Thanks for letting me know Connor! All of my structural videos are split between three playlists: statics, mechanics of materials, and structural analysis so if you can't find something in one of them do check the others. Cheers!
#wooow!_AMAZING! I AM CIVIL ENGINEERING student in ~~ETHIOPIA~~ I'm so glad to you thanku sir! I know more with few minuties!......you are best lecturer!
#WOOOW! THANKYOU!!.. WE Ethiopians inviting you to our country! @@Engineer4Free! thanku for your reply sir! yes I AM sudying at ADAMA SICIENCE AND TECHNOLOGY UNIVERSITY!..most of our university students follows your vedio.Again I would like to say thanku a lot! WISH YOU ALL THE BEST IN YOUR LIFE!.....
How come fy is 300n and not fx , since “ the hinge “ is attached on the wall ( on its side ) wouldn’t that make the fx component the y and the fy component the x ( fx)
The sum of the vertical forces must sum to zero on their own, regardless of any external horizontal forces (which there aren’t any actually in this case anyway). So the rest of the vertical applied forces sum to 300N down, so Fy must be 300N up to achieve static equilibrium in the vertical direction.
Hey Ryan, not necessarily. It depends on if you're looking at the opposite or adjacent side of the triangle. Construct a right angle triangle such that the hypotenuse is the diagonal, and then label the opp and adj side based on the known angle. Then determine which function you need. Basically if your using the hyo and adj, you use cos, and if you're using the hyp and opp, you use sin. Be careful to recognize which side your using and which trig function you need to use. Review basic trig / SOH CAH TOA if you need to, as that's all this is!
I don’t, I just draw assume a direction and go with it while they are unknowns. When doing this, pick your assumed direction in the positive direction of x and y, so that if you get a negative value, it’s clear that it must point in the negative direction.
@@asad5986 yeah, the negative sign just tells you that its preally pointing the other way. It's helpful to write (tension) or (compression) in brackets beside your answer to make it very clear to your prof that you understand which one it is.
Why is it that at the pinned point at joint F where you have Fx and Fy you just automatically make Fx=0? I know there’s no horizontal forces but if you take the moment at A wouldn’t you get two unknowns one for Fx and one for Fy?
This structure has only one horizontal reaction. That is Fx. So because there are no horizontally applied loads, then ΣFx = Fx = 0 (I apologize that "sum of forces in x direction" and "horizontal reaction force at point F" have the same syntax here...). But we find Fx to be zero in that first step. Because we know it's zero, it's no longer an unknown. So later in the sum of moments about A eapreasion, we don't need to write anything for the impact of Fx on the moment about A, because it's magnitude is zero, so it would cause a moment.
Why did you SUBTRACT BC??? When you calculated about point B?? Its a summation and there was no explanation. I've been searching the internet and textbook for 8 hours for that exact answer. waste of time
Ok so when I draw the fbd of joint B, I draw all unknown forces acting on the join in tension. Assume tension if you don't know. Everybody assumes tension, so that if you slget a positive magnitude, you know it is indeed in tension, and if you get a negsrive magnitude it is in fact in compression (so would actually push on that joint rather than pull). So on the Fbd I draw it in tension, this means it pulls, or otherwise points down. Draw a coordinate axis of x and y. Positive x is to the righrx and positive y is up. Vectors that point up have positive y magnitude and vectors that point down have negative y magnitude. So in the sum of y expression, BC is negative because it has a negative y value (because it points down in the FBD). The y component of AB is pointing upward, so it has a positive value in the sum of forces in y expression. Please watch this video that I troduces the method: www.engineer4free.com/4/truss-analysis-by-method-of-joints-explained it will help. And also I recommend watching videos 42 - 51 here www.engineer4free.com/statics as well as review any vector math from videos 1 - 13 if needed! I hope that clears it up for you
+Mickel nickey ahaha thanks!! Don't forget to check the rest of the statics videos here engineer4free.com/statics I should be finishing them up this week!
only channel that has helped me with my structural analysis unit.....thanks
Thanks for letting me know Connor! All of my structural videos are split between three playlists: statics, mechanics of materials, and structural analysis so if you can't find something in one of them do check the others. Cheers!
Amazing content. Easy to follow and understand. 🙏
Helped me with last minute cramming ;)
Glad I could help!! Check out engineer4free.com/statics for the rest of the videos if you haven’t already 🤜🤛
#wooow!_AMAZING! I AM CIVIL ENGINEERING student in ~~ETHIOPIA~~ I'm so glad to you thanku sir! I know more with few minuties!......you are best lecturer!
Thanks!! I'm hoping to visit Ethiopia 👌👌. Are you studying in Addis?
#WOOOW! THANKYOU!!.. WE Ethiopians inviting you to our country! @@Engineer4Free! thanku for your reply sir! yes I AM sudying at ADAMA SICIENCE AND TECHNOLOGY UNIVERSITY!..most of our university students follows your vedio.Again I would like to say thanku a lot! WISH YOU ALL THE BEST IN YOUR LIFE!.....
wow . brilliant channel , where have you been all this while. thkns a lot . you are helping me
Haha online since 2012. Welcome to the party! 🥳
@@Engineer4Free thank you very much .
Best channel for education purposes 🥳🥳🥳
love you bruh.
luv ya too
thanks you a looooooooooooooooooooooot
wish u all the best in ur hole life
Thankyou so much. I have two textbooks and they both solve graphically and couldn't explain anything properly
How come fy is 300n and not fx , since “ the hinge “ is attached on the wall ( on its side ) wouldn’t that make the fx component the y and the fy component the x ( fx)
The sum of the vertical forces must sum to zero on their own, regardless of any external horizontal forces (which there aren’t any actually in this case anyway). So the rest of the vertical applied forces sum to 300N down, so Fy must be 300N up to achieve static equilibrium in the vertical direction.
When solving for the components of a diagonal force, will it always be sine for the Y direction and will it always be Cosine for the X direction?
Hey Ryan, not necessarily. It depends on if you're looking at the opposite or adjacent side of the triangle. Construct a right angle triangle such that the hypotenuse is the diagonal, and then label the opp and adj side based on the known angle. Then determine which function you need. Basically if your using the hyo and adj, you use cos, and if you're using the hyp and opp, you use sin. Be careful to recognize which side your using and which trig function you need to use. Review basic trig / SOH CAH TOA if you need to, as that's all this is!
How do you know Fx and Fy are going up and to the right?
I don’t, I just draw assume a direction and go with it while they are unknowns. When doing this, pick your assumed direction in the positive direction of x and y, so that if you get a negative value, it’s clear that it must point in the negative direction.
@@Engineer4Free Ok thank you. Then it's just a matter of adjustment if the assumption turns out to be incorrect right?
@@asad5986 yeah, the negative sign just tells you that its preally pointing the other way. It's helpful to write (tension) or (compression) in brackets beside your answer to make it very clear to your prof that you understand which one it is.
@@Engineer4Free ok thank you
I dont understand why the angle is 45 dgrees..could you please explain or send a video related to it pls?
nvm😅😅
Why is it that at the pinned point at joint F where you have Fx and Fy you just automatically make Fx=0? I know there’s no horizontal forces but if you take the moment at A wouldn’t you get two unknowns one for Fx and one for Fy?
This structure has only one horizontal reaction. That is Fx. So because there are no horizontally applied loads, then ΣFx = Fx = 0 (I apologize that "sum of forces in x direction" and "horizontal reaction force at point F" have the same syntax here...). But we find Fx to be zero in that first step. Because we know it's zero, it's no longer an unknown. So later in the sum of moments about A eapreasion, we don't need to write anything for the impact of Fx on the moment about A, because it's magnitude is zero, so it would cause a moment.
I dont understand the directions of Fx or Fy .
Can you record a extensive video about free body diagrams please?
Yeah, try this one: www.engineer4free.com/4/how-to-draw-good-free-body-diagrams-fbds
I think the angle in point C is 90 degrees not 45
Member CD is definitely angled at 45 degrees from the horizontal
Thankyou!!
You're welcome!! =)
200/sin(45) = 282.84. You still are correct with everything else though
Why did you SUBTRACT BC??? When you calculated about point B?? Its a summation and there was no explanation. I've been searching the internet and textbook for 8 hours for that exact answer. waste of time
Ok so when I draw the fbd of joint B, I draw all unknown forces acting on the join in tension. Assume tension if you don't know. Everybody assumes tension, so that if you slget a positive magnitude, you know it is indeed in tension, and if you get a negsrive magnitude it is in fact in compression (so would actually push on that joint rather than pull). So on the Fbd I draw it in tension, this means it pulls, or otherwise points down. Draw a coordinate axis of x and y. Positive x is to the righrx and positive y is up. Vectors that point up have positive y magnitude and vectors that point down have negative y magnitude. So in the sum of y expression, BC is negative because it has a negative y value (because it points down in the FBD). The y component of AB is pointing upward, so it has a positive value in the sum of forces in y expression. Please watch this video that I troduces the method: www.engineer4free.com/4/truss-analysis-by-method-of-joints-explained it will help. And also I recommend watching videos 42 - 51 here www.engineer4free.com/statics as well as review any vector math from videos 1 - 13 if needed! I hope that clears it up for you
@@Engineer4Free god bless your soul thanks!!
You are a champion, thank you for your videos, they're so much clearer and easy to understand then the lectures uni uploads.
Thanks Khentse!! ☺️
u are the best!!!!~~~~~~ ty
+Mickel nickey ahaha thanks!! Don't forget to check the rest of the statics videos here engineer4free.com/statics I should be finishing them up this week!