Part 2/2: kzbin.info/www/bejne/q3TLdZxtfLaJZ9U Tricky beam question: kzbin.info/www/bejne/pYSvipeJd7KdhKM
@boonkeatwong42435 жыл бұрын
hi, may i know how to get reactions by using virtual work instead of using equilibrium method?
@AFMathandEngineering5 жыл бұрын
Hello, please search our videos we have a few virtual work problems solved. Thanks!
@aifydzri34524 жыл бұрын
this video is perfectly splendid
@huangweifeng63644 жыл бұрын
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
@AFMathandEngineering4 жыл бұрын
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.
@chesterpauldaguro73923 жыл бұрын
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?
@Tracks7777 жыл бұрын
Awesome! Keep it up!
@ugurbaksi60745 жыл бұрын
Hey! You are terrific man!
@AFMathandEngineering5 жыл бұрын
Thanks alot!
@ash54924 жыл бұрын
Is the virtual moment for DC not 1/30 X x^2/2?
@az938506 жыл бұрын
You are doing great thanks.
@AFMathandEngineering6 жыл бұрын
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
@andreic57137 жыл бұрын
Excellent video. Could you explain why you multiply 1.5x with (x/2). Why do you use (x/2) ?
@AFMathandEngineering7 жыл бұрын
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. Afmatheng@gmail.com
@andreic57137 жыл бұрын
Got it ! Thank you !
@AFMathandEngineering7 жыл бұрын
Andrei Ciohodaru np!
@MegaHuda994 жыл бұрын
Hello, how to know the virtual rotation 1kn is clockwise?
@niruda6386 жыл бұрын
Please explain the Same problem using slop deflection method
@AFMathandEngineering6 жыл бұрын
kzbin.info/www/bejne/hZiaeJR9ir-Krs0
@locom16deen785 жыл бұрын
Why for dc real is m negative ie -m. isnt m pushing it clockwise??????
@AFMathandEngineering5 жыл бұрын
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.
@tonpanuwat42536 жыл бұрын
Mv (for virtual) at member CB is not zero. It should be 1.
@AFMathandEngineering6 жыл бұрын
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.