Chapter 10 MOMENTS of INERTIA for AREAS, RADIUS OF GYRATION 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. READING QUIZ 1. The definition of the Moment of Inertia for an area involves an integral of the form A) da. B) da. C) dm. D) m da.. Select the SI units for the Moment of Inertia for an area. A) m 3 B) m C) kg m D) kg m 3 APPLICATIONS 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? 10.1,10.,10.3 1
MOMENTS OF INERTIA 3cm 10cm 10cm 1cm 10cm 1cm (A) (B) 10cm (C) R 3cm P 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 internal stress and deflection? Why? The answer depends on the MoI of the beam about the -ais. It turns out that Section (A) has the highest MoI because most of the area is farthest from the ais. Hence, it has the least stress and deflection. 10.1 MOMENTS OF INERTIA FOR AREAS For the differential area da, shown in the figure: d I = y da, d I y = da, and, d J O = r da, where J O is the polar moment of inertia about the pole O or z ais. Moments of inertia for the entire area are obtained by integration. I = A y da ; I y = A da J O = A r da = A ( + y ) da = I + 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 or in ). 10. PARALLEL-AXIS THEOREM FOR AN AREA This theorem relates the moment of inertia (MoI) of an area about an ais passing through the area s centroid to the MoI of the area about a corresponding parallel ais. This theorem has many practical applications, especially when working with composite areas. Consider an area with centroid C. The ' and y' aes pass through C. The MoI about the -ais, which is parallel to, and located at distance d y from ' ais, is found by using the parallel-ais theorem. 10.1,10.,10.3
READING QUIZ 1. The parallel-ais theorem for an area is applied between A) an ais passing through its centroid and any corresponding parallel ais. B) any two parallel ais. C) two horizontal aes only. D) two vertical aes only.. 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 PARALLEL-AXIS THEOREM (continued) y = y' + d y Using the definition of the centroid: y' = ( A y' da) / ( A da). Now since C is at the origin of the ' y' aes, y' = 0, and hence A y' da = 0. Thus, Similarly, PARALLEL-AXIS THEOREM (eamples) Moment of inertia I T of a circular area with respect to a tangent to the circle, I 1 T = I + Ad = π r + 5 = π r ( π r ) r Moment of inertia of a triangle with respect to a centroidal ais, I AA = I BB + Ad ( 1 h) 3 I 1 1 BB = I AA Ad = bh bh 1 3 1 3 = bh 36 9-9 10.1,10.,10.3 3
10.3 - RADIUS OF GYRATION OF AN AREA Consider area A with moment of inertia I. Imagine that the area is concentrated in a thin strip parallel to the ais with equivalent I. I I k A k = = A k = radius of gyration with respect to the ais Similarly, I y = k y A k y = JO = ko A ko = k O = k + k y I y A JO A 9-10 10. MoI FOR AN AREA BY INTEGRATION For simplicity, the area element used has a differential size in only one direction (d or dy). This results in a single integration and is usually simpler than doing a double integration with two differentials, d 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 ais about which the MoI is to be determined usually results in an easier solution. For eample, we typically choose a horizontal strip for determining I and a vertical strip for determining I y. MoI FOR AN AREA BY INTEGRATION, (continue) b) If y is easily epressed in terms of (e.g., y = + 1), then choosing a vertical strip with a differential element d wide may be advantageous.. Integrate to find the MoI. For eample, given the element shown in the figure above: I y = da = y d and I = d I = (1 / 3) y 3 d (using the information for a rectangle about its base from the inside back cover of the tetbook). Since in this case the differential element is d, y needs to be epressed in terms of and the integral limit must also be in terms of. As you can see, choosing the element and integrating can be challenging. It may require a trial and error approach plus eperience. 10.1,10.,10.3
EXAMPLE 10.1 This result can also be obtained by direct integration (as in part a) but with y varies from (0 to h) EXAMPLE 10.1, (continue) (,y) Solution EXAMPLE 10. Given: The shaded area shown in the figure. Find: The MoI of the area about the - and y-aes. Plan: Follow the steps given earlier. I = y da da = ( ) dy = ( y /) dy I = O y ( y /) dy = [ (/3) y 3 (1/0) y 5 ] 0 = 3.1 in 10.1,10.,10.3 5
y (,y) EXAMPLE 10. (continued) I y = da = y d = ( ) d = 0.5 d = [ (/3.5) 3.5 ] 0 = 73.1 in In the above eample, it will be difficult to determine I y using a horizontal strip. However, I in this eample can be determined using a vertical strip. So, I = (1/3) y 3 d = (1/3) ( ) 3 d. 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)? M y M A) Smaller I B) Smaller I y C) Larger I D) Larger I y. In the figure to the right, what is the differential moment of inertia of the element with respect to the y-ais (di y )? A) y d B) (1/1) 3 dy C) y dy D) (1/3) y dy Pipe section y y = 3,y GROUP PROBLEM SOLVING Given: The shaded area shown. Find: I and I y of the area. Plan: Follow the steps described earlier. (,(,y) da Solution I = 0 (1/3) y 3 d = 0 (1/3) d = [ / 6 ] 0 = 10.7 in I Y = da = y d = ( (1/3) ) d = 0 (7/3) d = [(3/10) (10/3) ] 0 = 307 in 10.1,10.,10.3 6
ATTENTION QUIZ 1. When determining the MoI of the element in the figure, di y equals A) dy B) d y y = (,y) C) (1/3) y 3 d D).5 d. Similarly, di equals A) (1/3) 1.5 d B) y da C) (1 /1) 3 dy D) (1/3) 3 d 10.1,10.,10.3 7