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Answer:
a) The ductility = -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) the true stress at fracture is 658.26 Mpa
Explanation:
Given that;
Original diameter [tex]d_{o}[/tex] = 12.8 mm
Final diameter [tex]d_{f}[/tex] = 10.7
Engineering stress [tex]\alpha _{E}[/tex] = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4([tex]d_{o}[/tex] )² ; Ag = π/4([tex]d_{f}[/tex] )²
% = π/4 [ ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ( π/4 ([tex]d_{o}[/tex] )²) ]
= ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ([tex]d_{o}[/tex] )² × 100
we substitute
= [( (10.7)² - (12.8)²) / (12.8)² ] × 100
= [(114.49 - 163.84) / 163.84 ] × 100
= - 0.3012 × 100
= -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) The true stress at fracture;
True stress [tex]\alpha _{T}[/tex] = [tex]\alpha _{E}[/tex] ( 1 + [tex]E_{E}[/tex] )
[tex]E_{E}[/tex] is engineering strain
[tex]E_{E}[/tex] = dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
[tex]E_{E}[/tex] = 0.431
so we substitute the value of [tex]E_{E}[/tex] into our initial equation;
True stress [tex]\alpha _{T}[/tex] = 460 ( 1 + 0.431)
True stress [tex]\alpha _{T}[/tex] = 460 (1.431)
True stress [tex]\alpha _{T}[/tex] = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa