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Stress, Strain, Hooke's Law, and Stress-Strain Curve



Stress And Strain

To quantify the deformation of a solid material under the action of forces, we use the concepts of stress and strain. These quantities are independent of the size and shape of the object to some extent, allowing us to compare the mechanical properties of different materials.


Stress ($\sigma$ or $S$)

Stress is a measure of the internal restoring forces acting within a deformed body per unit cross-sectional area. When an external deforming force is applied, internal restoring forces are set up. In equilibrium, the magnitude of the restoring force is equal to the magnitude of the applied deforming force.

Stress is defined as the restoring force per unit area across the surface where the force is acting.

$ \text{Stress} = \frac{\text{Restoring Force}}{\text{Area}} $

In equilibrium, $\text{Restoring Force} = \text{Applied Deforming Force}$. So, the magnitude of stress is often calculated as:

$ \text{Stress} = \frac{\text{Applied Force}}{\text{Area}} = \frac{F}{A} $

where $F$ is the magnitude of the applied force perpendicular or parallel to the surface and $A$ is the area of the surface.

The SI unit of stress is the same as that of pressure, which is Newton per square metre (N/m$^2$). This unit is also called Pascal (Pa). $1 \text{ Pa} = 1 \text{ N/m}^2$. Other units like megapascals (MPa) or gigapascals (GPa) are commonly used for materials. $1 \text{ MPa} = 10^6 \text{ Pa}$, $1 \text{ GPa} = 10^9 \text{ Pa}$.

Stress can be of different types depending on how the deforming force is applied:


Strain ($\epsilon$)

Strain is a measure of the deformation produced in the body as a result of stress. It is defined as the ratio of the change in dimension to the original dimension.

Strain is a dimensionless quantity, as it is a ratio of two lengths or volumes. It has no units.

Different types of strain correspond to different types of stress:

Diagram illustrating tensile stress and longitudinal strain.

(Image Placeholder: A cylindrical rod or wire being pulled by forces F at both ends. Show the cross-sectional area A. Show original length L and extended length L+delta L. Label tensile stress and longitudinal strain.)

Diagram illustrating shear stress and shear strain.

(Image Placeholder: A rectangular block fixed at the bottom. A tangential force F is applied to the top surface of area A. Show the height h of the block. The top surface shifts horizontally by delta x. The vertical side rotates by an angle phi. Label shear stress and shear strain.)

Example 1. A steel wire of length 2.0 m and uniform cross-sectional area $0.5 \times 10^{-4}$ m$^2$ is stretched by a force of 1000 N. Calculate the tensile stress in the wire.

Answer:

Applied force, $F = 1000$ N.

Cross-sectional area, $A = 0.5 \times 10^{-4}$ m$^2$.

The force is applied perpendicular to the area (pulling), so it causes tensile stress. Assuming the force is uniformly distributed over the cross-section, the tensile stress is:

$ \text{Tensile Stress} = \frac{F}{A} $

$ \text{Tensile Stress} = \frac{1000 \text{ N}}{0.5 \times 10^{-4} \text{ m}^2} $

$ \text{Tensile Stress} = \frac{10^3}{0.5 \times 10^{-4}} \text{ N/m}^2 = \frac{1}{0.5} \times 10^{3 - (-4)} \text{ Pa} = 2 \times 10^7 \text{ Pa} $

The tensile stress in the wire is $2 \times 10^7$ Pa, or 20 MPa.

Example 2. A wire of length 4 m is stretched by 0.5 mm by a certain load. Calculate the longitudinal strain produced in the wire.

Answer:

Original length, $L = 4$ m.

Change in length, $\Delta L = 0.5$ mm.

Convert the change in length to metres: $0.5 \text{ mm} = 0.5 \times 10^{-3}$ m.

Longitudinal strain is the ratio of the change in length to the original length:

$ \text{Longitudinal Strain} = \frac{\Delta L}{L} $

$ \text{Longitudinal Strain} = \frac{0.5 \times 10^{-3} \text{ m}}{4 \text{ m}} $

$ \text{Longitudinal Strain} = \frac{0.0005}{4} = 0.000125 $

The longitudinal strain is 0.000125. It is a dimensionless quantity.



Hooke’s Law

Through his experiments, Robert Hooke (1635-1703) discovered a fundamental relationship between stress and strain for elastic materials. This relationship is known as Hooke's Law.


Statement of Hooke's Law

Within the elastic limit, stress is directly proportional to strain.

Mathematically, this can be written as:

$ \text{Stress} \propto \text{Strain} $

To turn this proportionality into an equation, we introduce a constant of proportionality. This constant is called the modulus of elasticity or elastic modulus of the material.

$ \text{Stress} = \text{Modulus of Elasticity} \times \text{Strain} $

$ \text{Modulus of Elasticity} = \frac{\text{Stress}}{\text{Strain}} $

The modulus of elasticity is a property of the material and depends on the type of stress and strain. It represents the stiffness of the material – a higher modulus means the material is stiffer and requires more stress to produce a given amount of strain.

Since strain is dimensionless, the units of the modulus of elasticity are the same as those of stress (N/m$^2$ or Pascal).


Types of Modulus of Elasticity

Depending on the type of stress and the resulting strain, there are different moduli of elasticity:

Hooke's Law is an empirical law, based on observation. It is valid only within the proportional limit, which is usually the initial, linear part of the stress-strain curve. The proportional limit is slightly below or coincides with the elastic limit.

Graph illustrating Hooke's Law: Linear stress-strain relationship.

(Image Placeholder: A graph with Strain on the x-axis and Stress on the y-axis. A straight line from the origin with positive slope is shown, labelled "Elastic Region" or "Hooke's Law Region". The line extends up to a point labelled "Proportional Limit" or "Elastic Limit".)

Example 3. A brass wire of length 1.5 m and cross-sectional area $1 \times 10^{-5}$ m$^2$ is stretched by 0.3 mm when a weight of 100 N is suspended from it. Calculate the Young's modulus of brass.

Answer:

Original length, $L = 1.5$ m.

Cross-sectional area, $A = 1 \times 10^{-5}$ m$^2$.

Change in length, $\Delta L = 0.3$ mm $= 0.3 \times 10^{-3}$ m.

Applied force (equal to suspended weight), $F = 100$ N.

Tensile Stress: $ \sigma = \frac{F}{A} = \frac{100 \text{ N}}{1 \times 10^{-5} \text{ m}^2} = 100 \times 10^5 \text{ N/m}^2 = 1 \times 10^7 $ Pa.

Longitudinal Strain: $ \epsilon = \frac{\Delta L}{L} = \frac{0.3 \times 10^{-3} \text{ m}}{1.5 \text{ m}} = \frac{0.0003}{1.5} = 0.0002 $.

Young's Modulus, $ Y = \frac{\text{Stress}}{\text{Strain}} $

$ Y = \frac{1 \times 10^7 \text{ Pa}}{0.0002} = \frac{10^7}{2 \times 10^{-4}} = \frac{1}{2} \times 10^{7 - (-4)} = 0.5 \times 10^{11} = 5 \times 10^{10} $ Pa.

The Young's modulus of brass is $5 \times 10^{10}$ Pa or 50 GPa.



Stress-Strain Curve

The relationship between stress and strain for a material can be graphically represented by a stress-strain curve. This curve is obtained by applying an increasing load to a material sample (e.g., stretching a wire) and measuring the resulting strain at each level of stress. The shape of the stress-strain curve provides valuable information about the mechanical properties of the material.

Different materials (ductile, brittle) have characteristic stress-strain curves.


Stress-Strain Curve for a Ductile Material (like Steel)

Let's consider the stress-strain curve for a typical ductile material like steel under tensile stress:

Stress-strain curve for a ductile material, showing different points and regions.

(Image Placeholder: A stress-strain graph. X-axis is Strain (dimensionless). Y-axis is Stress (Pa or N/m^2). The curve starts from the origin, goes up as a straight line, then curves, reaches a peak, drops slightly, then rises again to another peak before dropping sharply to break. Label the following points/regions: - Origin (O) to Proportional Limit (A): Linear region, Hooke's Law valid. - Proportional Limit (A) to Elastic Limit (B): Elastic region, non-linear but recoverable deformation. - Elastic Limit (B) or Yield Point (C): Point where plastic deformation begins. - Yield Strength/Stress: Stress at the Yield Point. - Yielding Region (C to D): Significant increase in strain with little or no increase in stress. - Ultimate Tensile Strength (E): Maximum stress the material can withstand. - Necking Region: Region after UTS where cross-sectional area starts to decrease locally. - Fracture Point (F): Point where the material breaks. )

Key points and regions on the stress-strain curve:

  1. Proportional Limit (Point A): This is the point up to which stress is directly proportional to strain. Hooke's Law ($ \text{Stress} \propto \text{Strain} $) is valid in the region from the origin (O) to A. The slope of this linear region is the Young's Modulus ($Y$).
  2. Elastic Limit (Point B): This is the maximum stress the material can withstand and still return to its original size and shape after the load is removed. If the stress is increased beyond the elastic limit, the material starts to undergo permanent, plastic deformation. For some materials, the proportional limit and elastic limit are very close or coincide.
  3. Yield Point (Point C): For many ductile materials like mild steel, after the elastic limit, the stress drops slightly, and the material begins to yield. This is the point where significant plastic deformation occurs with little or no increase in stress. The stress at the yield point is called the Yield Strength or Yield Stress.
  4. Plastic Region (from C to F): Beyond the elastic limit/yield point, the material enters the plastic region. Deformation is permanent.
  5. Ultimate Tensile Strength (UTS) (Point E): This is the maximum stress the material can withstand before it begins to neck (local reduction in cross-sectional area). This point represents the maximum load the material can support.
  6. Necking (Region after E): Beyond the UTS, the material starts to neck down locally, and the stress needed to cause further deformation actually decreases (when calculated using the original cross-sectional area - engineers sometimes use "true stress" based on instantaneous area, which keeps increasing).
  7. Fracture Point (Point F): The point at which the material breaks or fractures.

Stress-Strain Curve for a Brittle Material

Brittle materials (like glass, ceramics, cast iron) fracture very soon after the elastic limit is reached, with little or no plastic deformation. Their stress-strain curve is typically steeper (higher modulus) and breaks at a lower strain compared to ductile materials.

Stress-strain curve for a brittle material.

(Image Placeholder: A stress-strain graph. X-axis is Strain. Y-axis is Stress. The curve is steeper and stops abruptly shortly after the proportional/elastic limit, at the fracture point. Show the linear elastic region and the fracture point shortly after.)

There is usually no distinct yield point or significant plastic region. The fracture point is very close to the ultimate tensile strength.


Factors Affecting the Stress-Strain Curve

The stress-strain curve of a material can be affected by factors such as:

Understanding the stress-strain curve is essential for engineers to predict how a material will behave under different loads and to ensure the safety and reliability of structures and components.