Lecture 5 - Page 1 of 8

Lecture 5 – Moment of Inertia of Non-symmetric Shapes

In general, most cross-sectional shapes of structural members are symmetric

(i.e., mirror image on both sides of both neutral axes). The determination of

section properties for these symmetric shapes involves plugging in numbers into

the formulas as discussed in Lecture 4.

Non-symmetric shapes are those that are NOT mirror image on both sides of one

or both of the neutral axes. Examples of non-symmetric shapes are as follows:

The procedure for determining the moment of inertia of these non-symmetric

shapes involves two steps:

1) Determine location of centroid of entire shape, y and x

piece

piecepiece

A

yA

y

Σ

Σ

=

)(

where:

A

piece

= area of the individual piece

y

piece

= dist. to centroid of the individual piece

piece

piecepiece

A

xA

x

Σ

Σ

=

)(

2) Determine the transformed moment of inertia,

I

t

by using the “Parallel

Axis Theorem

”

piece

t

AdII

∑

+= )(

2

where:

I

= moment of inertia of the piece

A

= area of the piece

d

=

y – y

piece

Angles

T-Sections

Channels

Sections w/ unsymmetric

holes

Lecture 5 - Page 2 of 8

Example 1

GIVEN

: A “T” shaped beam using a nominal wood 2x8 web with a 2x4 top flange

REQUIRED

: Determine

a) The location of the neutral axis

y

b) The transformed moment of inertia about the strong axis

c) The moment of inertia about the weak axis

a) Location of neutral axis y

:

Piece

Area

y

(Area)y

A

(1.5”)(7.25”) = 10.88 in

2

½(7.25”) = 3.625”

(10.88)(3.625”) = 39.44 in

3

B (1.5”)(3.5”) = 5.25 in

2

7.25” + ½(1.5”) = 8”

(5.25 in

2

)(8”) = 42 in

3

Totals:

16.13 in

2

81.44 in

3

piece

piecepiece

A

yA

y

Σ

Σ

=

)(

= 81.44 in

3

16.13 in

2

y = 5.05 in.

1½”

y

A

1½”

7¼”

3½”

Datum

Piece “B”

y

y

B

Piece “A” (centered

under piece “B”)

Lecture 5 - Page 3 of 8

b) Determine transformed moment of inertia about the strong “x” axis

:

Piece

Area

y

(Area)y

I

d = y-y

piece

Ad

2

A

10.88 in

2

3.625” 39.44 in

3

4

3

63.47

12

)"25.7)("5.1(

in=

3.625 - 5.05

= -1.425”

(10.88)(-1.425)

2

= 22.09 in

4

B

5.25 in

2

8”

42 in

3

4

3

98.0

12

)"5.1)("5.3(

in=

8 - 5.05 = 2.95”

(5.25)(2.95)

2

= 45.69 in

4

Totals:

16.13 in

2

81.44 in

3

48.61 in

4

67.78 in

4

piece

x

AdII

∑

+= )(

2

= 48.61 in

4

+ 67.78 in

4

I

x

= 116.39 in

4

c) Determine the moment of inertia about the weak “y” axis

:

Since the neutral axis of both pieces line–up over each other, the total

moment of inertia is the sum of the moment of inertias of the pieces.

I

y

= I

A

+ I

B

12

)5.3)(5.1(

12

)5.1)(25.7(

33

+=

y

I

= I

y

= 7.4 in

4

1½”

7¼”

3½”

1½”

Piece “B”

Piece “A” (centered

under piece “B”)

Neutral axis of

piece “A” and “B”

Lecture 5 - Page 4 of 8

Example 2

GIVEN

: A steel W18x35 “I” beam reinforced with a ½” x 8” steel plate welded to

the bottom flange of the beam as shown below

REQUIRED

: Determine:

a) The location of the neutral axis “

y”

b) The transformed moment of inertia about the strong axis

c) The moment of inertia about the weak axis

d) The section modulus about the strong axis

e) The radius of gyration about the strong axis

Make a Table as shown below:

Piece

Area

y

(Area)y

I

d = y-y

piece

Ad

2

1 4 in

2

0.25”

1 in

3

4

3

08.0

12

)"5.0)("8(

in

=

6.80”-0.25” = 6.55”

(4)(6.55)

2

=

171.6 in

4

2

10.3 in

2

9.35”

96.3 in

3

510 in

4

6.80”–9.35” = -2.55”

(10.3)(-2.55)

2

=

66.98 in

4

Totals:

14.3 in

2

97.3 in

3

510.08 in

4

238.58 in

4

a) Determine location of neutral axis “y”

:

piece

piecepiece

A

yA

y

Σ

Σ

=

)(

=

2

3

3.14

3.97

in

in

y = 6.80”

W18x35 steel beam (Area = 10.3 in

2

)

(

I

x

= 510 in

4

)

(

I

y

=15.3 in

4

)

Datum

Y

2

0.5”

Y

1

= 0.25”

8.85”

17.70”

8.85”

½” x 8” steel plate welded to center

of bottom flange of beam

y

Piece 1

Piece 2

From

textbook

Lecture 5 - Page 5 of 8

b) Determine Transformed Moment of Inertia about Strong Axis

:

piece

x

AdII

∑

+= )(

2

= 510.08 in

4

+ 238.58 in

4

I

x

= 748.66 in

4

c) Determine Moment of Inertia about Weak Axis

:

Since the neutral axis of both pieces line–up over each other, the total

moment of inertia is the sum of the moment of inertias of the pieces.

I

y

= I

1

+ I

2

4

3

3.15

12

)"8)("5.0(

inI

y

+=

I

y

= 36.63 in

4

4”

4”

8”

W18x35 steel beam (Area = 10.3 in

2

)

(

I

x

= 510 in

4

)

(

I

y

=15.3 in

4

)

Lecture 5 - Page 6 of 8

d) Determine the Section Modulus about Strong Axis

:

strong

strong

strong

y

I

S =

"80.6

66.748

4

in

=

S

strong

= 110.1 in

3

e) Determine Radius of Gyration about Strong Axis

:

tot

trong

s

strong

A

I

r =

2

4

3.14

66.748

in

in

r

strong

=

r

strong

= 7.24”

Lecture 5 - Page 7 of 8

Example 3

GIVEN

: Repeat Example 1, using a 2x4 top flange and a 2x8 web.

REQUIRED

: Using AutoCAD, determine the following:

a) The location of the neutral axis

y

b) The transformed moment of inertia about the strong axis

d) The moment of inertia about the weak axis

1½”

y

A

1½”

7¼”

3½”

Datum

Piece “B”

y

y

B

Piece “A” (centered

under piece “B”)

Lecture 5 - Page 8 of 8

Using AutoCAD, draw the shape shown and make a “REGION” out of it:

Results are exactly the same as in Example 1

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