How do you make coke for steel?

When do you know if you start volly ball

igomit [66]

Answer You ask your coach

7 0

3 years ago

What do u call a bad bird

statuscvo [17]

Answer:

A buzzerd

Explanation:

4 0

2 years ago

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

Nataly [62]

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

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making Y the subject

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}$

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, $\sigma_c$ is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682$

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making K the subject

$K=\sigma_c Y \sqrt {a\pi}$ and substituting 260 MPa for $\sigma_c$ while a is taken as 0.003m and Y is already known

$K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa$

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

6 0

3 years ago

Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape

Taya2010 [7]

Answer: ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

l₀l₀l₀=l₁l₂l₃

$\frac{lo*lo*lo}{l1*l2*l3}=1.0$

taking natural log on both sides

$ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)$

Considering the logarithmic Laws of division and multiplication :

ln(AB) = ln(A)+ln(B)

ln(A/B) = ln(A)-ln(B)

$ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0$

Use the image attached to see the definition of true strain defined as

ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0

8 0

2 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$