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40 30 10 10 D=15 15 30 55 40 FIG. EMS - PS6 002

40 30 10 10 D=15 15 30 55 40 FIG. EMS - PS6 002

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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1.) Total Area AT

2.) Centroid about the x-axis ȳ

3.) Centroid about the y – axis ?̅

4.) Moment of Inertia About the x-axis Ix

note:  In locating the centroid of the entire section, Varignon's Theorem. Varignon's Theorem locates the centroid of a composite body.

note: I hope you will use the formulas I will be giving.

STEP 4:VARIGNON'S THEOREM 6 Right triangle (Origin of axes at centroid) 1 bh h Qyr = 2@yn Arỹr = > Anyn 2 3 3 |3D C B- n bl hb³ 1 36 36 72 Qxt Qxn AnXn %3D %3D bl bh (h² + b³) I, 36 n n 12 Quarter circle (Origin of axes at center of circle) πι A = 4 4r x = y В- B C p4 I, = 1, = I. 16 (9n² – 64)r4 I BB = 0.05488,4 8 1447 Circle (Origin of axes at center) d = 2r nd² I, = nd4 A = tr² 4 4 64 C ndª I BB B- -B 5ar4 5nd4 Ixy =0 Ip = %3D %3D ху 2 32 4 64 1 |y Rectangle (Origin of axes at centroid) A = bh x y = 2 h C bh³ hb3 bh I 12 I, = 0 Ip 12 (h² + b?) 12 xy II
Transcribed Image Text:STEP 4:VARIGNON'S THEOREM 6 Right triangle (Origin of axes at centroid) 1 bh h Qyr = 2@yn Arỹr = > Anyn 2 3 3 |3D C B- n bl hb³ 1 36 36 72 Qxt Qxn AnXn %3D %3D bl bh (h² + b³) I, 36 n n 12 Quarter circle (Origin of axes at center of circle) πι A = 4 4r x = y В- B C p4 I, = 1, = I. 16 (9n² – 64)r4 I BB = 0.05488,4 8 1447 Circle (Origin of axes at center) d = 2r nd² I, = nd4 A = tr² 4 4 64 C ndª I BB B- -B 5ar4 5nd4 Ixy =0 Ip = %3D %3D ху 2 32 4 64 1 |y Rectangle (Origin of axes at centroid) A = bh x y = 2 h C bh³ hb3 bh I 12 I, = 0 Ip 12 (h² + b?) 12 xy II
40 30 10 10 D=15 15 30 55 FIG. EMS - PS6 002 40
Transcribed Image Text:40 30 10 10 D=15 15 30 55 FIG. EMS - PS6 002 40
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