Robert Soley
Structural
I posit it's the right side, you find the percentages & then calculate from total load; but I have no way to test this.
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Constituent components of an inclined load, hypothesis
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I posit it's the right side, you find the percentages & then calculate from total load; but I have no way to test this.
Seems like a simple statics problem. Transform the force vector into constituent force vectors orthogonal to your member to get shear and axial loads. It's the left side. Nothing to test - it's covered in first year physics/kinematics and then again in statics. What are you trying to figure out?
For the sloping cantilever beam in the middle diagram, the axial and shear components are shown below. Arc length has nothing to do with the problem. The applied force is resolved into components parallel and perpendicular to the beam which gives axial and shear components respectively.
BA
BA
It's interesting to think about - based on my experience from Calc 2 back in the day, the concept of arc length will haunt my dreams forever. However, I don't believe that arc length has much to do with this.
Regardless of where it exists in space, a force is a force. Changing arc length does not increase the magnitude applied force. It's important to note that the reaction moment at the base of the beam will be proportional to the length of the beam though. That is, the longer the cheater bar, the bigger the cheater force. The reaction forces will remain the same though (if your applied force is Fx = 2 kip at 5 ft out or 10 ft out, the reaction force would still be Fx = -2 kip).
I guess what I'm saying is that reaction moment is proportional to length, and arc length is proportional to length, therefore the reaction moment is indirectly proportional to arc length.
Determining the component forces for any given set of axes is a function of the angles that the X and Y (or X' and Y') axes make with respect to the given force. In other words, all you need to do is draw a right triangle where the force vector is the hypotenuse. Then, use trig (remember SohCahToa) and figure out the lengths of sides A and B. Component forces are essentially just projections of vectors onto other sets of axes.
Regardless of where it exists in space, a force is a force. Changing arc length does not increase the magnitude applied force. It's important to note that the reaction moment at the base of the beam will be proportional to the length of the beam though. That is, the longer the cheater bar, the bigger the cheater force. The reaction forces will remain the same though (if your applied force is Fx = 2 kip at 5 ft out or 10 ft out, the reaction force would still be Fx = -2 kip).
I guess what I'm saying is that reaction moment is proportional to length, and arc length is proportional to length, therefore the reaction moment is indirectly proportional to arc length.
Determining the component forces for any given set of axes is a function of the angles that the X and Y (or X' and Y') axes make with respect to the given force. In other words, all you need to do is draw a right triangle where the force vector is the hypotenuse. Then, use trig (remember SohCahToa) and figure out the lengths of sides A and B. Component forces are essentially just projections of vectors onto other sets of axes.
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Robert Soley
Structural
There's no such thing as the conservation of forces. Momentum, yes. Energy, certainly. Forces? Not so much. Equal and opposite, yes, but not conservation in a literal sense and definitely not in a a way you can do a scalar sum of vectors in multiple directions.
If you take the moment of the 5kip original force about the fixed end of the cantilever, the distance e you would use is the shortest distance between the line of action and the point. So Pe = M = 3kips*L. And then you also have the axial load.
100% for the left.
Curious - what's you level of formal education in engineering?
If you take the moment of the 5kip original force about the fixed end of the cantilever, the distance e you would use is the shortest distance between the line of action and the point. So Pe = M = 3kips*L. And then you also have the axial load.
100% for the left.
Curious - what's you level of formal education in engineering?
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