lecture material week 6

Section Properties of Structural Members

2

Center of Gravity & Center of Mass

The center of gravity G is a point which locates the resultant weight of a system of particles.

The weights of the particles are considered to be a parallel force system.

The system of weights can be replaced by a single (equivalent) resultant weight acting at the Center of Gravity.

3

Total weight, WR = ∑=

n

iiW

1

Thus, using the moment equation, we can write

x location:

nnRR Wx~...Wx~Wx~Wx~Wx +++= 332211

y location:

nnRR Wy~...Wy~Wy~Wy~Wy +++= 332211

z location:

nnRR Wz~...Wz~Wz~Wz~Wz +++= 332211

4

The coordinate ( )GGG z,y,x , is called the center of gravity for

the system.

5

Since W = mg, we can immediately deduce that

∑

∑

∑

∑

∑

∑

=

=

=

=

=

= === n

ii

n

iii

n

ii

n

iii

n

ii

n

iii

m

mz~

zm

my~

ym

mx~

x

1

1

1

1

1

1

=z,y,x coordinates of the center of mass

=iii z~,y~,x~ coordinates of the ith particle

mi = mass of the ith particle

6

Center of Gravity Let us imagine that a continuous body is made up

of an infinite number of particles, each possesses an infinitesimal weight, δWi → 0 but δWi ≠ 0.

Following the previous derivation, we can write

• This equation is also an equation for the center of gravity of a body, where γ is the specific weight of the body (weight per unit volume). The same goes for y and z coordinates.

7

Centroid The centroid C is a point which defines the

geometric center of an object.

The geometric center of an object is independent of the force or weight of the body.

It depends entirely on the complexities of the shape of the object.

Also, the location of the geometric center does not necessarily have to be within the object; it can be located off the object in space.

8

Centroid of a Volume

9

Centroid of an Area

10

Centroid of a Line

11

Example:

Locate the centroid of the line shown.

Solution:

12

= 0.410 m= 0.574 m

xy

13

Example:

Locate the centroid of the area shown

Solution:

14

15

Example: Locate the centroid of the area shown in the figure.

16

Example:Locate the yc centroid for the paraboloid of the revolution, which is generated by revolving the shaded area shown about the y-axis.

17

Composite Bodies Real engineering structures are very

complicated. How do we find the centroid?

We can cut the body into several pieces of simpler “bodies” of standard shapes, like square, rectangle, triangle, circle etc, and analyze them separately, and finally put all the result together.

18

19

Example:Locate the centroid C of the cross-sectional area for the T-beam shown.

Solution: The section is symmetry about the y-axis so that x = 0. The area is segmented into two rectangles

20

Example: Locate the centroid of the composite area.

21

Segment A (ft2) x y xA yA

1 4.5 1 1 4.5 4.5

2 6 -1 1.5 -6 9

3 1 -2.5 0.5 -2.5 0.5

∑A = 11.5 ∑xA = -4 ∑yA = 14

ft348.05.11

4~ −=−== ∑

A

Axx

ft22.15.11

14~ === ∑

A

Ayy

y

x 1

2

3

1ft 2ft 3ft

2ft

1ft

22

Moments of Inertia of an Area

In the preceding sections, we compute the centroid for an area by using the first moment of the area about an axis, i.e., ∫xdA.

Here, we are considering the second moment of an area given as

• These formulas are sometimes referred to as the moment of inertia for the area.

23

Instead of computing the second moment of area about an axis, we can also consider the second moment of area about the pole O, and is referred to as the polar moment of inertia, given mathematically as

• The unit for the second moment of area is [length]4.

24

Parallel-Axis Theorem If the second moment of area is known

about an axis passing through its centroid, we can use that information to compute the second moment of area about a corresponding parallel axis.

This is known as the Parallel-Axis Theorem.

We now derive the formula.

25

It is known that the second moment of δA about the X-axis is:

26

Integrating for the entire area, we have

Notice that ∫y’2dA is the second moment of area about centroidal axis, Ix’.

.

y 0=y∫dA = 0 since ∫y’dA – zero because ∫y’dA =

Finally, we conclude that

and

27

Example:Determine the moment of inertia for the rectangular area shown with respect to (a) the centroidal x’-axis, and (b) the xb-axis.

xb

28

(b) the moment of inertia passing about an axis passing through the base of the rectangle can be obtained by applying the parallel-axis theorem:

29

Moment of Inertia for Simple Areas

12

3bdI xx =1. Rectangle:

12

3dbI yy =

2. Triangle: 36

3hbI yy =

36

3bhI xx =

3. Circle:4

4rII yyxx

π==

4. Semicircle: 411.0 rI xx =8

4rI yy

π=

5. Quarter Circle: 4055.0 rII yyxx ==

30

Example:Determine the moments of inertia for the cross-section shown about the x-axis.

∫ ∫ −==A A

x dy)x(ydAyI 10022

dyy

yI x

−= ∫ 400

1002200

0

2

4610107 mmI x ×=

dxydyyI x ∫∫ −=200

0

4200

0

2

400

1100 dy

31

Example:Determine the moments of inertia for the cross-section shown about its centroidal axes.

A

B

D

100 mm

32

Solution: We segment the beam into 3 rectangles.

33

Example:Determine Ixx and Iyy. 6 in 3 in

6 in

6 in 1 2

3

x

y

( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )( )4

232323

in 630

2962

169

36

1236

2

163

36

136666

12

1

=

++

++

+=xxI

Ixx1 Ixx2 Ixx3

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4

232323

in 1971

6962

196

36

1736

2

136

36

136666

12

1

=

++

++

+=yyI

Iyy1 Iyy2 Iyy3