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I have a configuration with two bolts and a point of force application (see figure). I understand that, if perfectly rigid, the two parts depicted in the figure will behave identically in how they will transfer moment on the bolts from the point of force application.

In this specific case, there would be a torsion, that is moment on the upper bolt going forward and moment on the lower bolt going backward.

The question is: what options are there to keep the configuration (bolts, point of application of the force), but straighten the moment on the bolts?

Would having a shape like the left one made of a flexible material change anything?

enter image description here

edit: to clarify, in the picture below i added, in red the resulting forces on the bolts using either of the two depicted rigid shapes, and the desired resulting forces on the bolts

enter image description here

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  • $\begingroup$ If it is flexible then of course it will change things, unless the system is statically determinate (and the things you have drawn are probably not). However you seem to be using the word "momentum" in non-standard way, so it's hard to understand what you are really asking. $\endgroup$ – alephzero Aug 12 '19 at 10:53
  • $\begingroup$ thanks, I added a picture to clarify what is the desired result. $\endgroup$ – Pa_ Aug 12 '19 at 11:39
  • $\begingroup$ Your comments on the available answers is making me think this is an XY problem. Please edit your question to add a lot more context on what you want, what sort of support is "behind" those screws, etc. $\endgroup$ – Wasabi Aug 12 '19 at 19:39
  • $\begingroup$ Sorry for being picky, but: "momentum" is the mass-velocity product of a particle or body. You mean "moment". It would be nice if you'd edit your question. $\endgroup$ – TimWescott Aug 12 '19 at 19:51
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First a review of how to determine the direction of the supports

In these cases, it's always useful to imagine what would happen if there were no supports. In this case, what would happen if there were no bolts.

As you noticed, one thing that'd happen is that the piece would rotate counter-clockwise. We don't want that to happen, so we can already tell that the supports will need to resist that rotation.

But the piece would obviously also move to the left. So the supports will need to resist that motion as well.

In general, we want to satisfy three conditions for any 2D object:

$$\begin{align} \sum F_x &= 0 \\ \sum F_y &= 0 \\ \sum M_0 &= 0 \end{align}$$

That is, the sum of external forces (including reactions) in the horizontal direction must be zero, as must the sum of vertical forces and the sum of moments around any arbitrary point.

Now, let's start with the first equation. Subscripts 1 and 2 describe the two screws, and I start by assuming both reactions point to the right, and are therefore positive:

$$\begin{align} \sum F_x &= R_{1,x} + R_{2,x} - F = 0 \\ \therefore F &= R_{1,x} + R_{2,x} \end{align}$$

This is pretty obvious: the two reactions must cancel out the applied horizontal force. But how the screws divide that load is still unknown (maybe they split it 50-50, maybe the top screw absorbs it all by itself, maybe the bottom one has a reaction greater than the applied force and the top one then has to cancel out that excess reaction... we don't know yet).

Now for the second equation (now assuming both forces point up and are positive):

$$\begin{align} \sum F_y &= R_{1,y} + R_{2,y} = 0 \\ \therefore R_{1,y} &= -R_{2,y} \end{align}$$

The applied force has no vertical component, so we only need to balance out the vertical components of the reactions. The difference in sign tells us they must be equal and opposite: maybe they're both zero, maybe they're a billion tons each (in different directions). Still to be determined.

Now for the last equation, calculating the moment around the bottom screw (remembering that moment is equal to force times perpendicular distance from the force to the selected point). I'll adopt $d$ as the distance between the screws and $L$ as the vertical distance from the force to the bottom screw. Forces which would cause a counter-clockwise rotation are positive.

$$\begin{align} \sum M_2 &= -R_{1,x} \cdot d + R_{2,x} \cdot 0 + R_{1,y} \cdot 0 + R_{2,y} \cdot 0 + F \cdot L = 0 \\ -R_{1,x} \cdot d + F \cdot L &= 0 \\ \therefore R_{1,x} = \dfrac{F \cdot L}{d} \end{align}$$

Notice that $R_{2,x}$ and $R_{2,y}$ have a distance of zero and therefore are discarded for this calculation. This makes sense, since a force on a point doesn't cause rotation around that point. Likewise, $R_{1,y}$ also has zero distance since we care about the perpendicular distance, and that force's "line of action" (it's projection) crosses through the other screw.

So we now know the direction of the force on the top screw: the opposite of what you drew, it will point to the right, countering the applied force.

Using $\sum F_x$, we can then see that

$$\begin{align} R_{2,x} &= F - \dfrac{F \cdot L}{d} \\ &= F\left(1 - \dfrac{L}{d}\right) \end{align}$$

As drawn, $L$ is obviously greater than $d$, which would make this force "negative", meaning our initial hypothesis that it was pointing to the right was incorrect: it actually points to the left, the same direction as $F$. This makes sense: the previous calculation told us the top screw's reaction was greater than the applied force, so the bottom screw now needs to compensate for that excess reaction.

And now to answer your actual question

Notice that the entire calculation above didn't take the structure's shape into consideration in any way. Whether the beam was a diagonal or a U-shape or an M.C. Escher painting was irrelevant. No matter the shape of your beam, or whether you add more beams, the reaction will always be the same.

You can even add more screws between the existing ones and the conclusion won't change: more screws will reduce the load on each screw, but you'll still have some screws reacting to the right and others to the left.

But notice the caveat in that previous paragraph: "You can even add more screws between the existing ones". That's a hint to our eventual solution. And that hint points us back to the equation for $R_{1,x}$:

$$R_{1,x} = \dfrac{F \cdot L}{d}$$

As mentioned, the screws have reactions in different directions because $R_{1,x}$ is actually greater than $F$. So, how can we fix that?

Well, we just need to increase the spacing $d$ between the screws, such that it is greater than $L$. If that happens, then $R_{1,x} < F$, therefore

$$R_{2,x} = F\left(1 - \dfrac{L}{d}\right) > 0$$

If $R_{2,x}$ is greater than zero, that means our initial hypothesis that it also points to the right was correct.

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  • $\begingroup$ so if i have the constraint that the point of application of the force must be minimum 5cm above the top screw, then L is always > d, implying no solution with a rigid bracket, right? $\endgroup$ – Pa_ Aug 12 '19 at 18:53
  • $\begingroup$ @Pa_, I unfortunately don't know what you mean by "a rigid bracket". Rigid has nothing to do with the direction of the reaction forces. You can have a very stiff base for the two screws which is virtually unaffected by the moment between the screws. Or you can have a very flexible base which breaks even with the same horizontal forces in each screw. $\endgroup$ – Wasabi Aug 12 '19 at 19:38
  • $\begingroup$ I mean like if, for example, a bracket of the first shape would flex on the bottom curve. would this somehow help until of course the flex is "over" and then the bracket would be fully "pushed" and behaving like a rigid one? $\endgroup$ – Pa_ Aug 12 '19 at 19:51
  • $\begingroup$ @Pa_: As described in my answer, the shape of the beam is irrelevant to the support reactions. You can complicate the shape as much as you want, adding as many curves and new beams as you want, and the reactions will stay the same. The only way to have both reactions pointing in the same direction is to have $d > L$, that is, to increase the spacing of the bolts such that the force is applied "between" them. And as my recent comment to your question requested, please edit your question to describe why you care about the direction of the reactions being equal in the first place. $\endgroup$ – Wasabi Aug 12 '19 at 22:01
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If you can have the bases extend up a little so that the center of the two holes is at level with the point of application of the force you won't have torsion.

Lets call the lateral force on the screws $ F_A \text{ and } F_B \text{ when the load on the handle is } F_H$:

Let's call the up screw A and down screw B and the line section between them $L$. And let's call the projection of the point of force on this line x.

$$ F_A= F_H\cdot\dfrac{L-x}{L} \text{ and } F_B= F_H * \dfrac{x}{L} \\ \text{and if } x = L-x \text{ then } x = \dfrac{L}{2} \\ \text{therefore } F_A = F_B \text{ and there is no torque.}$$

Regardless of the lateral loads on the bolts/screws, you always have a pull-out force on the screws.

EDIT

After OP's comment saying they can not change the handle geometry but they could possibly add some bolts to the right of the point of application of force:

Let's say we add two more bolt with the order of sizes you have in mind at the right-hand side such that together with the original bolts they make a square of 10cmx10cm.

Now let's calculate what is the maximum$ \ F_H $ which is not going to create a negative reaction ( reaction pointing to the left).

By what you said the torque of the force on the handle is F times(5cm height of the handle offset+ 5cm from side of the base square to it CG=10cm,

Torque = $ F_H*10cm$

Horizontal bolt force without the effect of torque $V= F_H/4 $.

The resistance of the bolts to the torque is the sum of the squares of the bolts distances to the center of the bracket: $ I_{baseplate}= \Sigma (F_{bolt}* r^2 )=4*F_B*50cm^2 $ and each bolt contribution to resisting torque is $$ F_B= F_H *10*7.07/ 4*F_B*50cm^2 $$

The critical bolts are the upper right-hand bolt and the lower left-hand bolts.

In those two bolts as long as $F_B \text{ due to torque }< \frac{F_H}{4} $ we are ok.

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  • $\begingroup$ Right, that's also the obvious solution. Unfortunately, the point of force application can't go so down, at least not without adding a piece that would then be practically part of the bracket. The geometry is therefore more or less fixed. I can add more bolts, and are free to do so also on the right of the point of force application, but they must all be like ~5cm below that point. And i understand that with a rigid bracket there is no solution, so that's why i was wondering if the desired result could be approached, if not reached, with a bracket that would be flexible in specific points $\endgroup$ – Pa_ Aug 12 '19 at 18:50
  • $\begingroup$ @Pa_, check my edit to my original answer, hopefully this clarifies the situation. $\endgroup$ – kamran Aug 12 '19 at 20:44

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