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1- A steel rod has a transverse area of 0.50 cm^2 and is hung at one end to a structure to support a frame with a mass of 600kg. The frame hangs from the bottom end...

Question

1- A steel rod has a transverse area of 0.50 cm^2 and is hung at one end to a structure to support a frame with a mass of 600kg. The frame hangs from the bottom end of the rod. If the Young module is 2.0x10^11 Nm^2, determine the rod effort.2- A steel rod has a transverse area of 0.50 cm^2 and is hung at one end to a structure to support a frame with a mass of 600kg. The frame hangs from the bottom end of the rod. If the Young module is 2.0x10^11 Nm^2, what extension has the rod?3- A steel rod h

1- A steel rod has a transverse area of 0.50 cm^2 and is hung at one end to a structure to support a frame with a mass of 600kg. The frame hangs from the bottom end of the rod. If the Young module is 2.0x10^11 Nm^2, determine the rod effort.2- A steel rod has a transverse area of 0.50 cm^2 and is hung at one end to a structure to support a frame with a mass of 600kg. The frame hangs from the bottom end of the rod. If the Young module is 2.0x10^11 Nm^2, what extension has the rod?3- A steel rod has a transverse area of 0.50 cm^2 and is hung at one end to a structure to support a frame with a mass of 600kg. The frame hangs from the bottom end of the rod. If the Young module is 2.0x10^11 Nm^2 and its length is 5.0m, which elongation has the rod?Tranverse area is 0.50cm^2, the problem say it



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A metal rod, $0.5 \mathrm{~m}$ long, has a rectangular cross section of area $2 \mathrm{~mm}^{2}$ (a) With the rod vertical and a mass of $60 \mathrm{~kg}$ hung from the bottom, there is an extension of $0.25 \mathrm{~mm}$. What is Young's modulus $\left(\mathrm{N} / \mathrm{m}^{2}\right)$ for the material of the rod? (b) The rod is firmly clamped at the bottom as shown in the sketch, and at the top a force $F$ is applied in the $y$ direction as shown (parallel to the edge of length $b$ ). The result is a static deflection, $y$, given by $$ y=\frac{4 L^{3}}{Y a b^{3}} F $$ If the force $F$ is removed and a mass $m$, which is much greater than the mass of the rod, is attached to the top end of the rod, what is the ratio of the frequencies of vibration in the $y$ and $x$ directions (i.e., parallel to edges of length $b$ and $a$ )? (c) The mass is pulled aside in a certain transverse direction and released. It then traces a path like the one sketched. What is the ratio of $a$ to $b$ ?

For this problem on the topic of sound waves were told that a forgiven type of steel with the density listed in the given table, it bends permanently if subjected to a compressive stress that is greater than the elastic limit of 400 mega paschal's. This is called its yield strength. If we have a rod that is a length of 80 centimeters made from this type of steel and projected at 12 m per second at a hard wall. We want to find the speed of the waves moving along the rod and then we want to find the time that elapses before the back end of the rod receives the force that is exerted on the front end of the rod. We want to find how far the back end has moved in. This time. We want to find the strain and also the stress. And lastly, we want to show that if the Iraq is not to fail at the maximum impact speed that a rod can, can have is given by sigma over the square root of road times. Why now the speed of the compression wave in a bar V is given as the square root of the youngs modulators of that bar y over its density role. And this for this type of material is the square root of the youngs module is is given as 20 Times 10 to the power 10 newton's per square meter, divided by the density Of the steel, which is 7860 kilograms per cubic meter. This gives the speed of the waves in the spot to be five 04 Times 10 to the power three meters per second. For part B, the signal to stop, passes between layers of atoms as a sound wave reaching the back end of the bar in a time T, which is equal to the length of the bar, divided by the speed of the waves V. and so this is the length of the bar 0.8 m, divided by The speed at which the waves travel along the bar 5.04 times 10 to the three meters per second, which we calculated above. This gives a time of 1.5, 9 Times 10 to the -4 seconds. Now, as described by Newton's first law, the very most layer of steel has continued to move forward with its original speed V. I. For this time, and hence the bis compressed by a distance delta L, which is the speed V times T. And this is 12 m/s Near time of 1.5, 9 times 10 to the -4 seconds, Which causes an a compression of the bar of 1.9 Times 10 to the -3 m Which is 1.9 millimeters. The strain in the rod is delta L. Over L. The compression over the original length, Which is 1.9 times 10 to the -3 m, divided by 0.8 m. This gives the strain to be 2.38 Times 10 to the -3. The stress in the rod is sigma and sigma is equal to the youngs, modulates y times the strain delta L. Of L. And so this is equal to 20 times 10 to the power 10 newtons per square meter times two point 38 times 10 to the -3. And so the stress in the Rod is 476 mega paschal's. No, since sigma is greater than 400 Mega Paschal's, which is the elastic limit. This will mean that the rod is permanently distorted. So for part E we want to know if the Iraqis not to fail. You want to find the maximum impact speed that the rod can have. Now. We'll go to the same steps as in part A to E. But use algebraic expressions instead of numbers. And so the speed of sound in the rod. We know V is equal to the square root of why overall. And the back end of the road continues to move forward with speed V. I for a time T. Which is our of V. Oh. L. Times the square root of row over why it travels a distance delta L. Which is V. I. Times T. Which is also the compression of the rod. After the front end hit the wall. The strain in the rod is equal to data AL over L. Which is V. I. T. Over L. Which is V. I. Times the square root overall over why. Then the stress sigma is equal to the youngs modulates Y times the strain delta al overall. Which is why times V. I. Times the square root of roll over. Why? Which becomes V. I. Times the square root overall times Why? Now, for this to be less than the yield stressed sigma sigma. Y. We require that V. I times the square root. Overall times Y is less than sigma. Why? Or the initial speed of Tyrod. V. I must be less than the yield stress sigma Y, divided by the square root of row times Y as required.

In this problem, we're told that we have a piece of pre stressed concrete. So we have is a slab of concrete. What happens is we take a metal rod, we stretch it and hold it, and then we pour concrete around it that concrete cure and then let release the rod. And then we have big plates here that are very stiff. So the rod will try to spring back to its normal length and in doing so will compress the concrete because the concrete concrete new sister, so in the and you'll have a rod through the concrete. That's intention and the country in compression, which is what you want, because medals are very good in intention and just concrete is very good and compression, but very does very poorly in attention because cracks brought me easily. So this piece of concrete. This conflict block is 1/2 1 1/2 meters long. It has a cross sectional area, 50 centimeters. The steel rod that goes through it has a cross sectional area, 1.5 centimeters squared. Um, it's steel. So Chung's module is is 200 acre Pascal's, their country. We're told it has a young marginalised of 30 Guica Pascal and we're told that the stress in the concrete after the after the Rod is concrete is cured and the rod is released. Is minus eight Mecca Paschal's. Now I've assumed that this 1.5 meters is the final link of the concrete. No, what we'll see is that it really doesn't matter. And then we've given in a linear approximation. We have very small deformation is whether we use the final length of the initial length. You'll see really makes no difference at all. Um, but I used the final in character kind of show that why that is so we know that the strain is the change in length over the initial length, which is the stress divided by the young's modules. Um, and we know that the change in length is the final length minus the initial all over. The initial is the strength. So if we have, if we have the initial link, it's easy to find the if you have the initial length, the stress and the young marginalised season to find the change in length. However, if you have their final link, the stress than the young evangelist. We need to do a little bit of algebra and we see that what we get is that the change in length change in length to equal the final length the times of stress all over the module is time plus this stress. And so that's what I have here for the concrete. Now why it makes very, very little difference is if you look at the magnitude of these two terms, Um, if this time gets anywhere close to being this, then you have strains on the order of one, which means that you're long past any kind of many. Your material behaviour on pretty much everything in this chapter is is no longer could assumption. So you can see that, really, this term had better be very small compared to this. And in fact, we had When we have here, um, basically, you know, orders of three orders of magnitude difference here. So we can pretty much neglect this term, which means that whether this is the initial length of the final thing is kind of material. So what we're asked for how much does the concrete compress after the after the tension in the rod is released, and we can pluck all our numbers in here, and we get that it compresses a 0.4 millimeters. So from the from the cured length to its final link, it is now. It's your point formula short, Um, and again, you can kind of compare that to this. You know, the initial length of the final link are almost indistinguishable from now. We're asked what the tension in the rod is. Well, if we look at a free by diagram of, say, the plate that the rod the rod is attached to when it's pulling the rod is pulling the plate this way, and the concrete is pressing on. It has a pressure that's equal to stress. That's pushing it out. And so the net to keep this an equilibrium, the net force from the concrete, which is this must equal the tension in the rod so we can plug in our numbers and we get that. That's 40 kilometers now. If no one asked for how how much? How long is the rod after it's been released? How much has it stretched from it? So there's there's kind of Ford in three different states of the rod. It's on stretched length. It's stretched length when you're waiting for the concrete secure. And then it's second stretch length after it's been released, and now the concrete is stretching. So it's this last stretch length that we want to find. So we know the stress in the in the Rod, his t two, the tension divided by its cross sectional area. And again we can use the formula like this and again. This term is negligible compared to this, and we can plug all our values. And then we get that this final value is is two millimeters. And so what that tells us is, is that the you need to stretch the rod 2.4 millimeters and while the concrete rich jury and then once we let it go, it rebounded back 0.4 millimeters by compressing the concrete

Start by the equation. Years she scientist types not equals mg times are not over to. So from here you find that the tension these mg over do sorry data. Yeah. So that's trace. He was effort one a forces G cause I intend over a substituting be substituting the value off be here you get stress It was mg over to a Was I indeed now young smarter Thus he's if as not over you know why are dead end now? Suster This value off the stress for f over a So you have I m g and not over to a why times cause cocoa. Daniel instead of the school time laid off taker Now mass is given by density times volume which is that tax area So massive area comes up to the density times So this mass and the area over here can be replaced in both stress on till the end. So if we does it over here, massive area comes out to be royal. So this becomes half rule and G who abandoned off tender and there's the l, comes out to be one of our two. Why massive area is go not squared g o tangent off. No, if you put the numbers over here so that stress, if you put the numbers in you have, you would have stress. It was one foreign four times started a five baskets. And if you put the numbers over here, he would have delta l equals 2.2 times. 10 to the minus six. Made it. So what we find is when, as gets doubled, stress gets doubled as well. However, the desert is dubs ing means Triscuits. Quotable.

Three bars. Trust ABC. Constructed off. Still pipes span off 3 m identical rods. Act both vertically and horizontally at at the joint. See eso first. Um, I can't let us calculate the horizontal displacement off joint be when the Lord be acting is equal toe for 75 killanin tents. Applying moment electoral Librium equation about point. Uh, submission off is zero. So I m a c b l upon to plus B l upon to minus are the one. Why multiplied by Ellis zero. So are the one. Why l is ableto pay alto plus p l upon to so r B one y is equal to p now applying horizontal force equilibrium equation Submission off if it is zero. If fetches are one are a one explicitly is equal to zero. So are a one x equal to minus the end. Applying vertical force equilibrium equation. Some vertical is the road. So are Avon Wife Plus are given Why is a call to zero? Yeah, called p Sorry. Yeah, So are a one y equal to zero. Mhm. Mm. Now consider the equilibrium off joint at a and let us throw the free body diagram this at this point? A. Here's a force acting. That is, from a C Another force from B. There is Force A B and course a one. Why are a one X? Yeah. Now applying vertical force equilibrium equation off. Joint a summation off. Submission off. F Y is zero. That is, um f a C When she replied by sign, 45 is at Evan. Why so F a c is zero and along the horizontal is f A B minus f a one x equals zero f A b is equal toe f a one x so f a B difficult to minus P. Now let us find the horizontal deflection at point B is Delta B is equal toe F A B L upon e. Yeah, yeah. We have forced along a B s for 75 into 10. 24 3 into l s 3000 off one it is 202 10 to deport three and 8, 3900. So, delta, there is 1.8 to 9. Mm. Is the horizontal displacement off joined beef. Now let us calculate the maximum possible Lord. That is Delta. It did be Max help upon. He mhm. So P max is equal to the limit he a upon and we have Delta is 1.5 is 3900 and it is 200 court and to the power three upon 3000, there is the 3 90 kg Newton. Hence the maximum armies of a lord. Value is 3 90 kilometers. Now, um, the three bar space trust ABC constructed of steel pipe has a span off three. Mm. No applying equilibrium equation in extra addiction. Submission off affects is zero. So a X minus B is zero. So x tickle toe Be now Let us find the length off a be using triangle. Hey, Obie, no diagram is this is B See? Yeah. The triangle. Thank you. This is a l l upon to you see, using the tangle, it will be He is a cultural route over a square plus at square This desecration one using triangle. Yeah, C dash b is equal toe right over hell Square, my national upon to whole square. So we have ls three square minus three upon to whole square That is three through three upon now using triangles O C B is a girl toe route over three, the trade upon to whole square minus three upon to whole square. So it is 3 to 2 upon. So So, um, substituting the value of A and L in equation one length of Member A. B s sequel toe route over, please. A little too upon three whole square. Last three square His three Route six upon do now considering joint a off the dress and applying my third off section ah force equilibrium equation at joint. Hey, is permission off effects is zero. So I x minus f a b I replied by a upon say is zero familiar? We confined f a b a X see upon? Yeah, we have X. It's for 75 into three. Duke six upon to upon 3212 So ever baby is 8 22.72 kg Newtons. Now let us calculate the angle off or a or B. A angle off or B is equal to cause a one C. So it is three route to upon to what three road six upon to that is 54.7 36 degree again. The deflection at Point B is F A b l Fun e. We have a 22.7 2010, 24 3 into 3000 in two road six upon to 1, 202 10 30 Power three into 39. It is 3.87 life and on the deflection in direction Ex success ISS There be access there will be upon cause I'll be here. 3.875 upon because 54.73 Yes, 6.711 amendment the horizontal. The placement off joint bees is the horizontal displacement as the displacements are linearly related to the Lord. For this linearly lasting tip like in displacement, the maximum Lord will be Pemex will be 1.5 upon 6.71 254 to 4. 75. That is 106.1 45 kg. Newtown. Handsome maximum lord. Value P is 106.145 Cool, uneaten


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