Calculation of the second moments of area

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 = 66.6cm², respectively. Furthermore, the second moments of area are I_xx = 21400cm⁴ and I_yy = 645cm⁴.

The solution:

a)       i) y₁ and y₂

y₁ = 360 + 7.5 – 273.75 = 93.75mm

y₂ = 273.75 – 180 = 93.75mm

ii) I_xx

I_xx = (240 x 15³)/12 + 240 x 15 x 93.75² + (10 x 360³)/12 + 360 x 10 x 93.75² = 102.23 x 10⁶ mm⁴

iii) I_yy

I_yy = (15 x 240³)/12 + (360 x 10³)/12 = 17.31 x 10⁶ mm⁴

b)           i) y₁, y₂ and y₃

y₁ = 15 + 400 + 7.5 – 234.66 = 187.84mm

y₂ = 234.66 – (15+200) = 19.66mm

y₃ = 234.66 – 7.5 = 227.16mm

ii) I_xx

I_xx = (240 x 15³)/12 + 240 x 15 x 187.84² + (8 x 400³)/12 + 400 x 8 x 19.66² + (180 x 15³)/12 + 180 x 15 x 227.16² = 310.37 x 10⁶ mm⁴

iii) I_yy

I_yy = (15 x 240³)/12 + (400 x 8³)/12 + (15 x 180³)/12 = 24.59 x 10⁶ mm⁴

c)           i) y₁, y₂, y₃ and y₄

y₁ = 15 + 300 + 12 + 5 - 216.63 = 112.37mm

y₂ = 12 + 300 + 6 - 216.63 = 101.37mm

y₃ = 216.63 – (12+150) = 54.63mm

y₄ = 216.63 – 6 = 210.63mm

ii) I_xx

I_xx = (350 x 10³)/12 + 350 x 10 x 112.37² + (200 x 12³)/12 + 200 x 12 x 101.37² + (8 x 300³)/12 + 8 x 300 x 54.63² + (200 x 12³)/12 + 200 x 12 x 210.63²= 200.58 x 10⁶ mm⁴

iii) I_yy

I_yy = (10 x 350³)/12 + 2 x (12 x 200³)/12 + (300 x 8³)/12 = 51.74 x 10⁶ mm⁴

d)           i) x₁, x₂, y₁, y₂ and y₃

x₁ = 110 – 7.5 = 102.50mm

x₂ = 110 – 7.5 = 102.50mm

y₁ = 449.8 + 4 -303.51 = 150.29mm

y₂ = 449.8 - 41 -303.51 = 105.29mm

y₃ = 303.51 – (449.8/2) = 78.61mm

ii) I_xx

I_xx = (220 x 8³)/12 + 220 x 8 x 150.29² + 2 x (15 x 82³)/12 + 15 x 82 x 105.29² + (21400 x10⁴) + 6660 x 78.61²= 323.57 x 10⁶ mm⁴

iii) I_yy

I_yy = (8 x 220³)/12 + 2 x [(82 x 15³)/12 + (82 x 15 x 102.5²)] + 645 x 10⁴ = 39.44 x 10⁶ mm⁴

e)              i) x₁, x₂, y₁, y₂ and y₃

x₁ = 1200/2 + 5 = 605mm

x₂ = 1200/2 + 5 = 605mm

y₁ = 12 + 500 + 10 – 330.56 = 191.44mm

y₂ = 330.56 – (12 +250) = 68.56mm

y₃ = 330.56 – 6 = 324.56mm

ii) I_xx

I_xx = (1420 x20³)/12 + 1420 x 20 x 191.44² + 2 x (10 x 500³)/12 + 10 x 500 x 68.56² + (1220 x12³)/12 + 1220 x 12 x 324.56²= 2839.47 x 10⁶ mm⁴

iii) I_yy

I_yy = (20 x 1420³)/12 + 2 x [(500 x 10³)/12 + (10 x 500 x 605²)] + (12 x 1220³)/12 = 10248.30 x 10⁶ mm⁴