Combined stresses in columns are critical in civil engineering design, especially in Kenya where columns often support multi-storey buildings, bridges, and other infrastructure under complex loading. Understanding how axial loads interact with bending moments helps engineers ensure structural safety and serviceability. This chapter focuses on computing combined stresses, addressing direct and bending stresses, and the effects of eccentric loading on columns.

4.1 Stress Types and Their Implications in Columns

In civil engineering, columns carry loads primarily through axial compression but frequently experience bending due to eccentric loads or lateral forces. Accurate calculation of combined stresses enables design that prevents failure modes such as buckling or crushing.

4.1.1 Direct Stress and Bending Stress in Columns

Direct stress arises from axial loads acting uniformly across the cross-section, while bending stress results from moments causing tension and compression on opposite sides of the section. The combined effect influences the maximum and minimum stresses in the column.

The total stress at any fiber in the column cross-section is the sum of direct and bending stresses:

$$ \sigma = \sigma_{\text{direct}} \pm \sigma_{\text{bending}} $$

where

$$ \sigma_{\text{direct}} = \frac{P}{A} $$

and

$$ \sigma_{\text{bending}} = \frac{M y}{I} $$

• \(P\) = Axial load (N)
• \(A\) = Cross-sectional area (m²)
• \(M\) = Bending moment (Nm)
• \(y\) = Distance from neutral axis to fiber (m)
• \(I\) = Moment of inertia of cross-section (m⁴)

Worked Examples

Example 1: A square concrete column 300 mm × 300 mm carries an axial load of 500 kN with no bending moment. Calculate the direct stress.

Given:
\(P = 500 \times 10^3\) N,
\(A = 0.3 \times 0.3 = 0.09\) m²

$$ \sigma_{\text{direct}} = \frac{P}{A} $$

$$ \sigma_{\text{direct}} = \frac{500 \times 10^3}{0.09} $$

$$ \sigma_{\text{direct}} = 5,555,555.56 \text{ Pa} = 5.56 \text{ MPa} $$

Answer: 5.56 MPa (compression)

Example 2: The same column experiences a bending moment of 20 kNm about the centroidal axis. Calculate the maximum bending stress.

Given:
\(M = 20 \times 10^3\) Nm,
\(y = \frac{0.3}{2} = 0.15\) m,
\(I = \frac{(0.3)(0.3)^3}{12} = \frac{0.3 \times 0.027}{12} = 6.75 \times 10^{-5}\) m⁴

$$ \sigma_{\text{bending}} = \frac{M y}{I} $$

$$ \sigma_{\text{bending}} = \frac{20 \times 10^3 \times 0.15}{6.75 \times 10^{-5}} $$

$$ \sigma_{\text{bending}} = \frac{3,000}{6.75 \times 10^{-5}} = 44,444,444.44 \text{ Pa} = 44.44 \text{ MPa} $$

Answer: 44.44 MPa (tension on one side, compression on the other)

Example 3: Calculate the maximum and minimum stresses in the column subjected to axial load and bending moment.

Given:
Direct stress = 5.56 MPa (compression),
Bending stress = 44.44 MPa (tension/compression)

$$ \sigma_{\max} = \sigma_{\text{direct}} + \sigma_{\text{bending}} = -5.56 + 44.44 = 38.88 \text{ MPa} \quad (\text{tension}) $$

$$ \sigma_{\min} = \sigma_{\text{direct}} - \sigma_{\text{bending}} = -5.56 - 44.44 = -50.00 \text{ MPa} \quad (\text{compression}) $$

Answer: Maximum tensile stress = 38.88 MPa, Minimum compressive stress = 50.00 MPa

4.1.2 Combined Axial and Bending Stress for Eccentrically Loaded Columns

Eccentric loading produces both axial and bending stresses because the load does not act through the centroidal axis. This is common in structural columns supporting beams or subjected to lateral loads.

The bending moment due to eccentricity \(e\) is:

$$ M = P e $$

The combined stress at any fiber is:

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

Worked Examples

Example 1: A circular column of diameter 400 mm carries an axial load of 600 kN with an eccentricity of 50 mm. Calculate the maximum and minimum stresses at the extreme fibers.

Given:
\(P = 600 \times 10^3\) N,
Diameter \(d = 0.4\) m,
\(A = \frac{\pi d^2}{4} = \frac{3.142 \times 0.4^2}{4} = 0.1257\) m²,
\(I = \frac{\pi d^4}{64} = \frac{3.142 \times 0.4^4}{64} = 3.14 \times 10^{-3}\) m⁴,
\(e = 0.05\) m,
\(y = \frac{d}{2} = 0.2\) m

Calculate direct stress:

$$ \sigma_{\text{direct}} = \frac{600 \times 10^3}{0.1257} = 4,774,648.29 \text{ Pa} = 4.77 \text{ MPa} $$

Calculate bending moment:

$$ M = P e = 600 \times 10^3 \times 0.05 = 30,000 \text{ Nm} $$

Calculate bending stress:

$$ \sigma_{\text{bending}} = \frac{M y}{I} = \frac{30,000 \times 0.2}{3.14 \times 10^{-3}} = 1,911,4642.03 \text{ Pa} = 19.11 \text{ MPa} $$

Calculate max and min stresses:

$$ \sigma_{\max} = \sigma_{\text{direct}} + \sigma_{\text{bending}} = 4.77 + 19.11 = 23.88 \text{ MPa} $$

$$ \sigma_{\min} = \sigma_{\text{direct}} - \sigma_{\text{bending}} = 4.77 - 19.11 = -14.34 \text{ MPa} $$

Answer: Maximum stress = 23.88 MPa (compression), Minimum stress = 14.34 MPa (tension)

Example 2: A rectangular column 400 mm by 500 mm carries a load of 800 kN with an eccentricity of 30 mm about the shorter side. Calculate the combined stress at the extreme fiber.

Given:
\(P = 800 \times 10^3\) N,
\(b = 0.4\) m,
\(h = 0.5\) m,
\(A = 0.4 \times 0.5 = 0.2\) m²,
\(I = \frac{b h^3}{12} = \frac{0.4 \times 0.5^3}{12} = \frac{0.4 \times 0.125}{12} = 0.004167\) m⁴,
\(e = 0.03\) m,
\(y = \frac{h}{2} = 0.25\) m

Calculate direct stress:

$$ \sigma_{\text{direct}} = \frac{800 \times 10^3}{0.2} = 4,000,000 \text{ Pa} = 4.0 \text{ MPa} $$

Calculate bending moment:

$$ M = P e = 800 \times 10^3 \times 0.03 = 24,000 \text{ Nm} $$

Calculate bending stress:

$$ \sigma_{\text{bending}} = \frac{M y}{I} = \frac{24,000 \times 0.25}{0.004167} = 1,440,000 \text{ Pa} = 14.4 \text{ MPa} $$

Calculate max and min stresses:

$$ \sigma_{\max} = 4.0 + 14.4 = 18.4 \text{ MPa} $$

$$ \sigma_{\min} = 4.0 - 14.4 = -10.4 \text{ MPa} $$

Answer: Maximum compressive stress = 18.4 MPa, Minimum tensile stress = 10.4 MPa

Example 3: A square steel column 250 mm by 250 mm carries 400 kN load with eccentricity 20 mm about both axes equally. Calculate max and min combined stresses considering bending in both directions.

Given:
\(P = 400 \times 10^3\) N,
\(a = 0.25\) m,
\(A = 0.25 \times 0.25 = 0.0625\) m²,
\(I_x = I_y = \frac{a^4}{12} = \frac{0.25^4}{12} = 3.255 \times 10^{-4}\) m⁴,
\(e_x = e_y = 0.02\) m,
\(y = \frac{a}{2} = 0.125\) m

Calculate direct stress:

$$ \sigma_{\text{direct}} = \frac{400 \times 10^3}{0.0625} = 6,400,000 \text{ Pa} = 6.4 \text{ MPa} $$

Calculate bending moments:

$$ M_x = P e_x = 400 \times 10^3 \times 0.02 = 8,000 \text{ Nm} $$

$$ M_y = P e_y = 8,000 \text{ Nm} $$

Calculate bending stresses:

$$ \sigma_{b,x} = \frac{M_x y}{I_x} = \frac{8,000 \times 0.125}{3.255 \times 10^{-4}} = 3,070,000 \text{ Pa} = 3.07 \text{ MPa} $$

$$ \sigma_{b,y} = \frac{M_y y}{I_y} = 3.07 \text{ MPa} $$

Calculate total bending stress:

$$ \sigma_{\text{bending}} = \sigma_{b,x} + \sigma_{b,y} = 3.07 + 3.07 = 6.14 \text{ MPa} $$

Calculate max and min stresses:

$$ \sigma_{\max} = 6.4 + 6.14 = 12.54 \text{ MPa} $$

$$ \sigma_{\min} = 6.4 - 6.14 = 0.26 \text{ MPa} $$

Answer: Maximum compressive stress = 12.54 MPa, Minimum compressive stress = 0.26 MPa

Chapter Summary

This chapter focused on understanding the different types of stresses that act on columns, primarily direct stress and bending stress, and their significance in structural analysis. It examined how these stresses combine in columns subjected to eccentric loading, resulting in both axial and bending stresses that must be evaluated together for safe design. The chapter also discussed critical assumptions and considerations necessary when analyzing combined stresses, emphasizing the importance of accurately accounting for load eccentricity. Load eccentricity influences the magnitude and distribution of stresses within a column, affecting its capacity and stability. By integrating these concepts, the chapter provided a comprehensive approach to calculating combined stresses in columns, ensuring that structural elements are assessed realistically under practical loading conditions. This foundation is essential for designing columns that can safely resist complex stress states encountered in the field.