2- Handouts Lecture-44 45
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Transcript of 2- Handouts Lecture-44 45
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COURSE:COURSE: CE 201 (STATICS)CE 201 (STATICS)
LECTURE NO.:LECTURE NO.: 44 & 4544 & 45
FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD
DEPARTMENT:DEPARTMENT: CIVIL ENGINEERINGCIVIL ENGINEERING
UNIVERSITY:UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA& MINERALS, DHAHRAN, SAUDI ARABIA
TEXT BOOK:TEXT BOOK: ENGINEERING MECHANICSENGINEERING MECHANICS--STATICS STATICS by R.C. HIBBELER, PRENTICE HALLby R.C. HIBBELER, PRENTICE HALL
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LECTURE NO. 44 & 45LECTURE NO. 44 & 45MOMENT OF INERTIAMOMENT OF INERTIA
Objectives:Objectives: To develop a method for determining the moment To develop a method for determining the moment
of inertia for an areaof inertia for an area
To explain the ParallelTo explain the Parallel--Axis TheoremAxis Theorem
To show how to determine the moment of inertia To show how to determine the moment of inertia for composite areasfor composite areas
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MOMENT OF INERTIA FOR AREAS MOMENT OF INERTIA FOR AREAS
Moment of inertia for an area is determined by taking integral ofthe second moment of area about an axis, as follows.
2
2
2
x A
y A
o x yA
I y dA
I x dA
J r dA I I
=
=
= = +
Where:
Ix = moment of inertia about x-axis Iy = moment of inertia about y-axis Jo = polar moment of inertia
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PARALLEL AXIS THEOREMPARALLEL AXIS THEOREM
The parallel-axis theorem is useful for determination ofthe moment of inertia for composite bodies.
For an area, as shown in the figure below, the parallel-axis theorem may be stated as:
2'
2'
2
x x y
y y x
o C
I I Ad
I I Ad
J J Ad
= +
= +
= +
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1Determine the moment of inertia of the triangular area about the y-axis.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
Let us consider a differential element of thickness dx at a distance x from the y-axis, as shown in the following figure:
dA=ydx= (b-x)dxhb
( )
( )
2 2
3 42 3
4 4 4
3
3 4
3 4 12
Ans.12
b
yo
bb
o o
y
hI x dA x b x dxb
h h bx xbx x dxb b
h b b hbb b
hbI
= =
= =
= =
=
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
Determine the moment of inertia ofthe shaded area about the y-axis.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2y
x
dxx
h
b
y
y'=h-y
y
x
dxx
h
b
y
y'=h-y
( )
'
22
2 22
( )dA y dx h y dxhdA h x dxb
hdA b x dxb
= =
=
=
( ) ( )2 2 2 2 2 2 42 22 3 5 5 5
2 2
3
3 5 3 5
2 Ans.15
b b
yo o
b
o
y
h hI x dA x b x dx b x x dxb b
h b x x h b bb b
b hI
= = =
= =
=
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3Locate the centroid x of the abovecomposite section. Then find the moment of inertia of the compositearea about y-axis passing through thecentroid (i.e., about y-axis).
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
1in
4in
4in
2in
1in
B
A
x
yy'
x'
x
Centroid1in
4in
4in
2in
1in
B
A
x
yy'
x'
x
Centroid
The x coordinate of the centroid (i.e., x ) can be determined as:Rectangle A(in2) (in)x xA A 4 2 = 8 1 8 B 8 2 = 16 4 64 A = 24 x A = 72
72 3 in24
xAxA
= = =
For rectangle A: ( ) ( )'
32 44 2 4 2 2 34.66 in
12yI = + =
For rectangle B: 3
2 4'
2 8 8 2 1 101.33 in12y
I = + =
Summation: 4' 34.66 101.33 136 inyI = + = Ans.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
Determine the moment of inertia of thefollowing composite section about the x-axis.
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4 y
x
y'
x'C
x = 68mm
1
2
3
80mm
80mm
y = 80mm
80mm80mm
60mm20mm
48mm
12mm
y
x
y'
x'C
x = 68mm
1
2
3
80mm
80mm
y = 80mm
80mm80mm
60mm20mm
48mm
12mm
The x coordinate of the centroid (i.e., x ) can be determined as:
( )mmx ( )3mmxASegment A (mm2)1 40 80 = 3200 20 640002 160 40 = 6400 80 5120003 160 40 = 6400 80 512000
2 3 = 16000 mm ; 1088000 mmA xA = 1088000 68 mm16000
xAxA
= = =
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4 y
x
y'
x'C
x = 68mm
1
2
3
80mm
80mm
y = 80mm
80mm80mm
60mm20mm
48mm
12mm
y
x
y'
x'C
x = 68mm
1
2
3
80mm
80mm
y = 80mm
80mm80mm
60mm20mm
48mm
12mm
Summation 6 4
'6 4
'
1706666.66 2 23893333.33 49.49 10 mm A
7799466.66 2 14574933.33 36.94 10 mm Ax
y
I
I
= + =
= + =
Rectangle 1: 3
4'
32 4
'
40 80 1706666.66 mm12
80 40 80 40 48 7799466.66 mm12
x
y
I
I
= =
= + =
Rectangle 2: 3
2 4'
32 4
'
160 40 160 40 60 23893333.33 mm12
40 160 160 40 12 14574933.33 mm .12
x
y
I
I
= + =
= + =
Rectangle 3: 3
2 4'
32 4
'
160 40 160 40 60 23893333.33 mm12
40 160 160 40 12 14574933.33 mm .12
x
y
I
I
= + =
= + =
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5Determine the moment of inertia of thefollowing shaded area about the x and yaxes.
-
PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
( )4 2'x x yI in Ad= +I ( )4 2'y y xI in Ad= +ISegment A (in2) Ix (in4) Iy (in4)dy
(in)dx
(in)1 18 54 13.5 7 1.5 936 54.02 9 18 4.5 6 4.0 342 148.53 24 32 72 2 3.0 128 288.0
4 12.566
12.566
12.566 4 3.0 213.628 126.663
Ans. Ix= 1192.37 in4 Iy= 364.83 in4
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6Determine the moment of inertia of thefollowing composite section about the x
axis. y =154.4 mm
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PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
1
2
3
50mm
d y3=
60.6
mm
100m
m
d y2=
64.4
mm
y=154.4mm
15m
m
150mm
d y1=
146.
9mm
150m
m
15mm
x
1
2
3
50mm
d y3=
60.6
mm
100m
m
d y2=
64.4
mm
y=154.4mm
15m
m
150mm
d y1=
146.
9mm
150m
m
15mm
x
32 6 4
'150 15Segment 1: 150 15 146.9 48.596 10 mm
12xI = + =
32 6 4
'15 150Segment 2: + 15 150 64.4 = 13.550 10 mm
12xI =
( )4 2 2 6 4'Segment 3: 50 + (50) 60.6 = 33.751 10 mm4xI =
Summation: Ix = 95.898 106 mm4 Ans.
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Multiple Choice Problems1. 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:
3 342 28 12 6 12 5760 in8 12 6 6 12 2
12 12xI = + =+ +
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Multiple Choice Problems
2. The moment of inertia of the composite section (shown in the followingfigure) about the x-axis is
(a) 290.66 in4 (b) 190.66 in4 (c) 90.66 in4 (d) 85.33 in4 Ans: (a) Feedback:
3 3
42 2'
2 8 8 2 290.66 in2 8 2.5 8 2 2.512 12x
I = + =+ +
6.5 iny =