Please check ,may be mistake in using formula in calculating equivalent nodal load,in left beam bc used pb/L shoud be pb^2(L+2a)/L^3 and the right beam pa^2(L+2b)/L^3
You are correct, the expressions for the left and right member-end shear forces @4:50 are written incorrectly. Thanks for noticing it and bringing it to our attention.
For time 2.20, why did we draw our member end forces according to global instead of local coordinate which produce a problem with changing f1 and f2. Can I solve it by taking local coordinate?
Each member has its own local coordinate system. To assemble the system stiffness matrix and solve for the unknown displacements we need to use a single coordinate system, which we call global/system coordinate system. So, the properties of all the members have to be transformed into that coordinate system. As a result, the system displacement vector (which is the main result of the analysis) is going to be determined in the global coordinate system. From there, we can determine the forces at the ends of each member. These can be done in the global coordinate system or the local coordinate system. But at the end, we usually prefer having the members forces written down in local coordinate system since that makes it easier to draw the shear and moment diagrams for the member. From the computational standpoint, we can view the displacement method consisting of three main steps: pre-processing, main processing, and post-processing. In the pre-processing step, we determine the joint loads and transform everything into the global/system coordinate system, and assemble the system of equations to be solved. In the main processing step, we solve that system of equations for the unknown displacements. In the post-processing step, we use the computed displacements to determine the member-end forces.
Hi ! Please can you indicate the name of an app or software that we can use to solve matrices, i have downloaded (M)atrix which is a free app on appstore but still it misses a couple of things. Thank you :)
We use Mathematica for solving such problems. There is a free online version of it available under the name WolframAlpha. Here is the website: www.wolframalpha.com/
Hi @3:16, it's stated that member AB adds to the system stiffness in directions 1, 2, and 3. But the System Stiffness Matrix shows this contribution as K44, K45, K46, K55, K56, K66. Can you elaborate on this?
The system directions 1, 2, and 3 correspond to the member AB directions 4, 5, and 6. So, the stiffness contribution of AB in system directions 1, 2, and 3 comes from member stiffness directions 4, 5, and 6.
@@DrStructure Yes, the sketch at 2:18 illustrates these directions for member AB. Thank you. I came into another query. At 2:50, the sketch shows the degrees of freedom for member BC, which is the same coding for system (structure) degrees of freedom. I can't catch this. Why the matrix at 2:40 isn't the same when it comes to 2:50 for member BC?
@@hoshangmustafa9672 It depends on how we number the joints of the frame and assign DOF indices. For example, we can number the system DOFs at joint C as 1, 2, 3. And number the DOFs at joint B as 4, 5, 6. Then, 1,2,3 of member BC would correspond to 4,5,6 of the system. And, 4,5,6 of member AB corresponds to 4,5,6 of the system.
@2:40 the CZcams video shows an incorrect stiffness matrix for member BC. Please see this same (but updated) lecture in the online course (courses.structure.education/courses/structural-analysis-ii ) for the correct matrix representation. K12 (system) = k45 (AB @2:12) + k12 (BC @2:35) = 0 + 0 = 0. K22 (system) = k55 (AB @2:12) + k22 (BC @2:35) = 475000 + 3641 = 478641
Yes, there are prerequisites for understanding the displacement method. You need to have a basic understanding of statics and structural analysis including but not limited to the slope-deflection method. Please see the previous lectures, starting from SA01, for a more methodical approach to grasping the underlying ideas and methods of structural analysis.
this is very helpful, thanks!
Nicely done; looking forward to your upcoming airplane hangar video.
thank's very much. can you do videos about FEM ?
Thanks for your note. Yes, eventually we'll get to the generalized finite element method. But we have a lot of grounds to cover before we get there.
Dear Sir, Please make videos on arches and suspension cables
Please check ,may be mistake in using formula in calculating equivalent nodal load,in left beam bc used pb/L shoud be pb^2(L+2a)/L^3 and the right beam pa^2(L+2b)/L^3
You are correct, the expressions for the left and right member-end shear forces @4:50 are written incorrectly. Thanks for noticing it and bringing it to our attention.
Thanks for the video.have your channel any intention to make a video about arches?
Yes, eventually we'll get to arches.
please! i'm eagerly waiting
For time 2.20, why did we draw our member end forces according to global instead of local coordinate which produce a problem with changing f1 and f2. Can I solve it by taking local coordinate?
Each member has its own local coordinate system. To assemble the system stiffness matrix and solve for the unknown displacements we need to use a single coordinate system, which we call global/system coordinate system. So, the properties of all the members have to be transformed into that coordinate system. As a result, the system displacement vector (which is the main result of the analysis) is going to be determined in the global coordinate system. From there, we can determine the forces at the ends of each member. These can be done in the global coordinate system or the local coordinate system. But at the end, we usually prefer having the members forces written down in local coordinate system since that makes it easier to draw the shear and moment diagrams for the member.
From the computational standpoint, we can view the displacement method consisting of three main steps: pre-processing, main processing, and post-processing. In the pre-processing step, we determine the joint loads and transform everything into the global/system coordinate system, and assemble the system of equations to be solved. In the main processing step, we solve that system of equations for the unknown displacements. In the post-processing step, we use the computed displacements to determine the member-end forces.
Hi ! Please can you indicate the name of an app or software that we can use to solve matrices, i have downloaded (M)atrix which is a free app on appstore but still it misses a couple of things. Thank you :)
We use Mathematica for solving such problems. There is a free online version of it available under the name WolframAlpha. Here is the website: www.wolframalpha.com/
Hi
@3:16, it's stated that member AB adds to the system stiffness in directions 1, 2, and 3. But the System Stiffness Matrix shows this contribution as K44, K45, K46, K55, K56, K66. Can you elaborate on this?
The system directions 1, 2, and 3 correspond to the member AB directions 4, 5, and 6. So, the stiffness contribution of AB in system directions 1, 2, and 3 comes from member stiffness directions 4, 5, and 6.
@@DrStructure
Yes, the sketch at 2:18 illustrates these directions for member AB. Thank you.
I came into another query. At 2:50, the sketch shows the degrees of freedom for member BC, which is the same coding for system (structure) degrees of freedom. I can't catch this.
Why the matrix at 2:40 isn't the same when it comes to 2:50 for member BC?
@@hoshangmustafa9672 It depends on how we number the joints of the frame and assign DOF indices.
For example, we can number the system DOFs at joint C as 1, 2, 3. And number the DOFs at joint B as 4, 5, 6. Then, 1,2,3 of member BC would correspond to 4,5,6 of the system. And, 4,5,6 of member AB corresponds to 4,5,6 of the system.
@@DrStructure
It's clear now. Thanks.
How K12=0 for system?
How K22=478641 for system?
@2:40 the CZcams video shows an incorrect stiffness matrix for member BC. Please see this same (but updated) lecture in the online course (courses.structure.education/courses/structural-analysis-ii ) for the correct matrix representation.
K12 (system) = k45 (AB @2:12) + k12 (BC @2:35) = 0 + 0 = 0.
K22 (system) = k55 (AB @2:12) + k22 (BC @2:35) = 475000 + 3641 = 478641
i am a beginner..
i am unable to understand it.
any basics video for me.
Yes, there are prerequisites for understanding the displacement method. You need to have a basic understanding of statics and structural analysis including but not limited to the slope-deflection method. Please see the previous lectures, starting from SA01, for a more methodical approach to grasping the underlying ideas and methods of structural analysis.
Dr. Structure .
thanks for your support.
.
You are welcome.