Respuesta :
Answer:
a) %RA = 22.3 %
b) σT = 5.92 *10^8 N/m² (= 592 MPa)
Explanation:
Step 1: Data given
The original diameter = 11.8 mm
The original cross-sectional area A0= radius² * π
⇒ (11.8/2)² *π
Engineering fracture strength = 460 MPa
The cross-sectional diameter at fracture = 10.4 mm
The cross-sectional area at the point of fracture Afrac = radius² * π
⇒ (10.4/2)² *π
Step 2: The ductility in terms of percent reduction in area.
%RA = ((A0 - Afrac)/A0) *100%
⇒ with A0 = 11.8 mm
⇒ with Afrac = 10.4 mm
%RA = (((11.8mm/2)² *π -(10.4mm/2)² * π)/(11.8mm/2)² * π) *100 %
%RA = (109.36-84.95)/109.36 *100 %
%RA = 22.3 %
Step 3: The true stress at fracture:
True stress = the load F divided by the cross-sectional area over where deformation is occurring.
F = σf * A0
⇒ with σf = 460 *10^6 N/m²
F = 460 *10^6 * 109.36 mm² *(1m²/10^6 mm²)
F = 50305.6 N = 50.3 *10³ N
True stress σT = F/Afrac
σT = 50305.6 N / (84.95mm² * (1m²/10^6 mm²))
σT = 50305.6 N / 0.00008495 m²
σT = 592178928.78 = 5.92 *10^8 N/m² (= 592 MPa)
