Elongation from the true stress–true strain flow curve: 22-Mec-B8 Engineering Materials, December 2019, Q1

Question 1: Elongation from the true stress–true strain flow curve (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A metal deforming plastically in uniaxial tension, described by the Hollomon (power-law) flow curve $\sigma_{T} = K\varepsilon_{T}^{\,n}$, in which $\sigma_{T}$ is the true stress, $\varepsilon_{T}$ the true plastic strain, $K$ the strength coefficient and $n$ the strain-hardening exponent.

Given data for Question 1
Quantity	Symbol	Value
First (calibration) true stress	$\sigma_{T1}$	395 MPa
True plastic strain at that stress	$\varepsilon_{T1}$	0.425
Strain-hardening exponent	$n$	0.285
Second (applied) true stress	$\sigma_{T2}$	310 MPa
Original gauge length	$l_{0}$	300 mm

Find. The elongation $\Delta l = l_{i} - l_{0}$ of the specimen when the true stress is 310 MPa — that is, the extension the 300 mm gauge length has undergone at that point on the flow curve.

The single given pair of values fixes the strength coefficient $K$; the required strain is then read back off the same curve at 310 MPa.

Approach. Use the one given stress–strain pair to evaluate $K$, invert the flow curve at the second stress to obtain the true strain, and convert that true (logarithmic) strain into a length change through $\varepsilon_{T} = \ln(l_{i}/l_{0})$.

Evaluate the strength coefficient from the calibration point. The flow curve has two constants and the question supplies $n$, so a single measured $(\sigma_{T},\varepsilon_{T})$ pair is enough to fix the other:$$K = \frac{\sigma_{T1}}{\varepsilon_{T1}^{\,n}} = \frac{395\ \text{MPa}}{(0.425)^{0.285}} = \frac{395}{0.78359} = \boxed{504.1\ \text{MPa}}$$The value is physically sensible: $K$ is the true stress the material would carry at unit true strain, and for a work-hardening engineering alloy it sits a little above the ultimate tensile strength.
Invert the flow curve at the second stress. With $K$ and $n$ both known, the strain that accompanies any stress on the curve follows directly:$$\varepsilon_{T2} = \left(\frac{\sigma_{T2}}{K}\right)^{1/n} = \left(\frac{310}{504.1}\right)^{1/0.285} = (0.61497)^{3.5088} = 0.1816$$The lower stress necessarily returns the smaller strain, because the power-law curve rises monotonically.
Convert true strain to a length ratio. True strain is defined as the integral of $dl/l$, so it is the natural logarithm of the length ratio rather than the ratio itself. Solving for the instantaneous length,$$\varepsilon_{T} = \ln\!\left(\frac{l_{i}}{l_{0}}\right) \quad\Longrightarrow\quad l_{i} = l_{0}\,e^{\varepsilon_{T2}}$$Substituting the strain from Step 2,$$l_{i} = 300\,e^{0.1816} = 300(1.19917) = 359.75\ \text{mm}$$
Report the elongation. Subtracting the original gauge length,$$\Delta l = l_{i} - l_{0} = 359.75 - 300 = \boxed{59.8\ \text{mm}}$$which is an engineering elongation of $\Delta l/l_{0} = 19.9\,\%$ — comfortably inside the uniform-elongation range of a ductile alloy with $n = 0.285$, since necking would not begin until $\varepsilon_{T} = n = 0.285$.

Question 1 — final results
Quantity	Symbol / basis	Value
Strength coefficient	$K = \sigma_{T1}/\varepsilon_{T1}^{n}$	504.1 MPa
True plastic strain at 310 MPa	$\varepsilon_{T2}$	0.1816
Instantaneous gauge length	$l_{i} = l_{0}e^{\varepsilon_{T2}}$	359.75 mm
Elongation	$\Delta l$	59.8 mm
Engineering elongation	$\Delta l/l_{0}$	19.9%

Check: the calculation treats the quoted true strain as entirely plastic and neglects the elastic part. For a steel at 310 MPa the elastic true strain is about $310/207\,000 = 0.0015$, under one per cent of the plastic strain, so the omission changes the answer by less than 0.1 mm. It also assumes the specimen is still straining uniformly, which is guaranteed here because $\varepsilon_{T2} = 0.182$ is well below the instability strain $\varepsilon_{u} = n = 0.285$.

Concept & Theory

Topic. The Hollomon power-law description of plastic flow, and the difference between true (logarithmic) and engineering strain.

Key relations. $\sigma_{T} = K\varepsilon_{T}^{\,n}$; $\varepsilon_{T} = \ln(l_{i}/l_{0})$; $\sigma_{T} = \sigma(1+e)$ and $\varepsilon_{T} = \ln(1+e)$ relating true to engineering measures; necking at $\varepsilon_{u} = n$ (Considère).

Why this works. Once a metal yields, the flow stress it can carry rises with the accumulated plastic strain because dislocations multiply and tangle, and over the uniform-elongation range that rise is described remarkably well by a straight line on log–log axes. That straight line is the power law: its intercept is $K$ and its slope is $n$. Because the relation is a two-constant fit, one measured point plus a quoted exponent determines the whole curve, which is exactly what this question exploits. The second half of the answer rests on the definition of true strain. Engineering strain refers every extension to the original length, but the specimen is getting longer as it deforms, so the increment $dl$ should be referred to the current length $l$. Integrating $d\varepsilon = dl/l$ from $l_{0}$ to $l_{i}$ gives the logarithm, and that is why the length is recovered with an exponential rather than by simple multiplication. For the small strains of elastic design the two measures are numerically indistinguishable; at the 18 per cent strain of this problem the difference is already about two percentage points, which is why metal-forming calculations are always done in true measures.

Common pitfall. Two slips are common. The first is to treat the given 0.425 as an engineering strain and divide, which corrupts $K$. The second is to write $l_{i} = l_{0}(1+\varepsilon_{T})$, that is, to use the true strain as though it were an engineering strain; that gives 354.5 mm and a 54.5 mm elongation, about nine per cent low. A third, subtler one is to assume the answer is the elongation at fracture — it is not, it is the extension at one particular point on the loading path.

Source. Dieter, Mechanical Metallurgy, 3rd ed., §8-3 to §8-4 (true stress–strain and the flow curve); Callister & Rethwisch, 10th ed., §6.7 (true stress and strain, Eq. 6.19); Askeland & Wright, Ch. 6.

Question 1 of 7: Elongation from the true stress–true strain flow curve

Question 1: Elongation from the true stress–true strain flow curve (20 marks)