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nasty-shy [4]
2 years ago
7

Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s

tress states at various critical locations. Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using. These problems are designed to test your understanding of out-of-plane principal stresses!
Engineering
1 answer:
suter [353]2 years ago
3 0

Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”

Stress states at various critical locations are f= 2.662.

Read more about strain:

brainly.com/question/6390757

#SPJ1

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4 0
3 years ago
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Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012
devlian [24]

The pressure drop of air in the bed is  14.5 kPa.

<u>Explanation:</u>

To calculate Re:

R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}

From the tables air property

\mu_{394 k}=2.27 \times 10^{-5}

Ideal gas law is used to calculate the density:

ρ = \frac{2.2}{2.83 \times 10^{-3} \times 394.3}

ρ = 1.97 Kg / m^{3}

ρ = \frac{P}{RT}

R = \frac{R_{c} }{M} = 8.2 × 10^{-5} / 28.97×10^{-3}

R = 2.83 × 10^{-3} m^{3} atm / K Kg

q is expressed in the unit m/s

q=\frac{2.45}{1.97}

q = 1.24 m/s

Re = \frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}

Re = 2278

The Ergun equation is used when Re > 10,

\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}

\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24 +\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0
3 years ago
What’s the difference between engineering stress and strain and true stress and strain
Nana76 [90]

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:

Explanation:

4 0
3 years ago
Read 2 more answers
Which 1 of the following did women NOT do during WWII?
Anastaziya [24]

Answer:

Didn't do what, there are no women as answers. Contact me and give me the women.

Explanation:

3 0
3 years ago
Read 2 more answers
A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t
Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

tan\phi=\frac{\bigtriangleup l}{h}

tan\phi=\frac{0.08}{1.38}

tan\phi= 0.05797

For small angle of \phi, tan\phi can take as\phi.

\phi = 0.05797 rad.

Thus, the shear strain is 0.05797 rad.

7 0
3 years ago
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