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?

 

[CanEngJohn]

summation(M1t) = 0 = h*F2t - 0*(F1t + F3t) - (h/3)*Pc

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.

 

[vonlueke]

CanEngJohn: If the problem has a minute eccentricity generating a positive moment about line 1-3, then when cable 4 is removed, the plate will rotate, say, 90 degrees. And the force in cable 2 will be zero, because cable 2 goes slack.

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).

 

[btrueblood]

CanEngJohn: If the problem has a minute eccentricity generating a positive moment about line 1-3, then when cable 4 is removed, the plate will rotate, say, 90 degrees. And the force in cable 2 will be zero, because cable 2 goes slack.
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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.

 

[desertfox]

Hi All


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