High School

During a tensile test, the starting gauge length is 100.0 mm, and the cross-sectional area is 150 mm\(^2\). During testing, the following force and gauge length data are collected:

1. 17,790 N at 100.2 mm
2. 23,040 N at 103.5 mm
3. 27,370 N at 110.5 mm
4. 28,910 N at 122.0 mm
5. 27,400 N at 130.0 mm
6. 20,460 N at 135.1 mm

The final data point (6) occurred immediately prior to failure. Yielding occurred at a load of 19,390 N (0.2% offset value), and the maximum load (4) was 28,960 N.

Tasks:
(a) Plot the engineering stress-strain curve.
(b) Determine the yield strength.

Answer :

a. the engineering stress strain curve is attached.

b. the yield strength of the material is 129.27 MPaa.

Engineering stress is described as the force applied divided by the original cross-sectional area.

engineering_stress = force / 150 mm²

Engineering strain on the otherhand is described as the change in length divided by the original length.

engineering_strain = (gage_length - 100.0 mm) / 100.0 mm

The 0.2% offset yield strength is the stress at which the engineering strain is 0.2% greater than the strain at the yield point and the yield point is the point on the stress-strain curve where the material begins to deform plastically.

From the curve attached, the yield point appears to be around 19390 N.

the strain at the yield point =

(19390 N / 150 mm²) / 100.0 mm

= 0.12927.

So we have that 0.2% of the strain at the yield point

= 0.12927 * 0.002

= 0.00025854.

0.12927 + 0.00025854

= 0.12952854.

From the stress-strain curve, we see that the stress corresponding to a strain of 0.12952854 is 129.27 MPa.

Other Questions