1. Problem 7.6 Locate the centroid of the plane area shown. y x 20 mm

2. Solving Problems on Your Own
20 mm
30 mm
Solving Problems on Your Own
Locate the centroid of the plane area shown.
36 mm
Several points should be emphasized when solving these types of problems.
24 mm x
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.

3. Decide how to construct the given area from common shapes.y
Problem 7.6 Solution
Decide how to construct the given area from common shapes. C1 C2 30 x 10
Dimensions in mm

4. y
Problem 7.6 Solution
Construct a table containing areas and respective coordinates of the centroids. C1 C2 30 x 10
Dimensions in mm
A, mm x, mm y, mm xA, mm yA, mm3
x 60 = , ,000
2 (1/2) x 30 x 36 = , ,440
S , ,440

5. XS A = S xA X (1740) = 28,200 X = 16.21 mm YS A = S yA
Problem 7.6 Solution
Then
XS A = S xA
X (1740) = 28,200
X = mm C1 or C2
and
YS A = S yA 30
Y (1740) = 55,440 x 10
Y = 31.9 mm
Dimensions in mm or
A, mm x, mm y, mm xA, mm yA, mm3
x 60 = , ,000
2 (1/2) x 30 x 36 = , ,440
S , ,440

6. Problem 7.7 a
24 kN
30 kN
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.
0.3 m A B wA wB
1.8 m

7. Solving Problems on Your Own a
24 kN
30 kN
0.3 m
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. A B wA wB
1.8 m
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.

8. RII = (1.8 m)(wB kN/m) = 0.9 wB kN
Problem 7.7 Solution a
24 kN
30 kN
0.3 m
Replace the distributed
load by a pair of
equivalent forces. C A B
20 kN/m wB
0.6 m
0.6 m RI
RII 1 2
We have
RI = (1.8 m)(20 kN/m) = 18 kN 1 2
RII = (1.8 m)(wB kN/m) = 0.9 wB kN

9. SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0 or wB = 40 kN/m
Problem 7.7 Solution a
24 kN
30 kN
0.3 m C A B wB
0.6 m
0.6 m
RI = 18 kN
RII = 0.9 wB kN
(a) +
SMC = 0: (1.2 - a)m x 24 kN m x 18 kN
- 0.3m x 30 kN = 0
or a = m
(b) +
SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0
or wB = 40 kN/m

10. Problem 7.8 y For the machine element shown, locate the z coordinatex
0.75 in z
1 in
2 in
3 in
r = 1.25 in
For the machine element
shown, locate the z coordinate
of the center of gravity.

11. X S V = S x V Y S V = S y V Z S V = S z V
0.75 in z
1 in
2 in
3 in
r = 1.25 in
Problem 7.8
Solving Problems on Your Own
For the machine element
shown, locate the z coordinate
of the center of gravity.
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
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.

12. Determine the center of gravity of composite body.x
0.75 in z
1 in
2 in
3 in
r = 1.25 in
Problem 7.8 Solution
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.

13. II (p/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3p)] = 7.8488 36.987 y x z I II
III IV V y x
0.75 in z
1 in
2 in
3 in
r = 1.25 in
V, in3 z, in z V, in4
I (4)(0.75)(7) =
II (p/2)(2)2 (0.75) = [(4)(2)/(3p)] =
III p(11.25)2 (0.75)=
IV (1)(2)(4) =
V (p/2)(1.25)2 (1) = S
Z S V = S z V : Z ( in3 ) = in Z = 3.47 in