Engg-Mechanics-Centroid.ppt

Useful for btech 1st semester students

1. CENTROID
CENTRE OF GRAVITY
Centre of gravity : of a body is the pointat
which the whole weight of the body may be
assumed to be concentrated.
It is represented by CG. or simply Gor C.
A body is having only onecenter of gravity for
all positions of the body.
1
Contd.
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2. CENTRE OF GRAVITY
Consider a three dimensional
body of any size and shape,
having a mass m.
If we suspend the body as shown in
figure, from any point such as A, the
body will be in equilibrium under the
action of the tension in the cord and
the resultant Wof the gravitational
forces acting on all particles of the
body.
2
Contd.
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3. Resultant Wis collinear with
the Cord
Assume that we mark its
position by drilling a
hypothetical hole of negligible
size along its line of action
Cord
Resultant
CENTRE OF GRAVITY
3
Contd.
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4. To determinemathematically the location of the
centre of gravityof any body,
Centre of gravity is that point about whichthe
summation of the first moments of the weights of
the elements of the body is zero.
we apply the principle of momentsto the parallel
system of gravitational forces.
CENTRE OF GRAVITY
6
Contd.
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5. We repeat the experiment by
suspending the body from other
points such as Band C, and in
each instant we mark the line of
action of the resultant force.
For all practical purposes theselines of actionwill be
concurrentat a single point G, which is called the
centre of gravityof the body.
CENTRE OF GRAVITY
4
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6. w
A
A w
B
A
G
B
A w
A
B
C
G
B
A
C C
B
Example:
CENTRE OF GRAVITY
5
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7. if, we apply principle of moments, (Varignon’s Theorem)
about y-axis, for example,
The moment of the
resultant gravitational
force W, about any
axis
=
thealgebraicsumofthe
momentsaboutthesame
axisofthegravitational
forcesdWactingonall
infinitesimalelementsof
thebody.
dWx
Where W =
dW Wx
The moment of the
resultant about y-axis
=
The sum of moments of its
components about y-axis
CENTRE OF GRAVITY
7
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8. where = x-coordinate of centre of gravityx x x W
dWx
x



Similarly, y and z coordinates of the centre of gravity areW
dWy
y


 W
dWz
z



and ----(1)
CENTRE OF GRAVITY
8
8

9. x W
dWx
x


 W
dWy
y


 W
dWz
z


 With the substitution of W= m g and dW= g dmm
dmx
x


 m
dmy
y


 m
dmz
z



----(1)
----(2)
,,
,
,
(if ‘g’is assumed constant for all particles, then )
the expression for the coordinates of centre of gravity become
CENTRE OF MASS
9
Contd.
9

10. 




dV
dVx
x

 




dV
dVy
y

 




dV
dVz
z

 and ----(3)
If ρis not constant throughout the body, then we may
write the expression as
,
CENTRE OF MASS
The density ρof a body is mass per unit volume. Thus,
the mass of a differential element of volume dV
becomes dm = ρdV.
10
Contd.
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11. m
dmx
x


 m
dmy
y


 m
dmz
z


 ----(2)
,,
This point is called the centre of massand clearly
coincides with the centre of gravity as long as the gravity
field is treated as uniform and parallel.
CENTRE OF MASS
Equation 2 is independent of g and therefore define a
unique point in the body which is a function solely of the
distribution of mass.
11
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12. When the density ρof a body is uniform throughout,
it will be a constant factor in both the numeratorsand
denominatorsof equation (3) and will therefore
cancel.
The remaining expression defines a purely
geometrical propertyof the body.




dV
dVx
x

 




dV
dVy
y

 




dV
dVz
z


and, ----(3)
CENTROID
12
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13. When speaking of an actual physical body, we use the
term “centre of mass”.
Calculation of centroid falls within three distinct
categories, depending on whether we can model the
shape of the body involved as a line, an areaor a
volume.
The term centroidis used when the calculation concerns
a geometrical shape only.
13
Contd.
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14. LINES: for a slender rod or a wire of length L, cross-
sectional area A, and density ρ, the body approximates a
line segment, and dm = ρA dL. If ρ and A are constant over
the length of the rod, the coordinates of the centre of mass
also becomes the coordinates of the centroid, Cof the line
segment, which may be written as L
dLx
x


 L
dLy
y


 L
dLz
z



The centroid “C”of the line segment,
,,
14
Contd.
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15. AREAS:when the density ρ,is constant and the
body has a small constant thickness t,the body can be
modeled as a surface area.
The mass of an element becomes dm = ρ t dA.
If ρ and tare constant over entire area, the
coordinates of the ‘centre of mass’ also becomes
the coordinates of the centroid, Cof the surface
areaand which may be written asA
dAx
x


 A
dAy
y


 A
dAz
z



The centroid “C”of the Area segment,
,,
15
Contd.
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16. VOLUMES: for a general body of volume V and density ρ,
the element has a mass dm = ρ dV.
If the densityis constant the coordinates of the centre of
mass also becomes the coordinates of the centroid,Cof the
volume and which may be written asV
dVx
x


 V
dVy
y


 V
dVz
z



The centroid “C”of the Volume segment,
, ,
16
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17. Centroid of Simple figures:using method of
moment ( First moment of area)
Centroid of an area may or may not lie on the
area in question.
It is a unique point for a given area
regardless of the choice of the origin and the
orientation of the axes about which we take
the moment.
17
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18. Moment of
Total area ‘A’
about y-axis
=
Algebraic Sum of
moment of elemental
‘dA’ about the same
axis
where (A = a
1+ a
2+ a
3+ a
4+ ……..+ a
n)
(A) x= (a
1
) x
1
+ (a
2
) x
2
+ (a
3
) x
3
+
……….+(a
n
) x
n
= First moment of area
The coordinates of the centroid of the surface area
about any axis can be calculated by using the equn.
18
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19. If an area has an axis of symmetry, then the centroid
must lie on that axis.
If an area has two axes of symmetry, then the centroid
must lie at the point of intersection of these axes.
AXIS of SYMMETRY:
It is an axis w.r.t. which for an elementary area on one
side of the axis , there is a corresponding elementary
area on the other side of the axis (the first moment of
these elementary areas about the axis balance each
other)
19
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20. For example:
The rectangular shown in
the figure has two axis of
symmetry, X-X and Y-Y.
Therefore intersection of
these two axes gives the
centroid of the rectangle.
x
x
da
da
da ×x = da ×x
Moment of areas,da
about y-axis cancel
each other
da ×x + da ×x = 0
20
Contd.
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21. AXIS of SYMMETYRY
‘C’ must lie at the intersection
of the axes of symmetry
‘C’ must lie
on the axis
of symmetry
‘C’ must lie on
the axis of
symmetry
21
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22. EXERCISE PROBLEMS
Locate the centroid of the shaded area shown
Problem No.1:50
40
10
10
Ans: x=12.5, y=17.5
22
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23. Locate the centroid of the shaded area shown
Problem No.2:300
300
5001000 mm
1000 mm
500
r=600
D=600
Ans: x=474mm, y=474mm23
EXERCISE PROBLEMS
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24. Locate the centroid of the shaded area w.r.t. to the
axes shown
Problem No.3:x-axis
y-axis
90
20
20
60
120
r=40
Ans: x=34.4, y=40.324
EXERCISE PROBLEMS
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25. Locate the centroid of the shaded area w.r.t. to the
axes shown
Problem No.4:250 mm
20
10
380
10
200 mm x-axis
y-axis
Ans: x= -5mm, y=282mm
10
25
EXERCISE PROBLEMS
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26. Locate the centroid of the shaded area w.r.t. to the
axes shown
Problem No.530
50
40
20 20
40
x
y
r=20
Ans:x =38.94, y=31.46
30
26
EXERCISE PROBLEMS
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27. Locate the centroid of the shaded area w.r.t. to the
axes shown
Problem No.62.4 m
1.0
1.5
1.5
1.0
1.0
r=0.6
x
y
Ans: x=0.817, y=0.2427
EXERCISE PROBLEMS
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28. Problem No.7
Locate the centroid of the shaded area w.r.t. to the
axes shown
Ans: x= -30.43, y= +9.5828
EXERCISE PROBLEMS
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29. Problem No.8
Locate the centroid of the shaded area.
Ans: x= 0, y= 67.22(about base)
20
29
EXERCISE PROBLEMS
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30. Problem No.9
Locate the centroid of the shaded area w.r.t. to the
base line.
Ans: x=5.9, y= 8.17
2
30
EXERCISE PROBLEMS
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31. Problem No.10
Locate the centroid of the shaded area w.r.t. to the
axes shown
Ans: x=21.11, y= 21.11 31
EXERCISE PROBLEMS
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32. Problem No.11
Locate the centroid of the shaded area w.r.t. to the
axes shown
Ans: x= y= 22.22 32
EXERCISE PROBLEMS
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