Structural bending theory is fundamental to understanding how beams behave under loads in building construction. In Kenya’s building industry, accurate computation of bending stresses ensures safety and durability of structural elements such as floor beams, lintels, and roof supports. This chapter introduces the theory of simple bending, which forms the basis for analyzing bending stresses and strains in beams subjected to transverse loads.

6.1 Definition and basic principles of bending in beams

Bending occurs when an external load causes a beam to curve, inducing internal stresses that resist deformation. The theory of simple bending assumes the beam is initially straight, has a constant cross-section, and is subjected to pure bending without shear forces. Understanding bending behavior is crucial for designing beams that can safely carry imposed loads in buildings such as schools, hospitals, and office blocks.

6.1.1 Nature of bending in beams

Bending in beams results from moments created by transverse loads acting at some distance from the beam’s neutral axis. The beam’s fibers above the neutral axis are compressed while those below are stretched in tension, producing a stress distribution across the section.

The fundamental bending equation is:

$$\sigma = \frac{My}{I}$$

where

• \(\sigma\) = bending stress at distance \(y\) from the neutral axis (Pa)
• \(M\) = bending moment at the section (Nm)
• \(y\) = distance from the neutral axis to the fiber (m)
• \(I\) = moment of inertia of the section about the neutral axis (m\(^4\))

Worked Examples

Example 1: A simply supported timber beam of rectangular cross-section 150 mm wide and 300 mm deep carries a bending moment of 5 kNm at mid-span. Calculate the maximum bending stress.

Given:
Width \(b = 0.15\) m, Depth \(d = 0.30\) m, Moment \(M = 5,000\) Nm, Maximum \(y = d/2 = 0.15\) m

Moment of inertia for rectangle:$$I = \frac{bd^3}{12} = \frac{0.15 \times (0.30)^3}{12}$$$$I = \frac{0.15 \times 0.027}{12} = 0.0003375 \text{ m}^4$$

Bending stress:$$\sigma = \frac{My}{I} = \frac{5,000 \times 0.15}{0.0003375}$$$$\sigma = \frac{750}{0.0003375} = 2,222,222 \text{ Pa} = 2.22 \text{ MPa}$$

Answer: 2.22 MPa (tensile or compressive depending on fiber position)

Example 2: A concrete lintel with a rectangular cross-section 200 mm wide and 400 mm deep experiences a bending moment of 12 kNm. Find the maximum bending stress.

Given:
\(b=0.2\) m, \(d=0.4\) m, \(M=12,000\) Nm, \(y = 0.2\) m

Calculate moment of inertia:$$I = \frac{bd^3}{12} = \frac{0.2 \times (0.4)^3}{12} = \frac{0.2 \times 0.064}{12} = 0.001067 \text{ m}^4$$

Calculate stress:$$\sigma = \frac{My}{I} = \frac{12,000 \times 0.2}{0.001067} = \frac{2,400}{0.001067} = 2,248,836 \text{ Pa} = 2.25 \text{ MPa}$$

Answer: 2.25 MPa

Example 3: A steel beam with I-section has a moment of inertia \(I = 8.5 \times 10^{-6}\) m\(^4\) and is subjected to a bending moment of 25 kNm. The distance from neutral axis to extreme fiber is 75 mm. Calculate the bending stress.

Given:
\(I = 8.5 \times 10^{-6}\) m\(^4\), \(M=25,000\) Nm, \(y=0.075\) m

Calculate stress:$$\sigma = \frac{My}{I} = \frac{25,000 \times 0.075}{8.5 \times 10^{-6}} = \frac{1,875}{8.5 \times 10^{-6}} = 220,588,235 \text{ Pa} = 220.6 \text{ MPa}$$

Answer: 220.6 MPa

6.1.2 Assumptions of the theory of simple bending

The theory of simple bending is based on key assumptions that simplify analysis and design of beams:

• Plane sections before bending remain plane after bending, meaning no distortion occurs.
• The material of the beam is homogeneous and isotropic, with linear elastic behavior.
• The beam’s cross-section is constant along its length.
• Bending occurs about one principal axis only.
• Stresses due to shear and axial loads are negligible compared to bending stresses.

These assumptions enable the use of the bending formula to calculate stresses efficiently in building members such as beams and joists.

Worked Examples

Example 1: A rectangular beam section 100 mm by 200 mm is bent about the horizontal axis. Verify if the section is symmetrical and if the neutral axis passes through the centroid.

Given:
Width \(b=0.1\) m, Depth \(d=0.2\) m

Since the cross-section is symmetrical about the horizontal axis, the neutral axis passes through the centroid at mid-depth \(d/2=0.1\) m.

Answer: Neutral axis at 0.1 m from top fiber, assumptions valid

Example 2: A beam with a variable cross-section is loaded. Explain why the simple bending theory is not applicable.

Given:
Cross-section varies along length

Because the cross-section is not constant, the moment of inertia \(I\) changes along the beam, violating the assumption; thus, simple bending theory does not apply.

Answer: Simple bending theory invalid for variable cross-section beams

Example 3: A beam made of composite materials with different elastic moduli is loaded. Can simple bending theory be used?

Given:
Material is non-homogeneous

Since the material is not homogeneous, stress distribution is not linear and simple bending theory assumptions fail.

Answer: Simple bending theory is not applicable

6.1.3 Neutral axis and its significance

The neutral axis is the line within the beam’s cross-section where bending stress is zero during bending. Fibers above the neutral axis undergo compression, while those below experience tension. Determining the neutral axis location is essential for calculating bending stresses and designing safe beam sections.

The neutral axis passes through the centroid of the cross-section for symmetric, homogeneous beams under pure bending.

Worked Examples

Example 1: Find the neutral axis location of a rectangular beam 150 mm wide and 300 mm deep.

Given:
\(b=0.15\) m, \(d=0.3\) m

For symmetrical section, neutral axis is at centroid:$$y_{NA} = \frac{d}{2} = \frac{0.3}{2} = 0.15 \text{ m}$$

Answer: Neutral axis is 0.15 m from top

Example 2: Determine the neutral axis for a T-beam composed of a flange 100 mm wide by 20 mm thick and a web 200 mm high by 100 mm wide.

Given:
Flange area \(A_f = 0.1 \times 0.02 = 0.002\) m\(^2\)
Web area \(A_w = 0.1 \times 0.2 = 0.02\) m\(^2\)

Calculate centroid:$$y_{NA} = \frac{A_f y_f + A_w y_w}{A_f + A_w}$$

Where \(y_f = 0.01\) m (half flange thickness), \(y_w = 0.12\) m (flange thickness + half web height)

$$y_{NA} = \frac{0.002 \times 0.01 + 0.02 \times 0.12}{0.002 + 0.02} = \frac{0.00002 + 0.0024}{0.022} = \frac{0.00242}{0.022} = 0.11 \text{ m}$$

Answer: Neutral axis is 0.11 m from top of flange

Example 3: A beam with two different materials joined together has areas \(A_1 = 0.003\) m\(^2\) at \(y_1 = 0.05\) m, \(A_2 = 0.005\) m\(^2\) at \(y_2 = 0.15\) m. Find neutral axis location assuming modular ratio \(n=2\).

Given:
Transform area \(A_2\) into equivalent \(A_2' = n \times A_2 = 2 \times 0.005 = 0.01\) m\(^2\)

Calculate neutral axis:$$y_{NA} = \frac{A_1 y_1 + A_2' y_2}{A_1 + A_2'} = \frac{0.003 \times 0.05 + 0.01 \times 0.15}{0.003 + 0.01} = \frac{0.00015 + 0.0015}{0.013} = \frac{0.00165}{0.013} = 0.127 \text{ m}$$

Answer: Neutral axis at 0.127 m

6.1.4 Moment of inertia and its role in bending

The moment of inertia \(I\) of a beam’s cross-section measures its resistance to bending deformation. A larger \(I\) indicates greater stiffness and lower bending stresses for the same moment. Calculating \(I\) accurately is essential for beam design in buildings, especially for sections like rectangular, I, and T shapes.

For a rectangular section:

$$I = \frac{bd^3}{12}$$

where \(b\) is width and \(d\) is depth.

Worked Examples

Example 1: Calculate the moment of inertia of a rectangular beam 150 mm wide and 300 mm deep.

Given:
\(b=0.15\) m, \(d=0.3\) m

$$I = \frac{bd^3}{12} = \frac{0.15 \times (0.3)^3}{12} = \frac{0.15 \times 0.027}{12} = 0.0003375 \text{ m}^4$$

Answer: \(I = 0.0003375\) m\(^4\)

Example 2: Find the moment of inertia of an I-section with flange width 200 mm, flange thickness 20 mm, web height 300 mm, and web thickness 10 mm.

Given:
Flange area moment of inertia:$$I_f = 2 \times ( \frac{b_f t_f^3}{12} + A_f d_f^2 )$$

Where \(b_f=0.2\) m, \(t_f=0.02\) m, \(A_f = b_f \times t_f = 0.004\) m\(^2\), \(d_f\) = distance from flange centroid to neutral axis (assumed 0.16 m)

Web moment of inertia:$$I_w = \frac{b_w h_w^3}{12} = \frac{0.01 \times (0.3)^3}{12} = 0.0000225 \text{ m}^4$$

Calculate flange inertia:$$\frac{b_f t_f^3}{12} = \frac{0.2 \times (0.02)^3}{12} = 0.00000133 \text{ m}^4$$$$A_f d_f^2 = 0.004 \times (0.16)^2 = 0.0001024 \text{ m}^4$$$$I_f = 2 \times (0.00000133 + 0.0001024) = 2 \times 0.00010373 = 0.00020746 \text{ m}^4$$

Total moment of inertia:$$I = I_f + I_w = 0.00020746 + 0.0000225 = 0.00022996 \text{ m}^4$$

Answer: \(I = 0.00023\) m\(^4\) (approx.)

Example 3: Calculate the moment of inertia of a circular column with diameter 400 mm.

Given:
Diameter \(d=0.4\) m

Moment of inertia for circle:$$I = \frac{\pi d^4}{64} = \frac{3.1416 \times (0.4)^4}{64} = \frac{3.1416 \times 0.0256}{64} = 0.001256 \text{ m}^4$$

Answer: \(I = 0.001256\) m\(^4\)

Chapter Summary

This chapter introduced the fundamental concept of bending in beams, explaining how external loads cause bending stresses within structural elements. It covered the relationship between bending moment and shear force, emphasizing their roles in determining internal forces in beams. The derivation and explanation of the simple bending equations were presented, providing the mathematical foundation for calculating bending stresses. The concept of the neutral axis was described alongside the section modulus, highlighting their importance in assessing beam strength and behavior under bending. The chapter also explored how stress distribution varies across different cross-sectional shapes, demonstrating how geometry influences the performance of beams under load. Together, these topics establish a comprehensive understanding of the theory of simple bending essential for structural analysis.