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For the beam in Question No 5 , the maximum bending moment :ST69 " € auuoj metre;A .3 . 8 tonne metre4 . 1 tonne metre;7 .2 tonne metre_D .300874/1...

Question

For the beam in Question No 5 , the maximum bending moment :ST69 " € auuoj metre;A .3 . 8 tonne metre4 . 1 tonne metre;7 .2 tonne metre_D .300874/1

For the beam in Question No 5 , the maximum bending moment :ST 6 9 " € auuoj metre; A . 3 . 8 tonne metre 4 . 1 tonne metre; 7 .2 tonne metre_ D . 300874/1



Answers

If $L=9 \mathrm{m},$ the beam will fail when the maximum shear force is $V_{\max }=5 \mathrm{kN}$ or the maximum bending moment is $M_{\max }=2 \mathrm{kN} \cdot \mathrm{m} .$ Determine the magnitude $M_{0}$ of the largest couple moments it will support.

Here were given the equation for capital M, which stands for the bending moment of a being. And it's equal to one half of W times L A times x minus one half, uh, W times X squared. So X is our independent variable here, because that is a distance from the support, W and L are both some constant. So the question is, we want to find a point on the beam where the moment is greatest. So, meaning we're looking for maximum them. So any time we're looking for a maximum one minimum, we're basically Looking for the Derivative 1st. So let's actually rewrite this. So we can factor out a couple of things. So let's do one half W because we can factor that out. Um and then I know we can factor out an X as well, but let's not do that. So we have L x minus X squared. Okay, so the reason I didn't want to factor it out is because once when you do the derivative later, if you factor out the eggs, you would have to do the product rule. Whereas here it's you don't need product rule. Okay, So let's go ahead and find the derivative of them with respect to X. So this would be D. And the X. So this is equal to well, what we factor out on the front, that's just a constant. So let's leave that as it is. And then now we need to take the derivative of this inside part. So this is going to be just l minus two X. Thanks. So we can see that this is just um this this expression here um exists everywhere for for all X. Right. So there's nothing here that will actually break this expression. So which means our only critical points would be when the derivative is equal to zero. So let's go ahead and set equal to zero. So that's going to be copy that down. Okay. And we can see that. Um So this w is the uniform load per unit length. So it's some constant Which means this can only equal to zero. Yes. This thing here in parentheses is equal to zero, which means L. Has to equal to two X. So if we isolate X, we get X is equal to L over to. Okay, so now we found our critical value. We just need to double check and make sure that this is a maximum and not a minimum. Okay, so let's go ahead and find our second derivative. So this would be D squared over the glared. And this is going to equal to so one half W. That stays in the french and then derivative of L. Well that's a constant. So that zero and derivative of negative two X. It's just negative too. So this is equal to negative W. The W again is the uniform vote for unit length, which has to be some positive constant. So if we have negative W, then that means this is less than zero. So what this means is it's concave now at this critical value. So concave down means that this here must be a local maximum. So therefore we found where the point on the scene where the moment is greatest. So we can say um and the greatest at X equal to L over two.

This question we have to find the minimum depth edge of the beam that will certainly support the Lord. Uh And from the diagram in the textbook, if we throughout this year and the moment diagrams, you have that the maximum moment of um the beam is 72 K. P. Feet and maximum shear for three is 24 K. P. So first if we assume that branding moment controlled the design, know that as is equal to and bye bye thrice and msn max and threats here is the reliable stress but best is also equal to I oversee and I is one of which killed three. I'm H. to the power of three. What? By sea, which is H power to no have that Mm Max is 72 12 for cappie injures And read by 21 reliable stories. And we all have that edge Is 9.7". Oh You can use AS. nine aged of interest. That's your answer. And I have to shake for the maximum She's where is that? It is okay to use it is a greater or lower than the Lagos twist. So first we have to five Q. Which is Q. Max. Why I bought that that weight area. And if uh H. Is nine and an eight of inches it is one point 9.1252 x two x 4 times 9.125 by about two times three. Well by by that and ate at it is 31-2 cubic interest and I is 1 12 3 edged The power of three. It is 100 89.95 interest to the powerful. Now we can find a town max which is remax Q. Max. Do I buy item T. And it is 24 31.22. um 1918 9.95 Thickness is three and it is 1.315 Ks I. And it is lower the Allowable. She has raised 10 kids. I saw. It is okay to use the edge as 9.125".


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