True Stress Formula vs. Engineering Stress: Key Differences
Understanding the behavior of materials under load is fundamental to mechanical engineering. Two primary ways to quantify this behavior areengineering stressandtrue stress. While both describe the force acting on a material divided by its area, they differ significantly in how they account for changes in that area during deformation. This distinction is critical for accurate analysis and design, especially when dealing with materials undergoing significant plastic deformation. This article will delve into the true stress formula and its differences from engineering stress, providing real-world examples and practical guidance for engineers.
Defining Engineering Stress and True Stress
Engineering stress, often denoted by σ (sigma), is the force applied to a material divided by itsoriginalcross-sectional area. It’s a straightforward calculation:
σ = F/A₀
Where: σ = Engineering stress
F = Applied force
A₀ = Original cross-sectional area
Engineering stress is convenient because it relies on a readily available initial measurement. However, it becomes less accurate as a material deforms, particularly when the cross-sectional area changes considerably, such as during tensile testing beyond the yield point.
True stress, denoted by σt (sigma sub-t), addresses this limitation. It is defined as the force applied to a material divided by theinstantaneouscross-sectional area. This area continuously changes as the material deforms. The formula for true stress is:
σt = F/A
Where: σt = True stress
F = Applied force
A = Instantaneous cross-sectional area
The crucial difference lies in the area used for the calculation. Engineering stress assumes the area remains constant, while true stress accounts for the actual, changing area. As a material necks (localized reduction in cross-sectional area) during tensile testing, the true stress will be significantly higher than the engineering stress.
Deriving the Relationship Between True Stress and Engineering Stress in Plastic Deformation
Assuming constant volume during plastic deformation (a reasonable assumption for many metals), we can relate the instantaneous area (A) to the original area (A₀) using the following:
A₀L₀ = AL
Where:
A₀ = Original cross-sectional area
L₀ = Original length
A = Instantaneous cross-sectional area
L = Instantaneous length
Rearranging this equation gives:
A = A₀(L₀/L)
Substituting this into the true stress equation:
σt = F / [A₀(L₀/L)] = (F/A₀) (L/L₀)
Since F/A₀ is the engineering stress (σ), and L/L₀ is equal to (1 + ε), where ε is the engineering strain, we get:
σt = σ (1 + ε)
This equation relates true stress to engineering stress and engineering strain, under the assumption of constant volume.
Similarly, we can relate true strain (εt) to engineering strain (ε):
εt = ln(L/L₀) = ln(1 + ε)
Therefore, the true stress-strain relationship can be expressed as:
σt = K(εt)n
Where:
K is the strength coefficient.
n is the strain hardening exponent.
This power-law relationship is commonly used to model material behavior in finite element analysis and other advanced calculations.
Real-World Applications Illustrating the Importance of True Stress
The difference between true stress and engineering stress becomes critical in several engineering applications, especially those involving large plastic deformations: Metal Forming Processes: In processes like forging, rolling, and extrusion, metals undergo significant plastic deformation. Accurately predicting the forces required and the final shape of the component necessitates using true stress-strain relationships in simulations and calculations. Engineering stress would underestimate the stress levels, leading to inaccurate predictions and potential process failures. Fracture Mechanics: When analyzing crack propagation, the stress at the crack tip is highly localized and involves significant plastic deformation. True stress provides a more realistic representation of the stress state and is crucial for predicting fracture behavior. Pressure Vessels: While thin-walled pressure vessel calculations often rely on simplified stress formulas (like hoop stress = pr/t), in situations where the material undergoes significant yielding due to high pressure or temperature, true stress analysis becomes necessary to ensure structural integrity and prevent failure. Consider, for example, a pressure vessel undergoing autofrettage, a process that induces compressive residual stresses at the inner wall through controlled plastic deformation. This process is designed using true stress-strain relationships. Finite Element Analysis (FEA): FEA software often allows users to input either engineering stress-strain or true stress-strain data. When modeling materials undergoing large plastic deformations, using true stress-strain data will result in more accurate simulations. Rotating Machinery:In components like turbine blades or high-speed rotors, centrifugal forces can cause significant stresses. If these stresses lead to yielding, true stress analysis is necessary to accurately predict the component's long-term behavior and prevent creep or fatigue failures.
Worked Examples
Example 1: Tensile Test of a Steel Bar
A steel bar with an initial diameter of 10 mm is subjected to a tensile test. At a certain point, the applied force is 25,000 N, and the instantaneous diameter is measured to be 9 mm. Calculate the engineering stress and the true stress at this point.
Solution
1.Calculate the original area (A₀):
A₀ = π (d₀/2)² = π (10 mm/2)² = 78.54 mm²
2.Calculate the engineering stress (σ):
σ = F/A₀ = 25,000 N / 78.54 mm² =
318.3 MPa
3.Calculate the instantaneous area (A):
A = π (d/2)² = π (9 mm/2)² = 63.62 mm²
4.Calculate the true stress (σt):
σt = F/A = 25,000 N / 63.62 mm² =
393.0 MPa
In this example, the true stress (393.0 MPa) is significantly higher than the engineering stress (318.3 MPa), demonstrating the importance of using true stress when the material undergoes significant deformation.
Example 2: Calculating True Strain and True Stress from Engineering Values
A copper wire is subjected to a tensile test. At one point in the test, the engineering stress is measured to be 200 MPa, and the engineering strain is 0.15. Calculate the true stress and true strain.
Solution
1.Calculate the true stress (σt):
σt = σ (1 + ε) = 200 MPa (1 + 0.15) = 230 MPa
2.Calculate the true strain (εt):
εt = ln(1 + ε) = ln(1 + 0.15) =
0.1398
Common Pitfalls and Misconceptions
Assuming Engineering Stress is Always Sufficient: A common mistake is to assume that engineering stress is always adequate for design calculations. While it's suitable for situations with small elastic deformations, it becomes increasingly inaccurate when plastic deformation is significant. Forgetting the Constant Volume Assumption: The relationship between true stress, engineering stress, and strain (σt = σ (1 + ε)) is derived based on the assumption of constant volume. This assumption may not hold for all materials, especially those undergoing significant void formation or cavitation during deformation. Confusion with Nominal Stress: Engineering stress is sometimes referred to as "nominal stress." While technically correct, it’s important to remember that “nominal” doesn’t imply that it's always an accurate representation of the stress state. Incorrectly Applying True Stress in Elastic Region: True stress calculations are most relevant in the plastic deformation region. In the elastic region, where deformations are small, the difference between true stress and engineering stress is negligible, and using either will yield similar results.
When to Use True Stress vs. Engineering Stress
Here's a general guideline to help you decide when to use true stress versus engineering stress: Use Engineering Stress:
When dealing with small elastic deformations.
For initial estimations and preliminary calculations.
When material data is only available in terms of engineering stress and strain. Use True Stress:
When dealing with large plastic deformations (e.g., metal forming, fracture analysis).
When performing finite element analysis requiring accurate material modeling in the plastic region.
When analyzing material behavior at high temperatures, where creep and plastic deformation are significant.
When designing components that undergo significant yielding, such as pressure vessels under extreme conditions.
People Also Ask
How do you calculate hoop stress in thin-walled cylinders?
Hoop stress (σh) in a thin-walled cylinder is calculated using the formula: σh = (p r) / t, where p is the internal pressure, r is the radius of the cylinder, and t is the wall thickness. This formula assumes that the stress is uniformly distributed across the wall thickness and that the deformation is small enough that engineering stress is a reasonable approximation. However, for high pressures that cause significant yielding, a more complex analysis using true stress and strain would be necessary.
What is the difference between true stress and engineering stress?
The key difference lies in the area used in the calculation. Engineering stress uses theoriginalcross-sectional area, while true stress uses theinstantaneouscross-sectional area. True stress accounts for the reduction in area that occurs during deformation, especially in plastic deformation, making it a more accurate representation of the stress state.
When should principal stress formulas be applied in design?
Principal stress formulas are used to determine the maximum and minimum normal stresses at a point in a material, along with the orientation of the planes on which these stresses act. They are crucial when analyzing components subjected to multiaxial loading, such as beams under bending and torsion, or pressure vessels with complex geometries. Principal stresses are used to predict yielding and fracture based on failure criteria such as the von Mises or Tresca criteria. While engineering stress can be used to calculate principal stresses, for materials undergoing significant plastic deformation, true stress values might be needed to input into these failure criteria to get accurate predictions.
Conclusion
Understanding the difference between true stress and engineering stress is essential for accurate stress analysis and design in mechanical engineering. While engineering stress provides a convenient approximation for small elastic deformations, true stress is necessary for situations involving significant plastic deformation. By understanding the concepts, formulas, and applications discussed in this article, engineers can make informed decisions and ensure the safety and reliability of their designs. Furthermore, remember that the use of true stress requires material models (such as the power-law model) that are more advanced and require careful selection of material parameters, such as the strength coefficient (K) and strain hardening exponent (n). Proper use of these concepts, along with FEA tools, is critical for robust engineering designs.