This is not a man, this is a god of engineering tutorials! Thank you so much man, you have no idea how many lives you have just saved just by doing these kinds of videos. Keep it up man!
🙌🙌 Thanks Korajark!! If you haven't already, do check out the whole collection of statics vids I made over at engineer4free.com/statics and tell some friends!! =)
I'm sincerely grateful for you giving literally a free lecture on RUclips..! Thank you so much. It helped me not only to solve questions in assignments but also to understand the entire truss analysis..!
GOOD effort I often see your videos to learn structure analysis. I have learnt much from ur videos. Simple and effective way of teaching. Thanks for making this effort
Honestly I don’t know how people don’t learn from you, you literally made this the easiest method ever, I use to hate method of joints/sections but this is a great guide. Thank you
I know everyone on here is thanking you, but I must say how much I appreciate your videos. They are by far the most concise and helpful of statics on YT.
i like to thank you from the bottom of my heart for making this super simple to understand. being in this pandemic, online classes have been rough. ur vids gave me hope in succeeding in my first semester in degree. thank you so much and i hope you have a great life for helping us all.
This video has been so helpful. I really enjoyed the fact that you speed up the video during the time you were writing out the equations. Once again thank you.
Hey awesome thanks for writing Peter. Yeah I figured no one needs to sit and watch me write all that stuff out in real time =) =), glad you like the way I do it!
Question: when you were finding the sum of moments at A you assumed that anti clockwise moments were positive. Why the momements then expressed as negatives @4:31.
@@Engineer4Free could you please explain what ou mean? if you take moment about point f wouldnt the forces we want to find be acting in the point and not formed into an equation
Yeah, I was thinking the same. But finding out all the unknowns can be useful for checking the answer by summing all the moments about an arbitrary point and see if that equals to zero.
Hey. If you are unsure, of which to use, draw a right angle triangle overtop of the truss that includes the angle you're working with, and two of the sides along the members that make up that angle. From there, label the angle, opposite, adjacent, and hypotenuse sides. From there you can determine whether you need to use sin or cos. See this image: i.imgur.com/Dvdqf2t.jpg
When in doubt, draw a right angle triangle over top of the truss, using the angle in question, and both members that touch it as 2 of the 3 sides of the triangle. Then label on the opposite, adjacent, and hypotenuse sides, and from there you can determine if you need to use sin or cos to solve the unknown in question. See this image: i.imgur.com/Dvdqf2t.jpg
It's because Ax is can be determined to be equal to zero by inspection. There are no horizontally applied loads and the structure is simply supported, so Ax must = 0. Because its zero, I just don't bother writing it in the Fx expression. Hope that makes sense!
Hey, EG shouldn't be multiplied by 0.866 for any of these calculations. 0.866 is the perpendicular distance from A to the line of action of FH and and the x component of FG.
Can anyone explain why there is no EG in the equation at 4:05 ? is that because that is supposed to be EG * (0) ? I got exam tomorrow, thanks already :)
It's because the line of action of EG passes through A, so EG does not cause a moment about A. No need to write a term for a force in the moment about A eq if it's known to not be a part of it anyways 🙂
Yeah so in the first line of the sum of moments about A equation, notice that there are two terms that include FG. One has an FGsin60 and the other has an FGcos60. The FGsin60 is the y component of FG, and it's also multiplied to 2.5, which is the perpendicular distance from A to the line of action of the force FGsin60. The FGcos60 is the x component of FG, and it's also multiplied to 0.866, which is the perpendicular distance from A to the line of action of the force FGcos60. Notice that the line of action of the x component of FG (FGcos60) is a horizontal line, in line with the top of the truss, which is 0.866m above the bottom of the truss. Both x and y components of FG cause a moment about point A, so both need to be accounted for in that equation.
Yes. You would need to include it in the FBD of the whole structure, when calculating the reactions at A and B. It will cause A and B to be different than they are now. After that, proceed in the exact same way. You can make the same cut, and inspect the left side of the structure if you want, but you will get different answers, because Ay will be different than it is now.
@@Engineer4Free but with the problem I’m doing, it asks my to find EG but joint G has a downward force acting on it. But when I split the truss in half, it gets cut off. What do I do?
@@SHIZZLE71 That's ok, it's not a problem. Go to 1:40 where I take the FBD of the whole structure. Include your extra downward force, at G. So you'll have an extra term in the sum of y expression, and also the sum about A expression. You will fin that Ay will be bigger than if your downward force did not exist. Once you found Ay, then do the cut, and analyze the left side. The downward force at G will not be present, because it is on the left side of the cut. But the influence of that force is already accounted for. This is why Ay is bigger than if that force at G was not there. It's because Ay knows its there from the calcs at 1:40. When you solve for FG, FH, and EG, their magnitudes depend on Ay. And Ay depends on all of the external forces, including the one at G. Hope that helps.. If it's still not making sense, then I just recommend watching vides 42-51 here: engineer4free.com/statics After doing several examples you might start noticing the pattern that they all follow, especially that Ay and By depend on the location and magnitude of the externally applied forces.
I’m confused sir, on moment at A why did you use FG twice (horizontal and vertical) isn’t you just choose one or use one FG in he equation, and why not put EG on the equation?
FG is diagonal so has both x and y conponent. Both of their lines of action do not pass through point A, so they will cause a moment about A if their magnitude is not zero. That's why we use both. The line of action of EG passes through A so it cannot create a moment about A, so it just gets omitted from the equation.
it's wonderful, however, i'd like to ask about when the method of section is used, in terms of x direction calculation, why the force derived from pin end in the left doesn't be incorporated?(the example is at 3:08)
Thanks for the question! At 3:08, I write the horizontal force balance for the left hand side of the truss after it has been virtually cut in half. Ax (the horizontal component of the reaction force caused by the pin at A) was determined to be zero at 1:38 so I opted not to write it in the expression at 3:08. I could have included it at 3:08 and written 0 = EG + FH + FGcos60 + Ax, but that could immediately be reduced to 0 = EG + FH + FGcos60 because we know that Ax = 0. If you have already determined that a force is 0, then it's typically ok to drop it from other expressions that it is part of. If you or your prof prefers that you always write every variable in every expression to start, even if they are zero, then you should still do that as it's proof that you are aware of every term that is supposed to be in each equation. Hope that clears it up, let me know if you have any other questions!
If you're cutting the truss in half to completely separate one side from the other, then make sure to only pass your virtual cut through a maximum of 3 individual members. If you for some reason pass through 4 or more, you won't have enough equations to work with.
Excellent information but since all are pin joints we can directly take moment about F = 0 and get the force value of EG in single step after cutting the section
When finding the moments at A... when we're looking at the y component of FG acting at 2.5 metres, should it not be cos30? It's acting at 30 degrees to the y axis. Or am I misunderstanding? Thanks!
Hey, the equation for moment is M=Fd, where MN is the moment caused about the point, F is the magnitude of the force, and d is the perpendicular distance from the line of action of the force to the point. 2.5m is the perpendicular distance to the line of action of the y component of FG only. 0.866m is the perpendicular distance to the line of action of the x component of the force. Neither 2.5m or 0.866m are the perpendicular distances to the line of action of the force FG. It’s longer to try to find out what that distance is, than it is to just separate it into x and y components and solve by parts.
If everytime you have negative value and plug it in to another equation, should I plug it with remaining negative sign or can I change to positive sign? as long as the final value would tell you if its in tension or compression? I'm confused, please enlighten me. Thanks
In this example, you have FH = -3kN then plugged in the Sum. Fy equation to get have a value of EG = +2.885kN. What If I plug it with +3kN since I know that FH is in tension and changed the graph. The value is different I guess.
@@shariffnurrmohammad1314 Hey yeah. So just to clarify, FH is actually in compression. I drew it initially in tension, and found it to have a negative magnitude, which means it's true sense is compression, not tension. Once you know that a member is truly in compression you can do two things. Method #1) Do not update the arrow on any FBD (ie, keep it drawn in the tension direction) and then any time you analyze a FBD that involves that force, use it's direction to determine whether you add or subtract in the sum of force equation, but then substitute a negative value for the letter. That's what I did in this case. Calculated FH to be 3kN Compression in the moment eq, and then in the sum of force in x equation, I left the arrow for FH as if it were in tension, and because FH points to the right on this FBD, I give the FH a + sign in front because it points in the positive x direction. But then irrelevant to that + sign, I substitute "FH" for "-3kN" and "+FH" becomes "+(-3kN)" which is equivalent to "-3kN". Method #2) Update the arrow on all FBD's that it interacts with to show that it "pushes" on the joint rather than "pulling" and assign it a positive value. So if we did this, then we'd have to update FH at joint F to push on it (to the left) rather than pull (to the right). When we take the sum of force in x direction, then we'd give the FH a - sign in front because it points in the negative x direction. So we would have a "-FH" already in the expression, and then we would substitute "FH" for "(positive)3kN" and "-FH" would become "-(3kN)" which is equivalent to "-3kN". Hope that helps to clarify. We in a way have two layers of +/- rules going on here. One is that when solving for the unknown internal force of a member assumed to be in tension, that + means it is in tension and - means that it is in compression, and the other sort of unrelated thing when we are taking sum of force in x or y direction, that + refers to whether a force in a FBD is pointing toward the positive direction, and - refers to a force that is pointing toward the negative direction.
If you draw the forces going out of the joint and get a positive number, that means that the force is "pulling on the joint", making the joint in tension. But this means that the actual member is in compression right???
Ahh not quite. You got the first part right. If a member is pulling on a joint, then that member is in tension. Think about rope. Ropes can only be in tension. If you have a rope with a joint attached at each end, and that rope has been tightened somehow such that it is pulling on each joint, then the rope is in tension. The rope that pulls on support reactions is not in compression, and nether would a steel or wood member that is similarly "pulling" on each joint. If you're still not convinced, switch the perspective that you observing from. We're talking about the member pulling or pushing on the joint; so this what's going on from the joint's perspective. Try thinking from the member's perspective. If a member in static equilibrium is pulling on a joint, then the joint can be thought of as pulling on the member, just in the opposite direction. So if you're a two force truss member, and a joint is pulling on each side of you, then you're getting stretched out. Getting stretched out mean's that you're in tension. Conversely, if you're a two force truss member and your pushing on each joint, then those joints are pushing back on you. You're getting pushed on both ends. You're getting squished. Getting squished means that you're in compression. Hope that helps!
There are no externally applied forces with any horizontal component. Ax is the only reaction force that can withstand horizontal loads, so the sum of forces in x direction equations is just Ax = 0. I skipped it, because it's almost always zero in these problems. So you could label it on, but it equals zero anyways and has no affect.
Either works. If doing method of sections, make the virtual cut through members JH, JI, JK, and if doing method of joints, find the reactoins, then solve joint k first. Next you would solve for joint J and would find the answer.
The angle might have been different on your test, or you may have needed to use cos instead because of the certain context. You have to take care when constructing the right angle triangle that you will use for your trig functions, here is an example of how to do one,: imgur.com/a/YCT6I1a sin is usually used for the vertical component, and cos is usually used for the horizontal component, but not always, it depends how you draw the right angle triangle
Instead of equating forces, can we solve the entire question by using moment? Like if we first take moment about F, then we will be able to find F(EG) as the moment of the other two unknown forces will be zero and then we can take moment about B to find F(FG), and then solve for the remaining unknown force.
when you do the moment equation- why do you include the moment of FH but not EG? they're both purely horizontal forces so I don't understand why one is included and not the other
Hey I just have a few friction problems here: engineer4free.com/statics that are at the basic level of statics. I'll be getting around to friction problems in dynamics soon, and eventually may cover angle of repose if I ever do a serious on geotechnical engineering. Thanks for the suggestions =)
Because the line of action of EG passes through A. a force can only cause a moment about a point if it's line of action does not pass through that point
Not necessarily, but it's recommended that you draw unknowns in tension. If you do that, positive magnitude means it's in tension, and a negative magnitude means it's in compression. That's the convention, and is less confusing if everyone does it the same way ✌️
Hi I just wanna say THANK YOU SO MUCH FOR THIS. I have been actually struggling with statics because i dont know what i should do first like should i get the summation of horizontal forces first? Or should i do a moment on a member likeee i was so confused but with this technique just writing it down first then just substituting it later really makes sense to me. My finals is coming up in 2 days I guess i still have time to review from sections to space trusses
If you are talking about when he is doing the moments about A (time 3:30 ) then the answer is cause EG is pulling straight in the X direction with both Ax and EG having the same Y coordinates. If you are talking about the horizontal forces summing, he does have EG in there. If you are talking about the vertical forces summing, it is not in there because EG is *not* a vertical force.
Because there are no horizontally applied forces on this structure. So if you wrote the expression for sum of forces in x direction, you would just have Ax = 0. Ax is the only thing that can apply a horizontal load to the overall structure (ie, not an internal force), so it just equals zeor. Because of that, I opted not to write the expression. If you're in a test, it's worth writing, but after making so many of these videos I started leaving it out. The rest of the playlist is here btw: engineer4free.com/statics =)
Hey Dinh. It's because those are the perpendicular distances from point A to the lines of action of FGy and FGx. FGy's line if action is vertical, passing through point F. That line is two and a half members away from A horizontally (being AC, CE, and half of EG} - draw a vertical line passing through F, and you'll see it I tersects member EG in the middle. That's where the 2.5m comes from, cuz they're all 1m. The line of action of FGx is horizontal, passing through F. It's in line with the OOP part of the truss. The top part of the truss is 0. 866m above the bottom part of the truss. You can find this because you know that member AB is Angles at 60 degrees (part of an equilateral triangle). So 1m * sin60 =0.866m from basic trig. Hope that helps clarify it all. Thanks for the good question! 😊
The convention is to draw unknown internal forces in tension. If you do that, then a positive value indicates it is in fact in tension, and a negative value means that the direction must be switched so it is actuakly in compression. If you're drawing unknowns in compression though, then the logic is flipped. I recommend following the standard convention as I have done in this video!
@@Engineer4Free I had this same question. when you made the cut, you drew arrows for the members and the members are shown to be acting in compression (pulling away from the joints). Also when you plugged it into the equations, you kept the same convention. Positive answers confirm your convention of the member being in compression. What do you think?
All members are the same length, so it's an equilateral triangle. All angles of an equilateral triangle are 60° (180/3). That is something you may be expected to realize on an exam so watch for it!
Yes but you have some typos. The sum of moments about A can be written as Ma= -2kN(1m) - 3kN(4m) + ky(5m). Your measurements to the 3kN force and ky were the typos, as well as the sign of ky (it’s pointing up so is with the method of sections we don’t want to take the sum of moments about A fro the whole structure, in the later part of the problem. You want to take a virtual cut and the. Do the sum of moments including the internal forces of the members that were cut. You should watch this video to understand the method: www.engineer4free.com/4/truss-analysis-by-method-of-sections-explained also watching videos 42 - 51 here www.engineer4free.com/statics.html would help in general to understand trusses better
@@Engineer4Free lol i know now why im confuse sir its because i thought from point C to point I it measured 4m hahaha its from point A to I measured 4m thanks for the reply sir it was a stupid question hahaha ur video is very helpful sir thanks
we can also take the moment at G (that eliminate 2 unknown forces FG & EG). This will take less work: ∑Mg=-FG(.866)-2.2(3)+2(2)=0 FG=(2.2(3)-2(2))/.866=-3.00KN
Thanks for the contribution Toan! It's a good sign if you are recognizing other ways to solve problems, shows that you are really understanding it. Thanks for contributing the discussion and showing others!
It is the sin of 60. The diagonal member is on a 60 degree angle because it is an equilateral triangle. 0.866 is the y component of that member, aka the height of the triangle, because each member is 1m long
Good video, one question though, when calculating the suction part , fx fy and moment at a, I see you are only including external force and the Ay, and the cutted members "FH, FG and EG" , my question though, why did you not include all the other previous members, for example CD, and DE, and BC, and for that matter also their y and x components and also AC and CE, there are more members in the cutted section that you choose, I know there's a reason for it, and I'm thinking that these members cancel each other out, but my question is, how a nub like me would know this, how to spot it, you havent mentioned it in the video thanks and sorry for the long question.
it's a method of sections problem, which is useful when you only need to find some of the forces, not all of them. This problem specifically only asks for FH FG & EG.
@@user-cj2qi7im8c Wow brother I already graduated and I am en Engineer now but thanks for the info. I needed this 4 yrs ago for an exam haha. Good news I got an A :)
@@user-cj2qi7im8c Wow nice!! goodluck...Ive been in your shoes 4 yrs ago and yes I watched every single video plus lots of tutorials/pdf files, so really goodluck brother I hope you get an A.
Hiee...in the last step as u have taken sum of moments about point A and calculated force in FH. For the same step I've taken the sum of moments about point E and calculated the force in FH and the answer was 3.23KN compression which is different from yours...pls correct me if I've calculated wrong..
No matter where you decide to take the moment, the internal force in each member will be the same. Did you take the perpendicular distance to the line of action of the vertical component of FG to be 0.5? If not that will cause errors for everything else.
couldn't we take the moment of F to find EG? and then do the sigma of X? (notice: i am not saying you are wrong i was just wandering if my suggestion is correct or no) ty.
Hey sorry I am not understanding your question. When we virtually cut the bridge in half, there are 3 unknowns. We need 3 equations to be able to solve them, so we must take sum of forces in x direction, sum of forces in y direction, and sum of moments about some point. Picking A as our point to take moments about is just a convenient place as it will eliminate Ay from the calculation, and the distances to the lines of action of the x and y components of each unknown force is easy to measure. Does that clear it up?
SO what I think he means, and also what I am wondering, for this question in particular, is that it seems like if you take the moment about F after you've split the truss into sections, you don't need to analyze and find FG and FH to find EG.
Oh cool yeah that is definitely a quick way to solve for just EG. Thanks for taking the time to notice and then check that it works! That would be a clever/sneaky shortcut on a statics exam if they only asked for the force in the one member, you could potentially save a ton of time by not bothering to solve for the other two unknowns from the virtual cut.. cool!
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wow you literally taught me what it took my teacher 3 hours to confuse me about lololol
😊😊 You should check out engineer4free.com/statics for all of the ither videos I did too!
Im lucky that my teachers are good. I just forgot this and cbf loggin in to find the tute. Vids good
Literally same here
This is not a man, this is a god of engineering tutorials! Thank you so much man, you have no idea how many lives you have just saved just by doing these kinds of videos. Keep it up man!
🙌🙌 Thanks Korajark!! If you haven't already, do check out the whole collection of statics vids I made over at engineer4free.com/statics and tell some friends!! =)
Fantastic Video. It took the professor 3 weeks to confuse me, and it took you 5 mins to explain it properly. Thank you.
Glad it helped! Check out the rest of the playlist here: engineer4free.com/statics =)
I'm sincerely grateful for you giving literally a free lecture on RUclips..! Thank you so much. It helped me not only to solve questions in assignments but also to understand the entire truss analysis..!
Glad I can help!! There are more vids here: engineer4free.com/statics =)
GOOD effort
I often see your videos to learn structure analysis. I have learnt much from ur videos. Simple and effective way of teaching.
Thanks for making this effort
Hey sorry this comment wen't without a reply for so long, somehow it slipped through. Thanks a lot for your feedback!! :D
Honestly I don’t know how people don’t learn from you, you literally made this the easiest method ever, I use to hate method of joints/sections but this is a great guide. Thank you
Thanks for the feedback Kammy, glad the videos are working =) =)
I know everyone on here is thanking you, but I must say how much I appreciate your videos. They are by far the most concise and helpful of statics on YT.
Thanks Johan, It means a lot to me just to get a simple thank you but I always really appreciate a comment with genuine substance :)
i like to thank you from the bottom of my heart for making this super simple to understand. being in this pandemic, online classes have been rough. ur vids gave me hope in succeeding in my first semester in degree. thank you so much and i hope you have a great life for helping us all.
Thanks for the nice comment, and glad I can give you hope. please share the vids with other students that could use them too
thanks hommie, helped a lot boss, passed my external engineering exam because of this video, legend.
How can I express my gratitude, such a brilliant lecture . To the point, short, excellent.
This video has been so helpful. I really enjoyed the fact that you speed up the video during the time you were writing out the equations. Once again thank you.
Hey awesome thanks for writing Peter. Yeah I figured no one needs to sit and watch me write all that stuff out in real time =) =), glad you like the way I do it!
Damn man you simplified a lot of work for me. Thanks. We owe you a lot
Glad to hear it bro. Just tell some classmate that could use the help too :)
Moment about F would be more practical for timed assessments, but this explanation is a 10/10
OMAIGOD, what kinda magic did you use that make me understand just with 5 mins vid instead of 2 hours lectures
Moment about F would also give EG in one step. Great video
You're an unbelievable teacher. Thank you for your work.
You are very welcome!!
Question: when you were finding the sum of moments at A you assumed that anti clockwise moments were positive. Why the momements then expressed as negatives @4:31.
I'm studying this before finals and these videos are very helpful 🌸
Keep it up and good luck!!!
Thank you very much! With love from Sri-Lanka!
You're very welcome!! Incase you haven't seen the full playlist yet, it's here: engineer4free.com/statics plz share with some other students 🙏
@@Engineer4Free Definetely !!!!!!!
If you had taken moment about Point F, it would have been one step to solve the problem. Nonetheless, very well explained
Yeah true. If you can see that, then you're doing alright =) =)
@@Engineer4Free could you please explain what ou mean? if you take moment about point f wouldnt the forces we want to find be acting in the point and not formed into an equation
@@nahilismail3144 the force EG isn't, its perpendicular distance from point F is 0.866m- so you can calculate EG and sub it into Fx equation 🙃
@@inyoungbaek6426 thank you so much for helping a stranger you know nothing about. i hope you have a good day!
Yeah, I was thinking the same. But finding out all the unknowns can be useful for checking the answer by summing all the moments about an arbitrary point and see if that equals to zero.
your videos are as great as a university itself and your website is way coole , I wish I could know about your website sooner ... you Got my respect
Thanks a lot! I hope you get lots of use out of it, and feel free to tell some friends so they can benefit too =)
Getting ready to take the PE, this was a great help to try and remember schooling from 4 years ago
at 3:18 ik trignometry but can please explain why you use sine or cosine? it's difficult to see when to use each...
Hey. If you are unsure, of which to use, draw a right angle triangle overtop of the truss that includes the angle you're working with, and two of the sides along the members that make up that angle. From there, label the angle, opposite, adjacent, and hypotenuse sides. From there you can determine whether you need to use sin or cos. See this image: i.imgur.com/Dvdqf2t.jpg
When in doubt, draw a right angle triangle over top of the truss, using the angle in question, and both members that touch it as 2 of the 3 sides of the triangle. Then label on the opposite, adjacent, and hypotenuse sides, and from there you can determine if you need to use sin or cos to solve the unknown in question. See this image: i.imgur.com/Dvdqf2t.jpg
i finally understood how to use method of sections properly. thank you
Glad it helped!!! There are some more examples in videos 42-51 here: engineer4free.com/statics
Thank you very helpful I didn't understand the cutting part but seeing the cut members turn into forces was a lightbulb moment
Thank you! I was stuck on a problem set up exactly like this for hours!
Glad I could help! Full playlist is here: engineer4free.com/statics including lots on trusses
One can on yearn to solve problems as fast as this man.
I couldn't understand this method until I saw this video, never realized it would be so simple :p
Glad to hear I could help!!! It doesn't need to be complicated =) =)
@@Engineer4Free didn't even extend it till 10:01 for midroll ads, and for that I am eternally grateful
How did you know It was Cos(60º) at 3:11
By drawing. right angle triangle that includes FG, FGx, and FGy and applying SOH CAH TOA. All angles are 60 degrees. See this: i.imgur.com/dTR8W7X.jpg
at 3:24, why did you put FGsin60 as a negative?
The production quality on this video is through the roof
Thanks Bryce
thank you so much you're a legend im in ISE but I still have to take statics so you are saving me big time
Thanks Nick, glad I can help.. you can make it through Statics!!! 🤜🤛
Thanks Nick, glad I can help.. you can make it through Statics!!! 🤜🤛
When finding FH 4:01 why are you measuring 2.5 shouldn’t it be just 2 cuz that’s the member where the force is applied?
Thank you!! I was so confused, but this video made it so easy!
Simple, precise, and great!
3.25 how come you are including Ay of 2.2 kN in the Fy expression, but you’re not including Ax in the Fx expression?
It's because Ax is can be determined to be equal to zero by inspection. There are no horizontally applied loads and the structure is simply supported, so Ax must = 0. Because its zero, I just don't bother writing it in the Fx expression. Hope that makes sense!
Sir...in 4:00 I think u left EG × 0866?? Is it
Hey, EG shouldn't be multiplied by 0.866 for any of these calculations. 0.866 is the perpendicular distance from A to the line of action of FH and and the x component of FG.
@@Engineer4Free ok..thanks
Mannn may God bless you🥺thank you very much🌼
You Sir are a legend
Thanks Munashe =)
Can anyone explain why there is no EG in the equation at 4:05 ?
is that because that is supposed to be EG * (0) ?
I got exam tomorrow, thanks already :)
It's because the line of action of EG passes through A, so EG does not cause a moment about A. No need to write a term for a force in the moment about A eq if it's known to not be a part of it anyways 🙂
@@Engineer4Free Ah I see, thank you so much!
You really helped me out :)
Hey how is the FG cos 60 (0.866) shouldn’t be half of 0.866? I was just wondering this since FG sin 60 is (2.5) because the 1m was cut in half?
Yeah so in the first line of the sum of moments about A equation, notice that there are two terms that include FG. One has an FGsin60 and the other has an FGcos60. The FGsin60 is the y component of FG, and it's also multiplied to 2.5, which is the perpendicular distance from A to the line of action of the force FGsin60. The FGcos60 is the x component of FG, and it's also multiplied to 0.866, which is the perpendicular distance from A to the line of action of the force FGcos60. Notice that the line of action of the x component of FG (FGcos60) is a horizontal line, in line with the top of the truss, which is 0.866m above the bottom of the truss. Both x and y components of FG cause a moment about point A, so both need to be accounted for in that equation.
bro you saved my life thx !!!!!!!!!!!!!!!!
Glad I could help!! Check out all the statics vids I did over at engineer4free.com/statics when you have a minute ✌️
would u have to do anything extra if point G had a force pulling downwards from it?
Yes. You would need to include it in the FBD of the whole structure, when calculating the reactions at A and B. It will cause A and B to be different than they are now. After that, proceed in the exact same way. You can make the same cut, and inspect the left side of the structure if you want, but you will get different answers, because Ay will be different than it is now.
@@Engineer4Free but with the problem I’m doing, it asks my to find EG but joint G has a downward force acting on it. But when I split the truss in half, it gets cut off. What do I do?
@@SHIZZLE71 That's ok, it's not a problem. Go to 1:40 where I take the FBD of the whole structure. Include your extra downward force, at G. So you'll have an extra term in the sum of y expression, and also the sum about A expression. You will fin that Ay will be bigger than if your downward force did not exist. Once you found Ay, then do the cut, and analyze the left side. The downward force at G will not be present, because it is on the left side of the cut. But the influence of that force is already accounted for. This is why Ay is bigger than if that force at G was not there. It's because Ay knows its there from the calcs at 1:40. When you solve for FG, FH, and EG, their magnitudes depend on Ay. And Ay depends on all of the external forces, including the one at G. Hope that helps.. If it's still not making sense, then I just recommend watching vides 42-51 here: engineer4free.com/statics After doing several examples you might start noticing the pattern that they all follow, especially that Ay and By depend on the location and magnitude of the externally applied forces.
I’m confused sir, on moment at A why did you use FG twice (horizontal and vertical) isn’t you just choose one or use one FG in he equation, and why not put EG on the equation?
FG is diagonal so has both x and y conponent. Both of their lines of action do not pass through point A, so they will cause a moment about A if their magnitude is not zero. That's why we use both. The line of action of EG passes through A so it cannot create a moment about A, so it just gets omitted from the equation.
This is great as compared to the confusion after 2hrs of mechanics+2 hours wasted of my life during online lectures of mechanics
Glad I'm able to help, you can see all of my statics videos here: engineer4free.com/statics there are quite a few truss videos!
it's wonderful, however, i'd like to ask about when the method of section is used, in terms of x direction calculation, why the force derived from pin end in the left doesn't be incorporated?(the example is at 3:08)
Thanks for the question! At 3:08, I write the horizontal force balance for the left hand side of the truss after it has been virtually cut in half. Ax (the horizontal component of the reaction force caused by the pin at A) was determined to be zero at 1:38 so I opted not to write it in the expression at 3:08. I could have included it at 3:08 and written 0 = EG + FH + FGcos60 + Ax, but that could immediately be reduced to 0 = EG + FH + FGcos60 because we know that Ax = 0. If you have already determined that a force is 0, then it's typically ok to drop it from other expressions that it is part of. If you or your prof prefers that you always write every variable in every expression to start, even if they are zero, then you should still do that as it's proof that you are aware of every term that is supposed to be in each equation. Hope that clears it up, let me know if you have any other questions!
Much detailed explanation, it really make sense. Thanx a lot with a great expectation for your more video.
can you cut them multiple times?
If you're cutting the truss in half to completely separate one side from the other, then make sure to only pass your virtual cut through a maximum of 3 individual members. If you for some reason pass through 4 or more, you won't have enough equations to work with.
tnx for your videos 😁
Excellent information but since all are pin joints we can directly take moment about F = 0 and get the force value of EG in single step after cutting the section
When finding the moments at A... when we're looking at the y component of FG acting at 2.5 metres, should it not be cos30? It's acting at 30 degrees to the y axis. Or am I misunderstanding? Thanks!
Hey, the equation for moment is M=Fd, where MN is the moment caused about the point, F is the magnitude of the force, and d is the perpendicular distance from the line of action of the force to the point. 2.5m is the perpendicular distance to the line of action of the y component of FG only. 0.866m is the perpendicular distance to the line of action of the x component of the force. Neither 2.5m or 0.866m are the perpendicular distances to the line of action of the force FG. It’s longer to try to find out what that distance is, than it is to just separate it into x and y components and solve by parts.
Can you just take the moment about point F to get EG?
If everytime you have negative value and plug it in to another equation, should I plug it with remaining negative sign or can I change to positive sign? as long as the final value would tell you if its in tension or compression? I'm confused, please enlighten me. Thanks
In this example, you have FH = -3kN then plugged in the Sum. Fy equation to get have a value of EG = +2.885kN. What If I plug it with +3kN since I know that FH is in tension and changed the graph. The value is different I guess.
@@shariffnurrmohammad1314 Hey yeah. So just to clarify, FH is actually in compression. I drew it initially in tension, and found it to have a negative magnitude, which means it's true sense is compression, not tension. Once you know that a member is truly in compression you can do two things.
Method #1) Do not update the arrow on any FBD (ie, keep it drawn in the tension direction) and then any time you analyze a FBD that involves that force, use it's direction to determine whether you add or subtract in the sum of force equation, but then substitute a negative value for the letter. That's what I did in this case. Calculated FH to be 3kN Compression in the moment eq, and then in the sum of force in x equation, I left the arrow for FH as if it were in tension, and because FH points to the right on this FBD, I give the FH a + sign in front because it points in the positive x direction. But then irrelevant to that + sign, I substitute "FH" for "-3kN" and "+FH" becomes "+(-3kN)" which is equivalent to "-3kN".
Method #2) Update the arrow on all FBD's that it interacts with to show that it "pushes" on the joint rather than "pulling" and assign it a positive value. So if we did this, then we'd have to update FH at joint F to push on it (to the left) rather than pull (to the right). When we take the sum of force in x direction, then we'd give the FH a - sign in front because it points in the negative x direction. So we would have a "-FH" already in the expression, and then we would substitute "FH" for "(positive)3kN" and "-FH" would become "-(3kN)" which is equivalent to "-3kN".
Hope that helps to clarify. We in a way have two layers of +/- rules going on here. One is that when solving for the unknown internal force of a member assumed to be in tension, that + means it is in tension and - means that it is in compression, and the other sort of unrelated thing when we are taking sum of force in x or y direction, that + refers to whether a force in a FBD is pointing toward the positive direction, and - refers to a force that is pointing toward the negative direction.
You sir.... are a badass. Thank you!
Thanks dowg.
If you draw the forces going out of the joint and get a positive number, that means that the force is "pulling on the joint", making the joint in tension. But this means that the actual member is in compression right???
Ahh not quite. You got the first part right. If a member is pulling on a joint, then that member is in tension. Think about rope. Ropes can only be in tension. If you have a rope with a joint attached at each end, and that rope has been tightened somehow such that it is pulling on each joint, then the rope is in tension. The rope that pulls on support reactions is not in compression, and nether would a steel or wood member that is similarly "pulling" on each joint. If you're still not convinced, switch the perspective that you observing from. We're talking about the member pulling or pushing on the joint; so this what's going on from the joint's perspective. Try thinking from the member's perspective. If a member in static equilibrium is pulling on a joint, then the joint can be thought of as pulling on the member, just in the opposite direction. So if you're a two force truss member, and a joint is pulling on each side of you, then you're getting stretched out. Getting stretched out mean's that you're in tension. Conversely, if you're a two force truss member and your pushing on each joint, then those joints are pushing back on you. You're getting pushed on both ends. You're getting squished. Getting squished means that you're in compression. Hope that helps!
FG is in compression not tension, right? Because the arrows down are positive. Thus, in compression
Why isn't EG calculated in the Sum of the Moments? Sorry I have jumped right into the middle of my engineering course and need to catch up
When doing the sun of forces in the X for the section why did you exclude Ax?
Sum of forces for the entire structure shows us that Ax = 0, so I just don't bother writing it
how do you find the forces in members AB and AC?
method of joints
Yes, method of joints. You can find tutorials for it in the trusses section of engineer4free.com/statics 🤙
What happened to Ax at the sum of x of the left section
There are no externally applied forces with any horizontal component. Ax is the only reaction force that can withstand horizontal loads, so the sum of forces in x direction equations is just Ax = 0. I skipped it, because it's almost always zero in these problems. So you could label it on, but it equals zero anyways and has no affect.
if I need to calculate forces on JH then will a method of section works or we need to use the method of joints?
Either works. If doing method of sections, make the virtual cut through members JH, JI, JK, and if doing method of joints, find the reactoins, then solve joint k first. Next you would solve for joint J and would find the answer.
Why did you take FGsin60? I used the same method on my test it happened to be wrong.
The angle might have been different on your test, or you may have needed to use cos instead because of the certain context. You have to take care when constructing the right angle triangle that you will use for your trig functions, here is an example of how to do one,: imgur.com/a/YCT6I1a sin is usually used for the vertical component, and cos is usually used for the horizontal component, but not always, it depends how you draw the right angle triangle
Instead of equating forces, can we solve the entire question by using moment?
Like if we first take moment about F, then we will be able to find F(EG) as the moment of the other two unknown forces will be zero and then we can take moment about B to find F(FG), and then solve for the remaining unknown force.
when you do the moment equation- why do you include the moment of FH but not EG? they're both purely horizontal forces so I don't understand why one is included and not the other
Hi, could you please make some videos explaining how to tackle engineering friction problems and the angle of repose please..
Hey I just have a few friction problems here: engineer4free.com/statics that are at the basic level of statics. I'll be getting around to friction problems in dynamics soon, and eventually may cover angle of repose if I ever do a serious on geotechnical engineering. Thanks for the suggestions =)
Great refresher. Thanks!
Glad it was helpful!! =)
why do we ignore eg when doing the moment about a
Because the line of action of EG passes through A. a force can only cause a moment about a point if it's line of action does not pass through that point
Does the direction of the forces matter when drawing the FBD ?
Not necessarily, but it's recommended that you draw unknowns in tension. If you do that, positive magnitude means it's in tension, and a negative magnitude means it's in compression. That's the convention, and is less confusing if everyone does it the same way ✌️
@@Engineer4Free oh really? I've been stressing about the direction for hours! Thank you!😩
Is there a reason as to why you made the cut from FH FG and EG? Why didn’t the cut go from FH FG and FE? Maybe it’s obvious but I’m just not seeing it
This man is good at what he’s doing..... reminds me of Khan
Thanks cielcloud!! 🙌 You should also check out the rest of the playlist here: ruclips.net/p/PLOAuB8dR35oeXMk2C5fjHP2K306hGfk_w
@@Engineer4Free I already have your website bookmarked :D
Hi I just wanna say THANK YOU SO MUCH FOR THIS. I have been actually struggling with statics because i dont know what i should do first like should i get the summation of horizontal forces first? Or should i do a moment on a member likeee i was so confused but with this technique just writing it down first then just substituting it later really makes sense to me. My finals is coming up in 2 days I guess i still have time to review from sections to space trusses
Hope you're exam went well!!! 🙌🙌
same
Hello sir...wat terms do u use to justify that forces are all the same in a continuous truss?
I don't believe I said that. Can you clarify what you mean?
can someone answer my question ? why did he not include the EG in summing the forces ? thank you
If you are talking about when he is doing the moments about A (time 3:30 ) then the answer is cause EG is pulling straight in the X direction with both Ax and EG having the same Y coordinates.
If you are talking about the horizontal forces summing, he does have EG in there.
If you are talking about the vertical forces summing, it is not in there because EG is *not* a vertical force.
@@joshman9757 Yup thanks dude. The line of action of EG passes through point A, so it doesn't cause a moment about point A =) =)
please why did you leave out Ax when you were finding the sum of X's
As in forces in the x direction
Because there are no horizontally applied forces on this structure. So if you wrote the expression for sum of forces in x direction, you would just have Ax = 0. Ax is the only thing that can apply a horizontal load to the overall structure (ie, not an internal force), so it just equals zeor. Because of that, I opted not to write the expression. If you're in a test, it's worth writing, but after making so many of these videos I started leaving it out. The rest of the playlist is here btw: engineer4free.com/statics =)
you did a good job explaining this, cheers :D
hey great vid I just wanted to ask why does FGsin60 multiplies by 2.5 and the FGcos60 multiplies by 0.866
Hey Dinh. It's because those are the perpendicular distances from point A to the lines of action of FGy and FGx. FGy's line if action is vertical, passing through point F. That line is two and a half members away from A horizontally (being AC, CE, and half of EG} - draw a vertical line passing through F, and you'll see it I tersects member EG in the middle. That's where the 2.5m comes from, cuz they're all 1m. The line of action of FGx is horizontal, passing through F. It's in line with the OOP part of the truss. The top part of the truss is 0. 866m above the bottom part of the truss. You can find this because you know that member AB is Angles at 60 degrees (part of an equilateral triangle). So 1m * sin60 =0.866m from basic trig. Hope that helps clarify it all. Thanks for the good question! 😊
@@Engineer4Free thanks, i was curious about that too.
@@2a1n71 glad I could help!
what is the onscreen whiteboard software used to make this tutorial?
You can find all of the hardware and software that I use here: engineer4free.com/tools
i thought if the ans is in negative its tensile? and if the answer is positive its compressive? is it the opposite?
The convention is to draw unknown internal forces in tension. If you do that, then a positive value indicates it is in fact in tension, and a negative value means that the direction must be switched so it is actuakly in compression. If you're drawing unknowns in compression though, then the logic is flipped. I recommend following the standard convention as I have done in this video!
@@Engineer4Free I had this same question. when you made the cut, you drew arrows for the members and the members are shown to be acting in compression (pulling away from the joints). Also when you plugged it into the equations, you kept the same convention. Positive answers confirm your convention of the member being in compression. What do you think?
how did you get the 60 degrees?
All members are the same length, so it's an equilateral triangle. All angles of an equilateral triangle are 60° (180/3). That is something you may be expected to realize on an exam so watch for it!
Thanks 😊...can you make more videos please
Simple and efficient👌🏻
Thanks Muhammad!!!!
where did you get the 5m for ky in the moment?
It’s the horizontal distance between A and K
how can we use method of section if their are 3 unknowns in the center that forms a right triangle?
Useful and short 👍
Glad you liked it =)
To all the men and women I had to refer to when I was having trouble understanding anything because of covid-19, I salute you
Thanks Abdullah, glad I can help ✌️
help sir im just really confuse, in summation of moments at A isn't that -2kn(1)-3kn(5)-ky(6)=0?
Yes but you have some typos. The sum of moments about A can be written as Ma= -2kN(1m) - 3kN(4m) + ky(5m). Your measurements to the 3kN force and ky were the typos, as well as the sign of ky (it’s pointing up so is with the method of sections we don’t want to take the sum of moments about A fro the whole structure, in the later part of the problem. You want to take a virtual cut and the. Do the sum of moments including the internal forces of the members that were cut. You should watch this video to understand the method: www.engineer4free.com/4/truss-analysis-by-method-of-sections-explained also watching videos 42 - 51 here www.engineer4free.com/statics.html would help in general to understand trusses better
@@Engineer4Free lol i know now why im confuse sir its because i thought from point C to point I it measured 4m hahaha its from point A to I measured 4m thanks for the reply sir it was a stupid question hahaha ur video is very helpful sir thanks
we can also take the moment at G (that eliminate 2 unknown forces FG & EG).
This will take less work:
∑Mg=-FG(.866)-2.2(3)+2(2)=0
FG=(2.2(3)-2(2))/.866=-3.00KN
Thanks for the contribution Toan! It's a good sign if you are recognizing other ways to solve problems, shows that you are really understanding it. Thanks for contributing the discussion and showing others!
Good I got the same answer!!
Great video! Thank you so much.
You're welcome, thanks for watching! Check out the rest of my statics videos here if you haven't already: engineer4free.com/statics =)
How do you find 'FGcos60' when taking the sum of the moment at A?
from constructing a right angle triangle that is the same basically as this: i.imgur.com/Dvdqf2t.jpg
wow... i now understand the concepts. i still have more questions though, how can i contact you.
Great! I recommend checking out the rest of the tutorials here: engineer4free.com/statics
ffs lmao best guide out here appreciate the help fam
hah thanks!
How did you get the height of 0.866 ?
It is the sin of 60. The diagonal member is on a 60 degree angle because it is an equilateral triangle. 0.866 is the y component of that member, aka the height of the triangle, because each member is 1m long
much love, very clear
Thanks Dmitry!!
Good video, one question though, when calculating the suction part , fx fy and moment at a, I see you are only including external force and the Ay, and the cutted members "FH, FG and EG" , my question though, why did you not include all the other previous members, for example CD, and DE, and BC, and for that matter also their y and x components and also AC and CE, there are more members in the cutted section that you choose, I know there's a reason for it, and I'm thinking that these members cancel each other out, but my question is, how a nub like me would know this, how to spot it, you havent mentioned it in the video
thanks and sorry for the long question.
it's a method of sections problem, which is useful when you only need to find some of the forces, not all of them. This problem specifically only asks for FH FG & EG.
@@user-cj2qi7im8c Wow brother I already graduated and I am en Engineer now but thanks for the info. I needed this 4 yrs ago for an exam haha. Good news I got an A :)
@@xanh350 sheesh I love to hear that! Congrats! 🎉 I have just one year left before I graduate so I’m watching all these videos 🤣
@@user-cj2qi7im8c Wow nice!! goodluck...Ive been in your shoes 4 yrs ago and yes I watched every single video plus lots of tutorials/pdf files, so really goodluck brother I hope you get an A.
@@xanh350 thanks brother I’ll do my best 🙌
Thank you! My professor skipped this lesson but I still have a quiz on it for some reason so you just saved my ass lol
Typical haha. Lemme know how it goes!
I got a completely different answer solving from the RHS? I got Rk=+2.8kN
I got EG from taking moments about F to be 2.87kN
Ok so you are all good? Can't tell if the text after the edit is still relevant 🤔
thank you! and you could have used the moment about point F and get the same result:
2.2x2.5-2x1.5-EGx0.866=0
EG=2.9
Wouldn't the height of the truss be 0.43m?
0.5sin60 = 0.43
ah no, it should be 1m*sin60, as each member is 1m long =)
Hiee...in the last step as u have taken sum of moments about point A and calculated force in FH. For the same step I've taken the sum of moments about point E and calculated the force in FH and the answer was 3.23KN compression which is different from yours...pls correct me if I've calculated wrong..
rajeev raj44 the moments about different points changes..
No matter where you decide to take the moment, the internal force in each member will be the same. Did you take the perpendicular distance to the line of action of the vertical component of FG to be 0.5? If not that will cause errors for everything else.
You invited to my graduation 🔥
couldn't we take the moment of F to find EG? and then do the sigma of X? (notice: i am not saying you are wrong i was just wandering if my suggestion is correct or no) ty.
Hey sorry I am not understanding your question. When we virtually cut the bridge in half, there are 3 unknowns. We need 3 equations to be able to solve them, so we must take sum of forces in x direction, sum of forces in y direction, and sum of moments about some point. Picking A as our point to take moments about is just a convenient place as it will eliminate Ay from the calculation, and the distances to the lines of action of the x and y components of each unknown force is easy to measure. Does that clear it up?
SO what I think he means, and also what I am wondering, for this question in particular, is that it seems like if you take the moment about F after you've split the truss into sections, you don't need to analyze and find FG and FH to find EG.
Actually I just did the math and I got the same answer as him. So if you need to find only EG specifically, it is best to take moment about F :)
Oh cool yeah that is definitely a quick way to solve for just EG. Thanks for taking the time to notice and then check that it works! That would be a clever/sneaky shortcut on a statics exam if they only asked for the force in the one member, you could potentially save a ton of time by not bothering to solve for the other two unknowns from the virtual cut.. cool!
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