The answer to question 1: The operator was not being unreasonable.
The answer to question 2: The rigging superintendent was completely wrong.

All lifts and especially critical lifts should have a firm base for a crane to work safely. One of the best ways to provide that is to use timber mats that are in good condition and free of rot or pests. One-inch thick steel plates, even stacked four-high are not comparable substitutes for timber mats. In fact, it would take more than (13) layers of one-inch thick steel plates to provide the same load spreading capacity as one layer of 12" thick timber mats. If the plates could be welded together on four sides, four layers of one-inch thick steel plates would be comparable, or a single layer of a solid four-inch thick plate, such as a SuperLift counterweight slab, would also be comparable.

Steel plates are useful when a crane is working on areas paved with asphalt or concrete mainly to avoid damage to the paved surface and for increased shear resistance to punching through thin paving. It should be noted, however that mats should still be used on paved surfaces if there are sewer lines, storm drains, or other underground structures that must be protected.

To illustrate, if we assume that the timber mat or steel plate acts like a cantilever beam fixed by the tracks and extending out on both sides of each track (refer to the rigging quiz no. 14 to the dimension Leff), and the crane is working on an area that has an allowable soil bearing capacity of 2,000 lb/sq ft, we have:

*Maximum bending moment supported by the "beam", Mmax = SFb , where Fb = 2,050 psi for a timber mat and 21,600 psi for the steel plate, and S = section modulus = (width)(thk2) / 6 = 288 in3 for a 12" wide x 12" thk timber and 2 in3 for 12" wide x 1" thk steel plate. Then Mmax = 590,400 in-lb for the timber and 43,200 in-lb for the steel.

*The effective length of the "beam", Leff = Sqr(2Mmax /w), where "w" is the load distributed by the "beam" and is 166.7 lb/in in this example. Then Leff = 84.2" for the timber and only 22.8", or only about one-fourth as long for the steel.
Adding another steel plate doubles the value of "S" as S = (2plates)(12"wide)(1"thk2) / 6 = 4 in3, but not the value of "Leff". The total thickness of the plates can be squared in the equation only if the plates are welded together to act as a unit. If you go through the math, it will take (14) layers of plates for "Leff" to equal 84.2".

*To illustrate the above, try supporting something on a cantilevered beam made up of (20) sheets of paper laying flat. Then try it again after stapling all of the edges together. Big difference!

    

Quiz List     

Home Page