Lecture 5 – Moment of Inertia of Non-symmetric Shapes

sentencecopyElectronics - Devices

Oct 13, 2013 (3 years and 7 months ago)

102 views

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