y •r = 100 mm 50 mm -180 mm 1.
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter8: Centroids And Distributed Loads
Section: Chapter Questions
Problem 8.7P: Using integration, locate the centroid of the area under the n-th order parabola in terms of b, h,...
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CENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines.
PROBLEM 1 ONLY
PROVIDE THE SAME PROCESS OF SOLUTION IN THE GIVEN EXAMPLE :) THANK YOU.
Transcribed Image Text:Problems 1 and 2. Locate the centroid of the shaded area
r = 100 mm
100 mm
140 mm
50 mm
140 mm
50 mm
-180 mm
- 200 mm
1.
ITY
2.
Transcribed Image Text:2. Locate the centroid of the shaded area shown.
Divide the area into regular geometric shapes. Make
sure the centroids of the geometric shapes can 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.
8" x 6" x 1"
8" x 6" x 1"
12"
16"
16" x 1"
5" x 3" x 1/2"
5" x 3" x 1/2"
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 area.
16"
6"
6"
12"-20.7 lb channel
Solution:
Since the cross section is symmetrical with respect to a vertical axis, the centroid of the
cross section lies on the axis of symmetry. Therefore, only the location of the centroid with
respect to a horizontal axis will be determined.
ay, in³
(36)(8)=288
(72)(3)=216
(-36) (2) = -72
Part
a, in?.
х, in
y, in
ах, inз
(1/2)(12)(6)=36
(12)(6)=72
-(1/2)(12)(6)=-36
A=72
(36)(4)=144
(72)(6) = 432
(-36) (6)=-216
4
6
3
6
With the base of the cross section as the reference:
Total
360
432
E ax
8" x 6" x 1"
3. Locate the centroid of the shaded area in the figure,
created by cutting a semicircle of diameter r from a
quarter circle of radius r.
360
8" x 6" x 1"
%3D
A
72
* = 5"
16" + 0.28"= 16.28"
Eay
y =
432
16" x 1"
%3D
A
72
ỹ = 6"
5" x 3" x 1/2"
5" x 3" x 1/2"
ах, in3
ar? (4r
ay, in3
ar2 (4r
3T
a, in
X, in
4r
Part
у, in
4r
Quarter
circle
Semicircle
3n
4r
4 37
ar (4r
4
tr3
12"-20.7 lb channel
8
12
8
16
2
8
ay, in³
380.38
a, in².
2(13)=26
(16)(1)=16
2(3.75)=7.5
6.03
A=55.53
Part
3r3
= 0.25r3
y, in
16.28 -1.65 = 14.63
Total
ar2
0.137r3
= 0.393r2
8
2-8"x6"x1"
A =
12
16" x 1"
8 + 0.28 = 8.28
132.48
2-5"x3"x1/2"
0.75 + 0.28 = 1.03
7.725
E ax
0.393r
0.25r
12"-20.7 Ib channel
0.70
4.221
i = 0.636r
Total
524.806
Σαν
524.806
0.137r
A
55.53
0.393r?
y = 0.349r
y = 9.45" above the base of the cross section
RSITY
OTINN
SAPIEY
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