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(20 p) Consider a constitutive equation of the form: 0 = Ac0.2(1 + 0.01l0ge)(1 - (T/Tm)13)07 where A is a constant with units of stress. Tm is the absolute melting ...

Question

(20 p) Consider a constitutive equation of the form: 0 = Ac0.2(1 + 0.01l0ge)(1 - (T/Tm)13)07 where A is a constant with units of stress. Tm is the absolute melting temperature Determine the strain at instability using the Considere criterion:

(20 p) Consider a constitutive equation of the form: 0 = Ac0.2(1 + 0.01l0ge)(1 - (T/Tm)13)07 where A is a constant with units of stress. Tm is the absolute melting temperature Determine the strain at instability using the Considere criterion:



Answers

Develop an expression for the variation in temperature with pressure in a constant entropy process, $(\partial T / \partial P)_{s},$ that only includes the properties $P-v-T$ and the specific heat, $C_{p} .$ Follow the deyelopment of Eq. 13.32.

Transferred to a gas mixture contained in a piston cylinder discussed in the previous problem with the total entropic change here. Also in Tripoli change. The ex energy destruction are to be determined for two places. So when we understand this case and totally generally during the process is determined by applying the intro ph balance on the extended system. Okay that includes the question cylinder device and its immediate surroundings so that the boundary temperature of the extended system of them, environment temperature at all times it remains the same. Therefore we can find out that. Yeah, As in -1 out. This question was to delta A. System that is from this weekend. Find it out like that. Q. In upon T boundary plus as jim Equals two Delta S. Yeah that is as jen equals to em off S. Two minus S. One minus You. M. seven. Then exerted destroyed during the process from its deficient. It's a girl strong it destroyed. That is caused your T. zero into estrogen. Now noting the total mixture pressure and thus the partial pressure of each gas remains constant. The entropy change of accompaniment in the mixture during the process is Delta as I equals two AM I. Into C. P. And then T. Two upon T. One minus R. L. N. P. Two upon P. One. I. It is equals two AM I. C. P. I. Ln T two upon T. One. So assuming the ideal gas behavior and the values of C. P. The delta S of H. Two and N. Two can be determined by Delta. As of H. two ideal was to 18.21 kg two kelvin & Delta s. n. two. I did. Is it calls to you? 4.87 you know, don't work. Okay. No. As gen Equals to 8.496. Hello to kelvin. So putting that in the excel energy destroyed. We get 2489. He looked to No. You think that Megan's law finding out there is that one is the best 1? Is it has to for the cases we get that. It's gin using a mega slow. We have done using. I'm like at slow we get as gentle B nine 9.390 kg joules per kelvin and X destroyed was 22751 kg two. Yes, So I hope you understood the question. Thank you.

Indiscretion. We have to find the temperature rights that will cause uh the failure at the joint at the glue joint of the metal block that has the allowable maximum shear stress at two Z. So the figure on the right it is committed of the um metal block. We can see that there is a confinement in why direction after block. And that's gonna limit the cheer strain at wind direction to be zero. And from that we can use hopes law to calculate the spring, the strength at wind direction. And you can see from this equation, the hook's law, we can see that there is no straight in X. And Z. Direction because there's no flaws applied. So um and we know that we know E. And the alpha because it's the property that uh they gave us from the problem. So we can calculate the uh stretch in y. Direction which is minus point oh six. There are the T. And from that we can use a more circle to let me in the cheer strength. Um Now to find the circle we have to determine the center of the circle and the radius of the circle. The center of the circle is just the average value of the switch and zero and the average is my last point over six. The other T. Which is um sigma Y plus sigma eggs which is zero but by two. Yeah so Cinda will be at minus point or three. Yeah the T. And C. E. O. Now to find the radius we have to have another reference point which is point A point A. Is A. Is the absalon. Why? And uh how X. Y. And we know that this to where you are zero. So the radius it's just the this time between A. And Z. A. And Z. Yes ceo my next point or three they are the T. Square plus still and it is 0.43 that the T. And from that we can draw the circle. This is the center 0.3 to the T. So and as this where it is. Okay. And now from from this diagram we can see that the X. That direction yes in this direction and this direction is X. Direction. So uh the value uh the angle here is 1 40 be great. Now that means the tao Xy will be at this orientation that has then go of 280 degree which is 1 40 time to And from that we can calculate. So is that why dad? Which is equal to five oh three? They're dirty. And so this angle will be 80 degree, right? It's going to be signed 80 degree. That equals point 02 nine 54 There. That T. And allowable shear straight have to be equal to extent why that cheer straight to cost failure. That means to equals this number. And you have that that C is 67.7 for and high and this is the answer.

In this problem, valuation of length with temperature is given by reality and it is equal to and not one plus alpha Dlt solving it for the I can write DLL by and not is equal to Alpha guilty, which is equal to eat and we know that is equal to side by. Capitally so mm size equal to alpha. Dirty capital E, which is the south distress of pressure. Now, using all the given data from the appendix and taken guilty, they call 200°C. We get the value of The stress is required to 2.2 multiplication 10 to the power three 80 M has the final answer.


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