Virtual Work Method to Solve for Rotation at a Joint (Frame) - Part 1/2

[@AFMathandEngineering]

Part 2/2: kzbin.info/www/bejne/q3TLdZxtfLaJZ9U Tricky beam question: kzbin.info/www/bejne/pYSvipeJd7KdhKM

[@boonkeatwong4243]

hi, may i know how to get reactions by using virtual work instead of using equilibrium method?

[@AFMathandEngineering]

Hello, please search our videos we have a few virtual work problems solved. Thanks!

[@aifydzri3452]

this video is perfectly splendid

[@huangweifeng6364]

Could you explain why is the virtual moment for segment CB is 0, you are taking origin from point C shouldn't the applied virtual load at point C be considered when cutting CB

[@AFMathandEngineering]

The question is asking us to calculate the rotation at point C. That means in the virtual system we apply the virtual moment at point C and simply calculate the reactions. This is shown around 5:30.

[@chesterpauldaguro7392]

But since you're taking C as the origin and there's an existing 1k.ft moment at point C, shouldn't it be considered?

[@Tracks777]

Awesome! Keep it up!

[@ugurbaksi6074]

Hey! You are terrific man!

[@AFMathandEngineering]

Thanks alot!

[@ash5492]

Is the virtual moment for DC not 1/30 X x^2/2?

[@az93850]

You are doing great thanks.

[@AFMathandEngineering]

BLACK WATER thank you so much for the great feedback. Are you currently an engineering student? Let us know if there's anything we can help with :D

[@andreic5713]

Excellent video. Could you explain why you multiply 1.5x with (x/2). Why do you use (x/2) ?

[@AFMathandEngineering]

Andrei Ciohodaru thanks a lot for the positive feedback, glad you enjoyed it! The reason why we multiply by x/2 is because we're multiplying by the moment arm of the resolved force. When we represent a member in terms of variable distance x, the resolved distributed load as a point becomes the value of the load * x. in this case, 1.5x. In order to find the moment caused by this point load, we must multiply by the distance from the point we're considering to the point load. Since we know the resultant of a distributed load acts at l/2, where l in our case is the length of the distributed load, we multiply by x/2. This is because we're representing the length l as a variable x. Hard to explain by messages, if you require more clarification feel free to email us. [e-mail]

[@andreic5713]

Got it ! Thank you !

[@AFMathandEngineering]

Andrei Ciohodaru np!

[@MegaHuda99]

Hello, how to know the virtual rotation 1kn is clockwise?

[@niruda638]

Please explain the Same problem using slop deflection method

[@AFMathandEngineering]

kzbin.info/www/bejne/hZiaeJR9ir-Krs0

[@locom16deen78]

Why for dc real is m negative ie -m. isnt m pushing it clockwise??????

[@AFMathandEngineering]

As shown in the FBD of the forces acting on member DC, if we cut just to the right of joint C, we have a clockwise moment, hence positive. When you start to work with frames, it's quite difficult to envision sometimes how they're acting. Best just to rely on the math.

[@tonpanuwat4253]

Mv (for virtual) at member CB is not zero. It should be 1.

[@AFMathandEngineering]

Thanks for the comment. Our moment is acting on section DC, thus AC only has axial forces acting on it. Mv = 0. For section DC, we cut just before C with D as our origin, ignoring the 1 kN-m point moment.