Question

1. The strength coefficient and strain-hardening exponent of brass are 100,000 lb/in2 and 0.35 respectively. A cylindrical specimen of the metal with starting diameter of 2.5 in. and height of 3.0 in. is compressed to a length of 1.5 in. Determine the flow stress at this compressed length and the average flow stress that the metal has experienced during deformation.

Answer 1

To solve this problem we will apply the concepts of strain, flow stress and average flow stress to find the required data. We will start by calculating the Strain which is the logarithmic relationship between the longitudinal change. Later we will find the flow stress through the strength coefficient, the strain and the strain-hardening exponent. Finally with the found values it will be possible to find the average flow stress,

Now the strain is calculated with the logaritmic relation of the lengths.

$\epsilon = ln(\frac{1.5}{3.0})$

$\epsilon = ln(0.5)$

$\epsilon = 0.69315in/in$

With this value we can calculate the flow stress,

$Y_f = K\epsilon^n$

Here,

K = Strneght coefficient

n = Strainhardening exponent of brass

$Y_f = (100000)(0.69315)^{0.35}$

$Y_f = 87961lb/in^2$

Finally the average flow stress will be given under the relation:

$\bar{Y_f} = \frac{K\epsilon^n}{1+n}$

$\bar{Y_f} = \frac{(100000)(0.69315)^{0.35}}{1.35}$

$\bar{Y_f} = 65156lb/in^2$