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
