1. 9.3 Composite Bodies
Consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular or semicircular
A body can be sectioned or divided into its composite parts
Provided the weight and location of the center of gravity of each of these parts are known, the need for integration to determine the center of gravity for the entire body can be neglected

2. 9.3 Composite Bodies Accounting for finite number of weights Where
represent the coordinates of the center of gravity G of the composite body
represent the coordinates of the center of gravity at each composite part of the body
represent the sum of the weights of all the composite parts of the body or total weight

3. 9.3 Composite Bodies
When the body has a constant density or specified weight, the center of gravity coincides with the centroid of the body
The centroid for composite lines, areas, and volumes can be found using the equation
However, the W’s are replaced by L’s, A’s and V’s respectively

4. 9.3 Composite Bodies Procedure for Analysis Composite Parts
Using a sketch, divide the body or object into a finite number of composite parts that have simpler shapes
If a composite part has a hole, or a geometric region having no material, consider it without the hole and treat the hole as an additional composite part having negative weight or size

5. 9.3 Composite Bodies Procedure for Analysis Moment Arms
Establish the coordinate axes on the sketch and determine the coordinates of the center of gravity or centroid of each part

6. 9.3 Composite Bodies Procedure for Analysis Summations
Determine the coordinates of the center of gravity by applying the center of gravity equations
If an object is symmetrical about an axis, the centroid of the objects lies on the axis

7. 9.3 Composite Bodies
Example 9.9
Locate the centroid of the wire.

8. 9.3 Composite Bodies Solution Composite Parts Moment Arms
Location of the centroid for each piece is determined and indicated in the diagram

9. 9.3 Composite Bodies Solution Summations Segment L (mm) x (mm) y (mm)
z (mm)
xL (mm2)
yL (mm2)
zL (mm2) 1
188.5 60
-38.2
11 310
-7200 2 40 20
800 3
-10
-200
Sum
248.5
-5600

10. 9.3 Composite Bodies
Solution
Summations

11. 9.3 Composite Bodies Example 9.10
Locate the centroid of the plate area.

12. 9.3 Composite Bodies Solution Composite Parts
Plate divided into 3 segments
Area of small rectangle considered “negative”

13. 9.3 Composite Bodies Solution Moment Arm
Location of the centroid for each piece is determined and indicated in the diagram

14. 9.3 Composite Bodies Solution Summations Segment A (mm2) x (mm) y (mm)
xA (mm3)
yA (mm3) 1
4.5 2 9
-1.5
1.5
-13.5
13.5 3 -2
-2.5 5 -4
Sum
11.5 14

15. 9.3 Composite Bodies
Solution
Summations

16. 9.3 Composite Bodies Example 9.11 Locate the center of mass of the
composite assembly. The conical
frustum has a density of
ρc = 8Mg/m3 and the hemisphere
has a density of ρh = 4Mg/m3.
There is a 25mm radius
cylindrical hole in the center.

17. 9.3 Composite Bodies Solution Composite Parts
View Free Body Diagram
Solution
Composite Parts
Assembly divided into 4 segments
Area of 3 and 4 considered “negative”

18. 9.3 Composite Bodies Solution Moment Arm
Location of the centroid for each piece is determined and indicated in the diagram
Summations
Because of symmetry,

19. 9.3 Composite Bodies Solution Summations Segment m (kg) z (mm)
zm (kg.mm) 1
4.189 50 2
1.047
-18.75 3
-0.524
125 4
-1.571
Sum
3.141
45.815

20. 9.3 Composite Bodies
Solution
Summations

21. 9.4 Theorems of Pappus and Guldinus
A surface area of revolution is generated by revolving a plane curve about a non-intersecting fixed axis in the plane of the curve
A volume of revolution is generated by revolving a plane area bout a nonintersecting fixed axis in the plane of area
Example
Line AB is rotated about
fixed axis, it generates
the surface area of a
cone (less area of base)

22. 9.4 Theorems of Pappus and Guldinus
Example
Triangular area ABC rotated
about the axis would
generate the volume of
the cone
The theorems of Pappus and Guldinus are used to find the surfaces area and volume of any object of revolution provided the generating curves and areas do not cross the axis they are rotated

23. 9.4 Theorems of Pappus and Guldinus
Surface Area
Area of a surface of revolution = product of length of the curve and distance traveled by the centroid in generating the surface area

24. 9.4 Theorems of Pappus and Guldinus
Volume
Volume of a body of revolution = product of generating area and distance traveled by the centroid in generating the volume

25. 9.4 Theorems of Pappus and Guldinus
Composite Shapes
The above two mentioned theorems can be applied to lines or areas that may be composed of a series of composite parts
Total surface area or volume generated is the addition of the surface areas or volumes generated by each of the composite parts

26. 9.4 Theorems of Pappus and Guldinus
Example 9.12
Show that the surface area of a sphere is
A = 4πR2
and its volume
V = 4/3 πR3

27. 9.4 Theorems of Pappus and Guldinus
View Free Body Diagram
Solution
Surface Area
Generated by rotating semi-arc about the x axis
For centroid,
For surface area,

28. 9.4 Theorems of Pappus and Guldinus
Solution
Volume
Generated by rotating semicircular area about the x axis
For centroid,
For volume,

29. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
Consider the flat plate subjected to the loading function ρ = ρ(x, y) Pa
Determine the force dF acting on the differential area dA m2 of the plate, located at the differential point (x, y)
dF = [ρ(x, y) N/m2](d A m2)
= [ρ(x, y) d A]N
Entire loading represented as
infinite parallel forces acting on
separate differential area dA

30. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
This system will be simplified to a single resultant force FR acting through a unique point on the plate

31. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
Magnitude of Resultant Force
To determine magnitude of FR, sum the differential forces dF acting over the plate’s entire surface area dA
Magnitude of resultant
force = total volume under
the distributed loading
diagram

32. 9.5 Resultant of a General Distributed Loading
Pressure Distribution over a Surface
Location of Resultant Force
Line of action of action of the resultant force passes through the geometric center or centroid of the volume under the distributed loading diagram