A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
1 answer:
Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above
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Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
Putting all the value to get the velocity of the flow
V = 0.5 m/s
Answer:
A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].
B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.
C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.
Compleated question:
1. Miller Indices:
a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.
b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?
c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.
Explanation:
A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:
1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]
2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).
3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.
4- Join the points.
<u>In this case, for [1 2 1]:</u>
<u>for </u><u>:</u>
B) The closest distance between planes with the same Miller indices can be calculated as:
With ,the distance is
with lattice constant a.
<u>In this case, for [1 2 1]:</u>
<u /><u />
<u>for </u><u>:</u>
C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:
dir₁=|0 0 1|
dir₂=|0 0.5 1.5|≡|0 1 3|
dir₃=|0 1 1|
dir₄=|0 1.5 0.5|≡|0 3 1|
dir₅=|0 0 1|
