Two-Leg Bridle Lift of an Off-Centre Compressor Skid
A 5,400 lb skid-mounted compressor is being lifted off its housekeeping pad onto a flatdeck trailer. The skid has two lifting lugs 6 ft apart along its length; the centre of gravity is 2.5 ft from lug A and 3.5 ft from lug B. The crane can place the hook 9 ft directly above the lug line, and the rigger intends to hang it directly over the centre of gravity. The slings are two 10 ft lengths of 5/8 in 6x19 IWRC wire rope, each tagged 6,400 lb vertical capacity, and both will be used in a choker hitch (80 percent of vertical rating). The lugs are 1 in plate with 7/8 in holes. The only shackles in the rigging box are 1-1/2 ton screw-pin alloy shackles. Wind is calm and the trailer deck is 3 ft below the pad.
1. Calculate the tension in each sling leg and state whether the two slings are adequate for the lift.
Read the worked answer
Vertical reaction at each lug (hook line over the C of G, 6 ft lug spacing):
- R_A = 5,400 x (3.5 / 6) = 3,150 lb
- R_B = 5,400 x (2.5 / 6) = 2,250 lb (sum checks to 5,400 lb)
Leg lengths (hook 9 ft above the lug line):
- L_A = sqrt(2.5^2 + 9^2) = sqrt(87.25) = 9.34 ft
- L_B = sqrt(3.5^2 + 9^2) = sqrt(93.25) = 9.66 ft
Tension = reaction x (leg length / hook height):
- T_A = 3,150 x 9.34 / 9 = 3,270 lb
- T_B = 2,250 x 9.66 / 9 = 2,415 lb
Check by horizontal components: 3,150 x 2.5/9 = 875 lb and 2,250 x 3.5/9 = 875 lb - they cancel, so the geometry is consistent.
Capacity check: choker hitch = 0.80 x 6,400 = 5,120 lb per leg. The heavily loaded leg is at 3,270 / 5,120 = 64 percent. The slings are adequate, provided the choker is on the sling body and not on the splice or a fitting, and the legs are identified and inspected.
2. Are the 1-1/2 ton screw-pin shackles acceptable at the lugs? Show the check and state what you would require instead.
Read the worked answer
The shackle at lug A carries the full leg tension, not the vertical share:
- Required: 3,270 lb
- 1-1/2 ton shackle WLL: 3,000 lb
3,270 > 3,000, so it is overloaded by about 9 percent and is not acceptable, even though the load is only 5,400 lb and the shackle "looks" big enough. The common error is comparing the shackle to the share of the load rather than to the sling tension.
Specify a 2 ton (4,000 lb WLL) minimum, then verify the hardware fits: pin diameter against the 7/8 in lug hole and jaw width against the 1 in plate (a pin too small in a large hole loads the lug in bending and can slip; a jaw too narrow will side-load the shackle). The load line must run through the shackle's centreline - a shackle side-loaded against the lug cheek is derated substantially. Also confirm the hook block and master link are rated for 5,400 lb plus rigging weight.
3. Because of a crane-access restriction, the crew proposes to choke the slings shorter so the hook sits only 6 ft above the lug line. What changes, and what is your recommendation?
Read the worked answer
Recompute at a 6 ft hook height:
- L_A = sqrt(2.5^2 + 36) = 6.50 ft; L_B = sqrt(3.5^2 + 36) = 6.95 ft
- T_A = 3,150 x 6.50 / 6 = 3,413 lb; T_B = 2,250 x 6.95 / 6 = 2,606 lb
Tension rises only about 4 percent (still well inside 5,120 lb), but the horizontal thrust on the lugs jumps from 875 lb to 3,150 x 2.5/6 = 1,313 lb - a 50 percent increase. That sideways pull tries to bend both lugs inward and loads their welds in a direction they were probably not designed for; it also makes the load harder to control on the swing.
Other consequences: the hook block is closer to the skid top and may foul it at pick-up; less headroom to clear the pad bolts; the slings will be choked tighter, so inspect for damage at the choke point.
Recommendation: keep the highest hook height the crane can give, or use a spreader bar sized for the 6 ft lug spacing. A spreader eliminates the horizontal thrust on the lugs entirely at any hook height, so it fixes the real problem rather than just accepting it.