Spring constant determination in Mechanical Shock
Spring constant determination in Mechanical Shock
(OP)
PREFACE: I'm new to shock and vibration analysis. This is not a homework problem. It is a simplification of the type of analysis I deal with. I simply need clarification of what is probably a simple step for those of you familiar with shock and vibration.
PROBLEM STATEMENT: I have a large steel box with four legs. The legs are simple, six-inch lengths of 3" X 3" X 0.25" angle iron. The legs are welded to the deck of a ship. The box is subjected to shock. Possible shock directions are in the x, y, and z directions only. Consider each direction separately.
Find the spring constant for each direction of shock.
MY SOLUTION: There can be only one spring constant applicable to all three directions of shock. The equation I use is:
K = (AE)/L
Where: A = 1.44 in^2, the cross sectional area of one length of angle iron;
E = 29 X 10^6 psi, young's modulus for steel
L = 6 in, the length of a leg
K = 6.96 X 10^6 lb/in
But, since there are four legs, all acting to resist the shock force together (acting in parallel), my actual spring constant will be Kt = 4(K) = 2.78 X 10^7 lb/in
QUESTION: I believe this same spring constant will hold no matter what direction the shock force comes from. A NAVSEA document I read said that normally the spring constant will be different for each direction of shock. Is this true? And, if so, how is this determined in the simple illustration I've provided?
PROBLEM STATEMENT: I have a large steel box with four legs. The legs are simple, six-inch lengths of 3" X 3" X 0.25" angle iron. The legs are welded to the deck of a ship. The box is subjected to shock. Possible shock directions are in the x, y, and z directions only. Consider each direction separately.
Find the spring constant for each direction of shock.
MY SOLUTION: There can be only one spring constant applicable to all three directions of shock. The equation I use is:
K = (AE)/L
Where: A = 1.44 in^2, the cross sectional area of one length of angle iron;
E = 29 X 10^6 psi, young's modulus for steel
L = 6 in, the length of a leg
K = 6.96 X 10^6 lb/in
But, since there are four legs, all acting to resist the shock force together (acting in parallel), my actual spring constant will be Kt = 4(K) = 2.78 X 10^7 lb/in
QUESTION: I believe this same spring constant will hold no matter what direction the shock force comes from. A NAVSEA document I read said that normally the spring constant will be different for each direction of shock. Is this true? And, if so, how is this determined in the simple illustration I've provided?





RE: Spring constant determination in Mechanical Shock
RE: Spring constant determination in Mechanical Shock
You should be looking for vibration modes. The flexibility of your legs is one problem, and possibly a minor one. Sheet metal walls have vibration modes too.
If your box is not on the deck, but up on a mast or some other structure, the vibration modes of the mast become interesting.
JHG
RE: Spring constant determination in Mechanical Shock
Loads which do not act axially on the legs will cause bending, which will have a different spring rate, and the spring rate will vary with the direction of these lateral loads, because the stiffness of the angles varies with the direction of the applied loads.
Also, it seems likely you will have to take into account any additional structure between the legs and the C. G. of the box and its contents.
You may find this to be a rather complicated problem when dealing with combinations of loads simultaneously applied in different directions.
RE: Spring constant determination in Mechanical Shock
That's an excellent point. If so, what would an equation for transverse or longitudinal shock look like? I guess my central question is how does one formulate the spring constant equation---from a physical standpoint.
If I take bending of a leg into account, do I then somehow apply
sigma = Mc/I
to the basic spring force equation F =-kx?
RE: Spring constant determination in Mechanical Shock
The shock would usually be from an explosion.
Lcubed,
Please refer to my reply to Electricpete.
Thank you both for the assistance.
RE: Spring constant determination in Mechanical Shock
I take it you are analyzing for stresses due to something similar to an impact load.
The equation Mc/I is a stress equation for a beam loaded in bending. M is the bending moment. I think your beam is way too short for a bending moment to be as significant as the shear loading in the x and y directions. Also, the standard equations are written on the assumption that your beam is long. They are probably wildly inaccurate for your case.
I agree with Lcubed that your primary problem is likely the structure of the box around where your legs attach.
JHG
RE: Spring constant determination in Mechanical Shock
Just as AE/L gives pounds of applied load per inch of axial deflection, you will need to find the formula for the load required to cause an inch of deflection of the leg in the direction of the applied load. End fixity will be a significant factor.
Here is one case:
Assume complete fixity at the bottom of the leg, where it is attached to the deck, and rotational fixity at the upper end where it is attached to the box. That is, when the box sways laterally the tops of the legs are displaced in the direction of the applied load and are no longer directly above the bottoms, but the top surface of the leg is still horizontal, as is the bottom surface, and the deflected leg is slightly “S” shaped.
In this case, the lateral displacement of one leg is WL^3/12EI, W being the applied load, and the formula for the stiffness is 12EI/L^3.
I’d suggest you get a book such as Roark, with many types of beams and loadings, and try to simulate your conditions as closely as possible.
Another problem which you may need to consider is that if the CG of the box and its contents is not centered with respect to the legs, there will be a tendency for the box to rotate in the horizontal plane as well as a tendency to translate. That is, in the case above, the ends of the legs are not displaced exactly in the direction of the load, but have been forced to rotate slightly about the vertical axis and also to bend simultaneously about a different axis; this type of phenomena, of torsion being reacted by a set of beams, is “differential bending”. This can happen even with the CG optimally located, because the legs have different stiffnesses in different directions, and if they do not react the applied load equally the effect can be the same as if the mass were eccentrically located.
Angles, being “open” sections, are not very efficient in torsion, and not very stable in bending; all things considered, if you have the option, you might wish to consider a different type of leg.
Optech is correct in being concerned about shear failure versus bending failure, and about the validity of beam equations when the beams are very short, but these legs are long enough to be critical in bending, rather than shear, and long enough for standard beam equations to be useful.
Please pardon the long-winded post. I hope I am not needlessly complicating your problem.
RE: Spring constant determination in Mechanical Shock
Rather than complicating things, you've clarified them. My difficulty was in seeing where the spring constant equations come from. Originally, I hadn't made the connection between beam deflection equations(axial and bending)and Hooke's law. An excellent point, too, on torsion. I believe I can attack these analyses more confidently now. Thankyou very much.
mvrod1
RE: Spring constant determination in Mechanical Shock
RE: Spring constant determination in Mechanical Shock
TTFN
RE: Spring constant determination in Mechanical Shock
You're absolutely right. We are using Algor FEA to do a majority of the analysis. In the past, however, my predecessors did much of it by hand calculation, so, determination of spring constants was necessary. I predominantly employ computer analysis, since taking over the job a short time ago, but there are still crunch times when hand calcs are useful. The question of proper determination of spring constants has been nagging me until now.
mvrod1
RE: Spring constant determination in Mechanical Shock
The four legs you mention will only disipate the actual weigh of the box, they will act as one spring in a shock situation. This is why shock mountings are weight based.
If your trying to protect what ever is in the box, good luck, the G force on certain areas of the ship can exceed 20g's and a hard mounted system like you explained..well...you know.
I hope this has helped in some way
Ther are lots of COTS items used to protect your box.
good luck
RE: Spring constant determination in Mechanical Shock
The barge test involves installing the equipment and a facsimile of the decking to which it's attached on a barge in a pond. Depth charges are then exploded. Quite spectacular, overall
The g-forces as measured by Clement's paper from the '70's shows forces exceeding several hundred g's.
TTFN