Edupanda » Strength of materials »   Eccentric Axial Load   »   Core of the Section   » Example 1

Example 1

Determine cross-section core

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Solution

Geometrical characteristics

\begin{aligned} &I_y=\frac{12\cdot 30^3}{12}=27000 cm^4\\ &I_z=\frac{30\cdot 12^3}{12}=4320 cm^4\\ &A=12\cdot 30=360 cm^2\\ &i_y^2=\frac{I_y}{A}=\frac{27000}{360}=75 cm^2\\ &i_z^2=\frac{I_z}{A}=\frac{4320}{360}=12 cm^2\\ \end{aligned}

Core section corner points

Neutral axis AB

\begin{aligned} &a_y=6 cm\\ &a_z=\infty\\ &y=-\frac{i_z^2}{a_y}=-\frac{12}{6}=-2 cm\\ &z=-\frac{i_y^2}{a_z}=-\frac{75}{\infty}=0 cm\\ \end{aligned}

Neutral axis BC

\begin{aligned} &a_y=\infty\\ &a_z=-15 cm\\ &y=-\frac{i_z^2}{a_y}=-\frac{12}{\infty}=0 cm\\ &z=-\frac{i_y^2}{a_z}=-\frac{75}{-15}=5 cm\\ \end{aligned}

Neutral axis CD

\begin{aligned} &a_y=-6 cm\\ &a_z=\infty\\ &y=-\frac{i_z^2}{a_y}=-\frac{12}{-6}=2 cm\\ &z=-\frac{i_y^2}{a_z}=-\frac{75}{\infty}=0 cm\\ \end{aligned}

Neutral axis DA

\begin{aligned} &a_y=\infty\\ &a_z=15 cm\\ &y=-\frac{i_z^2}{a_y}=-\frac{12}{\infty}=0 cm\\ &z=-\frac{i_y^2}{a_z}=-\frac{75}{15}=-5 cm\\ \end{aligned}

Core section