Glad I can help! Consider checking out the rest of the videos/courses that I've done at engineer4free.com 👌

yep - that because university professors who lecture know zilch about learning and cognition. I bet most of them dont have education or teaching qualifications

Glad to hear it! Thanks for watching 🙂

Excellent! Make sure to check out the rest of my statics videos here engineer4free.com/statics if you haven’t already 🤜🤛

Thanks, I love you too. Make sure you check out engineer4free.com/statics if you haven't already!

Good luck! Let me know how it goes! 🤞

Thanks Dhruv!!!

Thanks yo. There is a full list of my hardware and software that I use at engineer4free.com/tools 👌

Thanks Saranga =)

The reactions will be the same whether it is straight or bent, or even If it looks like a potato. Reaction forces are not the same as internal forces. Internal forces refer to internal axial force, shear, and bending moment. A straight two force member like a rod that is pin connected and experience it some tension or compression will ideally only experience said tension/compression. There will be no shear or bending moment because none of the internal force is normal to the axis. A bent two force member or a potato that’s pin connected will have some component of the overall internal force that is not parallel to the axis, because the axis is changing it’s orientation along the length. So you will have some shear force, and also some bending moment due to that. It will will be different everywhere. The more irregular the shape, the harder it will be to calculate what it actually is. That is beyond the scope of statics, which is the series that this video is from. It’s even beyond the next course, which would be mechanics of materials. You might find this type of problem in a “mechanics of materials 2” course, or maybe even after that in some structural design/engineering course. If you’re in a first or second year course, you don’t need to concern yourself with this issue for a very long time, if ever.

2 force members can exist in 3D as well if they are connected with ball joints on each end

The L shaped rod is a two force member. There are two forces acting on it and their lines of action are the same. Don’t worry about the internal forces in each part of the rod, they are not necessary to know for you to determine the reaction force. At this stage, I would just recommend identifying that you have a two force member, and then realizing that the reaction must be equal and opposite to the applied force. So the x component of the reaction will be equal and opposite to the x component of the applied force, and the y component of the reaction will be equal and opposite to the y component of the applied force. You can determine the internal forces in the rod if you want using shear force and bending moment diagrams, but that’s overkill and unnecessary. In statics, I wouldn’t do extra stuff that’s not being asked for.

Usually at this level of statics, members are assumed to be massless / have no self weight. IRL yeah it does

Thanks for watching! In frame problems like this one engineer4free.com/4/frame-example-problem-1 for example, I identify the two force member and then describe it's internal force with only one letter (T). However, I draw two equal and opposite instances of T in the FBD of the two force member. I could have called them T1 and T2, but because it is a two force member then we already know that T1 and T2 would be equal and opposite, so as long as we draw them in opposite directions then we can use a single letter T to represent each force. "each force" is kind of weird to talk about though for a member in pure tension or compression, because tension/compression is just pulling/pushing in both directions with the same magnitude and at the same time, not two different kinds of tension or two different kinds of compression, just a single internal force. I never really thought too much about this but I see how it can be confusing. Does this clear it up?

Engineer4Free I don’t really no how to thank you man💙 yeah It seems I got too much on thinking about it😂 you know sometimes when you feel satisfy and that you understood the material and then thinking a-lot about it will lead to a silly confusion, and I again thank you so much.

Haha I absolutely know how that goes. Glad I was able to help clear it up!

That is true, the forces acting on the L-shaped rod do cancel put but not the momentum on the bar if you change or the 10kN force or the angle calculated, therefore the rod will tend to rotate and wont be in complete equilibrium. This can be easily verified by calculating momentum about the reaction point, with the 10kN and 36.869 degrees, you will get that the sum of momentum equals zero, however if you change the angle by small amount, nevertheless your forces cancel out, your momentum wont. So, the bar will tend to rotate. In this case, in order to have static equilibrium you will have to introduce a third force (making it a 3 body member) which can counteract the 10kN force at an arbitrary angle.

@@umedina98 Why wouldn't the momentum cancel for the second example at around 1:35?

@@umedina98 Thanks for the response by the way. I appreciate it

@@asad5986 You can try it by yourself, try taking a pencil, put it on a table and apply equal but opposite forces at each end of it. The pencil will immediately start to rotate.

@@umedina98 Thanks for the response.

I recommend taking a few hours to review the "Moments" and "Trusses" sections of engineer4free.com/statics then, it will help you to better grasp two force members.

Whoops🙃