A metal specimen with an original diameter of 0.50 in. and a gauge length of 2.75 in. is tested in tension until a fracture occurs. At the point of fracture, the diameter of the specimen is 0.30 in. and the fractured gage length is 4.20 in. Calculate the ductility in terms of percent elongation and percent reduction in area.

Percent Elongation = 52.72%

Percent Reduction in Area = 64%

Explanation:

First we find percent elongation:

Percent Elongation = {Final Gage Length - Initial Gauge Length/Initial Guage Length} x 100%

Percent Elongation = {(4.20 in - 2.75 in)/2.75 in} x 100%

Percent Elongation = 52.72%

Now, for the percent reduction in area:

Percent Reduction in Area = {Final Cross Sectional Area - Initial Cross Sectional Area|/Initial Cross Sectional Area Length} x 100%

Percent Reduction in Area = {π(0.3 in)² - π(0.5 in)²/π(0.5 in)²} x 100%

Percent Reduction in Area = - 64%

here, negative sign shows a decrease in area.

how is this a question? in order for me to answer it i have to understand the question

explanation:

answer:

you cant but you can report the sender

and do you want to talk?

im xavier btw but i go by x

explanation: