1. Centre of gravity and centroid

2. Refers to masses or weights
Centre of gravity
Known as centre of mass
The centre of mass of an object is the point where the whole of the object is assumed to be concentrated
Refers to masses or weights
Center of gravity - center of mass in a uniform gravitational field
The centre of gravity of the leaning tower of Pisa lies above its base of support , so the tower is in stable equilibrium

4. Locating the centre of mass of a symmetrical object

5. Find the centre of mass of the system of particle shown

6. Centroid
Ignoring weight and considering only volume
Different densities of material= the centre of gravity and centroid do not always coincide
Homogeneous = the centre of gravity and centroid would be coincide.
Type of centroid
Centroid of lines (rope.wire)
Centroid of areas (x,y)
Simple areas
Composite areas
Centroid of volume (x,y,z)

7. Centroid of line
for one-dimension object such as rope, wire and cable.
measured in length only.
the curve object, the centre of gravity is not located in the object and can be determined using formula of derivation.
the centroid can be determined by:
The centroid of line whether straight, curve or composite lines can only be determined if the elements are connected and made up from a homogenous material. For the straight line, the centroid lies at a distance L/2 from reference axis. Normally, the centroid of line is always coincides with the center of gravity

8. Example
Determine the centroid of a thin homogeneous wire

10. Centroid of simple areas
Shape
Area A
1. Triangle
2. Semicircle
3. Quarter circle
4. Rectangle b h y x y x r y x r b h y x

11. Centroid of composite area
Composite areas are the combination of simple areas.
The determination of centroid is more easier if calculate in the table.
1st Step: Divide the areas into several parts.
2nd Step: State the reference axis in the figures.
3rd Step: Find x1, x2 and y1, y2 from the reference axis that you stated.
4th Step: Find the centroid.
Ax = A1x1 + A2x2
Ay = A1y1 + A2y2
Total area, ΣA = A1 + A2

12. Example 1 : centroid of simple areay
6 m
4.5 m

13. Centroids x y
6 m
4.5 m

14. Centroids x y
6 m
4.5 m

15. Example 2 : centroid of composite area
Calculation steps:
Divide the areas into simple shapes
State the reference axes
Find A1, A2, x1, x2, y1 and y2 based on the reference axes
Find the centroid y x
150 mm
20 mm
30 mm
120 mm 2 1
Section
Area A (mm2)
x (mm)
y (mm)
Ax (mm3)
Ay (mm3) 1 2
3600 75 60
270000
216000
3000 75
130
225000
390000

16. Example 3 y x 150 mm 20 mm 30 mm 120 mm 2 20 mm diameter hole 3 1
Section
Area A (mm2)
x (mm)
y (mm)
Ax (mm3)
Ay (mm3) 1 2 3
3600
3000 75 60
130
270000
225000
216000
390000
314.16 75 60  

17. Example 4 y x 150 mm 20 mm 30 mm 120 mm 2 20 mm diameter hole 3 1 60
Section
Area A (mm2)
x (mm)
y (mm)
Ax (mm3)
Ay (mm3) 1 2 3
3600
3000
314.16 75
60
270000  75 10
225000
30000 75
60 

18. Finding a centroid for 3D body diagram
Example

19. Determine centre of mass or centre of gravity
The difference between calculating the centroid using volume, mass, or weight is simply a scale factor, since mass and weight are proportional to volume as indicated below.
 (Greek letter rho) = mass density (typically expressed in kg/m3 or slug/ft3)
m = V
 (Greek letter gamma) = weight density or specific weight (typically in N/m3 or lb/ft3)
W = mg = gV = V
the centroids for mass and weight can be found as follows:

20. Centroid of common shape in 3D

21. Centroid of common shape in 3D (continue)

22. PROBLEM SOLVING
Given: Two blocks of different materials are assembled as shown. The weight densities of the materials are A = kN / m3 and B = kN / m3.
Find: The center of gravity of this assembly.
Plan: Follow the steps for analysis
Solution
1. In this problem, the blocks A and B can be considered as two segments.

23.  PROBLEM SOLVING Weight = w =  (Volume in cm3)
wA = (0.5) (6) (6) (2) / (10)3 = N
wB = (6) (6) (2) / (10)3 = N      
Segment
w (N)
x (cm)
y (cm)
z (cm)
w x (N.cm)
w y (N.cm)
w z (N.cm) A B
3.125
18.75 4 1 1 3 2 3
12.5
18.75
3.125
56.25
6.25
56.25 
21.88
31.25
59.38
62.5

24. PROBLEM SOLVING (continued)~
x = ( x w) / ( w ) = /21.88 = 1.47 cm
y = ( y w) / ( w ) = /21.88 = 2.68 cm
z = ( z w) / ( w ) = / = cm
Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 9.3

25. Application : Distributed load on Beam
Illustration: The lumber on the rack shown below distributes the weight evenly across the supporting beam. This uniform loading is represented by a load curve with equal length lines. The total weight (resultant) equals the area under the load curve and it acts at the centroid of the load curve.

26. Application: Water tank
To design the structure for supporting a water tank, we will need to know the weights of the tank and water as well as the locations where the resultant forces representing these distributed loads act.
How can we determine these weights and their locations?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section

27. Application : Tilt-slab construction
In tilt-slab construction, we have a concrete wall (with doors and windows cut out) which we need to raise into position. We don't want the wall to crack as we raise it, so we need to know the center of mass of the wall. How do we find the center of mass for such an uneven shape?
Solution:
1. find the centroid of an area with straight sides
2. concept to areas with curved sides where we'll use integration.