Eccentrically loaded columns| Kern of the section

When you load the column about its axis means the action line of load and the centroid of the column are coinciding then it is called the concentrically loaded column. But when these axes are not in one line this type of loading is called eccentrically loaded column.

What are the stresses in eccentrically loaded column?

The eccentric force creates moment and axial compression in the column. Stress generated in compression is direct compression and the stress due to moment is bending stress.

When column is eccentrically loaded following action will happen:

Direct Compression

$$\sigma = \frac{P}{A}$$

Bending Stress

$$\frac{M}{I} = \frac{\sigma}{y}$$

Combined Stress

$$\sigma=\frac{P}{A} \pm \frac{M}{I}y$$

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What do we mean by kern of a section?

Masonry structures cannot take tension. Hence we ensure that minimum stress is zero.

$$\frac{P}{A} \pm \frac{M}{I}y \ge 0$$

The region of load application which satisfies above condition is kern of section.

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Kern for Rectangular Section

$$\sigma = \frac{P}{A} \pm \frac{Pe_x}{Z_y} \pm \frac{Pe_y}{Z_x}$$

Corner stresses:

$$\begin{aligned} \sigma_1 &= \frac{P}{A} - \frac{Pe_x}{Z_y} - \frac{Pe_y}{Z_x} \\ \sigma_2 &= \frac{P}{A} - \frac{Pe_x}{Z_y} + \frac{Pe_y}{Z_x} \\ \sigma_3 &= \frac{P}{A} + \frac{Pe_x}{Z_y} + \frac{Pe_y}{Z_x} \\ \sigma_4 &= \frac{P}{A} + \frac{Pe_x}{Z_y} - \frac{Pe_y}{Z_x} \end{aligned}$$

Critical stress:

$$\sigma_1 = \frac{P}{A} - \frac{Pe_x}{Z_y} - \frac{Pe_y}{Z_x}$$

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Section Modulus

$$Z_x = \frac{I_{xx}}{y} = \frac{\frac{bd^3}{12}}{d/2} = \frac{bd^2}{6}$$ $$Z_y = \frac{I_{yy}}{x} = \frac{\frac{db^3}{12}}{b/2} = \frac{db^2}{6}$$

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Substitute

$$\sigma_1= \frac{P}{bd}-\frac{6Pe_x}{db^2}-\frac{6Pe_y}{bd^2} \ge0$$

Solving

$$\frac{P}{bd}-\frac{6Pe_x}{db^2}-\frac{6Pe_y}{bd^2}=0$$ $$\frac{6Pe_x}{b}+\frac{6Pe_y}{d}=1$$

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Middle Third Rule

$$x/a + y/b=1$$

Comparing

$$e_x \le b/6$$ $$e_y \le d/6$$ $$e_x= \frac{1}{3}\left(b/2\right)$$ $$e_y= \frac{1}{3}\left(d/2\right)$$

Hence kern is middle third area.

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Middle Fourth Rule (Circular Section)

$$Z=\frac{I}{y} =\frac{\frac{\pi d^4}{64}}{d/2} = \frac{\pi d^3}{32}$$ $$\frac{P}{\frac{\pi d^2}{4}} - \frac{Pe}{\frac{\pi d^3}{32}} =0$$

Solving

$$e=d/8$$ $$e=\frac{1}{4}\left(d/2\right)$$

Hence middle fourth rule.

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Conclusion

Kern of section ensures no tension stress.

Key Points

• Kern of section
• Middle third rule
• Middle fourth rule