The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.
Explanation:
Given:
tensile stress is applied parallel to the [100] direction
Shear stress is 0.5 MPA.
To calculate:
The magnitude of applied stress in the direction of [101] and [011].
Formula:
zcr=σ cosФ cosλ
Solution:
For in the direction of 101
cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)
cos λ = 1/√2
The magnitude of stress in the direction of 101 is 12.25 MPA
In the direction of 011
We have an angle between 100 and 011
cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)
cosλ = 0
Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.
Answer:
a) V_2 = 82.1 m/s
b) m = 0.298 Kg/s
Explanation:
from A-11 to A-13 we have the following data
P_1 = 600 kpa
V_1 = 0.033925 m^3/kg
h_1 = 262.52 kJ/kg
P_2 = 700 kpa
V_2 = 0.0313 m^3/kg
T_2 = 40°C = 313K
h_2 = 278.66 kJ/kg
Now, from the conversation of mass,
A_2*V_2/u_2 = A_1*V_1/u_1
V_2 = A_1/A_2*u_2/u_1*V_1
V_2 = A_1/1.8*A_1 * 0.0313 /0.033925*160
V_2 = 82.1 m/s
now from the energy balance equation
E_in = E_out
Q_in + m(h_1 + V_1^2/2) = m(h_2 + V_2^2/2)
m = 0.298 Kg/s
Answer:
1. Assumption
2.Truth Table
3.K-Map
4.Mapping and Final Expression
Answer:
See explanations for completed answers
Explanation:
Given that; The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diameter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa
See attachment for completed solvings