Truss analysis by method of sections: worked example #1

hey welcome back in this video just going over an example on using the method of sections to solve truss problems in statics so this is useful if you've been asked to solve for just a single or a couple internal forces in a dress for example if your professor asks you to solve for the internal force and member eg here if we were to use the method of joints it would take forever we'd have to go joint by joining the whole way across but with the method of sections we pretty much do it in two quick steps the first thing that we do is we want to just cut we want to take a virtual cut and separate our trust into two sides and we want to make sure that our cut is passing through the the member in question and also that it's only passing through a maximum of three members because in 2d statics problems we have a maximum of three unknowns so I'm just going to pass my section like that and I'm going to eventually cut this truss into the left-hand side and the right-hand side well the first thing that we want to do before we get to that I guess is we will just want to actually find what the reaction supports are so we'll go ahead and we'll draw a Freebody diagram for the whole structure and solve for those so we have our two applied external forces here and then we have a horizontal reaction at our hinge and a vertical reaction for a and we also have our other vertical reaction over here at the roller another thing I just forgot to mention was that all of these members are going to be one meter long and then when we have everything in one meters long and they're triangles that we get these as equilateral triangles so we get all these angles in here is 60 degrees all right so that gives us these distances that I've drawn on here in this pink color and then this is just the 1 times sine 60 so that's just 0.866 meters in height all right so now let's take the sum of forces in the x-direction there is only one force in the x-direction so when we set the sum of forces in X equal to zero we only have that one force that's ax so ax is just equal to zero the sum of forces in the y-direction is a Y plus KY minus two minus three kilonewtons and then the sum of moments about a has these three components so if we just simplify that we get FY is 2.8 and then a Y is 2.2 all right now that we have the reaction supports what we want to do is we want to redraw our truss we can bring it down here and our we had a section passing through it like this that we were considering so what we want to do is we want to pick the left hand side or the right hand side and cut it in half and just draw the free body diagram of that side so I'm going to pick the the left hand side you can think either but this is the free body diagram once we take once you remove the other side and now all we have to do is take the sum of forces in X direction the sum of forces in the Y direction and sum of moments about some point and then we can solve for these three unknowns you'll notice that these are the internal forces and I have drawn them in tension we use it as a sign convention so that when we solve for our answers if we get a positive answer we know that that member is in tension and if we get a negative answer then we know that that member is in compression so if you remember the original question was you know is asking what is the internal force in eg so we're about to find that out so let's start with the sum of forces in the X direction but there's not enough information here there's too many unknowns so then the next thing we're gonna have to do is write out the expression for the sum of forces in the Y direction and this can just easily be arranged to get F G is equal to 2.31 kilonewtons and you'll notice that is a positive number so that means that this member is in tension now at this point our unknowns are still eg and FH so we're going to have to do one more step by solving for the sum of moments about point a so you'll see I've always written the the magnitude of the force times the distance and the distance here in brackets so this is this 2 kilonewtons here that's 2 times 1 meter FG sine 60 is the y component of this force and the distance from a to the perpendicular distance to the line of action that force is 2.5 meters we also need to break out the X component of this force because it's line of action would come over here so we have FG coasts 60 and then times that perpendicular distance to that line of action of the force and then same for FH we have the magnitude the force here times the distance of that perpendicular distance to the line of action of force again you notice all of these because I've defined the positive sense of a moment to be counterclockwise all of these would tend to cause this object to rotate clockwise about point a so they're all giving us negative moments by my definition all right so we can go ahead and simplify this a little bit more we'll find out that F H is actually negative three kilonewtons which means it's in compression so that's why we have to see here in brackets for compression whereas here when it was intentionally just little little T there just to keep track alright so the last thing that we want to do now is just plug back in all of these valleys that we found into this original expression so we still are missing EG so we had F H this was about Li minus three and then F G Co 60 F G was zero point two three one and Co sixty is just zero point five so that's all equal to zero and when we just solve for e G we get that our eg is two point eight eight was a two point eight five two point eight eight five that comes out as a positive number so that was kilonewtons and because it's positive that means we've correctly assumed its intention and there we go we've actually solved the problem now that we're looking for so put the answer there this is a much faster method than how do we use method of joints and gone through and solve every joint up until that point so there you go I will catch you guys in the next video and we'll go over one more example on using the method of sections

Channel: Engineer4Free

Views: 262,604

Keywords: free, engineer, engineering, math, university, science, tutor, tutorial, lesson, online, learning, mathematics, engineer4free, worked, example, statics, mechanics, static, equilibrium, AP physics, truss, tension, compression, joint, member, section, method, of, sections, virtual, cut, side, fbd, free body diagram, pin, bridge, civil, mechanical, rigid, support, reaction, force, forces, moment, sum of, moments

Id: PXS9n-b5CA8

Length: 5min 52sec (352 seconds)

Published: Mon Sep 12 2016