Simple? Statics
Simple? Statics
(OP)
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?
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?





RE: Simple? Statics
Look at this thread:- Thread2-101606.
I am assuming your load is at right angles to the bolt heads is that correct?
If my assumption is correct both you and I have contributed
threads to the link above.
Regards Desertfox
RE: Simple? Statics
Regards,
Lcubed
RE: Simple? Statics
There are many solutions to your problem depending on the arrangement of the bolt/plate system. I think that most of the load can be ditributed between 2 bolts but the third bolt also carries load even though is negligible in comparison with the load carried by the other 2 bolts.
RE: Simple? Statics
Orientation of the load is important; ie if the load is at right angles to the bolt axis you have shear across thr bolt diameters, if the load is parallel to axis of the threads you have tension and a prying action.
regards desertfox
RE: Simple? Statics
Right. Glad to hear from you. The problem that I am having trouble with is that this load, on the line between the two most widely separated remaining bolts, does not cause a moment about the line, and therefore doesn't seem likely to induce a reaction in the third remaining bolt. Yet, using the approach I recommended in the thread you cited, one gets 1/3 of the load on each bolt because the bolts are equidistant from the load. Being a little more realistic, and considering the load eccentric from the centroid of the pattern, one gets a different answer. But I still don't see how there can be any load at all on that third bolt???
Regards,
Lcubed
RE: Simple? Statics
I should have been more clear. Think of this as a flat horizontal plate hanging on four nuts attached to vertical tension rods extending from the ceiling. Say that the rods are at the corners of a square pattern, one foot on each side. It is 1.414 feet from a rod to the rod diagonally opposite. A load hangs from the center of the plate; it is .707 foot from each rod. There is no shear. The load is parallel to the axes of the rods. There is no prying. Consider this a rigid body problem with no deformation-just pure statics. Hope this helps.
Regards,
Lcubed
RE: Simple? Statics
RE: Simple? Statics
Regards,
Lcubed
RE: Simple? Statics
As I suspected, there is stress on the third bolt directly perpendicular to the load. The deformation of the plate looks exactly as I suspected it would, the unsupported corner had the largest deflection.
The material is going to deform, and when it does, stress is going to occur at all restraining points.
The two holes opposite each other had stresses spread out in three different directions.
RE: Simple? Statics
There is no stress in the 3rd bolt if it is suspended by a cable (tensile only) since as we all know the moments about the line of centers of the two inline bolts must be zero.
Your model assumed a restraint at that third bolt which would depend on the actual structure of the third supporting member(a rod or a cable).
RE: Simple? Statics
I didn't read the "no deformation" part of the thread... My model assumed deformation, which was stated in my thread.
Then again, that's an unrealistic situation...
RE: Simple? Statics
Ben T
RE: Simple? Statics
Reread my thread. If there is only tensile support, the force in the offline bolt is zero. It doesn't care about the previous history of the suppot system.
RE: Simple? Statics
If all bolts are attached to an effectively rigid piece, then I think I would agree that only two bolts take the force. However, we don't know whether rigid is a reasonable assumption.
Brad
RE: Simple? Statics
It sounds like this might be purely an academic question to satisfy someone's curiosity. Either you've got the appropriate engineering background to solve this or you don't. Since the plate has some deflection all three bolts will have some loading.
- - -Dennyd, P.E.
RE: Simple? Statics
"Reread my thread. If there is only tensile support, the force in the offline bolt is zero. It doesn't care about the previous history of the suppot system."
I have read and re-read your posts. Please note that what at least three people here are telling you (multiple times, and two of them purport to be PE's) is not BS. Real world members have elasticity (they stretch).
If, before applying a load, all four cables are taut, then each cable MUST STRETCH when the load is applied. This stretch develops a certain amount of tensile force in each cable. When one cable is then severed or loosened, the remaining cables react to the change in load pattern. I can not envision any combination of cable patterns or distribution of tensile forces where you can say that the third cable will have zero force -- its residual tension reacts to removal of the opposite force, pulling the plate "out of plumb" and producing a moment (due to the change in relative position of the downwards pulling force).
If you are now going to say that two cables have slack before the load is applied (and thus have zero tension at the loaded condition), then you have changed the problem.
I suggest you set this problem up with some strings and a block of wood, and some weights. Let us all know the results of your experiment.
RE: Simple? Statics
ssk, PE
RE: Simple? Statics
RE: Simple? Statics
RE: Simple? Statics
RE: Simple? Statics
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.
RE: Simple? Statics
JStephen, P.E.
RE: Simple? Statics
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
RE: Simple? Statics
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.
RE: Simple? Statics
RE: Simple? Statics
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.
RE: Simple? Statics
RE: Simple? Statics
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.
RE: Simple? Statics
Imagine that....
RE: Simple? Statics
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
RE: Simple? Statics
Best regards,
Matthew Ian Loew
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: Simple? Statics
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)(y)/(h^2) = (3/2)(Pc*h/3)(y)/(h^2) = (Pc*h)(y)/(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.
RE: Simple? Statics
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
RE: Simple? Statics
I must be missing something - why can't a non-level plate be in equilibrium with no external moments?
RE: Simple? Statics
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.
RE: Simple? Statics
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
RE: Simple? Statics
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 - www.livejournal.com/users/robcampbell
RE: Simple? Statics
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 - www.livejournal.com/users/robcampbell
RE: Simple? Statics
Rob Campbell, PE
Finite Monkeys - www.livejournal.com/users/robcampbell
RE: Simple? Statics
Here sums up where I see the issue. If you remove bolt 4 can you truly assume the system is still static? In the above statement that is you are assuming that the system is still static.
In this case where the system is static AFTER the 4th bolt has been removed (and you have done something to stabilize it) then yes there cannot be any force in bolt 2 if the system is supported with cables.
The question of whether this assumption is reasonable is dependant on the type of support.
The system becomes dynamic when the forth bolt is removed. The removal of the forth bolt will cause a temporary unbalance in the moment equation which then may or may not stabilize depending on the loading in the #2 support. The system where a series of long pieces of threaded rod or bolts are the support will have the compressive resistance to load the plate in such a way that it will return to a static state. Hence the load in #2 is compressive.
The system using the cables removing the forth support will put the system into a dynamic state and will not return to a static state for us to analyze until the supports are no longer supporting the load.
It looks as though the problem has become -
Given that the system is in static equilibrium (which can only be true if the load in #2 bolt is zero) please solve for the load in the #2 bolt. In making the assumption that the system is static after the removal of the support you are automatically declaring the answer.
RE: Simple? Statics
If, on the other hand, the problem has an absolutely zero eccentricity, or, say, a minute eccentricity generating a negative moment about line 1-3, then when cable 4 is removed, the plate won't rotate a large amount, and, as the statics equations indicate, the cable 2 force drops from (P + m*g)/4 to zero or essentially zero.
In either case, when cable 4 is removed, the force in cable 2 will become zero, or nearly zero, and the force in cables 1 and 3 will double. This is the most conservative answer to the given question (posts 8 and 10), unless you want to consider a dynamic amplification factor for the case where the plate rotates, say, 90 degrees.
rjcjr9: Good assessment. You asked the rhetorical question, "But if it is, why is the member there in the first place?" Cable 2 is there because, before cable 4 is removed, the four cables create a "stable" system, and because having cables 2 and 4 present reduce the force in cables 1 and 3 by a factor 2 (if all four cables are exactly the same length).
RE: Simple? Statics
--------------------------------
That's an assumption. I can give you cases where the force in cable 2 won't go to zero. I depends on how springy the cable is, i.e. how long the cable (a real cable, not a fictitious perfectly rigid cable) is. Also, if the load is offset from the plate by a vertical hanger, or is held by a "rigidly" attached member, the pivoting of the plate about cables 1-3 causes the center of the applied load to shift, which produces a moment that is reacted by cable 2. All combined, the force in cable 2 may be less than in cables 1-3, but saying it is insignificant is specious. Especially if this experiment gets done in the real world - when cable 4 is cut, cable 2 sees a dynamic (time varying) load with alternating stress much higher than cables 1 and 3.
RE: Simple? Statics
If we go back to the earlier problem and consider that the plate is bolted down to another plate and lets assume that
the plate we are talking about is 14" square ie a 1" border
around the bolts, now if we lose a bolt ie bolt number 4 we can calculate the bolt loads as follows:-
The plate will tend to pivot at the opposite corner from which the bolt is lost ie that closest to bolt number 2.
calculate the distances to the bolts from the pivoting corner:-
(1^2 + 1^2)^0.5 = distance from pivot edge to bolt numer 2
for the remaining 2 bolts the distance
would be (7^2 + 7^2)^0.5 ie this would be a line from the centre of the plate to the pivoting corner perpendicular to the diagonal line drawn between bolt numbers 1 and 3.
let u = tensile load on a bolt at a unit distance from
the pivoting corner
therefore we can write that
P*(7^2+7^2)^2= u*la^2 + 2*u*(lc^2)
P= load at plate centre
la = distance from pivot to bolt 2
lc = distance from pivot to bolts 1 and 3
assume P = 100lb
therefore 100 * (98)^0.5 = u * (((1^2+1^2)^0.5)^2 + -------
-----------2*((7^2+7^2)^0.5))^2)
solving for u
u = 4.999744917 tensile load per unit distance
now to find the loads in the bolts multiply u by the distance from the pivot edge to the bolt centres
ie:- 4.949747468 * la = 7lb force
4.949747468 * lc = 49.49lb force
therefore bolts 1 and 3 carry 49.49lb force and bolt 2 only 7lb
clearly showing that bolt 2 doesn't do very much.
Now if you replace the bolts with cable's the pivot for the plate would be cable 2 and therefore would carry zero load or very close to zero leaving the cables 1 and 3 to do all the work, hence assuming that all the load is carried by cables 1 and 3 would be conserative and increase the safety margin.
regards
desertfox