SA49: Matrix Displacement Method: Frame Analysis (Joint Loads)

[@huseinfaisal3402]

I mean if element AB not 90 degree,of course global stiffness matrix of element AB cant used to get forces of element AB,We first must transformation global displacement of element AB to local displacement and used f = k d, k and d in local coordinate

[@DrStructure]

Not quite, we can calculate the member forces in the global system using the global displacement vector. Then, transform the global member forces to local forces. That way, we don’t need to directly calculate the displacement vector in the local system. Put it in matrix notation: F = K D where K is the member stiffness matrix in the global system, D is the member displacement vector in the global system, and F is the member force vector in the global system. After calculating F using the above equation, multiply in be the transformation matrix to get f (the member force vector in the local system).

[@huseinfaisal3402]

@@DrStructure in your explanation d = T D,while F = T f, capital letter in global system and the other local system, in this formula any difference in transformation between displacement and force,I just tried if We calculate members force from F = K D and then transform F to local system by f = T F,the result not correct if the members of frames not 90 and 0 degree.

[@DrStructure]

@@huseinfaisal3402 1) Let Q be the transformation matrix and T be the transpose of Q. 2) Starting from the local system formulation, let f = kd where f is the member force, k is the member stiffness matrix, and d is the member displacement vector all in the local coordinate system. 3) To map the local into global, we can write F = Qf and D = Qd. 4) If we pre-multiply both sides of each equation in (3) by T, the transpose of Q, we get: TF = TQf and TD=TQd. Since TQ equals the identity matrix, we get: f = TF and d = TD. Note the expression f = TF, which means if we multiply T by F, we get the forces in the local coordinate system. But let's complete the derivation. 5) The equation in (2) now can be written as f = k TD. Here, we replaced d with TD. 6) Pre-multiply the equation in (5) by Q. We get: Qf = QkTD. Or, F = QkTD. And since QkT = K, the equation becomes: F = KD. That is, the member forces in the global system equals to the member stiffness matrix times the displacement vector both given the global coordinate system. 7) Once F is determined using the above equation, we can use f = TF (given in Step 4 above) to transform F from global to local system. In this way, there is no need to directly transform D into d in order to calculate f. If you have your sample test solution written in a complete and clear manner, feel free to email it to: [e-mail] We'll take a look at your solution and offer feedback accordingly.

[@huseinfaisal3402]

@@DrStructure It should be F = Tf not Qf,by this expression not only change posisition of force but also the sign of force. Use F = Tf,more advantage in assemble force vektor without the plot the position of force like you do in example of members loads.We just use the code of dof in assemble force vektor and also stiffness matrix.

[@DrStructure]

@@huseinfaisal3402 If you define your T to be the transpose of Q (as is defined above and in the video), then yes. Otherwise your formulation comes out incorrect. I suggest you go through the derivation and formulation step by step, compare and contrast it with what is given in the lecture and see how the two differ. Even better, write it up so that you have a written version of it too. That way, you can better substantiate your idea for further discussion.

[@rugbyrocks123321]

Best structural analysis videos on KZbin.

[@leonacndanmark]

I think this channel has a a high quality of learning output through these videos, specifically the Matrix deflection videos since they can be tough to learn :)

[@mehdigerman7970]

i just like the way of explanation as well as the representation

[@christinatesfaye1507]

dr.structure!!!!!! its amaazazzzzzzzzzzzzzzzzzzzinggggggggggggggggggg.tnx a lot

[@marcusaurelius8047]

excellent presentation. thanks.

[@irfanullah8255]

Hi Ma'am hope you are fine. Ma'am plz share the lecture pdfs of structure(ll) as they are not availibke in the course website.

[@irfanullah8255]

Ma'am plz have a look at this. Im also enrolled in structure(ll) bit tgere is not availible. Structure(l) pdfs were helpful very much so will be the structure(ll)

[@DrStructure]

Thanks for the note. We are working on producing a written version of all our lectures in PDF format. We will make them available through our courses when they are ready.

[@irfanullah8255]

Sixtillions of thanks Ma'am

[@dilafrozakarim9461]

Can you kindly tell me at 07:49 in the multiplication of member stiffness matrix how the plus sign has come? i am not understanding that

[@DrStructure]

Are you referring to the signs attached to the elements of the stiffness matrix that are shown @7:49? Are you asking how/why QkT results in matrix coefficients that have plus and minus signs in them? Please clarify.

[@dilafrozakarim9461]

I am asking how the result is matrix coefficient have plus & minus basically how the result came as we are multiplying the values

[@DrStructure]

​@@dilafrozakarim9461 That is the result of pure matrix multiplication. We have three 6x6 matrices: Q, k, and T. First, we can multiply Q and k, then multiply the resulting matrix by T. What does the product of Q and k look like? It is going to be a 6x6 matrix. How do we get the elements of that matrix? We multiply each row of Q by each column of k. So, first row of Q by the first column of k, then first row of Q by the second column of k, and so forth and so on until we have multiplied each row of Q by every column of k. To give you an example, the first row of Q is: [c -s 0 0 0 0 0] where c stands for cosine of theta and s stands for sine of theta. The first column of k, written here as a row is: [EA/L 0 0 -EA/L 0 0]. How do we multiply these two vectors? We multiply their corresponding terms, then add them all up. So, we get: (c)(EA/L) + (-s)(0) + (0)(0) + (0)(-EA/L) + (0)(0) + (0)(0) = cEA/L. This is the element in the first row and first column of (Qk) matrix. When we multiply the i-th row of Q by the j-th column of k, we get the element of the product matrix in position i-j. That is the element that is in row i and column j. Once we have generated all the elements of Qk, we then multiply that matrix by T in a similar manner in order to get the final stiffness matrix. Matrix multiplication is a standard procedure in matrix algebra. There should be ample videos/notes available on KZbin, if you want to look into it more.

[@dilafrozakarim9461]

Thank you so much Sir/Mam. That's very helpful. Thank you for clearing THIS term❤

[@DrStructure]

@@dilafrozakarim9461 You're most welcome!

[@Leo-cs6sb]

Excuse me , why at 10:44 for segment AB the axial force is f2 and f5, should it also be f1 and f4?

[@Leo-cs6sb]

ah sorry now I understand, they are all in the global coordinate system now, so f1 should always in horizontal. Thank you.

[@DrStructure]

Correct, they are specified in the global coordinate system.

[@huseinfaisal3402]

Good explanation but in calculation each element forces ,displacement should trasformized in local coordinate,we get d = TD,next each element forced f = k d ,k is local stiffness matrix.the result is the same but your formula not for general frames especially understanding to plot axial and shear forces in each members.

[@DrStructure]

I shall mildly disagree with your assertion that the formulation is not general enough. It is true that the member forces need to be converted from the global to the local coordinate system, but transforming the displacement vector is not a precursor for it. Member forces, once determined in the global coordinate system, can be converted to the local coordinate system for determining thrust, shear and moment in each member.

[@chungken8496]

Excuse me, @09:07, what is the unit of K bc? For column 1, 2, 4, 5 , the unit is N/mm, since the unit of belta or alpha is N/mm. For column 3, 6, the row 1, 2, 4, 5, the unit is kN, since belta (N/mm) times length (m) but in row 3, 6, the unit is kNm, since belta (N/mm) times Length square (m^2). Do I correct?

[@DrStructure]

Yes, you are correct. Alpha and Beta each has unit kN/m. As you correctly indicated, some of the elements of the matrix have unit kN, and some have unit kN-m. More specifically, the elements in rows 1, 2, 4 and 5 end up in kN, and the elements in rows 3 and 6 end up in kN-m. These correspond to the member force vector, since f1, f2, f4, and f5 are forces in the x and y directions at the two ends of the member (the unit of force is kN), and f3 and f6 are bending moments at the ends of the member (the unit of moment is kN-m).

[@chungken8496]

@@DrStructure 1. For the stiffness K matrix, should the unit of row 1 column 1 is kN/m?? Unit of 12Beta*s^2+alpha *c^2 2. Also , i don't understand that why do whole column 3 and 6 does not share the same unit? Since they should multiply with D3 and D6 respectively to provide moment. 3. I think the unit of D3 and D6 is radian. Does it correct?

[@DrStructure]

1. Yes, the unit of k11 is kN/m. K11 is the multiplier of d1 (the displacement in direction 1). That is, if we multiply the first row of the matrix by the displacement vector, we get: f1 = k11 (d1) + k12 (d2) + k13(d3) + .... Since the unit of force (f1) is kN, then the product of k11 (d1) should yield kN. The unit of displacement (d1) is m, the unit of k11 is kN/m, so their product has unit: kN. 2. Note the third term in the equation above: k13 (d3). This product has to have unit kN. Since d3 (rotation) is unitless, then k13 must be in kN (Beta is in kN/m and L is in m). However, now consider equation 3: f3 = k31 (d1) + k32 (d2) + k33 (d3) + .... Here, f3 is bending moment so, it must have unit: kN-m. This means k33 times d3 must give kN-m. Since d3 in radian (unitless), then k31 must be in kN-m. That unit is different than the unit for k13 (above), because one is for calculating force (f1) and the other for calculating moment (f3). 3. Yes, d3 and d6 are rotations given in radians, which is a unitless coefficient as far as forces and moments are concerned.

[@chungken8496]

@@DrStructure 1. In the stiffness matrix what is the unit of element row 3 column 3 --- 4 Beta * L^2 ???? kNm? 2. In the stiffness matrix what is the unit of element row 1 column 3 --- -6Beta*L*s ???? kN? 3. Why they are different ? I think that they should be share the same unit, cause it should times d3 to provide kNm.

[@DrStructure]

1. Yes, the unit of k33 is kN-m 2. Yes, the unit of k13 is kN. 3. No, they should not be the same. It is true that they both are being multiplied by d3 (unitless rotation), but the elements of the matrix in row 1 are associated with f1 (force in the x direction having a unit of kN) whereas the elements of the matrix in row 3 are for calculating f3 (bending moment which has a unit of kN-m). So, k13 should have a unit of kN, and k33 should be in kN-m.

[@ineverlickyoghurtlid3903]

i would suggest a video that neglecting the axial deformation of the members. so it would be more clear.

[@sturtevantirma2332]

nice method

[@dilafrozakarim9461]

Hello Sir/Mam, I have an another confusion ,when shouldn't we consider horizontal force as degree of freedom in frame analysis?

[@DrStructure]

A degree of freedom tells us that a frame joint can move in a specific direction. So, if a joint can move in the horizontal direction, we assign a degree of freedom to the joint in that direction. This is not a function of the applied loads, it is a function of the ability of the frame to move in a direction whether or not a load is applied to the joint. So, unless there is a fixed or pin support at the joint preventing its horizontal displacement, we should assign a degree of freedom to the joint in the x direction.

[@aaronvillajuan9060]

why is the rotation at joint B in memBC clockwise?

[@DrStructure]

@12:05 the calculated displacement vector shows a negative D4 (a clockwise rotation). This is the rotation at joint B. Why did D4 come out to be negative? because the joint is subjected to a clockwise moment of 10 kN.m causing the joint to rotate in that direction.

[@aaronvillajuan9060]

@@DrStructure Ohhh yes, I understood that part and wrote down everything. What I don't get is that rotation of local d3, memberBC, I thought that we should always assign positive directions for our unknowns. I'm sorry, I forgot to put the timestamp, It's @12:24

[@DrStructure]

I see where the problem lies. That rotation (at the left end of BC) is drawn in the wrong direction, it should have been drawn in the counterclockwise direction. Thanks for pointing it out.

[@moviesynopsis001]

Hey Dr. Structure awsome video. Was the narrators voice in this video clip a text to speech voice reader or was it a real person? It sounds really good either way.

[@DrStructure]

This is a human voice not a computer-generated one. We have only one or two lectures/videos that use computer-generated voice, like this one: kzbin.info/www/bejne/i564nGyKdtGle8U The text to speech technology is not quite there yet...

[@moviesynopsis001]

@@DrStructure I was only confused because of the lady at the end of this video, her lips were moving in an animated way. Thanks for the prompt response btw!

[@mohamedthasneem7327]

Thank you very much

[@albertowusu1688]

Excellent material. Can we get your contact?

[@DrStructure]

Please visit Lab101.space for contact information for the person in charge of this project.

[@fsxdogflight187]

method is Good for everyone

[@markwatson2779]

Can someone tell me how the theta angle is 90 and 0 for the two beams ? I am not understanding.

[@DrStructure]

Theta is the angle that the member makes with the x-axis, which for the example problem discussed in the video, is the horizontal axis. Member AB, being vertical, makes a 90 degrees angle with the horizontal axis. Member BC is horizontal, so it’s theta = 0.

[@markwatson2779]

@@DrStructure Do I always consider with respect to the x axis in any problem ? I have a rectangular box frame with a force acting perpendicular to the Y axis. it is in 3D so whats the best way to simplify it to 2D? If so can I solve with respect to the x axis for all the bars thats in the frame ? Thank you.

[@DrStructure]

As far as the angle itself is concerned, yes, we can always measure it with respect to the x-axis. A box structure in which there are no out-of-plane applied forces can be simplified to a 2D (in-plane) system. If however out of plane loads are present, the system needs to be analyzed in 3D space using the 3D version of the displacement method. In 2D space, we usually label the horizontal axis as x and the vertical axis as y. But this labeling is rather arbitrary. We can label the vertical axis as x, if we wish to do so. Regardless of which axis we label x, angle theta is always measured with respect to that (x) axis.

[@markwatson2779]

@@DrStructure Thank you so much for your answer. That makes sense to me. Do you also have any material I can see for 3D structures ? Also , at 10.29 minutes I don't understand how that matrix come? for an example how is k11 relationship come. If there are 6 forces like in BC what is the matrix be for that ? Is that because it always considers the adjacent beam we are calculating as well? I am confused in that section can you please help. Thank you.

[@DrStructure]

@@markwatson2779 No, we don’t have any lectures on 3D structures, at least not yet. As for the question on the formation of the system stiffness matrix, Degree of freedom (DOF) 1 is shared by member AB only (it resides at joint A which is not a part of member BC). So, AB is the only member that contributes to K11. Similarly, degrees of freedom 1 and 2 reside on member AB only. Although DOF 2 is also a part of member BC, but since DOF 1 is not on BC, only member AB contributes to K12. So, we basically identify the member(s) that share both DOFs, then compute the system stiffness element using the corresponding stiffness of those members. In this case, the only pair of DOF that are shared between members AB and BC are: Pair 2 2 (for calculating K22), Pair 2 3 (for K23), Pair 2 4 (for K24), Pair 3 3 (K33), Pair 3 4 (K34), and Pair 4 4 (K44). These are the degrees of freedom at joint B which belongs the both members. So, when calculating these system stiffness elements, we need to add the corresponding member stiffness elements from AB and BC.

[@alexhermawans]

THANKS !

[@DOTA2MAJISTRATE]

Hi Sir, For the truss we have been researching on below ibb.co/nRggxMB Does it have 8 degrees of freedom ? or 10 degrees of freedom ? If it is a 8 degrees of freedom, means i have to have like a 8x8 matrix ? or a 10x10 if i have 10 degrees of freedom ? Is solving this by hand calculation possible ?

[@DrStructure]

If the structure is a truss, you would need only two degrees of freedom per joint; There is no joint rotation in trusses, there are only x and y displacements. However, modeling this structure as a truss is problematic. You would need another diagonal member (in the lower level) to make it stable. If there are two pins at the based of the truss, the structure would have 10 degrees of freedom (an x and a y displacement at each of the joints except for two at the base). Yes, a structure with N degrees of freedom yields an N by N matrix. It is possible to solve a 10x10 problem by hand, but it is not common.

[@DOTA2MAJISTRATE]

@@DrStructure well assuming i am not modeling it as a truss ( no diagonal member in lower level ), will it be a 10 degree freedom frame or 8 degree freedom frame ? ( does point 9 and 10 exist a degree of freedom ? or more since it might rotate to ? )

[@DrStructure]

If you model the structure as a frame, you need to define three degrees of freedom per joint, unless the joint is restrained by supports. In this case, since there is a pin at A and B, each of them needs to have a rotational degree of freedom. Since, the pin can rotate rotate. So that is two degrees of freedom at the base of the frame. The remaining joints each needs to have 3 degrees of freedom, which gives you 15 degrees of freedom for the upper joints. Hence, the frame has a total of 17 degrees of freedom, each joint can move in the x and y directions and it can rotate, except for A and B that can only rotate.

[@DOTA2MAJISTRATE]

@@DrStructure Dear Sir, assuming i am having fixed A&B, that makes 15 degree of freedom, and even if i change the lower beam member to a pinned connection with the column it will not impose any changes as the degree of freedom is still 3 so total still 15 degree off freedom ?

[@DrStructure]

@@DOTA2MAJISTRATE No quite. The 3 degrees of freedom per joint assumes a rigid joint. Instead of a rigid connection, if you connect two members using a hinge (internal pin), naturally the formulation is going to be different. The two members that are connected with the hinge, although share the same x and y displacements, they are going to rotate differently. That is, we end up having two rotational degrees of freedom at the hinge, one degree of freedom per member. So in total there would be 4 degrees of freedom: displacement in the x direction, displacement in the y direction, rotation at the end of member i, and rotation at the end of member j, assuming the pin connects members i and j. The analysis of such a frame is going to be more complicated. This topic has not been discussed in any of our lectures.

[@huilingwong3689]

thankssssss!!!!

[@DrStructure]

You're welcome!