Answer: LED have the lowest cost of operation.
Explanation:
If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by
Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.
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:
The answer is 0.4490Mpa
Explanation:
Given :
P in direction of 001 , P = 1.1MPa
slip plane = 111, slip plane normal direction of 111,
slip direction = 101
= δcosФcosλ
= (001 * 101) = 1 = cosλ
(111 * 001) 1 cosФ
= 1.1 MPa * *
1.1 Mpa * 1/1.4142 * 1/1.7320
1.1 Mpa * 0.7072 * 0.5773
= 0.4490 MPa
therefore the resolved shear stress = 0.4490MPa