1 SIMPLE STRESSES AND STRAINS

 For 4th Semester Polytechnic ME Students

Written by Garima Kanwar | Blog: Rajasthan Polytechnic

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

1. Simple Stresses and Strains

This section covers basic concepts of stress and strain, their significance, mechanical properties of materials, and related calculations for real-world applications. Here's a breakdown of each topic to help you with exam preparation:

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1.1 Types of Forces; Stress, Strain, and Their Nature

• Forces: Forces are external influences that cause a body to deform. They can be classified as:

  • Tensile Force: Pulls an object, stretching it.
  • Compressive Force: Pushes an object, compressing it.
  • Shear Force: Causes a material to slide over itself.
  • Bending Force: Causes bending by applying a force at an angle.
  • Torsional Force: Causes twisting in the material.

• Stress:
  Stress is the internal resistance of a material to deformation when subjected to an external force. It is defined as force per unit area and is expressed as:

  $$\text{Stress} (\sigma) = \frac{F}{A}$$

  Where:

  • $F$ = Force applied
  • $A$ = Cross-sectional area

  The units of stress are N/m² (Pascal, Pa).

• Strain:
  Strain is the deformation or displacement per unit length resulting from the application of stress. It is a dimensionless quantity (no units). It is expressed as:

  $$\text{Strain} (\epsilon) = \frac{\Delta L}{L}$$

  Where:

  • $\Delta L$ = Change in length
  • $L$ = Original length

  Types of strain include:

  • Tensile Strain: Deformation under tensile stress.
  • Compressive Strain: Deformation under compressive stress.
  • Shear Strain: Deformation due to shear stress.

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1.2 Mechanical Properties of Common Engineering Materials

The mechanical properties of materials describe their response to mechanical forces and stresses. Common properties include:

• Young's Modulus (E): The ratio of tensile stress to tensile strain in the elastic region. It represents stiffness.

  $$E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$$

• Poisson’s Ratio ($\nu$): The ratio of lateral strain to axial strain when a material is stretched or compressed.

• Yield Strength (σ_y): The stress at which a material begins to deform plastically (permanent deformation).

• Ultimate Strength (σ_u): The maximum stress a material can withstand before fracture.

• Fracture Toughness: The ability of a material to resist crack propagation.

Examples:

• Mild Steel (M.S.): High tensile strength, moderate ductility.
• Cast Iron (C.I.): Brittle material, low tensile strength, good compressive strength.

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1.3 Significance of Various Points on Stress-Strain Diagram for M.S. and C.I. Specimens

A stress-strain curve shows the relationship between the stress applied to a material and the resulting strain. Key points on this curve are:

• Proportional Limit: The point up to which stress and strain are proportional (i.e., Hooke's Law holds).
• Elastic Limit: Beyond this point, permanent deformation occurs.
• Yield Point: The stress at which plastic deformation begins.
• Ultimate Strength: The maximum stress the material can withstand.
• Fracture Point: The point where the material ultimately breaks.

For Mild Steel (M.S.):

• The stress-strain curve is relatively linear up to the yield point, showing good ductility and strength.

For Cast Iron (C.I.):

• The curve is steep and brittle, showing little plastic deformation before fracture.

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1.4 Significance of Factor of Safety

The Factor of Safety (FoS) is the ratio of the material's ultimate or yield strength to the actual applied load or stress. It ensures that a material will not fail under normal working conditions. The formula is:

$$\text{FoS} = \frac{\text{Ultimate Strength or Yield Strength}}{\text{Working Stress}}$$

A higher FoS provides a greater safety margin, while a lower FoS can lead to failure if unexpected conditions arise.

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1.5 Relation Between Elastic Constants

Elastic constants relate different types of stresses and strains in materials. The primary elastic constants are:

1. Young’s Modulus (E): Describes the material's stiffness.
2. Poisson’s Ratio ($\nu$): Describes the ratio of lateral strain to longitudinal strain.
3. Shear Modulus (G): Describes the material’s response to shear stress.

The relationship between these constants is given by:

$$G = \frac{E}{2(1 + \nu)}$$

and

$$K = \frac{E}{3(1 - 2\nu)}$$

Where:

• $G$ = Shear Modulus
• $K$ = Bulk Modulus

These relations allow engineers to calculate other material properties when only one is known.

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1.6 Stress and Strain Values in Bodies of Uniform Section and Composite Section Under Normal Forces

• Uniform Section: A body with a constant cross-sectional area will experience uniform stress and strain when subjected to a normal force.

  • Stress = $\frac{F}{A}$
  • Strain = $\frac{\Delta L}{L}$

• Composite Section: A composite body consists of two or more materials with different properties. The total strain in the composite section is the sum of the strains in each material, assuming no slip between materials.

  $$\frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2} = \ldots = \frac{\Delta L_n}{L_n}$$

  The stresses are calculated based on the respective Young’s moduli of each material.

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1.7 Thermal Stresses in Bodies of Uniform Section and Composite Sections

Thermal stress arises when a material is subjected to a temperature change, causing it to expand or contract. If the material is constrained, thermal stresses develop. The formula for thermal stress is:

$$\sigma_{\text{thermal}} = E \cdot \alpha \cdot \Delta T$$

Where:

• $\alpha$ = Coefficient of linear expansion
• $\mathrm{\Delta }\mathrm{T\; =\; Change\; in\; temperature}$
• $E$ = Young’s Modulus

In composite sections, the thermal strain for each material will be proportional to its own coefficient of thermal expansion and Young's Modulus.

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1.8 Related Numerical Problems on the Above Topics

Numerical problems may include:

1. Calculating stress, strain, or Young’s Modulus given force, area, and change in length.
2. Determining the factor of safety for a given material under a specified load.
3. Calculating the thermal stress in a material subjected to a temperature change.
4. Solving for strain energy or determining the total strain in composite materials.

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1.9 Strain Energy and Its Significance

Strain Energy is the energy stored in a material when it is deformed due to applied forces. It can be calculated as the work done to deform the material. The formula for strain energy $U$ for a material under tensile stress is:

$$U = \frac{1}{2} \cdot \sigma \cdot \epsilon \cdot V$$

Where:

• $\sigma$ = Stress
• $\epsilon$ = Strain
• $V$ = Volume of the material

Strain energy is significant in understanding how much deformation a material can absorb before failure.

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1.10 Derivation of Strain Energy for Various Cases

1. Gradually Applied Load: When a load is applied gradually, the strain energy is given by:

   $$U = \frac{1}{2} \cdot F \cdot \Delta L$$
   

   Where:

   • $F$ = Applied force
   • $\Delta L$ = Change in length

2. Suddenly Applied Load: When the load is applied suddenly, the strain energy is:

   $$U = \frac{1}{2} \cdot F \cdot \Delta L$$
   

   Same as the gradually applied load, but the material may experience more stress due to the sudden application.

3. Impact/Shock Load: For impact or shock loading, strain energy is typically calculated using the dynamic load factor and is often more complex because of the high rate of loading.

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Summary of Key Concepts:

• Stress & Strain: Stress is the force per unit area, while strain is the deformation per unit length.
• Material Properties: Properties like Young’s Modulus, Poisson’s Ratio, and Yield Strength define material behavior.
• Factor of Safety: Ensures structures are designed with an adequate margin of safety.
• Elastic Constants: Relations between Young’s Modulus, Poisson’s Ratio, and Shear Modulus are fundamental in material calculations.
• Thermal Stress: Results from temperature changes and affects material deformation.
• Strain Energy: The energy stored in a material due to deformation.

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Important Questions for Exam Preparation:

1. Derive the relation between stress and strain for a material subjected to normal tensile force.
2. Explain the significance of the yield strength and ultimate strength on the stress-strain curve.
3. What is the factor of safety? How do you calculate it for a given material?
4. Discuss the thermal stresses in composite materials.
5. Explain the concept of strain energy and its significance.

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Numerical Problems to Practice:

1. A steel rod of 10 mm diameter and 2 m length is stretched by a force of 5 kN. Calculate the elongation of the rod (given E = 200 GPa).
2. A mild steel specimen is subjected to a gradually applied tensile load of 10 kN. If the elongation is 2 mm, calculate the strain energy stored.
3. Calculate the thermal stress developed in a steel bar of 3 m length when the temperature is increased by 50°C. Assume $\alpha =12\times {10}^{-6}\mathrm{/}\mathrm{°}\mathrm{C\; and }$$E=200\text{ GPa.}$

These notes should provide you with the theoretical understanding and practical tools to tackle simple stresses and strains topics effectively.

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