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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
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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
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
COSC321Haque (PPT_C7) 11
(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
COSC321Haque (PPT_C7) 16
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
