Shear lines from a FBD with distributed forces

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The discussion revolves around determining shear and moment lines for a half-wing of an airliner using a free body diagram (FBD). Participants are attempting to analyze the load distribution and share their FBD attempts, although there are issues with image visibility. The problem requires creating functions for shear and moment lines and organizing them in a table, starting from the wingtip. There is a debate about the direction of analysis, with some suggesting that working from the root to the tip may be easier than the instructed tip-to-root approach. Overall, the focus is on clarifying the methodology for solving the problem effectively.
Finn072
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Homework Statement
Formulate both the V and M diagrams for the following FBD.
Relevant Equations
How would I go about this? I've looked everywhere on youtube but couldn't find a single example that used distributed forces.
, this is the FBD in question
 
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Welcome!
I still can’t see the posted image.
 
Lnewqban said:
Welcome!
I still can’t see the posted image.
Hey thanks! this should work:
I'll also add my attempt at the 4 V(m) lines and the resultant graph (which isn't correct, I do know that much)
 

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Could you post the text of the problem?
It seems to be the load distribution on the half-wing of an airliner.
 
Lnewqban said:
Could you post the text of the problem?
It seems to be the load distribution on the half-wing of an airliner.
"Determine the functions for the Shear and Moment (“V & M”) lines and put these functions in a table. Choose x=0 at the wingtip, then start working from the wingtip to the wing root. Consequently use the right section of a separation. Create a complete FBD for each required separation."

My teacher mentioned during a lesson that working from tip to root was not mandatory, since the opposite is easier (apparently)
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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