SA02-A: Stability and Determinacy in Trusses
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- Опубліковано 9 лип 2016
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A pdf version of the lecture is available online at: Lab101.Space/pdf/lectures/SA02...
THANK YOUUUUUUU UGH LITERALLY NOBODY ELSE ON UA-cam IS COVERING THIS TOPIC THOROUGHLY!!!
This is so helpful! Thanks for posting this video.
Thank you. This is the best video on this topic.
I really like your videos! :)
When Subject of study is relevant, I always encourage my students to watch your videos!
Thank you very much!
Thanks for the comment. Glad to be of help to students.
Thank you for the videos, what are the answers for the practice problems of unstable / stable structures...
Amazing video. Thanks
Really wonderful
Thank you it's really helpful videos
thank you Sir ..very helpful video.
thank you dear!!!great man!!!go for top realy
11:50,I think By and Cy both are zero. since the vertical member is a zero force member it will lead to zero force in the inclined members so balancing the forces Bx and Cx will be equal and both Cy and By are zero. However for the left part of the structure if we write moment about A, Cx Ax and Ay are passing through the point A and Cy is zero so there will be a moment due to P which will make the structure unstable anyway
But then moment equilibrium will not be satisfied
Dear Dr. Structure, at 15:39, do we count 6 or 7 joints? I count 7, which means 2x7=14. 10 members. 3 support reactions. so, 14>13, which means unstable. But according to other answers below, seems I am wrong. Can you please help me?
Are you referring to Problem D? If so, you are not incorrect. It is an unstable truss.
I have one question sir??
"If stable then stable."
In the given example one end of the truss is supported with hinged and another end it has roller and both the support is connected with the horizontal member. If this horizontal member i.e., if this bottom chord doesn't exist then the structure would be unstable???
Correct, if that horizontal member connecting the pin to the roller is removed, the truss becomes unstable.
For the stability questions, can you explain how E, G, and J are unstable? Thank you
See the linked pdf for solutions to the three problems.
lab101.space/QA/SA02-A-Exercise%20Problem%20Solutions.pdf
Hi
What is the name of the program that you use for making such wonderful videos?
We use several software applications for creating videos.
Autodesk Graphic (on IPad) for creating text and drawings.
VideoScribe for handwriting simulation.
Adobe Animate for creating animations.
CrazyTalk for voice animation.
Camtasia Studio for compiling the components into a video.
+Dr. Structure
oh yes
thank you very much
Good day Dr Structure.
I don't know if I might have missed out the exact part for the solutions for this tutorial. I don't know how to locate them if they happen to be available already.
We don't have the pdf solutions for the exercise problems yet. But some explanations can be found in the comments section.
@@DrStructure ok thank you
Excuse me, at 07:17, it shows the method of determine a unstable truss, i want to ask if number of equations < number of unknowns, then is the truss stable?
If the number of equations is less than the number of unknowns, the truss could be stable. But it may also be unstable. For example, a beam that rests on many rollers, and nothing else, would give us many unknowns but only a few equations. Yet, the beam since it is not restrained against lateral movement is considered unstable.
Excuse me, at 10:39, does the truss is externally stable?
Yes, that truss is unstable.
How can there be a Cy force@ 12:00 ??
No reactions are present so how..?
Please explain..
The truss is subjected to a downward force of P.
If we cut the structure at C and examine the left(triangular) substructure, where there is Ay and P, then we must conclude that there is also a non-zero Cy at the right end of the substructure. Otherwise, the equilibrium conditions for the substructure cannot be satisfied. Let's assume that P is located half-way between A and C, then Ay = Cy = P/2.
The same Cy must be present in the free-body diagram of the right substructure, the one shown in the video. And if Cy = P/2, then By must have the same magnitude (but opposite direction) since the sum of the forces in the y direction must be zero... Yet, a non-zero By leads to an invalid moment equilibrium equation as stated in the video.
How do you create the fonts for your videos or maybe which fonts do you use. This one of my best fonts.
We have used different fonts, the latest one is called Verveine (regular): www.fonts.com/font/dalton-maag/verveine/regular
@@DrStructure ok thanks. And what is the font name that you used on video SA 02-A?
I believe the font is Times New Roman. But in that video, the written text/equations were traced manually using a stylus on an ipad.
@@DrStructure wow ok i see. I think you should also try the Dubai Font too.
At 11:48 why there is shear force Cy on the right segment?
Although the truss members carry an axial force only, at the hinges where two or more members connect, the sum of the member force could be an inclined vector with x and y components.
Why is he taking number of support reactions as 3 when it is clearly 2 (7:26)? What did i miss? can someone help me out..
The left support is a pin and the right one is a roller. The pin support has two reaction forces, an x-force and a y-force. The roller has only one reaction force, here, a y-force. So we get a total of three reaction forces.
@@DrStructure thanks a bunch☺️
At 9:02 question is there any other method beside method of joint?? Bcz it can be lengthy at times...
For larger structures we often use the matrix (stiffness) method for analyzing trusses. This involves formulating and then inverting a matrix. For an unstable structure the determinant of the matrix would be zero, its inverse does not exist. So, a faster way to do this analysis could be to formulate the stiffness matrix and see if it has a zero determinant.
Thxs for ans.
in this way of checking internal stability by cutting the structure as you done for rectangular shaped truss ....cant we prove even the simplest triangle truss is unstable?
You wrote: “... Can’t we prove even the simplest truss is unstable?”
Show (illustrate) what you mean.
@@DrStructure thank you for your reply... it was my miconception... extremely sorry
@@sidhanv.v7089 Glad to know you've resolved the misconception.
In the last part. Won't it be an unstable truss if you put a downward load at joint c? Need help.
Do you mean the truss @11:50?
The truss is unstable, and we can show its instability in various ways.
So, how do you prove its instability by placing a downward load at C?
@@DrStructure the truss at 18:09
@@luisgabrielcabico9389 No, we cannot show the system is unstable by placing a load at C. Here is a simple thought experiment: Remove all the members and joints to the right of C. What remains is a stable (cantilever) truss. The addition of the substructure to the right of C does not make that the stable system to the left unstable. However, the truss would have been unstable, if there was not support at either A or B. But the two supports, as well as the roller support at D, ensure the stability of the system.
Oh. Thank you😊
How is a truss, which undergoes rigid body translation for an arbitrary load,classified as
Options are
1 geometrically unstable
2 statically unslable
3 structurally unstable
Are you asking why a rigid body movement of a structural system renders it unstable? Because for such structures, the static equilibrium equations cannot be satisfied. If one or more equilibrium equations are violated, the system would not be able to carry the intended load.
@@DrStructure i m asking a rigid body translation of truss comes under which category of unstability given 3 option above by me ?
In the context of static structures, statically unstable and structurally unstable have the same meaning.
If the rigid body movement is local (i.e., only a part of the truss undergoes rigid body motion), the system is said to be geometrically unstable.
A geometrically unstable truss is statically and structurally unstable.
@@DrStructure if whole truss translate then ?
A structure with such a rigid body movement is also unstable. But we don’t refer to that as geometrical instability. We call it external instability.
Do we need to negate zero force members first
What do you mean by negating zero-force members? And for what purpose? Please elaborate.
I can't understand why b and c are unstable.please explain
For B, place a vertical load at the top middle joint. Now remove all the zero-force members from the truss. If we examine the lower left joint, we can conclude that the two members connected to the joint are indeed zero-force members. The same analysis can be applied to the right bottom joint, then the middle bottom joint, then the left middle joint and finally the right middle joint. This simple qualitative analysis shows that all the members except the top two (horizontal) members carry no force. But this remaining truss can be easily shown to be unstable since at the top middle joint there is a vertical applied load but no member force has any component in the y direction. So, the sum of the forces in the y direction at that joint is not zero, the condition of equilibrium cannot be satisfied, hence the truss is unstable.
To show C is unstable, place a horizontal load at the lower right joint of the truss where a roller is located. Then, since there is no load applied at the middle top joint, the two members connected to the joint carry no force. Remove them from the truss. What remains would be a pin, two rollers and two members. The right (inclined member) rests on two rollers, one at top and one at the bottom. That configuration is unstable since it can undergo rigid body motion.
Dr. Structure thank u...
You are welcome.
Dr structure
Can you plz check my answers
A) unstable B) stable C) unstable D) unstable E) unstable F) unstable G) stable H) stable I) stable J) unstable
Plz do explain me where i am wrong....
They are all unstable.
B) Place a vertical force at the middle joint on top of the truss, then identify and remove the zero-force members. What remains are the two horizontal members at the very top, with a vertical force applied in between the two. That configuration is unstable since the sum of the forces in the y direction at the middle joint is not zero.
G) Place a vertical load at the center joint of the truss, then cut the truss just to the right of that center joint thereby exposing the interval forces at the two inclined members. Now, considering the left side of the truss, take the sum of the moments about that middle joint. Since the sum of the moments about the joint must be zero, then the vertical support reaction at the left pin support must be zero, since it is the only force that could create a bending moment about the center point. Using the same line of reasoning we can show that the vertical reaction at the right support should also be zero for the equilibrium conditions to be met. But these vertical reactions cannot be zero, since their sum must add up to the applied load at the center joint.
H): Place a horizontal load at the lower right (roller) support, then identify and remove the zero-force members from the truss. That would result in an unstable truss configuration. The configuration is that of the three supports and two members only, one that connect the pin to the top roller , and one that connects the top and bottom rollers.
I) Place a vertical load at the right-most joint in the truss. Now, take the sum of the moments about the left pin. Since the reaction at the vertical roller passes through the pin connection, the moment sum equals to the applied load times its distance from the pin connection, that moment is not equal to zero, it should be, hence the truss is unstable.
at 5:37, how come there isn't a normal internal force?
Here we are dealing with a truss structure. Truss members carry axial loads only. A truss member is either in tension or compression. Therefore, when drawing free-body diagrams of a part of a truss, we end up having/showing axial member forces only (drawing either in the tension or compression direction). In the free-body diagram shown @5:37, we have cut through two members. The member forces are shown as Ay and Dy. These are the axial forces in the two members. No other forces are present/drawn on the free-body diagram.
@@DrStructure Got it. In the same figure, how come you don't treat it as a frame, but you treat it as a truss instead. In other words, how can you tell the difference between the two. Also, If the structure is a frame, it would have an internal shear and moment, is that correct?
@@nhilang9911 Since the lecture is about trusses, we treated the structure as a truss. In general, truss structures consist of long slender members arranged in such a away that external loads are applied at the joints only. Beams and frames, on the other hand, are bulkier and can be subjected to direct loads of all types. Yes, if the structure in question was a frame, an internal shear force and a bending moment, as well as an axial force, would have been drawn for each member.
Dr. Structure so if the problem is a truss, it will state it. Is that how one knows what type of structure it is?
@@nhilang9911 In textbook problems or in-class discussions, yes, we are often told whether we are dealing with a truss or a frame structure. In real life, however, when you come across a skeleton structure, you should be able to decide if it is a truss or a frame based on the size and arrangement of its members, the type of loads it is intended to carry, and how the loads are being transferred to the elements of the structure.
What is non zero force
A force vector with a magnitude different than zero.
In this vlog the non zero force is applying force or internal force
@1:39, the term is used to indicate an applied horizontal force.
please post the answers for the practice problems
Since these problems can be solved rather quickly, give it a serious try and post your solution here, I will then comment on your solution. That is definitely a more effective learning strategy than the alternative.
+Dr. Structure
i see, okay then.. here it goes:
equations > unstable because it does not prevent horizontal movement in the left part of the truss
(B): 16 = 16 --> something wrong in the middle, can't prove it though
(C): 8 = 8 --> [CAN"T SOLVE]
(D): 12 = 12 --> [CAN"T SOLVE]
(E): 12 > 13 --> concurrent reactions, or too many zero force members [NOT SURE]
(F): 12 = 12 --> unstable because when loaded at -say- the right middle joint, then separating the left part and taking the moment at the joint, the moment equation yields zero reaction which contradicts reaction values calculated earlier
(G): 14 = 14 --> same as F, when separating and taking moment at the joint it yields zero reaction
(H): 10 = 10 --> [CAN"T SOLVE]
(I): 8 = 8 --> concurrent support reactions at the pin support, moment@pin != zero when loaded
(J): same as F & G
degree of indeterminacy:
(A): ext: 1st
(B): int: 1st
(C): int: 1st
(D): ext: 3rd
(E): int: 2nd & ext: 2nd
For instability problems, the ones you missed:
B) Place a vertical load at the joint between the two pin supports, then identify and remove all the zero-force members. You should be able to easily see why the remaining truss is unstable. All the members except the horizontal ones between the two supports are zero-force members.
C) Place a horizontal load at C, then identify and remove all the zero-force members in the truss. What remains is unstable.
D) Cut the truss member attached to the pin support, and show the axial force in the member. Now, draw the free-body diagram of the truss where the pin is replaced by the axial force in the member. This force (which acts as a support reaction) is parallel to the reaction at the roller. Parallel reaction forces means unstable truss.
E) Similar to D above, cut the members attached to the two pins, replace each pin with the axial force in the member attached to the pin. These forces pass through the upper left pin (concurrent reactions), so the truss is unstable.
H) Place a horizontal force at the lower right roller support. Now, remove the zero-force members (there are four of them). What remains can be easily shown to be unstable.
On determinacy problems: There is only one incorrect answer, C is externally indeterminate to the first degree.
Dr. Structure
Thank you, i'm really glad for your support
You are welcome.
the pen is too big and annoying
I agree!