Understanding the Forces on a Beam

[Cyrus]

say you have a beam like this
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*---------------------------------------------------|
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Where the arrow is some downward acting force at any distance away from the end of the two *'s. The *'s represent bolts holding this thing in place. Each bolt is directly above the other bolt. If You sum the moments about a line in the vertical direction passing through both bolts, you will have a moment caused by the downard force times the distance to that line. But no counteracting moment? Whats going on here? I know that the sum of the moments about any point must be zero. Isn't it true for any line as well?

 

[Pyrrhus]

The support you describe, Cyrus does it constraint traslational and rotational movement? If it does then there will be a counter moment on the FBD. I guess it depends on the application. Another way to see it, if that the bolts will have a force with an unknown direction, but since there are no axial forces acting on our beam both of the x components on the bolts will have opposite directions and same magnitud, thus forming the counter moment (couple) to oppose your applied force's moment. Of course in practice for a cantilever beam, the correct support is usually embedded (like a metallic plate bolted to a concrete plate).

 

[Cyrus]

Well, what I am saying is that it matters not. because any force acting on the bold will pass through the line, thus have ZERO lever arm, no matter WHAT direction they are in.

 

[Pyrrhus]

Well the sum of moments equal 0 is only right about any point, what you mean is moment about an axis. Yes, if you take the moment of the forces acting on your beam with respect to the axis which goes throught both of the bolts, it will NOT BE ZERO, unless the force is parallel or intersects the axis. In your example your applied force will have a 0 moment about your chosen axis.

 

[Cyrus]

Why, it is a distance of L away from the axis passing through the bolts. I did say line, btw.

In your example your applied force will have a 0 moment about your chosen axis.

Not true.

 

[Pyrrhus]

Because of the definition of Moment about an axis

M_{OL} = \vec{\lambda} \cdot \vec{r} \times \vec{F}

The lambda unit vector goes in direction of the axis up or down, and the position vector can come out of any point on the axis to the point of application on the force.

 

[Pyrrhus]

The moment will be 0, let's be rigorous (using standard axis), where the origin is just above the beam left upper surface and our x-axis is the centroidal axis of the beam:

\vec{M}_{o} = -PL \vec{k}

Now if let's imagine the y-axis goes throught our both bolts centroids, then

\vec{\lambda}_{y} = \vec{j}

thus

M_{y} = \vec{j} \cdot -PL \vec{k} = 0

 

[Cyrus]

Huh? You are over complicating things. Look, just make an axis passing through both *'s. I don't care what you want to call that axis. Now take all the moments about that axis. That will be equal to the force, P, times the span of the beam, L, which is not zero.

 

[Cyrus]

OHHHHHHHHHHHHHHHHHHHHHH. Yeah you are right. That force does not tend to rotate that beam ABOUT that axis. It tends to rotate it about an axis comming out of the computer screen. I am an idiot. I forgot about the precise defintion of that. Thank you oh wise one. What the F' was I smoking. Thats just sad.

 

[Pyrrhus]

Don't worry, just thanks you didn't have this mental block on an exam.