Spreader Beam BLOCK E
Short Description
Download Spreader Beam BLOCK E...
Description
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
DESIGN OF SPREADER LIFTING BEAM of Hollow Circular Cross Section Belleli Energy srl, Dubai Branch prepared by R. Venkat Date
12/5/2013
Project: Hitachi Zosen, Arzew Plant, Algeria "BLOCK-E" Job No. 4776 Safe Working Load, SWL 112.6 Tons Spreading Length 10.000 metres Safety Factor in Compression 1.6 Yield stress of the Beam material 240 MPa Allowable Tensile Stress 150 MPa Elastic Modulus of the material 210000 MPa Geometry of Lifting, Solved using the equations of Static Equillibrium
SWL, W =
1109.7 kN
C
Tie Length = L =
10.000 metres a=
60.0 Degrees
Tensile Force, P =
637.9 kN
y x
A
Compressive Force, C =
318.9 kN
B
x
Spreading Length, L =
W/2 =
10.000 metres
552.4 kN
W/2 =
552.4 kN
As the Spreader beam is free for all its three planar DOF (x,y & R z Degrees 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). Hence, the STRUCTURAL analysis is made for (1) Direct Compressive Stress, and (2) Critical Load for Transverse Buckling. (A) Design for Normal Stress (Direct Compressive Stress) **** Selecting the Section initially based on this **** Considering the equillibrium of node B , 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 => P = kN Sin(180-a)/2
Tensile Force in the tie,
Prepared By: R.Venkat
P
=
637.9 kN
1 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
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
Minimum Area of Cross section required in mm2
A
Minimum Area of Cross section required in mm2
=
318.9 kN
=
Compressive Force in Newton X Safety Factor in Compression Allowable Stress in Mpa
=
C * FOS sallow
=
3401.91 mm2 **** CLICK **** TO INCREASE THE SECTION BY SELECTING NEXT PIPE SIZE
*****CLICK HERE TO SELECT THE PIPE****** [The Standard pipe (API 5L) selected shall be atleast 8" Sch20 with this cross section area and wall thickness is 219.1 minimum (i.e, maximum OD)] 8" Sch20
The Pipe selected is OD of the Pipe Tk of the Pipe Cross section Area, As
12" Sch20 323.80 mm 6.35 mm 2 6332.85 mm
Therefore the practical Safety Factor achieved
= =
nc
Buckling Stress is OK Buckling Stress is OK
Critical Load is OK Safety Margin is OK
Signal Box "OPTIMALITY"
as perAPI 5L
4 Moments of Inertia, Ixx=Iyy=I= 8.0E+07 mm Radii of Gyration, rx=ry=r= 112.26 mm Unit Weight of the Pipe 49.71 kg / m Compressive Strength of the pipe selected = Allowable Stress = sallow*As
Cactual
Signal Box "SAFE DESIGN" Load Factor is OK
Load Factor is OPTIMUM Buckling Stress is OPTIMUM Buckling Stress is OPTIMUM
Safety Margin is OPTIMUM
X
Cross section area
949.93 kN Cactual C
=
2.98 Nr.
SATISFACTORY OPTIMUM
(B) Design for Elastic Stability - Transverse Buckling [1] =
Slenderness Ratio (L/r)
Effective Length of the Spreader Beam Radius of gyration
For both ends HINGED members, the EFFECTIVE length equals the LENGTH of the member
The Crippling commencement factor,
L/r
=
Cc
=
Cc
=
89.08 Nr.
More than Euler's Critical Range 2p2 E sy
131.42 Nr.
*** Beam in the Intermediate-block, COMPRESSION & BUCKLING analyses needed *** Calculations for the Intermediate-block, Pls. ignore for the Slender Range FS Computing the factor, = (5/3) + (3/8) * [(L/r) / Cc] - (1/8) * [(L/r) / Cc]3
Allowable Stress in intermediate buckling
Prepared By: R.Venkat
FS
=
sallow(ib)
=
sallow(ib)
=
1.88 Nr. (sy/Fs) * [1 - (1/2) * {(L/r) / Cc}2] 98.23 MPa
2 of 23
OK OPTIMUM
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
Calculations for the Slender Range, Pls. ignore for the Intermediate-block Allowable Stress in slender buckling
sallow(sb)
=
(p2 * E) / [1.92 * (L/r)2]
sallow(sb)
=
136.04 MPa
OK OPTIMUM
Check for the Ultimate Buckling load Pu, which is the Euler's Critical load Pcr =
Euler's Critical Bucling load Pcr Buckling safety margin
[1]
1654.06 kN
= =
ncr
(p2 * E * I) / (L)2 OK
Euler's Critical Buckling Load Actual Compressive Force on Beam X Safety Factor in Compression
OK OPTIMUM This section is computed in accordance to the Manual Of Steel Construction , 9th edition, American Institute of Steel Construction, New Yark, 1959
Length Radius of the bracket Radius of the reinforcement Dia of the hole for Shackle Dia of the Shackle ring Yield stress of the material Allowable Tensile Stress
3.24 Nr.
=
('C) DESIGN OF ATTACHMENTS - (a) Design of eye bracket for strength L1 = 400 mm Height h1 = R1 = 125 mm Height h2 = R2 = 110 mm Thickness (Bkt & RF) t= d= 90 mm Weld joints' efficiency hj = ds = 76 mm Weld Fillet Size sw = syield = 248 MPa Elastic Modulus of the material sallow = 155 MPa
150 100 15 70 10 210000
mm mm mm % mm MPa
! Initially "assume" then "Iterate" these values with the help of following Signal Boxes Design Criterion (a-i) Tensile stress in the eye-bracket (a-ii) Bearing / Crushing stress in the eye-bracket (a-iii) Tearing stress in the eye-bracket (a-iv) Out-of plane buckling of the eye-bracket (a-v) Shearing stress in the shackle ring (b) Weld (set-on double fillet) size for eye-bracket with the beam Tensile Force, P = 3 * (t =
15
Signal Box "SAFE DESIGN" Signal Box "OPTIMALITY" OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM
637.9 kN R2 =
) mm
R1 =
110 mm
125 mm
b=
60.0 Degrees
h1 =
150 mm
d=
70 mm
h2 =
100 mm
Comp.Force, C= 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.
Prepared By: R.Venkat
3 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-i) Design of eye bracket for tensile strength - FAILURE MODE - 1 (allowable tensile stress is governing)
ds =
76 mm
~ 2R1 =
250 mm
3 * (t =
15
) mm
Resisting area for tension of the eye braket
= =
Resisting Area
At
=
Normal Force
P
=
Tensile Stress
=
Factor of safety with yield stress
Diameteral Difference (2R1 - ds) * 3t
X
Total Thickness
/
Resisting Area
2 7830 mm
637.9 kN Normal Force
st
=
81.5 MPa
OK
n1
=
3.0 Nr.
OPTIMUM
(a-ii) Design of eye bracket for bearing/crushing strength - FAILURE MODE -2 (80% of the yield stress is governing) ds =
3 * (t =
76 mm
15
) mm
Bearing area of the Sling hole for the Sling ring
= =
Bearing Area
Ab
=
Normal Force
P
=
Bearing Stress
Factor of safety with yield stress
Prepared By: R.Venkat
=
Diameter of the Sling ring ds * 3(t)
X
Total Thickness
/
Bearing Area
2 3420 mm
637.9 kN Normal Force
sbearing
=
186.5 MPa
OK
n2
=
1.3 Nr.
OPTIMUM
4 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-iii) Design of eye bracket for tearing strength - FAILURE MODE - 3 (50% of the yield stress is governing)
3 * (t =
15
) mm
Radii Difference
The bracket is in tearing due to shear along two planes against the sling ring, i.e., resisting it with the chordal sections Chordal (assumed to be Radial) area resisting the tearing shear
=
= Tearing area
At
=
Tearing Force
P
=
Tearing Stress
Factor of safety with yield stress
=
2X
Radii difference for the bracket + Radii difference for the reinfrmnt
X
Thickness
X
Total Thickness
2 * [(R1 - d/2) * t] + [(R2 - d/2) * (t*2)] 2 7200 mm
637.9 kN Tearing Force
/
Tearing Area
stearing
=
88.6 MPa
OK
n3
=
2.8 Nr.
OPTIMUM
(a-iv) Design of eye bracket for out-of plane buckling - FAILURE MODE - 4 (as per David T. Ricker
[2]
)
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
Factor of safety with thickness provided
[2]
=
0.25 * d
treq
=
22.5 mm
OK
n4
=
2.0 Nr.
OPTIMUM
This section is computed in accordance to David T. Ricker, " Design and Construction of Lifting Beams ", Engineering Journal, 4th Quarter, 1991
Prepared By: R.Venkat
5 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-v) Design of shackle ring for shearing strength - FAILURE MODE - 5 (50% of the yield stress is governing)
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
=
Shearing Force
P
=
Shearing Stress
=
Factor of safety with yield stress
2 * [p/4 * ds2] 2 9073 mm
637.9 kN Shearing Force
/
Shearing Area
sshearing
=
70.3 MPa
OK
n3
=
3.5 Nr.
OPTIMUM
(b) Design of weld joint of the eye-bracket with the spreader beam for shear strength (50% of the allowable stress is governing) Tensile Force, P = 3 * (t =
15
637.9 kN R2 =
) mm
R1 =
110 mm
125 mm
b=
60.0 Degrees
h1 =
150 mm
70 mm
h2 =
100 mm
Comp.Force, C= 318.9
L1 =
Prepared By: R.Venkat
d=
6 of 23
400 mm
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
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 axis Lw1 Transverse load on these joints
Allowable Shear Stress on effective throat area
=
2100 mm
Effective throat thickness tw Minimum Fillet Size of the Welds
W/2
= =
tallow-w
Prepared By: R.Venkat
(2+2) * (L1 + R1)
= Pw1
Factor of safety with fillet size provided
=
552.4 kN (sallow / 2) * hj
=
52.5 MPa
=
Pw1 * 1000 / (sw * Lw1)
=
5.01 mm
=
2 * tw
sw
=
7.09 mm
OK
n4
=
1.4 Nr.
OPTIMUM
7 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
DESIGN OF ADJUSTABLE SPREADER LIFTING BEAM of Standard Profile "HEB Series" Belleli Energy srl, Dubai Branch prepared by Venkat Date 12/5/2013 Project: Hitachi Zosen, Arzew Plant, Algeria "BLOCK-E" Job No. 4776 Safe Working Load, SWL 552.4 kN Spreading Length, L 4.000 metres Length, L1 1.998 metres
Value brought from previous worksheet
After reaching Flexural Stress "safe", Iterate this dimension L2 sothat this agrees with computed L2
0.202 metres
Initial Assumpn. Cant.lvr. Length, L2
Yield stress of the Beam material 248 MPa Allowable Stress 155 MPa Elastic Modulus of the material 210000 MPa Geometry of Lifting at the each end of the HOLLOW PIPE Spreader Beam, Solved using the equations of Static Equillibrium
UDL, q =
0.1 kN/m
SWL, W =
552.9 kN
y
A
C
D
x
B
x
Lgth, L1 =
Lgth, L2 =
0.202 metres
1.998 metres
Spreading Length, L =
4.000 metres
P= 276.2 kN P= 276.2 kN As this Adjustable Spreader beam is constrained for all its three planar DOF (x,y & R z Degrees 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). Hence, the STRUCTURAL analysis is made for the FLEXURAL Stress (A) STRENGTH DESIGN - Design for FLEXURAL Bending Stress **** Selecting the Section initially based on this **** 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 C (W * L1) + ((q * L2) * L2/2))
Prepared By: R.Venkat
=
(P * L) + ((q * L) * L/2)
L2
=
(2/q) * [(P * L) + ((q * L) * L/2) - (W * L1)]
L2
=
4.312 m
8 of 23
PLS. CHANGE THE INITIAL ASSUMPTION
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept. Bending Moment
= M
We have, Maximum Flexural Stress
sf(max) Factor of Safety achieved on yield stress
=
1105.760 kN-m Bending Moment
=
M / Zx
= n1
(P * L) + ((q * L) * L/2)
=
=
Moments of Inertia, Ixx=
PLS. INCREASE THE SECTION
syield / sf(max) 0.01 Nr.
*********** CLICK HERE ********** TO SELECT THE SECTION [The Standard Section (EN 53-62) selected shall be atleast with this Moment of Inertia] HE 100 AA 12.20 kg/m 4 2370000 mm
Elastic Section Modulus, Zx=
4 921000 mm 38.90 mm 24.30 mm 3 52000 mm
Elastic Section Modulus, Zy= Depth of the section, h = Width of the section, b =
3 18400 mm 91 mm 100 mm
Moments of Inertia, Iyy= Radius of Gyration, rx= Radius of Gyration, ry=
Section Modulus of the section about the axis perpendicular to plane of bending
/
21264.6 MPa
=
Structural member section Unit Weight of the member
Spreader Beam Design
OPTIMUM
**** CLICK **** TO INCREASE THE SECTION BY SELECTING NEXT SECTION
Signal Block "SAFE DESIGN" Counter Weight is NOT OK
Flexural Stress is NOT OK Max. deflection is NOT OK
as per EN 53-62
Signal Block "OPTIMALITY" Bending Stress is OPTIMUM
(B) STIFFNESS DESIGN - Design for DEFLECTION
The maximum deflection of the beam between loaded nodes and
C
=
B
dmax
Prepared By: R.Venkat
=
W * L1* (L2 - L12)3/2 9 * 31/2 * L * E * I
1481.222 mm
9 of 23
PLS. INCREASE THE SECTION
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(C) DESIGN OF ATTACHMENTS - (a) Design of adjustable (CG location variations) suspension bracket for strength Length of the bracket Lb = 200 mm Radius of the lug end R= 120 mm Clearance above the beam c= 40 mm Diameter of the hole dh = 60 mm Total Height of the lug h1 = 300 mm Diameter of the pin dp = 50 mm Height of the cut in the lug h2 = 50 mm Thickness of the lug t1 = 30 mm Height of taper in the lug h3 = 145 mm Thk of all other plates t2 = 10 mm Total Width of the lug w1 = 350 mm Weld joints' efficiency hj = 70 % Width of the cut in the lug w2 = 120 mm Weld Fillet Size sw = 10 mm Yield stress (all matl ex. pin) syield = 248 MPa Elastic Modulus of the material 210000 MPa Allowable Stress sallow = 155 MPa Yield stress (pin matl) sy(pin) = 300 MPa
! Initially "assume" then "Iterate" these values with the help of following Signal Boxes Design Criterion (a-i) Tensile stress in the lug (a-ii) Bearing / Crushing stress in the lug (a-iii) Tearing stress in the lug (a-iv) Out-of plane buckling of the lug (a-v) Shearing stress in the pin (b) Weld (set-on double fillet) size for lug with other plates (c) Tensile stress in the end plate (d) Weld (set-on double fillet) size for the bottom plate with other plates
Signal Box "SAFE DESIGN" Signal Box "OPTIMALITY" OK NOT OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM W=
w1 =
w2 =
350
120
(h + c) =
R=
120
dp =
50
dh =
60
h3 =
145
h2 =
50
h1 =
300
552.9 kN t1 =
131
=
=
Lb = t2 =
10
b=
30
200
100
The load on the bracket is the straight forward lifting force acting vertically upwards against the load.
Prepared By: R.Venkat
10 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-i) Design of lugs on the bracket for tensile strength - FAILURE MODE - 1 (allowable tensile stress is governing) wcs =
t1 =
309
W=
R=
120
dh =
60
h3 =
145
h1 =
300
h2 =
50
552.9 kN t1 =
30
30
w1 =
350
h3 - R
The width of the critical section i.e., across the diameter of the hole
=
w1 - 2 *
h3 - R
R tan
sin-1
+ tan-1 (h3 - R)2 + (w1 / 2)2
wcs The effective width at the critical section
wef
The effective normal area at the critical section resisting the force per lug
=
309 mm
=
Width at the critical section
=
wcs - dh
=
249 mm
=
Effective width
= Aef
-
X
(w1 / 2)
Diameter of the hole
Thickness
wef * t1 2
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
st Factor of Safety achieved on yield stress
Force on the lug
=
(W/2) / Aef 36.96 MPa
= =
n2
Prepared By: R.Venkat
=
/
Effective normal area
OK
syield / st
=
6.71 Nr.
11 of 23
NOT OPTIMUM
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-ii) Design of lugs on the bracket for bearing/crushing strength - FAILURE MODE -2 (80% of the yield stress is governing) W=
t1 =
30
The effective area bearing the crushing force per lug
Ab The bearing / crushing stress
sb Factor of Safety achieved on yield stress
wb =
50
dp =
50
=
Bearing width
=
wb * t1
t1 =
X
30
Thickness
( Note: Bearing width equals the projected diameter 2
1500 mm
= =
Crushing force
=
(W/2) / Ab
/
Bearing area
184.30 MPa
=
OK
syield / sb
= n3
552.9 kN
=
1.35 Nr.
OPTIMUM
(a-iii) Design of lug on the bracket for tearing strength - FAILURE MODE - 3 (50% of the yield stress is governing) W=
552.9 kN t1 =
Area resisting lateral tension (tearing) per lug
t1 =
30
(R - dh/2)=
90
=
Radii difference
= At
Prepared By: R.Venkat
X
30
Thickness
(R - dh/2) * t1 2
2700 mm
=
12 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept. Tearing stress
stear Factor of Safety achieved on yield stress
=
Tearing force
=
(W/2) / At
/
Area resisting lateral tension (tearing)
( Note: Tearing force conservatively equals lifiting force
102.39 MPa
=
OK
syield / sbtear
= n4
Spreader Beam Design
=
2.42 Nr.
OPTIMUM
(a-iv) Design of lug on the bracket for out-of plane buckling - FAILURE MODE - 4 (as per David T. Ricker
[1]
)
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
Factor of safety with thickness provided
[1]
0.25 * dh
= treq
=
15.0 mm
OK
n5
=
2.0 Nr.
OPTIMUM
This section is computed in accordance to David T. Ricker, " Design and Construction of Lifting Beams ", Engineering Journal, 4th Quarter, 1991 (a-v) Design of pin for shearing strength - FAILURE MODE - 5 (50% of the yield stress of the pin is governing) W=
552.9 kN t1 =
dp =
30
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
=
Shearing area
As
=
Shearing Force
P
=
Shearing Stress
Factor of safety with yield stress
Prepared By: R.Venkat
=
2 * [p/4 * dp2] 2 3927 mm
552.9 kN Shearing Force
/
Shearing Area
sshearing
=
140.8 MPa
OK
n6
=
2.1 Nr.
OPTIMUM
13 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(b) Design of weld joint of the lug with other plates for shear strength (50% of allowable stress is governing) W=
t2 =
w1 =
350
w2 =
120
552.9 kN t1 =
10
h2 =
50
b=
100
h3 =
145
h4 =
116
h1 =
300
30
Fixing the height h4 at 75% of the straight height (h1 - h3) h4 Total weld-length provided per lug
= Lw1
Transverse force on the weld joint per lug
Allowable Shear Stress on effective throat area
Effective throat thickness tw1 Minimum Fillet Size of the Welds
Prepared By: R.Venkat
2 * [2 * (h2 + h4 + t2 + h4) + b] 1370 mm W/2
= =
tallow-w
116 mm
= =
Pw1
Factor of safety with fillet size provided
=
276.5 kN (sallow / 2) * hj
=
54.25 MPa
=
Pw1 * 1000 / (sw * Lw1)
= =
3.72 mm 2 * tw
sw1
=
5.26 mm
OK
n7
=
1.9 Nr.
OPTIMUM
14 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(c) Design of end plates of the adjustable bracket for tensile strength (allowable tensile stress is governing) W=
552.9 kN
=
=
Lb = t2 =
200
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
= =
Aep
Length of the bracket
X
Thickness
Lb * t2 2 2000 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
st(ep) Factor of Safety achieved on yield stress
Force on the end plate
=
(W/2) / Aef
= =
n8
Prepared By: R.Venkat
=
138.23 MPa
/
Normal area
OK
syield / st(ep)
=
1.79 Nr.
15 of 23
OPTIMUM
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(d) Design of weld joint for the bottom plate of the adjustable bracket for shear strength (50% of allowable tensile stress is governing) W=
h5 =
552.9 kN
67
=
=
Lb = b=
200
100
Fixing the height h5 at one-third of length of the bracket, Lb h5
=
67 mm
Total weld-length provided
=
Directly for bottom plate
Total weld-length provided
=
2 * (Lb + b) + 4 *4* h5
Lw2 Transverse force on the weld joint
= Pw2
Allowable Shear Stress on effective throat area
Effective throat thickness tw2 Minimum Fillet Size of the Welds
Prepared By: R.Venkat
W 552.9 kN (sallow / 2) * hj
=
54.25 MPa
=
Pw1 * 1000 / (sw * Lw1)
= =
Indirectly for ribs supporitng bottom plate
1567 mm
= =
tallow-w
Factor of safety with fillet size provided
=
X
6.51 mm 2 * tw
sw2
=
9.20 mm
OK
n9
=
1.1 Nr.
OPTIMUM
16 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(D) DESIGN OF ATTACHMENTS - (a) Design of adjustable (Span variations) suspension bracket for strength Length of the bracket Lb = 150 mm Radius of the lug end R= 120 Clearance above the beam c= 20 mm Diameter of the hole dh = 60 Total Height of the lug h1 = 200 mm Diameter of the pin dp = 48 Height of the cut in the lug h2 = 50 mm Thickness of the lug t1 = 15 Height of taper in the lug h3 = 150 mm Thk of all other plates t2 = 8 Total Width of the lug w1 = 350 mm Weld joints' efficiency hj = 70 Width of the cut in the lug w2 = 116 mm Weld Fillet Size sw = 8 Yield stress (all matl ex. pin) syield = 248 MPa Elastic Modulus of the material 210000 Allowable Stress sallow = 155 MPa Yield stress (pin matl) sy(pin) = 300
mm mm mm mm mm % mm MPa MPa
! Initially "assume" then "Iterate" these values with the help of following Signal Boxes Design Criterion (a-i) Tensile stress in the lug (a-ii) Bearing / Crushing stress in the lug (a-iii) Tearing stress in the lug (a-iv) Out-of plane buckling of the lug (a-v) Shearing stress in the pin (b) Weld (set-on double fillet) size for lug with other plates (c) Tensile stress in the end plate (d) Weld (set-on double fillet) size for the top plate with other plates t2 =
Signal Box "SAFE DESIGN" Signal Box "OPTIMALITY" OK NOT OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM OK OPTIMUM NOT OK OPTIMUM
8 b=
100
=
w2 =
116
w1 =
350
h2 =
50
h1 =
200
h3 =
150
dh =
60
dp =
48
R=
120
W=
Lb =
150
(h + c) =
111
t1 =
15
=
276.2 kN
The load on the bracket is the straight forward load acting vertically downwards.
Prepared By: R.Venkat
17 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-i) Design of lugs on the bracket for tensile strength - FAILURE MODE - 1 (allowable tensile stress is governing) w1 =
t1 =
350 h2 =
50
h1 =
200
h3 =
150
dh =
60
R=
120
15
t1 = wcs =
304
W=
15
276.2 kN h3 - R
The width of the critical section i.e., across the diameter of the hole
=
w1 - 2 *
h3 - R
R tan
sin-1
+ tan-1 (h3 - R)2 + (w1 / 2)2
wcs The effective width at the critical section
wef
The effective normal area at the critical section resisting the force per lug
=
304 mm
=
Width at the critical section
=
wcs - dh
=
244 mm
=
Effective width
= Aef
-
X
(w1 / 2)
Diameter of the hole
Thickness
wef * t1 2
3653 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
st Factor of Safety achieved on yield stress
Force on the lug
=
(W/2) / Aef 37.80 MPa
= =
n2
Prepared By: R.Venkat
=
/
Effective normal area
OK
syield / st
=
6.56 Nr.
18 of 23
NOT OPTIMUM
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(a-ii) Design of lugs on the bracket for bearing/crushing strength - FAILURE MODE -2 (80% of the yield stress is governing)
dp = t1 =
48
15
t1 = wb = The effective area bearing the crushing force per lug
Ab The bearing / crushing stress
sb Factor of Safety achieved on yield stress
=
Bearing width
=
wb * t1
W= X
Thickness
( Note: Bearing width equals the projected diameter 2
=
Crushing force
=
(W/2) / Ab
=
276.2 kN
720 mm
=
= n3
48
15
/
Bearing area
191.81 MPa
OK
syield / sb
=
1.29 Nr.
OPTIMUM
(a-iii) Design of lug on the bracket for tearing strength - FAILURE MODE - 3 (50% of the yield stress is governing)
(R - dh/2) =
90
t1 =
15
t1 = W= Area resisting lateral tension (tearing) per lug
=
Radii difference
= At
Prepared By: R.Venkat
X
15
276.2 kN Thickness
(R - dh/2) * t1 2
1350 mm
=
19 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Tearing stress
stear Factor of Safety achieved on yield stress
=
Tearing force
=
(W/2) / At
/
Area resisting lateral tension (tearing)
( Note: Tearing force conservatively equals lifiting force
102.30 MPa
=
OK
syield / sbtear
= n4
Spreader Beam Design
=
2.42 Nr.
OPTIMUM
(a-iv) Design of lug on the bracket for out-of plane buckling - FAILURE MODE - 4 (as per David T. Ricker
[1]
)
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
OK
n5 Factor of safety with thickness provided = 1.0 Nr. OPTIMUM This section is computed in accordance to David T. Ricker, " Design and Construction of Lifting Beams ", Engineering Journal, [1] 4th Quarter, 1991 (a-v) Design of pin for shearing strength - FAILURE MODE - 5 (50% of the yield stress of the pin is governing)
dp =
48
t1 = W=
15
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
=
Shearing area
As
=
Shearing Force
P
=
Shearing Stress
Factor of safety with yield stress
Prepared By: R.Venkat
=
2 * [p/4 * dp2] 2 3619 mm
276.2 kN Shearing Force
/
Shearing Area
sshearing
=
76.3 MPa
OK
n6
=
3.9 Nr.
OPTIMUM
20 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(b) Design of weld joint of the lug with other plates for shear strength (50% of allowable stress is governing) t2 =
8 b=
w2 =
116
w1 =
350
100
h2 =
50
h4 =
38
h1 =
200
h3 =
150
t1 = W=
15
276.2 kN
Fixing the height h4 at 75% of the straight height (h1 - h3) h4 Total weld-length provided per lug
= Lw1
Transverse force on the weld joint per lug
Allowable Shear Stress on effective throat area
Effective throat thickness tw1 Minimum Fillet Size of the Welds
Prepared By: R.Venkat
2 * [2 * (h2 + h4 + t2 + h4) + b] 732 mm W/2
= =
tallow-w
38 mm
= =
Pw1
Factor of safety with fillet size provided
=
138.1 kN (sallow / 2) * hj
=
54.25 MPa
=
Pw1 * 1000 / (sw * Lw1)
= =
3.48 mm 2 * tw
sw1
=
4.92 mm
OK
n7
=
1.6 Nr.
OPTIMUM
21 of 23
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(c) Design of end plates of the adjustable bracket for tensile strength (allowable tensile stress is governing) t2 =
8 Lb = =
W=
150
=
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
= =
Aep
Length of the bracket
X
Thickness
Lb * t2 2 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
st(ep) Factor of Safety achieved on yield stress
Force on the end plate
=
(W/2) / Aef
= =
n8
Prepared By: R.Venkat
=
115.08 MPa
/
Normal area
OK
syield / st(ep)
=
2.15 Nr.
22 of 23
OPTIMUM
Printed on: 12/5/2013
BELLELI ENERGY srl Dubai Technical dept.
Spreader Beam Design
(d) Design of weld joint for the top plate of the adjustable bracket for shear strength (allowable tensile stress is governing) b=
100 Lb = =
h5 =
150
=
50
W=
276.2 kN
Fixing the height h5 at one-third of length of the bracket, Lb h5
=
50 mm
Total weld-length provided
=
Directly for bottom plate
Total weld-length provided
=
2 * (Lb + b) + 2*4 * h5]
Lw2 Transverse force on the weld joint
= Pw2
Allowable Shear Stress on effective throat area
Effective throat thickness tw2 Minimum Fillet Size of the Welds
W 276.2 kN (sallow / 2) * hj
=
54.25 MPa
=
Pw1 * 1000 / (sw * Lw1)
= =
Indirectly for ribs supporitng bottom plate
800 mm
= =
tallow-w
Factor of safety with fillet size provided
=
X
6.36 mm 2 * tw
sw2
=
9.00 mm
! NOT OK, INCREASE THE FILLET SIZE
n9
=
0.9 Nr.
OPTIMUM
(E) DESIGN OF ATTACHMENTS - (a) Design of fixed suspension bracket for strength ***** 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 beam and will be well within the operating range of the beam selected. Prepared By: R.Venkat
23 of 23
Printed on: 12/5/2013
View more...
Comments
