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ns 1 and 2. Locate the centroid of the shaded area

ns 1 and 2. Locate the centroid of the shaded area

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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CENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines. PROBLEM 2 ONLY PROVIDE THE SAME PROCESS OF SOLUTION IN THE GIVEN EXAMPLE ? THANK YOU.
2. Locate the centroid of the shoded crea shown, Divide the area into regulor geometric shapes. Make sure the centroids of the geometric shapes con 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. 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 orea. 12 hal Solution: Since the cross section is symmetrical with respect to a vertical axis, the centroid of the cross section les on the axis of symmetry. Therefore, only the location of the centroid with respect to a horizontal axis will be determined. ay, ie (36)(8)-288 (72)3)-216 (-36)|2) =-72 432 Part a in. x in in 1/2126)-36 12)(6)-72 F1/2)12|6)-36 A=72 ax, in (36(4)-144 2|(6)= 432 (36)(6)-216 360 4 3 2 With the base of the cross section as the reference: Total 3. Locate the centroid of the shoded orea in the fgure. created by cutfing a semiccle of diometer rom a Quarter cicie of rod 16+ 0.28" 16.28" Pert Quater cce Semicrce rarh dannel,. Pert 2-8"x6"x" 16"x 1" 2-5x3x1/2" 12-20.7 b channel Total a in 2(13)-26 (16)-16 213.75)-75 603 A-55.53 yin 16 28 -1.65- 14.63 8+0.28-8.28 0.75 + 0.28 = 1.03 0.70 ay, i 380.38 132.48 7.725 4221 fotal 524.806 9-945" above the base of the cross section 10TIN
Transcribed Image Text:2. Locate the centroid of the shoded crea shown, Divide the area into regulor geometric shapes. Make sure the centroids of the geometric shapes con 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. 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 orea. 12 hal Solution: Since the cross section is symmetrical with respect to a vertical axis, the centroid of the cross section les on the axis of symmetry. Therefore, only the location of the centroid with respect to a horizontal axis will be determined. ay, ie (36)(8)-288 (72)3)-216 (-36)|2) =-72 432 Part a in. x in in 1/2126)-36 12)(6)-72 F1/2)12|6)-36 A=72 ax, in (36(4)-144 2|(6)= 432 (36)(6)-216 360 4 3 2 With the base of the cross section as the reference: Total 3. Locate the centroid of the shoded orea in the fgure. created by cutfing a semiccle of diometer rom a Quarter cicie of rod 16+ 0.28" 16.28" Pert Quater cce Semicrce rarh dannel,. Pert 2-8"x6"x" 16"x 1" 2-5x3x1/2" 12-20.7 b channel Total a in 2(13)-26 (16)-16 213.75)-75 603 A-55.53 yin 16 28 -1.65- 14.63 8+0.28-8.28 0.75 + 0.28 = 1.03 0.70 ay, i 380.38 132.48 7.725 4221 fotal 524.806 9-945" above the base of the cross section 10TIN
Problems 1 and 2. Locate the centroid of the shaded area y -r = 100 mm 100 mm 140 mm 50 mm 50 mm 140 mm -180 mm -200 mm- 1. 2. ITY
Transcribed Image Text:Problems 1 and 2. Locate the centroid of the shaded area y -r = 100 mm 100 mm 140 mm 50 mm 50 mm 140 mm -180 mm -200 mm- 1. 2. ITY
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