45 0 202KB
DESIGN OF PADEYE AS PER AISC 360-2005 Section Properties
Py
P 20 THK
a
Px
b
w
Y X
Height of the padeye Height of the eye from top of the beam,h Width of the padeye, w Thickness of the padeye Radius of the padeye Hole diameter Shackle diameter Cross-sectional area Padeye Cross-sectional area Ag Material Properties Yield Strength Ultimate tensile strength
Fy Fu
STAAD input Maximum sling force Sling angle X component of P Px Y component of P Py Code checks Check for Dimensions a b beff w
= = = = = = =
= Selected grade of padeye = =
P
= = = =
= = =
122.5 mm 122.5 mm 2t+16
=
56 mm
=
280 120 320 20 160 75 50
320
mm mm mm mm mm mm mm
6400 mm2 S 275 275 MPa 400 MPa
244.84 66.38 98.099768 224.32802
kN degrees kN kN
Condition a>1.33beff
Fig. C-D5.1 O.K
beff2beff+d
O.K
a) Tensile rupture on the net effective area
Clause
Pn=2×t×beff ×Fu Фt Design strength
=
896 kN
=
0.75 [LRFD] Фt*Pn
=
= 672 kN Pass ratio = 0.36 SELECTED SECTION IS SAFE IN TENSILE RUPTURE
b) For shear rupture on the effective area
Clause Asf
Pn=0 . 6× Fu × Asf = Design strength
1416 kN Фt*Pn
= = =
Pass ratio
= =
1062 kN 0.23
SELECTED SECTION IS SAFE IN SHEAR
c) For bearing on the projected area of the pin Apb
Rn=1 . 8×Fy× Apb = Design strength
495 kN Фt*Pn
=
= = Apb - Projected bearing area
= 371.25 kN Pass ratio = 0.66 SELECTED SECTION IS SAFE IN BEARING
d) For tensile yielding in the gross section
Clause
Pn= Fy × Ag =
Фt
1435.6993350048 kN
=
0.9 [LRFD]
Фt*Pn = 1292.1294015043 kN Pass ratio = 0.19 SELECTED SECTION IS SAFE IN YIELDING
Design strength
=
e)Check for Bending stress Moment developed ,M =Px*h
=
11771972 Nmm
Moment of Inertia,Ixx y Section Modulus,Z = Ixx/y Bending Stress
= = =
fb = M ÷ Z =
Allowable Bending Stress Fb fb
= = < Fb
86603300 mm4 160 mm 541270.63 mm3
21.748773 N/mm2 0.66 * Fy 181.5 N/mm2
SELECTED SECTION IS SAFE IN BENDING Check for Weld size Thickness of weld Moment of Inertia,Ixx y Section Modulus, Z =Ixx/y Moment , M M/Z Area,A P/A P/A + M/Z
= = = = = = = = =
1 11855000 160 74093.75 11771972 158.87942 1340 167.40897 326.28839
mm mm4 mm3 Nmm mm2 N/mm X
Let required size of weld be S Allowable stress in weld
=
100 N/mm2
S×0. 707×100=( P÷ A )+( M ÷Z ) Reqd.Size S = Selected Size =
4.62 mm 6 mm
>S
HENCE O.K
Fig. C-D5.1
D5-1
D5-2 2t(a+d/2) 5900
Section J7
dpin x t 1000 ected bearing area
D2-1
Weld 20
Stiffener plates 10 85
60 X 320
10
Padeye