You can take moments about anywhere and the structure must still obey equilibrium, but the whole idea is eliminate as many unknown (forces C and D) as possible.
Hi Tim, Great video, enjoyed it. I do have a question on the last question where we have to select to delete the additional diagonals shown which may not carry a force. At 28.30 second you start to delete the diagonals until the mid point of the truss in same direction and reverse from there. this would be correct for the symmetric loading or resultant load is at center or near center. when we do have a significant resultant force away from center the symmetricity will be gone. appreciate you explain this scenario to identify the tension members. Thank you
Sorry, I just don't understand how the adjacent of the triangle in the beginning could be 50kN. Could you explain how that works? Also, is this method of joints?
If you take a freebody cut the whole structure still needs to be in equilibrium. There are 300kN going down and 350kN going up from the reaction. The diagonal member is the only member that can carry any vertical load, and the vertical component must equal 50kN downwards (350-300kN)
Useful videos. In terms of the maximum force, I don't calculate the tension, diagonal, and compression force as it is shown. Because I know that the maximum force is in the compression in the bottom/top cord. I simply calculate the maximum moment as the truss it is a simple beam. Then because applies M = F*h you can get straight away the compression force. For example in the first example: M_max = 4.5*350 - 100*6 = 975 kNm. F_max = 975 / (√3/2) = 1125.83 kN - compression. You have to know the reaction for the max. moment, the rest can be skipped. :)
Edit: there is not always maximum compression. It depends on the direction of the diagonals where the max. moment is. Basically, the maximum force is where you don't have any other diagonals connected to the top/bottom chord.
Hi Tim, thank you for all your videos, I really learn a lot! I have query on the crane example, I saw that at point B we have vertical member, is that member will not share taking the load (100kN) with diagonal AB? May I know why?
It would, but when we are making free body diagrams those internal axial forces of the members we cut become external forces. And in any statics problem we balance out the external forces first, before we start to resolve internal system forces
These videos are most helpful. Thanks for all. Waiting for more !!!
Glad you like them!
At 33:00 could you take moments at the pin opposite from x instead of y?
You can take moments about anywhere and the structure must still obey equilibrium, but the whole idea is eliminate as many unknown (forces C and D) as possible.
Hi Tim, Great video, enjoyed it. I do have a question on the last question where we have to select to delete the additional diagonals shown which may not carry a force. At 28.30 second you start to delete the diagonals until the mid point of the truss in same direction and reverse from there. this would be correct for the symmetric loading or resultant load is at center or near center. when we do have a significant resultant force away from center the symmetricity will be gone. appreciate you explain this scenario to identify the tension members. Thank you
Sorry, I just don't understand how the adjacent of the triangle in the beginning could be 50kN. Could you explain how that works? Also, is this method of joints?
If you take a freebody cut the whole structure still needs to be in equilibrium. There are 300kN going down and 350kN going up from the reaction. The diagonal member is the only member that can carry any vertical load, and the vertical component must equal 50kN downwards (350-300kN)
@@TheStructuralExam haha, of course! Brain malfunction there. Thanks for the content.
This channel is godsend.
In the crane example does this mean the diagonal AB takes all the vertical load from the 100kN and the others take 0?
AB does indeed take all the vertical load from the 100kN, but because it is inclined there will still be forces in the horizontal members beside it.
@@TheStructuralExam Thank you. What about the other inclined members?
Useful videos. In terms of the maximum force, I don't calculate the tension, diagonal, and compression force as it is shown. Because I know that the maximum force is in the compression in the bottom/top cord. I simply calculate the maximum moment as the truss it is a simple beam. Then because applies M = F*h you can get straight away the compression force. For example in the first example: M_max = 4.5*350 - 100*6 = 975 kNm. F_max = 975 / (√3/2) = 1125.83 kN - compression. You have to know the reaction for the max. moment, the rest can be skipped. :)
Edit: there is not always maximum compression. It depends on the direction of the diagonals where the max. moment is. Basically, the maximum force is where you don't have any other diagonals connected to the top/bottom chord.
Hi Tim, thank you for all your videos, I really learn a lot!
I have query on the crane example, I saw that at point B we have vertical member, is that member will not share taking the load (100kN) with diagonal AB? May I know why?
It would, but when we are making free body diagrams those internal axial forces of the members we cut become external forces. And in any statics problem we balance out the external forces first, before we start to resolve internal system forces
@@TheStructuralExam That's make sense. Thank you so much :)
my query is you are not providing the actual hand calculations for questions 1 and 2. if you could do so that would be very helpful
Thank you.....💓