Answer:
The time taken will be "1 hour 51 min". The further explanation is given below.
Explanation:
The given values are:
Number of required layers:
= 
= 
Diameter (d):
= 1.25 mm
Velocity (v):
= 40 mm/s
Now,
The area of one layer will be:
= 
= 
The area covered every \second will be:
= 
= 
= 
The time required to deposit one layer will be:
= 
= 
The time required for one layer will be:
= 
∴ Total times required for one layer will be:
= 
= 
So,
Number of layers = 152
Therefore,
Total time will be:
= 
= 
= 
The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.
<u>Explanation</u>:
<u>Given</u>:
tensile stress is applied parallel to the [100] direction
Shear stress is 0.5 MPA.
<u>To calculate</u>:
The magnitude of applied stress in the direction of [101] and [011].
<u>Formula</u>:
zcr=σ cosФ cosλ
<u>Solution</u>:
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)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa
Answer:
% reduction in area==PR=0.734=73.4%
% elongation=EL=0.42=42%
Explanation:
given do=12.8 mm
df=6.60
Lf=72.4 mm
Lo=50.8 mm
% reduction in area=((
*(do/2)^2)-(*(df/2)^2)))/*(do/2)^2
substitute values
% reduction in area=73.4%
% elongation=EL=((Lf-Lo)/Lo))*100
% elongation=((72.4-50. 8)/50.8)*100=42%
