Impact strips may be designed into a bumper cover. True False
1 answer:
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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 = 12.8 mm
Final diameter = 10.7
Engineering stress = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4( )² ; Ag = π/4(
)²
% = π/4 [ ( ( )² - (
)²) / ( π/4 (
)²) ]
= ( ( )² - (
)²) / (
)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 =
( 1 +
)
is engineering strain
= dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
= 0.431
so we substitute the value of into our initial equation;
True stress = 460 ( 1 + 0.431)
True stress = 460 (1.431)
True stress = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa
Answer:
Explanation:
Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to , where m is the mass and M the molar mass of the gas, and the density is
.
For air and
So,