Cantilever equilibrium problem

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The discussion revolves around the complexities of achieving both rotational and translational equilibrium in a cantilever beam. It highlights the necessity for the torques from the normal force and gravitational force to be equal for rotational equilibrium, while also emphasizing that the net forces must be zero for translational equilibrium. Participants explore the implications of treating the attachment point as both a force and a torque, clarifying that the torque from the attachment is independent of the net force. The conversation also addresses the concept of point forces and torques, discussing how they serve as useful approximations in analyzing the system. Ultimately, the thread underscores the intricate balance required for a cantilever beam to maintain equilibrium under various forces and torques.
  • #31
Do you understand the concept of a limit?
 
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  • #32
Yes, I understand the concept of the limit. But I don't understand how it is being applied in this situation.
 
  • #33
OK. Do you understand the idea that a point force is the limit of a pressure over an area as the pressure goes to infinity and the area goes to zero? This limiting function is sometimes called the delta function.
 
  • #34
Not exactly. I understand that if pressure were a function of area, then the limit as the area approached zero, pressure would approach infinity. I don't understand how that relates to point forces though.
 
  • #35
Say you have a 1 N force applied at a point. What is the pressure?
 
  • #36
Since a point by definition has virtually no area, the pressure would be infinite, right?
 
  • #37
Yes. Exactly.

Now, how realistic do you think that is? Do you think that you can really have an infinite pressure at a single point?
 
  • #38
No, the force must be spread across a small area. So how exactly would that work in the case of the cantilever?
 
  • #39
UMath1 said:
No, the force must be spread across a small area.
Yes. So the model of a point force is just a limiting approximation. It is not realistic, but if the pressure is relatively high over a relatively small area, then it can be a useful simplifying assumption.

Do you now understand the idea that a point force is a limit of a large pressure over a small area, and it is just a useful approximation?
 
  • #40
Assuming you understand the above, now consider what would happen if the pressure were not uniform. Suppose that it varies linearly from one side to the other. Do you see that the average of the pressure still gives you the applied force as before, but the linear variation also gives you an applied torque?
 
  • #41
I think I understand. But wouldn't the location the torque and applied force coincide?
 
  • #42
UMath1 said:
I think I understand. But wouldn't the location the torque and applied force coincide?
Yes, that is the idea. We are modeling the force as a point force and the torque as a point torque.
 
  • #43
So how would you do that in this case. The weld force and normal force must be treated as one point force producing one point torque?
 
  • #44
Yes.

For clarity, you have a cantilever force and a cantilever torque. If you choose an axis through the attachment point then the torque for the cantilever force is 0 so only the cantilever torque (bending moment) remains.
 
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  • #45
Not sure wht I did wrong.
Screenshot_2015-08-18-15-57-52.png
 
  • #46
Do the torques about the joint so that the lever arm of ##F_{joint}## is 0. Then you can easily calculate the point torque applied at the joint, call it ##\tau_{joint}## in order to satisfy the torque equilibrium condition. Note that ##\tau_{joint}## is a completely separate independent degree of freedom and is not related to ##F_{joint}##
 
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  • #47
Is the location of Tjoint the same as Fjoint, though? Does it have a zero moment arm?
 
  • #48
UMath1 said:
Is the location of Tjoint the same as Fjoint, though?
Yes.
UMath1 said:
Does it have a zero moment arm?
It is not a force, it is a torque (technically a bending moment). It doesn't have a moment arm, the concept is irrelevant for it.
 
  • #49
So then in this case, based on how I set it up, Tjoint would be 195 N*m, correct?

And what is a bending moment?
 
  • #50
UMath1 said:
And what is a bending moment?
Tension breaks a beam by pulling it apart. Compression breaks a beam by crushing or buckling. A shear moment breaks a beam by slicing it. A bending moment breaks a beam by snapping it. And torsion breaks a beam by twisting it.

The math is in the right ballpark. I can't check exactly since you didn't label the distances.
 
  • #51
So is bending moment a force or a torque? Tension is a force if I understand correctly.
 
  • #52
Neither. They are all stresses. The units for stress are the same as the units for pressure, but in solids they can point in directions other than perpendicular to the surface.
 
  • #53
So Tjoint is not a torque but a stress? How does it cancel out the other torques then?
 
  • #54
If you integrate a stress over an area you can get a torque or a force.

This is related to what we discussed earlier about the point force and point torque approximations.
 
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