A tension test is carried out on an Al alloy specimen which has an original diameter of 0.505 in and an original gauge length of
2 in. After fracture, the diameter at the fractured neck is measured to be 0.37 in and the gauge length after fracture is 2.32 in. The maximum load during the test was 37 kN whereas at fracture the load was 34 kN.
a. What is the tensile strength?
b. What are the engineering and true stresses at fracture?
c. What are the engineering and true strain at fracture?
d. What is the ductility of this metal?
a. What is the tensile strength?
b. What are the engineering and true stresses at fracture?
c. What are the engineering and true strain at fracture?
d. What is the ductility of this metal?
1 answer:
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This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ =
/ t₂
t₂ = t₁
t₂ = t₁ /
t₂ = ( /
)t₁
t₂ =
/
× t₁
so we substitute
t₂ = 0.0049 / 0.0018
× 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Answer:
A) About newtons
B) 76.518 newtons
C) 111.834 newtons
Explanation:
A) , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:
B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.
C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.
Hope this helps!