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.
A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.
A broken white line on the pavement show that the adjacent lane is traveling in the same direction and passing is permitted.
<h3>What does pavement markings show?</h3>
Pavement markings are known to be tools that are used to pass infor or messages to roadway users.
Note that they tell the part of the road that one need to use, give information about conditions ahead, and others
Note that A broken yellow line on the pavement tells that the adjacent lane is traveling in the opposite direction and passing is permitted.
Learn more about pavement markings from
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Answer:
(a) Precipitation hardening - 1, 2, 4
(b) Dispersion strengthening - 1, 3, 5
Explanation:
The correct options for each are shown as follows:
Precipitation hardening
From the first statement; Dislocation movement is limited by precipitated particles. This resulted in an expansion in hardness and rigidity. Precipitates particles are separated out from the framework after heat treatment.
The aging process occurs in the second statement; because it speaks volumes on how heated solutions are treated with alloys above raised elevated temperature. As such when aging increases, there exists a decrease in the hardness of the alloy.
Also, for the third option for precipitation hardening; This cycle includes the application of heat the alloy (amalgam) to a raised temperature, maintaining such temperature for an extended period of time. This temperature relies upon alloying components. e.g. Heating of steel underneath eutectic temperature. Subsequent to heating, the alloy is extinguished and immersed in water.
Dispersion strengthening
Here: The effect of hearting is not significant to the hardness of alloys hardening by the method in statement 3.
In statement 5: The process only involves the dispersion of particles and not the application of heat.
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that 
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
