centre of gravity and centroid. centre of gravity –known as centre of mass the centre of mass of...
TRANSCRIPT
• 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
The centre of gravity of the leaning tower of Pisa lies above its base of support , so the tower is in stable equilibrium
Center of gravity - center of mass in a uniform gravitational field
i
ii
m
rmR
• 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)
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
3b
3h
bh21
34r
2
2r
34r
34r
4
2r
x y
2b
2h
bh
yx
b
h
y
x
y
y
xr
yx
y
x
r
x
yb
h
y
x
Shape Area A
1. Triangle
2. Semicircle
3. Quarter circle
4. Rectangle
Centroid of simple areas
0
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
A
Ayy
A
Axx ,
Example 2 : centroid of composite area
150 mm
20 mm
30 mm
120 mm
y
x
Calculation steps:
1. Divide the areas into simple shapes
2. State the reference axes
3. Find A1, A2, x1, x2, y1 and y2
based on the reference axes
4. Find the centroid
i
ii
A
xAx
i
ii
A
yAy1
2
Section Area A (mm2) x (mm) y (mm) Ax (mm3) Ay (mm3)
1
2
6600 iA
3600
3000
75
75
60
130
270000
225000
216000
390000
495000 iixA 606000 iiyA
mm 75x mm 8291.y
Example 3
150 mm
20 mm
30 mm
120 mm
y
x
i
ii
A
xAx
i
ii
A
yAy
1
2
Section Area A (mm2) x (mm) y (mm) Ax (mm3) Ay (mm3)
1
2
3
846285.Ai
3600
3000
75
75
60
130
270000
225000
216000
390000
1471438.xA ii 4587150.yA ii
mm 75x mm 4193.y 20 mm
diameter hole
3
314.16 75 60 23561.9 18849.6
Example 4
150 mm
20 mm
30 mm
120 mm
y
x
i
ii
A
xAx
i
ii
A
yAy
1
2
Section Area A (mm2) x (mm) y (mm) Ax (mm3) Ay (mm3)
1
2
3
846285.Ai
75
75
60
10
270000 216000
225000 30000
1471438.xA ii 167150 iiyA
mm 75x mm 5926.y 20 mm
diameter hole
3
3600
3000
314.16 75 60 23561.9 18849.6
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
W = mg = gV = V
(Greek letter gamma) = weight density or specific weight (typically in N/m3 or lb/ft3)
the centroids for mass and weight can be found as follows:
i ii i ii
i i i
x m y m z mx y z
m m m
i ii i ii
i i i
x W y W z Wx y z
W W W
Determine centre of mass or centre of gravity
PROBLEM SOLVING
Given: Two blocks of different materials are assembled as shown. The weight densities of the materials are A = 86.8 kN / m3 and B = 260.4 kN / m3.
Find: The center of gravity of this assembly.
Plan: Follow the steps for analysisSolution
1. In this problem, the blocks A and B can be considered as two segments.
PROBLEM SOLVING
Weight = w = (Volume in cm3)wA = 86.8 (0.5) (6) (6) (2) / (10)3 = 3.125 N
wB = 260.4 (6) (6) (2) / (10)3 = 18.75 N
62.559.3831.2521.88
6.2556.25
3.12556.25
12.518.75
23
13
41
3.12518.75
AB
w z (N.cm)
w y (N.cm)
w x (N.cm)
z (cm)y (cm)x (cm)w (N)Segment
PROBLEM SOLVING (continued)
~x = ( x w) / ( w ) = 31.25/21.88 = 1.47 cmy = ( y w) / ( w ) = 59.38/21.88 = 2.68 cmz = ( z w) / ( w ) = 62.5 /21.88 = 2.82 cm
~
~
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.
Application : Distributed load on Beam
How can we determine these weights and their locations?
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.
Application: Water tank
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 sides2. concept to areas with curved sides where we'll use integration.
Application : Tilt-slab construction