22 Moment of Inertia Problem Set
MOMENT OF INERTIA | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: An I-beam is used to construct an offshore oil platform in the Gulf of Mexico. The thickness of the web and flanges of the I-beam is said to be the same, with a constant thickness βπ‘β. If the moment of inertia about the parallel x-axis is 245.833 cm4 and the cross-sectional area is 250 cm2, what is the moment of inertia about the centroidal axis most close to:
A. 2.1 π₯ 104 B. 8.0 π₯ 104 C. 1.5 π₯ 105 D. 2.5 π₯ 105 Made with
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SOLUTION 1: The FORMULAS for the MOMENT OF INERTIA can be referenced under the SUBJECT of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The first step in this problem is to look at the DIAGRAM of the I-beam and the GEOMETRY relating the CENTROILDAL X-AXIS of the I-beam to the PARALLEL XAXIS about which the moment of inertia is being calculated. We need to remember that the CENTROIDAL AXIS will line up with the CENTROID of the BODY being ANALYZED. As we are looking to solve for the moment of inertia about the PARALLEL AXIS, given the moment of inertia about the PARALLEL AXIS, we will use the PARALLEL-AXIS THEOREM to relate the two moment of inertia based on the PERPENDICULAR DISTANCE "d" between the TWO AXES. The FORMULAS for the MOMENT OF INERTIA PARALLEL-AXIS THEOREM can be referenced under the SUBJECT of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
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The moment of inertia about an area about any axis defined as the moment of inertia of the area about a parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance (d) from the centroidal axis to the axis in question: Ixβ² = Ix c + d2x A Iyβ² = Iy + d2y A c
Where: β’ ππ₯ , ππ¦ is the distance between the two axes in question and is also known as the TRANSFER DISTANCE β’ πΌπ₯π , πΌπ¦π is the moment of inertia about the centroidal axis β² β² β’ πΌπ₯π , πΌπ¦π is the moment of inertia about the new or parallel axis
β’ π΄ is the cross-sectional area β’ π is the perpendicular distance between the centroidal axis and the parallel axis, and is also referred to as the TRANSFER DISTANCE β’ π2 π΄ is often referred to as the TRANSFER COEFFICIENT or TRANSFER TERM As we are concerned with the INERTIA about the X-AXIS, we will use the following formula: Ixβ² = Ix c + d2x A
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Looking at this formula we are given a value for the moment of inertia about the x-axis and the cross-sectional area, but are not explicitly given a value for the perpendicular distance between the centroidal axis and the parallel axis. Going back to the geometry given in the original problem, we find that the perpendicular distance between the centroidal axis and the parallel axis is 30 cm.
Re-arranging the original equation for the PARALLEL AXIS THEOREM about the XAXIS, we can plug in the values to solve for the moment of inertia bout the centroidal axis as: Ixc = Ixβ² β d2x A = (245,833 cm4 ) β (250 cm2 )(30 cm)2 = 20,833 cm4
Therefore, the correct answer choice is A. π. π π πππ πππ Made with
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PROBLEM 2: The Moment of Inertia of the beamβs cross-sectional area about the π₯ βcentroidal axis is closest to (ππ.4 ):
A. 2.9 π₯ 109 B. 5.0 π₯ 109 C. 7.2 π₯ 109 D. 8.6 π₯ 109
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SOLUTION 2: The TOPIC of MOMENT OF INERTIA can be referenced under the SUBJECT of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS The first step we need to do is classify this object as a discrete region or continuous region by quickly observing that it can be broken up into several sub-regions composed of simple shapes (triangles, circles and rectangles are the most common shapes).
It is preferable to work with discrete regions as this means we are on the right track, and the NCEES Reference Manual, Version 9.4, confirms this by providing tables and formulas to determine the moment of inertia for discrete regions. A discrete region is a type of region made up of a combination of shapes referred to as Made with by Prepineer | Prepineer.com
sub-regions. Each sub-region has its own individual area, centroid, and moment of inertia calculation that is usually fairly simple to compute.
We then combine these sub-region properties to compute a single moment of inertia for the composite shape we are given. We will use a table (See below) to help us organize the different terms associated with moment of inertia calculations.
Iβx - Moment of Inertia Table Region
Area (A)
dy
dy2 A
Ixc
Iβx
We construct a table for each axis we are considering, with one value for the moment of inertia with respect to the π₯ β ππ₯ππ or one with respect to the π¦ β ππ₯ππ . Each sub-region has its own unique area and inertia values, as we have listed out in our Made with
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moment of inertia table below. From looking at the object given in the problem, we can break it in three sub-regions by identifying sub-region 1, our yellow region, as a rectangle (vertical orientation), sub-region 2, our green region, as a rectangle (horizontal orientation). and sub-region 3, our blue region, as a rectangle (vertical orientation).
Iβx - Moment of Inertia Table Region
Area (A)
dy
dy2 A
Ixc
Iβx
1 (rectangle) 2 (rectangle) 3 (rectangle) STEP 2: CALCULATE the AREA for each REGION The TABLE for the PROPERTIES OF COMMON SHAPES can be referenced under the SUBJECT of STATICS on pages 69 to 71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The next step is to calculate the area for each region using the tables in the Reference Handbook. If we grab our NCEES Reference Manual, Version 9.4, and flip back to pages 69 through 71, we will find a few tables defining the characteristics of a number of simple shapes.
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The first column defines the shape, whether itβs a triangle, rectangle, trapezoid, circle etc. The second column provides the area and centroid formulas for the particular shape and shape orientation, while the third column will give us the general formulas to determine the AREA MOMENT OF INERTIA of that particular shape.
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For sub-region 1 we are given a rectangle, which is a simple shape whose properties are provided in the tables found in the NCEES Reference Handbook, Version 9.4. It is important that you are able to easily identify the geometry of a shape, and identify which formulas to use to calculate the area, centroid, and moment of inertia of the shape.
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The FORMULA for the AREA OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Using the formula for the area of a rectangle, the area of sub-region π΄1 is defined as the base times the height. π΄1 = πβ = (100)(300) = 30,000ππ. With a base of 100ππ. and a height of 300ππ., we plug these numbers in to the formula for the area of a rectangle and get an area of 30,000ππ., which we then place into our moment of inertia table that we set up in step 1.
Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
dy
dy2 A
Ixc
Iβx
2 (rectangle) 3 (rectangle) As our second and third regions are a rectangle as well, there is no need to flip back to the tables, we can just move forward with using the same area formula base times height for the area of a rectangle. π΄2 = (600)(100) = 60,000in. Made with
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π΄3 = (100)(300) = 30,000ππ. Plugging in the dimensions, we define the area as 60,000ππ. for sub-region 2 and 30,000ππ. for sub-region 3, and list these values in the coordinate table as shown below.
Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
2 (rectangle)
60,000
3 (rectangle)
30,000
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dy
dy2 A
Ixc
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Iβx
STEP 3: CALCULATE the DISTANCE from the NEUTRAL AXIS to the CENTROID of each SHAPE We can now move on to Step 3, where we will calculate the distance from the NEUTRAL AXIS to the CENTROID of each shape, or sub-region. The problem is specifically asking us to determine the moment of inertia with respect to the π₯ β ππ₯ππ , such that the π₯ β ππ₯ππ will be our neutral axis, and reference axis to which we will be measuring back to.
Starting with region 1, we know that the centroid will be some distance π¦π from the bottom of its shape.
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The FORMULA for the CENTROID OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We know that a rectangle is symmetrical, and that the formula for the π¦ βcentroid of a rectangle is the vertical midpoint of the shape, as represented by the formula:
π¦π =
β 2
We can plug in the height of 300 ππ. and define this regionβs centroid as 150 ππ. from the bottom of its region.
π¦π1 =
β 300 = = 150in. 2 2
It is important that we realize that we need to add 50 ππ. to this measurement to take it all the way back to the π₯ β ππ₯ππ : ππ¦ = 150 + 50 = 200in.
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Doing so gives us 200 ππ., and we plug that value into our table:
Iβx - Moment of Inertia Table Region
Area (A)
dy
1 (rectangle)
30,000
200
2 (rectangle)
60,000
3 (rectangle)
30,000
dy2 A
Ixc
Iβx
Now moving to our second region, the centroidal axis of the overall object lines up with the centroidal axis for sub-region 2. As there is no distance between the two axes, we do not need to use the parallel-axis theorem. Therefore, we can go ahead and just put a tick mark to indicate a distance of 0, for when we do our calculations.
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Iβx - Moment of Inertia Table Region
Area (A)
dy
1 (rectangle)
30,000
200
2 (rectangle)
60,000
-
3 (rectangle)
30,000
dy2 A
Ixc
Iβx
Now moving to our third region, we know that the centroid will be some distance π¦π from the top of its region, which happens to reside on the π₯ β ππ₯ππ .
We can plug in the height of 300 ππ. and define this regionβs centroid as 150 ππ. from the top of its region.
π¦π3 =
β 300 = = 150in. 2 2 Made with
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It is important that we realize that we need to add 50ππ. to this measurement to take it all the way back to the π₯ β ππ₯ππ :
Doing so gives us 200ππ., and we plug that value in to our table: ππ¦ = 150 + 50 = 200in.
Iβx - Moment of Inertia Table Region
Area (A)
dy
1 (rectangle)
30,000
200
2 (rectangle)
60,000
-
3 (rectangle)
30,000
200
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dy2 A
Ixc
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Iβx
STEP 4: DEFINE the TRANSFER COEFFICIENT for each region by MULTIPLYING the value of the AREA by the SQUARE of its DISTANCE from the NEUTRAL AXIS Moving on to STEP 4, for each individual region, we now need to define what is called a TRANSFER COEFFICIENT. To determine this value, we take the AREA of each REGION and multiply it by the square of the distance.
STEP 5: CALCULATE the MOMENT OF INERTIA for each REGION about its own CENTROIDAL AXIS Now for step 5, we CALCULATE the MOMENT OF INERTIA for each individual REGION about its own CENTROIDAL AXIS. The formulas for the MOMENT OF INERTIA OF COMMON SHAPES can be referenced in the tables for the PROPERTIES OF COMMON SHAPES under the topic of STATICS on pages 69-71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Made with
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Referencing the tables in the Reference Handbook, we are able to determine that the moment of inertia for a rectangleβs centroidal π₯ β ππ₯ππ is given by the formula:
πΌπ₯π
πβ3 = 12
We plug in our values and get a calculate the moment of inertia for sub-region π΄1 as:
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πΌπ₯π1
πβ3 (100in. )(300in. )3 = = = 2.25 x 108 ππ.4 12 12
With our second and third sub-regions being a rectangle as well, we can use the same formula, and calculate the moment of inertia for each sub-region as:
πΌπ₯π2
πβ3 (600in. )(100in. )3 = = = 5.0 x 107 ππ.4 12 12
πΌπ₯π3
πβ3 (100in. )(300in. )3 = = = 2.25 x 108 ππ.4 12 12
We then list the calculated values into our TABLE.
STEP 6: Use the PARALLEL AXIS THEOREM to DEFINE the MOMENT OF INERTIA of each REGION about the NEUTRAL AXIS So moving on with our problem, the first value we are concerned with is the MOMENT OF INERTIA (MOI) of the individual shape about its CENTROIDAL axis, which we just defined in STEP 5 and this is highlighted in the table. We then the TRANSFER COEFFICIENT, or TRANSFER TERM which we defined in STEP 4, to the MOI. Made with
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Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 1 as: πΌ β² π₯1 = (1.88 π₯ 109 ) + (2.25 π₯ 108 ) = 1.43 π₯ 109 As the centroidal axis of sub-region 2 is the same as the centroidal axis for the overall object, the first term using the parallel-axis theorem will be 0, as there is no distance between the two axes: πΌ β² π₯2 = (0) + (5.0 π₯ 107 ) = 5.0 π₯ 107 Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 3 as: πΌ β² π₯3 = (1.88 π₯ 109 ) + (2.25 π₯ 108 ) = 1.43 π₯ 109
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STEP 7: SUM all the VALUES for the adjusted MOMENT OF INERTIA about the NEUTRAL AXIS to obtain the MOMENT OF INERTIA for the COMPOSITE SHAPE Summing the shapes that represent the composite T-section, we calculate the total moment of inertia as: Ixβ² = Ix1 + Ix2 + Ix 3 = (1.43 x 109 ) + (5.0 x 107 ) + (1.43 x 109 ) = 2.9 x 109 in4
Therefore, the correct answer choice is A. π. π π πππ
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PROBLEM 3: The Moment of Inertia of the beamβs cross-sectional area about the π¦ βcentroidal axis is closest to (ππ.4 ):
A. 2.4 π₯ 109 B. 3.8 π₯ 109 C. 4.1π₯ 109 D. 5.6 π₯ 109
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SOLUTION 3: The TOPIC of MOMENT OF INERTIA can be referenced under the SUBJECT of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS The first step we need to do is classify this object as a discrete region or continuous region by quickly observing that it can be broken up into several sub-regions composed of simple shapes (triangles, circles, and rectangles are the most common shapes).
It is preferable to work with discrete regions as this means we are on the right track, and the NCEES Reference Manual confirms this by providing us tables and formulas to determine the moment of inertia for discrete regions. A discrete region is a type of region made up of a combination of shapes referred to as Made with by Prepineer | Prepineer.com
sub-regions. Each sub-region has its own individual area, centroid, and moment of inertia calculation that is usually fairly simple to compute.
We then combine these sub-region properties to compute a single moment of inertia for the composite shape we are given. We will use a table to help us organize the different terms associated with moment of inertia calculations, as shown below.
Iβx - Moment of Inertia Table Region
Area (A)
dy
dy2 A
Ixc
Iβx
We construct a table for each axis we are considering, with one value for the moment of inertia with respect to the π₯ β ππ₯ππ or one with respect to the π¦ β ππ₯ππ .
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Each sub-region has its own unique area and inertia values, as we have listed in our moment of inertia table below. Looking at the object given in the problem, we can break it in three sub-regions by identifying sub-region 1, our yellow region, as a rectangle (vertical orientation), sub-region 2, our green region, as a rectangle (horizontal orientation), and sub-region 3, our blue region, as a rectangle (vertical orientation).
Iβy - Moment of Inertia Table Region
Area (A)
dx
dx 2 A
Iyc
Iβy
1 (rectangle) 2 (rectangle) 3 (rectangle) STEP 2: CALCULATE the AREA for each REGION The TABLES for the PROPERTIES OF COMMON SHAPES can be referenced under the SUBJECT of STATICS on pages 69 to 71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The next step is to calculate the area for each region using the tables in the Reference Handbook. If we grab our NCEES Reference Manual, Version 9.4, and flip back to pages 69 through 71, we will find a few tables defining the characteristics of a number of simple shapes.
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The first column defines the shape, whether itβs a triangle, rectangle, trapezoid, circle etc. The second column provides the area and centroid formulas for the particular shape and shape orientation, while the third column will give us the general formulas to determine the AREA MOMENT OF INERTIA of that particular shape.
For sub-region 1 we are given a rectangle, which is a simple shape whose properties are provided in the tables found in the NCEES Reference Handbook. It is important that you are able to easily identify the geometry of a shape, and identify which formulas to Made with
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use to calculate the area, centroid, and moment of inertia of the shape.
The FORMULA for the AREA OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Using the formula for the area of a rectangle, the area of sub-region 1 is defined as the base times the height.
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π΄1 = πβ = (100)(300) = 30,000 With a base of 100ππ. and a height of 300ππ., we plug these numbers into the formula for the area of a rectangle and get an area of 30,000ππ.2 , which we then place into our moment of inertia table that we set up in step 1.
Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
dy
dy2 A
Ixc
Iβx
2 (rectangle) 3 (rectangle) As our second and third regions are a rectangle as well, there is no need to flip back to the tables, we can just move forward with using the same area formula base times height for the area of a rectangle. π΄2 = (600in. )(100in. ) = 60,000in.2 π΄3 = (100ππ. )(300ππ. ) = 30,000ππ.2 Plugging in the dimensions, we define the area as 60,000ππ.2 for sub-region 2 and 30,000ππ.2 for sub-region 3, and list these values in the coordinate table as shown below.
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Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
2 (rectangle)
60,000
3 (rectangle)
30,000
dy
dy2 A
Ixc
Iβx
STEP 3: CALCULATE the DISTANCE from the NEUTRAL AXIS to the CENTROID of each SHAPE We can now move on to Step 3, where we will calculate the distance from the NEUTRAL AXIS to the CENTROID of each shape, or sub-region. The problem is specifically asking us to determine the moment of inertia with respect to the π¦ β ππ₯ππ , such that the π¦ β ππ₯ππ will be our neutral axis, and reference axis to which we will be measuring back to.
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Starting with region 1, we know that the centroid will be some distance π₯π from the bottom of its shape.
The FORMULA for the CENTROID OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We know that a rectangle is symmetrical, and that the formula for the π₯ βcentroid of a rectangle is the horizontal midpoint of the shape, as represented by the formula:
π₯π =
π 2
We can plug in the base of 100ππ. and define this regions centroid as 50ππ. from the left side of its region. Made with
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π₯π1 =
π 100 = = 50 2 2
It is important that we realize that we need to add 200ππ. to this measurement to take it all the way back to the y-axis: ππ₯ = 200 + 50 = 250
Doing so gives us 250ππ., and we list that value in our table:
Iβy - Moment of Inertia Table Region
Area (A)
dx
1 (rectangle)
30,000
250
2 (rectangle)
60,000
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dx 2 A
Iyc
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Iβy
3 (rectangle)
30,000
Now moving on to our second region, the centroidal axis of the overall object lines up with the centroidal axis for sub-region 2. As there is no distance between the two axes, we do not need to use the parallel-axis theorem. Therefore, we can go ahead and just put a tick mark to indicate a distance of 0, for when we do our calculations.
Iβy - Moment of Inertia Table Region
Area (A)
dx
1 (rectangle)
30,000
250
2 (rectangle)
60,000
-
3 (rectangle)
30,000
dx 2 A
Iyc
Iβy
Now moving to our third region, we know that the centroid will be some distance π₯π from the right side of its region:
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We can plug in the base of 100ππ. and define this regions centroid as 50ππ. from the right side of its region.
π₯π3 =
π 100 = = 50 2 2
It is important that we realize that we need to add 50ππ. to this measurement to take it all the way back to the π¦ β ππ₯ππ :
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Doing so gives us 250ππ., and we can list that value in our table: ππ₯ = 200 + 50 = 250in.
Iβy - Moment of Inertia Table Region
Area (A)
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dx
dx 2 A
Iyc
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Iβy
1 (rectangle)
30,000
250
2 (rectangle)
60,000
-
3 (rectangle)
30,000
250
STEP 4: DEFINE the TRANSFER COEFFICIENT for each region by MULTIPLYING the value of the AREA by the SQUARE of its DISTANCE from the NEUTRAL AXIS Moving on to STEP 4, for each individual region, we now need to define what is called a TRANSFER COEFFICIENT. To determine this value, we take the AREA of each REGION and multiply it by the square of the distance.
STEP 5: CALCULATE the MOMENT OF INERTIA for each REGION about its own CENTROIDAL AXIS Now for step 5, we CALCULATE the MOMENT OF INERTIA for each individual REGION about its own CENTROIDAL AXIS. Made with
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The FORMULAS for the MOMENT OF INERTIA OF COMMON SHAPES can be referenced in the TABLES for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on pages 69-71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
Referencing the tables in the Reference Handbook, we are able to determine that the moment of inertia for a rectangleβs centroidal π¦ β ππ₯ππ is given by the formula:
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πΌπ¦π
π3 β = 12
We plug in our values and calculate the moment of inertia for sub-region 1 as: π 3 h (100in. )3 (300in. ) πΌπ¦π1 = = = 2.5 x 107 ππ.4 12 12 With our second and third sub-regions being rectangles as well, we can use the same formula, and calculate the moment of inertia for each sub-region as:
πΌπ¦π2
π 3 h (600in. )3 (100in. ) = = = 1.8 x 109 ππ.4 12 12
πΌπ¦π3
π 3 h (100in. )3 (300in. ) = = = 2.5 x 107 ππ.4 12 12
We then list the calculated values in our TABLE.
Iβy - Moment of Inertia Table Region
Area (A)
dx
dx 2 A
Iyc
1 (rectangle)
30,000
250
1.88E9
2.5E7
2 (rectangle)
60,000
-
-
1.8E9
3 (rectangle)
30,000
250
1.88E9
2.5E7
Iβy
STEP 6: Use the PARALLEL AXIS THEOREM to DEFINE the MOMENT OF Made with
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INERTIA of each REGION about the NEUTRAL AXIS So moving on with our problem, the first value we are concerned with is the MOMENT OF INERTIA (MOI) of the individual shape about its CENTROIDAL axis, which we just defined in STEP 5 and is highlighted in the table. We then add the TRANSFER COEFFICIENT, or TRANSFER TERM which we defined in STEP 4, to the MOI.
Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 1 as: πΌ β² π¦1 = (1.88 π₯ 109 ) + (2.5 π₯ 107 ) = 1.9 π₯ 109 As the centroidal axis of sub-region 2 is the same as the centroidal axis for the overall object, the first term using the parallel-axis theorem will be 0, as there is no distance between the two axes: πΌ β² π¦2 = (0) + (1.8 π₯ 109 ) = 1.8 π₯ 109
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Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 3 as: πΌ β² π¦3 = (1.88 π₯ 109 ) + (2.5 π₯ 107 ) = 1.9 π₯ 109 STEP 7: SUM all the VALUES for the adjusted MOMENT OF INERTIA about the NEUTRAL AXIS to obtain the MOMENT OF INERTIA for the COMPOSITE SHAPE Summing the shapes that represent the composite T-section, we calculate the total moment of inertia as: Iyβ² = Iy1 + Iy2 + Iy = (1.9 x 109 ) + (1.8 x 109 ) + (1.9 x 109 ) = 5.6 x 109 in4 3
Therefore, the correct answer choice is D. π. π π πππ ππ.π
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PROBLEM 1: An I-beam is used to construct an offshore oil platform in the Gulf of Mexico. The thickness of the web and flanges of the I-beam is said to be the same, with a constant thickness βπ‘β. If the moment of inertia about the parallel x-axis is 245.833 cm4 and the cross-sectional area is 250 cm2, what is the moment of inertia about the centroidal axis most close to:
A. 2.1 π₯ 104 B. 8.0 π₯ 104 C. 1.5 π₯ 105 D. 2.5 π₯ 105 Made with
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SOLUTION 1: The FORMULAS for the MOMENT OF INERTIA can be referenced under the SUBJECT of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The first step in this problem is to look at the DIAGRAM of the I-beam and the GEOMETRY relating the CENTROILDAL X-AXIS of the I-beam to the PARALLEL XAXIS about which the moment of inertia is being calculated. We need to remember that the CENTROIDAL AXIS will line up with the CENTROID of the BODY being ANALYZED. As we are looking to solve for the moment of inertia about the PARALLEL AXIS, given the moment of inertia about the PARALLEL AXIS, we will use the PARALLEL-AXIS THEOREM to relate the two moment of inertia based on the PERPENDICULAR DISTANCE "d" between the TWO AXES. The FORMULAS for the MOMENT OF INERTIA PARALLEL-AXIS THEOREM can be referenced under the SUBJECT of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
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The moment of inertia about an area about any axis defined as the moment of inertia of the area about a parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance (d) from the centroidal axis to the axis in question: Ixβ² = Ix c + d2x A Iyβ² = Iy + d2y A c
Where: β’ ππ₯ , ππ¦ is the distance between the two axes in question and is also known as the TRANSFER DISTANCE β’ πΌπ₯π , πΌπ¦π is the moment of inertia about the centroidal axis β² β² β’ πΌπ₯π , πΌπ¦π is the moment of inertia about the new or parallel axis
β’ π΄ is the cross-sectional area β’ π is the perpendicular distance between the centroidal axis and the parallel axis, and is also referred to as the TRANSFER DISTANCE β’ π2 π΄ is often referred to as the TRANSFER COEFFICIENT or TRANSFER TERM As we are concerned with the INERTIA about the X-AXIS, we will use the following formula: Ixβ² = Ix c + d2x A
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Looking at this formula we are given a value for the moment of inertia about the x-axis and the cross-sectional area, but are not explicitly given a value for the perpendicular distance between the centroidal axis and the parallel axis. Going back to the geometry given in the original problem, we find that the perpendicular distance between the centroidal axis and the parallel axis is 30 cm.
Re-arranging the original equation for the PARALLEL AXIS THEOREM about the XAXIS, we can plug in the values to solve for the moment of inertia bout the centroidal axis as: Ixc = Ixβ² β d2x A = (245,833 cm4 ) β (250 cm2 )(30 cm)2 = 20,833 cm4
Therefore, the correct answer choice is A. π. π π πππ πππ Made with
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PROBLEM 2: The Moment of Inertia of the beamβs cross-sectional area about the π₯ βcentroidal axis is closest to (ππ.4 ):
A. 2.9 π₯ 109 B. 5.0 π₯ 109 C. 7.2 π₯ 109 D. 8.6 π₯ 109
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SOLUTION 2: The TOPIC of MOMENT OF INERTIA can be referenced under the SUBJECT of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS The first step we need to do is classify this object as a discrete region or continuous region by quickly observing that it can be broken up into several sub-regions composed of simple shapes (triangles, circles and rectangles are the most common shapes).
It is preferable to work with discrete regions as this means we are on the right track, and the NCEES Reference Manual, Version 9.4, confirms this by providing tables and formulas to determine the moment of inertia for discrete regions. A discrete region is a type of region made up of a combination of shapes referred to as Made with by Prepineer | Prepineer.com
sub-regions. Each sub-region has its own individual area, centroid, and moment of inertia calculation that is usually fairly simple to compute.
We then combine these sub-region properties to compute a single moment of inertia for the composite shape we are given. We will use a table (See below) to help us organize the different terms associated with moment of inertia calculations.
Iβx - Moment of Inertia Table Region
Area (A)
dy
dy2 A
Ixc
Iβx
We construct a table for each axis we are considering, with one value for the moment of inertia with respect to the π₯ β ππ₯ππ or one with respect to the π¦ β ππ₯ππ . Each sub-region has its own unique area and inertia values, as we have listed out in our Made with
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moment of inertia table below. From looking at the object given in the problem, we can break it in three sub-regions by identifying sub-region 1, our yellow region, as a rectangle (vertical orientation), sub-region 2, our green region, as a rectangle (horizontal orientation). and sub-region 3, our blue region, as a rectangle (vertical orientation).
Iβx - Moment of Inertia Table Region
Area (A)
dy
dy2 A
Ixc
Iβx
1 (rectangle) 2 (rectangle) 3 (rectangle) STEP 2: CALCULATE the AREA for each REGION The TABLE for the PROPERTIES OF COMMON SHAPES can be referenced under the SUBJECT of STATICS on pages 69 to 71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The next step is to calculate the area for each region using the tables in the Reference Handbook. If we grab our NCEES Reference Manual, Version 9.4, and flip back to pages 69 through 71, we will find a few tables defining the characteristics of a number of simple shapes.
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The first column defines the shape, whether itβs a triangle, rectangle, trapezoid, circle etc. The second column provides the area and centroid formulas for the particular shape and shape orientation, while the third column will give us the general formulas to determine the AREA MOMENT OF INERTIA of that particular shape.
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For sub-region 1 we are given a rectangle, which is a simple shape whose properties are provided in the tables found in the NCEES Reference Handbook, Version 9.4. It is important that you are able to easily identify the geometry of a shape, and identify which formulas to use to calculate the area, centroid, and moment of inertia of the shape.
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The FORMULA for the AREA OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Using the formula for the area of a rectangle, the area of sub-region π΄1 is defined as the base times the height. π΄1 = πβ = (100)(300) = 30,000ππ. With a base of 100ππ. and a height of 300ππ., we plug these numbers in to the formula for the area of a rectangle and get an area of 30,000ππ., which we then place into our moment of inertia table that we set up in step 1.
Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
dy
dy2 A
Ixc
Iβx
2 (rectangle) 3 (rectangle) As our second and third regions are a rectangle as well, there is no need to flip back to the tables, we can just move forward with using the same area formula base times height for the area of a rectangle. π΄2 = (600)(100) = 60,000in. Made with
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π΄3 = (100)(300) = 30,000ππ. Plugging in the dimensions, we define the area as 60,000ππ. for sub-region 2 and 30,000ππ. for sub-region 3, and list these values in the coordinate table as shown below.
Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
2 (rectangle)
60,000
3 (rectangle)
30,000
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dy2 A
Ixc
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Iβx
STEP 3: CALCULATE the DISTANCE from the NEUTRAL AXIS to the CENTROID of each SHAPE We can now move on to Step 3, where we will calculate the distance from the NEUTRAL AXIS to the CENTROID of each shape, or sub-region. The problem is specifically asking us to determine the moment of inertia with respect to the π₯ β ππ₯ππ , such that the π₯ β ππ₯ππ will be our neutral axis, and reference axis to which we will be measuring back to.
Starting with region 1, we know that the centroid will be some distance π¦π from the bottom of its shape.
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The FORMULA for the CENTROID OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We know that a rectangle is symmetrical, and that the formula for the π¦ βcentroid of a rectangle is the vertical midpoint of the shape, as represented by the formula:
π¦π =
β 2
We can plug in the height of 300 ππ. and define this regionβs centroid as 150 ππ. from the bottom of its region.
π¦π1 =
β 300 = = 150in. 2 2
It is important that we realize that we need to add 50 ππ. to this measurement to take it all the way back to the π₯ β ππ₯ππ : ππ¦ = 150 + 50 = 200in.
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Doing so gives us 200 ππ., and we plug that value into our table:
Iβx - Moment of Inertia Table Region
Area (A)
dy
1 (rectangle)
30,000
200
2 (rectangle)
60,000
3 (rectangle)
30,000
dy2 A
Ixc
Iβx
Now moving to our second region, the centroidal axis of the overall object lines up with the centroidal axis for sub-region 2. As there is no distance between the two axes, we do not need to use the parallel-axis theorem. Therefore, we can go ahead and just put a tick mark to indicate a distance of 0, for when we do our calculations.
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Iβx - Moment of Inertia Table Region
Area (A)
dy
1 (rectangle)
30,000
200
2 (rectangle)
60,000
-
3 (rectangle)
30,000
dy2 A
Ixc
Iβx
Now moving to our third region, we know that the centroid will be some distance π¦π from the top of its region, which happens to reside on the π₯ β ππ₯ππ .
We can plug in the height of 300 ππ. and define this regionβs centroid as 150 ππ. from the top of its region.
π¦π3 =
β 300 = = 150in. 2 2 Made with
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It is important that we realize that we need to add 50ππ. to this measurement to take it all the way back to the π₯ β ππ₯ππ :
Doing so gives us 200ππ., and we plug that value in to our table: ππ¦ = 150 + 50 = 200in.
Iβx - Moment of Inertia Table Region
Area (A)
dy
1 (rectangle)
30,000
200
2 (rectangle)
60,000
-
3 (rectangle)
30,000
200
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Ixc
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Iβx
STEP 4: DEFINE the TRANSFER COEFFICIENT for each region by MULTIPLYING the value of the AREA by the SQUARE of its DISTANCE from the NEUTRAL AXIS Moving on to STEP 4, for each individual region, we now need to define what is called a TRANSFER COEFFICIENT. To determine this value, we take the AREA of each REGION and multiply it by the square of the distance.
STEP 5: CALCULATE the MOMENT OF INERTIA for each REGION about its own CENTROIDAL AXIS Now for step 5, we CALCULATE the MOMENT OF INERTIA for each individual REGION about its own CENTROIDAL AXIS. The formulas for the MOMENT OF INERTIA OF COMMON SHAPES can be referenced in the tables for the PROPERTIES OF COMMON SHAPES under the topic of STATICS on pages 69-71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Made with
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Referencing the tables in the Reference Handbook, we are able to determine that the moment of inertia for a rectangleβs centroidal π₯ β ππ₯ππ is given by the formula:
πΌπ₯π
πβ3 = 12
We plug in our values and get a calculate the moment of inertia for sub-region π΄1 as:
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πΌπ₯π1
πβ3 (100in. )(300in. )3 = = = 2.25 x 108 ππ.4 12 12
With our second and third sub-regions being a rectangle as well, we can use the same formula, and calculate the moment of inertia for each sub-region as:
πΌπ₯π2
πβ3 (600in. )(100in. )3 = = = 5.0 x 107 ππ.4 12 12
πΌπ₯π3
πβ3 (100in. )(300in. )3 = = = 2.25 x 108 ππ.4 12 12
We then list the calculated values into our TABLE.
STEP 6: Use the PARALLEL AXIS THEOREM to DEFINE the MOMENT OF INERTIA of each REGION about the NEUTRAL AXIS So moving on with our problem, the first value we are concerned with is the MOMENT OF INERTIA (MOI) of the individual shape about its CENTROIDAL axis, which we just defined in STEP 5 and this is highlighted in the table. We then the TRANSFER COEFFICIENT, or TRANSFER TERM which we defined in STEP 4, to the MOI. Made with
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Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 1 as: πΌ β² π₯1 = (1.88 π₯ 109 ) + (2.25 π₯ 108 ) = 1.43 π₯ 109 As the centroidal axis of sub-region 2 is the same as the centroidal axis for the overall object, the first term using the parallel-axis theorem will be 0, as there is no distance between the two axes: πΌ β² π₯2 = (0) + (5.0 π₯ 107 ) = 5.0 π₯ 107 Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 3 as: πΌ β² π₯3 = (1.88 π₯ 109 ) + (2.25 π₯ 108 ) = 1.43 π₯ 109
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STEP 7: SUM all the VALUES for the adjusted MOMENT OF INERTIA about the NEUTRAL AXIS to obtain the MOMENT OF INERTIA for the COMPOSITE SHAPE Summing the shapes that represent the composite T-section, we calculate the total moment of inertia as: Ixβ² = Ix1 + Ix2 + Ix 3 = (1.43 x 109 ) + (5.0 x 107 ) + (1.43 x 109 ) = 2.9 x 109 in4
Therefore, the correct answer choice is A. π. π π πππ
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PROBLEM 3: The Moment of Inertia of the beamβs cross-sectional area about the π¦ βcentroidal axis is closest to (ππ.4 ):
A. 2.4 π₯ 109 B. 3.8 π₯ 109 C. 4.1π₯ 109 D. 5.6 π₯ 109
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SOLUTION 3: The TOPIC of MOMENT OF INERTIA can be referenced under the SUBJECT of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. STEP 1: LIST each of the AREAS that make up the DISCRETE REGION, including any HOLES or SUBTRACTED REGIONS The first step we need to do is classify this object as a discrete region or continuous region by quickly observing that it can be broken up into several sub-regions composed of simple shapes (triangles, circles, and rectangles are the most common shapes).
It is preferable to work with discrete regions as this means we are on the right track, and the NCEES Reference Manual confirms this by providing us tables and formulas to determine the moment of inertia for discrete regions. A discrete region is a type of region made up of a combination of shapes referred to as Made with by Prepineer | Prepineer.com
sub-regions. Each sub-region has its own individual area, centroid, and moment of inertia calculation that is usually fairly simple to compute.
We then combine these sub-region properties to compute a single moment of inertia for the composite shape we are given. We will use a table to help us organize the different terms associated with moment of inertia calculations, as shown below.
Iβx - Moment of Inertia Table Region
Area (A)
dy
dy2 A
Ixc
Iβx
We construct a table for each axis we are considering, with one value for the moment of inertia with respect to the π₯ β ππ₯ππ or one with respect to the π¦ β ππ₯ππ .
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Each sub-region has its own unique area and inertia values, as we have listed in our moment of inertia table below. Looking at the object given in the problem, we can break it in three sub-regions by identifying sub-region 1, our yellow region, as a rectangle (vertical orientation), sub-region 2, our green region, as a rectangle (horizontal orientation), and sub-region 3, our blue region, as a rectangle (vertical orientation).
Iβy - Moment of Inertia Table Region
Area (A)
dx
dx 2 A
Iyc
Iβy
1 (rectangle) 2 (rectangle) 3 (rectangle) STEP 2: CALCULATE the AREA for each REGION The TABLES for the PROPERTIES OF COMMON SHAPES can be referenced under the SUBJECT of STATICS on pages 69 to 71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The next step is to calculate the area for each region using the tables in the Reference Handbook. If we grab our NCEES Reference Manual, Version 9.4, and flip back to pages 69 through 71, we will find a few tables defining the characteristics of a number of simple shapes.
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The first column defines the shape, whether itβs a triangle, rectangle, trapezoid, circle etc. The second column provides the area and centroid formulas for the particular shape and shape orientation, while the third column will give us the general formulas to determine the AREA MOMENT OF INERTIA of that particular shape.
For sub-region 1 we are given a rectangle, which is a simple shape whose properties are provided in the tables found in the NCEES Reference Handbook. It is important that you are able to easily identify the geometry of a shape, and identify which formulas to Made with
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use to calculate the area, centroid, and moment of inertia of the shape.
The FORMULA for the AREA OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Using the formula for the area of a rectangle, the area of sub-region 1 is defined as the base times the height.
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π΄1 = πβ = (100)(300) = 30,000 With a base of 100ππ. and a height of 300ππ., we plug these numbers into the formula for the area of a rectangle and get an area of 30,000ππ.2 , which we then place into our moment of inertia table that we set up in step 1.
Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
dy
dy2 A
Ixc
Iβx
2 (rectangle) 3 (rectangle) As our second and third regions are a rectangle as well, there is no need to flip back to the tables, we can just move forward with using the same area formula base times height for the area of a rectangle. π΄2 = (600in. )(100in. ) = 60,000in.2 π΄3 = (100ππ. )(300ππ. ) = 30,000ππ.2 Plugging in the dimensions, we define the area as 60,000ππ.2 for sub-region 2 and 30,000ππ.2 for sub-region 3, and list these values in the coordinate table as shown below.
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Iβx - Moment of Inertia Table Region
Area (A)
1 (rectangle)
30,000
2 (rectangle)
60,000
3 (rectangle)
30,000
dy
dy2 A
Ixc
Iβx
STEP 3: CALCULATE the DISTANCE from the NEUTRAL AXIS to the CENTROID of each SHAPE We can now move on to Step 3, where we will calculate the distance from the NEUTRAL AXIS to the CENTROID of each shape, or sub-region. The problem is specifically asking us to determine the moment of inertia with respect to the π¦ β ππ₯ππ , such that the π¦ β ππ₯ππ will be our neutral axis, and reference axis to which we will be measuring back to.
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Starting with region 1, we know that the centroid will be some distance π₯π from the bottom of its shape.
The FORMULA for the CENTROID OF A RECTANGLE can be referenced in the TABLE for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on page 69 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. We know that a rectangle is symmetrical, and that the formula for the π₯ βcentroid of a rectangle is the horizontal midpoint of the shape, as represented by the formula:
π₯π =
π 2
We can plug in the base of 100ππ. and define this regions centroid as 50ππ. from the left side of its region. Made with
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π₯π1 =
π 100 = = 50 2 2
It is important that we realize that we need to add 200ππ. to this measurement to take it all the way back to the y-axis: ππ₯ = 200 + 50 = 250
Doing so gives us 250ππ., and we list that value in our table:
Iβy - Moment of Inertia Table Region
Area (A)
dx
1 (rectangle)
30,000
250
2 (rectangle)
60,000
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dx 2 A
Iyc
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Iβy
3 (rectangle)
30,000
Now moving on to our second region, the centroidal axis of the overall object lines up with the centroidal axis for sub-region 2. As there is no distance between the two axes, we do not need to use the parallel-axis theorem. Therefore, we can go ahead and just put a tick mark to indicate a distance of 0, for when we do our calculations.
Iβy - Moment of Inertia Table Region
Area (A)
dx
1 (rectangle)
30,000
250
2 (rectangle)
60,000
-
3 (rectangle)
30,000
dx 2 A
Iyc
Iβy
Now moving to our third region, we know that the centroid will be some distance π₯π from the right side of its region:
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We can plug in the base of 100ππ. and define this regions centroid as 50ππ. from the right side of its region.
π₯π3 =
π 100 = = 50 2 2
It is important that we realize that we need to add 50ππ. to this measurement to take it all the way back to the π¦ β ππ₯ππ :
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Doing so gives us 250ππ., and we can list that value in our table: ππ₯ = 200 + 50 = 250in.
Iβy - Moment of Inertia Table Region
Area (A)
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dx
dx 2 A
Iyc
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Iβy
1 (rectangle)
30,000
250
2 (rectangle)
60,000
-
3 (rectangle)
30,000
250
STEP 4: DEFINE the TRANSFER COEFFICIENT for each region by MULTIPLYING the value of the AREA by the SQUARE of its DISTANCE from the NEUTRAL AXIS Moving on to STEP 4, for each individual region, we now need to define what is called a TRANSFER COEFFICIENT. To determine this value, we take the AREA of each REGION and multiply it by the square of the distance.
STEP 5: CALCULATE the MOMENT OF INERTIA for each REGION about its own CENTROIDAL AXIS Now for step 5, we CALCULATE the MOMENT OF INERTIA for each individual REGION about its own CENTROIDAL AXIS. Made with
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The FORMULAS for the MOMENT OF INERTIA OF COMMON SHAPES can be referenced in the TABLES for the PROPERTIES OF COMMON SHAPES under the SUBJECT of STATICS on pages 69-71 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
Referencing the tables in the Reference Handbook, we are able to determine that the moment of inertia for a rectangleβs centroidal π¦ β ππ₯ππ is given by the formula:
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πΌπ¦π
π3 β = 12
We plug in our values and calculate the moment of inertia for sub-region 1 as: π 3 h (100in. )3 (300in. ) πΌπ¦π1 = = = 2.5 x 107 ππ.4 12 12 With our second and third sub-regions being rectangles as well, we can use the same formula, and calculate the moment of inertia for each sub-region as:
πΌπ¦π2
π 3 h (600in. )3 (100in. ) = = = 1.8 x 109 ππ.4 12 12
πΌπ¦π3
π 3 h (100in. )3 (300in. ) = = = 2.5 x 107 ππ.4 12 12
We then list the calculated values in our TABLE.
Iβy - Moment of Inertia Table Region
Area (A)
dx
dx 2 A
Iyc
1 (rectangle)
30,000
250
1.88E9
2.5E7
2 (rectangle)
60,000
-
-
1.8E9
3 (rectangle)
30,000
250
1.88E9
2.5E7
Iβy
STEP 6: Use the PARALLEL AXIS THEOREM to DEFINE the MOMENT OF Made with
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INERTIA of each REGION about the NEUTRAL AXIS So moving on with our problem, the first value we are concerned with is the MOMENT OF INERTIA (MOI) of the individual shape about its CENTROIDAL axis, which we just defined in STEP 5 and is highlighted in the table. We then add the TRANSFER COEFFICIENT, or TRANSFER TERM which we defined in STEP 4, to the MOI.
Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 1 as: πΌ β² π¦1 = (1.88 π₯ 109 ) + (2.5 π₯ 107 ) = 1.9 π₯ 109 As the centroidal axis of sub-region 2 is the same as the centroidal axis for the overall object, the first term using the parallel-axis theorem will be 0, as there is no distance between the two axes: πΌ β² π¦2 = (0) + (1.8 π₯ 109 ) = 1.8 π₯ 109
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Using the parallel-axis theorem, we are able to calculate the moment of inertia about the neutral axis for sub-region 3 as: πΌ β² π¦3 = (1.88 π₯ 109 ) + (2.5 π₯ 107 ) = 1.9 π₯ 109 STEP 7: SUM all the VALUES for the adjusted MOMENT OF INERTIA about the NEUTRAL AXIS to obtain the MOMENT OF INERTIA for the COMPOSITE SHAPE Summing the shapes that represent the composite T-section, we calculate the total moment of inertia as: Iyβ² = Iy1 + Iy2 + Iy = (1.9 x 109 ) + (1.8 x 109 ) + (1.9 x 109 ) = 5.6 x 109 in4 3
Therefore, the correct answer choice is D. π. π π πππ ππ.π
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