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Lug design for local loading

Lug design for local loading

Lug design for local loading

(OP)
Hi,

First confession - I am not using compress.

What I am doing is designing lugs for local loading on a heat exchanger shell and I am trying to find a good worked example of resolving lifting forces and moments for chain angles of 30 degrees and 60 degrees etc. I have already designed the lug itself - I am simply wanting to see a good visual example on a website or paper of the manner in wqhich the forces and moments are resolved.

Can anyone help me?

Many thanks


RE: Lug design for local loading

Seeker,

Sometimes a picture is worth 1000 words. I wish I could provide you with one. Instead, here are 1000 words :)

I suggest obtaining a textbook or other reference on engineering mechanics (or "statics") and read how forces can be resolved into components in other directions. A text on machinery design should address this as well; my old copy of Machinery's Handbook has a confusing chapter on "resolution of force systems". The TEMA manual has some nice pictures of lifting lugs, I wish they also showed the forces acting on the lugs.

In general, the tensile force in the lifting cable or chain can be applied to the lug in force components that are "orthogonal" to the surface to which the lug is attached. Forces can be resolved into components in any direction, why resolve them into these "orthogonal" components? The point of resolving them into force components parallel and/or perpendicular to the lug's attachment surface is that the various types of stress acting in the lug will generally be a function of such components. eg: the attachment surface must transfer the tensile force (perpendicular to the surface) and the shear force (parallel to the surface).

As an example, consider the heat exchanger shell in a horizontal position with two lugs on the top surface. Let the chain or cable at each lug be oriented at 60° from the horizontal. For this example let each cable support the same weight and the total weight is F = 10,000 lb. Then the tensile force in each cable is found by writing the equation of equilibrium for forces in vertical direction (from those books on engineering mechanics):

   sum of forces in vertical direction = 0 (because there is no acceleration)
     = lifted weight - 2 * ((tension in cable)*sin(60))
     = 10,000 - 2 * T * sin(60)

   solve for T:

      T = 10,000 / (2 * sin(60))
        = 5773.5 lb

Thus each lug is subject to a 5773.5 lb force applied at 60° from the horizontal.

The shear force acting on the lug is the horizontal component of the cable tension, or T * cos(60) = (5773.5)*cos(60) = 2886.75 lb.

The tensile force acting on the lug attachment surface is the vertical component of the cable tension, or T * sin(60) = (5773.5)*sin(60) = 5,000 lb. This checks out because it has to be 1/2 of the total lifted weight.

Of course, there are many more details to be worked out for lifting lugs. But hopefully this gives you some idea of how to resolve the forces in the different directions. If not, c'mon back and let me know.


Tom Barsh
Codeware Technical Support

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