Area, Centroid, Moment of Inertia, Radius of Gyration
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COSC321Haque (PPT_C7) 1 Area, Centroid, Moment of Inertia, Radius of Gyration Dr. Mohammed E. Haque, P.E. Professor Department of Construction science
description
Area, Centroid, Moment of Inertia, Radius of Gyration. Dr. Mohammed E. Haque, P.E. Professor Department of Construction science. y. A = b h I x = b h 3 /12 I y = h b 3 /12. x. h. b. Area, Moment of Inertia. Centroid. Area, Moment of Inertia. y. A = 0.5 b h I x = b h 3 /36 - PowerPoint PPT Presentation
Transcript of Area, Centroid, Moment of Inertia, Radius of Gyration
COSC321Haque (PPT_C7) 1
Area, Centroid, Moment of Inertia, Radius of Gyration
Dr. Mohammed E. Haque, P.E.
Professor
Department of Construction science
COSC321Haque (PPT_C7) 2
Area, Moment of Inertia
A = b h
Ix = b h3 /12
Iy = h b3 /12
x
y
b
hCentroid
COSC321Haque (PPT_C7) 3
Area, Moment of Inertia
A = 0.5 b h
Ix = b h3 /36
Iy = h b3 /36
x
y
b
h
b/3
h/3Centroid
COSC321Haque (PPT_C7) 4
Area, Moment of Inertia
R
A = π R2
Ix = Iy = π R4 /64
X
Y
Centroid
COSC321Haque (PPT_C7) 5
Radius of Gyration
rx =(Ix /A)
ry =(Iy /A)
COSC321Haque (PPT_C7) 6
b
h Ixc = b h3 /12
A = b h
dy
Ix-x = Ixc + A dy2
= b h3 /12 + b h dy2
x x
Centroid
h/2
b/2
Moment of Inertia about an axis parallel to centroidal axis
COSC321Haque (PPT_C7) 7
Area and Centroid
10’-0”
14’-0”3’-0”3’-0”
20’-0”
7’-0”
4’-0”
4’-0”
3’-0”
4’-0”
X
Y
Q1: A pre-cast concrete wall panel as shown in fig. Determine
(a) Wall Area
(b) Centroid (x and y axes referenced from the lower left corner).
COSC321Haque (PPT_C7) 8
Section A (ft2) X (ft) xA (ft3) Y (ft) yA (ft3)
1 (20x10)=200
10 2000 5 1000
2 (Door) -(7x3) = -21
(3+1.5)= 4.5
-94.5 3.5 -73.5
3 (Window) -(4x4)= -16
3+3+4+2=12
-192 3+2=5
-80
Total 163 1713.5 846.5
X = 1713.5 /163 = 10.512 ft
Y = 846.5 /163 = 5.193 ft
A = 163 Sqft
COSC321Haque (PPT_C7) 9
5”
2”
3”3”
2”
x
y
Y
X
Q2: Determine(a) Area(b) Centroid (c ) Moment of Inertia about x and y axes
5”
2”
3”3”
2”
x
y
1
2
COSC321Haque (PPT_C7) 10
(a) Area; (b) Centroid
Section A (in2) x xA y yA
1 2x5=10 4 40 4.5 45
2 2x8=16 4 64 1 16
Total 26 104 61
(a) AREA, A = 26 Sqin.
(b)X = 104 /26 = 4 in
Y = 61 /26 = 2.346 in
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(c ) Moment of Inertia about the centroidal axes
Section A (in2) Ixc
(in4)
dy
(in)
Ady2
(in4)
Iyc
(in4)
dx
(in)
Adx2
(in4)
1 10 2(5)3/12=20.833
4.5-2.346=2.154
46.397 5(2)3/12=3.333
0 0
2 16 8(2)3/12=5.333
2.346 -1=1.346
28.987 2(8)3/12=85.333
0 0
Total 26 26.167 75.384 88.667 0
Ixcg = 26.167 + 75.384 = 101.55 in4
Iycg = 88.667 + 0 = 88.667 in4
COSC321Haque (PPT_C7) 12
X
1”
1”
4”
2”2”2”
Y Q3: Determine
(a) Area
(b) Moment of Inertia, Ixc, Iyc
(c) Radius of Gyration, rx, ry
COSC321Haque (PPT_C7) 13
Section A (in2) Ixc
(in4)
dy
(in)
Ady2
(in4)
Iyc
(in4)
dx
(in)
Adx2
(in4)
1 6x1=6 6(1)3/12=0.5
2.5 37.5 1(6)3/12=18
0 0
2 6x1=6 6(1)3/12=0.5
2.5 37.5 1(6)3/12=18
0 0
3 2x4=8 2(4)3/12=10.667
0 0 4(2)3/12=2.667
0 0
Total 20 11.667 75.0 38.667 0
A= 20 in2
Ix = 11.667 + 75.0 = 86.667 in4
Iy = 38.667 + 0 = 38.667 in4
rx = (86.667/20) = 2.08 inry = (38.667/20) = 1.39 in
X
1”
1”
4”
2”2”2”
Y
1
2
3
COSC321Haque (PPT_C7) 14
X
1”
1”
4”
2”2”2”
Y
1
2
Q4: Determine
(a) Area
(b) Moment of Inertia, Ixc, Iyc
(c) Radius of Gyration, rx, ry
COSC321Haque (PPT_C7) 15
A= 28 in2
Ix = 97.333 in4
Iy = 105.333 in4
rx = (97.333/28) = 1.86 inry = (105.333/28) = 1.94 in
X
1”
1”
4”
2”2”2”
Y
1
2
Section A (in2) Ixc (in4) Ixy (in4)
1 (Ignoring hole) 6x6 = 36 6(6)3 /12
=108
6(6)3 /12
=108
2 (Hollow) -(2x4) = -8 -2(4)3 /12
= -10.667
-4(2)3 /12
= - 2.667
Total 28 97.333 105.333
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Q5: Determine
(a)Area
(b)Centroid
(c) Moment of Inertia, Ixc, Iyc
(d)Radius of Gyration, rx, ryX1”
4”
4”2”
Y
X1”
4”
4”2”
Y
1
2
COSC321Haque (PPT_C7) 17
(a) Area; (b) Centroid
Section A (in2) x xA y yA
1 2x4=8 1 8 3 24
2 1x6=6 3 18 0.5 3
Total 14 26 27
(a) AREA, A = 14 Sqin.
(b)X = 26 /14 = 1.86 in
Y = 27 /14 = 1.93 in
COSC321Haque (PPT_C7) 18
(c ) Moment of Inertia; (d) Radius of gyration
Section A (in2) Ixc
(in4)
dy
(in)
Ady2
(in4)
Iyc
(in4)
dx
(in)
Adx2
(in4)
1 8 2(4)3/12=10.667
3 -1.93=1.07
9.159 4(2)3/12=2.667
1-1.86= -0.86
5.92
2 6 6(1)3/12=0.5
0.5 -1.93=-1.43
12.26 1(6)3/12=18.0
3-1.86= 1.14
7.80
Total 14 11.167 21.419 20.667 13.72
Ix = 11.167 + 21.419 = 32.586 in4
Iy = 20.667 + 13.72 = 34.39 in4
rx = (32.586/14) = 1.53 inry = (34.39/14) = 1.57 in
