Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3
1 answer:
Answer:
The answer is NO. applied stress if 12MPa will not cause the single crystal to yield.
Explanation:
Given ;
Ф =
∧ =
critical resolved shear stress = 6.2MPa
Applied stress = 12MPa
= cosФ cos∧ = 12 * cos
* cos
= 12* 0.5 * 0.8191
= 4 . 9146MPa
The resolved shear stress (4.9146MPa) is lee than the critical resolved shear stress which is 6.2MPa. therefore, the single critical will not yield.
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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.
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
