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Vlada [557]
3 years ago
8

You are given a body with no body forces and told that the stress state is given as: ⎡ ⎣ 3αx 5βx2 + αy γz3 5βx2 + αy βx2 0 γz3 0

5 ⎤ ⎦ psi, where (α, β, γ) are constants with the following values: α = 1 psi/in, β = 1 psi/in2, and γ = 1 psi/in3. Does this represent an equilibrium state of stress? Assume the body occupies the domain Ω = [0, 1] × [0, 1] × [0, 1] (in inches).

Engineering
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer:

This doesn't represent an equilibrium state of stress

Explanation:

∝ = 1 , β = 1 ,  y = 1

x = 0 , y = 0 , z = 0 ( body forces given as 0 )

Attached is the detailed solution is and also the conditions for equilibrium

for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution

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Answer:

15.64 MW

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x = 15.64 MW

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8 0
3 years ago
Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They c
Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

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Effect AC

Effect AD

Effect BC

Effect BD

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Answer :

A  = significant

 B  = significant

C  = Non-significant

D  = Non-significant

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AD  = Non-significant

BC  = Non-significant

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Effect of A  = significant

Effect of B  = significant

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Effect of AD  = Non-significant

Effect of BC  = Non-significant

Effect of BD  = Non-significant

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8 0
3 years ago
Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w
shusha [124]

Answer:

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5 0
3 years ago
The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u
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Answer:

F = 0.0022N

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u = (5y-0.5y²) mm/s

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F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

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8 0
2 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
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Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

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∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

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Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

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4 0
3 years ago
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