Thank u very much.....So amazingly cleared hidden points....May God bless you with lots of happinesses and victories in ur life.Again thank u very much.
Those are fixed-end moments. A beam of length L with a concentrated load of P applied at its midpoint, the fixed-end moments are PL/8. So, here we get: (16 kN)(10 m)/8 = 20 kN.m
Thanks , for this video on Matrix Displacement Method.How one can solve continuous beam with all fix supports by stiffness matrix method, For example beam with 3 span of each 4m , all supports are fixed with uniformly distributed load on beam 2.5,3.5,3.9 kn/m
A beam with three spans has 4 supports, two at the ends and two intermediate supports. If by all supports you mean all four supports are fixed, then you basically have three independent beams each fixed at its ends. They can be analyzed independently, and there is no need to resort to displacement method to analyze them. The fixed-end moments for each beam are: w L^2/12. Knowing the end moments, you can then apply the equilibrium equations to determine the vertical reaction at the ends of each beam. If by all supports being fixed, you mean the end supports only, then the system has two degrees of freedom only, the joint rotations at the two interior supports. In this case, we can use the displacement method to calculate the two rotations, and then the support reactions.
The direction of the shear force is not defined by the direction of the force arrow, rather it is defined by how the force tends to rotate the beam segment. An upward force at the left end of the segment (wanting to turn the segment in the clockwise direction) signifies a positive shear. However, an upward shear force at the right end of the segment (wanting to turn the segment in the counter-clockwise direction) indicates a negative shear. In the example problem, the matrix solution gives us two positive (vertical) forces for member BC. The left force having a magnitude of 13.63 kN (since it want to turn BC in the clockwise direction) is considered a positive shear force. The force at the right end of BC is also upward with a magnitude of 10.37 kN. This force however wants to turn the beam in the counter-clockwise direction, therefore, it is considered a negative shear. Unrelated to the sign of shear, but related to static equilibrium, note that the sum of these two upward forces must be equal to the total downward (applied) load on BC.
Once again thank you much. I dont have enough words to thank this channel for these videos. Is there a way to know which direction the moment to be shown ma'am?
In our formulation of the matrix method, we always assume the unknown member-end moments to be in the counterclockwise direction. After the calculations are done, if we end up with a negative moment magnitude, then we know the moment is actually acting in the clockwise direction.
@@DrStructure Ok Thank you. I am looking at the member forces. In your question both members have a force ( either distributed or a concentrated) what if only one member has a force. In that case how to solve ma'am?
The same way, just right the equilibrium equations based on the member-end forces, without any applied loads. The member-end shear forces and moments would be sufficient to keep the member in equilibrium.
@@DrStructure Thank you . In lecture SA50 at 3.49 minutes in B corner the forces are 20 and 8. so if its equilibrium on the BC bar, does the B corner will have the same forces ? Can you guide me how to find the equilibrium forces in that case ? Is it by the sum of forces and moments ?
When we cut a member, as we have done at 3:49, the forces at the left and right faces of the cut need to balance each other out. The two shear forces need to have the same magnitude but act in opposite directions. The two moments too, need to act in opposite directions but have the same magnitude. @3:49, these forces/moments are shown in black. We calculate the shear force and moment using one of the free-body diagrams (say, beam BC), then place the same force and moment (but in opposite directions) at the other face of the cut.
There are no exercise problems for this lecture at the present time, but you may want to checkout this interactive web page: lab101.space/iexamples/SA51.html
Excuse me, @02:37, member AB, i have a question. I need to draw a picture to describe my question. The link of picture is attached as follow: facebook.com/photo.php?fbid=10157522690582710&set=a.10150380104392710&type=3&theater The question is around this lecture SA50 @02:37 & previous lecture SA49 @10:27, Both lecture teach same configuration of structure with different load types - joint load vs member load. For member AB, they both have theta = 90 degree; But the sequence of vector direction is different, I dont know why!??? In this lecture SA50, the force vector f2 is counted as horizontally in global view, but in previous lecture SA49, the force vector f2 is counted as vertically in global view., I dont get it!???
In SA50, you need to keep in mind that member AB is vertical, even though at 2:37 it is drawn horizontally. In that horizontal diagram, the shear force is f1 (in global coordinate system), the axial force is f2, ... Perhaps it would have been less confusing if that diagram would have been drawn vertically.
@@chungken8496 Vector p is defined in the local (member) coordinate system. When we pre-multiply it by Q (the transformation matrix), we get the member forces in global coordinate system. So, when writing p we consider the x-axis to be along the length of the member. When then transform the local coordinate system to the global one by multiplying p by Q.
Don't see any errors in those coefficients. How did you arrive at your numbers? Take k22. It equals k11 of member BC added to k44 of member AB. k11 of BC = EA/L = (200 x 10^6)(5000 x 10^-6)/8 = 125000 and k44 of AB = 72. When the two numbers added together, we get: 125072.
Thank you for your time to making this video! this is really helpful!
Great work.. keep uploading more such videos ..really appreciate your work
Just amazing, made it very simple to understand and solve
Loved it sir 💛
I dont understand anything in class lectures but ur videos have enlightened me.
Loved it ❤️
I should tell , This method is easiest method ever.
Thanks sir for making us understood.
Thanks sir nice and easy concept video uploaded..
very helpful, thank you.
Thank u very much.....So amazingly cleared hidden points....May God bless you with lots of happinesses and victories in ur life.Again thank u very much.
Thank you!
Now I understand better. thanks
Love and Respect sir ❤️ 💛 💚🙏🙏🙏
perfect!
Thanks you so much. In 2:18 how did yo end up having moments of 20knm at both ends of A-B member?
Those are fixed-end moments. A beam of length L with a concentrated load of P applied at its midpoint, the fixed-end moments are PL/8. So, here we get: (16 kN)(10 m)/8 = 20 kN.m
thank you so much. do you have an interactive web page for member loads using the displacement method? really helpful!
Currently, we have one such web page, for SA51. A similar page for SA50 has been in the works. We'll link to the page here when it is available.
Thanks , for this video on Matrix Displacement Method.How one can solve continuous beam with all fix supports by stiffness matrix method, For example beam with 3 span of each 4m , all supports are fixed with uniformly distributed load on beam 2.5,3.5,3.9 kn/m
A beam with three spans has 4 supports, two at the ends and two intermediate supports. If by all supports you mean all four supports are fixed, then you basically have three independent beams each fixed at its ends. They can be analyzed independently, and there is no need to resort to displacement method to analyze them. The fixed-end moments for each beam are: w L^2/12. Knowing the end moments, you can then apply the equilibrium equations to determine the vertical reaction at the ends of each beam.
If by all supports being fixed, you mean the end supports only, then the system has two degrees of freedom only, the joint rotations at the two interior supports. In this case, we can use the displacement method to calculate the two rotations, and then the support reactions.
Thank you very much.....looking for more such videos on stiffness matrix method
At end of rhe results how the shear force has in same direction in the segment left end and right end?
The direction of the shear force is not defined by the direction of the force arrow, rather it is defined by how the force tends to rotate the beam segment. An upward force at the left end of the segment (wanting to turn the segment in the clockwise direction) signifies a positive shear. However, an upward shear force at the right end of the segment (wanting to turn the segment in the counter-clockwise direction) indicates a negative shear.
In the example problem, the matrix solution gives us two positive (vertical) forces for member BC. The left force having a magnitude of 13.63 kN (since it want to turn BC in the clockwise direction) is considered a positive shear force. The force at the right end of BC is also upward with a magnitude of 10.37 kN. This force however wants to turn the beam in the counter-clockwise direction, therefore, it is considered a negative shear.
Unrelated to the sign of shear, but related to static equilibrium, note that the sum of these two upward forces must be equal to the total downward (applied) load on BC.
Once again thank you much. I dont have enough words to thank this channel for these videos.
Is there a way to know which direction the moment to be shown ma'am?
In our formulation of the matrix method, we always assume the unknown member-end moments to be in the counterclockwise direction. After the calculations are done, if we end up with a negative moment magnitude, then we know the moment is actually acting in the clockwise direction.
@@DrStructure Ok Thank you. I am looking at the member forces. In your question both members have a force ( either distributed or a concentrated) what if only one member has a force. In that case how to solve ma'am?
The same way, just right the equilibrium equations based on the member-end forces, without any applied loads. The member-end shear forces and moments would be sufficient to keep the member in equilibrium.
@@DrStructure Thank you . In lecture SA50 at 3.49 minutes in B corner the forces are 20 and 8. so if its equilibrium on the BC bar, does the B corner will have the same forces ? Can you guide me how to find the equilibrium forces in that case ? Is it by the sum of forces and moments ?
When we cut a member, as we have done at 3:49, the forces at the left and right faces of the cut need to balance each other out. The two shear forces need to have the same magnitude but act in opposite directions. The two moments too, need to act in opposite directions but have the same magnitude.
@3:49, these forces/moments are shown in black. We calculate the shear force and moment using one of the free-body diagrams (say, beam BC), then place the same force and moment (but in opposite directions) at the other face of the cut.
pleaaase , are there any exercises for this lecture ?
There are no exercise problems for this lecture at the present time, but you may want to checkout this interactive web page: lab101.space/iexamples/SA51.html
is member A a fix support?
In the example, the frame is pin connected at point A.
Excuse me, @02:37, member AB, i have a question. I need to draw a picture to describe my question. The link of picture is attached as follow:
facebook.com/photo.php?fbid=10157522690582710&set=a.10150380104392710&type=3&theater
The question is around this lecture SA50 @02:37 & previous lecture SA49 @10:27,
Both lecture teach same configuration of structure with different load types - joint load vs member load.
For member AB, they both have theta = 90 degree;
But the sequence of vector direction is different, I dont know why!???
In this lecture SA50, the force vector f2 is counted as horizontally in global view,
but in previous lecture SA49, the force vector f2 is counted as vertically in global view.,
I dont get it!???
In SA50, you need to keep in mind that member AB is vertical, even though at 2:37 it is drawn horizontally. In that horizontal diagram, the shear force is f1 (in global coordinate system), the axial force is f2, ... Perhaps it would have been less confusing if that diagram would have been drawn vertically.
@@DrStructure If f1 is shear force -- 8kN in global system, why do first row of vector P is equal to 0?
@@chungken8496 Vector p is defined in the local (member) coordinate system. When we pre-multiply it by Q (the transformation matrix), we get the member forces in global coordinate system. So, when writing p we consider the x-axis to be along the length of the member. When then transform the local coordinate system to the global one by multiplying p by Q.
İ think threre is a mistake in the stifness matrices for exampla k22 is equal to 12572 not 125072 also k33 is equal to 10140.6 not 100140,6
Don't see any errors in those coefficients. How did you arrive at your numbers?
Take k22. It equals k11 of member BC added to k44 of member AB.
k11 of BC = EA/L = (200 x 10^6)(5000 x 10^-6)/8 = 125000
and k44 of AB = 72.
When the two numbers added together, we get: 125072.
Hi doc. may I please get contact you directly via mail? I have an example (beam with two different EI values) that's making me pull my hairs out.
Yes, you can send email to:
Dr.structure@educativetechnologies.net