Determine the force in members C D, C J, K J, and D J of the truss which serves to support the d

step 1 We'll be looking at structural analysis. For this problem, we want to determine the magnitude of

Determine the force in members C D, C J, K J, and D J of the...

Key Concepts

Tension and Compression

In structural analysis, members in a truss experience either tension or compression, which refers to the type of axial force a member undergoes. Tension occurs when the member is being stretched, causing forces to act away from the joint, while compression occurs when the member is being squashed, leading to forces directed towards the joint. Identifying these forces is crucial for assessing the stability and performance of a truss.

Method of Sections

The method of sections is another approach in truss analysis that involves cutting through the truss to isolate a section. By applying equilibrium equations to this section, it becomes easier to directly solve for the forces in several members at once, rather than analyzing the entire truss sequentially.

Truss Analysis

Truss analysis involves evaluating the forces present in the individual members of a truss—a structure composed of connected struts arranged in interconnected triangles. This analysis determines how loads applied to the truss are transmitted through its members, helping ensure the overall integrity and stability of the structure.

Method of Joints

The method of joints is an analytical technique used in truss analysis where free-body diagrams are drawn for each joint. By applying the equilibrium conditions that the sum of forces in both the horizontal and vertical directions must equal zero at each joint, one can solve for the unknown forces in the connecting members.

Transcript

00:01 We'll be looking at structural analysis.

00:03 For this problem, we want to determine the magnitude of the forces and the members of the structure as well as their nature.

00:11 But to, and we'll apply the method of joint.

00:15 But before we do that, we'll have to look at determine the support reactions, okay? so to do that, let's take a summation of moment about g is equal to zero.

00:31 So we have at a that support at a there.

00:38 Okay, so let's call the vertical one a substitute y times 54.

00:47 Use the information given in the diagram minus 40 or 4 ,000 rather, okay? times 45 minus 8 ,000 times 36, minus 5.

01:26 5 ,000 times 9 equals 0.

01:35 So when you do that, you are going to arrive at arrow, okay, vertical component of the supporter to be equal to 9 ,500 ib.

01:56 So we can then, that means at g, you can also get the value at g, okay, from there.

02:08 So let us look at, let us look at joint a, let's call it j .a.

02:18 So for joint a, we have, we have this force, it's going to look like this, okay? these are our 9 ,500 going up there at, from that, okay, 9 ,500 at that support going up, i -b.

02:54 We have this force, we said the forces should be coming out of the joints.

02:58 By assumption f l, f, sosp, a, b, 1 .j.

03:09 A, it's joint a to b.

03:13 And this one is coming like this.

03:16 F, subscript, e, l going to form joint, a to l.

03:21 This angle is 33 .1 degree.

03:27 Use the information given a diagram to find, figure that out.

03:31 All right, we take summation of forces, this vertical direction, you know, have 900, okay? sorry, 9 ,500 that we got, okay, minus f, so shall be al, sign, sign 33 .1 degrees, is equal to zero.

04:15 So if you solve this out, you have, you have f source at force in that member, al to be equal to one, one, 1879 that's 117 .179 .179.

04:34 179 .6 .8.

04:46 So that's what you're gonna have.

04:48 0 .6 .8.

04:50 I can see that it is positive.

04:56 So it's a member under tension.

05:01 Okay.

05:03 So we can look at, we can also look at some form of forces in the horizontal direction.

05:09 If you do that, you are going to have a zero.

05:17 So that we have our f source with a, ab plus f source with a, al costs, cost, cost to 3 .1 degree, okay, to be equal to zero.

05:45 So plugging in values into this one.

05:48 Of course, we've known f, such with a, al initial.

05:55 So we just put values in.

05:56 And you have this one to be negative 7132, 7 ,132 .8 .1.

06:10 Sorry, 0 .8 id.

06:13 Okay? see that this is a member under compression because of the negative sign that is, okay? it's under compression.

06:29 All right, we can also go further look at some other joint.

06:36 Let's look at joint l.

06:39 Let's consider joint l.

06:41 That's called it jl.

06:46 So we have for that joint, we have, remember, forces configure this way.

06:58 We have this one is f, so with l k, going from the joint we are considering l to k.

07:06 These are l, these are joint l.

07:16 Then there's this other force, going from the joint to a.

07:21 Okay now this is 36 .9 degree point nine use information given in the diagram the question to solve to determine the angles you can always do that all right so this is f subscribes lb going from join l to b now for take summation of forces in the in the vertical direction 0, we are going to have f, substitute, lb, plus fl, s off with la, f sosf l, la, cost 36 .9 degrees, to be equal to zero.

08:32 Now, you sub this art, you are going to have this value, f sub with lb.

08:37 Of course, we know the ones we've known, just plug it in there.

08:41 We're going to have it to be equal to, negative 9, 9 ,0004 and 99 .9 .99, 0 .99, i'd say 0 .9 id, okay? so you can see that this member is under compression because of the negative sign, but the magnitude is that, okay? so we can look at summation of forces in the horizontal direction.

09:36 If we do that, we'll have, so just follow the same procedure.

09:46 It's minus fl that, sign 36 .9 degrees, okay? plus flk, struck with lk, this will be equal to zero.

10:19 So when you solve this out, putting in values you've gotten already into this equation.

10:27 And you are going to have 770132.

10:42 So 7 .132 .8 id.

10:50 Okay, you can see that this is a member under tension.

10:57 So we can also look at joint b.

11:05 Consider joint b, you are going to have then we have something that look like this.

11:13 That's just related joint.

11:16 Then b, j .b.

11:19 So we have that force coming down, 4 ,000 ib there, coming down there...