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Compress0d cylindcr i> uscd flan accumula or (or Uir-opbraled actualor. The accumulator is 20 inchos diameter and 18 Inches Iom Upon complation of the 0l ga5 tra...

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Compress0d cylindcr i> uscd flan accumula or (or Uir-opbraled actualor. The accumulator is 20 inchos diameter and 18 Inches Iom Upon complation of the 0l ga5 transter tho accumu"ator md a usiure 100 psig 4 qus lmpetulere dogrees The ambien: lemporalure build ng 112 degreds Whal is the approxi mate accumulator pressure once the accurnilalor len putatero equalizos wilh the building lemperaturo? Polnt}125 psh1105 pslg1270 palg

compress0d cylindcr i> uscd flan accumula or (or Uir-opbraled actualor. The accumulator is 20 inchos diameter and 18 Inches Iom Upon complation of the 0l ga5 transter tho accumu"ator md a usiure 100 psig 4 qus lmpetulere dogrees The ambien: lemporalure build ng 112 degreds Whal is the approxi mate accumulator pressure once the accurnilalor len putatero equalizos wilh the building lemperaturo? Polnt} 125 psh 1105 pslg 1270 palg



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A boiler is constructed of 8 -mm thick steel plates that are fastened together at their ends using a butt joint consisting of two 8 -mm cover plates and rivets having a diameter of $10 \mathrm{mm}$ and spaced $50 \mathrm{mm}$ apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress in the boiler's plate apart from the seam, (b) the circumferential stress in the outer cover plate along the rivet line $a-a,$ and (c) the shear stress in the rivets.

This question you have to find the temperature, the final temperature T. two at which the average nor most ways In each magnesium or still become 12. Make a pass out inform. Last question. We have already calculated um We have already know that this relationship which is um the compatibility equation, the displacement from magnesium is equal to the displacement of the still. So we substitute this fire. Um So first we can find the force F. F. Is from the it's great time area. So stress is the wealth 10 5 of six. And the area you can find still because it has the smallest cross sectional area. That is true By all of four. Um The diameter of steel and that is 1885 newton. Now if we use this where you are F. In this equation we can find a value of delta T. Where technology is 229 degree & T. two is does the tea when asked T. One Past the one which is 15 degree Which is 244 degree. That is a disaster.

Chinese solution when directed with stool and avoided give some right sector own additional access this compound. So could you inform general which hard place here's the correct terms.

And this question we have to find largest room with al that the broad can spend. And um there is a figure in the textbook and if we take a look at this figure and draw the sheer at the moment diagrams, We'll have that. The b. max is 225 pound and max is 75 time the with L. So first we have to shake under here stress for this um Lord which is 1.53 or a. For this structure and We is 2-5 Made by area 1.5 times eight. That is to the 8.1 c. And We will see that it is less than allowable. She swears over 180 g. p. s. i. Um So for bending women, you know that um max it is 75 owl And I. is 1 12 1.5 eight. The power three 64 um injures to the powerful and S. Is I. O. C. 64 divided by four days, 16 GB interest. And in my it is the allowable stress. S and M. S. Is 25 L like most West is on two 1000. Um, as is 16 so al is 35.6 feet. And that is the answer.

Hi friends pressure difference inside the film from outside can be written as peace get too and far Born upon urban press been upon hard here to R one because of suitable equal to edge and hard to. Is the radius of the the radius of tablet and can be neglected. Yeah. The total force exerted by mercury drop on the upper glass plate to buy our square and for because of twitter upon edge typically we should put mhm mhm Hedge upon and four edge because that ability is compressed and time. Mm let compressed and times. Mhm. Then since at G is nearly in compressive it so pioneer square edge would be constant. So are is our route over. So total force you will get mm Why are square and for Yeah course of Ceta any squared divided by edge. Part of the force needed to okay. Part of the force needed to keep mercury in the same bob. David. Yeah. Yeah. Or rather than in the shape of tin suit this part can be calibrated. Yeah but putting and it's got a but so you can write MG plus by our square and for more of costs that are up on edge that is proved by our square and far model. Of course it upon urging to any square. So um you will get two pi R square and for mode of course of theater upon urgent to G any square minus man. Subsequently all parameters you will get it is to be 0.7 cases that's all thanks for watching it


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