Question

Consider uniaxial extension of a test specimen. It has gauge length L = 22 cm (the distance between where it is clamped in the testing fixture) with a cross section of diameter D = 1.8 mm. You expect the material to have a Young’s modulus E = 205 GPa and ultimate tensile strength of σu = 420 MPa. You would like to conduct a tensile test such that the test takes 2 minutes to go from initial loading to fracture. In completing the calculations below, assume the material’s stress-strain curve is perfectly linear until fracture.
Required:
a. At what force do you expect the specimen to fracture?
b. At what strain rate should you pull the specimen, in mm/minute? Assume the material undergoes brittle fracture and its stress-strain curve is perfectly linear to fracture.
c. What is the elongation (in inches) of the specimen in the moment just prior to fracture?

Answer 1

Answer:

A) 4800.6 N

B) 0.2255 mm/min

C) 8.6792 inches

Explanation:

Given data :

gauge Length = 22 cm = 220 mm

cross section diameter/thickness = 1.8 mm

young's modulus ( E ) = 205 GPa

Ultimate tensile strength ( σu ) = 420MPa

Time = 2 mins

A) what force would fracture the specimen

σU = Ff /A

where A (area) = thickness * width ( unknown )  hence we assume a width of 6.35 mm

Ff = σU × A =  420 * 1.8 * 6.35 = 4800.6 N

B) calculate the strain rate

Ultimate tensile strength ( σU ) = εE

ε = σU / E = 420/205000 = 0.00205

also  ε ​​​​​​ =  ΔL / Li = ( Lf - Li) / Li = (Lf / Li) - 1

 therefore  ε ​​​​​​ + 1 = (Lf / Li) = 1 + 0.00205 = 1.00205

 Li = gauge length = 220 mm

Hence :  Lf = 1.00205 × gauge length =  1.00205 * 220 =  220.451 mm

  strain rate

(Lf - Li) / Time = (0.451) / 2 min = 0.2255 mm/min

C) Elongation in inches prior to fracture

ΔL = ε × Li = 0.451 mm

Lf = 220 mm + 0.451 mm = 220.451 mm =  8.6792 inches