Write a matrix, that is a lower triangular matrix.
1 answer:
Answer:
Explanation:
A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.
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Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence
Therefore, tension in the cable,
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then
Similarly,
Therefore, both reactions at A and D are 73.575 N
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