13.1 Centroids Concept Overview
CENTROIDS | CONCEPT OVERVIEW The TOPIC of CENTROIDS of MASSES, AREAS, LENGTHS, and VOLUMES can be referenced on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: The CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM. There are a number of ways that CENTROIDS can be derived in real world applications including, but not limited to, the use of complex INTEGRAL CALCULUS. We will see the basis of such INTEGRATION, however, for the purposes of the FE EXAM and in an effort to streamline the process of establishing CENTROIDS for various OBJECTS, simple and complex alike, we will focus on configurations that require only the use of established GEOMETRIC RELATIONSHIPS. CENTROIDS are a fundamental concept that we learn in STATICS, but will be seen as applicable in many SUBJECTS down the line, most commonly, in VECTOR MECHANICS problems that require the student to LOCATE an equivalent CENTRAL POINT OF APPLICATION of a DISTRIBUTED LOAD, or any other load for that matter, that is not APPLIED at SINGLE POINT. An example of this would be a CANITLEVER BEAM loaded with a UNIFORMLY
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DISTRIBUTED LOAD across the GEOMETRICAL LENGTH. As we will become accustomed to when working these types of problems, it will be paramount to realize that when we develop EQUIVALENT LOADS and asked to assess the affects of that LOAD in relation to a particular point, that we are using the GEOMETRIC LOCATION of that load, the CENTROID, as our POINT OF APPLICATION…otherwise all measurements, MOMENTS, EQUILIBRIUM, etc, will result in a incorrect result. This would be catastrophic in real world applications.
CENTER OF GRAVITY: The CENTER OF GRAVITY is the AVERAGE LOCATION at which the SELF WEIGHT of the OBJECT is assumed to ACT, or in more simple terms, the LOCATION on the OBJECT where the GRAVITATION FORCE is ACTING. A RIGID BODY will always have both a CENTER OF GRAVITY and CENTROID, but the LOCATION of these two points will NOT ALWAYS be the SAME; the driving factor being whether or not the BODY is of HOMOGENEOUS or NONHOMOGENEOUS DENSITY. A BODY of HOMOGENEOUS, or rather UNIFORM DENSITY, has properties that are unchanging throughout the ENTIRETY of the BODY. In this case, the CENTROID will be located at the GEOMETRIC CENTER of the AREA, per the standard definition, and due to the UNFIORM DENSITY that exists within that
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AREA, the CENTER OF GRAVITY will also be LOCATED at this SAME POINT. On the other hand, when dealing with a BODY that is NONHOMOGENEOUS in DENSITY, the CENTROID would be located in the same GEOMETRIC CENTER of the AREA, but the CENTER OF GRAVITY would be located based on the DISTRIBUTION of DENSITY throughout that AREA. There may be an instance where the CENTER OF GRAVITY will be LOCATED at the SAME POINT as the CENTROID of a NONHOMOGENOUS BODY, but that is completely DEPENDENT on the DISTRIBUTION of the PROPERTIES that surround it. It all comes down to BALANCING all PROPERTIES around a CENTER OF GRAVITY, whereas, the CENTROID is completely DEPENDENT on GEOMETRY alone. A few key things to NOTE before moving on: • SYMMETRY: An object is said to be SYMMETRICAL about an AXIS if we are able to establish that AXIS and confirm that the GEOMETRY of which resides on one side is a REFLECTION, or MIRRORED IMAGE, of what resides on the other side of the AXIS. Most shapes that will be encountered on the FE EXAM will be of SYMMETRICAL nature. • SELF WEIGHT: The SELF WEIGHT of a RIGID BODY will always ACT at the CENTER OF GRAVITY of that BODY. The SELF WEIGHT is the FORCE resulting from the GRAVITATIONAL EFFECTS due to the MASS of the OBJECT.
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CENTER OF MASS: The GENERAL FORMULA for defining the CENTER OF MASS 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 CENTER OF MASS is the SINGLE LOCATION at which the TOTALITY of the OBJECT’S MASS is ASSUMED to ACT, or in more simple terms, the LOCATION where we can compile and simplify the overall properties of the SYSTEM to an EQUIVALENT SINGLE ACTION. The CENTER OF MASS isn’t always tied to GEOMETRIC RELATIONSHIPS, but rather the DISTRIBUTION of the MASS within the OBJECT. The CHARACERTISTICS of DENSITY and VOLUME drive the LOCATION of the CENTER OF MASS, and can be further developed depending on whether the AREA of note is a DISCRETE or CONTINUOUS REGION. Expanding on this further: • A DISCRETE REGION is an AREA that can be BROKEN UP into several SUBREGIONS composed of SIMPLE SHAPES, such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. Working with DISCRETE REGIONS will commonly be encountered when assessing COMPOSITE BODIES giving us a means to easily and efficiently
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express the various CHARACTERISTICS of the object as a whole (CENTROIDS, CENTER OF GRAVITY, CENTER OF MASS, etc) due to the fact that GEOMETRICAL DATA can be established. • A CONTINUOUS REGION is any REGION that isn’t classified as a DISCRETE REGION. These REGION is are typically represented by COMPLEX, or IRREGULAR SHAPED, BOUNDARIES. To establish the HOLISTIC CHARACTERISTICS of a BODY that encompasses a CONTINUOUS REGION, one would need to DEFINE the BOUNDARIES using COMPLEX MATHEMATICAL FUNCTIONS and then employ BASIC CALCULUS and INTEGRATION TECHNIQUES. In the context of the FE EXAM, we would expect that a student would encounter OBJECTS that can be considered to inherit the BOUNDARIES of a CONTINUOUS REGION, but the analysis would not be COMPLEX, but rather straight forward using the GENERAL FORMULAS that are provided in the NCEES Reference Handbook. For DISCRETE MASSES, AREAS, LENGTHS, and VOLUMES, the CENTER OF MASS can be derived using the GENERAL FORMULA:
𝑟" =
𝑚% 𝑟% 𝑚_𝑛
Where: • 𝑚% is the mass of each particle making up the system Made with by Prepineer | Prepineer.com
• 𝑟% is the radius vector to each particle from a selected reference point • 𝑟" is the radius vector to the CENTROID OF THE TOTAL MASS from the selected reference point This GENERAL FORMULA for defining the CENTER OF MASS, as well as all the associated TERMINOLGY that revolves around it, can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Analyzing COMPOSITE BODIES will be the way in which we will be tested on our understanding of DISCRETE REGIONS come the day of the exam. Most students will be presented a BODY made up of a number of different shapes, presenting itself as COMPLEX OBJECT. However, in most all cases, it will be a BODY made up of a number of DISCRETE REGIONS that can be BROKEN DOWN into several SUB-REGIONS composed of SIMPLE SHAPES, such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. When you’re working with these types of BODIES, it’s 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 when dealing with COMPOSITE BODIES, 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.
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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, and the CENTER OF MASS.
DEFINING THE COORDINATES OF A CENTROID: 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. Just as we used the MASS of a BODY to calculate the CENTER OF MASS, we use the AREA of a BODY to calculate the CENTROID OF AN AREA. The LOCATION of the CENTROID of an AREA is wholly a function of it’s GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
𝑥)" =
𝑀)+ = 𝐴
𝑥% 𝑎% 𝐴
Where 𝑥)" is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
𝑦)" =
𝑀)/ = 𝐴
𝑦% 𝑎% 𝐴 Made with
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Where 𝑦)" which is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN Y-AXIS. The various other terms strewn throughout these two GENERAL FORMULAS can be defined as: • 𝑥% is the DISTANCE from the ORIGIN to the CENTROID of SUB-REGION “𝑖” measured PARALLEL to the CARTESIAN X-AXIS. • 𝑦% is the DISTANCE from the ORIGIN to the CENTROID of SUB-REGION “𝑖” measured PARALLEL to the CARTESIAN Y-AXIS • 𝑎% is the AREA of SUBREGION “𝑛” • 𝑛 is the NUMBER of SUBREGIONS that make up the DISCRETE REGION • 𝐴 =
𝑎% is the SUM of the SUBREGIONS for any “𝑛” NUMBER of
SUBREGIONS 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 on the TABLE with the CENTROID and AREA FORMULAS defined for us in the SECOND COLUMN. It’s important to NOTE that when we are working CENTROID problems, that the GEOMETRY, ORIENTATION and how the SHAPE lays out in our established
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COORDINATE SYSTEM is of the upmost IMPORTANTANCE.
COMPOSITE BODIES: A COMPOSITE BODY, in context of what we would expect to see on the FE EXAM, is 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 again would be expected to be 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, a COMPOSITE BODY will have CLEARLY defined locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. From there, with all the appropriate data available, we are able to determine the LOCATIONS of the CENTROID, CENTER OF GRAVITY, and the CENTER OF MASS of 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 or SUBTRACT REGIONS to establish the PROPERTIES of the COMPOSITE BODY as a whole.
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Generally speaking, the process of working with COMPOSITE BODIES plays out cleanly following the steps as identified below: 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 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
FIRST MOMENT OF AREA: The GENERAL FORMULA for the FIRST MOMENT OF 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.
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In most cases, the TABLES that we find in the NCEES Reference Handbook will suffice in getting us definition around the CENTROID of an OBJECT. However, there may be times where we will use the FIRST MOMENT OF AREA as seen earlier in the GENERAL FORMULAS provided to us define the CENTROIDAL COORDINATES. The expression: ∫ 𝑥 𝑑𝐴 Is known as the FIRST MOMENT OF THE AREA, or FIRST AREA MOMENT with respect to the Y-AXIS. The expression: ∫ 𝑦 𝑑𝐴 Is known as the FIRST MOMENT OF THE AREA, or FIRST AREA MOMENT with respect to the X-AXIS. Further, the MOMENT OF AREA 𝑀) is defined as:
𝑀)+ =
𝑥% 𝑎%
𝑀)/ =
𝑦% 𝑎%
And:
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The PRIMARY APPLICATION of the FIRST MOMENT OF AREA is to CALCULATE the COORDINATES of a CENTROID. While it may seem that this GENERAL FORMULA is mostly mathematical, one great APPLICATION of the FIRST MOMENT of AREA is to CHECK if an AREA is SYMMETRICALLY DISTRIBUTED about an AXIS…that is, that all points are EVENLY SPREAD out from that AXIS. This SYMMETRY has deep implications in STRUCTURAL ANALYSIS.
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CENTROIDS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The location of the centroid from the origin of the composite T-section illustrated below is closest to:
A. 45 B. 55 C. 65 D. 75
SOLUTION: 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.
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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 POINT O at the TOP of the COMPOSITE BODY that is presented. This can also be confirmed by tracing back the location where the X and Y AXES INTERSECT. A COMPOSITE BODY is 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.
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Although not directly delineated as presented to us, this COMPOSITE BODY CLEARLY has TWO locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. The COMPOSITE SHAPES here will be TWO RECTANGLES. 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 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
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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 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 TSECTION. 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. 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.
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Breaking up our T-SECTION COMPOSITE BODY, we have:
Each SUBREGION will have its own AREA and its own CENTROID. 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. 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: 𝑥)" = 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
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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. Let’s 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
yn an
1 (rectangle) 2 (rectangle) Come exam day, this table doesn’t 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
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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. 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 GREEN 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 = 100 mm h = 20 mm Taking these DIMENSIONS and plugging them in to this GENERAL FORMULA, we get: A = bh = 100 mm 20 mm = 2,000 mm@ With a base of 100 and a height of 20, we plug these numbers in to the formula and get an area of 2,000 mm@ …which we place in to our Y-COORDINATE table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
2,000
yn
yn an
2 (rectangle) 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.
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This time, we will tackle the REGION identified as the PURPLE RECTANGLE, focusing in on the information and formulas highlighted here in PURPLE:
Noting again 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 = 15 mm h = 150 mm Taking these DIMENSIONS and plugging them in to this GENERAL FORMULA, we get: A = bh = 15 mm 150 mm = 2,250 mm@ With a base of 15 and a height of 150, we plug these numbers in to the formula and get an area of 2,250 mm@ …which we place in to our Y-COORDINATE table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
2,000
2 (rectangle)
2,250
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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Starting with our GREEN REGION:
We know that the CENTROID will some DISTANCE 𝑦"A from the X-AXIS.
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Flipping back to our tables, we see that the formula for the Y-CENTROID is defined:
Taking this GENERAL FORMULA for the Y-CENTROID of this particular region is: 𝑦" = ℎ/2
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And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: h = 20 mm We calculate the CENTROID of region 1 using the formula for a horizontal rectangle:
yE =
h 20 mm = = 10 mm 2 2
Identifying this in our illustration:
Plugging this value in to our table we now have: Y-CENTROID COORDINATE TABLE
Region
Area (an)
yn
1 (rectangle)
2,000
10
2 (rectangle)
2,250
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yn an
We can do the same for our SECOND REGION, knowing that the CENTROID will be some distance 𝑦"A from the top of IT’S SHAPE…that’s an IMPORTANT DISTINCTION, we will explain moving forward. Illustrating this, here is what we are looking at:
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Flipping back to the TABLES, we will hone in on the section highlighted in PURPLE:
Taking this GENERAL FORMULA for the Y-CENTROID of this particular region is: 𝑦" = ℎ/2 And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: h = 150 mm
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We calculate the CENTROID of region 2 using the formula for a vertical rectangle:
yE =
h 150 mm = = 75 mm 2 2
Some will call it day here, plug this value in to the table an move on…this would be a CATASTROPHIC mistake. The top of REGION 2 starts at the BOTTOM of our FIRST REGION, which is 20 mm high…meaning the top of region 2 is 20 mm from the ORIGIN. We must account for this offset by adding that height to get the full distance back to the origin and a TRUE distance to the CENTROID of this REGION as it relates to the COMPOSITE BODY it resides. We will calculate the distance as:
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Or in FORMULAIC TERMS: yE = yEF + hH Plugging in the values from the problem we find: yE = 20 mm + 75 mm = 95 mm We conclude that a distance to the CENTROID of REGION 2 as 95 𝑚𝑚 and we plug this value in to our coordinate table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
1 (rectangle)
2,000
10
2 (rectangle)
2,250
95
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 ORIGIN…and plug this product in to our final column in the coordinate table.
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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)
2,000
10
20,000
2 (rectangle)
2,250
95
213,750
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: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
2,000
10
20,000
2 (rectangle)
2,250
95
213,750
= 4,250
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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. This CALCULATION takes in to account the COLUMN we have highlighted GREEN: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
2,000
10
20,000
2 (rectangle)
2,250
95
213,750
= 4,250
= 233,750
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, or T-SECTION. 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
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as it resides relative to an established CARTESIAN COORDINATE SYSTEM. The LOCATION of the CENTROID of an AREA is wholly a function of it’s GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
𝑥)" =
𝑀)+ = 𝐴
𝑥% 𝑎% 𝐴
Where 𝑥)" is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
𝑦)" =
𝑀)/ = 𝐴
𝑦% 𝑎% 𝐴
Where 𝑦)" 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 let’s focus in on the GENERAL FORMULA for the Y-COORDINATE, which is:
𝑦)" =
𝑀)/ = 𝐴
𝑦% 𝑎% 𝐴
We much don’t care for the FIRST HALF of this FORMULA, but rather the SECOND HALF, such that:
𝑦)" =
𝑦% 𝑎% 𝐴 Made with
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From our COORDINATE TABLE we have developed in this problem, we know that: •
𝑦% 𝑎% = 233,750
• 𝐴 = 4,250 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:
𝑦)" =
233,750 4,250
Or: 𝑦)" = 55 𝑚𝑚 The correct answer choice is B. 𝟓𝟓
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CONCEPT INTRO: The CENTROID of a BODY is the GEOMETRIC CENTER of the AREA as it resides relative to an established CARTESIAN COORDINATE SYSTEM. There are a number of ways that CENTROIDS can be derived in real world applications including, but not limited to, the use of complex INTEGRAL CALCULUS. We will see the basis of such INTEGRATION, however, for the purposes of the FE EXAM and in an effort to streamline the process of establishing CENTROIDS for various OBJECTS, simple and complex alike, we will focus on configurations that require only the use of established GEOMETRIC RELATIONSHIPS. CENTROIDS are a fundamental concept that we learn in STATICS, but will be seen as applicable in many SUBJECTS down the line, most commonly, in VECTOR MECHANICS problems that require the student to LOCATE an equivalent CENTRAL POINT OF APPLICATION of a DISTRIBUTED LOAD, or any other load for that matter, that is not APPLIED at SINGLE POINT. An example of this would be a CANITLEVER BEAM loaded with a UNIFORMLY
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DISTRIBUTED LOAD across the GEOMETRICAL LENGTH. As we will become accustomed to when working these types of problems, it will be paramount to realize that when we develop EQUIVALENT LOADS and asked to assess the affects of that LOAD in relation to a particular point, that we are using the GEOMETRIC LOCATION of that load, the CENTROID, as our POINT OF APPLICATION…otherwise all measurements, MOMENTS, EQUILIBRIUM, etc, will result in a incorrect result. This would be catastrophic in real world applications.
CENTER OF GRAVITY: The CENTER OF GRAVITY is the AVERAGE LOCATION at which the SELF WEIGHT of the OBJECT is assumed to ACT, or in more simple terms, the LOCATION on the OBJECT where the GRAVITATION FORCE is ACTING. A RIGID BODY will always have both a CENTER OF GRAVITY and CENTROID, but the LOCATION of these two points will NOT ALWAYS be the SAME; the driving factor being whether or not the BODY is of HOMOGENEOUS or NONHOMOGENEOUS DENSITY. A BODY of HOMOGENEOUS, or rather UNIFORM DENSITY, has properties that are unchanging throughout the ENTIRETY of the BODY. In this case, the CENTROID will be located at the GEOMETRIC CENTER of the AREA, per the standard definition, and due to the UNFIORM DENSITY that exists within that
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AREA, the CENTER OF GRAVITY will also be LOCATED at this SAME POINT. On the other hand, when dealing with a BODY that is NONHOMOGENEOUS in DENSITY, the CENTROID would be located in the same GEOMETRIC CENTER of the AREA, but the CENTER OF GRAVITY would be located based on the DISTRIBUTION of DENSITY throughout that AREA. There may be an instance where the CENTER OF GRAVITY will be LOCATED at the SAME POINT as the CENTROID of a NONHOMOGENOUS BODY, but that is completely DEPENDENT on the DISTRIBUTION of the PROPERTIES that surround it. It all comes down to BALANCING all PROPERTIES around a CENTER OF GRAVITY, whereas, the CENTROID is completely DEPENDENT on GEOMETRY alone. A few key things to NOTE before moving on: • SYMMETRY: An object is said to be SYMMETRICAL about an AXIS if we are able to establish that AXIS and confirm that the GEOMETRY of which resides on one side is a REFLECTION, or MIRRORED IMAGE, of what resides on the other side of the AXIS. Most shapes that will be encountered on the FE EXAM will be of SYMMETRICAL nature. • SELF WEIGHT: The SELF WEIGHT of a RIGID BODY will always ACT at the CENTER OF GRAVITY of that BODY. The SELF WEIGHT is the FORCE resulting from the GRAVITATIONAL EFFECTS due to the MASS of the OBJECT.
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CENTER OF MASS: The GENERAL FORMULA for defining the CENTER OF MASS 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 CENTER OF MASS is the SINGLE LOCATION at which the TOTALITY of the OBJECT’S MASS is ASSUMED to ACT, or in more simple terms, the LOCATION where we can compile and simplify the overall properties of the SYSTEM to an EQUIVALENT SINGLE ACTION. The CENTER OF MASS isn’t always tied to GEOMETRIC RELATIONSHIPS, but rather the DISTRIBUTION of the MASS within the OBJECT. The CHARACERTISTICS of DENSITY and VOLUME drive the LOCATION of the CENTER OF MASS, and can be further developed depending on whether the AREA of note is a DISCRETE or CONTINUOUS REGION. Expanding on this further: • A DISCRETE REGION is an AREA that can be BROKEN UP into several SUBREGIONS composed of SIMPLE SHAPES, such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. Working with DISCRETE REGIONS will commonly be encountered when assessing COMPOSITE BODIES giving us a means to easily and efficiently
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express the various CHARACTERISTICS of the object as a whole (CENTROIDS, CENTER OF GRAVITY, CENTER OF MASS, etc) due to the fact that GEOMETRICAL DATA can be established. • A CONTINUOUS REGION is any REGION that isn’t classified as a DISCRETE REGION. These REGION is are typically represented by COMPLEX, or IRREGULAR SHAPED, BOUNDARIES. To establish the HOLISTIC CHARACTERISTICS of a BODY that encompasses a CONTINUOUS REGION, one would need to DEFINE the BOUNDARIES using COMPLEX MATHEMATICAL FUNCTIONS and then employ BASIC CALCULUS and INTEGRATION TECHNIQUES. In the context of the FE EXAM, we would expect that a student would encounter OBJECTS that can be considered to inherit the BOUNDARIES of a CONTINUOUS REGION, but the analysis would not be COMPLEX, but rather straight forward using the GENERAL FORMULAS that are provided in the NCEES Reference Handbook. For DISCRETE MASSES, AREAS, LENGTHS, and VOLUMES, the CENTER OF MASS can be derived using the GENERAL FORMULA:
𝑟" =
𝑚% 𝑟% 𝑚_𝑛
Where: • 𝑚% is the mass of each particle making up the system Made with by Prepineer | Prepineer.com
• 𝑟% is the radius vector to each particle from a selected reference point • 𝑟" is the radius vector to the CENTROID OF THE TOTAL MASS from the selected reference point This GENERAL FORMULA for defining the CENTER OF MASS, as well as all the associated TERMINOLGY that revolves around it, can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Analyzing COMPOSITE BODIES will be the way in which we will be tested on our understanding of DISCRETE REGIONS come the day of the exam. Most students will be presented a BODY made up of a number of different shapes, presenting itself as COMPLEX OBJECT. However, in most all cases, it will be a BODY made up of a number of DISCRETE REGIONS that can be BROKEN DOWN into several SUB-REGIONS composed of SIMPLE SHAPES, such as RECTANGLES, CIRCLES, TRIANGLES, and PARABOLIC SEGMENTS, with established, or easily determined, GEOMETRIC RELATIONSHIPS. When you’re working with these types of BODIES, it’s 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 when dealing with COMPOSITE BODIES, 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.
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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, and the CENTER OF MASS.
DEFINING THE COORDINATES OF A CENTROID: 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. Just as we used the MASS of a BODY to calculate the CENTER OF MASS, we use the AREA of a BODY to calculate the CENTROID OF AN AREA. The LOCATION of the CENTROID of an AREA is wholly a function of it’s GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
𝑥)" =
𝑀)+ = 𝐴
𝑥% 𝑎% 𝐴
Where 𝑥)" is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
𝑦)" =
𝑀)/ = 𝐴
𝑦% 𝑎% 𝐴 Made with
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Where 𝑦)" which is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN Y-AXIS. The various other terms strewn throughout these two GENERAL FORMULAS can be defined as: • 𝑥% is the DISTANCE from the ORIGIN to the CENTROID of SUB-REGION “𝑖” measured PARALLEL to the CARTESIAN X-AXIS. • 𝑦% is the DISTANCE from the ORIGIN to the CENTROID of SUB-REGION “𝑖” measured PARALLEL to the CARTESIAN Y-AXIS • 𝑎% is the AREA of SUBREGION “𝑛” • 𝑛 is the NUMBER of SUBREGIONS that make up the DISCRETE REGION • 𝐴 =
𝑎% is the SUM of the SUBREGIONS for any “𝑛” NUMBER of
SUBREGIONS 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 on the TABLE with the CENTROID and AREA FORMULAS defined for us in the SECOND COLUMN. It’s important to NOTE that when we are working CENTROID problems, that the GEOMETRY, ORIENTATION and how the SHAPE lays out in our established
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COORDINATE SYSTEM is of the upmost IMPORTANTANCE.
COMPOSITE BODIES: A COMPOSITE BODY, in context of what we would expect to see on the FE EXAM, is 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 again would be expected to be 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, a COMPOSITE BODY will have CLEARLY defined locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. From there, with all the appropriate data available, we are able to determine the LOCATIONS of the CENTROID, CENTER OF GRAVITY, and the CENTER OF MASS of 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 or SUBTRACT REGIONS to establish the PROPERTIES of the COMPOSITE BODY as a whole.
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Generally speaking, the process of working with COMPOSITE BODIES plays out cleanly following the steps as identified below: 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 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
FIRST MOMENT OF AREA: The GENERAL FORMULA for the FIRST MOMENT OF 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.
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In most cases, the TABLES that we find in the NCEES Reference Handbook will suffice in getting us definition around the CENTROID of an OBJECT. However, there may be times where we will use the FIRST MOMENT OF AREA as seen earlier in the GENERAL FORMULAS provided to us define the CENTROIDAL COORDINATES. The expression: ∫ 𝑥 𝑑𝐴 Is known as the FIRST MOMENT OF THE AREA, or FIRST AREA MOMENT with respect to the Y-AXIS. The expression: ∫ 𝑦 𝑑𝐴 Is known as the FIRST MOMENT OF THE AREA, or FIRST AREA MOMENT with respect to the X-AXIS. Further, the MOMENT OF AREA 𝑀) is defined as:
𝑀)+ =
𝑥% 𝑎%
𝑀)/ =
𝑦% 𝑎%
And:
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The PRIMARY APPLICATION of the FIRST MOMENT OF AREA is to CALCULATE the COORDINATES of a CENTROID. While it may seem that this GENERAL FORMULA is mostly mathematical, one great APPLICATION of the FIRST MOMENT of AREA is to CHECK if an AREA is SYMMETRICALLY DISTRIBUTED about an AXIS…that is, that all points are EVENLY SPREAD out from that AXIS. This SYMMETRY has deep implications in STRUCTURAL ANALYSIS.
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CENTROIDS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. The location of the centroid from the origin of the composite T-section illustrated below is closest to:
A. 45 B. 55 C. 65 D. 75
SOLUTION: 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.
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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 POINT O at the TOP of the COMPOSITE BODY that is presented. This can also be confirmed by tracing back the location where the X and Y AXES INTERSECT. A COMPOSITE BODY is 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.
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Although not directly delineated as presented to us, this COMPOSITE BODY CLEARLY has TWO locations where we are able to SECTION, or DIVIDE, the BODY in to these COMPOSITE SHAPES. The COMPOSITE SHAPES here will be TWO RECTANGLES. 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 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
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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 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 TSECTION. 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. 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.
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Breaking up our T-SECTION COMPOSITE BODY, we have:
Each SUBREGION will have its own AREA and its own CENTROID. 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. 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: 𝑥)" = 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
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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. Let’s 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
yn an
1 (rectangle) 2 (rectangle) Come exam day, this table doesn’t 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
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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. 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 GREEN 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 = 100 mm h = 20 mm Taking these DIMENSIONS and plugging them in to this GENERAL FORMULA, we get: A = bh = 100 mm 20 mm = 2,000 mm@ With a base of 100 and a height of 20, we plug these numbers in to the formula and get an area of 2,000 mm@ …which we place in to our Y-COORDINATE table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
2,000
yn
yn an
2 (rectangle) 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.
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This time, we will tackle the REGION identified as the PURPLE RECTANGLE, focusing in on the information and formulas highlighted here in PURPLE:
Noting again 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 = 15 mm h = 150 mm Taking these DIMENSIONS and plugging them in to this GENERAL FORMULA, we get: A = bh = 15 mm 150 mm = 2,250 mm@ With a base of 15 and a height of 150, we plug these numbers in to the formula and get an area of 2,250 mm@ …which we place in to our Y-COORDINATE table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
1 (rectangle)
2,000
2 (rectangle)
2,250
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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Starting with our GREEN REGION:
We know that the CENTROID will some DISTANCE 𝑦"A from the X-AXIS.
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Flipping back to our tables, we see that the formula for the Y-CENTROID is defined:
Taking this GENERAL FORMULA for the Y-CENTROID of this particular region is: 𝑦" = ℎ/2
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And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: h = 20 mm We calculate the CENTROID of region 1 using the formula for a horizontal rectangle:
yE =
h 20 mm = = 10 mm 2 2
Identifying this in our illustration:
Plugging this value in to our table we now have: Y-CENTROID COORDINATE TABLE
Region
Area (an)
yn
1 (rectangle)
2,000
10
2 (rectangle)
2,250
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yn an
We can do the same for our SECOND REGION, knowing that the CENTROID will be some distance 𝑦"A from the top of IT’S SHAPE…that’s an IMPORTANT DISTINCTION, we will explain moving forward. Illustrating this, here is what we are looking at:
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Flipping back to the TABLES, we will hone in on the section highlighted in PURPLE:
Taking this GENERAL FORMULA for the Y-CENTROID of this particular region is: 𝑦" = ℎ/2 And that per the ORIENTATION of how this RECTANGLE sits in our established COORDINATE SYSTEM, we have: h = 150 mm
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We calculate the CENTROID of region 2 using the formula for a vertical rectangle:
yE =
h 150 mm = = 75 mm 2 2
Some will call it day here, plug this value in to the table an move on…this would be a CATASTROPHIC mistake. The top of REGION 2 starts at the BOTTOM of our FIRST REGION, which is 20 mm high…meaning the top of region 2 is 20 mm from the ORIGIN. We must account for this offset by adding that height to get the full distance back to the origin and a TRUE distance to the CENTROID of this REGION as it relates to the COMPOSITE BODY it resides. We will calculate the distance as:
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Or in FORMULAIC TERMS: yE = yEF + hH Plugging in the values from the problem we find: yE = 20 mm + 75 mm = 95 mm We conclude that a distance to the CENTROID of REGION 2 as 95 𝑚𝑚 and we plug this value in to our coordinate table, such that: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
1 (rectangle)
2,000
10
2 (rectangle)
2,250
95
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 ORIGIN…and plug this product in to our final column in the coordinate table.
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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)
2,000
10
20,000
2 (rectangle)
2,250
95
213,750
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: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
2,000
10
20,000
2 (rectangle)
2,250
95
213,750
= 4,250
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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. This CALCULATION takes in to account the COLUMN we have highlighted GREEN: Y-CENTROID COORDINATE TABLE Region
Area (an)
yn
yn an
1 (rectangle)
2,000
10
20,000
2 (rectangle)
2,250
95
213,750
= 4,250
= 233,750
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, or T-SECTION. 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
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as it resides relative to an established CARTESIAN COORDINATE SYSTEM. The LOCATION of the CENTROID of an AREA is wholly a function of it’s GEOMETRY, and generically identified using coordinate NOMENCLATURE and the GENERAL FORMULAS:
𝑥)" =
𝑀)+ = 𝐴
𝑥% 𝑎% 𝐴
Where 𝑥)" is the is the DISTANCE from the ORIGIN to the CENTROID of the DISCRETE REGION measured PARALLEL to the CARTESIAN X-AXIS.
𝑦)" =
𝑀)/ = 𝐴
𝑦% 𝑎% 𝐴
Where 𝑦)" 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 let’s focus in on the GENERAL FORMULA for the Y-COORDINATE, which is:
𝑦)" =
𝑀)/ = 𝐴
𝑦% 𝑎% 𝐴
We much don’t care for the FIRST HALF of this FORMULA, but rather the SECOND HALF, such that:
𝑦)" =
𝑦% 𝑎% 𝐴 Made with
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From our COORDINATE TABLE we have developed in this problem, we know that: •
𝑦% 𝑎% = 233,750
• 𝐴 = 4,250 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:
𝑦)" =
233,750 4,250
Or: 𝑦)" = 55 𝑚𝑚 The correct answer choice is B. 𝟓𝟓
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