---
title: "Flow curve constants K and n"
subject: "Manufacturing and Production Processes"
chapter: 2
key: "semester_4/manufacturing_and_production_processes:flow-curve-strength-coefficient-numerical"
source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/flow-curve-strength-coefficient-numerical
site: BCT Engineering
---

# Flow curve constants K and n

Manufacturing and Production Processes · Chapter 2: Manufacturing Properties of Materials

- Asked in: year not given, 5 marks
- Video: [Understanding True Stress and True Strain](https://www.youtube.com/watch?v=AkX6JqlWRqc) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/flow-curve-strength-coefficient-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-2

## Question

For a metal the flow curve is $\sigma = K\varepsilon^n$. Two points on it are: true stress 300 MPa at true strain 0.05 and true stress 450 MPa at true strain 0.20. Determine K and n, the flow stress at true strain 0.40, and the average flow stress for a deformation from zero to 0.40 strain. Explain what the value of n indicates.

## Short answer

Divide stress ratios to get n = ln1.5/ln4 = 0.2925 and K = 720.5 MPa. Flow stress at strain 0.40 is 551 MPa; average flow stress from 0 to 0.40 is 426 MPa. n, the strain-hardening exponent, indicates the material's capacity for uniform straining before necking.

## Answer

The flow curve is the plastic part of the true stress-true strain curve: $\sigma = K\varepsilon^{n}$, where K is the **strength coefficient** (MPa) and n the **strain-hardening exponent**.

#### Finding n and K

Divide the two readings:

$$
\frac{450}{300} = \left(\frac{0.20}{0.05}\right)^n \quad\Rightarrow\quad 1.5 = 4^n
$$

$$
n = \frac{\ln 1.5}{\ln 4} = \frac{0.4055}{1.3863} = 0.2925
$$

$$
K = \frac{300}{(0.05)^{0.2925}} = \frac{300}{0.4164} = 720.5\ \text{MPa}
$$

#### Flow stress at $\varepsilon = 0.40$

$$
\sigma = 720.5\,(0.40)^{0.2925} = 720.5\times 0.7649 = 551.1\ \text{MPa}
$$

#### Average flow stress from 0 to 0.40

$$
\bar{Y}_f = \frac{K\varepsilon_{max}^{\,n}}{1+n} = \frac{551.1}{1.2925} = 426.4\ \text{MPa}
$$

This average value is used to compute forming forces and work in rolling, drawing and forging, because the metal strain hardens during the process.

#### Meaning of n

- n is the slope of the flow curve on a log-log plot. It lies between 0 (perfectly plastic) and 1 (perfectly elastic).
- A large n (about 0.2-0.5, e.g. annealed copper, stainless steel) means strong strain hardening; the metal stretches uniformly and is good for stretch forming and deep drawing.
- A small n (about 0.05-0.15, e.g. cold-worked steel) means little hardening and early necking.
- In a tensile test, the true strain at the onset of necking equals n.

**Answer:** $n = 0.2925$, $K = 720.5$ MPa, flow stress at 0.40 strain $= 551$ MPa, average flow stress $= 426$ MPa.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is n found from the two points (300 MPa at 0.05, 450 MPa at 0.20)?
   A. Subtract the stresses and divide by strain difference
   B. Divide the equations: 1.5 = 4^n, so n = ln1.5/ln4 = 0.292
   C. n = 450/300 = 1.5
   D. n = 0.20/0.05 = 4
   Answer: B. Dividing sigma = K epsilon^n for the two points eliminates K, giving 1.5 = 4^n and n = 0.292. (Key point: Finding n)

2. What is the formula for average flow stress from zero to maximum strain?
   A. K epsilon_max^n x (1+n)
   B. K epsilon_max^n / (1+n)
   C. K epsilon_max^n / n
   D. K epsilon_max^(n+1)
   Answer: B. Y_f(avg) = K epsilon^n / (1+n) = 551.1/1.292 = 426.6 MPa. (Key point: Average flow stress)

3. A large strain-hardening exponent n indicates
   A. Little hardening and early necking
   B. Strong strain hardening, uniform stretching, good for deep drawing
   C. A perfectly plastic metal
   D. A perfectly elastic metal with no plastic range
   Answer: B. A large n (0.2-0.5) means strong strain hardening and uniform stretching; small n means early necking. (Key point: Meaning of n)

4. In a tensile test the true strain at the onset of necking equals
   A. K
   B. 1/n
   C. n
   D. 0.5
   Answer: C. The true strain at necking onset equals the strain-hardening exponent n. (Key point: n and necking)

## Related in Manufacturing and Production Processes

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