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Padeye Calculation

This document summarizes the design of a padeye according to AISC 360-2005 standards. It provides the dimensions and material properties of the padeye. It then describes performing checks to ensure the padeye is safe based on its tensile rupture strength, shear rupture strength, bearing strength, tensile yielding strength, and bending stress. All checks indicate the selected padeye section is safe based on the given loading conditions and material properties.

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Karun Das
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3K views6 pages

Padeye Calculation

This document summarizes the design of a padeye according to AISC 360-2005 standards. It provides the dimensions and material properties of the padeye. It then describes performing checks to ensure the padeye is safe based on its tensile rupture strength, shear rupture strength, bearing strength, tensile yielding strength, and bending stress. All checks indicate the selected padeye section is safe based on the given loading conditions and material properties.

Uploaded by

Karun Das
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLS, PDF, TXT or read online on Scribd
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DESIGN OF PADEYE AS PER AISC 360-2005

Section Properties

Py P

20 THK

a
b Px

w
Y

Height of the padeye = 280 mm


Height of the eye from top of the beam,h = 120 mm
Width of the padeye, w = 320 mm
Thickness of the padeye = 20 mm
Radius of the padeye = 160 mm
Hole diameter = 75 mm
Shackle diameter = 50 mm
Cross-sectional area
Padeye
Cross-sectional area Ag = 6400 mm2

Material Properties Selected grade of padeye S 275


Yield Strength Fy = 275 MPa
Ultimate tensile strength Fu = 400 MPa

STAAD input
Maximum sling force P = 244.84 kN
Sling angle = 66.38 degrees
X component of P Px = 98.099768 kN
Y component of P Py = 224.32802 kN

Code checks
Check for Dimensions Condition Fig. C-D5.1
a = 122.5 mm a>1.33beff O.K
b = 122.5 mm
beff = 2t+16
= 56 mm beff<b O.K
w = 320 w>2beff+d O.K
a) Tensile rupture on the net effective area Clause

Pn=2tbeff Fu
= 896 kN
t = 0.75 [LRFD]

Design strength = t*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 =
= 1416 kN
Design strength = t*Pn
= 1062 kN
Pass ratio = 0.23

SELECTED SECTION IS SAFE IN SHEAR

c) For bearing on the projected area of the pin

Rn=1 . 8Fy Apb Apb =


= 495 kN =
Design strength = t*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
= 1435.6993350048 kN
t = 0.9 [LRFD]
Design strength = t*Pn
= 1292.1294015043 kN
Pass ratio = 0.19
SELECTED SECTION IS SAFE IN YIELDING

e)Check for Bending stress

Moment developed ,M =Px*h = 11771972 Nmm


Moment of Inertia,Ixx = 86603300 mm4
y = 160 mm
Section Modulus,Z = Ixx/y = 541270.63 mm3

Bending Stress fb = M Z
= 21.748773 N/mm2
Allowable Bending Stress Fb = 0.66 * Fy
= 181.5 N/mm2
fb < Fb
SELECTED SECTION IS SAFE IN BENDING

Check for Weld size


Thickness of weld = 1 mm
Moment of Inertia,Ixx = 11855000 mm4
y = 160
Section Modulus, Z =Ixx/y = 74093.75 mm3
Moment , M = 11771972 Nmm
M/Z = 158.87942
Area,A = 1340 mm2
P/A = 167.40897
P/A + M/Z = 326.28839 N/mm
X
Let required size of weld be S
Allowable stress in weld = 100 N/mm2

S0. 707100=( P A )+( M Z )


Reqd.Size S = 4.62 mm
Selected Size = 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

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