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Locate the core or kernel of I-section with the properties : top and bottom flange: (150 mm x 10 mm) web (280 mm x 10 mm). Overall depth=300 mm.
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[1] ˉy=10×150×5+280×10×(10+140)+150×10×(280+10+5)10×150+280×10+150×10

ˉy=150 mm OR

¯y1=3002=150 mm (Symmetrical about X – axis)

Moment Inertia about XX – axis and YY – axis is Ixx & Iyy

[2] Ixx=150×300312(15010)×280312=81.39×106 mm4

Iyy=2[10×150312]+280×10312=5.64×106 mm4

[3] ¯y2=1502=75 mm (Symmetrical about Y – axis)

[4] Now, Let ex and ey be the maximum eccentricities about the XX – axis and YY – axis in order to avoid tensile stress. Let P be the longitudinal load on section.

Hence,

PA=Px exZxx & PA=PeyZyy

(A=10×150×2+280×10)

i.e. A=5800 mm2

i.e. ex=ZxxA & ey=ZyyA

But Zxx=Ixxy1 & Zyy=Iyyy2

Zxx=81.39×106150 & Zyy=5.64×10675

Zxx=542.6×103 mm3 & Zyy=75.2×103 mm3

e_x = \frac{542.6 \times 10^3}{5800} & e_y = \frac{75.2 \times 10^3} {5800}

\therefore e_x = 933.55 \ mm & e_y = 12.96 \ mm

The shaded region in I – Section shows core/kernel of section.

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