This is a stylistic choice to reduce the size of the matrix coefficients, making the matrix more compact. This is analogous to factoring in algebra. Such manipulations do not affect the solution of the system.
Is there possible for the case like applied inclined load and inclined support? Is that mean the member stiffness matrix will required to include the axial deformation and x-axis forces with transformation matrix? Can you make some example problem and solution for them?
Yes, both forces and supports could be inclined. The axis transformation matrix needs to be applied to the inclined forces in order to have them represented in member coordinate system (see Lecture SA49 where the method is being used to analyze frame structures). In the case of inclined supports, a bit of more work is needed in order to analyze the structure. When a support, say, a roller is inclined, then the displacement in direction of the roller needs to be represented as two dependent degrees of freedom, one in the global x and one in the global y direction. We then need to establish the relationship between the two degrees of freedom in the matrix equations. For example, an inclined roller (at 30 degrees) ends up with a degree of freedom in the x-direction equals to: Dx = cos(30) D = 0.86 D, and a degree of freedom in the y-direction equals to: Dy = sin(30) D = 0.5 D where D represent the degree of freedom in the plane of the roller. When writing the system matrix equations, we need to specify the relationship between Dx, Dy and D in the form of the simple equations given above, then simplify the system of equations accordingly before solving it for the unknown global displacements. We will add this topic to the list of future videos.
Adding the member stiffness matrices to get the system stiffness matrix which in essence involves matching member indices (1 through 6) to system indices (here there are 3 of them).
Rollers do not have moment reactions. Bending moment at a roller is always zero. Our matrix formulation is done for a typical beam member detached from any supports. Such a general member can be assumed to carry moments at its ends. However, if the member happens to be connected to a roller or a pin, at the end, the moment comes out to be zero.
Do you mean why K11_system is equal to k22_AB? Member AB has 4 degrees of freedom (dof): dof 1 is displacement in the y direction at end A, dof 2 is rotation at end A, dof 3 is displacement at end B and dof 4 is rotation at end B. That makes the system dof 1 (i.e., rotation at joint A) corresponds to dof 2 for member AB.
What to do when there are multiple loads on a beam segment?
When a beam segment is subjected to multiple loads, we calculate the equivalent joint loads for each, then add them up.
Thanks, this is very helpful. We need more examples.
for the new matrix at 4:02 why did you divide everything by 12 to get EI/18(1 3 -1 3) etc. instead of just leaving it as EI/216 and so on
This is a stylistic choice to reduce the size of the matrix coefficients, making the matrix more compact. This is analogous to factoring in algebra. Such manipulations do not affect the solution of the system.
Is there possible for the case like applied inclined load and inclined support? Is that mean the member stiffness matrix will required to include the axial deformation and x-axis forces with transformation matrix?
Can you make some example problem and solution for them?
Yes, both forces and supports could be inclined. The axis transformation matrix needs to be applied to the inclined forces in order to have them represented in member coordinate system (see Lecture SA49 where the method is being used to analyze frame structures). In the case of inclined supports, a bit of more work is needed in order to analyze the structure. When a support, say, a roller is inclined, then the displacement in direction of the roller needs to be represented as two dependent degrees of freedom, one in the global x and one in the global y direction. We then need to establish the relationship between the two degrees of freedom in the matrix equations.
For example, an inclined roller (at 30 degrees) ends up with a degree of freedom in the x-direction equals to: Dx = cos(30) D = 0.86 D, and a degree of freedom in the y-direction equals to: Dy = sin(30) D = 0.5 D where D represent the degree of freedom in the plane of the roller. When writing the system matrix equations, we need to specify the relationship between Dx, Dy and D in the form of the simple equations given above, then simplify the system of equations accordingly before solving it for the unknown global displacements.
We will add this topic to the list of future videos.
@@DrStructure opps it is already in frame analysis, i am not there yet. And thank you very much for going for add-on topic.
Can you solve a problem with an internal hinge?
Soon :)
How do you get the fixed end forces?
Please see the following video lecture for an explanation on fixed-end forces:
th-cam.com/video/Z_jxaKQirC0/w-d-xo.html
What are you doing after 7:00
Adding the member stiffness matrices to get the system stiffness matrix which in essence involves matching member indices (1 through 6) to system indices (here there are 3 of them).
I got confused maam. How rollers have a reaction moment?
Rollers do not have moment reactions. Bending moment at a roller is always zero.
Our matrix formulation is done for a typical beam member detached from any supports. Such a general member can be assumed to carry moments at its ends. However, if the member happens to be connected to a roller or a pin, at the end, the moment comes out to be zero.
excuse me! how can i calculate kAB24 equalEI/3?
For member AB, K24 = (EI/L^3)(2L^2) = 2EI/L.
Since L = 6 m, K24 = 2EI/6 = EI/3.
How does K11=K22 ?
Do you mean why K11_system is equal to k22_AB?
Member AB has 4 degrees of freedom (dof): dof 1 is displacement in the y direction at end A, dof 2 is rotation at end A, dof 3 is displacement at end B and dof 4 is rotation at end B.
That makes the system dof 1 (i.e., rotation at joint A) corresponds to dof 2 for member AB.
Solution for Exercise ????
Click on the circled i at the upper right corner of the video screen.
good
👍🏻
Hey ,good teacher 😉
Can we see the answers
Yes, the solution to the two exercise problems will be posted shortly.
Thanks
Can u make us an example of beam with internal hinge also??
Thank's for help
Yes, we'll get to it soon.
please uploade next part.