Area Centroid Moment of Inertia Radius of Gyration

Area, Centroid, Moment of Inertia, Radius of Gyration Dr. Mohammed E. Haque, P. E. Professor Department of Construction science COSC 321 Haque (PPT_C 7) 1

Area, Moment of Inertia y A=bh x Centroid h Ix = b h 3 /12 Iy = h b 3 /12 b COSC 321 Haque (PPT_C 7) 2

Area, Moment of Inertia y A = 0. 5 b h h x Centroid h/3 Ix = b h 3 /36 Iy = h b 3 /36 b/3 b COSC 321 Haque (PPT_C 7) 3

Area, Moment of Inertia Y Centroid X R A = π R 2 Ix = Iy = π R 4 /64 COSC 321 Haque (PPT_C 7) 4

Radius of Gyration rx = (Ix /A) ry = (Iy /A) COSC 321 Haque (PPT_C 7) 5

Moment of Inertia about an axis parallel to centroidal axis Centroid b/2 h A=bh Ixc = b h 3 /12 dy b x Ix-x = Ixc + A dy 2 = b h 3 /12 + b h dy 2 COSC 321 Haque (PPT_C 7) x 6

Area and Centroid 20’-0” Y 4’-0” 10’-0” 4’-0” 7’-0” 3’-0” X 3’-0” 14’-0” Q 1: 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). COSC 321 Haque (PPT_C 7) 7

Section A (ft 2) X (ft) x A (ft 3) Y (ft) y A (ft 3) 1 (20 x 10) =200 10 2000 5 1000 2 (Door) -(7 x 3) = -21 (3+1. 5) = 4. 5 -94. 5 3. 5 -73. 5 3 (Window) -(4 x 4) = -16 3+3+4+2 =12 -192 3+2 =5 -80 Total 163 1713. 5 846. 5 A = 163 Sqft X = 1713. 5 /163 = 10. 512 ft Y = 846. 5 /163 = 5. 193 ft COSC 321 Haque (PPT_C 7) 8

y Q 2: Determine (a) Area (b) Centroid (c ) Moment of Inertia about x and y axes 2” 3” 3” 5” X Y 2” x y 2” 3” 1 2 3” 5” 2” x COSC 321 Haque (PPT_C 7) 9

(a) Area; (b) Centroid Section A (in 2) x x A y y A 1 2 x 5=10 4 40 4. 5 45 2 2 x 8=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 COSC 321 Haque (PPT_C 7) 10

(c ) Moment of Inertia about the centroidal axes Section A (in 2) Ixc (in 4) dy (in) Ady 2 (in 4) Iyc (in 4) dx (in) Adx 2 (in 4) 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 in 4 Iycg = 88. 667 + 0 = 88. 667 in 4 COSC 321 Haque (PPT_C 7) 11

Q 3: Determine Y (a) Area 1” (b) Moment of Inertia, Ixc, Iyc X (c) Radius of Gyration, rx, ry 4” 1” 2” 2” 2” COSC 321 Haque (PPT_C 7) 12

Y 1 1” X 4” 3 2 2” 1” 2” A= 20 in 2 Ix = 11. 667 + 75. 0 = 86. 667 in 4 Iy = 38. 667 + 0 = 38. 667 in 4 rx = (86. 667/20) = 2. 08 in ry = (38. 667/20) = 1. 39 in 2” Section A (in 2) Ixc (in 4) dy (in) Ady 2 (in 4) Iyc (in 4) dx (in) Adx 2 (in 4) 1 6 x 1=6 6(1)3/12 =0. 5 2. 5 37. 5 1(6)3/12 =18 0 0 2 6 x 1=6 6(1)3/12 =0. 5 2. 5 37. 5 1(6)3/12 =18 0 0 3 2 x 4=8 2(4)3/12 =10. 667 0 0 4(2)3/12 =2. 667 0 0 Total 20 75. 0 38. 667 11. 667 COSC 321 Haque (PPT_C 7) 0 13

Y 1 Q 4: Determine 1” (a) Area X 4” 2 (b) Moment of Inertia, Ixc, Iyc (c) Radius of Gyration, rx, ry 1” 2” 2” 2” COSC 321 Haque (PPT_C 7) 14

Y 1 1” X 4” 2 1” 2” 2” Section A= 28 in 2 Ix = 97. 333 in 4 Iy = 105. 333 in 4 rx = (97. 333/28) = 1. 86 in ry = (105. 333/28) = 1. 94 in 2” A (in 2) Ixc (in 4) Ixy (in 4) 1 (Ignoring hole) 6 x 6 = 36 6(6)3 /12 =108 2 (Hollow) -(2 x 4) = -8 -2(4)3 /12 = -10. 667 -4(2)3 /12 = - 2. 667 Total 28 97. 333 105. 333 COSC 321 Haque (PPT_C 7) 15

Y Q 5: Determine (a) Area (b) Centroid 4” 1” 2” (c) Moment of Inertia, Ixc, Iyc X (d) Radius of Gyration, rx, ry Y 4” 1 4” 2 2” COSC 321 Haque (PPT_C 7) 1” X 4” 16

(a) Area; (b) Centroid Section A (in 2) x x A y y A 1 2 x 4=8 1 8 3 24 2 1 x 6=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 COSC 321 Haque (PPT_C 7) 17

(c ) Moment of Inertia; (d) Radius of gyration Section A (in 2) Ixc (in 4) dy (in) Ady 2 (in 4) Iyc (in 4) dx (in) Adx 2 (in 4) 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 -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 in 4 Iy = 20. 667 + 13. 72 = 34. 39 in 4 rx = (32. 586/14) = 1. 53 in ry = (34. 39/14) = 1. 57 in COSC 321 Haque (PPT_C 7) 18