Question

A metal specimen with an original diameter of 0.500 in.and a gage length of 2.000 in. is tested in tension until fracture occurs.At the point of fracture, the diameter of the specimen is 0.260in. and the fractured gage length is 3.08 in. Calculate the ductilityin terms of percent elongation and percent reduction in area.

Answer 1

Answer:

percent elongation is 54%

reduction area is 73%

Explanation:

original diameter = 0.500 in

gauge length = 2.000 in

diameter of the specimen is 0.260 in

fractured gauge length = 3.08 in

solution

we get here ductility in terms of percent elongation that is express as

elongation = $\frac{\delta }{L}$    ........1

elongation = $\frac{3.08-2}{2}$  

elongation = 0.54

so  percent elongation is 54%

and

now we get cross section of this section is

Ao = $\frac{\pi }{4}$ × 0.5²  

Ao = 0.1963 in²

so now calculate final cross section area

A(f) =  $\frac{\pi }{4}$ × 0.26²  

A(f) = 0.0530 in²

so percent reduction in area will be

percent reduction  area = $\frac{Ao-A(f)}{Ao}$     ..............2

percent reduction  area  = $\frac{0.1963 - 0.0530 }{0.1963}$  

percent reduction  area  = 73%

there is reduction area is 73%