Chapter 2 - Properties of Steel Areas

7/26/2019 Chapter 2 - Properties of Steel Areas

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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 + …





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



9 - 5

• 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



9 - 6

• 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.



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



• 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





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.



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



• 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



Determine the centroid, moment of inertia, section

modulus and radius of gyration of the shaded area

with respect to the y axis.



9 - 15

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



9 - 16

• 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

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Chapter 2 - Properties of Steel Areas