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For the beam and loading shown, determine the reactions at the beam supports.16 kN/m5 k/m B9 m5 m...

Question

For the beam and loading shown, determine the reactions at the beam supports.16 kN/m5 k/m B9 m5 m

For the beam and loading shown, determine the reactions at the beam supports. 16 kN/m 5 k/m B 9 m 5 m



Answers

For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

Here we have a parabolic load, um, forced density that spread over the span of a six meter beam. We were asked to find the magnitude and location of the resultant load in terms of the total force and location of that load at the central oId. Our strategy will be to write a function lower case F of X that defied size, the applied load density or a forced s ity over, um X as a coordinate along the beam. Secondly, will find the total load by finding the total area under the force density curve here by integrating it over the span X Um, thirdly, we'll find the central using the relation shown here that is essentially the integral of the product of X and the differential force, which can also which we will express. As for calculating the product of X and the forest density function D X integrated over the length of the beam from 0 to 6 meters Ah, after the numerator they integral in the numerator is evaluated will divide that by the total force of the load in the denominator. The first step is to find the, uh function of the parabolic force density along coordinate X. We'll start with the general function for a parabola that is both translated along the X axis as well as, um offset from the, um why equals zero X axis. I'm defined by this be offset. The translation along the X axis is the negative of the location of the Vertex. In our case, that location is at six meters. So here we can use weaken substitute in six meters for X Avi doing this, we know that the, um when we evaluate the function at the Vertex, that would be X also equal six here this whole term goes to zero and we know that at the Vertex the function has a value of 0.9 Kila Newtons per meter. We can split that on as substitute that value for F of X when x is equal, Teoh Exelby and we can deduce in that case that are offset valuable will be 0.9 killing Newtons for meter. The, um second ah step will do in this, um of ah determination of the parabolic forest density function is to determine to determine our constant A. We can do that by getting evaluating the function at another point for which we have its value in that point is at the point a X equals zero where were given, uh, to killing Newtons per meter as the value of the function at that point. So weekends substitute that value in for our function up here Ah, for X equals zero we're left with a times the square root of negative six meters plus are offset value a 60.9 killing Newtons per meter is equal to our to kill a newtons per meter that is a given as we saw above. Therefore, our constant is the quotient 1.1 killing Newtons per meter divided by 36 meters squared. That gives our co efficient value 0.306 killing Newtons per meter cubed. That just happens to be the unit value of that constant And here we express it in Newtons per cubic meter so we can write fully right out our function that defines the parabolic low density along the being. Step two is to use that function we just defined to first find the total load and in the next step, using this well to find this century But before we get ahead of ourselves. We, um, can evaluate this function for the total load as integrating our functional load density. Uh, over the span of the beam from 0 to 6. Ah. Shortcut will be to substitute in for X minus six into a formula a dummy variable you, which makes the integration easier, as you can see. And, um, the integral of a U squared term turns out to be a you cubed over three. The integral of the B dy you term is be you. These air evaluated we had to change our integration limits when we switch to ah w dummy variable on those limits became from negative 6 to 0. Evaluating our, um, our anti derivative here, if you want to call it that. Ah, we have zero for the first part, and we're gonna subtract our, um, the result actually sub in negative six, which is as follows and be insert in our numerical values here and evaluate them, and we find we have a total load force of 7.60 killer Newton's in the downward direction. Step three. We will find the centrally by, um, this time rather than Ah, simply evaluating the integral of f of x dx. We're gonna multiply it by Are our distance variable X as well. So the integral in the numerator we will have to evaluate is the, um, integral of x times, the low density function f of X dx. We can use that same substitution and this scheme is well and ah, accompanied by a change of our integration boundaries. And you can if you follow the evaluation of the integral. As I've listed here, you can notice that once subbing in the numerical values given in this well given in this problem as well as the while me derived in the last step the total load, we find that the central aid evaluates to 2.56 meters. So given that we have both the total load and its location along the beam in the form of this centrally, we can now calculate the torques about the central oId that the two supports will have to or they are torques. However they're all we're concerned about is Thea um forces there generated since the moment arms or no. So we will find the forces the reaction forces that supports A and B. Given our information of these two previous steps, the total load, it's Willis the central of the load because it's a static system. The, um beam is not translating in space, so all of our linear forces must sum to zero. These forces include both the load forces as well as the reaction forces of the supports. Since the game is not rotating about any point, it's static. All of our torque must sum to zero. Those include the torque about the central aid from loading forces, as well as the opposing torques that opposed the loading toe. Works from the supports. Those some 20 So we know that our the supports must, um some the forces of those supports must there's some must equal the total load that we derived before so we can express that Assef total equals the opposing force and support a plus the opposing force at support be since, um, the load was downward. The some of these two forces of the reaction forces at the supports must be upward to oppose it. We also can derive from our, um, our torque, some that the torque exerted by the support about the central aid from the supported at point A must exactly counter that the opposing torque from the support at point B And since our coordinate origin is at zero where F A is its moment arm is simply the set. The location of the central oId the moment arm of the, uh support be must be the length of the beam minus the century. We can use equation one to write things in terms of F B, right f of a in terms of effort B and F total and substitute that into equation f of a in equation too. And we can't get things in. Terms of the givens are in terms of things we derive before this central area and the total force and F B only so we can solve for FCB doing that, we find that FCB is the product of the central eight times the total force load divided by six which comes to 3.25 killing mutants. We can use this value that we just calculated for FCB subtract it from our force total toe leave us with f survey the support of the, um the support forced reaction force at support a and that leaves us with 4.35 killing Newtons. But I forgot to do here was show that each of these vectors must oppose the load. They're in the upward, They're both in the upward direction. They are vectors, and that concludes our problem.

The agreement as soon the figure. We have to find the reaction. Adopt beam support support Here it is a right. This is m a more. Yeah, it will be urban. Yes, it will be our toe. And this is Are we calculation Urban? How 50 into 12 inch, 300 bucks. Article 50 into contains £1000. This will be eight inch 14 inch taking vertically. Oberto, be positive. Submission off our fight must read doodle Eva I minus £300 minus £1000 plus £400. So why you will get £900 in upper direction? This is your jalape concept not taking moment off and contract clockwise to be positive about the 0.8. Then you will get em a minus 302 8 inch minus 1000 in tow, don't you? Two inch plus 400 and to 38 on solving it. You will get a Toby £9200 into inch in counterclockwise direction. So we just got to answer for it. Thanks for watching it

Elephant as soon the figure. We have to calculate reactions at the support carefully. Reactions at beam support at this point. Support This is the right. Hey, this is he bites. This is our one after the paint off of it. This is article at that distance. Forfeit. This is our three have guarded stance, benefit value off urban reaction fools. Uh huh. Three in tow. Heretics. It would be 7. 24 are off six into 600. It will be 1804 and our three two into 600. £1200. Yeah. Moment off. Couple moment off boots, about three point. Must be zero. You will get two in 27 20 all in tow. 1800 six and to see. Right. They went in to 12 in days. Must be here. So from here, See? Right. Mhm. Okay. He will get 2360 for mhm permission off all forces around direction. Toby. Zero by the seven projects. But let's be right. Minus a candid. That's for three. Six of you, minus 1200. Must be a hero. Toby. Why? You will get 1360 court. So our answer is the reaction at the point is £1360 in a good direction. On direction at sea pointed 136 years but totally found in other directions. Oh, that's all for it. Thanks for watching it.

Calculate reaction. Actor being supposed so at the support eight supports. Reaction is eight. And here the reaction is B. The reaction are one. Is he here on this distended one? We did. The reaction is Artem on this distances. Quiet place. Let us start solving it. Value off urbanists. Ban by Tau 902 1.5 at 6. 75. And our to it. He ended Newton taking commission off momento course about 8.0. You will get minus 6. 75 in 21 point foot, 302 point night be in 22.5. It's called 20 So value are we will be to 17. Newton's in upward direction. No submission off Phi Tau zero. Okay, Vertical Albert, We are taking protective submission off by We're taking the zero. If you will get in my nest. 6 70 fights plus 300. The last 2 70 must be zero. So a reaction is one of Europe's five new death. So the dancer rejection at a 105 Newton on the reaction at these to 70 neutral in a direction. That's all for it. Thanks for budget.


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