Question

A metal is plastically deformed in a tensile test. The starting specimen had a gage length = 2.0 in and an area = 0.50 in2 . At one point during the test, the gage length = 2.3 in, and the corresponding engineering stress = 25,000 lb/in2 . At another point prior to necking, the gage length = 3.0 in, and the corresponding engineering stress = 28,000 lb/in2 . Determine the strength coefficient and the strain-hardening exponent for this metal.

Answer 1

Answer:

K = 57904.32 lb/in²

σt = 57904.32×$\epsilon^{0.3558}$

Explanation:

given data

gage length l1 = 2 in

area A1 = 0.50 in²

gage length l2 = 2.3 in

engineering stress σ1 = 25000 lb/in²

gage length l3 = 3 in

engineering stress σ2 = 28,000 lb/in²

to find out

strength coefficient and the strain-hardening exponent for this metal

solution

we know that volume v of work is constant so

v1 = v2 = v3

so we can say A1×l1 = A2×l2 = A3×l3     ............1

here v is volume , A is area and l is length

so

we find first area of cross section (A) at both state 1 and 2

use equation 1

A1×l1 = A2×l2 = A3×l3

A2 = $\frac{A1*l1}{l2}$ =  $\frac{0.5*2}{2.3}$

A2 = 0.43478 in²

and

A3 =  $\frac{A1*l1}{l3}$ =  $\frac{0.5*2}{3}$

A3 = 0.3333 in²

and

engineering stress is express as

engineering stress = $\frac{P}{A}$

here P is true load and A is area of cross section so

and P = engineering stress × A

so

true stress will be at both state 1 and 2

σt =  $\frac{P}{A}$     ...................2

σt 2 =  $\frac{25000*0.5}{0.43478}$

σt 2 = 28750.17 lb/in²

and

σt 3 =  $\frac{28000*0.5}{0.3333}$

σt 3 = 42000.42 lb/in²

and

true strain will be at both state 1 and 2

ε = $ln\frac{l}{l0}$      ....................3

ε 2 = $ln\frac{2.3}{2}$  

ε 2 = 0.139762

ε 3 = $ln\frac{3}{2}$  

ε 3 = 0.405465

and

we know relation between true stress and stain is

σt = K ×$\epsilon^{n}$      ........................4

put here value for both state

28750.17 = K ×$0.139762^{n}$       ...........................5

42000.42 = K ×$0.405465^{n}$     ............................6

now from equation 5 divide by  6

$\frac{28750.17}{42000.42} = \frac{K(0.139762)^n}{K(0.405465)^n}$

take ln both side

ln ( 0.68452) = n ln $(\frac{0.139762}{0.405465})$

n = 0.3558

so

K from equation 5 is

28750.17 = K ×$0.139762^{0.3558}$  

K = 57904.32 lb/in²

and

flow curve equation is from equation 4

σt = K ×$\epsilon^{n}$

σt = 57904.32×$\epsilon^{0.3558}$