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CENTROID OF AN AREA
The centroid of a area is analogous to thecenter of gravity of a homogenous body. Thecentroid s often described as the point at
which a thin homogenous plate wouldbalance.
The centroid of complex area can be foundby dividing the area into basic shapes
(rectangles, triangles, circles, etc.)AT Xc = a1x1 + a2x2 + a3x3 + …
AT Yc = a1y1 + a2y2 + a3y3 + …
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9 - 4
• Consider distributed forces whose magnitudes are
proportional to the elemental areas on which theyact and also vary linearly with the distance of
from a given axis.
F
A
A
• Example: Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from the
neutral axis which passes through the section centroid.
momentsecond
momentfirst022
dA ydA yk M
QdA ydA yk R
Aky F
x
•
Example: Consider the net hydrostatic force on asubmerged circular gate.
dA y M
dA y R
A y A p F
x
2
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• Second moments or moments of inertia of
an area with respect to the x and y axes,
dA x I dA y I y x22
• Evaluation of the integrals is simplified by
choosing d A to be a thin strip parallel to
one of the coordinate axes.
• For a rectangular area,
331
0
22 bhbdy ydA y I h
x
• The formula for rectangular areas may also
be applied to strips parallel to the axes,
dx y xdA xdI dx ydI y x223
31
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• The moment of inertia of a composite area A about a given axis is
obtained by adding the moments of inertia of the component areas
A1, A2, A3, ... , with respect to the same axis.
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In structural engineering, the section modulus of a
beam is the ratio of a cross section's second moment of area
to the distance of the extreme compressive fiber from the
neutral axis.
z or s =
For symmetrical sections this will mean the Zx max and
Zx min are equal. For unsymmetrical sections (T-beam forexample) the section modulus used will differ depending on
whether the compression occurs in the web or flange of the
section.
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• The polar moment of inertia is an important
parameter in problems involving torsion of
cylindrical shafts and rotations of slabs.
dAr J 20
• The polar moment of inertia is related to the
rectangular moments of inertia,
x y I I
dA ydA xdA y xdAr J
22222
0
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• Consider area A with moment of inertia
I x. Imagine that the area isconcentrated in a thin strip parallel to
the x axis with equivalent I x.
A
I k Ak I x x x x 2
k x = radius of gyration with respectto the x axis
• Similarly,
A
J k Ak J
A
I k Ak I
OOOO
y y y y
2
2
222 y xO k k k
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The strength of a W14x38 rolled steel
beam is increased by attaching a plate
to its upper flange.
Determine the moment of inertia and
radius of gyration with respect to an
axis which is parallel to the plate and
passes through the centroid of the
section.
SOLUTION:
• Determine location of the centroid ofcomposite section with respect to a
coordinate system with origin at the
centroid of the beam section.
•
Apply the parallel axis theorem todetermine moments of inertia of beam
section and plate with respect to
composite section centroidal axis.
• Calculate the radius of gyration from the
moment of inertia of the compositesection.
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SOLUTION:
• Determine location of the centroid of
composite section with respect to a
coordinate system with origin at the
centroid of the beam section.
12.5095.17
0011.20SectionBeam12.50425.76.75Plate
in,in.,in,Section 32
A y A
A y y A
in.792.2in17.95
in12.5023
A A yY A y AY
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• Apply the parallel axis theorem to determine
moments of inertia of beam section and plate
with respect to composite section centroidal
axis.
4
23
43
1212
plate,
4
22section beam,
in2.145
792.2425.775.69
in3.472
792.220.11385
Ad I I
Y A I I
x x
x x
• Calculate the radius of gyration from the
moment of inertia of the composite section.
2
4
in17.95
in5.617
A
I k x x
in.87.5 xk
2.1453.472 plate,section beam, x x x I I I
4
in618 x I
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Determine the centroid, moment of inertia, section
modulus and radius of gyration of the shaded area
with respect to the y axis.
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SOLUTION:
• Compute the moments of inertia of the bounding
rectangle and half-circle with respect to the x axis.
Rectangle:
46
313
31 mm102.138120240 bh I x
Half-circle:moment of inertia with respect to AA’,
464
814
81 mm1076.2590 r I A A
23
2
2
12
2
1
mm1072.12
90
mm81.8a-120 b
mm2.383
904
3
4
r A
r a
moment of inertia with respect to x’,
46
362
mm1020.7
1072.121076.25
Aa I I A A x
moment of inertia with respect to x,
46
2362
mm103.92
8.811072.121020.7
Ab I I x x
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• The moment of inertia of the shaded area is obtained by
subtracting the moment of inertia of the half-circle fromthe moment of inertia of the rectangle.
46mm109.45 x I
x I 46mm102.138 46mm103.92