Determine the amount of elastic strain induced.

[Winzer]

A steel alloy specimen having a rectangular cross section of dimensions 18.8 mm × 3.2 mm (0.7402 in. × 0.1260 in.) has the stress-strain behavior shown in the Figure. If this specimen is subjected to a tensile force of 108900 N (24480 lbf) then
(a) Determine the amount of elastic strain induced.
(b) Determine the amount of plastic strain induced.
(c) If its original length is 610 mm (24.02 in.), what will be its final length after the given strain is applied and then released?

Homework Equations

$$\sigma=\frac{F}{A}$$

The Attempt at a Solution

So I calculated at stress of about 1810.17 MPa. Which gives a strain of about 0.0225.
a) I drew a parallel line to represent the linear elastic recovery. The line intersects at about 0.0125(strain). So the elastic plastic strain is the recoverable strain right? so then this should be 0.0225-0.0125=0.01 right?
b) This is the permanent deformation which is not recoverable. 0.0125 right?

 

[PhanthomJay]

A steel alloy specimen having a rectangular cross section of dimensions 18.8 mm × 3.2 mm (0.7402 in. × 0.1260 in.) has the stress-strain behavior shown in the Figure. If this specimen is subjected to a tensile force of 108900 N (24480 lbf) then
(a) Determine the amount of elastic strain induced.
(b) Determine the amount of plastic strain induced.
(c) If its original length is 610 mm (24.02 in.), what will be its final length after the given strain is applied and then released?

Homework Equations

$$\sigma=\frac{F}{A}$$

The Attempt at a Solution

So I calculated at stress of about 1810.17 MPa. Which gives a strain of about 0.0225.

looks about right

a) I drew a parallel line to represent the linear elastic recovery. The line intersects at about 0.0125(strain).

OK

So the elastic plastic strain is the recoverable strain right? so then this should be 0.0225-0.0125=0.01 right?

No. The elastic strain portion of the 0.0225 total strain is determined from the value of the strain at the peak of the linear piece of the stress -strain curve.

b) This is the permanent deformation which is not recoverable. 0.0125 right?

it is the permanent strain that is not recoverable. It is not the plastic strain under the given load.

 

[Mene]

c) To find the final length, we can use the equation L = L0(1+e), where L0 is the original length and e is the strain induced. In this case, e = 0.0225. So the final length would be L = 610(1+0.0225) = 623.225 mm (24.526 in.).