Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
![Y=\frac {K}{\sigma_c \sqrt {a\pi}}](https://tex.z-dn.net/?f=Y%3D%5Cfrac%20%7BK%7D%7B%5Csigma_c%20%5Csqrt%20%7Ba%5Cpi%7D%7D)
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness,
is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
![Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682](https://tex.z-dn.net/?f=Y%3D%5Cfrac%20%7BK%7D%7B%5Csigma_c%20%5Csqrt%20%7Ba%5Cpi%7D%7D%3D%20%5Cfrac%20%7B40%7D%7B300%5Csqrt%20%7B%280.002%2A%5Cpi%29%7D%7D%3D1.682)
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for
while a is taken as 0.003m and Y is already known
![K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa](https://tex.z-dn.net/?f=K%3D260%2A1.682%2A%5Csqrt%20%7B0.003%2A%5Cpi%7D%3D42.455%20Mpa)
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
Answer: C.) John Herschel
Answer:
B) the liquid accelerated to high velocities.
<em>I</em><em> </em><em>hope</em><em> </em><em>this helps</em><em> </em>
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
![\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}](https://tex.z-dn.net/?f=%5Cfrac%7BV2%7D%7BV1%7D%20%3D%5Cfrac%7B%5Calpha1%20%7D%7B%5Calpha%202%7D)
Now we can use the equation that defines the percentage of increase, in this case for speed
![i=(\frac{V2}{V1} -1) 100](https://tex.z-dn.net/?f=i%3D%28%5Cfrac%7BV2%7D%7BV1%7D%20-1%29%20100)
Now we use the equation obtained in the previous step, and replace values
![i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65](https://tex.z-dn.net/?f=i%3D%28%5Cfrac%7B%5Calpha1%20%7D%7B%5Calpha%202%7D%20-1%29%20100%5C%5Ci%3D%28%5Cfrac%7B1.2%7D%7B0.955%7D%20-1%29%20100%3D25.65)
the percent increase in the velocity of air is 25.65%
Answer:
1.44 mm
Explanation:
Compute the maximum allowable surface crack length using
where E is the modulus of elasticity,
is surface energy and
is tensile stress
Substituting the given values
The maximum allowable surface crack is 1.44 mm