Truss analysis by method of joints: worked example #2

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Truss analysis by method of joints: worked example #2

Рет қаралды 24,556

Күн бұрын

This engineering statics tutorial goes over another full example using the method of joints for truss analysis. You first need to solve for the reaction forces by drawing a FBD of the entire structure. Once you have the reactions, you then need to draw a FBD of every joint. It’s the standard convention to draw unknown forces in tension, so that if you get a positive value for their magnitude, you know you have correctly assumed tension, and if you get a negative value, it just means that the member is in compression. Once you calculate an internal force in a member, it will be a known value for the next joint it touches. So if it’s found to be in compression, then just draw it in compression when you do the next FBD (meaning it’s pushing on the joint rather than pulling on it).
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Thanks for watching, I hope it helps!

Пікірлер: 40

@cnw9997 6 жыл бұрын

only channel that has helped me with my structural analysis unit.....thanks

@Engineer4Free 6 жыл бұрын

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!

@ARMY_RUGS 6 ай бұрын

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@Engineer4Free 6 ай бұрын

Haha online since 2012. Welcome to the party! 🥳

@ARMY_RUGS 6 ай бұрын

@@Engineer4Free thank you very much .

@meelba-tube8869 4 жыл бұрын

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@Engineer4Free 4 жыл бұрын

Thanks!! I'm hoping to visit Ethiopia 👌👌. Are you studying in Addis?

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@richardb6635 4 жыл бұрын

Amazing content. Easy to follow and understand. 🙏 Helped me with last minute cramming ;)

@Engineer4Free 4 жыл бұрын

Glad I could help!! Check out engineer4free.com/statics for the rest of the videos if you haven’t already 🤜🤛

@jewelhoaasin7403 2 жыл бұрын

Best channel for education purposes 🥳🥳🥳

@oliverscannone7838 9 ай бұрын

Thankyou so much. I have two textbooks and they both solve graphically and couldn't explain anything properly

@Dandamany 4 жыл бұрын

You are a champion, thank you for your videos, they're so much clearer and easy to understand then the lectures uni uploads.

@Engineer4Free 4 жыл бұрын

Thanks Khentse!! ☺️

@shivaramvalagadri9333 5 жыл бұрын

love you bruh.

@Engineer4Free 5 жыл бұрын

luv ya too

@wanted1729 7 жыл бұрын

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@rufaromanjalanz8219 4 жыл бұрын

200/sin(45) = 282.84. You still are correct with everything else though

@tiantiansumile623 5 жыл бұрын

Thankyou!!

@Engineer4Free 5 жыл бұрын

You're welcome!! =)

@ryankinloch8610 4 жыл бұрын

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?

@Engineer4Free 4 жыл бұрын

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!

@MGMGn 5 жыл бұрын

I dont understand the directions of Fx or Fy . Can you record a extensive video about free body diagrams please?

@Engineer4Free 5 жыл бұрын

Yeah, try this one: www.engineer4free.com/4/how-to-draw-good-free-body-diagrams-fbds

@jonathanagpaoa3937 4 жыл бұрын

I think the angle in point C is 90 degrees not 45

@Engineer4Free 4 жыл бұрын

Member CD is definitely angled at 45 degrees from the horizontal

@genesisr8851 4 жыл бұрын

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)

@Engineer4Free 4 жыл бұрын

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.

@connorfowler9327 4 жыл бұрын

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?

@Engineer4Free 4 жыл бұрын

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.

@toemyintnaing2141 7 жыл бұрын

u are the best!!!!~~~~~~ ty

@Engineer4Free 7 жыл бұрын

+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!

@asad5986 4 жыл бұрын

How do you know Fx and Fy are going up and to the right?

@Engineer4Free 4 жыл бұрын

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 4 жыл бұрын

@@Engineer4Free Ok thank you. Then it's just a matter of adjustment if the assumption turns out to be incorrect right?

@Engineer4Free 4 жыл бұрын

@@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.

@asad5986 4 жыл бұрын

@@Engineer4Free ok thank you

@lukecaetano8405 3 жыл бұрын

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

@Engineer4Free 3 жыл бұрын

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

@lukecaetano8405 3 жыл бұрын

@@Engineer4Free god bless your soul thanks!!