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

K = 57904.32 lb/in²

σt = 57904.32×
`\epsilon^(0.3558)`

Step-by-step 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 =
`(A1*l1)/(l2)` =
`(0.5*2)/(2.3)`

A2 = 0.43478 in²

and

A3 =
`(A1*l1)/(l3)` =
`(0.5*2)/(3)`

A3 = 0.3333 in²

and

engineering stress is express as

engineering stress =
`(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 =
`(P)/(A)` ...................2

σt 2 =
`(25000*0.5)/(0.43478)`

σt 2 = 28750.17 lb/in²

and

σt 3 =
`(28000*0.5)/(0.3333)`

σt 3 = 42000.42 lb/in²

and

true strain will be at both state 1 and 2

ε =
`ln(l)/(l0)` ....................3

ε 2 =
`ln(2.3)/(2)`

ε 2 = 0.139762

ε 3 =
`ln(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

`(28750.17)/(42000.42) = (K(0.139762)^n)/(K(0.405465)^n)`

take ln both side

ln ( 0.68452) = n ln
`((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)`