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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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5. A rectangle has a base of 8 in. and a height of 12 in. a) Determine the moment of inertia with respect to the base using the approximate method by dividing the area into six equal horizontal strips. b) Use the exact formula from Table 3 (inside the back cover of the textbook) to determine the moment of inertia with respect to the base. b=8" h = 12"
Transcribed Image Text:5. A rectangle has a base of 8 in. and a height of 12 in. a) Determine the moment of inertia with respect to the base using the approximate method by dividing the area into six equal horizontal strips. b) Use the exact formula from Table 3 (inside the back cover of the textbook) to determine the moment of inertia with respect to the base. b=8" h = 12"
TABLE 3 Properties of areas Shape |- h CG b CG LE YO Y₂ - Rectangle CG b Triangle Yo R Circle H 14 -X₂ X ·X₂ X ·Xo Area (A) A = bh A-b A=4² А = 0.7854² Moment of Inertia (1) Radius of Gyration (r) Polar Moment of Inertia (J) 64² 12 A6³ 12 Joo = (²+²) 32 = 4,- 64 = 0.04909d4 = 0.7854 1 Xo R Shape Yo CG 0 Semicircle d₂ CG Yo Hollow circle b b₁ H CG Yo Hollow rectangle Yo CG 4R AR 45 71 3xX 4.R 3x R Quarter-circle 4R 3m Xo -X X₂ ·X Area (A) A=TR² = 1.5718² A= (²-4) 4 -0.7854(²-) A=bd-bid, A = R² Moment of Inertia (1) 1,-0.1098R R.R 8 =0,3927 64 130 = 110 _bd³-byd 110² 12 40 12 1,-1,-0.05498 Radius of Gyration (r) ₁,-0.264R = \d² + d} 14 "%" bd³ - byd 12A db³ - dib 12A ==0.2644R 7,=0.5R Polar Moment of Inertia (J) Jo=1₁₂+1% = 0.5025R TR Joa Ridª 32 { Jco=1₁₂+1% Ja=0.1098R
Transcribed Image Text:TABLE 3 Properties of areas Shape |- h CG b CG LE YO Y₂ - Rectangle CG b Triangle Yo R Circle H 14 -X₂ X ·X₂ X ·Xo Area (A) A = bh A-b A=4² А = 0.7854² Moment of Inertia (1) Radius of Gyration (r) Polar Moment of Inertia (J) 64² 12 A6³ 12 Joo = (²+²) 32 = 4,- 64 = 0.04909d4 = 0.7854 1 Xo R Shape Yo CG 0 Semicircle d₂ CG Yo Hollow circle b b₁ H CG Yo Hollow rectangle Yo CG 4R AR 45 71 3xX 4.R 3x R Quarter-circle 4R 3m Xo -X X₂ ·X Area (A) A=TR² = 1.5718² A= (²-4) 4 -0.7854(²-) A=bd-bid, A = R² Moment of Inertia (1) 1,-0.1098R R.R 8 =0,3927 64 130 = 110 _bd³-byd 110² 12 40 12 1,-1,-0.05498 Radius of Gyration (r) ₁,-0.264R = \d² + d} 14 "%" bd³ - byd 12A db³ - dib 12A ==0.2644R 7,=0.5R Polar Moment of Inertia (J) Jo=1₁₂+1% = 0.5025R TR Joa Ridª 32 { Jco=1₁₂+1% Ja=0.1098R
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