4. Bending and Shear Stresses in Beams, CE 3003 notes in English, Mechanics of Materials notes in English

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4. Bending and Shear Stresses in Beams

4.1 Concept and Theory of Pure Bending, Assumptions, Flexural Equation (with Derivation)

Pure Bending:

Pure Bending occurs when a beam is subjected to a bending moment without any axial force or shear force. It is characterized by uniform curvature along the length of the beam. This condition typically arises when the beam is subjected to moments applied at its ends (like a cantilever beam under a moment or a simply supported beam with equal moments at both ends).

Assumptions in Pure Bending:

To derive the flexural equation, certain assumptions are made:

1. The material is homogeneous and isotropic.
2. The beam is subjected to pure bending without any axial or shear forces.
3. The cross-section of the beam remains plane before and after deformation (plane sections remain plane).
4. The beam follows Hooke's law, and material is within the elastic limit.
5. The radius of curvature is large compared to the dimensions of the beam.

Flexural Equation (Bending Equation):

The flexural equation relates the bending stress at a point in a beam to the applied moment and the geometry of the beam's cross-section. The equation is:

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

where:

• $\sigma$ = Bending stress at a point in the beam (N/m² or Pascals).
• $M$ = Bending moment at the section (Nm).
• $y$ = Distance from the neutral axis to the point where stress is calculated (m).
• $I$ = Moment of inertia of the beam's cross-section about the neutral axis (m⁴).

Bending Stress and Their Nature:

• The bending stress varies linearly with the distance from the neutral axis.
• Compressive stress occurs above the neutral axis (in the upper part of the beam when subjected to bending).
• Tensile stress occurs below the neutral axis (in the lower part of the beam).
• The maximum bending stress occurs at the farthest point from the neutral axis.

Bending Stress Distribution Diagram:

• The bending stress distribution in a beam is triangular (for symmetric loading) with maximum tensile and compressive stresses occurring at the top and bottom surfaces of the beam. The neutral axis experiences zero stress.
• The stress decreases linearly toward the neutral axis.

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4.2 Concept of Moment of Resistance and Simple Numerical Problems Using Flexural Equation

Moment of Resistance:

The Moment of Resistance is the internal moment developed in the beam to resist the applied bending moment. It is given by the bending stress at the extreme fiber multiplied by the distance from the neutral axis to the extreme fiber, and the section modulus $Z$.

$$M_R = \sigma_{\text{max}} \times Z$$

where:

• $M_R$ = Moment of resistance (Nm)
• $\sigma_{\text{max}}$ = Maximum bending stress (N/m²)
• $Z$ = Section modulus, which is given by $Z = \frac{I}{y}$, where $y$ is the distance from the neutral axis to the extreme fiber.

Numerical Example:

Given a simply supported beam with a bending moment of 500 Nm, a rectangular cross-section of width $b = 100 \, \text{mm}$ and height $h = 200 \, \text{mm}$, and a material with modulus of elasticity $E = 200 \, \text{GPa}$, calculate the bending stress at the extreme fiber.

• Moment of inertia for the rectangular section: $$I = \frac{b h^3}{12} = \frac{100 \times (200)^3}{12} = 1.33 \times 10^8 \, \text{mm}^4$$ $$I = 1.33 \times 10^{-2} \, \text{m}^4$$
• Section modulus $Z = \frac{I}{y} = \frac{I}{h/2} = \frac{1.33 \times 10^{-2}}{200/2} = 1.33 \times 10^{-4} \, \text{m}^3$
• Maximum bending stress: $$\sigma_{\text{max}} = \frac{M}{Z} = \frac{500}{1.33 \times 10^{-4}} = 3.76 \times 10^6 \, \text{N/m}^2 \, \text{or} \, 3.76 \, \text{MPa}$$

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4.3 Shear Stress Equation (Without Derivation)

The shear stress at any point in a beam is given by the equation:

$$\tau = \frac{V Q}{I b}$$

where:

• $\tau$ = Shear stress at a point (N/m² or Pa).
• $V$ = Shear force at the section (N).
• $Q$ = First moment of the area above or below the point where shear stress is being calculated (m³).
• $I$ = Moment of inertia of the entire beam cross-section about the neutral axis (m⁴).
• $b$ = Width of the beam at the point where shear stress is being calculated (m).

This equation gives the shear stress at any point within the beam.

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4.4 Relation Between Maximum and Average Shear Stress for Rectangular and Circular Sections

For a Rectangular Section:

For a rectangular beam with width $b$ and height $h$, the maximum shear stress occurs at the neutral axis and is given by:

$$\tau_{\text{max}} = \frac{3V}{2A}$$

where $A = b \times h$ is the area of the cross-section.

The average shear stress is given by:

$$\tau_{\text{avg}} = \frac{V}{A}$$

Thus, the maximum shear stress is 1.5 times the average shear stress for a rectangular section.

For a Circular Section:

For a solid circular section with diameter $d$, the maximum shear stress occurs at the neutral axis and is given by:

$$\tau_{\text{max}} = \frac{4V}{3A}$$

where $A = \frac{\pi d^2}{4}$ is the area of the cross-section.

The average shear stress is given by:

$$\tau_{\text{avg}} = \frac{V}{A}$$

Thus, the maximum shear stress is approximately 1.33 times the average shear stress for a circular section.

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4.5 Shear Stress Distribution Diagram

The shear stress distribution across the height of a beam varies depending on the type of cross-section.

1. Rectangular Section:

   • The shear stress is maximum at the neutral axis and decreases parabolically towards the top and bottom surfaces.

2. Circular Section:

   • The shear stress is maximum at the center and decreases as you move towards the perimeter.

3. I-Section:

   • The shear stress is maximum at the web and decreases as you move toward the flanges.

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4.6 Shear Stress Distribution for Different Sections

Square and Rectangular Sections:

• The shear stress is maximum at the neutral axis and follows a parabolic distribution. It is zero at the top and bottom surfaces.

Circular Section:

• The shear stress is maximum at the center and decreases radially towards the circumference, forming a parabolic distribution.

Hollow Sections:

• For a hollow section (like a pipe), the shear stress distribution is more complex, with maximum stress near the thinner section of the wall and reducing towards the outer edge.

Angle, Channel, I, and T Sections:

• The shear stress is maximum at the web and zero at the flanges for I and T sections. For angle and channel sections, the distribution is more complex due to the geometry of the sections.

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4.7 Simple Numerical Problems Based on Shear Equation

Example 1: Rectangular Beam

Given a simply supported beam with a rectangular cross-section of width $b = 100 \, \text{mm}$ and height $h = 200 \, \text{mm}$, subjected to a shear force $V = 10 \, \text{kN}$, calculate the maximum shear stress at the neutral axis.

• Area $A = b \times h = 100 \times 200 = 20000 \, \text{mm}^2 = 20 \times 10^{-3} \, \text{m}^2$
• Moment of inertia $I = \frac{b h^3}{12} = \frac{100 \times 200^3}{12} = 1.33 \times 10^8 \, \text{mm}^4 = 1.33 \times 10^{-2} \, \text{m}^4$

Using the shear stress equation:

$$\tau = \frac{V Q}{I b}$$

For a point at the neutral axis, $Q = \frac{A_{\text{top}} \times y_{\text{top}}}{2}$.