The engineering stress–strain curve does not accurately represent a material’s deformation behavior because it is calculated using the specimen’s original dimensions, which continuously change during testing. In metalworking operations such as wire drawing, for instance, the cross-sectional area of the workpiece decreases significantly. Therefore, it is more meaningful to define stress and strain based on the instantaneous dimensions. Since dimensional changes are minimal within the elastic range, this distinction was unnecessary in the previous discussion.
True Strain vs Engineering Strain
The equation
$ e=\frac{\Delta L}{L_0}=\frac{L-L_0}{L_0}\tag{1} $
where $L_0$ is the initial distance between two gage marks and $L$ is the current distance between them, describes the conventional concept of unit linear strain, namely, the change in length referred to the original unit length.
$ e = \frac{\Delta L}{L_0} = \frac{1}{L_0} \int_{L_0}^{L} dL $
This definition of strain, called engineering strain or nominal strain, is satisfactory for elastic strains where $\Delta L$ is very small. However, in plastic deformation the strains are frequently large, and during the extension the gage length changes considerably. Ludwik in 1909 first proposed the definition of true strain, or natural strain, $\epsilon$, which obviates this difficulty. In this definition of strain the change in length is referred to the instantaneous gage length, rather than to the original gage length.
$ \epsilon = \frac{L_1-L_0}{L_0} + \frac{L_2-L_1}{L_1} + \frac{L_3-L_2}{L_2} + \dots \tag{2} $
or
$ \epsilon = \int_{L_0}^{L} \frac{dL}{L} = \ln \frac{L}{L_0} \tag{3} $
The relationship between true strain and conventional linear strain follows from Eq. (1).
$ e = \frac{\Delta L}{L_0} = \frac{L-L_0}{L_0} = \frac{L}{L_0} - 1 $$ e+1 = \frac{L}{L_0} $$ \epsilon = \ln \frac{L}{L_0} = \ln(e+1) \tag{4} $
The two measurements of strain give nearly identical results up to strains of about 0.1.
Because the volume remains essentially constant during plastic deformation, Eq. (3) can be written in terms of either length or area.
$ \epsilon = \ln \frac{L}{L_0} = \ln \frac{A_0}{A} \tag{5} $
Also, because of constancy of volume, the summation of the three principal strains is equal to zero.
$ \epsilon_1 + \epsilon_2 + \epsilon_3 = 0 \tag{6} $
This relationship is not valid for the principal conventional strains.
The advantage of using true strain should be apparent from the following example:
Consider a uniform cylinder which is extended to twice its original length. The linear strain is then $e = (2L_0 - L_0)/L_0 = 1.0$, or a strain of 100 percent. To achieve the same amount of negative linear strain in compression, the cylinder would have to be squeezed to zero thickness. However, intuitively we should expect that the strain produced in compressing a cylinder to half its original length would be the same as, although opposite in sign to, the strain produced by extending the cylinder to twice its length. If true strain is used, equivalence is obtained for the two cases. For extension to twice the original length, $\epsilon = \ln (2L_0/L_0) = \ln 2$. For compression to half the original length,
$\epsilon = \ln [(L_0/2)/L_0] = \ln \frac{1}{2} = -\ln 2.$
True Stress vs Engineering Stress
When analyzing the mechanical behavior of materials, it is crucial to distinguish between two key definitions of stress. While these two measures are nearly identical in the elastic region where deformations are small, their values diverge significantly during plastic deformation.
1. Engineering Stress (s) [1]
Also known as nominal or conventional stress, this is the measure most commonly used in introductory mechanics. It is calculated by dividing the applied load (P) by the specimen's original, undeformed cross-sectional area (A0). This area is a constant value measured before the test begins.
2. True Stress (σ) [2]
True stress provides a more physically accurate measure of the stress within the material at any given moment. It is defined as the applied load (P) divided by the instantaneous, actual cross-sectional area (A) of the specimen. This area changes continuously as the material elongates and necks down during the test.
Engineering vs. True Curves: A Visual Comparison
The following figure compares the engineering and true stress-strain curves for two common structural alloys. Several features are worth noting.
The two curves for each material lie almost on top of each other in the elastic region (below roughly 1% strain), confirming that the engineering and true measures are interchangeable for small deformations. Once significant plastic flow begins, the curves diverge. For the steel, the engineering curve peaks at approximately 440 MPa (the UTS) near 6% engineering strain and then falls to about 330 MPa at fracture. The true stress at that same fracture point is approximately 545 MPa (nearly 65% higher) because the rapidly shrinking neck area drives σ = P/A upward even as the load P falls. The aluminum shows the same qualitative behavior, with the true fracture
stress (approximately 415 MPa) substantially exceeding the engineering UTS (~310 MPa).
Deriving the Relationship Between True and Engineering Stress
We can derive a direct mathematical relationship to convert the easily measured engineering stress into the more physically meaningful true stress. This conversion relies on a key assumption about material behavior during plastic deformation.
We begin with the definition of true stress and use a simple algebraic manipulation to introduce the term for engineering stress. We multiply the equation by $A_0/A_0$, which is equivalent to multiplying by one:
$ \sigma = \frac{P}{A} = \frac{P}{A_0} \cdot \frac{A_0}{A} $
Notice that the term $P/A_0$ is simply the definition of engineering stress, $s$. Therefore, we can write:
$ \sigma = s \left( \frac{A_0}{A} \right) $
Since the volume of the specimen remain constant through the tests, we have A0L0 = AL and thus we can write
$ \sigma=s \frac{L}{L_0} $
Because
$ e = \frac{L - L_0}{L_0} = \frac{L}{L_0} - 1 $$ \Rightarrow \frac{L}{L_0} = 1 + e $
we obtain
$ \sigma = s (1+e)\tag{9} $
Example: A tensile test is performed on a metal specimen with an initial diameter of 15 mm. The specimen reaches a maximum load of 125 kN and then fractures at a load of 105 kN. The diameter of the necked-down region at fracture is measured to be 12.5 mm. Determine the engineering stress at maximum load (the ultimate tensile strength), the true fracture stress, the true strain at fracture, and the engineering strain at fracture.
Engineering stress at maximum load
$ \begin{aligned} \text{Engineering stress} &= \frac{P_{\max}}{A_{0}} \\ &= \frac{125 \times 10^3 \text{ N}}{\pi(15 \times 10^{-3} \text{ m})^2 / 4} \\ &= \frac{125 \times 10^3}{1.77 \times 10^{-4}}\\[6pt] &= 706 \text{ MPa} \end{aligned} $True fracture stress
$ \begin{aligned} \text{True fracture stress} &= \frac{P_{f}}{A_{f}} \\ &= \frac{105 \times 10^3 \text{ N}}{\pi(12.5 \times 10^{-3} \text{ m})^2 / 4} \\ &= \frac{105 \times 10^3}{1.23 \times 10^{-4}} \\[6pt] &= 854 \text{ MPa} \end{aligned} $True strain at fracture
$ \begin{aligned} \epsilon_f &= \ln \frac{A_0}{A_f} = \ln \left(\frac{d_0}{d_f}\right)^2 \\ &= \ln \left(\frac{15^2}{12.5^2}\right) \\[9pt] &= 2 \ln (15/12.5) = 2(0.1823) \\ &= 0.365 =36.5\% \end{aligned}$Engineering strain at fracture
$ \epsilon = \ln(1 + e)\Rightarrow \exp(\epsilon) = (1 + e)$$e = \exp(\epsilon) - 1 $$ e_f = \exp(0.365) - 1 = 1.44 - 1.00 = 0.44 $