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azamat
2 years ago
15

If the Poisson’s ratio of a 5 mm X 5 mm titanium alloy pin is 0.31 and it is elastically loaded

Engineering
1 answer:
leonid [27]2 years ago
7 0

The new dimensions of the titanium alloy pin will be that the width is 0.0775 mm and the length is 4.9225m.

<h3>What is Poisson's ratio?</h3>

The Poisson's ratio is the proportion of a material's change in width per unit width to its change in length per unit length due to strain. In order for a stable, isotropic, linear elastic material to have a positive Young's modulus, shear modulus, and bulk modulus, the Poisson's ratio must be between 1.0 and +0.5. Poisson's ratio values for the majority of materials fall between 0.0 and 0.5.

The formula for the longitudinal strain is:

= Change in length / Initial length

Based on the information, the longitudinal strain will be:

= 105 - 100 / 100

= 0.05

Poisson ratio will be illustrated as the change in the width divided by the longitudinal strain. :

0.31 = ∆w/5 / 0.05

∆w = 0.0775 mm

New side length will be the difference in the changes in the dimensions:

= w - ∆w

= 5 - 0.0775

= 4.9225m

Learn more about Poisson on:

brainly.com/question/7879375

#SPJ1

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Explanation:

7 0
3 years ago
In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)
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Answer:

(b) Constant (minimum) volume

Explanation:

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So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

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3 years ago
Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i
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Answer:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

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Explanation:

The internal energy is equal to:

dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV

The entropy is equal to:

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If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }

Replacing:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

4 0
3 years ago
The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight
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Answer:

hello your question lacks the required question attached below is the missing diagram

Forces in GJ = -4.4444 i.e. 4.4444 tons

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Explanation:

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

attached below is the detailed solution

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3 years ago
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