Padeye Design Cheek [PDF]

Zitiervorschau

Discipline: Calc'd by:

TITLE: LOCATION:

DESIGN OF PADEYE USING CHEEK PLATES

Reference:

Rev C1

Material Properties: Yield strength of plate Material performance factor

Fy phi

275 MPa 0.95

Loads (using factors for Lift Points): Dynamic Amplification Factor Skew Load Factor Contingency Factor Load applied to padeye Maximum angle from vertical Factored Vertical Load Factored Horizontal Load

DAF SLF CF SWL Alpha Pf Hf

1.1 1.4 1 423 60 548 27

Padeye Geometry: Diameter of Pin Diameter of Hole Diameter of Cheek Pl. Thickness of Cheek Pl. Radius of Main Plate Height to Center of Hole Length of Main Plate Thickness of Main Plate a) Capacity at Base of the Padeye Case 1: Load acts Vertically Factored Moment at Base Moment Resistance at Base Tension Resistance at Base Interaction

c) Pullout Capacity in Main Plate Length in tension (1/4 cheek) Length in shear Area for pull out Direct Tension Resistance d) Bearing Bearing Capacity

File: M:\btucker\Padeye-Design-Cheek-xls.xls

kN degrees kN kN

= (1.05 + 1.35 * (DAF*SLF*CF-1))*SWL = 0.05 * Pf (see Reference) Dr Rm

Dp Dh Dc Tc Rm Hh Lp Tp

Mf Mr Pr Mf/Mr Pf/Pr

Case 2: Load acts 60 degrees from Vertical Shear at Base Vfy Shear Resistance at Base Vry Tension at Base Pf_2 Moment in Plane of Plate Mfy Moment Resistance Mry Interaction Mf/Mr Pf_2/Pr 0.455 Vfy/Vry 0.727 Mfy/Mry b) Shear Pullout Capacity of Ring Tear out height Area subject to shear Shear strength Shear Resistance

(1.1 for onshore lift, 1.2 for offshore) (generally 1.4) (generally 1.35)

45 72 120 6 79 77 250 50

mm mm mm mm mm mm mm mm

3.1 40.8 3266 0.08 0.17 0.24

kN.m kN.m kN

475 2155 274 54 204 0.08 0.08 0.10 0.19 0.45

kN kN kN kNm kNm

CHEEK PL. MAIN PLATE

Hh

Dh

Lp

= Hf * (Hh+Dh/2)/1000 = phi * (Lp * Tp^2/4) * Fy * 1e-6 = phi* Lp * Tp * Fy * 1e-3

< 1.0 OK =Pf*sin(alpha*pi()/180) = phi * Lp * Tp * 0.66 Fy * 1e-3 = Pf*cos(alpha*pi()/180) = Vfy * (Hh+Dh/2)/1000 = phi * (Tp*Lp^2/4) *Fy * 0.000001

(Interaction from 13.4.6 of S16.1) < 1.0 OK

hv 43 mm 2 Aw 6316 mm Fs 182 MPa Vr 1089 kN Vr >= Pf OK

= Rm-Dh/2

Lt 94 Ls 197 At 11214 TFr 2930 Tfr >= Pf OK

= PI() * Dc * 1/4 = PI() * Dc * 1/4 + 2*SQRT(Rm^2-(Dc/2)^2) = Tp * (Lt + 0.66*Ls) = phi*At*Fy * 1e-3

Br Br >= Pf OK

mm mm mm2 kN

729 kN

= 2* (hv * Tp + 2*(Dc-Dh/2)*Tc) = 0.66*Fy = phi*Aw*Fs * 1e-3

= phi*Fy*Dp*(Tp+2*Tc)/1000 Page 1 of 1 Printed: 16/06/2022 at 12:47