Consider the equalflange steel section in Figure 3 given tha

Consider the equal-flange steel section in Figure 3, given that h = 1300mm, b = 320mm, t_w = 30mm. and t_f = 50mm. determine: The position of neutral axes for bending about the major and minor axes of the section. The second moment of area and the radius of gyration about both the major and minor axes of the section.

Solution

a.Position about major axis;

X*=(320/2)=160, since symmetrical about minor axis

Position about minor axis:

A1=320*50=16000;A2=30*1200=36000;A3=50*320=16000

Y*=(A1Y1+A2Y2+A3Y3)/(A1+A2+A3)

Y1=50/2=25mm; Y2=50+(1200/2)=650mm; Y3=50+1200+(50/2)=1275mm

so Y*=650mm

b. Second moment of inertia about minor axis=0

Second moment of inertia about major axis=(I₁+I₂+I₃)=(2I₁+I₂) since I₁=I₃

  Now section is rectangular section so I₁=(1/12)bt_f³=(1/12)(320*50³)=3.3*10⁶mm⁴

  I₂=I₁+A₁d².....from parallel axis theorem

so I₂=3.3*10⁶+(320*50)(25+600)=13.3*10⁶mm⁴