Two force members explained (statics)

[@lamborghinicentenario2497]

3 years into engineering, and this is the first time I get a good explanation for 2-force members. Thanks! It was about time

[@Engineer4Free]

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

[@johnmifsud6814]

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

[@noahdangerio9540]

usually i am slow at understanding things, for this one I knew its definition, but was not sure why it was important. This makes things so much clearer, thank you so much!

[@Engineer4Free]

Glad to hear it! Thanks for watching 🙂

[@aryan-singh]

Thanks Sir, you explained the cases very clearly.

[@Engineer4Free]

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

[@akhmadjonovakromjon3669]

I love you man. You are true hero.

[@Engineer4Free]

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

[@furqonmuhammad8232]

exam tomorrow and yer video is truly useful thanks sir

[@Engineer4Free]

Good luck! Let me know how it goes! 🤞

[@hakankosebas2085]

What about in 3d, are them pinned or have a ball joint? Is it matter type of connection of two force member?

[@Engineer4Free]

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

[@bhakemono346]

Hey man, amazing tutorial. Just wanted to ask what program you are doing this on? Much appreciated keep it up champ.

[@Engineer4Free]

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

[@ravigautam8685]

Thank you for the tutorial. As I can see L-shaped rod is acting as two force member. But is it necessary that force acting on the free end should be along of line joining rod's end points in order to have L-shaped rod as two force member. How do find equilibrium force is in horizontal direction (it may lie in the plane of the rod).

[@Engineer4Free]

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.

[@DarkGT]

Hello sir, i find out your tutorials about C++ and Linear Algebra for very helpful, thank you !

[@shahzad5162]

quick question what about the mass, doesn't that exert a force

[@Engineer4Free]

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

[@asad5986]

For the second case(around 1:35), I still am not getting why that member would not be in static equilibrium if the forces cancel out. Also, how were you able to conclude that the angle just below the 10 kN force is 36.869 degrees?

[@umedina98]

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.

[@asad5986]

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

[@asad5986]

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

[@umedina98]

@@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.

[@asad5986]

@@umedina98 Thanks for the response.

[@dhruvbhattofficial]

very nice video. u explained very well

[@Engineer4Free]

Thanks Dhruv!!!

[@johnpatrickaguilar9442]

Hi there, isn't it supposed to be artan because sin is opposite over hypotenuse?

[@thomasjefferson5727]

He used sine. The opposite over hyp.

[@ravigautam8685]

Dear Engineers4free, Thank you for reply. I am new to this subject. thank you for the explanation. I still have one question regrading your statement " identifying that you have a two force member, and then realizing that the reaction must be equal and opposite to the applied force. " Suppose I have frame with one non-straight member. Then its seems that shape of the curved member has no bearing on the equilibrium of the frame (curved member can be parabolic arc, circular arc, L-shaped rod of different sized arms). It actually seems to be same as that of a straight element! Also axial stress (or normal stress) in the curved element will be same as that of straight element. Please correct me I may be most probably wrong some where!

[@Engineer4Free]

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.

[@InfinityBladeRecollection]

Thanks very helpful

[@devsaranga]

Well explained

[@Engineer4Free]

Thanks Saranga =)

[@bbcooper17]

Well explained! thanks too much

[@j.s.s9108]

thank you for the great explanation, I just wonder why do we represent the 2 force member in a frame with one force while if we to analyze the FBD of the 2force member alone we have opposite and equal forces, so in frame we shouldn’t show any force represents it ?

[@Engineer4Free]

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?

[@j.s.s9108]

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.

[@Engineer4Free]

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

[@wiktorkawala6298]

Thought this looked complicated but then realized its just Components of a force problems from M1 OCR A-level. Nice explanation though :)

[@namsaying]

Thank u sir

[@raphaelgamba4016]

What a shame for the statics world to not use a 3/4/5 triangle ratio but the angle

[@alonkulungrai3796]

I did not get much concept.

[@Engineer4Free]

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.

[@xuewenyu1434]

can i have your email so i can send my question to you

[@Penitent946]

Why u talking so fast xD

[@Engineer4Free]

Whoops🙃

[@3vv2.]

❤❤