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Announcements Moments of Inertia - Chapter 10 Today s Objectives: Students will be able to: a) Define the moments of inertia (MoI) for an area. b) Determine the MoI for an area by integration. In-Class

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Announcements Moments of Inertia - Chapter 10 Today s Objectives: Students will be able to: a) Define the moments of inertia (MoI) for an area. b) Determine the MoI for an area by integration. In-Class Activities: Reading quiz Applications MoI: concept and definition MoI by integration Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 10 1 Reading Quiz 1. The definition of the Moment of Inertia for an area involves an integral of the form A) x da. B) x 2 da. C) x 2 dm. D) m da. 2. Select the SI units for the Moment of Inertia for an area. A) m 4 B) m 2 C) kg m 2 D) kg m 3 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? Engr222 Spring 2004 Chapter 10 2 Applications - continued Many structural members are made of tubes rather than solid squares or rounds. Why? What parameters of the cross sectional area influence the designer s selection? How can we determine the value of these parameters for a given area? Moment of Inertia of an Area Consider a plate submerged in a liquid. The pressure of a liquid at a distance z below the surface is given by p = γ z, where γ is the specific weight of the liquid. The force on the area da at that point is df = p da. The moment about the x-axis due to this force is z (df). The total moment is A z df = A γ z 2 da = γ A ( z 2 da). This sort of integral term also appears in solid mechanics when determining stresses and deflection. This integral term is referred to as the moment of inertia of the area of the plate about an axis. Engr222 Spring 2004 Chapter 10 3 1cm (A) Moment of Inertia of an Area - continued 10cm 10cm 1cm 3cm (B) 10cm (C) 3cm 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 internal stress and 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 stress and deflection. x R P S Moment of Inertia - Definition For the differential area da, shown in the figure: d I x = y 2 da, d I y = x 2 da, and d J O = r 2 da, where J O 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. I x = A y 2 da ; I y = A x 2 da J O = A r 2 da = A ( x 2 + y 2 ) da = I x + I y The MoI is also referred to as the second moment of an area and has units of length to the fourth power (m 4 or in 4 ). Engr222 Spring 2004 Chapter 10 4 Radius of Gyration of an Area - Section 10.3 y k x A x For a given area A and its MoI, I x, imagine that the entire area is located at distance k x from the x axis. 2 Then, I x = k x A or k x = ( I x / A). This k x is called the radius of gyration of the area about the x axis. Similarly; k Y = ( I y / A ) and k O = ( J O / 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. Moment of Inertia by Integration - Section 10.4 The step-by-step procedure is: 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. 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 I x and a vertical strip for determining I y. Engr222 Spring 2004 Chapter 10 5 Moment of Inertia by Integration - continued b) If y is easily expressed in terms of x (e.g., y = x 2 + 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: I y = x 2 da = x 2 y dx and I x = d I x = (1 / 3) y 3 dx (using the information for a rectangle about its base from the inside back cover of the textbook and from the parallel-axis theorem). 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 I x = y 2 da da = (4 x) dy = (4 y 2 /4) dy 4 I x = O y2 (4 y 2 /4) dy = [ (4/3) y 3 (1/20) y 5 ] 0 = 34.1 in 4 4 Engr222 Spring 2004 Chapter 10 6 Example - continued y (x,y) I y = x 2 da = x 2 y dx = x 2 (2 x) dx 4 = 2 0 x 2.5 dx = [ (2/3.5) x 3.5 ] 0 4 = 73.1 in 4 In the above example, it will be difficult to determine I y using a horizontal strip. However, I x in this example can be determined using a vertical strip. So, I x = (1/3) y 3 dx = (1/3) (2 x) 3 dx. Concept Quiz 1. A pipe is subjected to a bending moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)? y x M M Pipe section A) Smaller I x B) Smaller I y C) Larger I x D) Larger I y 2. In the figure to the right, what is the differential moment of inertia of the element with respect to the y-axis (di y )? y y=x 3 x,y A) x 2 y dx B) (1/12) x 3 dy C) y 2 x dy D) (1/3) y dy x Engr222 Spring 2004 Chapter 10 7 Group Problem Solving (x,y) Given: The shaded area shown. Find: I x and I y of the area. Plan: Follow the steps described earlier. Solution I x = (1/3) y 3 dx 8 = 0 (1/3) x dx = [x 2 / 6 ] 0 = 10.7 in 4 8 Group Problem Solving - continued (x,y) I Y = x 2 da = x 2 y dx = x 2 ( x (1/3) ) dx 8 = 0 x (7/3) dx = [(3/10) x (10/3) ] 0 = 307 in 4 8 Engr222 Spring 2004 Chapter 10 8 Attention Quiz 1. When determining the MoI of the element in the figure, di y equals A) x 2 dy B) x 2 dx C) (1/3) y 3 dx D) x 2.5 dx y 2 = x (x,y) 2. Similarly, di x equals A) (1/3) x 1.5 dx B) y 2 da C) (1/12) x 3 dy D) (1/3) x 3 dx Textbook Problem 10- Engr222 Spring 2004 Chapter 10 9 Announcements HW #4 and #5 due Friday Moment of Inertia for Composite Areas Today s Objectives: Students will be able to: 1. Apply the parallel-axis theorem. 2. Determine the moment of inertia (MoI) for a composite area. In-Class Activities: Reading quiz Applications Parallel-axis theorem Method for composite areas Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 10 10 Reading Quiz 1. The parallel-axis theorem for an area is applied between A) an axis passing through its centroid and any corresponding parallel axis. B) any two parallel axis. C) two horizontal axes only. D) two vertical axes only. 2. The moment of inertia of a composite area equals the of the MoI of all of its parts. A) vector sum B) algebraic sum (addition or subtraction) C) addition D) product Applications Cross-sectional areas of structural members are usually made of simple shapes or combinations of simple shapes. Is there a simpler method for determining the MoI of such cross-sectional areas as compared to the integration method? If yes, can you describe the method? Engr222 Spring 2004 Chapter 10 11 Applications - continued This is another example of a structural member with a composite cross-area. Design calculations typically require use of the MoI for these cross-sectional areas. Can you describe a simple method to calculate MoI? Parallel-Axis Theorem for an Area - Section 10.2 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 d y from the x 'axis, is found by using the parallel-axis theorem. Engr222 Spring 2004 Chapter 10 12 Parallel-Axis Theorem for an Area - continued I X = A y 2 da = A (y'+ d y ) 2 da = A y' 2 da + 2 d y A y'da + d 2 y A da Using the definition of the centroid: y = y'+ d y 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 I X = I X ' + A d 2 y Similarly, I Y = I Y '+ A d X 2 and J O = J C + A d 2 Parallel-axis Theorem for a Rectangle See inside back cover of your textbook Engr222 Spring 2004 Chapter 10 13 Moment of Inertia for a Composite Area Section 10.5 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 I X = I X + A ( d y ) 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). Engr222 Spring 2004 Chapter 10 14 Example Given: The beam s cross-sectional area. Find: Plan: The moment of inertia of the area about the y-axis and the radius of gyration k y. Follow the steps for analysis. [1] [2] [3] 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. [1] [2] [3] Example - continued 3. From the inside back cover of the book, the MoI of a rectangle about its centroidal y axis is (1/12) hb 3. I y[1] = (1/12) (25mm) (300mm) 3 = (10 6 ) mm 4 Using the parallel-axis theorem, I Y[2] = I Y[3] = I Y + A (d X ) 2 = (1/12) (100) (25) 3 + (25) (100) ( 87.5 ) 2 = (10 6 ) mm 4 Engr222 Spring 2004 Chapter 10 15 Example - continued 4. I y = I y1 + I y2 + I y3 = 94.8 (10 6 ) mm 4 k y = ( I y / A) A = 300 (25) + 25 (100) + 25 (100) = 12,500 mm 2 k y = ( 94.79) (10 6 ) / (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. d 3 d 2 d 1 A C Axis 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. Engr222 Spring 2004 Chapter 10 16 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 d 3 d 2 d 1 A C Axis (a) (b) (c) Group Problem Solving Given: The shaded area as shown in the figure. Find: Plan: Solution The moment of inertia for the area about the x-axis and the radius of gyration k X. Follow the steps for analysis. 1. The given area can be obtained by subtracting both the circle (b) and triangle (c) from the rectangle (a). 2. Information about the centroids of the simple shapes can be obtained from the inside back cover of the book. The perpendicular distances of the centroids from the x-axis are: d a = 5 in, d b = 4 in, and d c = 8 in. Engr222 Spring 2004 Chapter 10 17 Group Problem Solving - continued 3. I Xa = (1/12) 6 (10) (10)(5) 2 = 2000 in 4 I Xb = (1/4) π (2) 4 + π (2) 2 (4) 2 (a) (b) (c) = in 4 I Xc = (1 /36) (3) (6) 3 + (½) (3) (6) (8) 2 = 594 in 4 I X = I Xa I Xb I Xc = 1190 in 4 k X = ( I X / A ) A = 10 ( 6 ) π (2) 2 (½) (3) (6) = in 2 k X = (1192 / 38.43) = 5.57 in. Attention Quiz 1. For the given area, the moment of inertia about axis 1 is 200 cm 4. 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 d 2 d 1 C C A=10 cm 2 d 1 = d 2 = 2 cm The moment of inertia of the rectangle about the x-axis equals A) 8 cm 4 B) 56 cm 4 C) 24 cm 4 D) 26 cm 4 2cm 2cm 3cm x Engr222 Spring 2004 Chapter 10 18 Textbook Problem 10- Announcements WOP schedule next week. Class meets 10:45 11:30am. Engr222 Spring 2004 Chapter 10 19 Mass Moment of Inertia - Section 10.9 Today s Objectives: Students will be able to : a) Explain the concept of the mass moment of inertia (MMI). b) Determine the MMI of a composite body. In-Class Activities: Reading quiz Applications MMI: concept and definition Determining the MMI Concept quiz Group problem solving Attention quiz Reading Quiz 1. The formula definition of the mass moment of inertia about an axis is. A) r dm B) r 2 dm C) m dr D) m 2 dr 2. The parallel-axis theorem can be applied to determine. A) only the MoI B) only the MMI C) both the MoI and MMI D) None of the above. Note: MoI is the moment of inertia of an area and MMI is the mass moment inertia of a body. Engr222 Spring 2004 Chapter 10 20 Applications The large flywheel in the picture is connected to a large metal cutter. The flywheel is used to provide a uniform motion to the cutting blade. What property of the flywheel is most important for this use? How can we determine a value for this property? Why is most of the mass of the flywheel located near the flywheel s circumference? Applications - continued If a torque M is applied to a fan blade which is initially at rest, its angular speed begins to increase. On which property (P) of the fan blade does the angular acceleration (α) depend? How can we determine a value for P? What is the relationship between M, P, and α? Engr222 Spring 2004 Chapter 10 21 Concept of the Mass Moment of Inertia 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 = ma. The MMI is often used when analyzing rotational motion (done in dynamics). Definition of the Mass Moment of Inertia 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 r 2 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 m 2 or slug ft 2. Engr222 Spring 2004 Chapter 10 22 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 through the body s center of mass G. The formula is I p = I G + (m) (d) 2 (where m is the mass of the body). m G d p 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. q r p Example Given: The wheel consists of a thin ring with a mass 10 kg and four spokes (slender rods) with a mass of 2 kg each. Find: The wheel s MMI about an axis perpendicular to the screen and passing through point A. Plan: Follow steps similar to finding the MoI for a composite area. Solution: 1. The wheel can be divided into a thin ring (p) and two slender rods (q and r). Will both rods be treated the same? Engr222 Spring 2004 Chapter 10 23 Example - continued q r p 2. The center of mass for each of the three pieces is at point O, 0.5 m from Point A. 3. The MMI data for a thin ring and slender rod are given on the inside back cover of the textbook. Using those data and the parallel-axis theorem, calculate the following: I A = I O + (m) (d) 2 I Ap = 10 (0.5) (0.5) 2 = 5.0 kg m 2 I Aq = I Ar = (1/12) (4) (1) (0.5) 2 = kg m 2 4. Now add the three MMIs about point A. I A = I Ap + I Aq + I Ar = 7.67 kg m 2 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. A) 9 kg m 2 B) 16 kg m 2 C) 25 kg m 2 D) 36 kg m 2 2. Consider a rectangular frame made of four slender bars with four axes (z P, z Q, z R and z S ) 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? A) z P B) z Q C) z R D) z S E) Not possible to determine. x z P S P(3,4,6) Q y R Engr222 Spring 2004 Chapter 10 24 Group Problem Solving P Plan: R 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 - continued P R 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 on the inside cover of the textbook. Using those data and the parallel-axis theorem, I P = (1/12) (24/32.2) ( ) + (24/32.2) (3.5) 2 = slug ft 2 I R = (1/12) (8/32.2) (5) 2 + (8/32.2) (0.5) 2 = slug ft 2 4. I O = I P + I R = = slug ft 2 5. Total mass (m) equals (24+8)/32.2 = slug Radius of gyration k = I O / m = 3.15 ft Engr222 Spring 2004 Chapter 10 25 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 (z P, z Q, z R and z S ) 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? A) z P B) z Q C) z R D) z S E) Not possible to determine. z 1 m y P S Q R Textbook Problem 10- Engr222 Spring 2004 Chapter 10 26

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