61889691-17-Tons

June 16, 2018 | Author: Fernando Humberto | Category: Bending, Stress (Mechanics), Strength Of Materials, Physics, Physics & Mathematics
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Calculation SheetProject title Calc no Client Proj no Calculation title Phase/CTR Path Page of 1.0 Introduction The purpose of this calculation is to document the design of a padeye. This spreadsheet was developed primarily to satisfy the requirements of Reference 1, Edwin P Russo and Rudolph A Hall, " Systematic Approach to Lifting Eye Design" 1.1 Applied Loading The lifting loads were determined from a lift analysis which considered a load factor of 2.0 applied to the structure gravity loads, in order to determine the design load to be used for the padeye design. From SACS analysis of bridge lift: Padeye design load 72.2 kips 2.0 Padeye Configuration The principal features of the padeye are shown below, while details have been included on drawing INSERT DRAWING NUMBER. r R SWL SWL Sw  R 2r b b c B a B Rh g A A h t Tp t T where: SWL = safe working load b = angle from horizontal Rh = Hole radius r = Cheek plate radius R = Main plate radius Tp = Main plate thickness t = Cheek plate thickness T = Total plate thickness h = Base width b = Distance from edge of taper to center of hole c = Distance from base of plate to center of hole Sw = Cheek plate weld leg e = Main plate to cheek plate clearance a = Taper angle p = Shackle pin diameter The dimension T should equal 60 - 85% of shackle jaw width. 938 in Cheek plate radius = 2.563 in Main plate to cheek plate clearance a = 90 degrees = 1.094 in Say main plate radius = 2.047 radians Ph = 36.429 > and < 2. capacity 17 tons.thecrosbygroup.250 in Total plate thickness h = 14. determined at the base of padeye plate and at various points along padeye plate Taper Angle . shackle pin size 1. in-plane bending and out-of-plane bending 3.05 short tons Use G-2130 Crosby 2-inch shackle. Shackle load 36.5 kips Vertical In-Plane Component 3.1-1) b = 60 degrees = 1.1 kips Horizontal In-Plane Component Pv = 62.630 in Pin hole diameter should be 2.63 inches Refer to http://catalog.375 in Rh = 1.5t The cheek plate thickness (t) should be less than or equal to the plate thickness (Tp).0 Stress Checks The following stress checks are performed in this analysis Bearing Stress .750 in Base width b = 0.com/maininterface. Weld Stress . two planes from pin hole to plate edge in load direction.check on value of taper angle Out-of-plane bending . should be 1/8" greater than the selected shackle pin size The main plate radius is approximately R = 3 Rh Cheek plate radius is approximately r = R .031 in Hole radius r = 2. d.630 in 200 ton shackle Fy = 36 ksi Yield strength of padeye material 3.check combined tension.htm Shackle jaw width A = 2.1 Load P = 72. calculated around cheek plate Combined Stress .938 in Main plate radius Tp = 1.571 radians p = 1.1. Shear Stress . calculated on the net area through plate. determines the average bearing stress of pin on hole.2 kips Design Load (max from Table 1. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of The pin hole diameter.000 in Distance from edge of taper to center of hole c = 3.938 in Distance from base of plate to center of hole e = 0. Tensile Stress .1 kips = 18.382 in Total padeye width should be between 1.375 in Cheek plate radius R = 2.031 in so main plate radius should be approximately 3.025 in Shackle pin size D = 1.063 in Hole radius = 1.625 in Cheek plate thickness T = 2.2 Design Stresses .000 in Main plate thickness t = 0. 2 ksi Allowable = 0. the distance used in calculations.6 times the hole diameter.699 OK Tensile Stress From Section D3.2 of AISC.0.4C2) / TP h The maximum principal stress that exists on an element is given by: fmax = ((fa + fb)/2) + [((fa + fb)/2)2 + fv2]1/2 and the maximum principal shear stress on an element is given by: fmax = [((fa + fb)/2)2 + fv2]1/2 . is the minimum of 4 times the plate thickness at the pinhole or 1. ultimate strength UC4 = 0.45Fy UC3 = 0.707*Sw)) Sw = 0.5P cosb (1 .0 ksi E70XX Electrodes. Tensile stress = 9.500 in fillet weld size Allowable = 21.608 OK Shear Stress Shear = 10.3 in Allowable stress = 16.326 OK Combined Stress The axial stress due to uniform tension along a section is given by: fa = P sinb / TP h The elemental bending stress which is linearly distributed along the section B-B may be expressed as: fb = 12PC (b cosb .4 ksi Allowable = 0.600 OK Weld Stress Load on one weld = 36.0 ksi Allowable = 0. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of Bearing Stress Bearing = 19.9Fy UC1 = 0.7 ksi Bearing = P/(T*p) Allowable = 32. For Section A-A.3Fuw Fw = 70.7 ksi Tensile = P/(Effective width * plate thickness) Effective width = 3.8 ksi Weld Shear = F/(2pr*(0.3 in Plate thickness = 2.1 kips Weld load F = P/2 (assume each weld takes full load) Weld Shear = 6.1 ksi Shear = P/(4(r-Rh)*t+2(R-Rh)*Tp) Allowable = 14.4 ksi Allowable = 0. The shearing stress varies parabolically along the length of the padeye and is given as: fV = 1.4Fy UC2 = 0. across the hole.5 h sinb + R sinb) / TP h2 where C represents the distance of an element from the neutral axis. b should be replaced by c in the above equation. 0 3.10 4.27 0.8 ksi (M / S) Allowable = 27.0 ksi (0.2 -0. Consequently.4 0.5 5.6 0.40 4.2 -6. shear stress.6 Fy) Tensile stress / allowable = 0. Stresses calculated at Section A-A.7 0.6 ksi (0. is not critical.10 0. In addition as the padeyes are aligned with the slings.2 -3.75 Fy) Out-of-plane bending stress / allowable = 0.1 3.16 2.30 4.00 4.6 kips (P / 20) Lever arm.1 0. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of The following table shows the values of the axial stress.29 0.5 in3 (1/6 hT2p) Out-of-plane bending stress = 5.20 OK Maximum in-plane bending stress = 0.7 6. Out-of-plane load PL = 3.1 1.2 0.1 4.4Fy Stresses calculated at Section B-B. no analysis of this section is required.24 3.17 3.50 4.00 OK Sum of ratios = 0. principal stress and maximum shear stress at various points along the length of the plate.2 0.2 ksi Allowable = 21.29 4.5 3.1 0.2 -7.0 1.23 0.1 3.5 5.29 0.11 0.18 0.3 2.29 Out-of Plane Bending Check out-of plane bending at section A-A.7 0.41 OK For out-of-plane bending at the B-B axis.6 2.0 0.8 0.2 0.0 ksi Allowable = 23.10 1.02 MAX 0. M = 14.7 6.6 0.29 4. .5 3.4 0.03 1. take a minimum load of 5% of the total applied load.0 3.00 4.7 3.7 0.2 -4.4 0.29 0.2 in-kips (PL c) Section modulus S = 2.3 0.2 2.8 ksi (0.30 4.7 0.22 0.40 4.2 -2. Note that this out of plane stress is combined with the inplane bending stress and tensile stress calculated above.26 3.0 0.66 Fy) In-plane bending stress / allowable = 0.0 0. the out-of-plane bending moment is reduced by the ratio of the lever arms and therefore.3 0.9 0.00 1.10 4.26 0.29 0. which is the worst case scenario.21 OK Maximum tensile stress = 4.20 4.3 3.7 0. bending stress.10 2. c = 3.2 -1.5 0.7 0.1 0.5 0.2 -1.3 2. ignoring the gusset plates.6 0.02 0.2 0.2 0.22 3.6 ksi Allowable = 0.29 0.16 0.20 4.4 ksi Allowable = 0.4 0. The allowable stresses are: Tension Allowable = 21. Tension Shear Tension Shear C fa fb fv fmax UC fvmax UC C fa fb fv fmax UC fvmax UC (in) (ksi) (ksi) (ksi) (ksi) (ksi) (in) (ksi) (ksi) (ksi) (ksi) (ksi) 0.50 4.2 -3.9 in Bending moment.2 -3.12 0.7 0.6Fy Shear Allowable = 14. 0 Padeye Configuration The principal features of the padeye are shown below.0 applied to the structure gravity loads. .85% of shackle jaw width.04 kips 2. From SACS analysis of bridge lift: Padeye design load. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of 1. " Systematic Approach to Lifting Eye Design" 1.0 Introduction The purpose of this calculation is to document the design of a padeye. in order to determine the design load to be used for the padeye design. while details have been included on drawing INSERT DRAWING NUMBER.1 Applied Loading The lifting loads were determined from a lift analysis which considered a load factor of 2. Edwin P Russo and Rudolph A Hall. P 7. This spreadsheet was developed primarily to satisfy the requirements of Reference 1. r R SWL SWL Sw  R 2r b b c B a B Rh g A A h t Tp t T where: SWL = safe working load b = angle from horizontal Rh = Hole radius r = Cheek plate radius R = Main plate radius Tp = Main plate thickness t = Cheek plate thickness T = Total plate thickness h = Base width b = Distance from edge of taper to center of hole c = Distance from base of plate to center of hole Sw = Cheek plate weld leg e = Main plate to cheek plate clearance a = Taper angle p = Shackle pin diameter The dimension T should equal 60 . 0 kips Design Load (max from Table 1.950 > and < 2.thecrosbygroup.check on value of taper angle Out-of-plane bending .611 radians p = 2. calculated on the net area through plate.com/maininterface. in-plane bending and out-of-plane bending 3. should be 1/8" greater than the selected shackle pin size The main plate radius is approximately R = 3 Rh Cheek plate radius is approximately r = R .300 in Pin hole diameter should be 2.5 kips Horizontal In-Plane Component Pv = 6.htm Shackle jaw width A = 3.528 in Base width b = 1.1 Load P = 7.236 in Cheek plate thickness T = 0.472 in Main plate thickness t = 0.2 Design Stresses .433 in Hole radius r = 0. calculated around cheek plate Combined Stress .750 in Say main plate radius = 5.1 kips Vertical In-Plane Component 3.1.945 in Total plate thickness h = 4.1-1) b = 60 degrees = 1. shackle pin size 2.5t The cheek plate thickness (t) should be less than or equal to the plate thickness (Tp).000 in Cheek plate radius = 4.76 short tons Use G-2130 Crosby 2-inch shackle.000 in Rh = 0.763 in Shackle pin size D = 2.378 in Main plate radius Tp = 0. two planes from pin hole to plate edge in load direction. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of The pin hole diameter.0 Stress Checks The following stress checks are performed in this analysis Bearing Stress . Tensile Stress .250 in so main plate radius should be approximately 3.984 in Cheek plate radius R = 1. determines the average bearing stress of pin on hole.300 in 200 ton shackle Fy = 50 ksi Yield strength of padeye material 3.772 in Distance from edge of taper to center of hole c = 1.52 kips = 1. d. capacity 35 tons. determined at the base of padeye plate and at various points along padeye plate Taper Angle .250 in Total padeye width should be between 1. Weld Stress .check combined tension.772 in Distance from base of plate to center of hole e = 0.394 in Main plate to cheek plate clearance a = 35 degrees = 0.500 in Hole radius = 1. Shackle load SWL 3. Shear Stress .047 radians Ph = 3.3 inches Refer to http://catalog. 0 ksi Allowable = 0.0 ksi E70XX Electrodes.45Fy UC3 = 0.239 OK Weld Stress Load on one weld = 3.4 in Plate thickness = 0.5 ksi Allowable = 0.2 ksi Bearing = P/(T*p) Allowable = 45. For Section A-A.4Fy UC2 = 0.249 OK Tensile Stress From Section D3.0.5P cosb (1 .3Fuw Fw = 70.2 of AISC. b should be replaced by c in the above equation.5 kips Weld load F = P/2 (assume each weld takes full load) Weld Shear = 1.6 times the hole diameter.4C2) / TP h The maximum principal stress that exists on an element is given by: fmax = ((fa + fb)/2) + [((fa + fb)/2)2 + fv2]1/2 and the maximum principal shear stress on an element is given by: fmax = [((fa + fb)/2)2 + fv2]1/2 .5 h sinb + R sinb) / TP h2 where C represents the distance of an element from the neutral axis. across the hole.077 OK Combined Stress The axial stress due to uniform tension along a section is given by: fa = P sinb / TP h The elemental bending stress which is linearly distributed along the section B-B may be expressed as: fb = 12PC (b cosb . ultimate strength UC4 = 0. is the minimum of 4 times the plate thickness at the pinhole or 1.707*Sw)) Sw = 0.4 ksi Tensile = P/(Effective width * plate thickness) Effective width = 1.9 in Allowable stress = 22.0 ksi Allowable = 0.6 ksi Weld Shear = F/(2pr*(0.0 ksi Shear = P/(4(r-Rh)*t+2(R-Rh)*Tp) Allowable = 20.0 ksi Allowable = 0. Tensile stress = 5. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of Bearing Stress Bearing = 3. the distance used in calculations. The shearing stress varies parabolically along the length of the padeye and is given as: fV = 1.500 in fillet weld size Allowable = 21.072 OK Shear Stress Shear = 5.9Fy UC1 = 0. 3 0.3 0.1 2.5 0.4 4.4Fy Stresses calculated at Section B-B.6 in-kips (PL c) Section modulus S = 0.1 4. shear stress. no analysis of this section is required.6 3.5 ksi (0.66 Fy) In-plane bending stress / allowable = 0.0 2. In addition as the padeyes are aligned with the slings.14 2.09 0.5 4.14 0.7 0.14 2.14 0.2 2.9 0.14 2. Stresses calculated at Section A-A.4 0.9 0.3 0.75 Fy) Out-of-plane bending stress / allowable = 0.14 2. is not critical.9 0.3 1.4 0.0 ksi (0.14 2.5 0.1 4.13 0. which is the worst case scenario.14 0.2 0.9 0.50 2.5 0.02 OK Sum of ratios = 0. Out-of-plane load PL = 0.11 0.9 0.9 ksi Allowable = 30. principal stress and maximum shear stress at various points along the length of the plate. take a minimum load of 5% of the total applied load.20 2.0 ksi (0.9 0.11 1.9 0.14 0.0 2. Calculation Sheet Project title Calc no Client Proj no Calculation title Phase/CTR Path Page of The following table shows the values of the axial stress.13 2.5 0.14 2.9 0.9 0.10 OK Maximum in-plane bending stress = 0.09 0.12 1.11 0. Note that this out of plane stress is combined with the inplane bending stress and tensile stress calculated above.6 3. Tension Shear Tension Shear C fa fb fv fmax UC fvmax UC C fa fb fv fmax UC fvmax UC (in) (ksi) (ksi) (ksi) (ksi) (ksi) (in) (ksi) (ksi) (ksi) (ksi) (ksi) 0.5 ksi Allowable = 33. bending stress.14 0.9 0. ignoring the gusset plates.4 0.21 OK For out-of-plane bending at the B-B axis.9 0.1 2.30 2.12 1.7 0.00 2.9 0.40 2.00 2.0 ksi Allowable = 0.9 3.8 0.2 2.6 Fy) Tensile stress / allowable = 0.50 2. Consequently.8 in Bending moment.9 3.14 0. .6 0.5 4.40 2.8 0.1 0.08 MAX 0.9 0.20 2.30 2.9 0.8 0.4 0.9 0.6Fy Shear Allowable = 20. M = 0.4 kips (P / 20) Lever arm.9 0. The allowable stresses are: Tension Allowable = 30.3 0. the out-of-plane bending moment is reduced by the ratio of the lever arms and therefore.11 1.0 3.1 0.08 0.10 OK Maximum tensile stress = 2.14 Out-of Plane Bending Check out-of plane bending at section A-A.10 2.10 2.2 in3 (1/6 hT2p) Out-of-plane bending stress = 3.0 ksi Allowable = 0.13 2.6 0. c = 1.3 1.8 0.7 ksi (M / S) Allowable = 37.2 0.4 4.14 0.13 0.0 3.


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