36 mm 24 mm Locate the centroid of the plane area shown 2 Solving Problems on Your Own Several points should be emphasized when solving these types of problems Locate the centroid of the plane area shown ID: 783714
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Slide1
1
y
x
20 mm
30 mm
Problem 7.6
36 mm
24 mm
Locate the centroid of the plane
area shown.
Slide22
Solving Problems on Your Own
Several points should be emphasized when solving these types of problems.
Locate the centroid of the plane area shown.
y
x
20 mm
30 mm
36 mm
24 mm
1.
Decide how to construct the given area from common shapes
.
2.
It is strongly recommended that you
construct a table
containing areas or length and the respective coordinates of
the centroids
.
3.
When possible,
use symmetry to help locate the centroid
.
Problem 7.6
Slide33
Problem 7.6 Solution
y
x
24 + 12
20 + 10
10
30
Decide how to construct the given area from common shapes.
C
1
C
2
Dimensions in mm
Slide44
Problem 7.6 Solution
y
x
24 + 12
20 + 10
10
30
C
1
C
2
Dimensions in mm
Construct a table containing areas and respective coordinates of the centroids
.
A
, mm
2
x
, mm
y
, mm
xA
, mm
3
yA
, mm
3
1 20 x 60 =1200 10 30 12,000 36,000
2 (1/2) x 30 x 36 =540 30 36 16,200 19,440
S
1740 28,200 55,440
Slide55
Problem 7.6 Solution
y
x
24 + 12
20 + 10
10
30
C
1
C
2
Dimensions in mm
X
S
A
=
S
xA
X (1740) = 28,200
Then
or
X = 16.21 mm
and
Y
S
A
=
S
yA
Y (1740) = 55,440
or
Y = 31.9 mm
A
, mm
2
x
, mm
y
, mm
xA
, mm
3
yA
, mm
3
1 20 x 60 =1200 10 30 12,000 36,000
2 (1/2) x 30 x 36 =540 30 36 16,200 19,440
S
1740 28,200 55,440
Slide66
Problem 7.7
The beam
AB
supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b
) the corresponding value wB.
w
A
w
B
A
B
a
0.3 m
24 kN
30 kN
1.8 m
Slide77
The beam
AB
supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b
) the corresponding value wB.
w
A
w
B
A
B
a
0.3 m
24 kN
30 kN
1.8 m
Solving Problems on Your Own
1.
Replace the distributed load by a single equivalent force.
The magnitude of this force is equal to the area under the
distributed load curve and its line of action passes through
the centroid of the area.
2.
When possible,
complex distributed loads should be
divided into common shape areas
.
Problem 7.7
Slide88
Problem 7.7 Solution
w
B
A
B
a
0.3 m
24 kN
30 kN
20 kN/m
C
0.6 m
0.6 m
R
I
R
II
We have
R
I
= (1.8 m)(20 kN/m) = 18 kN
1
2
R
II
= (1.8 m)(
w
B
kN/m) = 0.9
w
B
kN
1
2
Replace the distributed
load by a pair of
equivalent forces.
Slide99
Problem 7.7 Solution
w
B
A
B
a
0.3 m
24 kN
30 kN
C
0.6 m
0.6 m
R
I
= 18 kN
R
II
= 0.9
w
B
kN
(a)
S
M
C
= 0: (1.2 -
a
)m x 24 kN - 0.6 m x 18 kN
- 0.3m x 30 kN = 0
or
a
= 0.375 m
(b)
S
F
y
= 0: -24 kN + 18 kN + (0.9
w
B
) kN - 30 kN= 0
or
w
B
= 40 kN/m
+
+
Slide1010
Problem 7.8
y
x
0.75 in
z
1 in
2 in
2 in
3 in
2 in
2 in
r
= 1.25 in
r
= 1.25 in
For the machine element
shown, locate the
z
coordinate
of the center of gravity.
Slide1111
Solving Problems on Your Own
y
x
0.75 in
z
1 in
2 in
2 in
3 in
2 in
2 in
r
= 1.25 in
r
= 1.25 in
X
S
V
=
S
x V Y
S
V
=
S
y V Z
S
V
=
S
z V
where
X, Y, Z
and
x, y, z
are the coordinates of the centroid of the body and the components, respectively.
For the machine element
shown, locate the
z
coordinate
of the center of gravity.
Problem 7.8
Determine the center of gravity of composite body.
For a homogeneous body
the center of gravity coincides
with the centroid of its volume. For this case the center of gravity can be determined by
Slide1212
Problem 7.8 Solution
y
x
0.75 in
z
1 in
2 in
2 in
3 in
2 in
2 in
r
= 1.25 in
r
= 1.25 in
Determine the center of gravity
of composite body.
First assume that the machine
element is homogeneous so
that its center of gravity will
coincide with the centroid of
the corresponding volume.
y
x
z
I
II
III
IV
V
Divide the body into five common shapes.
Slide1313
y
x
z
I
II
III
IV
V
y
x
0.75 in
z
1 in
2 in
2 in
3 in
2 in
2 in
r
= 1.25 in
r
= 1.25 in
V, in
3
z
, in.
z
V, in
4
I (4)(0.75)(7) = 21 3.5 73.5
II (
p
/2)(2)
2
(0.75) = 4.7124 7+ [(4)(2)/(3
p
)] = 7.8488 36.987
III -
p
(11.25)
2
(0.75)= -3.6816 7 -25.771
IV (1)(2)(4) = 8 2 16
V -(
p
/2)(1.25)
2
(1) = -2.4533 2 -4.9088
S
27.576 95.807
Z
S
V
=
S
z V
:
Z
(27.576 in
3
) = 95.807 in
4
Z
= 3.47 in