Calculate the moments of inertia of the area in question 2 about the X-X and Y-Y centroidal axes.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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Calculate the moments of inertia of the area in question 2 about the X-X and Y-Y centroidal axes.
2.) Locate the X-X and Y-Y centroidal axes for the area shown. Assume the "circle"
is aligned horizontally with the centerline of the 2" wide "arm". Neglect any tiny
amount of overlap of the two in your calculations.
12"
15"
(2
12
6" Diam.
1) Square
2)rectangle
3) circle
seclion
Area
K inches
Vinches
(12")-144 12/2 =6
12/2=6
12+52
= 19.5
10+2=1
2
I5x2=30
(32)
= 28.274
12+15+2
= 30
3
8+3=1
2-144(6)+30 (19.5)+2 8.274 (30)
144+36+28-.274
X-x = 12".
+ 122
X11.356in
+ 30x (11.356 -7.440)2
+ TL 97(11-7.440)
64
スー -2838.65in
y=144(6)+30 (11)+ 28.274 (1)
144+ 30+ 28.274
Y= 7.440in
Y-Y= 12"
144 (11.356-69-2x 153
+
(7, 5) (1. 35€ ,7.40")
12
+ 30(19.5 -11.356)²+T 64 9x(30-11.354)2
64
Y-Y=18302.87in"
3.) Calculate the moments of inertia of the area in question about the X-X and Y-
Y centroidal axes.
2.
Transcribed Image Text:2.) Locate the X-X and Y-Y centroidal axes for the area shown. Assume the "circle" is aligned horizontally with the centerline of the 2" wide "arm". Neglect any tiny amount of overlap of the two in your calculations. 12" 15" (2 12 6" Diam. 1) Square 2)rectangle 3) circle seclion Area K inches Vinches (12")-144 12/2 =6 12/2=6 12+52 = 19.5 10+2=1 2 I5x2=30 (32) = 28.274 12+15+2 = 30 3 8+3=1 2-144(6)+30 (19.5)+2 8.274 (30) 144+36+28-.274 X-x = 12". + 122 X11.356in + 30x (11.356 -7.440)2 + TL 97(11-7.440) 64 スー -2838.65in y=144(6)+30 (11)+ 28.274 (1) 144+ 30+ 28.274 Y= 7.440in Y-Y= 12" 144 (11.356-69-2x 153 + (7, 5) (1. 35€ ,7.40") 12 + 30(19.5 -11.356)²+T 64 9x(30-11.354)2 64 Y-Y=18302.87in" 3.) Calculate the moments of inertia of the area in question about the X-X and Y- Y centroidal axes. 2.
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