written 5.5 years ago by | • modified 5.4 years ago |
[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−(150–10)×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.