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Spreader Beam Design
Date
Project:
Job No.
Safe Working Load, SWL Tons
Spreading Length metres
Safety Factor in Compression
Yield stress of the Beam material MPa
Allowable Tensile Stress MPa
Elastic Modulus of the material MPa
SWL, W = 1109.7 kN
10.000 metres
a = 60.0 Degrees 637.9 kN
y
x 318.9 kN
x
10.000 metres
W/2 = 552.4 kN W/2 = 552.4 kN
Considering the equillibrium of node , We have
1) the vertical component of resolved P is balanced with W/2, hence
P * Sin a = (W/2)
=> P * Sin a = 552.4
552.4
Sin(180-a)/2
Tensile Force in the tie, P = 637.9 kN
Tie Length = L =
112.6
150
Geometry of Lifting, Solved using the equations of Static Equillibrium
12/5/2013
(A) Design for Normal Stress (Direct Compressive Stress) **** Selecting the Section initially based on this ****
240
Belleli Energy srl, Dubai Branch
10.000
1.6
prepared by
DESIGN OF SPREADER LIFTING BEAM of Hollow Circular Cross Section
Spreading Length, L =
Compressive Force, C =
Tensile Force, P =
H itachi Zosen, Arzew Plant, Algeri a " BLOCK-E"
4776
R. Venkat
=> P
Hence, the STRUCTURAL analysis is made for (1) Direct Compressive Stress, and (2) Critical Load for Transverse Buckling.
210000
kN
As the Spreader beam is free for all its three planar DOF (x,y & RzDegrees Of Freedom) at the nodes of application of load, The bar behaves like
a TRUSS member and it will resist only the AXIAL force (here, Compression) and it will NOT resist BENDING in the plane.
This Spreader Beam is a typical case of Timoshenko's BEAM-COLUMN(Horizontal members having axial loads in addition to lateral loads)
with both the ends HINGED. The Elastic Instability in the lateral direction causes the Spreader beam to BUCKLE due to the SLENDERNESS.
This imposes the limitation on the Compressive load. The load at which the TRANSVERSE BUCKLING commenced is the CRITICAL load
(Pcr).
=
A B
C
B
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Spreader Beam Design
2) the horizontal component of resolved P induces a compressive force C in spreader beam, hence
P * Cos a = C
Compressive Force in Beam C = 318.9 kN
C * FOS
sallow
= 3401.91 mm2
8" Sch20
219.1
8" Sch20
as perAPI 5L
323.80 mm
6.35 mm
6332.85 mm
8.0E+07 mm4
112.26 mm
49.71 kg / m
Compressive Strength of the pipe selected = X
= sallow*AsCactual = 949.93 kN
Cactual
C
nc = 2.98 Nr.
For both ends HINGED members, the EFFECTIVE length equals the LENGTH of the member
L/r = 89.08 Nr.
2p2 Esy
Cc = 131.42 Nr.
Computing the factor, FS = (5/3) + (3/8) * [(L/r) / Cc] - (1/8) * [(L/r) / Cc]3
FS = 1.88 Nr.
Allowable Stress in intermediate buckling sallow(ib) = (sy/Fs) * [1 - (1/2) * {(L/r) / C c}2]sallow(ib) = 98.23 MPa
More than Euler's Critical Range
Calculations for the Intermediate-block, Pls. ignore for the Slender Range
Compressive Force in Newton X Safety Factor in Compression
Cross section area
OD of the Pipe
A
Moments of Inertia, Ixx=Iyy=I=
Allowable Stress
Allowable Stress in Mpa
Tk of the Pipe
SATISFACTORY
=CcThe Crippling commencement factor,
OK
*** Beam in the Intermediate-block, COMPRESSION & BUCKLING analyses needed ***
=
=Minimum Area of Cross section required in mm2
Load Factor is OK
Signal Box "SAFE
DESIGN"
Cross section Area, As
Radii of Gyration, rx=ry=r=
Safety Margin is OK
Buckling Stress is OPTIMUM
Safety Margin is OPTIMUM
12" Sch20
Minimum Area of Cross section required in mm2
=Slenderness Ratio (L/r)
OPTIMUM
(B) Design for Elastic Stability - Transverse Buckling[1]
OPTIMUM
Effective Length of the Spreader Beam
Radius of gyration
Buckling Stress is OPTIMUM
Load Factor is OPTIMUM
Buckling Stress is OK
Buckling Stress is OK
Critical Load is OK
Therefore the practical Safety Factor achieved =
The Pipe selected is Signal Box
"OPTIMALITY"
Unit Weight of the Pipe
*****CLICK HERE TO SELECT THE PIPE******
[The Standard pipe (API 5L) selected shall be atleast
with this cross section area and wall thickness is
minimum (i.e, maximum OD)]
**** CLICK ****
TO INCREASE THE
SECTION BY
SELECTING NEXT
PIPE SIZE
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Allowable Stress in slender buckling sallow(sb) = (p2* E) / [1.92 * (L/r)2]sallow(sb) = 136.04 MPa
Euler's Critical Bucling load = p2
* E * I) / (L)
2
Pcr = 1654.06 kN
ncr = 3.24 Nr.
[1]
Length L1 = mm Height h1 = mm
Radius of the bracket R 1 = mm Height h2= mmRadius of the reinforcement R2 = mm Thickness (Bkt & RF) t = mm
Dia of the hole for Shackle d = mm Weld joints' efficiency hj= %Dia of the Shackle ring ds = mm Weld Fillet Size sw= mm
Yield stress of the material syield= MPa Elastic Modulus of the material MPaAllowable Tensile Stress sallow= MPa
637.9 kN
3 * (t = 15 ) mm R2= 110 mm
d = 70 mm
R1= 125 mm h2= 100 mmb = 60.0 Degrees
h1 = 150 mm 318.9
L1= 400 mm
As the geometry reveals the criticality of the obligue tensile force from the tie is significant than the horizontal compressive force from the beam.
Hence, the design for the tensile stress ensures the design for the compressive stress also.
Tensile Force, P =
Comp.Force, C=
(a-v) Shearing stress in the shackle ring OK OPTIMUM
(b) Weld (set-on double fillet) size for eye-bracket with the beam OK OPTIMUM
248
155
210000
Design Criterion Signal Box "SAFE DESIGN" Signal Box "OPTIMALITY"
! Initially "assume" then "Iterate" these values with the help of following Signal Boxes
15
Calculations for the Slender Range, Pls. ignore for the Intermediate-block
OK
Check for the Ultimate Buckling load Pu, which is the Euler's Critical load Pcr
OK
110
150
100
400
This section is computed in accordance to theManual Of Steel Construction , 9th edition, American Institute of Steel Construction,
New Yark, 1959
125
('C) DESIGN OF ATTACHMENTS - (a) Design of eye bracket for strength
OPTIMUM
Euler's Critical Buckling LoadBuckling safety margin
OPTIMUM
=
10
90 70
OPTIMUM
(a-iv) Out-of plane buckling of the eye-bracket OK OPTIMUM
OK(a-i) Tensile stress in the eye-bracket OPTIMUM
OPTIMUM(a-ii) Bearing / Crushing stress in the eye-bracket OK
76
(a-iii) Tearing stress in the eye-bracket
OK
Actual Compressive Force on Beam X Safety Factor in Compression
OK
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Spreader Beam Design
ds= 76 mm
~ 2R1= 250 mm
3 * (t = 15 ) mm
Resisting area for tension of the eye braket = Diameteral Difference X Total Thickness
=
Resisting Area At = 7830 mm2
Normal Force P = 637.9 kN
Tensile Stress = Normal Force / Resisting Area
st = 81.5 MPaFactor of safety with yield stress n1 = 3.0 Nr.
ds= 76 mm
3 * (t = 15 ) mm
Bearing area of the Sling hole for the Sling ring = Diameter of the Sling ring X Total Thickness
= ds * 3(t)
Bearing Area Ab = 3420 mm
Normal Force P = 637.9 kN
Bearing Stress = Normal Force / Bearing Area
sbearing = 186.5 MPaFactor of safety with yield stress n2 = 1.3 Nr.
OK
(2R1- ds) * 3t
OPTIMUM
(a-ii) Design of eye bracket for bearing/crushing strength - FAILURE MODE -2 (80% of the yield stress is governing)
OPTIMUM
OK
(a-i) Design of eye bracket for tensile strength - FAILURE MODE - 1 (allowable tensile stress is governing)
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Spreader Beam Design
3 * (t = 15 ) mm
The bracket is in tearing due to shear along two planes against the sling ring, i.e., resisting it with the chordal sections
X Thickness
X Total Thickness
=
Tearing area At = 7200 mm
Tearing Force P = 637.9 kN
Tearing Stress = Tearing Force / Tearing Area
stearing = 88.6 MPaFactor of safety with yield stress n3 = 2.8 Nr.
Requirement is the minimum thickness of the eye-bracket shall be ensured for 13 mm and 0.25 times the hole diameter d.
Required thickness = 0.25 * d
treq = 22.5 mm
Factor of safety with thickness provided n4 = 2.0 Nr.
[2]
Chordal (assumed to be Radial) area resisting the
tearing shear
(a-iv) Design of eye bracket for out-of plane buckling - FAILURE MODE - 4 (as per David T. Ricker[2]
)
This section is computed in accordance to David T. Ricker, "Design and Construction of Lifting Beams ", Engineering Journal,
4th
Quarter, 1991
OK
OPTIMUM
Radii Difference
= 2 X+
Radii difference for the bracket
OPTIMUM
OK
(a-iii) Design of eye bracket for tearing strength - FAILURE MODE - 3 (50% of the yield stress is governing)
Radii difference for the reinfrmnt
2 * [(R1 - d/2) * t] + [(R2 - d/2) * (t*2)]
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Spreader Beam Design
3 * (t = 15 ) mm
The curved shackle ring is under double shear along two parallel planes of the faces of the bracket, i.e., resisting it with the cross section area
The cross section area of the curved sling ring =
Shearing area As = 9073 mm
Shearing Force P = 637.9 kN
Shearing Stress = Shearing Force / Shearing Area
sshearing = 70.3 MPaFactor of safety with yield stress n3 = 3.5 Nr.
637.9 kN
3 * (t = 15 ) mm R2= 110 mm
d = 70 mm
R1= 125 mm h2= 100 mmb = 60.0 Degrees
h1 = 150 mm 318.9
L1= 400 mm
OPTIMUM
(a-v) Design of shackle ring for shearing strength - FAILURE MODE - 5 (50% of the yield stress is governing)
Tensile Force, P =
(b) Design of weld joint of the eye-bracket with the spreader beam for shear strength (50% of the allowable stress is governing)
2 * [p/4 * ds2]
OK
Comp.Force, C=
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Spreader Beam Design
A) Shear Stress on the Weld joints between the eye-bracket and the beam (Set-on double fillet without any grooving)
Total length of the weld joint parallel to the beam axi = (2+2) * (L1+ R1)
Lw1 = 2100 mm
Transverse load on these joints = W/2
Pw1 = 552.4 kN
Allowable Shear Stress on effective throat area = sallow / 2)* hjtallow-w = 52.5 MPa
Effective throat thickness = Pw1* 1000 / (sw* Lw1)tw = 5.01 mm
Minimum Fillet Size of the Welds = 2 * tw
sw = 7.09 mm
Factor of safety with fillet size provided n4 = 1.4 Nr. OPTIMUM
OK
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Dubai Technical dept.
Spreader Beam Design
Date
Project:
Job No.
Safe Working Load, SWL kN
Spreading Length, L metres
Length, L1 metres
Initial Assumpn. Cant.lvr. Length, L2 metres
Yield stress of the Beam material MPa
Allowable Stress MPa
Elastic Modulus of the material MPa
UDL, q = 0.1 kN/m SWL, W = 552.9 kN
y
x
x
Lgth, L1= 1.998 metres
Lgth, L2 0.202 metres 4.000 metres
P = 276.2 kN P = 276.2 kN
Ensuring the Translational equillibrium along y axis, We have to equate the downward forces with upward reactions
=> W = (2 * P) + q * (L + L2)
W = 552.90 kN
Ensuring the Rotational equillibrium about z axis, We have to equate the clockwise moments with counter clockwise moments
Taking the moments about the node
(W * L1) + ((q * L2) * L2/2)) = (P * L) + ((q * L) * L/2)
L2 = (2/q) * [(P * L) + ((q * L) * L/2) - (W * L1)]
2 = 4.312 m
(A) STRENGTH DESIGN - Design for FLEXURAL Bending Stress **** Selecting the Section initially based on this ****
PLS. CHANGE THE INITIAL ASSUMPTION
210000
1.998
prepared by Venkat
248
0.202
4.000
After reaching Flexural Stress "safe", Iterate this
dimension L2sothat this agrees with computed L
2
155
Value brought from previous worksheet
Geometry of Lifting at the each end of the HOLLOW PIPE Spreader Beam, Solved using the equations of Static Equillibrium
Spreading Length, L =
Hence, the STRUCTURAL analysis is made for the FLEXURAL Stress
As this Adjustable Spreader beam is constrained for all its three planar DOF (x,y & RzDegrees Of Freedom) at the nodes of application of load,
The bar behaves like a FRAME member and it will resist BOTH the AXIAL force and BENDING in the x-y plane.
This Spreader Beam is a typical case of Both the ends fixed with a Cantilever for Counter-weight, having a point load at an offset and
UDL for the entire length. The governing stress for such a configuration is the FLEXURAL STRESS (sb).
DESIGN OF ADJUSTABLE SPREADER LIFTING BEAM of Standard Profile "HEB Series"
H itachi Zosen, Arzew Plant, Algeri a " BLOCK-E"
4776
552.4
12/5/2013Belleli Energy srl, Dubai Branch
A BD
C
C
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Spreader Beam Design
Bending Moment = (P * L) + ((q * L) * L/2)
M = 1105.760 kN-m
= M / Zx
sf(max) = 21264.6 MPaFactor of Safety achieved on yield stress = syield/sf(max)
n1 = 0.01 Nr.
as per EN 53-62
12.20 kg/m
2370000 mm
921000 mm
38.90 mm
24.30 mm
52000 mm
18400 mm
91 mm
100 mm
dmax = 1481.222 mm
Depth of the section, h =
Width of the section, b =
Elastic Section Modulus, Zx=
Elastic Section Modulus, Zy=
(B) STIFFNESS DESIGN - Design for DEFLECTION
Structural member section
=W * L1* (L
2- L12)3/2
Radius of Gyration, rx=
Moments of Inertia, Ixx=
Unit Weight of the member
Bending Stress is OPTIMUM
9 * 31/2
* L * E * I
The maximum deflection of the beam between
loaded nodes and
Radius of Gyration, ry=
Signal Block
"SAFE DESIGN"
Flexural Stress is NOT OK
Counter Weight is NOT OK
Moments of Inertia, Iyy=
PLS. INCREASE THE SECTION
Max. deflection is NOT OK
Signal Block
"OPTIMALITY"
We have, Maximum Flexural Stress
OPTIMUM
PLS. INCREASE THE SECTION
HE 100 AA
Section Modulus of the section about the
axis perpendicular to plane of bending/Bending Moment=
*********** CLICK HERE **********
TO SELECT THE SECTION
[The Standard Section (EN 53-62) selected
shall be atleast with this Moment of Inertia]
**** CLICK ****
TO INCREASE THE
SECTION BY
SELECTING NEXT
SECTION
C B
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Length of the bracket Lb = mm Radius of the lug end R = mm
Clearance above the beam c= mm Diameter of the hole dh = mm
Total Height of the lug h1 = mm Diameter of the pin dp = mm
Height of the cut in the lug h2 = mm Thickness of the lug t1 = mm
Height of taper in the lug h3 = mm Thk of all other plates t2 = mm
Total Width of the lug w1 = mm Weld joints' efficiency hj= %Width of the cut in the lug w2 = mm Weld Fillet Size sw= mm
Yield stress (all matl ex. pin) syield= MPa Elastic Modulus of the material MPaAllowable Stress sallow= MPa Yield stress (pin matl) sy(pin)= MPa
W = 552.9 kN
w1= 350 R = 120 t1= 30
dp= 50
w2= 120 dh= 60
h3= 145
h2= 50
h1= 300
(h + c) = 131
= =
Lb= 200
t2= 10 b = 100
The load on the bracket is the straight forward lifting force acting vertically upwards against the load.
300
(c) Tensile stress in the end plate OK OPTIMUM
(d) Weld (set-on double fillet) size for the bottom plate with other plates OK OPTIMUM
(C) DESIGN OF ATTACHMENTS - (a) Design of adjustable (CG location variations) suspension bracket for strength
200
40
50
120
60
50
30
300
145
350
120
10
70
10
! Initially "assume" then "Iterate" these values with the help of following Signal Boxes
Design Criterion Signal Box "SAFE DESIGN" Signal Box "OPTIMALITY"
248 210000155
(a-i) Tensile stress in the lug OK NOT OPTIMUM
(a-ii) Bearing / Crushing stress in the lug OK OPTIMUM
(a-iii) Tearing stress in the lug OK OPTIMUM
(a-iv) Out-of plane buckling of the lug OK OPTIMUM
(a-v) Shearing stress in the pin OK OPTIMUM
(b) Weld (set-on double fillet) size for lug with other plates OK OPTIMUM
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Spreader Beam Desig
wcs= 309 W = 552.9 kN
R = 120 t1= 30
dh= 60
t1= 30
h3= 145
h1= 300
w1= 350 h2= 50
h3- R
tan sin-1
+ tan-1
(w1/ 2)
wcs = 309 mm
= Width at the critical section - Diameter of the hole
= wcs- dh
wef = 249 mm
= wef* t1
Aef = 7480 mm
As these two lugs are placed, with the lifting lug in-between, at a closer clearance always, the bending effects on the lug and the pin are ignored.
And also, it could be reasonably assumed that the entire force of lifting is shared equally without any moments on them.
The tensile stress on this critical section = Force on the lug / Effective normal area
= (W/2) / Aef
st = 36.96 MPaFactor of Safety achieved on yield stress = syield/st
n2 = 6.71 Nr.
= Effective width
OK
NOT OPTIMUM
The effective width at the critical section
X ThicknessThe effective normal area at the critical section
resisting the force per lug
(a-i) Design of lugs on the bracket for tensile strength - FAILURE MODE - 1 (allowable tensile stress is governing)
h3- R
R
(h3- R)2+ (w1/ 2)
2
The width of the critical section i.e., across the
diameter of the hole= w1- 2 *
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Dubai Technical dept.
Spreader Beam Design
W = 552.9 kN
wb= 50 t1= 30
t1= 30 dp= 50
The effective area bearing the crushing force per lug = Bearing width X Thickness
= wb* t1 ( Note: Bearing width equals the projected diameter
Ab = 1500 mm2
The bearing / crushing stress = Crushing force / Bearing area
= (W/2) / Ab
sb = 184.30 MPaFactor of Safety achieved on yield stress = syield/sb
n3 = 1.35 Nr.
W = 552.9 kN
t1= 30
t1= 30
(R - dh/2)= 90
Area resisting lateral tension (tearing) per lug = Radii difference X Thickness
=
At = 2700 mm
(R - dh/2) * t1
OK
OPTIMUM
(a-iii) Design of lug on the bracket for tearing strength - FAILURE MODE - 3 (50% of the yield stress is governing)
(a-ii) Design of lugs on the bracket for bearing/crushing strength - FAILURE MODE -2 (80% of the yield stress is governing)
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Spreader Beam Design
Tearing stress = Tearing force / Area resisting lateral tension (tearing)
= (W/2) / At ( Note: Tearing force conservatively equals lifiting force
stear = 102.39 MPaFactor of Safety achieved on yield stress = syield/sbtear
n4 = 2.42 Nr.
Requirement is the minimum thickness of the lug on the bracket shall be ensured for 13 mm and 0.25 times the hole diameter d.
Required thickness = 0.25 * dh
treq = 15.0 mm
Factor of safety with thickness provided n5 = 2.0 Nr.
[1]
W = 552.9 kN
t1= 30
dp = 50
The pin under double shear along two parallel planes of the inner faces of the lugs, i.e., resisting it with the cross section area
The cross section area of the pin = 2 * [p/4 * dp2]
Shearing area As = 3927 mm
Shearing Force P = 552.9 kN
Shearing Stress = Shearing Force / Shearing Area
sshearing = 140.8 MPaFactor of safety with yield stress n6 = 2.1 Nr.
OK
OPTIMUM
(a-v) Design of pin for shearing strength - FAILURE MODE - 5 (50% of the yield stress of the pin is governing)
(a-iv) Design of lug on the bracket for out-of plane buckling - FAILURE MODE - 4 (as per David T. Ricker[1]
)
OK
This section is computed in accordance to David T. Ricker, " Design and Construction of Lifting Beams ", Engineering Journal,
4thQuarter, 1991
OK
OPTIMUM
OPTIMUM
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Spreader Beam Design
W = 552.9 kN
w1= 350 t1= 30
w2= 120
h3= 145
h4= 116
h1= 300
h2= 50
t2= 10 b = 100
Fixing the height h4at 75% of the straight height (h1- h3)
h4 = 116 mm
Total weld-length provided per lug = 2 * [2 * (h2+ h4+ t2+ h4) + b]
Lw1 = 1370 mm
Transverse force on the weld joint per lug = W/2
Pw1 = 276.5 kN
Allowable Shear Stress on effective throat area = (sallow / 2)*hj
tallow-w = 54.25 MPaEffective throat thickness = Pw1* 1000 / (sw* Lw1)
tw1 = 3.72 mm
Minimum Fillet Size of the Welds = 2 * tw
sw1 = 5.26 mm
Factor of safety with fillet size provided n7
= 1.9 Nr.
(b) Design of weld joint of the lug with other plates for shear strength (50% of allowable stress is governing)
OK
OPTIMUM
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Spreader Beam Design
W = 552.9 kN
= =
Lb= 200
t2= 10
The load on the end plates of the bracket is the straight forward lifting force acting vertically against the load.
The normal area per end plate resisting tensile force = Length of the bracket X Thickness
= Lb* t2
Aep = 2000 mm2
As these two end plates are fabricated as box and the thickness is sufficiently large, the membrane effects and bending effects are ignored.
And also, it could be reasonably assumed that the entire force of lifting is shared equally without any moments on them.
The tensile stress on the cross section = Force on the end plate / Normal area
= (W/2) / Aef
st(ep) = 138.23 MPaFactor of Safety achieved on yield stress = syield/st(ep)
n8 = 1.79 Nr.
(c) Design of end plates of the adjustable bracket for tensile strength (allowable tensile stress is governing)
OK
OPTIMUM
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Spreader Beam Design
W = 552.9 kN
h5= 67
= =
Lb= 200
b = 100
Fixing the height h5at one-third of length of the bracket, Lb
h5 = 67 mm
Total weld-length provided = Directly for bottom plate X Indirectly for ribs supporitng bottom plate
Total weld-length provided = 2 * (Lb+ b) + 4 *4* h5
Lw2 = 1567 mm
Transverse force on the weld joint = W
Pw2 = 552.9 kN
Allowable Shear Stress on effective throat area = (sallow / 2)*hj
tallow-w = 54.25 MPaEffective throat thickness = Pw1* 1000 / (sw* Lw1)
tw2 = 6.51 mm
Minimum Fillet Size of the Welds = 2 * tw
sw2 = 9.20 mm
Factor of safety with fillet size provided n9 = 1.1 Nr.
OK
OPTIMUM
(d) Design of weld joint for the bottom plate of the adjustable bracket for shear strength (50% of allowable tensile stress is governing)
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Spreader Beam Design
dp= 48
t1= 15
t1= 15
wb= 48 W = 276.2 kN
The effective area bearing the crushing force per lug = Bearing width X Thickness
= wb* t1 ( Note: Bearing width equals the projected diameter
Ab = 720 mm2
The bearing / crushing stress = Crushing force / Bearing area
= (W/2) / Ab
sb = 191.81 MPaFactor of Safety achieved on yield stress = syield/sb
n3 = 1.29 Nr.
(R - dh/2) = 90
t1= 15
t1= 15
W = 276.2 kN
Area resisting lateral tension (tearing) per lug = Radii difference X Thickness
=
At = 1350 mm2
(R - dh/2) * t1
(a-ii) Design of lugs on the bracket for bearing/crushing strength - FAILURE MODE -2 (80% of the yield stress is governing)
OK
OPTIMUM
(a-iii) Design of lug on the bracket for tearing strength - FAILURE MODE - 3 (50% of the yield stress is governing)
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Dubai Technical dept.
Spreader Beam Design
Tearing stress = Tearing force / Area resisting lateral tension (tearing)
= (W/2) / At ( Note: Tearing force conservatively equals lifiting force
stear = 102.30 MPaFactor of Safety achieved on yield stress = syield/sbtear
n4 = 2.42 Nr.
Requirement is the minimum thickness of the lug on the bracket shall be ensured for 13 mm and 0.25 times the hole diameter d.
Required thickness = 0.25 * dh
treq = 15.0 mm
Factor of safety with thickness provided n5 = 1.0 Nr.
[1]
dp = 48
t1= 15
W = 276.2 kN
The pin under double shear along two parallel planes of the inner faces of the lugs, i.e., resisting it with the cross section area
The cross section area of the pin = 2 * [p/4 * dp2]
Shearing area As = 3619 mm2
Shearing Force P = 276.2 kN
Shearing Stress = Shearing Force / Shearing Area
sshearing = 76.3 MPaFactor of safety with yield stress n6 = 3.9 Nr.
4th
Quarter, 1991
(a-v) Design of pin for shearing strength - FAILURE MODE - 5 (50% of the yield stress of the pin is governing)
OK
OPTIMUM
OK
OPTIMUM
(a-iv) Design of lug on the bracket for out-of plane buckling - FAILURE MODE - 4 (as per David T. Ricker[1]
)
OK
OPTIMUM
Prepared By: R.Venkat 20 of 23 Printed on: 12/5/2013
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Dubai Technical dept.
Spreader Beam Design
t2= 8
b = 100
h2= 50
h4= 38
h1= 200
h3= 150
w2= 116
w1= 350
t1= 15
W = 276.2 kN
Fixing the height h4at 75% of the straight height (h1- h3)
h4 = 38 mm
Total weld-length provided per lug = 2 * [2 * (h2+ h4+ t2+ h4) + b]
Lw1 = 732 mm
Transverse force on the weld joint per lug = W/2
Pw1 = 138.1 kN
Allowable Shear Stress on effective throat area = (sallow / 2)*hj
tallow-w = 54.25 MPaEffective throat thickness = Pw1* 1000 / (sw* Lw1)
tw1 = 3.48 mm
Minimum Fillet Size of the Welds = 2 * tw
sw1 = 4.92 mm
Factor of safety with fillet size provided n7 = 1.6 Nr. OPTIMUM
(b) Design of weld joint of the lug with other plates for shear strength (50% of allowable stress is governing)
OK
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Dubai Technical dept.
Spreader Beam Design
t2= 8
Lb= 150
= =
W = 276.2 kN
The load on the end plates of the bracket is the straight forward lifting force acting vertically against the load.
The normal area per end plate resisting tensile force = Length of the bracket X Thickness
= Lb* t2
Aep = 1200 mm
As these two end plates are fabricated as box and the thickness is sufficiently large, the membrane effects and bending effects are ignored.
And also, it could be reasonably assumed that the entire force of lifting is shared equally without any moments on them.
The tensile stress on the cross section = Force on the end plate / Normal area
= (W/2) / Aef
st(ep) = 115.08 MPaFactor of Safety achieved on yield stress = syield/st(ep)
n8 = 2.15 Nr.
(c) Design of end plates of the adjustable bracket for tensile strength (allowable tensile stress is governing)
OK
OPTIMUM
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Dubai Technical dept.
Spreader Beam Design
b = 100
Lb= 150
= =
h5= 50
W = 276.2 kN
Fixing the height h5at one-third of length of the bracket, Lb
h5 = 50 mm
Total weld-length provided = Directly for bottom plate X Indirectly for ribs supporitng bottom plate
Total weld-length provided = 2 * (Lb+ b) + 2*4 * h5]
Lw2 = 800 mm
Transverse force on the weld joint = W
Pw2 = 276.2 kN
Allowable Shear Stress on effective throat area = (sallow / 2)*hj
tallow-w = 54.25 MPaEffective throat thickness = Pw1* 1000 / (sw* Lw1)
tw2 = 6.36 mm
Minimum Fillet Size of the Welds = 2 * tw
sw2 = 9.00 mm
Factor of safety with fillet size provided n9 = 0.9 Nr.
***** Design procedure for the adjustable (Span variations) suspension bracket for strength shall be followed but for the additional
consideration of welding with the adjustable cross beam *****
*** IMPORTANT NOTES ***
1) All the basic assumptions about material properties and their linear beaviour, as made in elementary STRENGTH OF MATERIAL and
ELASTICITY THEORY will hold good
2) The adjustments for the load carrying brackets considered throughout this computation are of small quantities compared to the span of the
(d) Design of weld joint for the top plate of the adjustable bracket for shear strength (allowable tensile stress is governing)
! NOT OK, INCREASE THE FILLET SIZE
OPTIMUM
(E) DESIGN OF ATTACHMENTS - (a) Design of fixed suspension bracket for strength