SA04U: Truss Analysis (Method of Joints)
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- เผยแพร่เมื่อ 24 ก.ค. 2024
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.structure.education/
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Thank you for breaking this complicated process down into an easy to follow process.
Best explanation
I liked this video.
I like to watch this type of videos.
I don't like those videos in which a teacher/professor teach because in that there is a problem of blur vision on board.
Thanks for your sharing
Thank you for the video, I'm at a loss how you how Fac & Fbc sadly could this be broken down a little teeny bit more for slowpokes such as myself :)
Its excellent in conciseness and lucidity. Is it possible to get the slides in .ppt ?
Thanks for the note. These lectures are not generated from ppt slides, there are no such slides.
Thank you for your videos! They are helping me study for the ARE exams. I was hoping you could explain at about 11 mins in, why in the sum of forces you add Fac x cos45. I understand that the forced of Fac is in the x direction, but I don't understand how you decided to multiply Fac by cos45. Is it because the adjacent and hypotenuse are the only forces acting on the joint?
As an inclined force, Fac has a x component and a y component.
Imagine a right triangle having Fac as its hypotenuse. That is, the length of the hypotenuse represents the magnitude of Fac. Then the x component of Fac would be the base of the triangle and the y component of the force would the height of the triangle.
Since the hypotenuse makes a 45 degree angle with the base of the triangle, then:
cos(45) = Base/Hypotenuse
sin(45) = Height/Hypotenuse
Since Base is the x component of Fac, Height is the y component of Fac, and Hypotenuse is Fac, we can rewrite the above relationships as:
cos(45) = Fac_x/Fac
sin(45) = Fac_y/Fac
Or,
Fac_x = Fac cos(45)
Fac_y = Fac sin(45).
When writing the equilibrium equations, summing the forces in the x direction requires using Fac_x, or Fac cos(45). And, summing the forces in the y direction requires using Fac_y, or Fac sin(45).
Dr. Structure Thank you for the response! I completely understand! So, cos and sin are interchangeable in this regard. I actually worked all of this on my own actually which I am proud to say but thank you for clarifying! Glad to know I am on the correct path! What was throwing me off was that I wanted to make the equation cos45=Fad/Fac, which gave me Fac cos(45)=Fad and made my equilibrium equation Ax + Fad + Fad.
I noticed the pin and roller support swapped places. Was this intended?
Good observation. :) No, it was done accidentally; the swapping has no significance.
Hello. I have the same question. But is the first one (@2:45) correct? Because solving reaction forces won't be possible because roller can only have upward normal force.
@@luisgabrielcabico9389 The position of the roller/pin support is irrelevant in this case. We can still calculate the support reactions. There is one reaction force at the roller and two reaction forces at the pin. So we have a total of three reactions forces and three static equilibrium equations.
For the diagram shown @2:45, if we set the sum the moments about point B to zero, we get the reaction force at the roller support (at A). Suppose the big triangle has the same height and base, say, 3 meters. The moment equilibrium equation about B can be written as: (5 N)(3 m) + R (3 m) = 0 where R is vertical reaction force at A. Solving the equation for R, we get: R = -5 N. That is, the vertical reaction at the roller is downward with a magnitude of 5 N.
I have always thought that roller supports can only create an upward reaction. Thank you Dr. Structure.
how do you put this animation face? is it in the program you are using
what is it ?
thank you
very great work
Check out CrazyTalk Animator.
@@DrStructure I downloaded it , after making the progect I could not find a way to put it into powerpoint
I ll be very greatful if you answer
You can use CrazyTalk to create a video clip of a talking head/avatar. Is that what you did? Did you generate a video file using CrazyTalk?
You should be able to insert a video file, any video file, into Powerpoint. That is pretty straight forward. And you can play that video from inside PowerPoint. But I suspect that is not what you want to do. What are using PowerPoint for? and why do you want to have the talking-head video inside PowerPoint?
If you are using this to create lecture videos, then you need more than PowerPoint to make it work. You can create the presentation slides using Powerpoint, and publish it as a video (without voice). And use CrazyTalk to create a talking Avatar. Then, use a video editing software like Camtasia Studio to synchronize the audio and video files, and publish the final video.
I am preparing to my phd and thought it would be interesting if I add a face in some slide who says smth 😅
but I cant find a way
I cant save crasytalk progect as vedio
is that because I am using the free trial
anyway
thank u so much
you were very generous
Yes, you need the non-trial version of the software to be able to save.... You're welcome.
You covered a joint method only in 18 minutes .
If I learn this from book then it will take more time
you made this video earlier...??
Yes. This is an updated version of the original video (SA04). A few minor corrections had to be made to the original one in order to make the math more precise.
OK
so good
Excuse me, how to analysis a truss if it need to resist a moment at joint?
If a joint is to be designed to resist moment, then the structure is no longer a pure truss. A moment-resisting joint means there are moment-resisting members which makes the structure a hybrid system, a mix of beam/column and truss elements. A good technique for analyzing such systems is the displacement method (see video lectures SA45 thru SA50).
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In the beginning, it said that the load shouldn't place directly at the truss member. But in the real case, the load is applied at the truss member such as the self-weight of the truss member and another kind of uniform distributed load. I got a case like this imgur.com/GiYFSJv How to translate those uniform distributed load at the member of trusses into a pointed load at the trusses joint. I hope you can share your wisdom, thanks.
You can multiply the magnitude of the distributed load by the distance between the two truss joints. That gives the total load on the span to be applied to its end joints. Divide the total load by 2 to get the load per joint. Example: Suppose the bottom chord of a bridge truss consists of five joints. Let's say the distance between two consecutive joints in the truss is 3 meters, and the distributed load applied to the bottom chord of the truss is 2.6 kN/m.
Since the bottom chord of the truss consists of 4 spans (5 joints), the total load per span is: 2.6x3 = 7.8 kN. Dividing the total load by 2, we get: 3.9 kN per joint.
Let's number the joints along the bottom chord of the truss as 1 through 5.
For the load in the first span, place a concentrated load of 3.9 kN at joint 1 and a load of 3.9 kN at joint 2.
For the load in the second span, place a concentrated load of 3.9 at joint 2, and a load of 3.9 kN at joint 3.
For the load in the third span, place a concentrated load of 3.9 at joint 3, and a load of 3.9 kN at joint 4.
For the load in the fourth span, place a concentrated load of 3.9 at joint 4, and a load of 3.9 kN at joint 5.
Add the joint loads for each span load to get the actual joint loads:
Joint 1: 3.9 kN
Joint 2: 7.8 kN
Joint 3: 7.8 kN
Joint 4: 7.8 kN
Joint 5: 3.9 kN
@@DrStructure Thanks for your wisdom, hope we still connect through this channel.