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Lorry chassisWhich component(se would be expected experience torsion? Load applied vertically downwards, perpendicular t0 {he lory chassis.attemptApplied loadSeleci...

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Lorry chassisWhich component(se would be expected experience torsion? Load applied vertically downwards, perpendicular t0 {he lory chassis.attemptApplied loadSeleci ono: MembersMernbor DMembers & CMomberhete t0 searchEi

Lorry chassis Which component(se would be expected experience torsion? Load applied vertically downwards, perpendicular t0 {he lory chassis. attempt Applied load Seleci ono: Members Mernbor D Members & C Momber hete t0 search Ei



Answers

Rod $A B$ is fixed to a smooth collar $D,$ which slides freely along the vertical guide. Determine the internal normal force, shear force, and moment at point $C,$ which is located just to the left of the 60 -lb concentrated load.

Have another beam that ISS this time simply supported both in. So we have been joint here, a ball joint support here they radically there could be court Donald Reaction force here, but because there are no or forces acting on the beam, external forces acting on being this was zero so weak now do force and moment balances for the entire team. And what we'll find is that, um, this is eight Kips on. This is 40 feet. Apply torque moment here. We'll find that get two equations into unknowns for people will find it. P one is seven. MP three is one kid, and that makes make some sense. Number one, they need to add up to eight, which is this and the load is culture to this side. And plus this moment is basically trying to push this bar down against this reaction. So it's not surprising he's looking reasonable, and so they want us. Teoh, check what the The internal sheer force and moment are at a couple points one point here at one point here. So this is 80 in from the end, and then this one is just Now I'm gonna take a look at the left hand side when I cut it. So when we cut here, we have p one and p two. I'm assuming that this, um and actually, I guess it just it's actually says that this is just to the right, um, Pete to because it doesn't jump in sheer force across this point, we have p one and out because this is the right of p two that also acts. And then we have one. And then moments, um, we have, um, the one and P one active practically the same point. So a feet in from here. And then we have someone acting. And so this gives us two more equations for V one and m one, and we confined that V one for six skips and, um, want equals 56 feet weekend draw a free body diagram for this section here, and we have p one p two and P to acting. And then we again have p to creating a moment be to creating a moment and then to the internal and two equations and 21 knows, and we get that B two is one. Kip and M two is 48

In this problem where given this power transmission line power, I've just drawn the top of it here because we actually don't need way. Don't need the bottom of it because we don't need the reaction forces between the ground so we can just draw free by Don agree with cop. What we're looking for is there's the cross numbers throughout the structure and we're told that they can only be intention. And so where asked to look at the ones and in the section of the second section from the top here to see which one of these what forces are in these two members and knowing that they have to vote more that they cannot be negative so we can write force factors. Here is these are the external loads acting from the part of the transmission lines and the internal loads here, from where we cut across their structure and weaken some the forces and then some of the moments and I picked This is my origin here and again, I can write the these vectors just like you get in all the previous problems and we wind up with these three equations if we some forces in some moments. And so what we need to do? We have 44 unknowns of these equations. So the problem is, is obviously can't find for gnomes with only three equations. But we could make the assumption that that one of these is that zero and then check with the other way. And so if we assume that this members seek from Cedar J zero, we find that the member from the age is has a negative forces, which means it's in compression. So that is not the correct assumption. So we can assume that the remember either h has zero attention in it, Onda and find the tension in C J with force. And if he is positive and it's 1.3 killing Newton's. So this is our answer here there is zero. Um, there was this Perry is carrying no load, and this one is carrying his intention

And this problem we have, I'll be him. This will and let's see. Here we are looking that is pin and it's left here and the roller support here as an applied load angle at the end and then has distributed load That's uniform over one section and various Linda Early, down zero over another section, we are asked to find the Internal Forces moment that, um, this point just to the left of the roller support. There's also an external couple that acts here, um, and then were asked to find that the Internal Forces moments at this point, if we take a look at the whole structure here we have in the extraction F one X plus 4 55 in the why we have f one Why plus F two minus of three minus four with these are the point the equivalent white loads of used to distributed loads and then minus 3 55 for moments. We take moments about this point Here we have minus 1.5 this distance here, three plus three F two minus four at four and minus six times the vertical component of five, which is 3 55 and then we also have to remember about the point moment here, and that's called them six. This gives us three equations, birthday unknowns, these three forces and read. Here we find the F one axis minus four, killin it Lose F one. Why is minus three? Killing Newton's in F two is 15 kilometers. This makes sense just because there's a lot This this point here is carrying a lot of load. Also basically counteracting this applied woman here. Counteracted this here. No, if we want to look at the Internal Forces and moments we have at this point here we have in the X direction we have f one acts plus and one equals zero and I'm looking at this is n one here one I am one. So then f one y minus f three plus 10 You don't have to look, we don't have to mess around with on changing this f three because we're looking at the entire region of this distributed load. So in who have once we cut it that we have the same equivalent force here and then we have minus for 1.5 after me minus three one and plus M one, which is the internal moment here on F two doesn't come in because we're told it's to the left of this. And likewise, m six doesn't because cut is before is to the left of that. Also, these three equations give us one is nigh killing. It was in one is four filming M one is minus 18. Now we can look at take a cut here way. Look at this section here. So have and, um, Internal forces and woman's because we're cutting through this distributed load. Um, you have to find out what the equivalent reaction is. Cool point for us, for just for this section. And it turns out that that's f prime is 4/3. What? There is 3/4 of killing four was three. Killing em's enough 66 and a 36 Calculate those And again, what are these things that is common in these problems is we cut a linearly very low. Well, that means that I want the triangular section that we still have three equipment loaded his quarter because we cut the distance in half. Only cut the peak and half so way basically get two factors are too. So get 1/4 of this. So if we go down here in the X direction, we have, uh 4 55 minus and to And then we have minus 3 55 minus f four, prime minus two. And then we have this distance. Turns out to be one here because this is 1.5 answer. We have one times. That's positive times four prime plus one in the head, out on the to and then minus. I am, too. And that has to be zero. So these equations, three equations we can then use to get to and to m two and two is minus 3.75 minutes into his for killing it. And m two is minus 4.875 killing two meters.

Now we're ask Teoh. Look at the counter forces in these counters in a section Two more down on this power transmission line tower. So again, I cut it through that section and you can see we have now to more powered power lines acting on external structure in this top part of it. And again we could draw great force factors for all of these loads and right position vectors to all of them. I haven't written them all out here on day. We can obviously some of the forces and then some of the moments and I some of the moments about his point here. And if you do all that thes three equations which I won't go through the whole deprivation off, but you can check them. And so what happens is we have again four unknowns and only three equations. But what we want to do is see, you can assume that one of these cross counter tension and one with the force and one of those counters zero. So if we assume that the force in this one is zero, we get that the force in this one is negative. So that means that we made the wrong assumption about which 10 So if we assume that this one is zero, then we get the force. This one is 2.5 killing. So this one is intention. Just like in the previous case, this was intention. But now we have about twice twice the load on that one. Because again, it's that this section, as we move down is carrying a larger, more external load. Something slowed. This is in much more attention this one year.


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