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y •r = 100 mm 50 mm -180 mm 1.

y •r = 100 mm 50 mm -180 mm 1.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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CENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines.

PROBLEM 1 ONLY

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Problems 1 and 2. Locate the centroid of the shaded area r = 100 mm 100 mm 140 mm 50 mm 140 mm 50 mm -180 mm - 200 mm 1. ITY 2.
Transcribed Image Text:Problems 1 and 2. Locate the centroid of the shaded area r = 100 mm 100 mm 140 mm 50 mm 140 mm 50 mm -180 mm - 200 mm 1. ITY 2.
2. Locate the centroid of the shaded area shown. Divide the area into regular geometric shapes. Make sure the centroids of the geometric shapes can be identified. 5. Locate the centroid of the built-up section shown in the figure. Refer to Table 6-6.2 on page 199 of your textbook for the properties of the elements. 8" x 6" x 1" 8" x 6" x 1" 12" 16" 16" x 1" 5" x 3" x 1/2" 5" x 3" x 1/2" The area can be divided into three parts as shown, with the unshaded triangle as a negative area since it is not part of the shaded area. 16" 6" 6" 12"-20.7 lb channel Solution: Since the cross section is symmetrical with respect to a vertical axis, the centroid of the cross section lies on the axis of symmetry. Therefore, only the location of the centroid with respect to a horizontal axis will be determined. ay, in³ (36)(8)=288 (72)(3)=216 (-36) (2) = -72 Part a, in?. х, in y, in ах, inз (1/2)(12)(6)=36 (12)(6)=72 -(1/2)(12)(6)=-36 A=72 (36)(4)=144 (72)(6) = 432 (-36) (6)=-216 4 6 3 6 With the base of the cross section as the reference: Total 360 432 E ax 8" x 6" x 1" 3. Locate the centroid of the shaded area in the figure, created by cutting a semicircle of diameter r from a quarter circle of radius r. 360 8" x 6" x 1" %3D A 72 * = 5" 16" + 0.28"= 16.28" Eay y = 432 16" x 1" %3D A 72 ỹ = 6" 5" x 3" x 1/2" 5" x 3" x 1/2" ах, in3 ar? (4r ay, in3 ar2 (4r 3T a, in X, in 4r Part у, in 4r Quarter circle Semicircle 3n 4r 4 37 ar (4r 4 tr3 12"-20.7 lb channel 8 12 8 16 2 8 ay, in³ 380.38 a, in². 2(13)=26 (16)(1)=16 2(3.75)=7.5 6.03 A=55.53 Part 3r3 = 0.25r3 y, in 16.28 -1.65 = 14.63 Total ar2 0.137r3 = 0.393r2 8 2-8"x6"x1" A = 12 16" x 1" 8 + 0.28 = 8.28 132.48 2-5"x3"x1/2" 0.75 + 0.28 = 1.03 7.725 E ax 0.393r 0.25r 12"-20.7 Ib channel 0.70 4.221 i = 0.636r Total 524.806 Σαν 524.806 0.137r A 55.53 0.393r? y = 0.349r y = 9.45" above the base of the cross section RSITY OTINN SAPIEY
Transcribed Image Text:2. Locate the centroid of the shaded area shown. Divide the area into regular geometric shapes. Make sure the centroids of the geometric shapes can be identified. 5. Locate the centroid of the built-up section shown in the figure. Refer to Table 6-6.2 on page 199 of your textbook for the properties of the elements. 8" x 6" x 1" 8" x 6" x 1" 12" 16" 16" x 1" 5" x 3" x 1/2" 5" x 3" x 1/2" The area can be divided into three parts as shown, with the unshaded triangle as a negative area since it is not part of the shaded area. 16" 6" 6" 12"-20.7 lb channel Solution: Since the cross section is symmetrical with respect to a vertical axis, the centroid of the cross section lies on the axis of symmetry. Therefore, only the location of the centroid with respect to a horizontal axis will be determined. ay, in³ (36)(8)=288 (72)(3)=216 (-36) (2) = -72 Part a, in?. х, in y, in ах, inз (1/2)(12)(6)=36 (12)(6)=72 -(1/2)(12)(6)=-36 A=72 (36)(4)=144 (72)(6) = 432 (-36) (6)=-216 4 6 3 6 With the base of the cross section as the reference: Total 360 432 E ax 8" x 6" x 1" 3. Locate the centroid of the shaded area in the figure, created by cutting a semicircle of diameter r from a quarter circle of radius r. 360 8" x 6" x 1" %3D A 72 * = 5" 16" + 0.28"= 16.28" Eay y = 432 16" x 1" %3D A 72 ỹ = 6" 5" x 3" x 1/2" 5" x 3" x 1/2" ах, in3 ar? (4r ay, in3 ar2 (4r 3T a, in X, in 4r Part у, in 4r Quarter circle Semicircle 3n 4r 4 37 ar (4r 4 tr3 12"-20.7 lb channel 8 12 8 16 2 8 ay, in³ 380.38 a, in². 2(13)=26 (16)(1)=16 2(3.75)=7.5 6.03 A=55.53 Part 3r3 = 0.25r3 y, in 16.28 -1.65 = 14.63 Total ar2 0.137r3 = 0.393r2 8 2-8"x6"x1" A = 12 16" x 1" 8 + 0.28 = 8.28 132.48 2-5"x3"x1/2" 0.75 + 0.28 = 1.03 7.725 E ax 0.393r 0.25r 12"-20.7 Ib channel 0.70 4.221 i = 0.636r Total 524.806 Σαν 524.806 0.137r A 55.53 0.393r? y = 0.349r y = 9.45" above the base of the cross section RSITY OTINN SAPIEY
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