1312 - Ch10 - 1209059858
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View 1312 - Ch10 - 1209059858 as PDF for free.
More details
- Words: 2,859
- Pages: 44
Chapter 10: MOMENTS OF INERTIA
APPLICATIONS
Many structural members like beams and columns have cross sectional shapes like I, H, C, etc.. Why do they usually not have solid rectangular, square, or circular cross sectional areas? What primary property of these members influences design decisions? How can we calculate this property?
THE CONCEPT OF THE MOMENT OF INERTIA (MoI) OF AN AREA
Consider a beam of uniform cross section which is subjected to two equal and opposite couples applied at each end of the beam. It is shown in mechanics of materials that the internal forces in any section of the beam are distributed forces whose magnitudes ΔF=ky ΔA vary linearly with the distance y between the element of area ΔA and an axis passing through the centroid of the section. The moment about the x-axis due to this force is y (ΔF). The total moment is ∫A y dF = ∫ A ky2 dA = k ∫A( y2 dA). This integral term is referred to as the moment of inertia of the area about x-axis.
THE CONCEPT OF THE MoI (continued) 10cm 1cm
3cm 10cm
10cm
(A)
x (B)
1cm
P
3cm
(C)
R
S
Consider three different possible cross sectional shapes and areas for the beam RS. All have the same total area and, assuming they are made of same material, they will have the same mass per unit length. For the given vertical loading P on the beam, which shape will develop less deflection? Why? The answer depends on the MoI of the beam about the x-axis. It turns out that Section A has the highest MoI because most of the area is farthest from the x axis. Hence, it has the least deflection.
MoI – DEFINITION For the differential area dA, shown in the figure: d Ix = y2 dA , d Iy = x2 dA , and, d JO = r2 dA , where JO is the polar moment of inertia about the pole O or z axis. The moments of inertia for the entire area are obtained by integration. Ix = ∫A y2 dA ; JO = ∫A r2 dA =
Iy =
∫A x2 dA
∫A ( x2 + y2 ) dA =
Ix +
Iy
The MoI is also referred to as the second moment of an area and has units of length to the fourth power (m4 or in4).
MoI FOR AN AREA BY INTEGRATION For simplicity, the area element used has a differential size in only one direction (dx or dy). This results in a single integration and is usually simpler than doing a double integration with two differentials, dx·dy. The step-by-step procedure is: 1. Choose the element dA: There are two choices: a vertical strip or a horizontal strip. Some considerations about this choice are: a) The element parallel to the axis about which the MoI is to be determined usually results in an easier solution. For example, we typically choose a horizontal strip for determining Ix and a vertical strip for determining Iy.
MoI BY INTEGRATION b) If y is easily expressed in terms of x (e.g., y = x2 + 1), then choosing a vertical strip with a differential element dx wide may be advantageous. 2. Integrate to find the MoI. For example, given the element shown in the figure above: Iy = ∫ d Iy = ∫ x2 dA = ∫ x2 y dx
and
Ix = ∫ d Ix = ∫ (1 / 3) y3 dx (using the information for a rectangle about its base from the EX 10-1a). Since in this case the differential element is dx, y needs to be expressed in terms of x and the integral limit must also be in terms of x. As you can see, choosing the element and integrating can be challenging. It may require a trial and error approach plus experience.
EXAMPLE (x,y)
Given: The shaded area shown in the figure. Find: The MoI of the area about the x- and y-axes. Plan: Follow the steps given earlier. Solution
Ix
=
dA = Ix
=
∫ y2 dA (4 – x) dy = (4 – y2/4) dy 4
2 (4 – y2/4) dy ∫ y O 4
= [ (4/3) y3 – (1/20) y5 ] 0 = 34.1 in4
EXAMPLE (continued) y (x,y)
Iy
= ∫ x2 dA
=
∫ x2 y dx
= ∫ x2 (2 √ x) dx = 2
4
0∫
x 3.5 dx 4
= [ (2/3.5) x 3.5 ] 0 = 73.1 in 4 In the above example, it will be difficult to determine Iy using a horizontal strip. However, Ix in this example can be determined using a vertical strip. So, Ix = ∫ (1/3) y3 dx = ∫ (1/3) (2√x)3 dx .
CONCEPT QUIZ 1. A pipe is subjected to a bending y moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)? A) Smaller Ix
B) Smaller Iy
C) Larger Ix
D) Larger Iy
2. In the figure to the left, what is the differential moment of inertia of the element with respect to the y-axis (dIy)? A) x2 y dx
B) (1/12) x3 dy
C) y2 x dy
D) (1/3) y dy
M
M
x Pipe section
y
y=x3 x,y
x
GROUP PROBLEM SOLVING Given: The shaded area shown.
(x,y)
Find: Ix and Iy of the area. Plan: Follow the steps described earlier.
Solution Ix
= ∫ (1/3) y3 = =
0∫
8
dx
(1/3) x dx
10.7 in 4
8
= [x 2 / 6 ]0
GROUP PROBLEM (continued) (x,y)
IY = ∫ x 2 dA
∫ x 2 y dx
=
= ∫ x 2 ( x (1/3) dx =
0∫
8
x (7/3) dx
= [(3/10) x = 307 in 4
(10/3)
8 ]0
ATTENTION QUIZ 1. When determining the MoI of the element in the figure, dIy equals A) x 2 dy
B) x 2 dx
C) (1/3) y 3 dx D) x 2.5 dx
2. Similarly, dIx equals A) (1/3) x 1.5 dx
B) y 2 dA
C) (1 /12) x 3 dy
D) (1/3) x 3 dx
(x,y)
y2 = x
HOMEWORK
• 10-1, 10-10
RADIUS OF GYRATION OF AN AREA A
y
For a given area A and its MoI, Ix , imagine that the entire area is located at distance kx from the x axis.
kx x
Then, Ix = k2xA or kx = √ ( Ix / A). This kx is called the radius of gyration of the area about the x axis. Similarly; kY = √ ( Iy / A ) and kO = √ ( JO / A )
The radius of gyration has units of length and gives an indication of the spread of the area from the axes. This characteristic is important when designing columns.
PARALLEL-AXIS THEOREM FOR AN AREA This theorem relates the moment of inertia (MoI) of an area about an axis passing through the area’s centroid to the MoI of the area about a corresponding parallel axis. This theorem has many practical applications, especially when working with composite areas. Consider an area with centroid C. The x' and y' axes pass through C. The MoI about the x-axis, which is parallel to, and distance dy from the x ' axis, is found by using the parallel-axis theorem.
PARALLEL-AXIS THEOREM (continued) IX = ∫A y 2 dA =
∫A (y' + dy)2 dA
= ∫A y' 2 dA + 2 dy ∫A y' dA + dy 2 ∫A dA Using the definition of the centroid: y = y' + dy
y'
=
(∫A y' dA) / (∫A dA) . Now
since C is at the origin of the x' – y' axes, y' = 0 , and hence ∫A y' dA = 0 . Thus IX = IX' +
A dy 2
Similarly, IY = IY' + A dX 2 and JO = JC +
Ad2
A=10 cm2
POP QUIZ 1. For the given area, the moment of inertia about axis 1 is 200 cm4 . What is the MoI about axis 3 (the centroidal axis)? A) 90 cm 4
B) 110 cm 4
C) 60 cm 4
D) 40 cm 4
2. The moment of inertia of the rectangle about the x-axis equals A) 8 cm
4.
C) 24 cm 4 .
4
B) 56 cm . D) 26 cm 4 .
d2
C C •
3 2
•
d1
1
d1 = d2 = 2 cm 3cm 2cm 2cm
x
MOMENT OF INERTIA FOR A COMPOSITE AREA A composite area is made by adding or subtracting a series of “simple” shaped areas like rectangles, triangles, and circles. For example, the area on the left can be made from a rectangle minus a triangle and circle.
The MoI of these “simpler” shaped areas about their centroidal axes are found in most engineering handbooks as well as the inside back cover of the textbook. Using these data and the parallel-axis theorem, the MoI for a composite area can easily be calculated.
STEPS FOR ANALYSIS 1. Divide the given area into its simpler shaped parts. 2. Locate the centroid of each part and indicate the perpendicular distance from each centroid to the desired reference axis. 3. Determine the MoI of each “simpler” shaped part about the desired reference axis using the parallel-axis theorem ( IX = IX’ + A ( dy )2 ) . 4. The MoI of the entire area about the reference axis is determined by performing an algebraic summation of the individual MoIs obtained in Step 3. (Please note that MoI of a hole is subtracted).
3
bh Ix' = 12
3
bh Ix' = 36 πa Ix = 4
4
EXAMPLE Given: The beam’s cross-sectional area.
[1] [2] [3]
Find:
The moment of inertia of the area about the y-axis and the radius of gyration ky.
Plan:
Follow the steps for analysis.
Solution
1. The cross-sectional area can be divided into three rectangles ( [1], [2], [3] ) as shown. 2. The centroids of these three rectangles are in their center. The distances from these centers to the y-axis are 0 mm, 87.5 mm, and 87.5 mm, respectively.
EXAMPLE (continued) 3. From the Table in the book, the MoI of a rectangle about its centroidal axis is (1/12) b h3. Iy[1] = (1/12) (25mm) (300mm)3 = 56.25 (106) mm4 [1] [2] [3] Using the parallel-axis theorem, IY[2] = IY[3] = IY’ + A (dX)2 = (1/12) (100) (25)3 + (25) (100) ( 87.5 )2 =
19.27 (106) mm 4
EXAMPLE (continued)
4.
Iy = Iy1 + Iy2 + Iy3 = 94.8 ( 106) mm 4
ky = √ ( Iy / A) A = 300 (25) + 25 (100) + 25 (100) = 12,500 mm 2 ky = √ ( 94.79) (106) / (12500) = 87.1 mm
CONCEPT QUIZ 1. For the area A, we know the centroid’s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem ___ . A) directly between the axes 1 and 2. B) between axes 1 and 3 and then between the axes 3 and 2. C) between axes 1 and 4 and then axes 4 and 2. D) None of the above.
Axis A
d3 d2 d1
C
•
4 3 2 1
CONCEPT QUIZ (continued) 2. For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number? A) Axis 1 B) Axis 2 C) Axis 3 D) Axis 4 E) Can not tell.
Axis A
d3 d2 d1
C
•
4 3 2 1
GROUP PROBLEM SOLVING Given: The shaded area as shown in the figure. Find: The moment of inertia for the area about the x-axis and the radius of gyration kX. Plan: Follow the steps for analysis.
(a) (b) (c)
Solution
1. The given area can be obtained by subtracting both the circle (b) and triangle (c) from the rectangle (a). 2. The perpendicular distances of the centroids from the x-axis are: da = 5 in , db = 4 in, and dc = 8 in.
GROUP PROBLEM SOLVING (continued) 3. IXa = (1/12) 6 (10)3 + 6 (10)(5)2 = 2000 in 4 IXb = (1/4) π (2)4 +
π (2)2 (4)2
= 213.6 in 4 IXc
(a) (b) (c) IX = IXa kX =
(½) (3) (6) (8)2 = 594 in 4 –
IXb
– IXc
=
1190 in 4
√ ( IX / A )
A = 10 ( 6 ) – kX =
= (1 /36) (3) (6)3 +
π (2)2 – (½) (3) (6) =
√ (1192 / 38.43) =
5.57 in.
38.43 in 2
Homework
• 10-33,10-36,10-39
MASS MOMENT OF INERTIA (Moment of Inertia of a Body)
CONCEPT OF THE MMI T G
·
Consider a rigid body with a center of mass at G. It is free to rotate about the z axis, which passes through G. Now, if we apply a torque T about the z axis to the body, the body begins to rotate with an angular acceleration α.
T and α are related by the equation T = I α . In this equation, I is the mass moment of inertia (MMI) about the z axis. The MMI of a body is a property that measures the resistance of the body to angular acceleration. This is similar to the role of mass in the equation F = m a. The MMI is often used when analyzing rotational motion (done in dynamics).
DEFINITION OF THE MMI p
Consider a rigid body and the arbitrary axis p shown in the figure. The MMI about the p axis is defined as I = ∫m r2 dm, where r, the “moment arm,” is the perpendicular distance from the axis to the arbitrary element dm. The MMI is always a positive quantity and has a unit of kg ·m2 or slug · ft2.
RELATED CONCEPTS Parallel-Axis Theorem: Just as with the MoI for an area, the parallel-axis theorem can be used to find the MMI about a parallel axis p’ that is a distance d from the axis m through the body’s center of mass G. The formula is Ip’ = IG + (m) (d)2 (where m is the mass of the body).
d
G·
The radius of gyration is similarly defined as k = √(I / m) Finally, the MMI can be obtained by integration or by the method for composite bodies. The latter method is easier for many practical shapes.
p’
CONCEPT QUIZ 1. Consider a particle of mass 1 kg located at point P, whose coordinates are given in meters. Determine the MMI of that particle about the z axis. B) 16 kg·m2 A) 9 kg·m2 C) 25 kg·m2 D) 36 kg·m2 2. Consider a rectangular frame made of four slender bars with four axes (zP, zQ, zR and zS) perpendicular to the screen and passing through the points P, Q, R, and S respectively. About which of the four axes will the MMI of the frame be the largest? B) zQ C) zR A) zP D) zS E) Not possible to determine.
z ·P(3,4,6) y x P • S•
Q • •R
GROUP PROBLEM SOLVING R
P
Plan:
Given: The pendulum consists of a 24 lb plate and a slender rod weighing 8 lb. Find: The radius of gyration of the pendulum about an axis perpendicular to the screen and passing through point O. Determine the MMI of the pendulum using the method for composite bodies. Then determine the radius of gyration using the MMI and mass values (check units!!).
Solution 1. Separate the pendulum into a square plate (P) and a slender rod (R).
GROUP PROBLEM SOLVING R P
2. The center of mass of the plate and rod are 3.5 ft and 0.5 ft from point O, respectively.
3. The MMI data on plates and slender rods are given in the Table. Using those data and the parallel-axis theorem, IP = (1/12) (24/32.2) (12 + 12) + (24/32.2) (3.5)2 = 9.254 slug·ft2 IR = (1/12) (8/32.2) (5)2 + (8/32.2) (0.5)2 = 0.5797 slug·ft2 4. IO = IP + IR = 9.254 + 0.5797 = 9.834 slug·ft2 5. Total mass (m) equals (24+8)/32.2 = 0.9938 slug Radius of gyration k = √IO / m = 3.15 ft
ATTENTION QUIZ 1. A particle of mass 2 kg is located 1 m down the y-axis. What are the MMI of the particle about the x, y, and z axes, respectively? x A) (2, 0, 2) B) (0, 2, 2) C) (0, 2, 2) D) (2, 2, 0) 2. Consider a rectangular frame made of four slender bars and four axes (zP, zQ, zR and zS) perpendicular to the screen and passing through points P, Q, R, and S, respectively. About which of the four axes will the MMI of the frame be the lowest? B) zQ C) zR A) zP D) zS E) Not possible to determine.
z 1m •
P • S•
y
Q • •R
Homework
• 10-92, 10-101, 10-110
APPLICATIONS
Many structural members like beams and columns have cross sectional shapes like I, H, C, etc.. Why do they usually not have solid rectangular, square, or circular cross sectional areas? What primary property of these members influences design decisions? How can we calculate this property?
THE CONCEPT OF THE MOMENT OF INERTIA (MoI) OF AN AREA
Consider a beam of uniform cross section which is subjected to two equal and opposite couples applied at each end of the beam. It is shown in mechanics of materials that the internal forces in any section of the beam are distributed forces whose magnitudes ΔF=ky ΔA vary linearly with the distance y between the element of area ΔA and an axis passing through the centroid of the section. The moment about the x-axis due to this force is y (ΔF). The total moment is ∫A y dF = ∫ A ky2 dA = k ∫A( y2 dA). This integral term is referred to as the moment of inertia of the area about x-axis.
THE CONCEPT OF THE MoI (continued) 10cm 1cm
3cm 10cm
10cm
(A)
x (B)
1cm
P
3cm
(C)
R
S
Consider three different possible cross sectional shapes and areas for the beam RS. All have the same total area and, assuming they are made of same material, they will have the same mass per unit length. For the given vertical loading P on the beam, which shape will develop less deflection? Why? The answer depends on the MoI of the beam about the x-axis. It turns out that Section A has the highest MoI because most of the area is farthest from the x axis. Hence, it has the least deflection.
MoI – DEFINITION For the differential area dA, shown in the figure: d Ix = y2 dA , d Iy = x2 dA , and, d JO = r2 dA , where JO is the polar moment of inertia about the pole O or z axis. The moments of inertia for the entire area are obtained by integration. Ix = ∫A y2 dA ; JO = ∫A r2 dA =
Iy =
∫A x2 dA
∫A ( x2 + y2 ) dA =
Ix +
Iy
The MoI is also referred to as the second moment of an area and has units of length to the fourth power (m4 or in4).
MoI FOR AN AREA BY INTEGRATION For simplicity, the area element used has a differential size in only one direction (dx or dy). This results in a single integration and is usually simpler than doing a double integration with two differentials, dx·dy. The step-by-step procedure is: 1. Choose the element dA: There are two choices: a vertical strip or a horizontal strip. Some considerations about this choice are: a) The element parallel to the axis about which the MoI is to be determined usually results in an easier solution. For example, we typically choose a horizontal strip for determining Ix and a vertical strip for determining Iy.
MoI BY INTEGRATION b) If y is easily expressed in terms of x (e.g., y = x2 + 1), then choosing a vertical strip with a differential element dx wide may be advantageous. 2. Integrate to find the MoI. For example, given the element shown in the figure above: Iy = ∫ d Iy = ∫ x2 dA = ∫ x2 y dx
and
Ix = ∫ d Ix = ∫ (1 / 3) y3 dx (using the information for a rectangle about its base from the EX 10-1a). Since in this case the differential element is dx, y needs to be expressed in terms of x and the integral limit must also be in terms of x. As you can see, choosing the element and integrating can be challenging. It may require a trial and error approach plus experience.
EXAMPLE (x,y)
Given: The shaded area shown in the figure. Find: The MoI of the area about the x- and y-axes. Plan: Follow the steps given earlier. Solution
Ix
=
dA = Ix
=
∫ y2 dA (4 – x) dy = (4 – y2/4) dy 4
2 (4 – y2/4) dy ∫ y O 4
= [ (4/3) y3 – (1/20) y5 ] 0 = 34.1 in4
EXAMPLE (continued) y (x,y)
Iy
= ∫ x2 dA
=
∫ x2 y dx
= ∫ x2 (2 √ x) dx = 2
4
0∫
x 3.5 dx 4
= [ (2/3.5) x 3.5 ] 0 = 73.1 in 4 In the above example, it will be difficult to determine Iy using a horizontal strip. However, Ix in this example can be determined using a vertical strip. So, Ix = ∫ (1/3) y3 dx = ∫ (1/3) (2√x)3 dx .
CONCEPT QUIZ 1. A pipe is subjected to a bending y moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)? A) Smaller Ix
B) Smaller Iy
C) Larger Ix
D) Larger Iy
2. In the figure to the left, what is the differential moment of inertia of the element with respect to the y-axis (dIy)? A) x2 y dx
B) (1/12) x3 dy
C) y2 x dy
D) (1/3) y dy
M
M
x Pipe section
y
y=x3 x,y
x
GROUP PROBLEM SOLVING Given: The shaded area shown.
(x,y)
Find: Ix and Iy of the area. Plan: Follow the steps described earlier.
Solution Ix
= ∫ (1/3) y3 = =
0∫
8
dx
(1/3) x dx
10.7 in 4
8
= [x 2 / 6 ]0
GROUP PROBLEM (continued) (x,y)
IY = ∫ x 2 dA
∫ x 2 y dx
=
= ∫ x 2 ( x (1/3) dx =
0∫
8
x (7/3) dx
= [(3/10) x = 307 in 4
(10/3)
8 ]0
ATTENTION QUIZ 1. When determining the MoI of the element in the figure, dIy equals A) x 2 dy
B) x 2 dx
C) (1/3) y 3 dx D) x 2.5 dx
2. Similarly, dIx equals A) (1/3) x 1.5 dx
B) y 2 dA
C) (1 /12) x 3 dy
D) (1/3) x 3 dx
(x,y)
y2 = x
HOMEWORK
• 10-1, 10-10
RADIUS OF GYRATION OF AN AREA A
y
For a given area A and its MoI, Ix , imagine that the entire area is located at distance kx from the x axis.
kx x
Then, Ix = k2xA or kx = √ ( Ix / A). This kx is called the radius of gyration of the area about the x axis. Similarly; kY = √ ( Iy / A ) and kO = √ ( JO / A )
The radius of gyration has units of length and gives an indication of the spread of the area from the axes. This characteristic is important when designing columns.
PARALLEL-AXIS THEOREM FOR AN AREA This theorem relates the moment of inertia (MoI) of an area about an axis passing through the area’s centroid to the MoI of the area about a corresponding parallel axis. This theorem has many practical applications, especially when working with composite areas. Consider an area with centroid C. The x' and y' axes pass through C. The MoI about the x-axis, which is parallel to, and distance dy from the x ' axis, is found by using the parallel-axis theorem.
PARALLEL-AXIS THEOREM (continued) IX = ∫A y 2 dA =
∫A (y' + dy)2 dA
= ∫A y' 2 dA + 2 dy ∫A y' dA + dy 2 ∫A dA Using the definition of the centroid: y = y' + dy
y'
=
(∫A y' dA) / (∫A dA) . Now
since C is at the origin of the x' – y' axes, y' = 0 , and hence ∫A y' dA = 0 . Thus IX = IX' +
A dy 2
Similarly, IY = IY' + A dX 2 and JO = JC +
Ad2
A=10 cm2
POP QUIZ 1. For the given area, the moment of inertia about axis 1 is 200 cm4 . What is the MoI about axis 3 (the centroidal axis)? A) 90 cm 4
B) 110 cm 4
C) 60 cm 4
D) 40 cm 4
2. The moment of inertia of the rectangle about the x-axis equals A) 8 cm
4.
C) 24 cm 4 .
4
B) 56 cm . D) 26 cm 4 .
d2
C C •
3 2
•
d1
1
d1 = d2 = 2 cm 3cm 2cm 2cm
x
MOMENT OF INERTIA FOR A COMPOSITE AREA A composite area is made by adding or subtracting a series of “simple” shaped areas like rectangles, triangles, and circles. For example, the area on the left can be made from a rectangle minus a triangle and circle.
The MoI of these “simpler” shaped areas about their centroidal axes are found in most engineering handbooks as well as the inside back cover of the textbook. Using these data and the parallel-axis theorem, the MoI for a composite area can easily be calculated.
STEPS FOR ANALYSIS 1. Divide the given area into its simpler shaped parts. 2. Locate the centroid of each part and indicate the perpendicular distance from each centroid to the desired reference axis. 3. Determine the MoI of each “simpler” shaped part about the desired reference axis using the parallel-axis theorem ( IX = IX’ + A ( dy )2 ) . 4. The MoI of the entire area about the reference axis is determined by performing an algebraic summation of the individual MoIs obtained in Step 3. (Please note that MoI of a hole is subtracted).
3
bh Ix' = 12
3
bh Ix' = 36 πa Ix = 4
4
EXAMPLE Given: The beam’s cross-sectional area.
[1] [2] [3]
Find:
The moment of inertia of the area about the y-axis and the radius of gyration ky.
Plan:
Follow the steps for analysis.
Solution
1. The cross-sectional area can be divided into three rectangles ( [1], [2], [3] ) as shown. 2. The centroids of these three rectangles are in their center. The distances from these centers to the y-axis are 0 mm, 87.5 mm, and 87.5 mm, respectively.
EXAMPLE (continued) 3. From the Table in the book, the MoI of a rectangle about its centroidal axis is (1/12) b h3. Iy[1] = (1/12) (25mm) (300mm)3 = 56.25 (106) mm4 [1] [2] [3] Using the parallel-axis theorem, IY[2] = IY[3] = IY’ + A (dX)2 = (1/12) (100) (25)3 + (25) (100) ( 87.5 )2 =
19.27 (106) mm 4
EXAMPLE (continued)
4.
Iy = Iy1 + Iy2 + Iy3 = 94.8 ( 106) mm 4
ky = √ ( Iy / A) A = 300 (25) + 25 (100) + 25 (100) = 12,500 mm 2 ky = √ ( 94.79) (106) / (12500) = 87.1 mm
CONCEPT QUIZ 1. For the area A, we know the centroid’s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem ___ . A) directly between the axes 1 and 2. B) between axes 1 and 3 and then between the axes 3 and 2. C) between axes 1 and 4 and then axes 4 and 2. D) None of the above.
Axis A
d3 d2 d1
C
•
4 3 2 1
CONCEPT QUIZ (continued) 2. For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number? A) Axis 1 B) Axis 2 C) Axis 3 D) Axis 4 E) Can not tell.
Axis A
d3 d2 d1
C
•
4 3 2 1
GROUP PROBLEM SOLVING Given: The shaded area as shown in the figure. Find: The moment of inertia for the area about the x-axis and the radius of gyration kX. Plan: Follow the steps for analysis.
(a) (b) (c)
Solution
1. The given area can be obtained by subtracting both the circle (b) and triangle (c) from the rectangle (a). 2. The perpendicular distances of the centroids from the x-axis are: da = 5 in , db = 4 in, and dc = 8 in.
GROUP PROBLEM SOLVING (continued) 3. IXa = (1/12) 6 (10)3 + 6 (10)(5)2 = 2000 in 4 IXb = (1/4) π (2)4 +
π (2)2 (4)2
= 213.6 in 4 IXc
(a) (b) (c) IX = IXa kX =
(½) (3) (6) (8)2 = 594 in 4 –
IXb
– IXc
=
1190 in 4
√ ( IX / A )
A = 10 ( 6 ) – kX =
= (1 /36) (3) (6)3 +
π (2)2 – (½) (3) (6) =
√ (1192 / 38.43) =
5.57 in.
38.43 in 2
Homework
• 10-33,10-36,10-39
MASS MOMENT OF INERTIA (Moment of Inertia of a Body)
CONCEPT OF THE MMI T G
·
Consider a rigid body with a center of mass at G. It is free to rotate about the z axis, which passes through G. Now, if we apply a torque T about the z axis to the body, the body begins to rotate with an angular acceleration α.
T and α are related by the equation T = I α . In this equation, I is the mass moment of inertia (MMI) about the z axis. The MMI of a body is a property that measures the resistance of the body to angular acceleration. This is similar to the role of mass in the equation F = m a. The MMI is often used when analyzing rotational motion (done in dynamics).
DEFINITION OF THE MMI p
Consider a rigid body and the arbitrary axis p shown in the figure. The MMI about the p axis is defined as I = ∫m r2 dm, where r, the “moment arm,” is the perpendicular distance from the axis to the arbitrary element dm. The MMI is always a positive quantity and has a unit of kg ·m2 or slug · ft2.
RELATED CONCEPTS Parallel-Axis Theorem: Just as with the MoI for an area, the parallel-axis theorem can be used to find the MMI about a parallel axis p’ that is a distance d from the axis m through the body’s center of mass G. The formula is Ip’ = IG + (m) (d)2 (where m is the mass of the body).
d
G·
The radius of gyration is similarly defined as k = √(I / m) Finally, the MMI can be obtained by integration or by the method for composite bodies. The latter method is easier for many practical shapes.
p’
CONCEPT QUIZ 1. Consider a particle of mass 1 kg located at point P, whose coordinates are given in meters. Determine the MMI of that particle about the z axis. B) 16 kg·m2 A) 9 kg·m2 C) 25 kg·m2 D) 36 kg·m2 2. Consider a rectangular frame made of four slender bars with four axes (zP, zQ, zR and zS) perpendicular to the screen and passing through the points P, Q, R, and S respectively. About which of the four axes will the MMI of the frame be the largest? B) zQ C) zR A) zP D) zS E) Not possible to determine.
z ·P(3,4,6) y x P • S•
Q • •R
GROUP PROBLEM SOLVING R
P
Plan:
Given: The pendulum consists of a 24 lb plate and a slender rod weighing 8 lb. Find: The radius of gyration of the pendulum about an axis perpendicular to the screen and passing through point O. Determine the MMI of the pendulum using the method for composite bodies. Then determine the radius of gyration using the MMI and mass values (check units!!).
Solution 1. Separate the pendulum into a square plate (P) and a slender rod (R).
GROUP PROBLEM SOLVING R P
2. The center of mass of the plate and rod are 3.5 ft and 0.5 ft from point O, respectively.
3. The MMI data on plates and slender rods are given in the Table. Using those data and the parallel-axis theorem, IP = (1/12) (24/32.2) (12 + 12) + (24/32.2) (3.5)2 = 9.254 slug·ft2 IR = (1/12) (8/32.2) (5)2 + (8/32.2) (0.5)2 = 0.5797 slug·ft2 4. IO = IP + IR = 9.254 + 0.5797 = 9.834 slug·ft2 5. Total mass (m) equals (24+8)/32.2 = 0.9938 slug Radius of gyration k = √IO / m = 3.15 ft
ATTENTION QUIZ 1. A particle of mass 2 kg is located 1 m down the y-axis. What are the MMI of the particle about the x, y, and z axes, respectively? x A) (2, 0, 2) B) (0, 2, 2) C) (0, 2, 2) D) (2, 2, 0) 2. Consider a rectangular frame made of four slender bars and four axes (zP, zQ, zR and zS) perpendicular to the screen and passing through points P, Q, R, and S, respectively. About which of the four axes will the MMI of the frame be the lowest? B) zQ C) zR A) zP D) zS E) Not possible to determine.
z 1m •
P • S•
y
Q • •R
Homework
• 10-92, 10-101, 10-110
