Estimating Weight and CG of a Fabricated Skid Without a Tag
A maintenance crew must lift a welded steel skid with no weight tag. The skid is a 10 ft × 4 ft base of 1/2 in. plate with four 6 in. × 6 in. × 1/2 in. wall square columns, each 8 ft tall, welded at the corners. A 500 lb pump sits on the base, centered 3 ft from the left end. The crew estimates "about a ton" and plans a 2-leg bridle from a single hook using a 4,000 lb chain sling. A 2-ton overhead crane is available. Steel weighs about 490 lb per cubic foot. The crew wants to know whether the crane is adequate and where to place the hook.
1. Estimate the skid's steel weight and total weight, showing your method. Is the 2-ton crane adequate?
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Base plate: 10 ft × 4 ft × 0.5 in. Volume = 10 × 4 × (0.5/12) = 1.667 cu ft, weight ≈ 1.667 × 490 ≈ 817 lb. Columns: each is 6 in. × 6 in. × 1/2 in. wall, 8 ft tall. The steel cross-section area is the outer square minus the inner void: outer 6 × 6 = 36 sq in., inner (6 − 2×0.5) = 5 in. square = 25 sq in., so steel area = 11 sq in. = 11/144 = 0.0764 sq ft. Volume per column = 0.0764 × 8 = 0.611 cu ft, weight ≈ 300 lb each, 1,200 lb for four. Steel total ≈ 817 + 1,200 = 2,017 lb. Add the 500 lb pump: total ≈ 2,517 lb, about 1.26 tons. The 2-ton crane is not adequate with any margin once you add the rigging (chain sling, hooks, shackles, perhaps 100–150 lb) and the dynamic allowance for a starting or stopping load, which commonly adds 10–25 percent. The crew's "about a ton" estimate is off by more than 50 percent. Use a larger crane or confirm the actual weight by scale before lifting.
2. Where should the hook be placed, and how does the pump's position shift the CG?
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Compute the CG along the 10 ft length, taking the left end as zero. The base plate is uniform, so its CG is at 5 ft. The four columns are symmetric at the corners, so their combined CG is also at 5 ft. The pump is centered at 3 ft. Weighted average: (2,017 × 5 + 500 × 3)/2,517 = (10,085 + 1,500)/2,517 = 11,585/2,517 ≈ 4.6 ft from the left end. So the CG sits about 4.6 ft from the left end, slightly left of center, not at the geometric center. The hook must be plumb over that point, or the skid will hang tilted and the bridle legs will load unevenly. A single-hook bridle cannot guarantee the hook stays over the CG unless the pick points are placed symmetrically about the CG; otherwise use a spreader or two hooks positioned so the resultant passes through the CG. Mark the CG on the skid and confirm it after the pump is mounted, since the pump's position is the largest variable.