3. THEORY OF SIMPLE BENDING AND DEFLECTION OF BEAMS

For 4th Semester Polytechnic ME Students
Written by Garima Kanwar | Blog: Rajasthan Polytechnic

📢 🔔 Important:
👉 Full PDF available in our WhatsApp Group | Telegram Channel
👉 Subscribe YouTube Channel: BTER Polytechnic Classes

Course Code ME 4002 (Same in MA/MP/MT 4002)
Course Title STRENGTH OF MATERIALS

3. Theory of Simple Bending and Deflection of Beams

The theory of simple bending is essential in understanding how beams respond to loads and how bending stresses and deflections are calculated. This topic includes the analysis of bending stresses, deflection of beams, and related numerical problems. Here’s a breakdown of each sub-topic:

---

3.1 Explanation of Terms:

1. Neutral Layer:

   • The neutral layer (or neutral axis) is the layer of the beam that experiences no strain during bending. In simple bending, the fibers above the neutral layer are in compression, while those below are in tension.
   • It is located at the point where the bending stress is zero.

2. Neutral Axis:

   • The neutral axis is the axis within the beam where the bending stress is zero.
   • The beam bends around this axis, with compression occurring above it and tension below it.

3. Modulus of Section (S):

   • The modulus of section (also known as the section modulus) is a geometric property of the beam's cross-section.
   • It is defined as the ratio of the moment of inertia (I) of the section about the neutral axis to the distance (y) from the neutral axis to the furthest point of the section.
   • Formula: $$S = \frac{I}{y}$$

   where $I$ is the moment of inertia, and $y$ is the distance from the neutral axis to the outermost fiber.

4. Moment of Resistance:

   • The moment of resistance (or resisting moment) is the moment a beam can resist before failing due to bending.
   • It is calculated using the bending stress and the section modulus.
   • Formula: $$M_r = \sigma \times S$$
     

   where $\sigma$ is the bending stress and $S$ is the modulus of section.

5. Bending Stress:

   • Bending stress is the stress developed in the beam due to the bending moment. It varies linearly across the section from maximum compression at the top to maximum tension at the bottom.
   • Formula: $$\sigma = \frac{M \times y}{I}$$

   where $M$ is the bending moment at a particular section, $y$ is the distance from the neutral axis to the point where the stress is being calculated, and $I$ is the moment of inertia of the cross-section.

6. Radius of Curvature (R):

   • The radius of curvature is the radius of the arc that the beam takes when bent.
   • Formula: $$R = \frac{EI}{M}$$

   where $E$ is the modulus of elasticity of the material, $I$ is the moment of inertia of the beam, and $M$ is the bending moment at the section.

---

3.3 Problems Involving Calculations of Bending Stress, Modulus of Section, and Moment of Resistance

1. Bending Stress Calculation:

   • Given a beam with a specific bending moment and dimensions, the bending stress at a point can be calculated.
   • Example:
     • A simply supported beam with a rectangular cross-section (width $b = 100 \, \text{mm}$, height $h = 200 \, \text{mm}$) carries a bending moment $M = 150 \, \text{Nm}$.
     • To calculate the bending stress at the topmost and bottommost fibers:
       • First, find the moment of inertia $I$ of the section: $$I = \frac{b h^3}{12} = \frac{100 \times (200)^3}{12} = 1.33 \times 10^7 \, \text{mm}^4$$
         
       • The distance from the neutral axis to the topmost fiber is $y = \frac{h}{2} = 100 \, \text{mm}$.
       • Now, calculate the bending stress at the topmost fiber: $$\sigma = \frac{M \times y}{I} = \frac{150 \times 100}{1.33 \times 10^7} = 1.13 \, \text{MPa}$$

2. Modulus of Section:

   • The modulus of section $S$ is calculated using the formula: $$S = \frac{I}{y}$$
     • Example: Using the same beam as above, the modulus of section is: $$S = \frac{1.33 \times 10^7}{100} = 1.33 \times 10^5 \, \text{mm}^3$$
       

3. Moment of Resistance:

   • The moment of resistance $M_r$ is calculated by multiplying the bending stress $\sigma$ by the modulus of section $S$: $$M_r = \sigma \times S = 1.13 \times 1.33 \times 10^5 = 150 \, \text{Nm}$$
     

---

3.4 Calculation of Safe Loads, Safe Span, and Dimensions of Cross-Section

1. Safe Load Calculation:

   • The safe load that a beam can carry is determined using the bending stress and the material's allowable stress.
   • Example:
     • If the allowable bending stress of the material is 150 MPa, the safe load is: $$\text{Safe Load} = \frac{M_r}{\text{distance from neutral axis to extreme fiber}}$$
     • For the same beam as above, the safe load is: $$\text{Safe Load} = \frac{150}{100} = 1.5 \, \text{kN}$$
       

2. Safe Span Calculation:

   • The safe span can be calculated based on the bending stress, material strength, and load conditions. The formula for safe span depends on the bending moment and loading type.

3. Calculation of Cross-Section Dimensions:

   • For a given maximum bending moment and allowable stress, the dimensions of the cross-section (e.g., height and width) can be calculated. Typically, this involves iterating on the dimensions that yield a safe design.

---

3.5 Definition and Explanation of Deflection as Applied to Beams (Standard Cases Only)

1. Deflection:

   • Deflection is the displacement of a point on the beam under the action of a load. It is a measure of the beam's flexibility.
   • In bending, deflection occurs as the beam bends, and it is crucial to ensure that the deflection does not exceed permissible limits (as per standards) to avoid structural damage or failure.

2. Standard Cases of Beam Deflection:

   • Simply Supported Beam with a Point Load at the Center:

     • Formula for maximum deflection ($\delta$) at the center: $$\delta = \frac{P L^3}{48 E I}$$

     where $P$ is the point load, $L$ is the length of the beam, $E$ is the modulus of elasticity, and $I$ is the moment of inertia of the beam cross-section.

   • Simply Supported Beam with Uniformly Distributed Load:

     • Formula for maximum deflection at the center: $$\delta = \frac{5 w L^4}{384 E I}$$

     where $w$ is the uniform load per unit length.

3. Example for a Simply Supported Beam with a Point Load at the Center:

   • A simply supported beam with length $L = 4 \, \text{m}$, subjected to a central point load $P=1000\text{N.}$
   • For a beam made of steel with modulus of elasticity $E = 200 \, \text{GPa}$, and a moment of inertia $I=4\times {10}^{-6}{\text{m}}^{\mathrm{4:}}$$$\delta = \frac{1000 \times 4^3}{48 \times 200 \times 10^9 \times 4 \times 10^{-6}} = 0.024 \, \text{m} = 24 \, \text{mm}$$
     

   This is the deflection at the center of the beam.

---

3.6 Related Numerical Problems

1. Problem 1: A simply supported beam of 5 m length carries a uniform load of 3 kN/m along its entire length. Find the maximum deflection and bending stress in the beam if its cross-sectional dimensions are 150 mm by 200 mm.

2. Problem 2: A cantilever beam with a length of 4 m carries a 5 kN point load at its free end. Calculate the maximum deflection and bending stress in the beam, given the beam's cross-sectional dimensions are 200 mm by 300 mm, and the modulus of elasticity of the material is 210 GPa.

---

Conclusion:

The theory of simple bending provides the foundation for analyzing beams subjected to various loads. The calculation of bending stress, moment of resistance, and deflection is essential for safe structural design. By understanding the relationships between load, moment, stress, and deflection, engineers can design beams that are both efficient and safe. Numerical problems help reinforce these concepts and are key to mastering the topic.

📢 🔔 Download PDF & Join Study Groups:
📥 WhatsApp Group: Join Now
📥 Telegram Channel: Join Now
📺 Watch Lecture on YouTube: BTER Polytechnic Classes
📍 Stay connected for more study materials! 🚀