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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Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
The Biot number is given as;
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
The time for the heating process is given as;
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.