Simple? Statics 2

[Lcubed]

A load is imposed at the center of a square 4-bolt pattern. One bolt is removed. How is the load distributed to the remaining 3 bolts?

I know the location of the "new" centroid of the fastener pattern, and I have calculated the moments about that point and the resulting reactions. I have also looked at the problem using "rigid body" methods, per Maleev and others.

However, I think that the load lies on a line between two fasteners, and there is no moment about that line, and so those two fasteners take all of the load while the third remaining bolt takes no load at all. I need a reference to show that I am right or wrong, as the case may be. Can anyone help?

 

[zekeman]

I just realized that if the plate has uniform weight density, then the CM Of the plate will also be in the line of centers so I can state unequivocally, no force in the remaining bolt and I will stand by that conclusion no matter how many PEs state otherwise or how many experiments you perform.

 

[ivymike]

if you number the bolts 1 thru 4 moving clockwise around the 4-bolt pattern, load the plate dead-center, and remove bolt 4 from the equation, and if all components are inelastic, you will have zero force in bolt 2 (diag from the removed bolt). All of the load will be reacted by bolts 1 and 3.

If you let the cables stretch, but retain an inelastic plate, and assume the cables cannot ever transmit a moment to the plate, then the plate rotates about the axis between connections 1 and 3 until cable 2 is unloaded, and maintains that position (or a position with greater rotation from the original) with zero load in cable 2.

If you let the plate bend and make the cables inelastic again, then the plate will want to "taco" about a line between points 2 and 4. It will again have to rotate to a position that allows the moments to equal zero - but this time that will require some amount of load on cable 2 because the load application will no longer be on a line between points 1 and 3 after the plate "tacos" and rotates. Cable 2 will end up with a tiny load to balance the moment due to the offset force application point.

If you let everything be elastic, then you will increase the load in cable 2 slightly because the loaded cables (1 and 3) will stretch slightly and make the plate rotation angle a little bit higher, further offsetting the force application point.

 

[JStephen]

Zeke, if you're hanging stuff off the ceiling, there's no significant load carried by the third cable or bolt. If you're designing a space shuttle, maybe there is.

JStephen, P.E.

 

[brad]

Why the heck are we talking about cables? These are bolts, folks--each able to react 6 degrees of freedom.

6 dof x 4 bolts = 24 degrees of freedom to solve.

Statics = 6 equations.

How do you solve 24 unknowns with only 6 equations available? It's not statics!

You can simplify all you want, and try to extract some meaning from your cables, but the fact is the REAL answer is very much dependent on the stiffnesses of the whole system.

What if I keep all four bolts, and attach three of those bolts to a very thin piece of plastic, while I attach the fourth to a very thick steel plate? Intuitively we all know that the forces will be driven through this fourth bolt. How come--your claims to statics say that they should all distribute the forces equally. This is static indeterminacy--too many equations to solve with basic statics assumptions. Additional assumptions must be included (such as mechanics of solids).

Brad

 

[CoryPad]

Brad,

To answer your first question, this post's originator stated in the 10th post of this thread that s/he wanted to change the bolts to cables. I do agree with everything you wrote, although this P.E.'s opinion doesn't account for much. ;-)


Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[JStephen]

Brad, the REAL answer is also very much dependent on the temperature that the whole system is installed at, taking into account the actual operating temperature and the differing coefficients of expansion, and of course radiation heat transfer from the lights in the building must be included in these effects. And it's also dependent on the exact bolt tension and coefficents of friction and the stiffness of the support material. Also, air currents in the room will vary the stresses slightly, and vibration in the support structure must be taken into account. And don't forget that small torsional force that results from the earth's rotation. But a part of engineering is deciding what factors can be neglected for a given application. Otherwise, you'd never finish analyzing anything.

 

[CanEngJohn]

In my limited experience ...

If you have bolts which are holding a plate in place creating a shelf scenario then the third bolt should have some sort of compressive force on it as the center of support is not in line with the center of mass. You have a triangle supporting a square and all your second moments just changed. In order to compensate between the I values of your moments there will be some force requirement.

Once you remove the first bolt the centers of area no longer line up with your centers of mass. You either have a new force to balance the imblance or you now have a dynamic situation. Cables usually result in a dynamic situation that results in excessive effort to clean up the static mess.

Many first hand accidental experiments on this concept.

 

[profengmen]

Funny how you can balance the plate with just two cables...

 

[CoryPad]

profengmen,

With imaginary conditions like atomically precise plate thickness/mass distribution, atomically precise cable positions, etc. You find these in the same place as massless pans and frictionless pulleys used in physics experiments.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[profengmen]

Corypad,
Imagine that....

 

[brad]

JStephen,
YOur post, though sarcastic, is correct. However, most of us can likely agree that many of those assumptions are second-order effects for most use-cases. (Note that if we were talking of an engine, I likely would've cited thermal expansion as a point of concern).

Nobody who has taken basic statics and mechanics of solids classes should presume that torque reactions due to bolts are second-order for such a configuration. Suggesting that these are second-order effects (which is exactly what is implicit in this line of reasoning) can lead to severe errors.

As others have noted, the basic conclusion of the cable scenario is nonsense--you've got a perfectly balanced system between the two cables in this line of reasoning. Why did we ever need the removed bolt in the first place, since by this logic neither that bolt ("cable") nor its diagonal ever would've taken any load.

This line of reasoning is truly frightening to me.

The original 4-bolt pattern can be reduced only by the fact that we have dual planes of symmetry, and we have made some implicit assumptions about adjoining stiffnesses. That original problem was not statics, but rather relied on such reasonable assumptions which often fall out for symmetric problems. If one is not able to appreciate that, and thus understand WHY these same assumptions do not hold for a no-longer symmetric configuration, perhaps one should not give advice for designing such a system.

Brad

 

[mloew]

[pedantic mode=on]The plate, even with conditions like atomically precise plate thickness/mass distribution, as Cory stated will still not have uniform weight density as not all points on the are the same distsance from the center of the gravitational field.[/pedantic]



Best regards,

Matthew Ian Loew


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[vonlueke]

Reading through the first ten posts, the original poster is asking for a rigid-body mechanics answer. A rigid, horizontal, square plate, having a vertical point load at its center, is suspended by four inextensional, vertical cables. One cable is removed. How is the load redistributed to the remaining three cables? The original poster also indicates he wants to consider the load to be slightly eccentric from the center of the plate.

Let a = length of plate side, h = half of diagonal length = a*cos(45 deg), P = applied load at center of plate, m = plate mass, and g = gravitational acceleration. Let F1, F2, F3, F4 = force in cables 1, 2, 3, and 4, respectively.

Before cable number 4 is removed, moment equilibrium (Newton's first law), and symmetry, dictates that cable number 2 must carry one-fourth of the load, as follows. Summing moments about connection 2 gives, summation(M2) = 0 = h*F1 + h*F3 + (2*h)*F4 - h*(P + m*g). But, by symmetry, F1 = F2 = F3 = F4, reducing four unknowns to one unknown. Therefore, substituting and solving for F2 gives, F2 = (P + m*g)/4.

As soon as cable 4 is removed, moment equilibrium, and symmetry, dictates that the force in cable 2 must now change to zero, as follows.

Summing moments about connection 1 gives, summation(M1) = 0 = 0*(P + m*g) - 0*F1 - 0*F3 + h*F2. Therefore, solving for F2 gives, F2 = 0.

Now let's say you instead want to solve this problem like a "bolt pattern," after cable 4 is removed. Let A = cross-sectional area of each cable, and yb = distance of "bolt pattern" centroid from line 1-3. Therefore, yb = summation(A*y)/summation(A) = (0 + 0 + A*h)/(3*A) = h/3. Now translate the applied loads to the centroid. The equivalent applied loads at the centroid are, Pc = (P + m*g) and Mc = (h/3)(P + m*g) = Pc*h/3. Summing moments about connection 1 for applied tensile load Pc gives, summation(M1t) = 0 = h*F2t - 0*(F1t + F3t) - (h/3)*Pc. Solving for F2t gives, F2t = Pc/3. Summation of vertical forces gives, summation(Ft) = 0 = F1t + F2t + F3t - Pc. By symmetry, F1t = F3t. Therefore, substituting and solving for F1t gives, F1t = F3t = (Pc - Pc/3)/2 = Pc/3. To distribute the applied centroidal moment Mc, sigma = M*y/I; therefore, F = M*y*A/I (Eq. 1). But I = integral[(y^2)(dA)] = A*summation(y^2); therefore, F = M*y/summation(y^2). But summation(y^2) = 2(yb^2) + (h-yb)^2 = 2*(h/3)^2 + (h - h/3)^2 = 2(h^2)/3. Therefore, substituting, Eq. 1 becomes, F = (3/2)(Mc)/(h^2) = (3/2)(Pc*h/3)/(h^2) = (Pc*h)/(2*h^2). Therefore, the force in cable 1 due to applied moment Mc is, F1m = (Pc*h)(h/3)/(2*h^2) = Pc/6. And, likewise, the force in cable 2 due to applied moment Mc is, F2m = (Pc*h)(h/3 - h)/(2*h^2) = -Pc/3. Thus, by superposition, the total force in cable 1 is, F1 = F1t + F1m = Pc/3 + Pc/6 = Pc/2 = (P + m*g)/2. And the total force in cable 2 is, F2 = F2t + F2m = Pc/3 - Pc/3 = 0.

Both approaches give exactly the same answer; F2 = 0. Notice in the second approach, I distribute the applied tensile load and moment to the "bolt pattern" separately, using superposition. I.e., I distribute the applied centroidal moment using the familiar linear, elastic beam flexure formula, sigma = M*y/I. (This formula assumes a rigid, planar lamina and assumes a linear stress distribution.) For the applied tensile load, I can't use the flexure formula because the tensile load generates zero moment at the centroid. Therefore I use basic moment equilibrium to distribute the applied tensile load about one axis -- the same method you would use for calculating the reactions on a simply supported beam. This is how you distribute an applied tensile load to, say, a bolt pattern or structure in rigid-body mechanics -- sum moments about the orthogonal, principal axes, and solve for the unknowns. In your problem, we didn't need to sum moments about the other principal axis, because a symmetry condition reduced the number of unknowns. The problem was statically determinate with only two equations and two unknowns. But when the number of unknowns exceeds the number of statics equations, then Hooke's law (component stiffnesses) and compatibility equations must also be utilized to add additional relations, to, e.g., write the plate rotation angle in terms of component elongations.

The assumption of rigid-body mechanics can sometimes be reasonable, depending on the problem, depending on the stiffnesses (and deflections) of the components relative to the magnitude of the applied load. And it's realistic to assume there will be imperfect mass distribution in the plate, imperfect location of the cable attachment points, and/or slight eccentricity in the applied load. Therefore, it's reasonable to assume there will be a finite, minute eccentricity in this problem that generates either a positive or negative minute moment about a line from cable 1 to 3. If this slight eccentricity generates a minute positive moment about line 1-3, then when cable 4 is removed, the plate will rotate almost 90 degrees before coming to rest. And the force in cable 2 will be zero. If, on the other hand, this minute moment about line 1-3 (due to eccentricities) is negative, then when cable 4 is removed, the plate will not move at all (in rigid-body mechanics). And the force in cable 2 will be only a minute tensile force that balances the minute eccentricity moment. I.e., the force in cable 2 will be virtually zero.

The force in this problem is carried by only two of the three cables because the weight of the plate is almost perfectly balanced by gravity about a line from cable 1 to 3, and load P is applied at the midpoint of line 1-3. Therefore, moment equilibrium dictates there can be no force applied to cable 2 (except for a minute tensile force to balance the minute eccentricity moment). And notice that the force in cable 2, F2 = 0, doesn't change if you change the gravitational field, whereas the force in cable 1 does change, because F1 = (P + m*g)/2.

Lcubed, if you now want an answer different from the rigid-body mechanics answer you requested, let us know, because it would give a slightly different answer (though perhaps not significantly different). Sometimes rigid-body mechanics is applicable and is a reasonable approximation, depending on the problem; other times it's not. In your given problem (posts 8 and 10), it's a very reasonable, close approximation (even if the cables are elastic), provided the plate doesn't bend much. If the cables are elastic (and the plate is still relatively rigid), then when you remove cable 4, as ivymike said, you'll actually see corner 2 rise slightly as the force in cable 2 changes from (P + m*g)/4 to zero; and you'll see corners 1 and 3 move downward slightly as the force in cables 1 and 3 doubles. In other words, the plate will rotate slightly and now be "crooked." But notice, even then, the force in cable 2 is still zero. The plate is "crooked" merely because cable 2 is now too short to let the plate hang perfectly level.

But also notice, once you remove cable 4, you are now in a state of unstable equilibrium; so the plate could rotate about line 1-3 to almost anywhere between, say, zero and 90 degrees and still satisfy equilibrium.

 

[brad]

Vonlueke:
you state "In other words, the plate will rotate slightly and now be "crooked." But notice, even then, the force in cable 2 is still zero."

This illustrates a fundamental error in this line of reasoning. If there is absolutely no force in cable 2, then what causes this rigid plate to be anything but level? If the sum of the moments on the plate is zero, and the force in cable 2 is also zero, then the plate MUST be level, with or without cable 2 there. Since cable 2 has no force per your contention, its effect on the rest of the system must be absolutely nothing.

Thus, there is no force (in your set of assumptions) to "pull" the plate out of level. It can only be out of level due to some amount of force in cable 2.

You are trying to combine what you intuitively know to be true (the plate being made not level for this scenario) with what you observe from your calculations. Upon reflection, these two observations do not jibe. Please reconsider your statement as to the plate being out of level. What could cause this?

It is caused by the fact that the only way in which such a construct is statically stable is by the fact that there is some non-zero moment reaction which is taking place (and is not accounted for in the "cable" assumptions). There IS force in bolt 2 (thus the plate can be non-level), and it is reacted by moments at bolts 1 and 3. Whether these moments are small or large is very much problem dependent.

Some keep trying to introduce cables to reduce this to a statics problem. Granted, the original requestor suggested this as an appropriate assumption. My contention is that we do not have enough information to presume that the original set of assumptions is valid. Lcubed could not figure out on his own that per cable assumptions, the force in bolt 2 is (theoretically) 0. Given that lack of insight, on what basis can we conclude that his framing of this problem as cables is appropriate?

Brad

 

[ivymike]

It can only be out of level due to some amount of force in cable 2.

I must be missing something - why can't a non-level plate be in equilibrium with no external moments?

 

[vonlueke]

Brad: Thanks for your comments. To be more accurate, I should have said, "But notice, even then, the force in cable 2 is still zero, or essentially zero."

It depends on whether we assume the problem has a minute eccentricity generating a negative moment about line 1-3, or whether we assume a theoretical zero eccentricity. Either way, the force in cable 2 is either zero or nearly zero. But for sake of clarity, let's assume a theoretical zero eccentricity. In that case, when cable 4 is removed, you are now no longer in stable equilibrium but instead in a state of neutral equilibrium. (The reason I said "unstable equilibrium" before, is because, realistically, the system could be in unstable equilibrium. But for the theoretical case of absolutely zero eccentricity, the system will now be in neutral equilibrium.) In neutral equilibrium, the force in cable 2 will be exactly zero, and the plate will be level for inelastic cables and nonlevel for elastic cables.

Neutral equilibrium means if you push on a system then remove the force, the system will not return to its original location. For elastic cables, when the cable 2 force drops from (P + m*g)/4 to zero, it shortens, whereas cables 1 and 3 lengthen (as their force doubles). The plate must now be rotated. The momentary, slight pull of cable 2 to cause the plate to rotate in neutral equilibrium is nearly zero. But that slight force is only momentary, then the force in cable 2 drops to exactly zero.

 

[brad]

ivymike--
The basic assumption that people have started with is that this was an originally-level plate. Something must have caused this plate to become "out-of-level" upon the removal of cable 4. Some net moment must cause this to rotate, then must be reacted by an opposing moment to cause it to cease rotation (and in the world of idealized cables, this is impossible). I can reasonably accept vonlueke's justification immediately above. Just please note that we are now devolving from the concepts of idealized tension-only members and "pure" static equilibrium.

My fundamental concern still holds: A set of assumptions have been employed in order to generate the "elegant" solution that bolt 2 has zero force. It is questionable whether these assumptions are appropriate. For most real-world problems, it is important to seriously consider the assumptions which were posed by Lcubed before concluding that bolt 2 has zero force. When is a plate sufficiently rigid? When are the bolt bending stiffnesses sufficiently negligible?

vonlueke--
I've read your posts before, and I think you can appreciate my concerns. I expect that there are people reading this thread who could take the advice given at face value without appreciating the fundamental assumptions employed. It is these people for whom I write these cautions.

Brad

 

[imagitec]

Brad wrote:

The basic assumption that people have started with is that this was an originally-level plate. Something must have caused this plate to become "out-of-level" upon the removal of cable 4. Some net moment must cause this to rotate, then must be reacted by an opposing moment to cause it to cease rotation (and in the world of idealized cables, this is impossible).

It is clear you are missing the assumption implicit in the problem statement that the janitor bumped the plate when your back was turned.

Rob Campbell, PE
Finite Monkeys -

http://www.livejournal.com/users/robcampbell

 

[imagitec]

I'm assuming this was a contrived example and not a real world problem. As people pointed out weaknesses, lcubed tried to address them, which only raised more questions. If it was cables from the get go, I imagine he would have noted that.

Can you solve the original problem using statics? It all depends on the assumptions, and it won't be exact in any case. But that's OK because we're engineers, not scientists, right?

An anaogy is zero force members in a truss. Since I can't paraphrase it any better, I'll quote MERM:

A third member framing into a joint already connecting two collinear members carries no internal forces unless there is a load applied at that joint. Similarly, both members forming an apex at the truss are zero-force members unless there is a load applied at the apex.

Is the force in those members really zero? No. Is it a valid assumption in the solution of the problem? Probably.

Is there a set of conditions for which a zero force in the third member of the original problem is valid? Yes. But if it is, why is the memeber there in the first place? Again, it's a contrived example. I don't think it was meant to be a real design problem.

Rob Campbell, PE
Finite Monkeys -

http://www.livejournal.com/users/robcampbell

 

[imagitec]

Statics is crucial, but it's relatively straightforward because of all the idealizations. At Virginia Tech, people didn't start switching to business majors until after statics.

Rob Campbell, PE
Finite Monkeys -

http://www.livejournal.com/users/robcampbell