same exact TABLE we just used as well as the same ROW and COLUMN for each of our SUB-REGIONS. For repetition purposes and to become familiar with the NCEES Reference Handbook, we will not illustrate this table, but rather challenge you to crosscheck as we move through each measurement. Starting with REGION 1, we know that the X-CENTROID will fall at the CENTER of the HALF CIRCLE as it lies on the X-AXIS, this point is located at 60mm from the Y-AXIS, giving us: π₯=> = 60 ππ
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For REGION 2, same as with the HALF CIRCLE, the X-CENTROID will fall at the CENTER of the CIRCLE as it lies on the X-AXIS, this point is located at 60ππ from the Y-AXIS as well: π₯=? = 60 ππ For REGION 3 is our RECTANGLE, again, the X-CENTROID will fall at the HALF WAY point of the BASE WIDTH, this point is located at 60ππ from the Y-AXIS: π₯=@ = 60 ππ Lastly, for REGION 4, our TRIANGLE, this one is different from the others, as itβs XCENTROID coordinate is located at a THIRD of its BASE WIDTH, or:
π₯=A =
π 120 = = 40 ππ 3 3
Plugging all of these values for each SUB-REGION in to our COORDINATE TABLE, we find the table is written as: X-CENTROID COORDINATE TABLE
Region
Area (an)
xn
1 (half circle)
5,645
60
2 (circle)
-5,026
60
3 (rectangle)
9,600
60
4 (triangle)
3,600
40
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xn an
STEP 4: MULTIPLY THE VALUE OF THE AREA BY ITS DISTANCE FROM THE ORIGIN FOR EACH REGION Moving on to STEP 4, for each individual region, we now need to take our SECOND COLUMN, the AREA, and MULTIPLY it by the THIRD COLUMN, the DISTANCE FROM THE ORIGINβ¦and plug this product in to our final column in the coordinate table. Highlighting the TWO COLUMNS in which we are MULTIPLYING as GREEN, we carry out this calculation and plug our RESULTS in to the last column, giving us:
X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle)
5,645
60
338,700
2 (circle)
-5,026
60
-301,560
3 (rectangle)
9,600
60
576,000
4 (triangle)
3,600
40
144,000
It is important to note that the NEGATIVE VALUE for the CIRCLE carries through in the calculation to represent the holeβ¦this is our ADJUSTMENT to the OVERESTIMATION of the FIRST and THIRD REGIONS.
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STEP 5: SUM UP THE VALUES OF ALL THE AREAS AND NOTE THAT VALUE AT THE BOTTOM OF THE AREA COLUMN Now for step 5, we SUM all the AREAS that we have defined up to this point. This CALCULATION takes in to account the AREA of ALL REGIONS that reside in the COLUMN we have highlighted GREEN: X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle)
5,645
60
338,700
2 (circle)
-5,026
60
-301,560
3 (rectangle)
9,600
60
576,000
4 (triangle)
3,600
40
144,000
= 13,819
STEP 6: SUM UP THE VALUES IN THE LAST COLUMN AND NOTE THE VALUE AT THE BOTTOM OF THAT COLUMN For step 6, we do the same, SUMMING all the products that we calculated in step 4 and provide the SUM at the bottom of our table.
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This CALCULATION takes in to account the COLUMN we have highlighted GREEN: X-CENTROID COORDINATE TABLE Region
Area (an)
xn
xn an
1 (half circle)
5,645
60
338,700
2 (circle)
-5,026
60
-301,560
3 (rectangle)
9,600
60
576,000
4 (triangle)
3,600
40
144,000
= 13,819
= 757,140
STEP 7: COMPUTE THE CENTROID COORDINATE In our final step, we now have all the data we need to define the X-COORDINATE for the CENTROID of this COMPOSITE BODY. The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Remember, that the CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM.
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The LOCATION of the CENTROID of an AREA is wholly a function of itβs GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
π₯B= =
πBD = π΄
π₯E πE π΄
Where π₯B= is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
π¦B= =
πBH = π΄
π¦E πE π΄
Where π¦B= which is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN Y-AXIS. We are only concerned with the X-COORDINATE in this problem, so letβs focus in on that GENERAL FORMULA:
π₯B= =
πBD = π΄
π₯E πE π΄
We much donβt care for the FIRST HALF of this FORMULA, but rather the SECOND HALF, such that:
π₯B= =
π₯E πE π΄
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From our COORDINATE TABLE we have developed in this problem, we know that: β’
π¦E πE = 13,819
β’ π΄ = 757,140 We can just plug and chug with our calculators to find that the X-COORDINATE of the CENTROID for this COMPOSITE BODY is calculated as:
π¦B= =
757,140 13,819
Or: π¦B= = 54.8 ππ Putting this value in to our SCHEMATIC, we have:
The correct answer choice is B. ππ
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SOLUTION 2: The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM. In this problem, we are presented that BODY as:
The ORIGIN is located at the location where the X and Y AXES INTERSECT. It isnβt IDENTIFIED using any NOMENCLATURE, so if we hadnβt known the basic DEFINITION of where an ORIGIN resides in your typical CARTESIAN COORDINATE SYSTEM, then this problem would be impossible. We are analyzing a COMPOSITE BODY, which refers generally to any BODY that can be BROKEN DOWN in to smaller SUB REGIONS composed of more SIMPLE SHAPES,
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such as such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. These GEOMETRIC RELATIONSHIPS are present in the TABLES provided to us under the subject of STATICS on page 69 through 71 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Although not directly delineated as presented to us, this COMPOSITE BODY CLEARLY has FIVE locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. The COMPOSITE SHAPES here will be a number of RECTANGLES with HORIZONTAL and VERTICAL ORIENTATIONS. Once we SECTION this BODY out, and with all the appropriate data defined, we will determine the LOCATIONS of the CENTROID for each SHAPE individually, and as it relates to the COORDINATE AXIS in which the COMPOSITE BODY resides. We will then bring everything back together by ADDING them to establish the PROPERTIES of the COMPOSITE BODY as a whole. Generally speaking, the process of working through this COMPOSITE BODY problem will lay out like this: STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS
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STEP 2: CALCULATE the AREA for each REGION STEP 3: CALCULATE the DISTANCE from the ORIGIN to the CENTROID of each SHAPE STEP 4: MULTIPLY the value of the AREA by its DISTANCE from the ORIGIN for each REGION STEP 5: SUM UP the VALUES of all the AREAS and NOTE that value at the BOTTOM of the AREA column STEP 6: SUM UP the VALUES in the LAST COLUMN and NOTE the value at the BOTTOM of that column STEP 7: COMPUTE the CENTROID COORDINATE Lets run through this problem STEP BY STEP:
STEP 1: LIST EACH OF THE AREAS THAT MAKE UP THE DISCRETE REGION, INCLUDING ANY HOLES OR SUBTRACTED REGIONS Our first step is to list all of the areas that make up the DISCRETE REGION of the COMPOSITE BODY. For illustration purposes, we will be color coding each shape that we will evaluate in the problem. Each SUB-REGION will have its OWN INDIVIDUAL AREA and CENTROID, as they have DIFFERENT GEOMETRY and DIMENSIONS.
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When combined, we will use the GEOMETRY and CENTROID of each area, relative to the ORIGIN as established, to calculate the CENTROID of the COMPOSITE STRUCTURE, object, or body of interest, as provided in the problem statement. Breaking up our T-SECTION COMPOSITE BODY, we have:
Each SUBREGION will have its own AREA and its own CENTROID such that:
To ORGANIZE our calculations for defining the COMPOSITE CENTROIDAL COORDINATES, we will need to develop a SEPARATE TABLE for each X and Y CENTROIDAL DIMENSION.
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Through OBSERVATION, we can conclude that SYMMETRY exists along the Y-AXIS, or in other terms, the X-COORDINATE of our CENTROID will fall in line with the ORIGIN or rather: π₯B= = 0 Recall that an OBJECT is said to be SYMMETRICAL about an AXIS if we are able to establish that AXIS and confirm that the GEOMETRY that resides on one side is a REFLECTION, or MIRRORED IMAGE, of what resides on the opposite side. Most shapes that will be encountered on the FE EXAM will be of SYMMETRICAL nature. Letβs move on to defining our Y-COORDINATE. To determine the Y-CENTROID LOCATION, we will start with creating a table with the column headings as shown: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
1 (rectangle) 2 (rectangle) 3 (rectangle) 4 (rectangle) 5 (rectangle)
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yn an
Come exam day, this table doesnβt have to be fancy, it can be scratched out in its most rudimentary form, the point is to keep everything organized and in the forefront for use as we progress through each of the steps.
STEP 2: CALCULATE THE AREA FOR EACH REGION The second step of our process requires us to CALCULATE the AREA for EACH REGION, and populate our coordinate table that we set up in step 1. The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, provides a series of TABLES on pages 69 through 71 that include GENERAL FORMULAS for defining the CENTROIDS, and various other CHARACTERISTICS of the most COMMON SHAPES we will encounter come exam day. Specifically, each SHAPE will be defined for us in the FIRST COLUMN of the TABLE with the CENTROID and AREA FORMULAS defined for us in the SECOND COLUMN. We will be heavily relying on the use of these TABLES as we move forward.
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Revisiting the COMPOSITE BODY as we have it broken up at this point, we have:
It is important to recognize the ORIENTATION of each SHAPE as well as the AXIS OF SYMMETRY, as the GENERAL FORMULAS are based on the GEOMETRY and DIMENSIONS of a given SHAPE.
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Flipping back to page 96 of the The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we will want to specifically hone in on the SECOND COLUMN titled AREA & CENTROID, which is highlighted:
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The first SUBREGION we will consider is the PURPLE RECTANGLE, focusing in on the information and formulas highlighted here in green:
Noting that the GENERAL FORMULA for the AREA of this particular region is: A = bh
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And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: b = 3 in h = 1 in Taking these DIMENSIONS and plugging them in to this GENERAL FORMULA, we get: A = bh = 3 in 1 in = 3 in1 With a base of 3 ππ and a height of 1 ππ, we plug these numbers in to the formula and get an area of 3 in1 β¦which we place in to our Y-COORDINATE table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
3
yn
yn an
2 (rectangle) 3 (rectangle) 4 (rectangle) 5 (rectangle) We can now repeat this same process of referencing the TABLE, getting the GENERAL FORMULA, and quantifying the AREA of each SUB-REGION. Flipping back to page 96 of the The NCEES Supplied Reference Handbook, Version 9.4 Made with by Prepineer | Prepineer.com
for Computer Based Testing, we will want to again hone in on the SECOND COLUMN titled AREA & CENTROID, which we highlighted previously, for each of the REGIONS. Doing so for our REGION 2, we get: π΄1 = π β = 1 3 = 3 For our REGION 3, we get: π΄7 = π β = 8 1 = 8 For our REGION 4, we get: π΄< = π β = 1 3 = 3 For our REGION 5, we get: π΄Q = π β = 3 1 = 3
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Plugging in the calculated values for each SUB-REGION area into our Y-CENTROID COORDINATE TABLE, we find get: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
3
2 (rectangle)
3
3 (rectangle)
8
4 (rectangle)
3
5 (rectangle)
3
yn
yn an
STEP 3: CALCULATE THE DISTANCE FROM THE ORIGIN TO THE CENTROID OF EACH SHAPE We can now move on to Step 3 where we will CALCULATE the DISTANCE from the ORIGIN TO the CENTROID of EACH SHAPE we have defined.
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We know that the CENTROID will some DISTANCE π¦B= from the X-AXIS. Starting with our GREEN REGION and flipping back to our tables, we see that the formula for the Y-CENTROID of a RECTANGLE is defined:
Taking this GENERAL FORMULA for the Y-CENTROID of this particular region is: π¦= = β/2
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And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: h = 1 in We calculate the CENTROID of region 1 using the formula for a horizontal rectangle:
yT =
h 1 in = = .5 in 2 2
Plugging this value in to our table we now have: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
1 (rectangle)
3
0.5
2 (rectangle)
3
2.5
3 (rectangle)
8
4.5
4 (rectangle)
3
2.5
5 (rectangle)
3
0.5
yn an
We can now repeat this same process of referencing the TABLE, getting the GENERAL FORMULA, and quantifying the AREA of each SUB-REGION. Flipping back to page 96 of the The NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing, we will want to again hone in on the SECOND COLUMN titled AREA & CENTROID, which we highlighted previously, for each of the REGIONS.
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Moving to REGION 2, we know that the CENTROID will be some distance yT from the bottom of its shape PLUS the HEIGHT of REGION 1 since it is between it and the π₯ βaxis. This gives us a CENTROID COORDINATE of:
yT ? = 1 +
h 3 = 1 + = 2.5 2 2
Moving to REGION 3, the centroid will still remain at half the height, as all rectangles, but itβs also elevated 4 inches from the π₯ βaxis, so this must be accounted for. Plugging in our dimensions, we find that this centroid is located at 4.5 inches from the π₯ βaxis:
yT @ = 1 + 3 +
h 1 = 1 + 3 + = 4.5 2 2
Now we could continue on to REGION 4 and 5 just as we have done up to this point, that will work, but you can HACK time off the solution by recognizing the SYMMETRY of the COMPOSITE OBJECT in relation to the Y-AXIS. The Y-COORDINATE for the CENTROIDS of REGION 4 and 5 must be identical to those of REGION 1 and 2.
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For REGION 4:
yT A = 1 +
h 3 = 1 + = 2. 5 2 2
For REGION 5:
yT V =
h 1 = = 0.5 2 2
Plugging in the derived values for each SUB-REGION area into our Y-CENTROID COORDINATE TABLE, we find get: Y-CENTROID COORDINATE TABLE
Region
Area (an)
yn
1 (rectangle)
3
0.5
2 (rectangle)
3
2.5
3 (rectangle)
8
4.5
4 (rectangle)
3
2.5
5 (rectangle)
3
0.5
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yn an
STEP 4: MULTIPLY THE VALUE OF THE AREA BY ITS DISTANCE FROM THE ORIGIN FOR EACH REGION Moving on to STEP 4, for each individual region, we now need to take our SECOND COLUMN, the AREA, and MULTIPLY it by the THIRD COLUMN, the DISTANCE FROM THE ORIGINβ¦and plug this product in to our final column in the coordinate table. Highlighting the TWO COLUMNS in which we are MULTIPLYING as GREEN, we carry out this calculation and plug our RESULTS in to the last column, giving us: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
3
0.5
1.5
2 (rectangle)
3
2.5
7.5
3 (rectangle)
8
4.5
36
4 (rectangle)
3
2.5
7.5
5 (rectangle)
3
0.5
1.5
STEP 5: SUM UP THE VALUES OF ALL THE AREAS AND NOTE THAT VALUE AT THE BOTTOM OF THE AREA COLUMN Now for step 5, we SUM all the AREAS that we have defined up to this point.
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This CALCULATION takes in to account the AREA of ALL REGIONS that reside in the COLUMN we have highlighted GREEN: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
3
0.5
1.5
2 (rectangle)
3
2.5
7.5
3 (rectangle)
8
4.5
36
4 (rectangle)
3
2.5
7.5
5 (rectangle)
3
0.5
1.5
= 20
STEP 6: SUM UP THE VALUES IN THE LAST COLUMN AND NOTE THE VALUE AT THE BOTTOM OF THAT COLUMN For step 6, we do the same, SUMMING all the products that we calculated in step 4 and provide the SUM at the bottom of our table.
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This CALCULATION takes in to account the COLUMN we have highlighted GREEN: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
3
0.5
1.5
2 (rectangle)
3
2.5
7.5
3 (rectangle)
8
4.5
36
4 (rectangle)
3
2.5
7.5
5 (rectangle)
3
0.5
1.5
= 20
= 54
STEP 7: COMPUTE THE CENTROID COORDINATE In our final step, we now have all the data we need to define the Y-COORDINATE for the CENTROID of this COMPOSITE BODY. The GENERAL FORMULAS for defining the COORDINATES of a CENTROID of a general AREA can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Remember, that the CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM.
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The LOCATION of the CENTROID of an AREA is wholly a function of itβs GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
π₯B= =
πBD = π΄
π₯E πE π΄
Where π₯B= is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
π¦B= =
πBH = π΄
π¦E πE π΄
Where π¦B= which is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN Y-AXIS. We know the X-COORDINATE already, due to SYMMETRY, so letβs focus in on the GENERAL FORMULA for the Y-COORDINATE, which is:
π¦B= =
πBH = π΄
π¦E πE π΄
We much donβt care for the FIRST HALF of this FORMULA, but rather the SECOND HALF, such that:
π¦B= =
π¦E πE π΄
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From our COORDINATE TABLE we have developed in this problem, we know that: β’
π¦E πE = 54
β’ π΄ = 20 We can just plug and chug with our calculators to find that the Y-COORDINATE of the CENTROID for the T-SECTION is calculated as:
π¦B= =
54 20
Or: π¦B= = 2.7 ππ Identifying this on our schematic, we have:
The correct answer choice is C.(π, π. π)
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