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4.) Moment of Inertia About the x-axis Ix 5.) Moment of Inertia About the y – axis Iy 6.) Polar Moment of Inertia about the axes Jo 7.) Moment of Inertia About the centroidal x-axis Ix

4.) Moment of Inertia About the x-axis Ix 5.) Moment of Inertia About the y – axis Iy 6.) Polar Moment of Inertia about the axes Jo 7.) Moment of Inertia About the centroidal x-axis Ix

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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Question

Given:

A1= 3400, A2=800, A3=706.85834, A4=176.71458

Y1=42.5, Y2=26.666, Y3=52.7324, Y4= 62.5

X1=20, X2=53.333, X3=52.7324, X4=17.5

Total Area: 4730.14376

ȳ=40.604

x̄ (x-bar) = 30.6224

 

Please, I want you to solve the:

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

5.) Moment of Inertia About the y – axis Iy

6.) Polar Moment of Inertia about the axes Jo

7.) Moment of Inertia About the centroidal 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. 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
Expert Answer Step 1 Dividing the given area into standard areas as shown below and showing the x and y axes 40 30 10 10 D=15 15 30 1 2 55 40 FIG. EMS - PS6 002
Transcribed Image Text:Expert Answer Step 1 Dividing the given area into standard areas as shown below and showing the x and y axes 40 30 10 10 D=15 15 30 1 2 55 40 FIG. EMS - PS6 002
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