Thank you. Your Videos are really helpful to understand the basics. I haven't seen anyone explaining these thing better then you did in all your videos. Thanks again.
*PLEASSSSEEEEEEE KEEP THIS CHANNEL ALIVE!!!!! AND PLEASE MAKE MORE VIDEOS, 'CAUSE YOU'RE DOING A GREAT JOB!! THANK YOU SO MUCH DOCTOR STRUCTURE!!!* 💖💖💖
Please review Lecture SA02-A for further discussion on this topic, and an answer to your question. Here is the link to the lecture: th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
Structure C has a a hinge at the bottom chord. Hinge does not transfer a moment. So, you can cut the a section that passes through the hinge and solve for the moment. You will find that the moment will not be zero. Since moment is not zero your structure will rotate or accelerate.
Thank you so much Dr. Structure, it's really sad that I found this channel now and courses are superb but it would have been great if you could provide such courses for other subjects as well (Design of Steel Structures, RCC Structures etc.) Please see if possible and seriously thanks a lot!!! 🙌🙌🙌
Beautiful video together with n. 1 clear, amazing, and remarkably comprehensible facilitated by a classic and clear technical language. CONGRATULATIONS truly a very good exhibition initiative in a truly current, important and conceptually difficult subject. I hope you will continue with other groups of videos related to the various topics of "Construction Science", a difficult, difficult and demanding subject, but with good understandable language, it becomes within everyone's reach. Thank you so much. I would think of carrying out a complete course of this material, video after video, starting from the positioning of a rigid body in a three-dimensional space (degrees of freedom), characteristics of the deformation, state of deformation, simple and compound stresses, resistance criteria, bonds tension deformations, the linear elastic problem, non-linear elastic problems, and arriving lesson after lesson with exercises applied at the end of the course. Do not forget ... the problem already exposed to you related to a really interesting course on bridge design. A greeting and thanks.
Solution to exercise problem: 1. Stable 2. Unstable (i)if we remove that horizontal member present in the mid of the truss then it will become stable ,this is one way of solution (ii) IN other way if we provide two more extra members to the given structure then it will become stable. 3. Unstable 4. Unstable ( as the roller doesn't resist vertical movement if we apply the force at any force) ( in other way , in that given truss series of triangles are missing as it seems like a rhombus in the second part of structure. 5. Unstable
@@rahatulislam6150 The number of unknown (member + reaction) forces is less than the number of equilibrium equations. Therefore, the truss is unstable. lab101.space/pdf/exercises/SA02-Exercises.pdf
We can easily guess whether the truss is stable or unstable by physical inspection itself. In the second example of truss there is one extra additional member i.e., member 9 which makes the structure unstable. If we remove that member then structure become stable. We can guess all the trusses using the same idea.A truss is made up of a series of triangle joined together to make it rigid.
Your videos are very superb i like it very much,but kindly i am requesting you to provide more videos on structural analysis which should cover all topics related to civil engineer subjects.thank u so much and congrats sir for your great job.
Thanks for this channel sir. I have a request could you please upload a video of analysis of G+2 building. I mean how load is transferring from slab to beam to column and finishing at foundations with the calculation as shown by you in of your video by moment distribution method. Please take a video like how much slab transferred to beam and how beam is transferring to column with animation. Please sir I want to really understand this basic concept slab-beam-culumn-foundation. Please upload sir
I never knew this kind of approach before I watched this video. This video is fantastic. However, my teacher wants me to use her way to do it. The other solution approach that I meant is by using rigid pieces an number of constraints do you have any video/ explanation on this approach? please reply fast I'm having my finals the day after tomorrow
The only other video we have on stability and determinacy is here: th-cam.com/video/w7rAiqzlanQ/w-d-xo.html But I doubt it matches your needs per your teacher's requirements.
i would like to know, if you can put a charge somewhere in the trellis that involve all the trellis to work (i mean by that that you don'T apply the load on the support, or too close to it that half of the trellis are not used), if you are able to find all the force with equilibrium equation with the joint method, it mean that it is stable? Or is it possible that a trellis is stable until a load is placed at a specific place?
+mrmadame28 A structure does not have to be subjected to any load to be classified as unstable. A structure is considered unstable if there exists a load (even though at the time it is not actually applied to the structure) such that if placed on the structure will cause rigid-body motion.
Dr. Structure ok. That's because my teacher ask me if some trellis are stable or not, i can find instability and prove them when it is unstable. But when it is stable, im not sure how i can prove that it is stable beside of simulating every load possible and resolving the trellis... it is kind of very very long and pretty boring... Yet i vaguely remember a teacher saying that when a load is wisely chosen and the trellis can be resolved, than it is stable, else it will not work. If i remember, the "wisely" refer to loading the trellis in a way that the stress would be distributed along the trellis and not localy.. But im not sure, it's 3 years ago so
Can you explain how B is unstable? Also, even though some of the trusses passed the instability test of 2J = M + R, that didn't mean they were unstable. We had to analyze the structure further to determine instability. However, it seems that process is more complicated since there isn't any rule we can use and we just simply put loads in random places and see if any instability occurs. Are there any simpler process to it?
There could be times that 2J = M + R, yet the truss is still unstable. But the reverse is not true. If the number of unknowns is less than the number of equations, we have instability. You are right, there are no encompassing fast and easy rules for deciding the stability of structures. In the case of Problem B, we can use the notion of zero-force members to prove the instability of the truss. Suppose we place a unit load at the middle right node (the one on the inclined member), and assume it to be the only force applied to the truss. Then, the vertical member is a zero-force member which we can remove from the truss. Also, the horizontal member connecting the two middle nodes is a zero-force member. If remove it from the truss, we end up with a triangle sitting on a pin and a roller, with the applied load acting at the middle of the right inclined member. By definition, such a truss is conisdered unstable.
Either way. As long as the force is not in the direction of the two members, it is going to have a component in the perpendicular direction. And since it would be the only force in that direction, then the sum of the forces in that direction cannot be zero.
Tell me your line of reasoning for solving this problem. What prevents you from reaching the conclusion , what confuses you, what is (where is) your starting point for solving the problem, and where do you get stuck?
A great lecture dr. Structure i am impressed with it... Please upload also about Rcc amd steel structure, geotechnical engineering... All civil engineering related course if it is possible please...
Hello Dr. Structure... This is great video I have a query that how to ensure after getting the mathematical stability(means by ensuring that R = N or R + B = 2J ) that the truss will be stable? as in your examples you have told that we should see the forces and moments equilibrium but it will not be easy to check this for large structure *IN THOSE CASES DO WE JUST NEED TO WORRY ABOUT THE UNSTABLE TYPE OF GEOMETRY(LIKE RECTANGULAR) ? OR THERE ARE SOME OTHER CRITERIAS AS WELL ??*
In the final analysis, it comes down to showing whether or not the equilibrium equations can be satisfied for ALL possible loading scenarios. If the answer is NO, the structure is considered unstable. Checking the R + B = 2J equation is not sufficient by itself. We have an updated lecture on this topic which you can found in our (free) online course. The link is provided in the video description field.
dr is it always necessary that a truss having rectangular pattern will always be unstable. consider the case of a truss having a rectangular member with roller support at one end and pin support at another end ,and its top member is restrained by fix support ,does this truss is unstable .
Generally speaking a rectangular pattern signifies instability. But there are always exceptions, like your example. If a rectangular configuration is supported by a pin and a roller at the bottom joints, and the top joints are also restrained by a pin and a roller (there is no fixed support in trusses), then no rigid-body movement can take place, meaning we have a stable configuration.
Think of a spring that we can pull or push. How much the spring elongates, when we apply a tension force to it is a function a property called spring stiffness. The higher the stiffness, the less it is going to elongate due to the applied force. We can write this relationship as f = k d where f is the force, d is the elongation and k is the stiffness of the member. Stiffness is a function of both material and section properties of the member. A structure can be viewed as a system of springs, that is, each structural member resists rotation and displacement the same way that a spring resists elongation. When we have a set of structural elements forming a system that resists loads, then the simple spring equation becomes a system of equations where f is a force vector, d is a displacement vector, and k is a two-dimensional matrix of stiffness coefficients which dictates/controls how much the systems deforms under the applied loads.
Thanks for the feedback. For this particular presentation, we traced the lecture notes using a stylus on an ipad and saved the results as a set of svg files, which were converted into video files using VideoScribe. The video files were then synchronized with audio using Camtasia Studio.
A pin/hinge like that is a source of instability. If a vertical external force is to be applied to that joint, then the sum of the forces in the y direction would not be zero, meaning the truss would not be in equilibrium.
excuse me Dr Structure, if i have a truss (2 pin supports, 10 members and 7 joints) externally indeterminate and internally determinate, so whats the overall status?
For the equation used, What does it mean by the no. Of members? If a member is connected at 90 degree with another member with no pin support in between, do I treat it as one member or two member
For trusses, the straight bars are always assumed to be connected to each other using pins. So if two members are connected a 90-degree angle, we assume there is a pin connecting the two at that intersection.
Bending moment at the ends of a beam is always zero, if the ends rest on pin or roller. However, within the member (away from the ends) bending moment is zero where there is an internal hinge. If the beam does not have an internal hinge, but it rests on an interior pin or roller support, the internal bending moment in the beam at that support point is not zero.
It is a good rule-of-thumb to suspect instability if a non-triangular configuration exists, but you need to further investigate it. Why? because we could have stable trusses that are not formed using simple triangles.
That is one way to do it, though I would not recommend it. Take a look the following video for another perspective on this topic: th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
Yes, you can find them in many bridges. For example, see: www.dot.state.mn.us/historicbridges/4700.html and commons.wikimedia.org/wiki/File:Roller_Bearing,_Jensen_Drive_(Hill_Street)_Bridge_over_Buffalo_Bayou,_Houston,_Texas_1310261116_(10577360794).jpg New bridges often utilize bearing pads, which for analysis purposes behave like rollers. see: www.archiexpo.com/prod/mageba/product-126411-1333289.html and vsl.com/home/technologies/bearings/
Why is A) unstable? It embodies only triangular patterns as you mention @ th-cam.com/video/Oj8hIdXukkE/w-d-xo.html. I don't understand exactly the "cut method". If I were to cut the triangular stable system somewhere near the midpoint, how are equilibrium conditions verified?
"A) embodies only triangles..." Not quite. By a stable triangle we mean three joints (vertices) connected together using three members. Consider the following process. We start with a stable triangle. Then add a joint and two members to the triangle to form a second stable triangle. Now we have two stable triangles. Keep adding a joint and two members (forming a stable triangle) until an entire truss is constructed. That truss is stable. A truss that cannot be constructed that way may be unstable. The truss in Problem A does not fall into that category. Please watch the following video for additional explanation on stability: th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
It is important to make a distinction between bending moment at a point (like at Point A) and the moment equation that we can write about that point. They are not the same thing. One refers to the actual bending moment the other is an algebraic equation which represents the sum of the moments about the point. It is true that bending moment at A is zero, but that does not mean we cannot sum the moments about the point. Regardless of the value of bending moment at A, if sum of the moments about A (caused by other forces acting on the structure) is not zero the structure would not be in equilibrium.
+Taha Mohammed As a general rule, stability of a structure can be decided by formulating the system of equations necessary for its analysis. If that system (often linear in nature) has a unique solution, then the structure is stable. If the system of (linear) equations has no unique solution. For example, if it has a zero determinant, then the structure is unstable.
Here unknowns are less, hence it can be solved using equilibrium equations. Hence it is a determinate structure. But it doesn't mean that it is stable. Actually here 11
when we consider the basic form of a truss that is a triangle, it will be having three joints right ? for that if we are using 2j, it will be 6. But actually it is having only three equations and the other three are one plus in all the members no ? but after that, if we are increasing the number of joints, exactly two equations we will be getting right ? that is why i said that 2j-3. Is my assertion right ?
Okay, I think I see where the misconception comes from. A simply supported truss, with a roller and pin support, has exactly three (3) support reactions. Say, the truss has j joints and m members. Then, we say: There are 2j equilibrium equations and m + 3 unknown forces (the truss has m unknown member forces and 3 unknown support reactions). For us to be able to solve for the unknowns, the number of equations must be equal to the number of unknowns, or: m + 3 = 2j. If the above condition holds true, then we have a determinate truss. We can rephrase (rewrite) the above condition as: m = 2j - 3 That is, if the number of members (in a simply supported truss with a roller and pin support only) equals to 2 times the number of joints minus 3, then we call the truss statically determinate. Here, m is the number of members, not the number of equations.
please excuse me if am annoying ! First of all Thank you for that explanation ! But i want to know if a frame is having end conditions not as a pinned and a roller, will that expression 2j-3 goes wrong while determining the number of members required for the structure to be determinate ? Asking this because i have seen in a book that the degree of internal indeterminacy can be found out using the formula m-(2j-3) for pin jointed plane frames !
You are not annoying at all. I would be delighted to help clarify issues and answer your questions. Number 3 that appears in the equation is a reference to the number of support reactions. Yes, if the number of unknown support reactions is different than 3, you should not use that expression to decide if the truss is determinate. Although in most cases determinate trusses have 3 reactions only, but there are exceptions.
Three (3) support reactions. Eight (8) members. Six (6) joints. Number of Equations = 2*6 = 12 Number of Unknowns = 3 + 8 = 11 Number of Unknowns < Number of Equations Therefore, the truss is unstable.
In both cases the number of equations is more than the number of unknowns, hence the instability. For A, we have: 8 joints = 16 equations 12 unknown member forces + 3 support reactions = 15 unknown forces For E, we have: 6 joints = 12 equations 8 unknown member forces + 3 support reactions = 11 unknown forces
We must have missed your question. I can only see your solution for the exercise problems that you posted recently. But do not see any questions posed, with a question mark at the end. :)
@@DrStructure oh I c . But I didn't asked any query. I had posted two comments but it seems missing. It's ok.. give me reply for my solution whether it is correct or not?
If a vertical load is applied to the middle bottom joint, the sum of the forces in the y direction at that joint does not vanish, the equilibrium condition at that joint cannot be maintained. Except for the applied load, there is no member force (or force component) in the y direction at that joint. And since the applied load is not zero, then the sum of the forces in the y direction is not zero either.
I like Structural Analysis by Russel Hibbeler, and Structural Analysis by Aslam Kassimali. Although you can also find many other good textbooks on the subject.
Dr. Structure Thanks a million. I can't get enough of your tutorials and i subcribed too. You are the best teacher. Nothing is better than putting real situation or examples like you do)
Please i want to be very good at structures and dynamics of structures, but am struggling with understanding and brainstorming that involves structures. Please what is your advice ??? I have this passion for it
You need to enroll in a civil/structural engineering program, if you haven't already. That gives you the best opportunity to take various classes and interact with professors and like-minded students. Although, structural dynamics is not something generally studies in depth in undergraduate programs; you need to be in a graduate program for that. At that present time, learning and interacting online is not a good substitute for traditional academic programs.
+au yinglun We can show the truss is externally unstable with rather ease. The structure has a pin support at the left end (say, point A) and a vertical roller at the right end (say, point B). So there is one (horizontal) support reaction at B and two (one horizontal and one vertical) reactions at A. Suppose we apply a downward load (P) on one of the top joints of the truss. So, the load is placed on the truss between A and B. Say, the distance from the point of application of the load to point B is L. Now, sum the moments about B. This sum must be zero if the structure is to remain in equilibrium. Here is the sum: L(P) = 0. Notice that all the support reactions pass through B, so they vanish in the equation. The only non-zero term is L times P. But neither L nor P is zero. Therefore, the equilibrium equation cannot be satisfied. This implies the structure is unstable.
On your explaining, sum the moments about B must be L(P) + or - (Ay)L' = 0, Ay is vertical reaction at A, L' is the distance from the point A to point B. But everything will be fine if we sum the moments about A. Otherwise, if the position of point B go up for a short distance then the sum moments about A or B will be satisfied, so how do we prove this truss is unstable?
A stable structure must remain stable under any and all loading conditions. So, if it can be shown to be unstable under only one loading case, then it is unstable. For Problem D, we can also prove instability since the truss has concurrent reactions, all the support reactions pass through the left pin support. This means if we place a vertical load at one of the top, or bottom nodes of the truss, then sum of the moments about the left support cannot be made to be zero.
How about the vertical reaction of Pin A? Wouldn't the vertical reaction of A be equal to whatever the downward load of P will be? So, the equation would be sum of moments on B is PL + Ay*(Distance from A to B)?
Yes, the equation that you mentioned works, it can be satisfied. But that is not enough for concluding that the structure is stable. For it to be in stable (in equilibrium), no matter which point we pick, the sum of the moments about that point must vanish. If there exists a point, any point, on or off the structure that makes any of the equilibrium equations invalid, the structure is considered unstable.
Thank you. Your Videos are really helpful to understand the basics. I haven't seen anyone explaining these thing better then you did in all your videos. Thanks again.
*PLEASSSSEEEEEEE KEEP THIS CHANNEL ALIVE!!!!! AND PLEASE MAKE MORE VIDEOS, 'CAUSE YOU'RE DOING A GREAT JOB!!
THANK YOU SO MUCH DOCTOR STRUCTURE!!!* 💖💖💖
Thanks for your comment and feedback.
I was having tough time with structural analysis 😀 thanks a million for taking huge time to make this presentations
A) r=3 b=12 j=8 so 15
Why are C and D unstable if they're both 0?
Please review Lecture SA02-A for further discussion on this topic, and an answer to your question. Here is the link to the lecture: th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
Structure C has a a hinge at the bottom chord. Hinge does not transfer a moment. So, you can cut the a section that passes through the hinge and solve for the moment. You will find that the moment will not be zero. Since moment is not zero your structure will rotate or accelerate.
for option D. If you take the moment at the center joint, the vertical force a the pin will cause moment.
QANIE ADNAN d is stable
Thank you so much Dr. Structure, it's really sad that I found this channel now and courses are superb but it would have been great if you could provide such courses for other subjects as well (Design of Steel Structures, RCC Structures etc.) Please see if possible and seriously thanks a lot!!! 🙌🙌🙌
Beautiful video together with n. 1 clear, amazing, and remarkably comprehensible facilitated by a classic and clear technical language. CONGRATULATIONS truly a very good exhibition initiative in a truly current, important and conceptually difficult subject.
I hope you will continue with other groups of videos related to the various topics of "Construction Science", a difficult, difficult and demanding subject, but with good understandable language, it becomes within everyone's reach. Thank you so much.
I would think of carrying out a complete course of this material, video after video, starting from the positioning of a rigid body in a three-dimensional space (degrees of freedom), characteristics of the deformation, state of deformation, simple and compound stresses, resistance criteria, bonds tension deformations, the linear elastic problem, non-linear elastic problems, and arriving lesson after lesson with exercises applied at the end of the course. Do not forget ... the problem already exposed to you related to a really interesting course on bridge design. A greeting and thanks.
I have my first SA exam tomorrow. Thanks for the awesome videos
Solution to exercise problem:
1. Stable
2. Unstable (i)if we remove that horizontal member present in the mid of the truss then it will become stable ,this is one way of solution (ii) IN other way if we provide two more extra members to the given structure then it will become stable.
3. Unstable
4. Unstable ( as the roller doesn't resist vertical movement if we apply the force at any force) ( in other way , in that given truss series of triangles are missing as it seems like a rhombus in the second part of structure.
5. Unstable
They are all unstable. See: lab101.space/pdf/exercises/SA02-Exercises.pdf
@@DrStructure Can you kindly explain why A is unstable?
@@rahatulislam6150 The number of unknown (member + reaction) forces is less than the number of equilibrium equations. Therefore, the truss is unstable.
lab101.space/pdf/exercises/SA02-Exercises.pdf
We can easily guess whether the truss is stable or unstable by physical inspection itself. In the second example of truss there is one extra additional member i.e., member 9 which makes the structure unstable. If we remove that member then structure become stable. We can guess all the trusses using the same idea.A truss is made up of a series of triangle joined together to make it rigid.
Again looking back into your videos to just memorize the basics...❤️
great work Dr. Structure. . .!!!!
i m very thankful to u. .
u make this subject very easy . .
thanks again. .
Its a awesome channel with great explanation! Please start with lectures related to Structural Engineering subjects !
so this video has nothing to do with structural engineering?
Your videos are very superb i like it very much,but kindly i am requesting you to provide more videos on structural analysis which should cover all topics related to civil engineer subjects.thank u so much and congrats sir for your great job.
very nicely explained in lesser amt of time.100 percent attention is a must
Amazing channel .. thanks a lot
Amazing explanation wow what a great work....
Thank you Dr. Structure it is so great videos...
great work Dr. Structure
thank you
this channel is just awesome
thanks a lot
by inspection, a and c are unstable and using the formulae for BDE to confirm they are also unstable, therefore, all are unstable. am i right?
Yes, they are all unstable.
Thanks for this channel sir. I have a request could you please upload a video of analysis of G+2 building. I mean how load is transferring from slab to beam to column and finishing at foundations with the calculation as shown by you in of your video by moment distribution method. Please take a video like how much slab transferred to beam and how beam is transferring to column with animation.
Please sir I want to really understand this basic concept slab-beam-culumn-foundation.
Please upload sir
Sir how to design a bicycle frame truss element. And how v can say that the design is safe?
I never knew this kind of approach before I watched this video. This video is fantastic. However, my teacher wants me to use her way to do it. The other solution approach that I meant is by using rigid pieces an number of constraints do you have any video/ explanation on this approach? please reply fast I'm having my finals the day after tomorrow
The only other video we have on stability and determinacy is here: th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
But I doubt it matches your needs per your teacher's requirements.
Guys check the description for the solution!
i would like to know, if you can put a charge somewhere in the trellis that involve all the trellis to work (i mean by that that you don'T apply the load on the support, or too close to it that half of the trellis are not used), if you are able to find all the force with equilibrium equation with the joint method, it mean that it is stable? Or is it possible that a trellis is stable until a load is placed at a specific place?
+mrmadame28 A structure does not have to be subjected to any load to be classified as unstable. A structure is considered unstable if there exists a load (even though at the time it is not actually applied to the structure) such that if placed on the structure will cause rigid-body motion.
Dr. Structure ok. That's because my teacher ask me if some trellis are stable or not, i can find instability and prove them when it is unstable. But when it is stable, im not sure how i can prove that it is stable beside of simulating every load possible and resolving the trellis... it is kind of very very long and pretty boring... Yet i vaguely remember a teacher saying that when a load is wisely chosen and the trellis can be resolved, than it is stable, else it will not work. If i remember, the "wisely" refer to loading the trellis in a way that the stress would be distributed along the trellis and not localy.. But im not sure, it's 3 years ago so
Can you explain how B is unstable?
Also, even though some of the trusses passed the instability test of 2J = M + R, that didn't mean they were unstable. We had to analyze the structure further to determine instability. However, it seems that process is more complicated since there isn't any rule we can use and we just simply put loads in random places and see if any instability occurs. Are there any simpler process to it?
There could be times that 2J = M + R, yet the truss is still unstable. But the reverse is not true. If the number of unknowns is less than the number of equations, we have instability.
You are right, there are no encompassing fast and easy rules for deciding the stability of structures. In the case of Problem B, we can use the notion of zero-force members to prove the instability of the truss. Suppose we place a unit load at the middle right node (the one on the inclined member), and assume it to be the only force applied to the truss. Then, the vertical member is a zero-force member which we can remove from the truss. Also, the horizontal member connecting the two middle nodes is a zero-force member. If remove it from the truss, we end up with a triangle sitting on a pin and a roller, with the applied load acting at the middle of the right inclined member. By definition, such a truss is conisdered unstable.
Are you assuming the unit load is placed perpendicularly to the truss or is it vertical?
Either way. As long as the force is not in the direction of the two members, it is going to have a component in the perpendicular direction. And since it would be the only force in that direction, then the sum of the forces in that direction cannot be zero.
Tell me your line of reasoning for solving this problem. What prevents you from reaching the conclusion , what confuses you, what is (where is) your starting point for solving the problem, and where do you get stuck?
A great lecture dr. Structure i am impressed with it... Please upload also about Rcc amd steel structure, geotechnical engineering... All civil engineering related course if it is possible please...
Hello Dr. Structure...
This is great video
I have a query that how to ensure after getting the mathematical stability(means by ensuring that R = N
or R + B = 2J ) that the truss will be stable?
as in your examples you have told that we should see the forces and moments equilibrium but it will not be easy to check this for large structure
*IN THOSE CASES DO WE JUST NEED TO WORRY ABOUT THE UNSTABLE TYPE OF GEOMETRY(LIKE RECTANGULAR) ? OR THERE ARE SOME OTHER CRITERIAS AS WELL ??*
In the final analysis, it comes down to showing whether or not the equilibrium equations can be satisfied for ALL possible loading scenarios. If the answer is NO, the structure is considered unstable. Checking the R + B = 2J equation is not sufficient by itself.
We have an updated lecture on this topic which you can found in our (free) online course. The link is provided in the video description field.
Super, Thanx , keep on your Project!!!
dr is it always necessary that a truss having rectangular pattern will always be unstable. consider the case of a truss having a rectangular member with roller support at one end and pin support at another end ,and its top member is restrained by fix support ,does this truss is unstable .
Generally speaking a rectangular pattern signifies instability. But there are always exceptions, like your example. If a rectangular configuration is supported by a pin and a roller at the bottom joints, and the top joints are also restrained by a pin and a roller (there is no fixed support in trusses), then no rigid-body movement can take place, meaning we have a stable configuration.
Thanx dr for u r quick reply except on the question of future video but their is always exception as u said.....
Can I ask a question outside the lecture.... what is the physical meaning of stiffness method?!
Think of a spring that we can pull or push. How much the spring elongates, when we apply a tension force to it is a function a property called spring stiffness. The higher the stiffness, the less it is going to elongate due to the applied force. We can write this relationship as f = k d where f is the force, d is the elongation and k is the stiffness of the member. Stiffness is a function of both material and section properties of the member.
A structure can be viewed as a system of springs, that is, each structural member resists rotation and displacement the same way that a spring resists elongation. When we have a set of structural elements forming a system that resists loads, then the simple spring equation becomes a system of equations where f is a force vector, d is a displacement vector, and k is a two-dimensional matrix of stiffness coefficients which dictates/controls how much the systems deforms under the applied loads.
Very good presentatiin
Clear concept
Thank yoy
May I know with what app do you prepare this presentation
Thanks for the feedback. For this particular presentation, we traced the lecture notes using a stylus on an ipad and saved the results as a set of svg files, which were converted into video files using VideoScribe. The video files were then synchronized with audio using Camtasia Studio.
In the given problem no.C,how to deal with that internal hinge????or how can we analyse that given structure????
A pin/hinge like that is a source of instability. If a vertical external force is to be applied to that joint, then the sum of the forces in the y direction would not be zero, meaning the truss would not be in equilibrium.
Dr. Structure thank u....
excuse me Dr Structure, if i have a truss (2 pin supports, 10 members and 7 joints) externally indeterminate and internally determinate, so whats the overall status?
That would be considered an indeterminate structure since we cannot analyze it completely using the static equilibrium equations only.
Dr, may you make a video about stability and indeterminacy in frames and composite structures, please ?!
It is really helpful
For the equation used, What does it mean by the no. Of members?
If a member is connected at 90 degree with another member with no pin support in between, do I treat it as one member or two member
For trusses, the straight bars are always assumed to be connected to each other using pins. So if two members are connected a 90-degree angle, we assume there is a pin connecting the two at that intersection.
Awesome video keep up the great work 👍
Awesome plz keep it up..
Thank you so much 😊😊😊😊😊
In which case we should not assume the bending moment zero at internal support? Plz i have confusion
Bending moment at the ends of a beam is always zero, if the ends rest on pin or roller. However, within the member (away from the ends) bending moment is zero where there is an internal hinge.
If the beam does not have an internal hinge, but it rests on an interior pin or roller support, the internal bending moment in the beam at that support point is not zero.
Dr structure, if I understood correctly, structures with non-triangular shapes are unstable, is this right?
It is a good rule-of-thumb to suspect instability if a non-triangular configuration exists, but you need to further investigate it. Why? because we could have stable trusses that are not formed using simple triangles.
So we investigate further using method of sections to check if each section is individually stable, right?
That is one way to do it, though I would not recommend it. Take a look the following video for another perspective on this topic:
th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
very good, thanks
Thank you for these videos, it is helpfull ^^
great explaining
procedure for sturucture is stable example
Is a "roller" type of support actually used on real structure, like a bridge?
Yes, you can find them in many bridges. For example, see:
www.dot.state.mn.us/historicbridges/4700.html
and
commons.wikimedia.org/wiki/File:Roller_Bearing,_Jensen_Drive_(Hill_Street)_Bridge_over_Buffalo_Bayou,_Houston,_Texas_1310261116_(10577360794).jpg
New bridges often utilize bearing pads, which for analysis purposes behave like rollers. see:
www.archiexpo.com/prod/mageba/product-126411-1333289.html
and
vsl.com/home/technologies/bearings/
@@DrStructure Thank you very much. I will look more closely the next time I see a truss style bridge.
what application do you use ?
For this specific video, VideoScribe and Camtasia Studio.
@@DrStructure Thankyou
You're welcome!
Why is A) unstable? It embodies only triangular patterns as you mention @ th-cam.com/video/Oj8hIdXukkE/w-d-xo.html.
I don't understand exactly the "cut method". If I were to cut the triangular stable system somewhere near the midpoint, how are equilibrium conditions verified?
"A) embodies only triangles..." Not quite. By a stable triangle we mean three joints (vertices) connected together using three members.
Consider the following process. We start with a stable triangle. Then add a joint and two members to the triangle to form a second stable triangle. Now we have two stable triangles. Keep adding a joint and two members (forming a stable triangle) until an entire truss is constructed. That truss is stable. A truss that cannot be constructed that way may be unstable. The truss in Problem A does not fall into that category.
Please watch the following video for additional explanation on stability:
th-cam.com/video/w7rAiqzlanQ/w-d-xo.html
You are a beast! You actually answered my question within hours, wow!!! Thank you! :)
can you please clarify why at 3:18 the force on the roller is Bx and not By in the vertical direction???
Because the roller is not acting on the horizontal plane, it has been turned 90 degrees, it acts on the vertical plane.
answer of the problems how we marked whether we are right or wrong
You should be able to find what you are looking for in the comment section.
thank you i just cleared this years upsc ese prilims and videos of yours helped me a lot that the burden of books doesnt
can you please make a video on three moment theorm and bowgirder analysis
As there is hinged support at A. so how can you get moment at point A(3:38)
It is important to make a distinction between bending moment at a point (like at Point A) and the moment equation that we can write about that point. They are not the same thing. One refers to the actual bending moment the other is an algebraic equation which represents the sum of the moments about the point. It is true that bending moment at A is zero, but that does not mean we cannot sum the moments about the point. Regardless of the value of bending moment at A, if sum of the moments about A (caused by other forces acting on the structure) is not zero the structure would not be in equilibrium.
hey ! can i ask doubts in structures which are not included in these videos ?
Yes!
Wow! this is great. keep it up Dr :)
thank you😍😍😍😍
you gave us two examples about the unstable case !... So how can know about stable???!
+Taha Mohammed As a general rule, stability of a structure can be decided by formulating the system of equations necessary for its analysis. If that system (often linear in nature) has a unique solution, then the structure is stable. If the system of (linear) equations has no unique solution. For example, if it has a zero determinant, then the structure is unstable.
when no. of unknows is less than no. of equations then the truss is stable i.e 11
Your assertion is not true. If the number of unknowns is less than the number of equations the structure is considered unstable.
Here unknowns are less, hence it can be solved using equilibrium equations. Hence it is a determinate structure. But it doesn't mean that it is stable. Actually here 11
yes, you are correct.
I m loving it
Why not
We are working on a pdf version of the solution to the exercise problems. It should be available in a few days.
hai,
Here it is mentioned that the number of equlibrium equations for a truss is 2j. Isn't it 2j-3?
If we the truss has j joints and we can write two equations per joint, then we get 2j for the number of equations. Why would you subtract 3 from that?
when we consider the basic form of a truss that is a triangle, it will be having three joints right ? for that if we are using 2j, it will be 6. But actually it is having only three equations and the other three are one plus in all the members no ? but after that, if we are increasing the number of joints, exactly two equations we will be getting right ? that is why i said that 2j-3. Is my assertion right ?
Okay, I think I see where the misconception comes from.
A simply supported truss, with a roller and pin support, has exactly three (3) support reactions. Say, the truss has j joints and m members. Then, we say:
There are 2j equilibrium equations and m + 3 unknown forces (the truss has m unknown member forces and 3 unknown support reactions).
For us to be able to solve for the unknowns, the number of equations must be equal to the number of unknowns, or:
m + 3 = 2j.
If the above condition holds true, then we have a determinate truss. We can rephrase (rewrite) the above condition as:
m = 2j - 3
That is, if the number of members (in a simply supported truss with a roller and pin support only) equals to 2 times the number of joints minus 3, then we call the truss statically determinate. Here, m is the number of members, not the number of equations.
please excuse me if am annoying ! First of all Thank you for that explanation ! But i want to know if a frame is having end conditions not as a pinned and a roller, will that expression 2j-3 goes wrong while determining the number of members required for the structure to be determinate ? Asking this because i have seen in a book that the degree of internal indeterminacy can be found out using the formula m-(2j-3) for pin jointed plane frames !
You are not annoying at all. I would be delighted to help clarify issues and answer your questions.
Number 3 that appears in the equation is a reference to the number of support reactions. Yes, if the number of unknown support reactions is different than 3, you should not use that expression to decide if the truss is determinate. Although in most cases determinate trusses have 3 reactions only, but there are exceptions.
amazing...!!!
why the (E) is unstable?
Three (3) support reactions.
Eight (8) members.
Six (6) joints.
Number of Equations = 2*6 = 12
Number of Unknowns = 3 + 8 = 11
Number of Unknowns < Number of Equations
Therefore, the truss is unstable.
Dr Structure, why are A and E unstable?
In both cases the number of equations is more than the number of unknowns, hence the instability.
For A, we have:
8 joints = 16 equations
12 unknown member forces + 3 support reactions = 15 unknown forces
For E, we have:
6 joints = 12 equations
8 unknown member forces + 3 support reactions = 11 unknown forces
Dr. Structure Thank you very much for your reply.
You're welcome.
Where can we find answers??
You should find them here, in the comment section.
What happened to dr. Structure team? Yet no reply.
We must have missed your question. I can only see your solution for the exercise problems that you posted recently. But do not see any questions posed, with a question mark at the end. :)
@@DrStructure oh I c . But I didn't asked any query. I had posted two comments but it seems missing. It's ok.. give me reply for my solution whether it is correct or not?
It is there after scrolling the comment section I have found it.
why C is unstable?
If a vertical load is applied to the middle bottom joint, the sum of the forces in the y direction at that joint does not vanish, the equilibrium condition at that joint cannot be maintained. Except for the applied load, there is no member force (or force component) in the y direction at that joint. And since the applied load is not zero, then the sum of the forces in the y direction is not zero either.
which book would you recommend for me???
thanks
I like Structural Analysis by Russel Hibbeler, and Structural Analysis by Aslam Kassimali. Although you can also find many other good textbooks on the subject.
Dr. Structure Thanks a million.
I can't get enough of your tutorials and i subcribed too. You are the best teacher.
Nothing is better than putting real situation or examples like you do)
Please i want to be very good at structures and dynamics of structures, but am struggling with understanding and brainstorming that involves structures.
Please what is your advice ???
I have this passion for it
You need to enroll in a civil/structural engineering program, if you haven't already. That gives you the best opportunity to take various classes and interact with professors and like-minded students. Although, structural dynamics is not something generally studies in depth in undergraduate programs; you need to be in a graduate program for that. At that present time, learning and interacting online is not a good substitute for traditional academic programs.
May I ask why Question D is unstable, how to cut the sturcture?
+au yinglun We can show the truss is externally unstable with rather ease. The structure has a pin support at the left end (say, point A) and a vertical roller at the right end (say, point B).
So there is one (horizontal) support reaction at B and two (one horizontal and one vertical) reactions at A.
Suppose we apply a downward load (P) on one of the top joints of the truss. So, the load is placed on the truss between A and B. Say, the distance from the point of application of the load to point B is L.
Now, sum the moments about B. This sum must be zero if the structure is to remain in equilibrium. Here is the sum:
L(P) = 0.
Notice that all the support reactions pass through B, so they vanish in the equation. The only non-zero term is L times P. But neither L nor P is zero. Therefore, the equilibrium equation cannot be satisfied. This implies the structure is unstable.
On your explaining, sum the moments about B must be L(P) + or - (Ay)L' = 0, Ay is vertical reaction at A, L' is the distance from the point A to point B. But everything will be fine if we sum the moments about A.
Otherwise, if the position of point B go up for a short distance then the sum moments about A or B will be satisfied, so how do we prove this truss is unstable?
A stable structure must remain stable under any and all loading conditions. So, if it can be shown to be unstable under only one loading case, then it is unstable.
For Problem D, we can also prove instability since the truss has concurrent reactions, all the support reactions pass through the left pin support. This means if we place a vertical load at one of the top, or bottom nodes of the truss, then sum of the moments about the left support cannot be made to be zero.
How about the vertical reaction of Pin A? Wouldn't the vertical reaction of A be equal to whatever the downward load of P will be? So, the equation would be sum of moments on B is PL + Ay*(Distance from A to B)?
Yes, the equation that you mentioned works, it can be satisfied. But that is not enough for concluding that the structure is stable. For it to be in stable (in equilibrium), no matter which point we pick, the sum of the moments about that point must vanish. If there exists a point, any point, on or off the structure that makes any of the equilibrium equations invalid, the structure is considered unstable.
You should post a pay pal or something if people want to give back.
really nice and cool presentation. (y) (y)
Where is the answer ??!!!
You should be able to find the answers in the comment section.
The link to solutions is given in the video description field.
👍👌
B, C, D, E are unstable