Determine the amount of elastic strain induced.

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SUMMARY

The discussion focuses on determining the elastic and plastic strain induced in a steel alloy specimen with a rectangular cross-section of 18.8 mm × 3.2 mm, subjected to a tensile force of 108900 N. The calculated stress is approximately 1810.17 MPa, resulting in a total strain of 0.0225. The elastic strain is determined to be 0.0125, while the plastic strain, representing permanent deformation, is clarified as not being recoverable and is also 0.0125. The final length of the specimen after applying and releasing the strain is not explicitly calculated in the discussion.

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Winzer
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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?
 

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Winzer said:
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.
 

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