Centroid of Area (Center of Gravity of an area): Point

73 Pages·2009·5.35 MB·English

Centroid or center of gravity is the point within an object from which the force
Centroid
of gravity appears to act.
Centroid of 3D objects often (but not always) lies somewhere along the lines
of symmetry.
Centroidal axis
or Neutral axis
Centroid of Area (Center of Gravity of an area):
( )
Point x , y that defines the geometric center of
the area
∫ yδA ∫ xδA
y = A x = A
A A
Q = ∫ yδA = y ⋅ A
First Moment of an Area with
x VQ
respect to the x-axis
A τ =
Ib
First Moment of an Area with Q = ∫ xδA = x ⋅ A
y
respect to the y-axis
A
Calculate the center of gravity of the 
rectangle:
(a)Without a hole
(b)With a hole of dimensions c and d
a) Without a hole
∫ yδA b
∫ ayδy The centroid of any area can be found by taking
ab2 b
y = A = 0 = = moments of identifiable areas (such as
A ab 2ab 2
rectangles or triangles) about any axis.
∫ xδA a The moment of an area about any axis is equal
∫ axδx
a2b a
to the algebraic sum of the moments of its
x = A = 0 = =
A ab 2ab 2 component areas.
b) With a hole
n ⎛ b ⎞ d
∫ yδA ∑ y A ⎜ ⎟⋅a⋅b −( f + )⋅c⋅d
i i
⎝ 2⎠ 2
y = A y = 1 y =
A A a⋅b −c⋅d
n ⎛ a ⎞ c
∫ xδA ∑ x A ⎜ ⎟⋅a⋅b −(e− )⋅c⋅d
i i
⎝ 2 ⎠ 2
x = A x = 1 x =
A A a⋅b −c⋅d
Moment of Inertia
also known as the Second Moment of the Area is a term used to describe the
capacity of a cross-section to resist bending.
It is a mathematical property of a section concerned with a surface area and how
that area is distributed about the reference axis. The reference axis is usually a
centroidal
Second Moment or Moment of Inertia of an Area
I = ∫ y2δA I = ∫ x2δA
x y
A A
Rectangular Moments of Inertia: Moments of 
inertia with respect to an axis
Polar moments of inertia: Moment of inertia with 
respect to a point
J = ∫ρ2δA = I + I
O x y
A
Simple rectangular shape
I = ∫ y2dA dA = bdy
x
A
h
h
2 y 3 2 b ⎡ h 3 ⎛ h 3 ⎞ ⎤
I = ∫ y 2bdy = b = − ⎜ − ⎟
⎢ ⎥
⎜ ⎟
x
3 3 8 8
h ⎣ ⎝ ⎠ ⎦
− h −
2
2
bh3 hb3
I = I =
y
x 12
12
I = ∫ x2dA dA = hdx
y
A
b
b
2 x 3 2 h ⎡ b 3 ⎛ b 3 ⎞⎤
I = ∫ x 2hdx = h = − ⎜ − ⎟
⎢ ⎥
⎜ ⎟
y
3 3 8 8
b ⎣ ⎝ ⎠⎦
−b −
2
2
Polar Moment of Inertia
The polar moment of inertia is an important parameter
in problems involving torsion of cylindrical shafts and
rotations of slabs.
2
J = ∫ r dA
0
The polar moment of inertia is related to the
rectangular moments of inertia,
( )
J = ∫r2dA = ∫ x2 + y2 dA = ∫ x2dA + ∫ y2dA J = I + I
0 0 y x
Simple rectangular shape
3 3
bh3 bh hb
I = I = I + I = +
x 12 Polar x y
12 12
bh
( )
hb3 I = h2 + b2
I = P
12
y
12
Moment of Inertia: Parallel Axis Theorem for an Area
If the moment of inertia of an area is known about its neutral axis (centroid
axis), we can determine the moment of inertia of area about a corresponding 
parallel axis using the parallel axis theorem
Consider moment of inertia of the shaded area
( )2
I = ∫ d + y dA
x 1
dA
I = ∫ y2dA + 2d ∫ ydA + d 2 ∫ dA
x 1 1
dA dA dA
First integral represents the moment of inertia 
of the area about the centroidal axis 2
I = I + A ⋅ d
Second integral is equal to zero,  since x passes  x xc 1
c
through the area’s centroid C
Third integral represents the total area A
2
I = I + A ⋅ d
Similarly: y yc 2
2
Simple rectangular shape ⎛ h ⎞ bh3 bh3
I = I + A⎜ ⎟ = +
BB x
3 ⎝ 2 ⎠ 12 4
bh
I =
bh3
x
12
I =
BB
3
Note:
2 2
I = r A I = r A
Radius of Gyration x x y y
J = r2A r2 = r2 + r2
O O O x y
Moment of Inertia of Composite Areas
Determine the moment of inertia of the area shown with
respect to the x-axis
The moment of inertia of the area shown can be
obtained by sustracting the circle from the rectangle
Circle
I = I + Ad 2 X-axis
x x' y
1
( )
= π(25 )4 + π(25 )2 (75 )2 = 11.4 10 6 mm 4
4
Rectangle
I = I + Ad 2
x x' y
1
( )
= (100)(150)3 + (100)(150)(75)2 =112.5 106 mm4
12
I = 112 .5 × 10 6 − 11.4 × 10 6 = 101 × 10 6 mm 4
x
Determine the moment of inertia of
the area shown with respect to the
centroids.
Rectangle A
I = I + Ad 2
x x' y
1
= (100 )(300 )3 + (100 )(300 )(200 )2 = 1.425 ×109 mm 4
12
I = I + Ad 2
y y' x
1
= (300 )(100 )3 + (100 )(300 )(250 )2 = 1.90 ×109 mm 4
12
Rectangle B
I = I + Ad 2
x x' y
1
= (600 )(100 )3 = 0.05 ×109 mm 4
12
I = I + Ad 2
y y' x
1
= (100 )(600 )3 = 1.8010 9 mm 4
12
Rectangle D
I = I + Ad 2
x x' y
1
= (100 )(300 )3 + (100 )(300 )(200 )2 = 1.425 ×109 mm 4
12
I = I + Ad 2
y y' x
1
= (300 )(100 )3 + (100 )(300 )(250 )2 = 1.90 ×109 mm 4
12
Adding each segment
( )
9
I = 1.425 + 0.05 + 1.425 × 10
x
9 4
I = 2.90 × 10 mm
x
( )
9
I = 1.90 + 1.80 + 1.90 × 10
y
9 4
I = 5.60 × 10 mm
y