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When mechanical components carry loads during service, they deform. To design safe and efficient machines, it is essential to understand how materials respond to forces. This response is described quantitatively by stress and strain. The stress-strain diagram graphically illustrates the relationship between applied load and resulting deformation. It reveals critical points that determine the material's behavior, strength, and failure modes. For engineers, mastering the stress-strain diagram is crucial for predicting performance and ensuring safety.
Before diving into stress-strain diagrams, let's understand what stress and strain mean.
Stress is defined as the internal force per unit area within a material that arises when an external load is applied. When the load acts axially (along the length), the stress is called normal stress. It is given by:
Here, \(F\) is the applied axial force in Newtons and \(A\) is the cross-sectional area of the specimen in square meters. Stress is measured in Pascals (Pa), where 1 Pa = 1 N/m². Commonly, we use Megapascal (MPa) with 1 MPa = \(10^{6}\) Pa.
Strain is the measure of deformation representing the change in length compared to the original length. It is a dimensionless quantity expressed as:
Since strain is a ratio, it has no units. It is often expressed in microstrains (\(\mu\varepsilon\)), where 1 microstrain = \(10^{-6}\).
The figure above shows a cylindrical rod with length \(L\) and cross-sectional area \(A\) subjected to equal and opposite axial tensile forces \(F\). This setup generates normal tensile stress and produces elongation causing strain.
Initially, when the axial load is applied, stress and strain relate linearly. This range is called the elastic region. Within this region, the material behaves like a spring-if the load is removed, the material returns to its original shape without permanent deformation.
This behavior is captured by Hooke's Law:
Here, Young's Modulus \(E\) is a material constant called the modulus of elasticity. It measures the stiffness of the material: the higher the \(E\), the stiffer the material.
In the linear region of the graph above, the stress and strain are proportional. The end of this linear region is called the proportional limit. Beyond this point, the linear relationship no longer holds exactly, yet the material may still be elastic.
The elastic limit is the maximum stress up to which the material returns to its original shape upon unloading. Note that the proportional limit and elastic limit are close but not always the same.
When the stress exceeds the elastic limit, the material no longer behaves elastically. Permanent deformation sets in. This is known as plastic deformation.
The yield point is a distinct stress level where the material begins to deform plastically at nearly constant stress, showing a horizontal flat region in the curve for some materials (like mild steel). Beyond the yield point, strain increases rapidly with little increase in stress.
After yielding, the material may get stronger as it is strained further, a phenomenon called strain hardening (or work hardening). The stress rises again to the ultimate tensile strength (UTS), the maximum stress the material can withstand.
Beyond the UTS, with continued deformation, the material eventually breaks or fractures at the fracture point, showing a sudden drop in stress.
So far, the stress-strain curve discussed uses engineering stress and strain, which assume the original cross-sectional area and length throughout the test.
However, during plastic deformation, the cross-sectional area decreases notably due to necking, which engineering stress does not account for. True stress refers to the actual load divided by the instantaneous cross-sectional area, and true strain accounts for incremental changes in length.
| Aspect | Engineering Stress-Strain | True Stress-Strain |
|---|---|---|
| Definition | \(\sigma = \frac{F}{A_0}\), \(\varepsilon = \frac{\Delta L}{L_0}\) | \(\sigma_{true} = \frac{F}{A_i}\), \(\varepsilon_{true} = \ln \left(\frac{L_i}{L_0}\right)\) |
| Area Consideration | Original area \(A_0\) | Instantaneous area \(A_i\) |
| Strain Calculation | Ratio of total elongation over original length | Natural logarithm of elongation ratio |
| Usefulness | Simple, suitable for elastic and early plastic regions; standard in most tests | More accurate in plastic region for precise material behavior |
Step 1: Convert Units
Cross-sectional area \(A = 100\; \text{mm}^2 = 100 \times 10^{-6} = 1 \times 10^{-4}\; \text{m}^2\)
Load \(F = 10\; \text{kN} = 10,000\; \text{N}\)
Length \(L = 500\; \text{mm} = 0.5\; \text{m}\)
Elongation \(\Delta L = 0.2\; \text{mm} = 0.0002\; \text{m}\)
Step 2: Calculate Normal Stress
\[ \sigma = \frac{F}{A} = \frac{10,000}{1 \times 10^{-4}} = 100,000,000\, \text{Pa} = 100\, \text{MPa} \]
Step 3: Calculate Strain
\[ \varepsilon = \frac{\Delta L}{L} = \frac{0.0002}{0.5} = 4 \times 10^{-4} \] (dimensionless)
Answer: Normal stress is 100 MPa, and strain is 0.0004 (or 400 microstrain).
Step 1: Recall Hooke's Law:
\[ \sigma = E \varepsilon \Rightarrow E = \frac{\sigma}{\varepsilon} \]
Step 2: Calculate \(E\):
\[ E = \frac{150 \times 10^6}{0.0015} = 1 \times 10^{11} \, \text{Pa} = 100\, \text{GPa} \]
Answer: Young's modulus is \(100\, \text{GPa}\).
Step 1: The yield point is identified as the stress at which the curve flattens after the linear rise. From the graph, this is approximately at \(130\, \text{MPa}\).
Step 2: The ultimate tensile strength is the maximum stress reached on the curve before it drops due to necking, approximately \(140\, \text{MPa}\) here.
Answer: Yield strength \(\sigma_y \approx 130\, \text{MPa}\), Ultimate tensile strength \(\sigma_{uts} \approx 140\, \text{MPa}\).
Step 1: Recall the formula connecting true stress \(\sigma_{true}\) to engineering stress \(\sigma_{eng}\) and strain \(\varepsilon_{eng}\):
\[ \sigma_{true} = \sigma_{eng} (1 + \varepsilon_{eng}) \]
Step 2: Substitute values:
\[ \sigma_{true} = 300 \times (1 + 0.1) = 300 \times 1.1 = 330\, \text{MPa} \]
Answer: True stress is \(330\, \text{MPa}\).
Step 1: Toughness is the energy absorbed per unit volume before fracture, equal to area under stress-strain curve.
Step 2: Approximate area of trapezoid:
\[ \text{Area} = \text{Average height} \times \text{Base} = 250 \times 0.15 = 37.5\, \text{MPa} \]
Step 3: Convert MPa to Pa and strain is dimensionless, so:
\[ 37.5\, \text{MPa} = 37.5 \times 10^6\, \text{Pa} = 37.5 \times 10^6\, \frac{\text{N}}{\text{m}^2} \]
Answer: Toughness is approximately \(37.5 \times 10^6 \, \text{J/m}^3\), representing the energy absorbed before failure.
When to use: Before solving all stress-strain problems to avoid unit-related errors.
When to use: For exam problems that provide graphical data.
When to use: For materials like aluminum where yield plateau is absent.
When to use: To avoid unnecessary complex calculations in exams.
When to use: Whenever interpreting or describing material behavior graphically.
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