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Method notes / Liftrix: Rigging and lugs

Method

Tail end: twin lugs, the tail beam and the skirt base ring

How two offset tailing lugs share the tail load and are resolved into their own planes, the Moss 10-5 tail beam check, and the base ring check with Table 10-7 in closed form.

Twin tailing lugs

Two plate lugs sit at ±β about the lift plane, each carrying half the tail load (T/2). They apply to a plate lug pointing up or down, not to a trunnion or a lug to the side. With a spreader the slings hang vertical. With slings converging to one hook at α from vertical, each lug sees a vertical share T/2, a pull T/2·tan α toward the lift plane (the two cancel between the lugs) and a sling load (T/2)/cos α. The side allowance S is added on top, in the sense that makes the across-plate load worse; the plate checks run both senses and keep the worse value per check.

Because each plate stands at β to the vertical, the load is resolved into its own plane. With X = ±T/2 (up or down) and Y = C + s·σ·S, where C is the convergence, s = ±1 the sense of S and σ the sign of X·sin β + C·cos β:

normal = |X·cos β − Y·sin β|, across = |X·sin β + Y·cos β|

The tail pin is then (R_o + h)·cos β from the axis, not R_o + h, and the tail offset in the statics should be set to that. The mesher also needs the two lugs to stay clear: the arc between centres, 2β·R_m, must hold the footprint of the pad (or section) plus six line spacings. The panel reports the minimum half-angle; a half-angle of 15° is refused for the default geometry.

Twin lugs, β = 30°, T/2 = 100 kN, side 10 kN, vertical slings, pointing down

X, YX = −100, C = 0, σ = −1; sense +1 gives Y = −10
Sense +1normal = |−86.60 + 5.00| = 81.6 kN; across = |−50.00 − 8.66| = 58.7 kN
Sense −1normal = 91.6 kN; across = 41.3 kN
Clearance, 2400 × 20 shell, 700 padR_m = 1210; λ = √(1210·20) = 155.6; h_line = 18.7 mm; footprint = 710 mm; need = 710 + 6·18.7 = 822 mm; minimum β = 822/(2·1210) rad = 19.5°

The check lists keep the worse of the two senses for each line.

Tail beam (Moss 10-5)

An alternative tail attachment is a beam across the base under the skirt, with the pin beyond the skirt at x from the beam's bolted face and y along the beam. On the factored tail load T at angle θ, the load along the vessel axis is T·sin θ and along the beam is T·cos θ.

M_b = x·T·cos θ + y·T·sin θ, σ_b = M_b/Z ≤ 0.66·F_y, f_t = M_b/y1, f_c = T·sin θ + f_ty1 bolt-group lever arm; f_c is spread over the skirt crippling length l_n = N + 2·l_r: σ_c = f_c/(t_sk·l_n) ≤ 0.66F_y. Bolt shear 0.5·T·cos θ/(n·A_b); bolt tension f_t/(n·A_b); skirt weld τ = T·cos θ/(π·D_sk·0.707·w4) against 108.9 MPa.

Default beam at 30°, T = 446.8 kN (W_L)

LoadsT·sin θ = 223.4 kN, T·cos θ = 386.9 kN; x = 150, y = 500 mm, Z = 900·10³ mm³
BendingM_b = 150·386940 + 500·223400 = 169.7·10⁶ N·mm; σ_b = 188.6 MPa against 171.6: ratio 1.10, over
Bolts4 bolts, 561 mm² each; shear 86.2 MPa against 68.9: ratio 1.25, over; tension 31.3 against 137.9
Skirtl_n = 25 + 2·16·20 = 665 mm; f_c = 293.5 kN; σ_c = 22.1 MPa

These are the panel's default beam values with F_y 260 MPa: they fail, which shows the checks respond. They are not a recommendation.

Skirt base ring (Moss 10-5, Table 10-7)

The tail lug pulls the skirt's base ring radially. The code models Moss's one-point case: a ring of radius R_B under a radial point load T, reacted by the skirt's tangential shear flow, with α the angle from the load. The table's coefficients are used in closed form:

K_r(α) = [2 + cos α − 2(π − α)·sin α] / (4π), K_T(α) = −[3·cos α + 2(π − α)·sin α] / (4π)
M = K_r·T·R_B, T_t = K_T·T, σ = |M|·c/I + |T_t|/AR_B is the neutral-axis radius of the ring section (plate, plus the skirt leg if it lies within the ring width), c the larger distance to a face.

It is evaluated every 0.5° from 0 to 180° and enveloped, since the vessel set down reverses the load. The checks are |M|·c/I ≤ 0.66·F_y and f_a/(0.6F_y) + f_b/(0.66F_y) ≤ 1 at the worst α. The test file records that the printed K_r at 125° and 130° in Table 10-7 appear transposed against the closed form.

Ring 300 × 50, inside diameter 2120, T = 446.8 kN, no skirt leg

SectionA = 15000 mm²; C1 = C2 = 150 mm; I = 50·300³/12 = 112.5·10⁶ mm⁴; R_B = 1060 + 150 = 1210 mm
At α = 0K_r = 3/(4π) = 0.2387; M = 0.2387·446800·1210 = 129.1·10⁶ N·mm
Bendingσ = 129.1·10⁶·150/112.5·10⁶ = 172.1 MPa against 0.66·260 = 171.6: ratio 1.00
Interactionf_a = 106.7 kN/15000 = 7.1 MPa; 7.1/156 + 172.1/171.6 = 1.05

|K_r| is largest at α = 0 for this load case (about −0.09 at 90°).

Limits

  • One-point stiffening is the unstiffened ring. Struts or a tail beam across the base only lower M, so the result is an upper bound for them.
  • The whole tail load T acts at the lug, for twin lugs as well; no share to each lug is taken in the ring.
  • The ring is a hand check. It is not in the FE mesh.

Related notes

Next: The FE model of the lug on the shell: what it holds and what it leaves out