- #1
Svelte1
- 9
- 2
- Homework Statement
- dV/dx=-W
- Relevant Equations
- dV/dx=-W
How is the slope of the shear force equal to the negative of the load?
Click For Summary
The discussion centers on the relationship between shear force and load in beam mechanics, specifically addressing the negative slope of the shear force diagram when a downward point load is applied. It highlights confusion regarding the shear force sign convention, noting that while the slope should be positive according to the equation, the actual representation is negative due to the downward direction of the load. Participants reference a paper that treats the load as positive downwards, clarifying the convention used. The conversation also touches on the importance of understanding shear force in relation to external forces acting on a beam. Overall, the forum serves as a platform for further questions and clarification on this topic.
- #2
erobz
Gold Member
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Personally, I don't think I've ever understood the shear convention myself. I thought it acts to oppose the external forces on the beam so that the beam remains static, but I'm not sure now. I look forward to hearing other's thoughts on the matter.
- #3
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At https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites/53/2016/10/OPTI_222_W8.pdf, they seem to take q as a magnitude, and the equation dV/dx=-q is assuming it represents a load. Hence, effectively, the load convention is positive down.
- #4
erobz
Gold Member
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So based on the convention given in the paper that @haruspex shared, the shear being positive from ##\rm{A}## until around ##x_1 = 5.7 \rm{m} ## where ##V=0## means the beam is deflecting like:
and after ##x_1## the beam in deforming like the mirror image of that:
and after ##x_1## the beam in deforming like the mirror image of that:
- #5
Svelte1
- 9
- 2
Ok, thanks guys. Assuming it's a load makes sens I suppose. Also I have a lot of questions around this level in the coming weeks, is this the correct forum for that? Thanks.
- #6
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Should be fine, but there is alsoSvelte1 said:
https://www.physicsforums.com/forums/engineering-and-comp-sci-homework-help/
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...