Answer:
Both Brass and 1040 Steel maintain the required ductility of 20%EL.
Explanation:
Solution:-
- This questions implies the use of empirical results for each metal alloy plotted as function of CW% and Yield Strength.
- So for each metal alloy use the attached figures as reference and determine the amount of CW% required for a metal alloy to maintain a Yield Strength Y = 345 MPa.
- Left Figure (first) at Y = 345 MPa ( y -axis ) and read on (x-axis):
1040 Steel --------> 0% CW
Brass ---------------> 22% CW
Copper ------------> 66% CW
The corresponding ductility (%EL) for cold Worked metal alloys can be determined from the right figure. Using the %CW for each metal alloy determined in first step and right figure to determine the resulting ductility.
- Right Figure (second) at respective %CW (x-axis) read on (y-axis)
1040 Steel (0% CW) --------> 25% EL
Brass (22% CW) -------------> 21% EL
Copper (66% CW) ----------> 4% EL
We see that both 1040 Steel and Brass maintain ductilities greater than 20% EL at their required CW% for Yield Strength = 345 MPa.
The Bidirectional Shift Register is the shift register that allows one to write a 5bit register if the series is shifted from left to right.
<h3>What is a value of a 5-bit register?</h3>
A 5-bit register is valuable because it can represent up to 32 items. It is to be noted that a bit is a digit that is binary that is indicative of two states.
While a 5-bit register can have a possible number of 32 values,
- 6 can hold 64;
- 7 can hold 128;
- 8 can hold 256 etc.
Learn more about registers at;
brainly.com/question/19091159
#SPJ1
Answer:
Tech A is correct
Explanation:
Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.
Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.
Answer:
The bending stress is 502.22 MPa
Explanation:
The diameter of the pinion is equal to:
Where
m = module = 5
Np = number of teeth of pinion = 26
= 0.13 m
The pitch line velocity is equal to:
Where
wp = speed of the pinion = 1800 rpm
The factor B is equal to:
The factor A is equal to:
A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82
The dynamic factor is:
The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables
JR = 0.41
Ka = 1
Kb = 1
Ks = 1
Ki = 1.42
Km = 1.7
The diametrical pitch is equal to:
The bending stress is equal to:
Answer:
77.2805 μF
Explanation:
Given data :
V = 2460 V
Q = 191 Kva
<u>Calculate the size of Each capacitor </u>
first step : calculate for the value of Xc
Q = V^2/ Xc
Xc ( capacitive reactance ) = V^2 / Q = 2460^2 / ( 191 * 10^3 ) = 31.683 Ω
Given that 1 / 2πFc = 31.683
∴ C ( size of each capacitor ) = 
= 77.2805 μF
