Thank you for not spending 10 minutes of the video drawing and correcting your drawing while you talked. The fact that you had it all prepared before hand made it much easier to follow and absorb the information!
Thanks for the video!! When you say " Tension" and "compression", is it with respect to the member or with respect to the joint?? I noticed that you determine the type of the force with respect to the joint. Am I right?
With respect to the member. The member is in tension, or the member is in compression. A member in tension will "pull" on the joints and a member in compression will "push" on the joints.
karishma chandini yaa coz there is no horizontal force anywhere to balance that horizontal force of pin G so it will become zero... most of the times in trusses horizontal force of pin is zero
You need to draw free body diagram first to see his point. Sum of X forces are supposed to be zero. Since A is sliding bearing it won't have an X force. Since there is only G bearing's X force , it equals to zero.
Because at this stage we have done no calculations, so the internal force in member AB is unknown. It is the convention to draw unknown forces in tension, so from the point of view of joint A, the unknown force is drawn pulling on it, ie, pointing away. In the next video, we solve this truss with some actual loads applied, and find that member AB to have a negative magnitude, which indicates it's pointing in the opposite direction as indicated on the fbd, ie, we find out later on that it is in fact pointing towards A in that fbd. Here's the link to the next video: th-cam.com/video/PXS9n-b5CA8/w-d-xo.html
When you work out each joint do the arrows always be drawn in tension before working it out. Ive seen some examples where they are drawn in compression before being solved.
It's not an absolute requirement. Often you can know by inspection if a member will be in tension or compresseion even when the actual magnitude is still unknown. If you draw an unknown in compression, just be careful that you remember that a positive value for that particular member means it's actually in compression and a negative means it's actually in tension. Always drawing unknowns in tension even if you expect them to be in compression usually decreases the chance of you making an error because you apply the exact same logic to every member every single time.
If a truss is only constructed with equilateral triangles like this then technically no, if there is a load somewhere then every member will have some amount of tension or compression in in it. Once you start getting trusses that resemble right angle triangles and stuff you might be finding ZFM's depending on the loading type.
There are no externally applied forces on the structure, and because G has the only reaction capable of providing horizontal support (Gx), Gx is equal to zero. You would see it if you write the Sum of Force in x expression for the whole structure, it would just be Gx = 0, so I just skipped it. In a test, it might be worth the few seconds to write it in to really show you know!
Just include that in the FBD of the whole structure. Ra and Rb will be different (bigger). Then once you know Ra, just proceed with the same method, joint by joint. See videos starting at #42 here: engineer4free.com/statics
Thank you so much for such great tutorials sir. May I ask if what kind of software or application are you using fir this kind of tutorial? Thank you in advance. 😇
Hey Edmil, yeah I use a few different pieces of software and hardware together. Best is to look through the list of tools that I use here: engineer4free.com/tools 🤙
Hey Daniel, I’ve got a list of all the hardware and software that I use at engineer4free.com/tools. The drawing program that I used for this video was sketchbook.
If you have two pins, then it is actually statically indeterminate, and is no longer in the realm of a statics course. If you are taking a statics course, everything must be statically determinate, so maybe it is a drawing typo. If you are in mechanics of materials or structural analysis course, then you need to use one of the methods for indeterminate structures. I made full playlists for both of those courses: engineer4free.com/mechanics-of-materials and engineer4free.com/structural-analysis
@@Engineer4Free Thank you, I am in a statics course. In the problem, a truss has two pinned supports and all applied forces act in the x direction. On closer inspection, of the two members attached to the support pins, one is perpendicular to the applied forces, while one meets the pin at a 45 degree angle. Does that make the problem statically determinate?
No it doesn't. If the lines of actions of the forces all both pass through both of the two supports then you can do some trickery to kind of skip the work that is actually involved in it being statically indeterminate, but I'm assuming that's not the case for you, as that's beyond the scope of statics. I think it is most likely a diagram typo. I would ask your professor if the truss is "simply supported" which means that one reaction is a pin and the other is a roller. Express your concern to them that two pins in the diagram implies indeterminacy and I because they give you 4 unknowns while you only have 3 equations (sum of forces in x, sum of forces in y, and sum of moments about some point). I'm pretty sure your prof will tell you it's a typo.
@@Engineer4Free Okay, my professor hasn't brought up static determinacy/indeterminacy in class so I was banging my head over the question for a while. Thank you very much for the prompt replies, I'll see what he has to say.
Hey Shubham. Finding forces is critical for real life structural enghineering. Structures support loads, and if the load is too big, the structure will break. Structures are generally made of smaller members (truss is a good example) , and if a certain member cannot support an internal force, it will break. Common internal forces include axial (tension and compression), shear, and bending moment, and torsion. This video is simplified to only be dealing with axial forces. Basically, a stronger material can support a greater load. Imagine 2 truss members of the same cross sectional area, but one is made of steel and the ither is made of wood. Long story short, the steel one is stronger, and could support a greater internal axial force compared to the wooden one. When you design a truss bridge, or any structure and its members, you need to select appropriate materials and sizes of members so that it will not break. Statics is just the super basics of structural engineering, with many simplifications in place, but finding internal forces is basically the core of all further studies and application of structural engineering.
This structure is made up of 11 pieces that are the same length, forming equilateral triangles connected by pinned joints. There are two horizontal members on the top, and three horizontal members on the bottom. The rest are diagonal @ 60 degrees.
not all heroes wear capes, real heroes explain the method of joints on youtube
🤜🤛
@Kyrie Trent no one caressssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss
@@Weminan why all the hate
@@shaquillchole8042 I think it was some guy who has now deleted his comment trying to scam people with his "free movies" stuff.
Instablaster
Thank you for not spending 10 minutes of the video drawing and correcting your drawing while you talked. The fact that you had it all prepared before hand made it much easier to follow and absorb the information!
The magic of video editing used for good! =)
Just wanted to say thankyou for explaining this so beautifully and in such a short time. Have a wonderful day!
Thanks a lot for the comment! Really glad you liked it and took the time to write. You have an awesome day too =)
@@Engineer4Free :))
currently reviewing for my statics exam! Thanks for helping create new engineers!
Thanks for explaining in short yet effective manner. 👍
exactly what i was looking for. Thanks homie
Awesome, glad to hear it!🤜🤛
Bruh that was so satisfying when you made all the fbds simultaneously haha
Magiccccc 💫
This guy right here is a life saver.
I gochu! More vids are here: engineer4free.com/statics =)
Thanks for the video!! When you say " Tension" and "compression", is it with respect to the member or with respect to the joint?? I noticed that you determine the type of the force with respect to the joint. Am I right?
With respect to the member. The member is in tension, or the member is in compression. A member in tension will "pull" on the joints and a member in compression will "push" on the joints.
you sir explained to me what no doctor managed to do
Thanks! The full playlist is here too: engineer4free.com/statics =)
Explaining all of that in 5mins
I would say thank you so much 🌹
shouldnt the fixed support at G have a reaction in the x direction?
Dude, you're an absolute legend. Thank you so much!
Thanks carter!! 🤜🤛
I would really appreciate it if you could make a video about how to convert uniformly distributed loading on the truss to point loads.
this actually makes it so simple! ty
You're welcome!! I've got a few more truss tutorials in videos # 42-51 here: engineer4free.com/statics =)
Amazing layout of the problem, super helpful!
Thanks Daniel!
I like this guy
I like u too
why didn't you take horizontal force at pin joint G?
karishma chandini because it's zero.
karishma chandini yaa coz there is no horizontal force anywhere to balance that horizontal force of pin G so it will become zero... most of the times in trusses horizontal force of pin is zero
RAHUL plz can you explain more in detail plsss
You need to draw free body diagram first to see his point. Sum of X forces are supposed to be zero. Since A is sliding bearing it won't have an X force. Since there is only G bearing's X force , it equals to zero.
But it would still be good practice to put a horizontal component, even though it’s zero
I dont get it, to me it looks like the forces you drew for ab would make the member be in compression. Why does it look like it should be flipped
How come you didn’t use the Gx doesn’t it have an X and a Y since it’s a pin?
Sum of forces for the entire structure shows us that Gx = 0, so I just don't bother writing it
why isnt the AB arrow pointing back towards Joint A?
Because at this stage we have done no calculations, so the internal force in member AB is unknown. It is the convention to draw unknown forces in tension, so from the point of view of joint A, the unknown force is drawn pulling on it, ie, pointing away. In the next video, we solve this truss with some actual loads applied, and find that member AB to have a negative magnitude, which indicates it's pointing in the opposite direction as indicated on the fbd, ie, we find out later on that it is in fact pointing towards A in that fbd. Here's the link to the next video: th-cam.com/video/PXS9n-b5CA8/w-d-xo.html
G is two supports there
When you work out each joint do the arrows always be drawn in tension before working it out. Ive seen some examples where they are drawn in compression before being solved.
It's not an absolute requirement. Often you can know by inspection if a member will be in tension or compresseion even when the actual magnitude is still unknown. If you draw an unknown in compression, just be careful that you remember that a positive value for that particular member means it's actually in compression and a negative means it's actually in tension. Always drawing unknowns in tension even if you expect them to be in compression usually decreases the chance of you making an error because you apply the exact same logic to every member every single time.
Thanks sir you are the great
Thanks Usman =) =)
Can you have 0 force members in this type of structure?
If a truss is only constructed with equilateral triangles like this then technically no, if there is a load somewhere then every member will have some amount of tension or compression in in it. Once you start getting trusses that resemble right angle triangles and stuff you might be finding ZFM's depending on the loading type.
Thank you it helped me a lot......
Awesome, glad I could help!! 🙂
why dont you include the horizontal force off the pin joint G?
There are no externally applied forces on the structure, and because G has the only reaction capable of providing horizontal support (Gx), Gx is equal to zero. You would see it if you write the Sum of Force in x expression for the whole structure, it would just be Gx = 0, so I just skipped it. In a test, it might be worth the few seconds to write it in to really show you know!
Thanks a bunch !! you saved me an hour of frustration XD
Thanks so much for this!!!!!!!!!!!!!!! 👌👌👌
You’re welcome!!!!!
how does it change if i have a weight hanging from point E ? please help me
Just include that in the FBD of the whole structure. Ra and Rb will be different (bigger). Then once you know Ra, just proceed with the same method, joint by joint. See videos starting at #42 here: engineer4free.com/statics
Thank you so much for such great tutorials sir. May I ask if what kind of software or application are you using fir this kind of tutorial? Thank you in advance. 😇
Hey Edmil, yeah I use a few different pieces of software and hardware together. Best is to look through the list of tools that I use here: engineer4free.com/tools 🤙
Hey. What program are you using to draw/edit the pictures?
Hey Daniel, I’ve got a list of all the hardware and software that I use at engineer4free.com/tools. The drawing program that I used for this video was sketchbook.
Engineer4Free legend. Thanks mate.
How do you go about solving these problems when none of the supports are designated as an expansion joint or rocker?
If you have two pins, then it is actually statically indeterminate, and is no longer in the realm of a statics course. If you are taking a statics course, everything must be statically determinate, so maybe it is a drawing typo. If you are in mechanics of materials or structural analysis course, then you need to use one of the methods for indeterminate structures. I made full playlists for both of those courses: engineer4free.com/mechanics-of-materials and engineer4free.com/structural-analysis
@@Engineer4Free Thank you, I am in a statics course.
In the problem, a truss has two pinned supports and all applied forces act in the x direction.
On closer inspection, of the two members attached to the support pins, one is perpendicular to the applied forces, while one meets the pin at a 45 degree angle. Does that make the problem statically determinate?
No it doesn't. If the lines of actions of the forces all both pass through both of the two supports then you can do some trickery to kind of skip the work that is actually involved in it being statically indeterminate, but I'm assuming that's not the case for you, as that's beyond the scope of statics. I think it is most likely a diagram typo. I would ask your professor if the truss is "simply supported" which means that one reaction is a pin and the other is a roller. Express your concern to them that two pins in the diagram implies indeterminacy and I because they give you 4 unknowns while you only have 3 equations (sum of forces in x, sum of forces in y, and sum of moments about some point). I'm pretty sure your prof will tell you it's a typo.
@@Engineer4Free Okay, my professor hasn't brought up static determinacy/indeterminacy in class so I was banging my head over the question for a while. Thank you very much for the prompt replies, I'll see what he has to say.
Ok cool!
can you answer my question ? i do it this question like in 3 days but didn't get the answer yet , can you help me ?
I am confused with sign convention please explain in details by solving a problem.
Yes I have examples on the truss method of joints. Please watch videos 44 and 45 here: engineer4free.com/statics
Good explanation thanks!
Thanks for commenting!
What is the use of finding forces? I mean where is it used in real life or in designing?
Hey Shubham. Finding forces is critical for real life structural enghineering. Structures support loads, and if the load is too big, the structure will break. Structures are generally made of smaller members (truss is a good example) , and if a certain member cannot support an internal force, it will break. Common internal forces include axial (tension and compression), shear, and bending moment, and torsion. This video is simplified to only be dealing with axial forces. Basically, a stronger material can support a greater load. Imagine 2 truss members of the same cross sectional area, but one is made of steel and the ither is made of wood. Long story short, the steel one is stronger, and could support a greater internal axial force compared to the wooden one. When you design a truss bridge, or any structure and its members, you need to select appropriate materials and sizes of members so that it will not break. Statics is just the super basics of structural engineering, with many simplifications in place, but finding internal forces is basically the core of all further studies and application of structural engineering.
Your two horizontal members:: Were they two long pieces??? Thanks. K.S.
This structure is made up of 11 pieces that are the same length, forming equilateral triangles connected by pinned joints. There are two horizontal members on the top, and three horizontal members on the bottom. The rest are diagonal @ 60 degrees.
why does it give me different values when i used method of section to find BD?
can you come teach my structures class, please??!!
What software did you use to generate the images?
I think I used sketchbook for this one. I have a full list of the hardware and software that I use for the videos here: engineer4free.com/tools
Glad to be the 100th comment😁
Welcome to the party! 🥳
I’m in love with you
👍🏻
👍
'di ko naman kasi pinangarap maging engineer eh 😭
where is the next video?
See videos 43-52 here: engineer4free.com/statics