LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Weight of component to be lifted = Component force acting on beam, F = Impact factor = 1) SKID LUG SIZING Distance from lug hole to edge of beam, k = Lug radius, rL = Lug thickness, tL = Lug base width, wL = Diameter of hole, d = Distance from lug hole to base, hL = Collar plate thickness, tcp = Collar ring diameter, Dcp = Clearance btw shackle & lug size Lug thickness, tL A= 42.9 mm 40 mm Lug radius, rL C= 95.5 mm 70 mm Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTABLE Per PTS Section 6.3 a) Lug hole diameter, d shall be Max of i) Dp + 3mm = ii) Dp X 1.05 = b) Lug hole diameter, d shall be less than < (Dp + 6mm) = Dp = 33 result a) = LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure b) = Hole,d Diameter of hole, d btw 31.70 33 34.70 No of lug eye, = Maximum combined force acting on lug eye, Fc = = LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure 2) LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = Specified yield stress, Sy = Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = Allowable tensile stress, St.all ( = 0.6Sy ) = Allowable bearing stress, Sbr.all ( = 0.9Sy ) = Allowable shear stress, Ss.all ( = 0.4Sy ) = SHACKLES Shackle rating ( S.W.L ) : Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = MAXIMUM SLING TENSION ON PADEYE Ts = FACTOR OF SAFETY F.O.S. = DESIGN LOAD: SLING TENSION P = FOS * Ts P = LIFTING ANGLE a = ACTUAL OUT OF PLANE ANGLE b = VERTICAL FORCE ON PADEYE Fz = P * sin a Fz = OUT OF PLANE FORCE Fyl = P * sin b Fyl = HORIZONTAL FORCE ON PADEYE Fx = P * cos a Fx = Horizontal dist.PIN CL to N.A. exl = 3) STRESS CHECK AT BASE a) Moment Calc at distance , H In Plane Moment My = ( Fx*H ) - ( Fz*ex l) My = Out of plane moment Mx = ( FyI*hL ) Mx = b) Tensile Stress Maximum tensile force, ft = Fz / [ tL * wL ] = Allowable tensile stress, St.all ( = 0.6Sy ) = Since ft < St.all, therefore the lug size is satisfactory. c) Bending stress (In Plane) Maximum bending stress , fbx = ( 6*Mx ) / ( wL * [(tL+tcp)^2] ) = Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = Since fbx < fbx.all,therefore the lug size is satisfactory. d) Bending stress (Out of Plane) LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Maximum bending stress , fby = ( 6*My ) / [ tL +(2*tcp)] * [ wL^2 ] ) = Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = Since fby < fby.all,therefore the lug size is satisfactory. LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure e) Combined stresses, U = St/St.all + fby/fby.all + fbx/fbx.all = Since U < 1, therefore the lug size is satisfactory. f) SHEAR stress (In Plane) Maximum SHEAR stress , fsx = Fx / [ wL * tL ] = Allowable shear stress, Ss.all ( = 0.4Sy ) = Since fsx < Ss.all,therefore the lug size is satisfactory. g) Bending stress (Out of Plane) Maximum SHEAR stress , fsy = Fyl / [ wL * tL ] = Allowable shear stress, Ss.all ( = 0.4Sy ) = Since fsx < Ss.all,therefore the lug size is satisfactory. 3.1) CHECKING VON-MISES CRITERIA a) Sum of stress in X-PLANE fx = St + fby = b) Sum of stress in Y-PLANE fy = St + fbx = c) Therefore, average Shear stress fxy = SQRT [ (fsx^2)+(fsy^2) ] = d) Maximum Combined stress Fcomb = SQRT [ (fx^2)+(fy^2)-(fx+fy+3fxy^2) ] = Allowable combined stress : Fcomb.all ( = 0.66Sy ) = Since fsx < Ss.all,therefore the lug size is satisfactory. 4) STRESS CHECK AT PIN HOLE a) Tensile Stress Maximum tensile force, P = Cross sectional area of lug eye, At = [ 2 * ( tL* ( rL - d/2 ))] + [ 2 * ( tcp* = (( Dcp/2) - d/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d/2 ))] Tensile stress, St = Allowable tensile stress, St.all ( = 0.6Sy ) = Since St < St.all, therefore the lug size is satisfactory. b) Bearing Stress Maximum bearing force, P = Cross sectional area of lug eye, Ab = Dp * ( tL + 2tcp ) = Bearing stress, Sbr = Fbr / Ab = Allowable bearing stress, Sbr.all ( = 0.9Sy ) = LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Since Sbr < Sbr.all,therefore the lug size is satisfactory. LIFTING LUG DESIGN CALCULATION (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure c) Shear Stress ` Maximum shear force, P = Cross sectional area of lug eye, At = [ 2 * ( tL* ( rL - d/2 ))] + [ 2 * ( tcp* = (( Dcp/2) - d/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d/2 ))] Shear stress, Ss = Allowable shear stress, Ss.all ( = 0.4Sy ) = Since Ss < Ss.all,therefore the lug size is satisfactory. d) Combined stresses, U = St/St.all + fby/fby.all + fbx/fbx.all = Since U < 1, therefore the lug size is satisfactory. 5) WELD SIZE CALCULATIONS Weld leg used, = Weld throat thickness used, tr = Filler metal material : Fillet weld joint efficiency, E = Welding stress for steel grade 43 ( E-43 ), = Allowable welding stress,Sw = (a) Tensile Stress Maximum tensile force,Ft = Area of weld, Aw = 2*(tL+wL)*tr = Tensile stress, St = [(Ft/Aw)] = Since St < Sw,therefore weld leg is satisfactory. (b) Shear stress Maximum shear force,Ft = Shear stress, Ss = (Ft/Aw) = Allowable welding stress for steel grade 43 ( E-43 ), Sw = Since Ss < Sw,therefore weld leg dimension is SATISFACTORY. (c) Bending stress Maximum bending force,Fb = Bending stress, Sb = [(Fb/Aw)] = Allowable welding stress for steel grade 43 ( E-43 ), Sw = Since Sb < Sw,therefore weld leg dimension is SATISFACTORY. nasrul: PLEASE ENABLE MACRO LINK 6,660 kg 130,675 N 2 100 mm 70 mm 20 mm 203 mm 33 mm 198 mm 10 mm 100 mm Result mm OK mm OK fore the Lug is is 31.70 mm 30.14 mm 34.70 mm satisfactory nasrul: PLEASE ENABLE MACRO LINK satisfactory OK 4 32669 N 3330 kg nasrul: PLEASE ENABLE MACRO LINK A 36 248.21 N/mm² 163.82 N/mm² 186.16 N/mm² 148.93 N/mm² 223.39 N/mm² 99.28 N/mm² 8.5 tons T Type Anchor shackle G2130 28.70 mm 16,334 N 2.00 32,669 N 90.00 Deg. 25.00 Deg. 32,669 N 13,806 N 0 N 0.00 0 N-mm 2,733,669 N-mm 4 N 148.93 N/mm² 50 N/mm² 163.82 N/mm² nasrul: PLEASE ENABLE MACRO LINK 0 N/mm² 186.16 N/mm² nasrul: PLEASE ENABLE MACRO LINK 0.34 0 N/mm² 99.28 N/mm² 3 N/mm² 99.28 N/mm² 9.39 N/mm² 59.89 N/mm² 3.40 N/mm² 59.75 N/mm² 163.82 N/mm² 32669 N 3480 mm² 9.39 N/mm² 148.926 N/mm² 32669 N 1148 mm² 28.46 N/mm² 223.389 N/mm² nasrul: PLEASE ENABLE MACRO LINK 32669 N 3480 mm² 9.39 N/mm² 99.28 N/mm² 0.22 10 mm 7 mm E-43 0.49 125 N/mm² 61.25 N/mm² 32669 N 3122 mm² 10.46 N/mm² 32669 N 10.46 N/mm² 61.25 N/mm² SFACTORY. 46723 N 3.41 N/mm² 61.25 N/mm² SFACTORY. LIFTING SPREADER PIPE SIZING CALCULATIONS (SKID) nasrul: PLEASE ENABLE ITEM : C.I. SKID (A-6810) MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Weight of component to be lifted ( + Spreader beam weight ) = 6,928 kg Component force acting on beam, F = 101,939 N Impact factor = 2 nasr INPU SPREADER LGTH = 1) PIPE SIZING LENGTH SKID + 280 Pipe size : 10 in SCH 80 Outer diameter of pipe, D = 273.1 mm Thickness of pipe, = 15.09 mm Outer Radius of pipe, R = 136.55 mm Inner radius of pipe, r = 121.46 mm Section modulus of pipe, Zx-x = 747,911 mm3 Second Moment of pipe, I = 6,371,592 mm4 Cross section area of pipe, A = 12231 mm2 Unbraced length of member, L = 2784 mm Modulus Of Elasticity , E = 2.0E+05 N/mm² Spreader Pipe Weight = 267.3 kg Material used = A 106.Gr.B Specified yield stress, Sy = 241.32 N/mm² a) Bending Stress F R1=R2=F/2 L R1 R2 Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam. (See Appendix) Total bending moment, M ( = F*L / 4 ) = 47,299,895 Nmm Bending stress, Sb ( = M / Zx-x ) = 63.24 N/mm² Max Bending stress = Fx * L / ( 60 * E * I ) 3 = 28.78 N/mm² Allowable bending stress, Sb.all ( = 0.66Sy ) = 159.27 N/mm² Since Sb < Sb.all,therefore the pipe size is satisfactory. OK b) Compressive Stress Compressive force, Fc = 101,939 N Compressive stress, Sc = Fc / A = 8.33 N/mm² Allowable compressive stress, Sc.all ( = 0.6Sy ) = 144.79 N/mm² c) Combined stresses, U = Sc + Sb = 0.45 Sc.all Sb.all Since U < 1, therefore the pipe size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (SKID) nasrul: PLEASE ENABLE ITEM : C.I. SKID (A-6810) MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure 2) LUG SIZING Fz P rL(s) tL(s) d(s) a Fx Fyl hL(s) wL(s) Lug radius, rL(s) = 70 mm Lug thickness, tL(s) = 25 mm Lug base width, wL(s) = 192 mm Diameter of hole, d(s) = 33 mm Distance from lug hole to base, hL(s) = 80 mm Collar plate thickness, tcp = 0 mm Collar ring diameter, Dcp = 0 mm Clearance btw shackle & lug size Result Lug thickness, tL(s) A= 42.9 mm 25 mm OK Lug radius, rL(s) C= 95.5 mm 70 mm OK Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTABLE Per PTS Section 6.3 Check a) Lug hole diameter, d shall be Max of i) Dp + 3mm = 31.70 mm ii) Dp X 1.05 = 30.14 mm b) Lug hole diameter, d shall be less than < (Dp + 6mm) = 34.70 mm Dp = 33 result a) = satisfactory b) = satisfactory Hole,d Diameter of hole, d(s) btw 31.70 33 34.70 OK No of lug eye, = 2 Maximum combined force acting on lug eye, Fc = 50970 N = 5196 kg LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = A 36 Specified yield stress, Sy = 248.21 N/mm² Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = 163.82 N/mm² Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = 186.16 N/mm² Allowable tensile stress, St.all ( = 0.6Sy ) = 148.93 N/mm² Allowable bearing stress, Sbr.all ( = 0.9Sy ) = 223.39 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² LIFTING SPREADER PIPE SIZING CALCULATIONS (SKID) nasrul: PLEASE ENABLE ITEM : C.I. SKID (A-6810) MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure SHACKLES Shackle rating ( S.W.L ) : 8.5 tons Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = 28.70 mm MAXIMUM SLING TENSION ON PADEYE Ts = 33,980 N FACTOR OF SAFETY F.O.S. = 2.00 DESIGN LOAD: SLING TENSION P = FOS * Ts P = 67,960 N LIFTING ANGLE a = 60.00 Deg. ACTUAL OUT OF PLANE ANGLE b = 0.00 Deg. VERTICAL FORCE ON PADEYE Fz = P * sin a Fz = 58,855 N OUT OF PLANE FORCE Fyl = 5% of P Fyl = 3,398 N HORIZONTAL FORCE ON PADEYE Fx = P * cos a Fx = 33,980 N Horizontal dist.PIN CL to N.A. exl = 0.00 3) STRESS CHECK AT BASE a) Moment Calc at distance , H In Plane Moment My = ( Fx*H ) - ( Fz*ex l) My = 2,718,385 N-mm Out of plane moment Mx = ( FyI*hL ) Mx = 271,838 N-mm b) Tensile Stress Maximum tensile force, ft = Fz / [ tL(s) * wL(s) ] = 12 N Allowable tensile stress, St.all ( = 0.6Sy ) = 148.93 N/mm² Since ft < St.all, therefore the lug size is satisfactory. c) Bending stress (In Plane) Maximum bending stress , fbx = ( 6*Mx ) / ( wL(s) * [(tL(s)+tcp)^2] ) = 14 N/mm² Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = 163.82 N/mm² Since fbx < fbx.all,therefore the lug size is satisfactory. d) Bending stress (Out of Plane) Maximum bending stress , fby = ( 6*My ) / [ tL(s) +(2*tcp)] * [ wL(s)^2 ] ) = 18 N/mm² Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = 186.16 N/mm² Since fby < fby.all,therefore the lug size is satisfactory. e) Unity Check : Combined stresses, U = St/St.all + fby/fby.all + fbx/fbx.all = 0.26 Since U < 1, therefore the lug size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (SKID) nasrul: PLEASE ENABLE ITEM : C.I. SKID (A-6810) MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure f) SHEAR stress (In Plane) Maximum SHEAR stress , fsx = Fx / [ wL(s) * tL(s) ] = 7 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. g) Bending stress (Out of Plane) Maximum SHEAR stress , fsy = Fyl / [ wL(s) * tL(s) ] = 1 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. 3.1) CHECKING VON-MISES CRITERIA a) Sum of stress in X-PLANE fx = St + fby = 43.10 N/mm² b) Sum of stress in Y-PLANE fy = St + fbx = 39.00 N/mm² c) Therefore, average Shear stress fxy = SQRT [ (fsx^2)+(fsy^2) ] = 7.11 N/mm² d) Maximum Combined stress Fcomb = SQRT [ (fx^2)+(fy^2)-(fx+fy+3fxy^2) ] = 56.08 N/mm² Allowable combined stress : Fcomb.all ( = 0.66Sy ) = 163.82 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. 4) STRESS CHECK AT PIN HOLE a) Tensile Stress Maximum tensile force, P = 67960 N Cross sectional area of lug eye, At = [ 2 * ( tL(s)* ( rL(s) - d(s)/2 ))] + [ 2 * ( tcp* = 2675 mm² (( Dcp/2) - d(s)/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d(s)/2 ))] Tensile stress, St = 25.41 N/mm² Allowable tensile stress, St.all ( = 0.6Sy ) = 148.926 N/mm² Since St < St.all, therefore the lug size is satisfactory. b) Shear Stress ` Maximum shear force, P = 67960 N Cross sectional area of lug eye, At = [ 2 * ( tL(s)* ( rL(s) - d(s)/2 ))] + [ 2 * ( tcp* = 2675 mm² (( Dcp/2) - d(s)/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d(s)/2 ))] Shear stress, Ss = 25.41 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since Ss < Ss.all,therefore the lug size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (SKID) nasrul: PLEASE ENABLE ITEM : C.I. SKID (A-6810) MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure c) Bearing Stress Maximum bearing force, P = 67960 N Cross sectional area of lug eye, Ab = Dp * ( tL + 2tcp ) = 825 mm² Bearing stress, Sbr = Fbr / Ab = 82.38 N/mm² Allowable bearing stress, Sbr.all ( = 0.9Sy ) = 223.39 N/mm² Since Sbr < Sbr.all,therefore the lug size is satisfactory. d) Unity check, Combine Stresses St Sbr Ss = 0.80 = ------------- + -------------- + -------------- is < than 1 St.all Sbr.all Ss.all Therefore, the lifting lug size is Satisfactory. 5) WELD SIZE CALCULATIONS Weld leg used, = 10 mm Weld throat thickness used, tr = 7 mm Filler metal material : E-43 Fillet weld joint efficiency, E = 0.49 Welding stress for steel grade 43 ( E-43 ), = 125 N/mm² Allowable welding stress,Sw = 61.25 N/mm² a) Tensile Stress Maximum tensile force,Ft = 67960 N Area of weld, Aw = 2*(tL+wL)*tr = 3038 mm² Tensile stress, St = [(Ft/Aw)] = 22.37 N/mm² Since St < Sw,therefore weld leg is satisfactory. (b) Shear stress Maximum shear force,Ft = 67960 N Shear stress, Ss = (Ft/Aw) = 22.37 N/mm² Allowable welding stress for steel grade 43 ( E-43 ), Sw = 61.25 N/mm² Since Ss < Sw,therefore weld leg dimension is SATISFACTORY. (c) Bending stress Maximum bending force,Fb = 46723 N Bending stress, Sb = [(Fb/Aw)] = 3.41 N/mm² Allowable welding stress for steel grade 43 ( E-43 ), Sw = 61.25 N/mm² Since Sb < Sw,therefore weld leg dimension is SATISFACTORY. SLING AND WIRE ROPE CALCULATION @ SPREADER PIPE (SKID) ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Selection of Sling's Safe Working Load (SWL) (SLING S5 & S6) - REFER DWG Design safety Factor = 2 Number of leg = 2 Vertical WLL = Weight of Load lifted / No. of legs = 14,683 lb = ( Load on each sling ) = 6.66 ton To calculate actual Sling capacity when lifting load at specified angle, a Sling angle factor will be used as shown in the calculation below : TABLE 1 Sling ratio = 92% hock: USE VALUE FROM TABLE Actual Sling Capacity = Factor x Rated Capacity of Sling being used Min Required Sling SWL = 6.66 ton (From calculation above) Sling angle = 60 (Advisable using btw 45° to 60, but not below 30° ) Rated Sling SWL used = 8.40 ton (As per table EN13414-1-Table 3) Actual Sling Capacity = 0.866 x 8.40 = 7.2744 ton OK (Advisable using below 80% : check!! Yee Loong Wire Rope Diameter calculation SEE NOTE Safe working load = (diameter)2 × 8 Wire rope diameter (inch) & SWL ( tonne ) The above formula can be used to estimate SWL of the wire rope to be used when lifting the loads. Therefore, the estimated diameter of wire ropes is : Wire rope diameter = SQRT (SWL of wire rope / 8) = 1.0247 in = 26 mm To Used ----> 28 mm (See note) (See note) NOTE : ( Estimated values only. Please consult wire rope manufacturer for confirmation on rated SWL of wire rope calculated) SUMMARY WIRE ROPE DIAMETER USED = 28 mm SWL OF SLING USED = 8.40 ton MIN. REQUIRED SWL OF SLING = 6.66 ton SLING AND WIRE ROPE CALCULATION @ (SKID) LIFTING LUG ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Selection of Sling's Safe Working Load (SWL) (SLING S1 / S2 / S3 /S4 ) - REFER DWG Design safety Factor = 2 Number of leg = 4 Vertical WLL = Weight of Load lifted / No. of legs = 7,342 lb = ( Load on each sling ) = 3.33 ton To calculate actual Sling capacity when lifting load at specified angle, a Sling angle factor will be used as shown in the calculation below : TABLE 1 Sling ratio = 88% hock: USE VALUE FROM TABLE Actual Sling Capacity = Factor x Rated Capacity of Sling being used Min Required Sling SWL = 3.33 ton (From calculation above) Sling angle = 60 (Advisable using btw 45° to 60, but not below 30° ) Rated Sling SWL used = 4.35 ton (As per table EN13414-1-Table 3) Actual Sling Capacity = 0.866 x 4.35 = 3.7671 ton OK Proceed to next calc Yee Loong Wire Rope Diameter calculation SEE NOTE Safe working load = (diameter)2 × 8 Wire rope diameter (inch) & SWL ( tonne ) The above formula can be used to estimate SWL of the wire rope to be used when lifting the loads. Therefore, the estimated diameter of wire ropes is : Wire rope diameter = SQRT (SWL of wire rope / 8) = 0.73739 in = 19 mm To Used ----> 20 mm (See note) (See note) NOTE : ( Estimated values only. Please consult wire rope manufacturer for confirmation on rated SWL of wire rope calculated) SUMMARY WIRE ROPE DIAMETER USED = 20 mm SWL OF SLING USED = 4.35 ton MIN. REQUIRED SWL OF SLING = 3.33 ton LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-6820 A/B PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure WARNING : DO NOT LIFT MORE THAN Fy 10.00 ° OF ANGLE U `U nasrul: PLEASE ENABLE Dcp w2 MACRO LINK tL rL d w2 A A tcp k tank Pad b plate Wp w3 a tp w1 Lp tM ( USED OPERATING Tank Weight) Weight of tank , We = 2,540 kg = 24,922 N Number of lifting lug, N = 4 1) LIFTING LUG Distance k = 80 mm Lug radius, rL = 42 mm Diameter of hole, d = 27 mm Lug thickness, tL = 12 mm Collar plate thickness, tcp = 0 mm Collar ring diameter, Dcp = 0 mm Length a = 84 mm Length b = 70 mm Pad length, Lp = 140 mm Pad width, Wp = 100 mm Pad thickness, tp = 6 mm Shell thickness, tM = 6 mm Angle, U = 10 ° Shackle S.W.L : 4.75 tons Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = 22.40 mm LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-6820 A/B PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure Per PTS Section 6.3 Check a) Lug hole diameter, d shall be Max of i) Dp + 3mm = 25.40 mm ii) Dp X 1.05 = 23.52 mm b) Lug hole diameter, d shall be less than < (Dp + 6mm) = 28.40 mm Diameter of hole, d 27 result a) = satisfactory b) = satisfactory Hole,d Diameter of hole, d btw 25.40 27 28.40 OK Clearance btw shackle & lug size Result lug thk A= 42.9 mm 24 mm OK lug radius C= 95.5 mm 42 mm OK Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTABLE 2) LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = A 36 Specified yield stress, Sy = 248.22 N/mm² Impact load factor, p = 2.00 Specified Tensile stress, Ts = 206.00 N/mm² 3) ALLOWABLE STRESSES Allowable tensile stress, St.all ( = 0.6 Sy ) = 148.93 N/mm² Allowable bearing stress, Sbr.all ( = 0.9 Sy ) = 223.40 N/mm² Allowable bending stress, Sbn.all ( = 0.66 Sy ) = 163.83 N/mm² Allowable shear stress, Ss.all ( = 0.4 Sy ) = 99.29 N/mm² 4) LIFTING LUG DESIGN - VERTICAL LIFTING 4.1 DESIGN LOAD Design load , Wt ( = p.We ) = 49843 N Design load per lug, W ( = Wt / N ) = 12461 N Vertical component force, Fy = 12461 N LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-6820 A/B PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure 4.2 STRESS CHECK AT PIN HOLE (a) Tensile Stress Vertical component force, Fy = 12461 N Cross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) = 684 mm² Tensile stress, St ( = Fy / Ae ) = 18.22 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. (b) Bearing Stress Vertical component force, Fy = 12461 N Cross-sectional area , Ae = Dp x (tL+2tcp) = 269 mm² Bearing stress, Sbr ( = Fy / Ae ) = 46.36 N/mm² Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory. (c) Shear Stress Vertical component force, Fy = 12461 N Cross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) = 684 mm² Shear stress, Ss ( = Fy / Ae ) = 18.22 N/mm² Since Ss < Ss.all,therefore the lifting lug size is satisfactory. (d) Unity check, Combine Stresses St Sbr Ss = 0.51 ------------- + -------------- + -------------- is < than 1 St.all Sbr.all Ss.all Therefore, the lifting lug size is Satisfactory. LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-6820 A/B PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure 5) STRESS CHECK AT SECTION A-A (a) Bending Stress Bending stress due to Pa ( = Fy x tan U ) = 2197 N Bending moment, Mb ( = Pa x k ) = 175,775 Nmm Section modulus, Z = ( 2*rL*tL / 6 ) + ( 2*(Dcp*tcp / 6) ) 2 = 2016 mm3 Bending stress, Sbn ( = Mb/Z ) = 87.19 N/mm² Since Sbn < Sb.all, therefore the lifting lug size is satisfactory. (b) Tensile Stress due to Fy Cross section area, Ae = 2rL x tL = 1008 mm² Tensile Stress, St = Fy / Ae = 12.36 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. (c) Unity check Combine Stress Ratio, CS = (St / St.all) + (Sb / Sbn.all) = 0.62 Since CS <= 1.0, therefore lifting lug size is satisfactory. 6) DESIGN OF WELD SIZE AT LUG TO PAD JOINT 6.1 GENERAL Weld leg , w1 = 6 mm Weld throat thickness, tr1 = 4.2 mm Weld leg , w2 = 4.2 mm Weld throat thickness, tr2 = 3.0 mm Fillet weld joint efficiency, E = 0.6 Allowable welding stress for steel grade 43 ( E-43 ) Sw = E x Ts = 123.6 N/mm² LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-6820 A/B PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure 6.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES Area of weld, Aw1 ( = tr1 ( a + 2b ) = 950 mm² Area of weld, Aw2 ( = tr2 ( (2a-tM) + (2M+tM ))) = 534 mm² Total area of weld, Aw ( = Aw1 + Aw2 ) = 1485 mm² 6.3 STRESS DUE TO FORCE Fy Component force, Fy = 12461 N Shear stress, Ssx ( = Fy / Aw ) = 8.39 N/mm² Allowable welding stress, Sa ( = E.Sw ) = 123.60 N/mm² Since Ssx < Sa, therefore the selected weld size is satisfactory . 7.0 DESIGN OF WELD SIZE AT PAD TO TANK JOINT 7.1 GENERAL Weld leg , w = 4.2 mm Weld throat thickness, tr = 3.0 mm Fillet weld joint efficiency, E = 0.6 Allowable welding stress for steel grade 43 ( E-43 ) Sw = E x Ts = 123.6 N/mm² 7.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES Area of weld, Aw ( = 2 tr ( Wp + Lp ) ) = 1425 mm² 7.3 STRESS DUE TO FORCE Fy Component force, Fy = 12461 N Shear stress, Ssx ( = Fy / Aw ) = 8.74 N/mm² Allowable welding stress, Sa ( = E.Sw ) = 74.16 N/mm² Since Ssx < Sa, therefore the selected weld size is satisfactory . LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-6820 A/B MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Weight of component to be lifted ( + Spreader beam weight ) = 2,575 kg Component force acting on beam, F = 37,890 N Impact factor = 2 SPREADER LGTH = 1) PIPE SIZING LENGTH SKID + 280 nasrul: INPUT -TablePipe Pipe size : 6 in SCH 40 Outer diameter of pipe, D = 168.3 mm Thickness of pipe, = 7.11 mm Outer Radius of pipe, R = 84.15 mm Inner radius of pipe, r = 77.04 mm Section modulus of pipe, Zx-x = 139,230 mm3 Second Moment of pipe, I = 730,961 mm4 Cross section area of pipe, A = 3600 mm2 Unbraced length of member, L = 1219 mm Modulus Of Elasticity , E = 2.0E+05 N/mm² Spreader Pipe Weight = 34.4 kg Material used = A 106.Gr.B Specified yield stress, Sy = 241.32 N/mm² a) Bending Stress F R1=R2=F/2 L R1 R2 Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam. (See Appendix) Total bending moment, M ( = F*L / 4 ) = 7,697,888 Nmm Bending stress, Sb ( = M / Zx-x ) = 55.29 N/mm² Max Bending stress = Fx * L / ( 60 * E * I ) 3 = 7.83 N/mm² Allowable bending stress, Sb.all ( = 0.66Sy ) = 159.27 N/mm² Since Sb < Sb.all,therefore the pipe size is satisfactory. OK b) Compressive Stress Compressive force, Fc = 37,890 N Compressive stress, Sc = Fc / A = 10.52 N/mm² Allowable compressive stress, Sc.all ( = 0.6Sy ) = 144.79 N/mm² c) Combined stresses, U = Sc + Sb = 0.40 Sc.all Sb.all Since U < 1, therefore the pipe size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-6820 A/B MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure 2) LUG SIZING Fz P rL(s) tL(s) d(s) a Fx Fyl hL(s) wL(s) Lug radius, rL(s) = 70 mm Lug thickness, tL(s) = 25 mm Lug base width, wL(s) = 192 mm Diameter of hole, d(s) = 33 mm Distance from lug hole to base, hL(s) = 80 mm Collar plate thickness, tcp = 0 mm Collar ring diameter, Dcp = 0 mm Clearance btw shackle & lug size Result Lug thickness, tL(s) A= 42.9 mm 25 mm OK Lug radius, rL(s) C= 95.5 mm 70 mm OK Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTABLE Per PTS Section 6.3 Check a) Lug hole diameter, d shall be Max of i) Dp + 3mm = 31.70 mm ii) Dp X 1.05 = 30.14 mm b) Lug hole diameter, d shall be less than < (Dp + 6mm) = 34.70 mm Dp = 33 result a) = satisfactory b) = satisfactory Hole,d Diameter of hole, d(s) btw 31.70 33 34.70 OK No of lug eye, = 2 Maximum combined force acting on lug eye, Fc = 18945 N = 1931 kg LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = A 36 Specified yield stress, Sy = 248.21 N/mm² Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = 163.82 N/mm² Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = 186.16 N/mm² Allowable tensile stress, St.all ( = 0.6Sy ) = 148.93 N/mm² Allowable bearing stress, Sbr.all ( = 0.9Sy ) = 223.39 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-6820 A/B MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure SHACKLES Shackle rating ( S.W.L ) : 8.5 tons Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = 28.70 mm MAXIMUM SLING TENSION ON PADEYE Ts = 12,630 N FACTOR OF SAFETY F.O.S. = 2.00 DESIGN LOAD: SLING TENSION P = FOS * Ts P = 25,260 N LIFTING ANGLE a = 60.00 Deg. ACTUAL OUT OF PLANE ANGLE b = 0.00 Deg. VERTICAL FORCE ON PADEYE Fz = P * sin a Fz = 21,876 N OUT OF PLANE FORCE Fyl = 5% of P Fyl = 1,263 N HORIZONTAL FORCE ON PADEYE Fx = P * cos a Fx = 12,630 N Horizontal dist.PIN CL to N.A. exl = 0.00 3) STRESS CHECK AT BASE a) Moment Calc at distance , H In Plane Moment My = ( Fx*H ) - ( Fz*ex l) My = 1,010,387 N-mm Out of plane moment Mx = ( FyI*hL ) Mx = 101,039 N-mm b) Tensile Stress Maximum tensile force, ft = Fz / [ tL(s) * wL(s) ] = 5N Allowable tensile stress, St.all ( = 0.6Sy ) = 148.93 N/mm² Since ft < St.all, therefore the lug size is satisfactory. c) Bending stress (In Plane) Maximum bending stress , fbx = ( 6*Mx ) / ( wL(s) * [(tL(s)+tcp)^2] ) = 5 N/mm² Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = 163.82 N/mm² Since fbx < fbx.all,therefore the lug size is satisfactory. d) Bending stress (Out of Plane) Maximum bending stress , fby = ( 6*My ) / [ tL(s) +(2*tcp)] * [ wL(s)^2 ] ) = 7 N/mm² Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = 186.16 N/mm² Since fby < fby.all,therefore the lug size is satisfactory. e) Unity Check : Combined stresses, U = St/St.all + fby/fby.all + fbx/fbx.all = 0.10 Since U < 1, therefore the lug size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-6820 A/B MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure f) SHEAR stress (In Plane) Maximum SHEAR stress , fsx = Fx / [ wL(s) * tL(s) ] = 3 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. g) Bending stress (Out of Plane) Maximum SHEAR stress , fsy = Fyl / [ wL(s) * tL(s) ] = 0 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. 3.1) CHECKING VON-MISES CRITERIA a) Sum of stress in X-PLANE fx = St + fby = 16.02 N/mm² b) Sum of stress in Y-PLANE fy = St + fbx = 14.49 N/mm² c) Therefore, average Shear stress fxy = SQRT [ (fsx^2)+(fsy^2) ] = 2.64 N/mm² d) Maximum Combined stress Fcomb = SQRT [ (fx^2)+(fy^2)-(fx+fy+3fxy^2) ] = 20.38 N/mm² Allowable combined stress : Fcomb.all ( = 0.66Sy ) = 163.82 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. 4) STRESS CHECK AT PIN HOLE a) Tensile Stress Maximum tensile force, P = 25260 N Cross sectional area of lug eye, At = [ 2 * ( tL(s)* ( rL(s) - d(s)/2 ))] + [ 2 * ( tcp* = 2675 mm² (( Dcp/2) - d(s)/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d(s)/2 ))] Tensile stress, St = 9.44 N/mm² Allowable tensile stress, St.all ( = 0.6Sy ) = 148.926 N/mm² Since St < St.all, therefore the lug size is satisfactory. b) Shear Stress ` Maximum shear force, P = 25260 N Cross sectional area of lug eye, At = [ 2 * ( tL(s)* ( rL(s) - d(s)/2 ))] + [ 2 * ( tcp* = 2675 mm² (( Dcp/2) - d(s)/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d(s)/2 ))] Shear stress, Ss = 9.44 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since Ss < Ss.all,therefore the lug size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-6820 A/B MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure c) Bearing Stress Maximum bearing force, P = 25260 N Cross sectional area of lug eye, Ab = Dp * ( tL + 2tcp ) = 825 mm² Bearing stress, Sbr = Fbr / Ab = 30.62 N/mm² Allowable bearing stress, Sbr.all ( = 0.9Sy ) = 223.39 N/mm² Since Sbr < Sbr.all,therefore the lug size is satisfactory. d) Unity check, Combine Stresses St Sbr Ss = 0.30 = ------------- + -------------- + -------------- is < than 1 St.all Sbr.all Ss.all Therefore, the lifting lug size is Satisfactory. 5) WELD SIZE CALCULATIONS Weld leg used, = 10 mm Weld throat thickness used, tr = 7 mm Filler metal material : E-43 Fillet weld joint efficiency, E = 0.49 Welding stress for steel grade 43 ( E-43 ), = 125 N/mm² Allowable welding stress,Sw = 61.25 N/mm² a) Tensile Stress Maximum tensile force,Ft = 25260 N Area of weld, Aw = 2*(tL+wL)*tr = 3038 mm² Tensile stress, St = [(Ft/Aw)] = 8.31 N/mm² Since St < Sw,therefore weld leg is satisfactory. (b) Shear stress Maximum shear force,Ft = 25260 N Shear stress, Ss = (Ft/Aw) = 8.31 N/mm² Allowable welding stress for steel grade 43 ( E-43 ), Sw = 61.25 N/mm² Since Ss < Sw,therefore weld leg dimension is SATISFACTORY. (c) Bending stress Maximum bending force,Fb = 46723 N Bending stress, Sb = [(Fb/Aw)] = 3.41 N/mm² Allowable welding stress for steel grade 43 ( E-43 ), Sw = 61.25 N/mm² Since Sb < Sw,therefore weld leg dimension is SATISFACTORY. SLING AND WIRE ROPE CALCULATION @ SPREADER PIPE @ (TANK ) ITEM : T-6820 A/B PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Selection of Sling's Safe Working Load (SWL) (SLING S5 & S6) - REFER DWG Design safety Factor = 1 Number of leg = 2 Vertical WLL = Weight of Load lifted / No. of legs = 4,059 lb = ( Load on each sling ) = 1.83 ton To calculate actual Sling capacity when lifting load at specified angle, a Sling angle factor will be used as shown in the calculation below : TABLE 1 Sling ratio = 136% Actual Sling Capacity = Factor x Rated Capacity of Sling being used Min Required Sling SWL = 1.83 ton (From calculation above) Sling angle = 60 (Advisable using btw 45° to 60, but not below 30° ) Rated Sling SWL used = 1.55 ton (As per table EN13414-1-Table 3) Actual Sling Capacity = 0.866 x 1.55 = 1.3423 ton NOT OK, USED OTHER ROPE SWL (Advisable using below 80% : check!! Wire Rope Diameter calculation Safe working load = (diameter)2 × 8 Wire rope diameter (inch) & SWL ( tonne ) The above formula can be used to estimate SWL of the wire rope to be used when lifting the loads. Therefore, the estimated diameter of wire ropes is : Wire rope diameter = SQRT (SWL of wire rope / 8) = 0.44017 in = 11.1803 mm To Used ----> 12 mm (See note) (See note) NOTE : ( Estimated values only. Please consult wire rope manufacturer for confirmation on rated SWL of wire rope calculated) SUMMARY WIRE ROPE DIAMETER USED = 12 mm SWL OF SLING USED = 1.55 ton MIN. REQUIRED SWL OF SLING = 1.83 ton SLING AND WIRE ROPE CALCULATION @ (TANK ) LIFTING LUG ITEM : T-6820 A/B PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Selection of Sling's Safe Working Load (SWL) (SLING S1 / S2 / S3 /S4 ) - REFER DWG Design safety Factor = 2 Number of leg = 4 Vertical WLL = Weight of Load lifted / No. of legs = 2,100 lb = ( Load on each sling ) = 0.95 ton To calculate actual Sling capacity when lifting load at specified angle, a Sling angle factor will be used as shown in the calculation below : TABLE 1 Sling ratio = 105% Actual Sling Capacity = Factor x Rated Capacity of Sling being used hock: Min Required Sling SWL = 0.95 ton (From calculation above) USE VALUE FROM TABLE Sling angle = 60 (Advisable using btw 45° to 60, but not below 30° ) Rated Sling SWL used = 1.05 ton (As per table EN13414-1-Table 3) Actual Sling Capacity = 0.866 x 1.05 = 0.9093 ton NOT OK, USED OTHER ROPE SWL (Advisable using below 80% : check!! Wire Rope Diameter calculation Safe working load = (diameter)2 × 8 Wire rope diameter (inch) & SWL ( tonne ) The above formula can be used to estimate SWL of the wire rope to be used when lifting the loads. Therefore, the estimated diameter of wire ropes is : Wire rope diameter = SQRT (SWL of wire rope / 8) = 0.36228 in = 9.20202 mm To Used ----> 10 mm (See note) (See note) NOTE : ( Estimated values only. Please consult wire rope manufacturer for confirmation on rated SWL of wire rope calculated) SUMMARY WIRE ROPE DIAMETER USED = 10 mm SWL OF SLING USED = 1.05 ton MIN. REQUIRED SWL OF SLING = 0.95 ton LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-2521 PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure WARNING : DO NOT LIFT MORE THAN Fy 10.00 ° OF ANGLE U `U nasrul: Dcp PLEASE ENABLE w2 MACRO LINK tL rL d w2 A A tcp k tank Pad b plate Wp w3 a tp w1 Lp tM Weight of skid , We = 0 kg = 0N Number of lifting lug, N = 4 1) LIFTING LUG Distance k = 80 mm Lug radius, rL = 42 mm Diameter of hole, d = 27 mm Lug thickness, tL = 12 mm Collar plate thickness, tcp = 0 mm Collar ring diameter, Dcp = 0 mm Length a = 84 mm Length b = 70 mm Pad length, Lp = 140 mm Pad width, Wp = 100 mm Pad thickness, tp = 6 mm Shell thickness, tM = 6 mm Angle, U = 10 ° Shackle S.W.L : 4.75 tons Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = 22.40 mm LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-2521 PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure Per PTS Section 6.3 Check a) Lug hole diameter, d shall be Max of i) Dp + 3mm = 25.40 mm ii) Dp X 1.05 = 23.52 mm b) Lug hole diameter, d shall be less than < (Dp + 6mm) = 28.40 mm Diameter of hole, d 27 result a) = satisfactory b) = satisfactory Hole,d Diameter of hole, d btw 25.40 27 28.40 OK Clearance btw shackle & lug size Result lug thk A= 42.9 mm 24 mm OK lug radius C= 95.5 mm 42 mm OK Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTABLE 2) LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = A 36 Specified yield stress, Sy = 248.22 N/mm² Impact load factor, p = 2.00 Specified Tensile stress, Ts = 206.00 N/mm² 3) ALLOWABLE STRESSES Allowable tensile stress, St.all ( = 0.6 Sy ) = 148.93 N/mm² Allowable bearing stress, Sbr.all ( = 0.9 Sy ) = 223.40 N/mm² Allowable bending stress, Sbn.all ( = 0.66 Sy ) = 163.83 N/mm² Allowable shear stress, Ss.all ( = 0.4 Sy ) = 99.29 N/mm² 4) LIFTING LUG DESIGN - VERTICAL LIFTING 4.1 DESIGN LOAD Design load , Wt ( = p.We ) = 0N Design load per lug, W ( = Wt / N ) = 0N Vertical component force, Fy = 0N LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-2521 PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure 4.2 STRESS CHECK AT PIN HOLE (a) Tensile Stress Vertical component force, Fy = 0N Cross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) = 684 mm² Tensile stress, St ( = Fy / Ae ) = 0.00 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. (b) Bearing Stress Vertical component force, Fy = 0N Cross-sectional area , Ae = Dp x (tL+2tcp) = 269 mm² Bearing stress, Sbr ( = Fy / Ae ) = 0.00 N/mm² Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory. (c) Shear Stress Vertical component force, Fy = 0N Cross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) = 684 mm² Shear stress, Ss ( = Fy / Ae ) = 0.00 N/mm² Since Ss < Ss.all,therefore the lifting lug size is satisfactory. (d) Unity check, Combine Stresses St Sbr Ss = 0.00 ------------- + -------------- + -------------- is < than 1 St.all Sbr.all Ss.all Therefore, the lifting lug size is Satisfactory. LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-2521 PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure 5) STRESS CHECK AT SECTION A-A (a) Bending Stress Bending stress due to Pa ( = Fy x tan U ) = 0N Bending moment, Mb ( = Pa x k ) = 0 Nmm Section modulus, Z = ( 2*rL*tL / 6 ) + ( 2*(Dcp*tcp / 6) ) 2 = 2016 mm3 Bending stress, Sbn ( = Mb/Z ) = 0.00 N/mm² Since Sbn < Sb.all, therefore the lifting lug size is satisfactory. (b) Tensile Stress due to Fy Cross section area, Ae = 2rL x tL = 1008 mm² Tensile Stress, St = Fy / Ae = 0.00 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. (c) Unity check Combine Stress Ratio, CS = (St / St.all) + (Sb / Sbn.all) = 0.00 Since CS <= 1.0, therefore lifting lug size is satisfactory. 6) DESIGN OF WELD SIZE AT LUG TO PAD JOINT 6.1 GENERAL Weld leg , w1 = 6 mm Weld throat thickness, tr1 = 4.2 mm Weld leg , w2 = 4.2 mm Weld throat thickness, tr2 = 3.0 mm Fillet weld joint efficiency, E = 0.6 Allowable welding stress for steel grade 43 ( E-43 ) Sw = E x Ts = 123.6 N/mm² LIFTING LUG DESIGN CALCULATION (TANK) ITEM : T-2521 PROJ NO : C.I Injection system (A-6810) - Permas Field Substructure 6.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES Area of weld, Aw1 ( = tr1 ( a + 2b ) = 950 mm² Area of weld, Aw2 ( = tr2 ( (2a-tM) + (2M+tM ))) = 534 mm² Total area of weld, Aw ( = Aw1 + Aw2 ) = 1485 mm² 6.3 STRESS DUE TO FORCE Fy Component force, Fy = 0N Shear stress, Ssx ( = Fy / Aw ) = 0.00 N/mm² Allowable welding stress, Sa ( = E.Sw ) = 123.60 N/mm² Since Ssx < Sa, therefore the selected weld size is satisfactory . 7.0 DESIGN OF WELD SIZE AT PAD TO TANK JOINT 7.1 GENERAL Weld leg , w = 4.2 mm Weld throat thickness, tr = 3.0 mm Fillet weld joint efficiency, E = 0.6 Allowable welding stress for steel grade 43 ( E-43 ) Sw = E x Ts = 123.6 N/mm² 7.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES Area of weld, Aw ( = 2 tr ( Wp + Lp ) ) = 1425 mm² 7.3 STRESS DUE TO FORCE Fy Component force, Fy = 0N Shear stress, Ssx ( = Fy / Aw ) = 0.00 N/mm² Allowable welding stress, Sa ( = E.Sw ) = 74.16 N/mm² Since Ssx < Sa, therefore the selected weld size is satisfactory . LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-2521 MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Weight of component to be lifted ( + Spreader beam weight ) = 34 kg Component force acting on beam, F = 668 N Impact factor = 2 SPREADER LGTH = 1) PIPE SIZING LENGTH SKID + 280 nasrul: INPUT -TablePipe Pipe size : 6 in SCH 40 Outer diameter of pipe, D = 168.3 mm Thickness of pipe, = 7.11 mm Outer Radius of pipe, R = 84.15 mm Inner radius of pipe, r = 77.04 mm Section modulus of pipe, Zx-x = 139,230 mm3 Second Moment of pipe, I = 730,961 mm4 Cross section area of pipe, A = 3600 mm2 Unbraced length of member, L = 1204 mm Modulus Of Elasticity , E = 2.0E+05 N/mm² Spreader Pipe Weight = 34.0 kg Material used = A 106.Gr.B Specified yield stress, Sy = 241.32 N/mm² a) Bending Stress F R1=R2=F/2 L R1 R2 Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam. (See Appendix) Total bending moment, M ( = F*L / 4 ) = 200,939 Nmm Bending stress, Sb ( = M / Zx-x ) = 1.44 N/mm² Max Bending stress = Fx * L / ( 60 * E * I ) 3 = 0.13 N/mm² Allowable bending stress, Sb.all ( = 0.66Sy ) = 159.27 N/mm² Since Sb < Sb.all,therefore the pipe size is satisfactory. b) Compressive Stress Compressive force, Fc = 668 N Compressive stress, Sc = Fc / A = 0.19 N/mm² Allowable compressive stress, Sc.all ( = 0.6Sy ) = 144.79 N/mm² c) Combined stresses, U = Sc + Sb = 0.01 Sc.all Sb.all Since U < 1, therefore the pipe size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-2521 MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure 2) LUG SIZING Fz P rL(s) tL(s) d(s) a Fx Fyl hL(s) wL(s) Lug radius, rL(s) = 70 mm Lug thickness, tL(s) = 25 mm Lug base width, wL(s) = 192 mm Diameter of hole, d(s) = 33 mm Distance from lug hole to base, hL(s) = 80 mm Collar plate thickness, tcp = 0 mm Collar ring diameter, Dcp = 0 mm Clearance btw shackle & lug size Result Lug thickness, tL(s) A= 42.9 mm 25 mm OK Lug radius, rL(s) C= 95.5 mm 70 mm OK Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTABLE Per PTS Section 6.3 Check a) Lug hole diameter, d shall be Max of i) Dp + 3mm = 31.70 mm ii) Dp X 1.05 = 30.14 mm b) Lug hole diameter, d shall be less than < (Dp + 6mm) = 34.70 mm Dp = 33 result a) = satisfactory b) = satisfactory Hole,d Diameter of hole, d(s) btw 31.70 33 34.70 OK No of lug eye, = 2 Maximum combined force acting on lug eye, Fc = 334 N = 34 kg LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = A 36 Specified yield stress, Sy = 248.21 N/mm² Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = 163.82 N/mm² Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = 186.16 N/mm² Allowable tensile stress, St.all ( = 0.6Sy ) = 148.93 N/mm² Allowable bearing stress, Sbr.all ( = 0.9Sy ) = 223.39 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-2521 MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure SHACKLES Shackle rating ( S.W.L ) : 8.5 tons Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = 28.70 mm MAXIMUM SLING TENSION ON PADEYE Ts = 167 N FACTOR OF SAFETY F.O.S. = 2.00 DESIGN LOAD: SLING TENSION P = FOS * Ts P = 334 N LIFTING ANGLE a = 60.00 Deg. ACTUAL OUT OF PLANE ANGLE b = 0.00 Deg. VERTICAL FORCE ON PADEYE Fz = P * sin a Fz = 289 N OUT OF PLANE FORCE Fyl = 5% of P Fyl = 17 N HORIZONTAL FORCE ON PADEYE Fx = P * cos a Fx = 167 N Horizontal dist.PIN CL to N.A. exl = 0.00 3) STRESS CHECK AT BASE a) Moment Calc at distance , H In Plane Moment My = ( Fx*H ) - ( Fz*ex l) My = 13,351 N-mm Out of plane moment Mx = ( FyI*hL ) Mx = 1,335 N-mm b) Tensile Stress Maximum tensile force, ft = Fz / [ tL(s) * wL(s) ] = 0N Allowable tensile stress, St.all ( = 0.6Sy ) = 148.93 N/mm² Since ft < St.all, therefore the lug size is satisfactory. c) Bending stress (In Plane) Maximum bending stress , fbx = ( 6*Mx ) / ( wL(s) * [(tL(s)+tcp)^2] ) = 0 N/mm² Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane = 163.82 N/mm² Since fbx < fbx.all,therefore the lug size is satisfactory. d) Bending stress (Out of Plane) Maximum bending stress , fby = ( 6*My ) / [ tL(s) +(2*tcp)] * [ wL(s)^2 ] ) = 0 N/mm² Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane = 186.16 N/mm² Since fby < fby.all,therefore the lug size is satisfactory. e) Unity Check : Combined stresses, U = St/St.all + fby/fby.all + fbx/fbx.all = 0.00 Since U < 1, therefore the lug size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-2521 MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure f) SHEAR stress (In Plane) Maximum SHEAR stress , fsx = Fx / [ wL(s) * tL(s) ] = 0 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. g) Bending stress (Out of Plane) Maximum SHEAR stress , fsy = Fyl / [ wL(s) * tL(s) ] = 0 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since fsx < Ss.all,therefore the lug size is satisfactory. 3.1) CHECKING VON-MISES CRITERIA a) Sum of stress in X-PLANE fx = St + fby = 0.21 N/mm² b) Sum of stress in Y-PLANE fy = St + fbx = 0.19 N/mm² c) Therefore, average Shear stress fxy = SQRT [ (fsx^2)+(fsy^2) ] = 0.03 N/mm² d) Maximum Combined stress Fcomb = SQRT [ (fx^2)+(fy^2)-(fx+fy+3fxy^2) ] = Err:502 N/mm² Allowable combined stress : Fcomb.all ( = 0.66Sy ) = 163.82 N/mm² Since fsx < Ss.all,therefore the lug size is Err:502 4) STRESS CHECK AT PIN HOLE a) Tensile Stress Maximum tensile force, P = 334 N Cross sectional area of lug eye, At = [ 2 * ( tL(s)* ( rL(s) - d(s)/2 ))] + [ 2 * ( tcp* = 2675 mm² (( Dcp/2) - d(s)/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d(s)/2 ))] Tensile stress, St = 0.12 N/mm² Allowable tensile stress, St.all ( = 0.6Sy ) = 148.926 N/mm² Since St < St.all, therefore the lug size is satisfactory. b) Shear Stress ` Maximum shear force, P = 334 N Cross sectional area of lug eye, At = [ 2 * ( tL(s)* ( rL(s) - d(s)/2 ))] + [ 2 * ( tcp* = 2675 mm² (( Dcp/2) - d(s)/2 ))] + [ 2 * ( tcp* (( Dcp/2) - d(s)/2 ))] Shear stress, Ss = 0.12 N/mm² Allowable shear stress, Ss.all ( = 0.4Sy ) = 99.28 N/mm² Since Ss < Ss.all,therefore the lug size is satisfactory. LIFTING SPREADER PIPE SIZING CALCULATIONS (TANK) nasrul: PLEASE ENABLE ITEM : T-2521 MACRO LINK PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure c) Bearing Stress Maximum bearing force, P = 334 N Cross sectional area of lug eye, Ab = Dp * ( tL + 2tcp ) = 825 mm² Bearing stress, Sbr = Fbr / Ab = 0.40 N/mm² Allowable bearing stress, Sbr.all ( = 0.9Sy ) = 223.39 N/mm² Since Sbr < Sbr.all,therefore the lug size is satisfactory. d) Unity check, Combine Stresses St Sbr Ss = 0.00 = ------------- + -------------- + -------------- is < than 1 St.all Sbr.all Ss.all Therefore, the lifting lug size is Satisfactory. 5) WELD SIZE CALCULATIONS Weld leg used, = 10 mm Weld throat thickness used, tr = 7 mm Filler metal material : E-43 Fillet weld joint efficiency, E = 0.49 Welding stress for steel grade 43 ( E-43 ), = 125 N/mm² Allowable welding stress,Sw = 61.25 N/mm² a) Tensile Stress Maximum tensile force,Ft = 334 N Area of weld, Aw = 2*(tL+wL)*tr = 3038 mm² Tensile stress, St = [(Ft/Aw)] = 0.11 N/mm² Since St < Sw,therefore weld leg is satisfactory. (b) Shear stress Maximum shear force,Ft = 334 N Shear stress, Ss = (Ft/Aw) = 0.11 N/mm² Allowable welding stress for steel grade 43 ( E-43 ), Sw = 61.25 N/mm² Since Ss < Sw,therefore weld leg dimension is SATISFACTORY. (c) Bending stress Maximum bending force,Fb = 46723 N Bending stress, Sb = [(Fb/Aw)] = 3.41 N/mm² Allowable welding stress for steel grade 43 ( E-43 ), Sw = 61.25 N/mm² Since Sb < Sw,therefore weld leg dimension is SATISFACTORY. SLING AND WIRE ROPE CALCULATION @ SPREADER PIPE @ (TANK ) ITEM : T-2521 PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Selection of Sling's Safe Working Load (SWL) (SLING S5 & S6) - REFER DWG Design safety Factor = 2 Number of leg = 2 Vertical WLL = Weight of Load lifted / No. of legs = 0 lb = ( Load on each sling ) = 0.00 ton To calculate actual Sling capacity when lifting load at specified angle, a Sling angle factor will be used as shown in the calculation below : TABLE 1 Sling ratio = 0% hock: USE VALUE FROM TABLE Actual Sling Capacity = Factor x Rated Capacity of Sling being used Min Required Sling SWL = 0.00 ton (From calculation above) Sling angle = 60 (Advisable using btw 45° to 60, but not below 30° ) Rated Sling SWL used = 2.12 ton (As per table EN13414-1-Table 3) Actual Sling Capacity = 0.866 x 2.12 = 1.83592 ton OK Proceed to next calc Wire Rope Diameter calculation Safe working load = (diameter)2 × 8 Wire rope diameter (inch) & SWL ( tonne ) The above formula can be used to estimate SWL of the wire rope to be used when lifting the loads. Therefore, the estimated diameter of wire ropes is : Wire rope diameter = SQRT (SWL of wire rope / 8) = 0.51478 in = 13.0755 mm To Used ----> 14 mm (See note) (See note) NOTE : ( Estimated values only. Please consult wire rope manufacturer for confirmation on rated SWL of wire rope calculated) SUMMARY WIRE ROPE DIAMETER USED = 14 mm SWL OF SLING USED = 2.12 ton MIN. REQUIRED SWL OF SLING = 0.00 ton SLING AND WIRE ROPE CALCULATION @ (TANK ) LIFTING LUG ITEM : T-6820 A/B PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Selection of Sling's Safe Working Load (SWL) (SLING S1 / S2 / S3 /S4 ) - REFER DWG Design safety Factor = 2 Number of leg = 4 Vertical WLL = Weight of Load lifted / No. of legs = 0 lb = ( Load on each sling ) = 0.00 ton To calculate actual Sling capacity when lifting load at specified angle, a Sling angle factor will be used as shown in the calculation below : TABLE 1 Sling ratio = 0% Actual Sling Capacity = Factor x Rated Capacity of Sling being used Min Required Sling SWL = 0.00 ton (From calculation above) Sling angle = 60 (Advisable using btw 45° to 60, but not below 30° ) Rated Sling SWL used = 1.05 ton (As per table EN13414-1-Table 3) Actual Sling Capacity = 0.866 x 1.05 = 0.9093 ton OK Proceed to next calc Wire Rope Diameter calculation Safe working load = (diameter)2 × 8 Wire rope diameter (inch) & SWL ( tonne ) The above formula can be used to estimate SWL of the wire rope to be used when lifting the loads. Therefore, the estimated diameter of wire ropes is : Wire rope diameter = SQRT (SWL of wire rope / 8) = 0.36228 in = 9.20202 mm To Used ----> 10 mm (See note) (See note) NOTE : ( Estimated values only. Please consult wire rope manufacturer for confirmation on rated SWL of wire rope calculated) SUMMARY WIRE ROPE DIAMETER USED = 10 mm SWL OF SLING USED = 1.05 ton MIN. REQUIRED SWL OF SLING = 0.00 ton SKID SUPPORT STRENGTH CALCULATIONS ITEM : C.I. SKID (A-6810) PROJECT NO. C.I Injection system (A-6810) - Permas Field Substructure Maximum Force Acting on Skid Support = 5,563 kg Component force acting on beam, Fx = 54,570 N 1) BEAM SIZING Assumption : The worst case scenario occur at the maximum span of Skid Support. nasrul: 11 PLEASE ENABLE MACRO LINK 203 EMPTY WEIGHT,TANK WEIGHT AND 8 MAX LIFT FORCE 203 Member size : UC 200 X 200X 56.2 kg/m3 Depth of section, D = 203 mm Flange width, = 203 mm Thickness of flange, = 11.0 mm Thickness of web, = 8 mm Unbraced length of member, L = 6,000 mm Cross sectional area, A = 6,136 mm² Section modulus , Zx-x = 58,905,600 mm³ Material used = A 36 Specified yield stress, Sy = 248.21 N/mm² a) Bending Stress Fx L R1 R2 Assume single point load at mid of beam, Total bending moment, M ( = Fx.L / 4 ) = 81,855,120 Nmm Bending stress, Sb ( = M / Zx-x ) = 1.39 N/mm² Allowable bending stress, Sb.all ( = 0.66Sy ) = 163.82 N/mm² Since Sb < Sb.all,therefore the section size is satisfactory. b) Compressive Stress Compressive force, Fc = 54,570 N Compressive stress, Sc = 8.89 N/mm² Allowable compressive stress, Sc.all ( = 0.6Sy ) = 148.93 N/mm² Combined stresses, Sc Sb U = + = 0.07 Sc.all Sbn.all Since U < 1, therefore the Section size is satisfactory. LIFTING LUG DESIGN CALCULATION (SKID) ITEM : T-6901 PROJECT NO. SKO Pipelines & Facilities Rejuvenation Project : WLK-A Fy WARNING : DO NOT LIFT MORE THAN `U 10.00 ° OF ANGLE U Dcp w2 tL nasrul: rL d w2 PLEASE ENABLE MACRO LINK A A tcp k I-Beam Pad plate b Wp w3 a tp w1 Lp Weight of skid , We = 29 kg = 283 N Number of lifting lug, N = 4 Number of tailing lug, Nt = 0 1.1 LIFTING LUG Distance k = 80 mm Lug radius, rL = 70 mm Diameter of hole, d = 33 mm Lug thickness, tL = 20 mm Collar plate thickness, tcp = 10 mm Collar ring diameter, Dcp = 100 mm Length a = 140 mm Length b = 53 mm Pad length, Lp = 206 mm Pad width, Wp = 180 mm Pad thickness, tp = 10 mm Angle, U = 10 ° Shackle S.W.L : 8.5 tons Type of shackle BOLT Type Anchor shackle G2130 Pin size, Dp = 28.70 mm Per PTS Section 6.3 Check a) Lug hole diameter, d shall be Max of i) Dp + 3mm = 31.70 mm ii) Dp X 1.05 = 30.14 mm b) Lug hole diameter, d shall be less than < (Dp + 6mm) = 34.70 mm Diameter of hole, d 33 result a) = satisfactory b) = satisfactory Hole,d Diameter of hole, d btw 31.70 33 34.70 OK Clearance btw shackle & lug size Result lug thk A= 42.9 mm 40 mm check lug radius C= 95.5 mm 70 mm check Since A & C clearance against Lug size , Therefore the Lug is is CHECK THK. LIFTING LUG DESIGN CALCULATION (SKID) ITEM : T-6901 PROJECT NO. SKO Pipelines & Facilities Rejuvenation Project : WLK-A Fy 2.0 LIFTING LUG MATERIAL & MECHANICAL PROPERTIES Material used = A 36 Specified yield stress, Sy = 248.22 N/mm² Impact load factor, p = 2.00 Specified Tensile stress, Ts = 206.00 N/mm² 3.0 ALLOWABLE STRESSES Allowable tensile stress, St.all ( = 0.6 Sy ) = 148.93 N/mm² Allowable bearing stress, Sbr.all ( = 0.9 Sy ) = 223.40 N/mm² Allowable bending stress, Sbn.all ( = 0.66 Sy ) = 163.83 N/mm² Allowable shear stress, Ss.all ( = 0.4 Sy ) = 99.29 N/mm² 4.0 LIFTING LUG DESIGN - VERTICAL LIFTING 4.1 DESIGN LOAD Design load , Wt ( = p.We ) = 565 N Design load per lug, W ( = Wt / N ) = 141 N Vertical component force, Fy = 141 N 4.2 STRESS CHECK AT PIN HOLE (a) Tensile Stress Vertical component force, Fy = 141 N Cross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) = 2140 mm² Tensile stress, St ( = Fy / Ae ) = 0.07 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. (b) Bearing Stress Vertical component force, Fy = 141 N Cross-sectional area , Ae = Dp x (tL+2tcp) = 1148 mm² Bearing stress, Sbr ( = Fy / Ae ) = 0.12 N/mm² Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory. (c) Shear Stress Vertical component force, Fy = 141 N Cross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) = 2140 mm² Shear stress, Ss ( = Fy / Ae ) = 0.07 N/mm² Since Ss < Ss.all,therefore the lifting lug size is satisfactory. (d) Unity check, Combine Stresses St Sbr Ss = 0.00 ------------- + -------------- + -------------- is < than 1 St.all Sbr.all Ss.all Therefore, the lifting lug size is Satisfactory. LIFTING LUG DESIGN CALCULATION (SKID) ITEM : T-6901 PROJECT NO. SKO Pipelines & Facilities Rejuvenation Project : WLK-A Fy 5.0 STRESS CHECK AT SECTION A-A (a) Bending Stress Bending stress due to Pa ( = Fy x tan U ) = 25 N Bending moment, Mb ( = Pa x k ) = 1,993 Nmm Section modulus, Z ( = (2*rL*tL2/6) + (2*(Dcp*tcp/6)) = 12667 mm3 Bending stress, Sbn ( = Mb/Z ) = 0.16 N/mm² Since Sbn < Sb.all, therefore the lifting lug size is satisfactory. (b) Tensile Stress due to Fy Cross section area, Ae (=2rL x tL) = 2800 mm² Tensile Stress, St (=Fy/Ae) = 0.05 N/mm² Since St < St.all, therefore the lifting lug size is satisfactory. (c) Unity check Combine Stress Ratio, CS (= St/St.all + Sb/Sbn.all) = 0.00 Since CS <= 1.0, therefore lifting lug size is satisfactory. 6.0 DESIGN OF WELD SIZE AT LUG TO PAD JOINT 6.1 GENERAL Weld leg , w1 = 10 mm Weld throat thickness, tr1 = 7.1 mm Weld leg , w2 = 7 mm Weld throat thickness, tr2 = 4.9 mm Fillet weld joint efficiency, E = 0.6 Allowable welding stress for steel grade 43 ( E-43 ) Sw = 0.3xTs = 61.8 N/mm² 6.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES Area of weld, Aw1 ( = tr1 ( a + 2b ) = 1739 mm² Area of weld, Aw2 ( = tr2 ( (2a-tM) + (2M+tM ))) = 1386 mm² Total area of weld, Aw ( = Aw1 + Aw2 ) = 3125 mm² 6.3 STRESS DUE TO FORCE Fy Component force, Fy = 141 N Shear stress, Ssx ( = Fy / Aw ) = 0.05 N/mm² Allowable welding stress, Sa ( = E.Sw ) = 61.80 N/mm² Since Ssx < Sa, therefore the selected weld size is satisfactory . 7.0 DESIGN OF WELD SIZE AT PAD TO SKID 7.1 GENERAL Weld leg , w = 7 mm Weld throat thickness, tr = 4.9 mm Fillet weld joint efficiency, E = 0.6 Allowable welding stress for steel grade 43 ( E-43 ) Sw = 0.3xTs = 61.8 N/mm² 7.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES Area of weld, Aw ( = 2 tr ( Wp + Lp ) ) = 3821 mm² 7.3 STRESS DUE TO FORCE Fy Component force, Fy = 141 N Shear stress, Ssx ( = Fy / Aw ) = 0.04 N/mm² Allowable welding stress, Sa ( = E.Sw ) = 37.08 N/mm² Since Ssx < Sa, therefore the selected weld size is satisfactory . ALLOWABLE STRESS REFERENCE ITEM : #REF! PROJECT NO. #REF! nasrul: PLEASE ENABLE MACRO LIN EMPTY WEIGHT,TANK WEIG MAX LIFT FORCE ALLOWABLE STRESS REFERENCE ITEM : #REF! PROJECT NO. #REF! CRO LINK NK WEIGHT AND