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) 1

Area, Centroid, Moment of Inertia, Radius of Gyration

Dr. Mohammed E. Haque, P.E.

Professor

Department of Construction science


Area, Moment of Inertia

A = b h

Ix = b h3 /12

Iy = h b3 /12

x

y

b

hCentroid



A = 0.5 b h

Ix = b h3 /36

Iy = h b3 /36

x

y

b

h

b/3

h/3Centroid



R

A = π R2

Ix = Iy = π R4 /64

X

Y

Centroid


Radius of Gyration

rx =(Ix /A)

ry =(Iy /A)


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


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).


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


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


(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


(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


X

1”

1”

4”

2”2”2”

Y Q3: Determine

(a) Area

(b) Moment of Inertia, Ixc, Iyc

(c) Radius of Gyration, rx, ry


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


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


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


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


(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


(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

COSC321Haque (PPT_C7)1 Area, Centroid, Moment of Inertia, Radius of Gyration Dr. Mohammed E. Haque, P.E. Professor Department of Construction science.

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