For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains prior to necking:
1 answer:
Answer:
<h2><em> </em>The engineering stress necessary to produce an engineering strain of 0.28 is 339 Mpa</h2>Explanation:
<em>On the basis of this information, we are going to compute the engineering stress necessary to produce an engineering strain of 0.28.</em>
<h2>Engineering Engineering </h2><h2> Stress (MPa) Strain</h2><h2>315 0.105 </h2><h2>340 0.220 </h2>For clarity and you will find the solving in the attached file for your reference.
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GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
=
Formula Used:
Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:
so,
= 75.39 rad/s
Using the given formula:
Damping is negligible, so,
will give the tranfer function
Therefore,
=
=
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
