SA26: Force Method (Truss Analysis)

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[@williamvisgauss9391]

It really is amazing that so many professors make this much more complicated than it has to be.

[@johnreagan4]

Mahn! So true

[@ThePriince786]

You are the best, keep the great content up.

[@accessuploads7834]

Always I am waiting the new video....you are great

[@DrStructure]

Thanks, I appreciate your input.

[@njwu4117]

At 12:37, the force in member BD might be incorrect. I get 15.5*5/sqrt(41)+0.84*10/sqrt(116)=12.88., not 14.

[@koasteroid6882]

Its a really helpful vidieo

[@niteshmeena9961]

How to judge the direction of unit load. For finding triangle X. You took in same direction of displacement. But in finding delta x you took at opposite direction of displacement. In question 1

[@DrStructure]

When a force or displacement is unknown, we can assume any direction for it. As long as we are consistent in our formulation, it will work out at the end. If we end up with a negative value for the assumed force/displacement, then we know that actual direction is opposite to what we've assumed.

[@greenjellymc6850]

Which lecture should I watch to determine how you got 4kN, -17.26kN, etc? 4:57 Thanks in advance

[@DrStructure]

We have a few lectures on truss analysis. I suggest watching: kzbin.info/www/bejne/ramkpJ2vrKeVg8k

[@karlcorrz]

regarding with the internally indeterminate case, what if my assumptions for the directions of the forces are opposite?

[@DrStructure]

You will then end up with a negative value for the magnitude of the force indicating that the force is acting in the opposite direction.

[@bhattaraibikash3805]

Sir need a force method of frame analysis plz

[@willrichards1454]

At 10:43, how did you find the force based on the virtual unit force (i.e method of joints, sections?). Also, are there any external forces for this?

[@DrStructure]

Correct, we can analyze the (statically determinate) truss using either the method of joints or the method of sections. When doing so, we treat the virtual unit load(s) as the applied loads. They are the only loads acting on the truss, all the real loads should be removed/ignored.

[@willrichards1454]

@@DrStructure Ok, but are the A and B forces removed, or just the 10 kN P-value?

[@DrStructure]

The truss shown @10:27 is the one we want to analyze, a statically determinate truss with a pin support at A, a roller support at B, an (upward) unit load at B, and a (downward) unit load at C. We are not removing any of the support reactions, rather, we are calculating them. If we apply the static equilibrium equations to the entire truss, we get zero for the support reactions at A and B. Knowing the reactions, now we use the method of joints to calculate the member forces.

[@youxizhang1171]

hi thanks very much for your video. i have a question : In the first example of internal statically indeterminacy , while you are calculating the deflection from a pair of virtual force, why is there 6 bars in total ? shouldn't it be 5 since one is cut already ? thanks

[@DrStructure]

It is true that one of the member has been cut, but then we are applying couple of virtual forces to the cut ends causing the two separated segments of the cut member to elongate by themselves. That elongation times the force in each segment contributes to the total internal work done by the virtual load.

[@FayazKhan-vc6is]

sir please tell me what is EA?? IN TRUSS {L/EA} E is modulus of elasticity but i dont know what is A please sir tell me

[@DrStructure]

A is the cross-sectional area of the member.

[@FayazKhan-vc6is]

thank you sir GOD BLESS YOU

[@raheelmalook5884]

Dr structures, don't we find external reactions at roller and hinge supports when we apply internal unit loads?

[@DrStructure]

If it is necessary to calculate the reactions, yes, we do it. But since we only need the member forces, sometimes calculating support reactions can be avoided.

[@CR-uh8yb]

hi Dr. Structure, I have a frame that is 2 bays, 2 story, 3 pins as the ground supports and 2 inverted V braces (one at bottom first bay first story & the other at the top second bay second story). There's a uniform load on the roof level, a lateral loads at the first and second floor far left joints . I am trying to find the force on the bottom left inv-V brace. I know I need to find the reactions and then make a cut to find the brace force but I'm not sure which force method to use: the one for trusses or the one for frames. Is this braced frame fall under the frame or truss category? . also is the force method the fastest method for this particular problem or should I use Moment distribution, slope-deflection, or stiffness matrix? which is best??

[@DrStructure]

The structure consists of beam-column and truss elements (the bracings). The classical techniques such as the force method and the slope-deflection method are not suitable for this kind of analysis. A more general technique such as the matrix/displacement method is the best option here.

[@CR-uh8yb]

@@DrStructure Is it safe and correct to make the assumption that the braces take the lateral only (since its the force of the bottom brace that is in question) therefore solving it using a shortcut approach like Brace force = story shear / (2cos theta) ? or do I absolutely have to use the matrix method

[@DrStructure]

@@CR-uh8yb The bracings effect joint displacements which in turn effect the member forces, including the force in each bracing element. Further, the shear force is going to be carried mainly by the columns, not by the bracings. By its very nature, a bracing is mainly a tension (truss) member carrying an axial force only. I would use the matrix method to analyze the system so as to better understand its behavior.

[@CR-uh8yb]

@@DrStructure Ok I will use the matrix method.Thank u Dr. !!

[@niteshmeena9961]

How to determine that truss is internally or externally inderminante

[@DrStructure]

There is a discussion on this here: kzbin.info/www/bejne/rWjVcpynr9GUpLM

[@joseh7612]

how do you know what is a redundant force and what is not?

[@DrStructure]

The number of redundant forces equals the degree of indeterminacy of the structure. Hence, a determinate system has zero/no redundant forces. An indeterminate system has one or more redundant forces.

[@cvanskarna8050]

does it apply when determinacy is more than 1

[@DrStructure]

Yes, the force method can be used to analyze a structure with N degrees of indeterminacy. Although as the degree of indeterminacy increases, the formulation becomes more complex.

[@yazanalmazaidah6642]

where is the solution for the proplems ?

[@DrStructure]

Solution for Exercise 1: Lab101.space/pdf/exercises/SA26-Exercise1.pdf Solution for Exercise 2: Lab101.space/pdf/exercises/SA26-Exercise2.pdf

[@huzeyfeyavuzyegit1894]

Thank you for making your videos.could u send that videos pdf please ?

[@DrStructure]

We are working on producing a pdf version of each video on the channel, but that will take some time.

[@niteshmeena9961]

Where is solution for exercise question.

[@DrStructure]

Probably lost in the outer space, we will look for it, or work on it. Thanks for the reminder.

[@mariasophiaenguito3899]

may i have a copy of this?

[@DrStructure]

Please let us know why you need a copy of the video file. What is the intended usage of the file?

[@chungken8496]

Excuse me, @0836, how to calculate the axial deformation by virutal method?

[@DrStructure]

We don't need the virtual work method for calculating axial deformation in a truss member. If we know the force in the member, then we can easily calculate its axial deformation. Let S be axial stress in the member, E: modulus of elasticity, U: axial strain A: cross-sectional area of the member L: Length of the member P: Axial force in the member D: Axial deformation (Axial Stress) = (Modulus of Elasticity)(Axial Strain) S = E U Since axial stress = axial force divided by cross-sectional area, S = P/A And, axial strain is axial deformation divided by member length: U = D/L Then, we can write: (P/A) = E(D/L) Solving the above equation for D (axial deformation), we get: D = (PL)/(EA) So, if know axial force P, length L, area A and modulus of elasticity E, we can use the above equation to calculate D.

[@chungken8496]

Which member should be calculate? Since AD is cut , how to calculate its axial force?

[@DrStructure]

I see where the confusion might be. I misunderstood your previous question and gave you a misleading answer. In the force method, how do we calculate an axial deformation caused by cutting a member using the virtual work method? The answer to that question is here: kzbin.info/www/bejne/qnWQh4yKhKh7jKM

[@chungken8496]

Actually I saw SA21 video before, but it does not have a case of cut the members and find the deformation. In your previous reply, I understand how to find the axial deformation, but I dont know which member force I need to find in @0836 example to find the cut member deformation. Let the point of intersection of BC and AD be a point E, if ED is cut, should I imagine ED is gone and find the member force? But after I find the member force, I cannot find the member force of ED , since it is not exist. Or should I find the member without imagine removed any members? But it is internally indeterminacy, so I cannot find the member force without removed one of the member. That's my confuse.

[@DrStructure]

Okay, the given truss is indeterminate. So, we need to cut a member. Here, we cut member AD. By the way, there is no need to define point E, since AD and BC are not connected, they just touch each other, overlap, at their mid point. Let’s cut the redundant member AD, And remove it form the truss. So, we end up with a statically determine truss subjected to load P at joint C. This truss does not have any member connecting A to D. Now, go to Lecture SA21 and calculate displacement between A and D. Don’t view it as the axial displacement in member AD, since the member is not there (we cut it and removed it), think of it as displacement along axis (line) AD. That displacement can be calculated using the virtual work method per SA21. We refer to this displacement as DELTA. Going back to the indeterminate truss, remove the applied load P. Then cut AD again, but this time, replace it with a unit axial force. This means a unit force at D point toward A, and a unit force at A pointing toward D. Again, remove member AD. This again gives us a statically determinate truss just like the above. Except that here the loads are different. In the case above the load was P. Here, there truss is subjected to couple of unit loads, one at A and one at D. But the truss has the same geometry as the case above, there is no member AD, the truss is statically determinate. We can use the virtual work method to calculate displacement between A and D due to the pair of unit loads, just like the above case. We refer to this displacement at delta. Then we can write DELTA = (Fab) delta. Or, Fad = DELTA/delta. Let me know if I need to explain a bit more.

[@puneetgarg3481]

this is displacement method

[@legspinnermuhammadikramkha3454]

I'm Iqra National University Peshawar Hayatabad students our teacher is too dull