Spreader Beam BLOCK E

8/13/2019 Spreader Beam BLOCK E

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Dubai Technical dept.

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

Prepared By: R.Venkat 1 of 23 Pr inted on: 12/5/2013


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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

Prepared By: R.Venkat 2 of 23 Printed on: 12/5/2013


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Dubai Technical dept.Spreader Beam Design

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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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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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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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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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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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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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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Dubai Technical dept.Spreader Beam Design

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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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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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


[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




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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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


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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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.


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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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



tw2 = 6.51 mm


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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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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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


[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




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



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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



tw1 = 3.48 mm


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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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.


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



23/23

BELLELI ENERGY srl



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



tw2 = 6.36 mm


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

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