A hub a signal that refreshes the signal strength.
1 answer:
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Answer:
a. 2x/3
b. 8
Explanation:
fundamental period can be defined to mean that at after every period of 2π radians or 360° the value of graph is repeated. For such functions the fundamental period is the period after which they repeat themselves.
It van also be looked as The fundamental period of cos(θ) is 2π. That is (for example) cos(0) to cos(2π) represents one full period.
Please see attachment for the step by step solution.
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