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__The problem:__

Determine the second moments of area, I_{xx} and I_{yy}, for the sections indicated in Figures 2a to 2e.

__Tip__: It is given that the total height and the area of the 457 x 152 x 52 UB section are: h = 449.8mm and A = 66.6cm^{2}, respectively. Furthermore, the second moments of area are I_{xx} = 21400cm^{4} and I_{yy} = 645cm^{4}.

__The solution:__

a) i) y_{1 }and y_{2}

y_{1 }= 360 + 7.5 – 273.75 = 93.75mm_{}

y_{2 }= 273.75 – 180 = 93.75mm

ii) I_{xx}

I_{xx} = (240 x 15^{3})/12 + 240 x 15 x 93.75^{2} + (10 x 360^{3})/12 + 360 x 10 x 93.75^{2}_{ }= 102.23 x 10^{6} mm^{4}

iii) I_{yy}

I_{yy} = (15 x 240^{3})/12 + (360 x 10^{3})/12 = 17.31 x 10^{6} mm^{4}

b) i) y_{1}, y_{2 }and y_{3}

_{}

y_{1 }= 15 + 400 + 7.5 – 234.66 = 187.84mm_{}

y_{2 }= 234.66 – (15+200) = 19.66mm

y_{3 }= 234.66 – 7.5 = 227.16mm

ii) I_{xx}

I_{xx} = (240 x 15^{3})/12 + 240 x 15 x 187.84^{2} + (8 x 400^{3})/12 + 400 x 8 x 19.66^{2}_{ }+ (180 x 15^{3})/12 + 180 x 15 x 227.16^{2}_{ }= 310.37 x 10^{6} mm^{4}

iii) I_{yy}

I_{yy} = (15 x 240^{3})/12 + (400 x 8^{3})/12 + (15 x 180^{3})/12 = 24.59 x 10^{6} mm^{4}

c) i) y_{1}, y_{2}, y_{3} and y_{4}

_{}

y_{1 }= 15 + 300 + 12 + 5 - 216.63 = 112.37mm_{}

y_{2 }= 12 + 300 + 6 - 216.63 = 101.37mm

y_{3 }= 216.63 – (12+150) = 54.63mm

y_{4 }= 216.63 – 6 = 210.63mm

ii) I_{xx}

I_{xx} = (350 x 10^{3})/12 + 350 x 10 x 112.37^{2} + (200 x 12^{3})/12 + 200 x 12 x 101.37^{2}_{ }+ (8 x 300^{3})/12 + 8 x 300 x 54.63^{2}_{ }+ (200 x 12^{3})/12 + 200 x 12 x 210.63^{2}= 200.58 x 10^{6} mm^{4}

iii) I_{yy}

I_{yy} = (10 x 350^{3})/12 + 2 x (12 x 200^{3})/12 + (300 x 8^{3})/12 = 51.74 x 10^{6} mm^{4}

^{}

d) i) x_{1}, x_{2}, y_{1}, y_{2} and y_{3}

_{}

x_{1 }= 110 – 7.5 = 102.50mm_{}

x_{2 }= 110 – 7.5 = 102.50mm_{}

y_{1 }= 449.8 + 4 -303.51 = 150.29mm_{}

y_{2 }= 449.8 - 41 -303.51 = 105.29mm_{}

y_{3 }= 303.51 – (449.8/2) = 78.61mm

ii) I_{xx}

I_{xx} = (220 x 8^{3})/12 + 220 x 8 x 150.29^{2} + 2 x (15 x 82^{3})/12 + 15 x 82 x 105.29^{2}_{ }+ (21400 x10^{4}) + 6660 x 78.61^{2}= 323.57 x 10^{6} mm^{4}

iii) I_{yy}

I_{yy} = (8 x 220^{3})/12 + 2 x [(82 x 15^{3})/12 + (82 x 15 x 102.5^{2})] + 645 x 10^{4} = 39.44 x 10^{6} mm^{4}

e) i) x_{1}, x_{2}, y_{1}, y_{2} and y_{3}

_{}

x_{1 }= 1200/2 + 5 = 605mm_{}

x_{2 }= 1200/2 + 5 = 605mm_{}

y_{1 }= 12 + 500 + 10 – 330.56 = 191.44mm_{}

y_{2 }= 330.56 – (12 +250) = 68.56mm

y_{3 }= 330.56 – 6 = 324.56mm

ii) I_{xx}

I_{xx} = (1420 x20^{3})/12 + 1420 x 20 x 191.44^{2} + 2 x (10 x 500^{3})/12 + 10 x 500 x 68.56^{2}_{ }+ (1220 x12^{3})/12 + 1220 x 12 x 324.56^{2}= 2839.47 x 10^{6} mm^{4}

iii) I_{yy}

I_{yy} = (20 x 1420^{3})/12 + 2 x [(500 x 10^{3})/12 + (10 x 500 x 605^{2})] + (12 x 1220^{3})/12 = 10248.30 x 10^{6} mm^{4}

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