COURSE:
CE 201 (STATICS)
LECTURE NO.:
44 & 45
FACULTY:
DR. SHAMSHAD AHMAD
DEPARTMENT:
CIVIL ENGINEERING
UNIVERSITY:
KING FAHD UNIVERSITY OF PETROLEUM
& MINERALS, DHAHRAN, SAUDI ARABIA
TEXT BOOK:
ENGINEERING MECHANICS-STATICS
by R.C. HIBBELER, PRENTICE HALL
LECTURE NO. 44 & 45
MOMENT OF INERTIA
Objectives:
►
To develop a method for determining the moment
of inertia for an area
►
To explain the Parallel-Axis Theorem
►
To show how to determine the moment of inertia
for composite areas
MOMENT OF INERTIA FOR AREAS
Moment of inertia for an area is determined by taking integral of
the second moment of area about an axis, as follows.
I x = ∫ y 2 dA
A
I y = ∫ x 2 dA
A
J o = ∫ r 2 dA = I x + I y
A
Where:
Ix = moment of inertia about x-axis
Iy = moment of inertia about y-axis
Jo = polar moment of inertia
PARALLEL AXIS THEOREM
• The parallel-axis theorem is useful for determination of
the moment of inertia for composite bodies.
• For an area, as shown in the figure below, the parallelaxis theorem may be stated as:
I x = I x ' + Ad
2
y
I y = I y ' + Ad
2
x
J o = J C + Ad
2
PROBLEM SOLVING: Example # 1
Determine the moment of inertia of the triangular area
about the y-axis.
PROBLEM SOLVING: Example # 1
Let us consider a differential element
of thickness dx at a distance x from
the y-axis, as shown in the following
figure:
b
h
dA=ydx= (b-x)dx
b
h
I y = ∫ x dA = ∫ x ⋅ ( b − x ) dx
b
o
2
2
b
h
h ⎡ bx x ⎤
2
3
= ∫ ( bx − x ) dx = ⎢
− ⎥
bo
b⎣ 3
4 ⎦o
b
3
h ⎡ b 4 b 4 ⎤ hb 4
= ⎢ − ⎥=
b ⎣ 3 4 ⎦ 12b
hb3
Ans.
∴Iy =
12
4
PROBLEM SOLVING: Example # 2
Determine the moment of inertia of
the shaded area about the y-axis.
PROBLEM SOLVING: Example # 2
y
dA = y ' dx = (h − y )dx
b
h 2⎞
⎛
⇒ dA = ⎜ h − 2 x ⎟ dx
b
⎝
⎠
h
⇒ dA = 2 ( b 2 − x 2 ) dx
b
h
y'=h-y
y
x
x
dx
b
h
h
I y = ∫ x 2 dA = ∫ x 2 ⋅ 2 ( b 2 − x 2 ) dx = 2
b
b
o
b
h ⎡ b 2 x3 x5 ⎤
h ⎡ b5 b5 ⎤
= 2⎢
− ⎥ = 2⎢ − ⎥
5 ⎦o b ⎣ 3 5 ⎦
b ⎣ 3
2b3 h
⇒ Iy =
15
Ans.
b
∫ (b x
2
o
2
− x 4 ) dx
PROBLEM SOLVING: Example # 3
Locate the centroid x of the above
composite section. Then find the
moment of inertia of the composite
area about y-axis passing through the
centroid (i.e., about y′-axis).
PROBLEM SOLVING: Example # 3
y
y'
x
2in
4in
A
Centroid
1in
x'
1in
B
x
4in
The x coordinate of the centroid (i.e., x ) can be determined as:
Rectangle
A(in2)
x (in)
A
B
4×2=8
8 × 2 = 16
ΣA = 24
1
4
xA
8
64
Σ x A = 72
∴x =
ΣxA
72
=
= 3 in
ΣA 24
4 ( 2)
2
I y' =
+ 4 × 2 × ( 2 ) = 34.66 in 4
12
2 × 83
I y' =
+ 8 × 2 × 12 = 101.33 in 4
12
3
For rectangle A:
For rectangle B:
Summation:
I y ' = 34.66 + 101.33 = 136 in 4 Ans.
PROBLEM SOLVING: Example # 4
Determine the moment of inertia of the
following composite section about the x′axis.
PROBLEM SOLVING: Example # 4
y
x = 68mm
y'
2
80mm
80mm
1
C
x'
20mm
48mm
12mm
60mm
y = 80mm
3
x
80mm
80mm
The x coordinate of the centroid (i.e., x ) can be determined as:
Segment
A (mm2)
x ( mm )
( mm 3 )
xA
1
2
3
40 × 80 = 3200
160 × 40 = 6400
160 × 40 = 6400
20
80
80
64000
512000
512000
ΣA = 16000 mm2 ;
= 1088000 mm3 x =
ΣxA
ΣxA
1088000
=
= 68 mm
ΣA
16000
PROBLEM SOLVING: Example # 4
y
x = 68mm
Rectangle 1:
y'
40 × 803
I x' =
= 1706666.66 mm 4
12
80 × 403
I y' =
+ 80 × 40 × 482 = 7799466.66 mm 4
12
2
80mm
1
C
x'
Rectangle 2:
48mm
20mm
80mm
12mm
60mm
y = 80mm
3
x
80mm
80mm
Rectangle 3:
160 × 403
Ix' =
+ 160 × 40 × 602 = 23893333.33 mm 4
12
40 ×1603
I y' =
+ 160 × 40 × 122 = 14574933.33 mm 4 .
12
160 × 403
I x' =
+ 160 × 40 × 602 = 23893333.33 mm 4
12
40 × 1603
I y' =
+ 160 × 40 × 122 = 14574933.33 mm 4 .
12
Summation
I x ' = 1706666.66 + 2 × 23893333.33 = 49.49 × 106 mm 4 A
I y ' = 7799466.66 + 2 × 14574933.33 = 36.94 × 106 mm 4 A
PROBLEM SOLVING: Example # 5
Determine the moment of inertia of the
following shaded area about the x and y
axes.
PROBLEM SOLVING: Example # 4
–
Segment
1
2
3
4
A
(in2)
18
9
24
–
12.566
Ix′
(in4)
54
18
32
–
12.566
Iy′
(in4)
13.5
4.5
72
–
12.566
dx
dy
(in) (in)
7
1.5
6
4.0
2
3.0
4
I x ( in 4 ) = I x ' + Ad y2
I y ( in4 ) = I y ' + Ad x2
936
342
128
54.0
148.5
288.0
3.0
–213.628
–126.663
Ans.
Ix= 1192.37 in4
Iy= 364.83 in4
PROBLEM SOLVING: Example # 6
Determine the moment of inertia of the
following composite section about the x′
axis. y =154.4 mm
PROBLEM SOLVING: Example # 6
150mm
15mm
2
x’
3
50mm
150 × 153
Segment 1: I x ' =
+ 150 × 15 × 146.92 = 48.596 × 106 mm 4
12
15 ×1503
Segment 2: I x ' =
+ 15 ×150 × 64.42 = 13.550 ×106 mm 4
12
Segment 3: I x ' =
Summation:
π
4
4
( 50 ) +π (50)2 × 60.62
= 33.751× 106 mm 4
Ix′ = 95.898 × 106 mm4
Ans.
100mm
dy3=60.6mm
dy2=64.4mm
y’=154.4mm
150mm
dy1=146.9mm
15mm
1
Multiple Choice Problems
1. The moment of inertia of the composite section (shown in the following
figure) about the x-axis is
(a) 288 in4
(b) 928 in4
(c) 2016 in4
(d) 5760 in4
Ans: (d)
Feedback:
⎡ 8 × 123
⎡ 6 × 123
⎤
4
2⎤
Ix = ⎢
+ 8 × 12 × 6 ⎥ + ⎢
+ 6 × 12 × 22 ⎥ = 5760 in
⎣ 12
⎦ ⎣ 12
⎦
Multiple Choice Problems
2. The moment of inertia of the composite section (shown in the following
figure) about the x′-axis is
y = 6.5 in
(a) 290.66 in4
(b) 190.66 in4
(c) 90.66 in4
Ans: (a)
Feedback:
⎡ 2 × 83
⎡ 8 × 23
⎤
4
2⎤
I x' = ⎢
+ 2 × 8 × 2.5 ⎥ + ⎢
+ 8 × 2 × 2.52 ⎥ = 290.66 in
⎣ 12
⎦ ⎣ 12
⎦
(d) 85.33 in4