ORganic chemistry was the🐐 before but now that I’m actually taking my civil engineering classes you are quickly taking his place👑 you’re a gift to the world 🙏 please don’t stop making videos
Thank you soo soo much. I was having so much trouble not with the problems themselves, but with the directions of the reaction forces. In class it wasn't made known that you had to draw equal and opposite reactions when you move from FBD to FBD, so I was incredibly confused as to why my answers were turning out incorrect. You clarified this in literally minutes!
Is there anyway to determine the orientation of reaction forces without solving the actual values of them? or is the only way to be sure of orientation is through actual calculations
In statics there is a distinction between trusses and frames. Trusses are made up entirely of straight 2 force members connected at their ends with pins. The structure in this video is not a truss because member ABC and FED do not meet the criteria; the structure is a frame. ABC and FED are not 2 force members because they are both experiencing a third applied force from BE that is not acting at the end (if it were acting at the end at a pin it would be combined with the other forces also acting at the same pin and be treated as the same force). Because trusses are straight two force members that are pin connected at the ends, we simplify the problem to assuming that the internal forces in any given member are axial, ie, no internal shear and internal bending moments within the members; pure tension or compression. If you have something that is not a straight two force member (like ABC or FED), there will be internal shear and bending forces developing in the member. So at point B or point E, members ABC and FED will have internal shear forces, and if you draw some virtual cuts around the points, you're going to see that for static equilibrium at the joint, that the force that BE applies on the joint will be equal and opposite to the internal shear force at that point. That's why it's not a zero-force member. It's important to realize that member ABC is a single member, not two separate members AB and BC. Same for member FED. Another giveaway that this is a frame and not a truss is the applied load F1 that is not acting at a joint. In truss problems, the applied loads ALWAYS act at joints. If you have a load not acting at a joint and not in line with the axis of the member, it's also going to be introducing internal shear and bending moment in that member, which takes us into the realm of frames that we can no longer assume to be in pure tension and compression. In statics you won't actually need to calculate those internal forces, but just knowing that they are present indicates we are talking frames.
ORganic chemistry was the🐐 before but now that I’m actually taking my civil engineering classes you are quickly taking his place👑 you’re a gift to the world 🙏 please don’t stop making videos
Thank you soo soo much. I was having so much trouble not with the problems themselves, but with the directions of the reaction forces. In class it wasn't made known that you had to draw equal and opposite reactions when you move from FBD to FBD, so I was incredibly confused as to why my answers were turning out incorrect. You clarified this in literally minutes!
Ah yeah if that detail is left out then frames make no sense. Glad I could help!!!
really saving me on my exam. thank you so much. these are quality videos.
Thanks for the comment! Please tell some friends about them :)
Short, clear, and simple explanation. Thanks!
Thanks for watching! Make sure to check out engineer4free.com/statics if you haven't already :)
I'm so grateful for you , i was in the class after i had listened ur video & magnificently i become like an expert.waw🥰🥰🥰🥰🥰🥰👏👏👏
fb.d. of a each member of the frame is solved by . equations of equilibrium
Is there anyway to determine the orientation of reaction forces without solving the actual values of them? or is the only way to be sure of orientation is through actual calculations
Does anyone know why force Fr doesnt act upon member CD in the exploded diagram?
Why is BEin the first diagram not a zero-force member? Why in this case is it a two force member?
In statics there is a distinction between trusses and frames. Trusses are made up entirely of straight 2 force members connected at their ends with pins. The structure in this video is not a truss because member ABC and FED do not meet the criteria; the structure is a frame. ABC and FED are not 2 force members because they are both experiencing a third applied force from BE that is not acting at the end (if it were acting at the end at a pin it would be combined with the other forces also acting at the same pin and be treated as the same force). Because trusses are straight two force members that are pin connected at the ends, we simplify the problem to assuming that the internal forces in any given member are axial, ie, no internal shear and internal bending moments within the members; pure tension or compression. If you have something that is not a straight two force member (like ABC or FED), there will be internal shear and bending forces developing in the member. So at point B or point E, members ABC and FED will have internal shear forces, and if you draw some virtual cuts around the points, you're going to see that for static equilibrium at the joint, that the force that BE applies on the joint will be equal and opposite to the internal shear force at that point. That's why it's not a zero-force member. It's important to realize that member ABC is a single member, not two separate members AB and BC. Same for member FED. Another giveaway that this is a frame and not a truss is the applied load F1 that is not acting at a joint. In truss problems, the applied loads ALWAYS act at joints. If you have a load not acting at a joint and not in line with the axis of the member, it's also going to be introducing internal shear and bending moment in that member, which takes us into the realm of frames that we can no longer assume to be in pure tension and compression. In statics you won't actually need to calculate those internal forces, but just knowing that they are present indicates we are talking frames.
Thank you!
You're welcome!!! More over @ engineer4free.com/statics =)
How to decide where to include F2..? in segment CD or FED..?
Can choose either, should result with the same final answer
nice ! can you make a video about arch? including internal force like shear force and axial force
oooooh yeah I don’t have any videos on arches. Good idea!
@@Engineer4Free yeaa! Thanks for concern about my comment 🥺
Brooo this guy is so goood!!!!
anyone from swinburne taking structural mechanics?? what is the answer
blesssssssssss up xx
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exam in 1 hr. ez lol
Are u canadian lol
haha how can you tell :p
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can i hae your email