ns 1 and 2. Locate the centroid of the shaded area
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
CENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines.
PROBLEM 2 ONLY
PROVIDE THE SAME PROCESS OF SOLUTION IN THE GIVEN EXAMPLE ? THANK YOU.
Transcribed Image Text:2. Locate the centroid of the shoded crea shown,
Divide the area into regulor geometric shapes. Make
sure the centroids of the geometric shapes con 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.
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 orea.
12 hal
Solution:
Since the cross section is symmetrical with respect to a vertical axis, the centroid of the
cross section les on the axis of symmetry. Therefore, only the location of the centroid with
respect to a horizontal axis will be determined.
ay, ie
(36)(8)-288
(72)3)-216
(-36)|2) =-72
432
Part
a in.
x in
in
1/2126)-36
12)(6)-72
F1/2)12|6)-36
A=72
ax, in
(36(4)-144
2|(6)= 432
(36)(6)-216
360
4
3
2
With the base of the cross section as the reference:
Total
3. Locate the centroid of the shoded orea in the fgure.
created by cutfing a semiccle of diometer rom a
Quarter cicie of rod
16+ 0.28" 16.28"
Pert
Quater
cce
Semicrce
rarh dannel,.
Pert
2-8"x6"x"
16"x 1"
2-5x3x1/2"
12-20.7 b channel
Total
a in
2(13)-26
(16)-16
213.75)-75
603
A-55.53
yin
16 28 -1.65- 14.63
8+0.28-8.28
0.75 + 0.28 = 1.03
0.70
ay, i
380.38
132.48
7.725
4221
fotal
524.806
9-945" above the base of the cross section
10TIN
Transcribed Image Text:Problems 1 and 2. Locate the centroid of the shaded area
y
-r = 100 mm
100 mm
140 mm
50 mm
50 mm 140 mm
-180 mm
-200 mm-
1.
2.
ITY
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