Ametal 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.

K = 57904.32 lb/in²

σt = 57904.32×

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 = =  

A2 = 0.43478 in²

and

A3 =   =  

A3 = 0.3333 in²

and

engineering stress is express as

engineering stress =

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 =       ...................2

σt 2 =  

σt 2 = 28750.17 lb/in²

and

σt 3 =  

σt 3 = 42000.42 lb/in²

and

true strain will be at both state 1 and 2

ε =      ....................3

ε 2 =  

ε 2 = 0.139762

ε 3 =  

ε 3 = 0.405465

and

we know relation between true stress and stain is

σt = K ×      ........................4

put here value for both state

28750.17 = K ×       ...........................5

42000.42 = K ×     ............................6

now from equation 5 divide by  6

take ln both side

ln ( 0.68452) = n ln

n = 0.3558

so

K from equation 5 is

28750.17 = K ×  

K = 57904.32 lb/in²

and

flow curve equation is from equation 4

σt = K ×

σt = 57904.32×