#754 v-3103 Spreader & Lifting 60 Degree
-
Upload
hafizi-zecky -
Category
Documents
-
view
93 -
download
0
description
Transcript of #754 v-3103 Spreader & Lifting 60 Degree
LIFTING DESIGN Unit: E -21221 SPREADER PIPE DESIGN Reference Drawing: D018-DWG-011A Lifting Arrangement DrawingPROPERTIES OF SPREADER
Type of spreader Size Pipe material Minimum specified yield strength, Modulus of elasticity, Pipe OD, Pipe thickness, Pipe ID, Cross sectional area, Moment of inertia, Setion modulus, Radius of gyration, Length of pipe, Effective end factor, Slenderness ratio Allowable compressive stress - AISC (for 0 < Kl/r < Cc)
: Pipe : 4" Sch.40 : A 106 GR B Fy = 241.32 N/mm2 E= 200000 N/mm2 D= 114.30 mm t= 5.30 mm d= 103.70 mm A= 1814.90 mm2 I = 2.70E+06 mm4 Z = 4.72E+04 mm3 r= 38.55 mm l= 3000 mm K= 1.0 K*l/r = 77.82
[1 - (K*l/r)/(2*Cc)]*Fy Fc = ---------------------------------------------- = 5/3+3(K*l/r)/(8*Cc)-(K*l/r)^3/(8*Cc^3) where,COMPRESSIVE STRESS IN PIPE
105.35 N/mm2
Cc = \/ (2*pi^2*E/Fy ) =
127.90
Sling angle to horizontal, Whole Structural Weight Design Lifting Factor Design sling force, Compressive force on pipe, Eccentric moment, Eccentric Compressive stress due to Fp, Bending stress due to Me, Stress interaction ratio, where,
= : : Fs = Fp = Fs*Cos = Me = e= Sp = Fp / A = Sb = Me / Z = Sp/Fc +Cm*Sb/[(1-Sp/Fe')*0.6*Fy] = Fe' = 12*pi^2*E/(23*(K*l/r)^2) = Cm =
60 4032 2.00 22829 11415 1793789 157.15 6.29 38.01 0.332
Deg. kg N N Nmm N/mm2 N/mm2 < 1.00 OK!
170 N/mm2 1
LIFTING LUG DESIGN AT COLUMNYZ Y
Material Minimum specified yield strength, Structural Lifting Weight Design Lift Impact Factor PAD Eye Impact Factor Lift Angle Vertical Force at Lug 9.81 x I x W/4 Sling Force at Lug Fv/Cos b Longitudinal Force at Lug Fv x Tanb Skew Force 0.05 x FsFv Fs b
: Sy = W I Ieb = = = = =
A 36 248.21 N/mm24032 kg 2 2 30 Deg
Fv
19771 N
Fs
=
22829 N
Fx
=
11415 N
Fz
=
1141 NFv
Fz Fx B Radius R C Diameter D T1
A
B
C
A
T2 H E E
L1
Lug Dimensions R D L1 H T1 T2 = = = = = = 50 28 150 65 16 16 mm mm mm mm mm mm
Pin Diameter of Shackle used for Lifting
Dp
= 25.4 mm (6.5 Tonne Shackle)
STRESSES IN LIFTING LUG Tensile Stress across A-A Fs/((2 x R - D) x T1)
ft
=
19.82 N/mm^2
SATISFACTORYShear Stress across B-B & C-C Fs/(2 x ((R^2)-(D/2)^2)^0.5 x T1)