SI 5211 Firstka Safira 25019325 PDF [PDF]

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SI-5211 PERILAKU STRUKTUR RANGKA BAJA TUGAS 5



oleh



25019325 NIM



Firstka Safira NAMA



DOSEN Dyah Kusumastuti, ST, MT, Ph.D.



PROGRAM STUDI TEKNIK SIPIL FAKULTAS TEKNIK SIPIL DAN LINGKUNGAN INSTITUT TEKNOLOGI BANDUNG 2020 1



Problem 8.1 For the SMF shown, design the beam-to-column connections for the first story beam using only the following types of connections prequalified per AISC 358. (a) WUFW connections (e) Bolted flange plate (BFP) connections Assume that, at the story under consideration, the beam is W30 × 173, and the columns are W14 × 311. All loads are shown below. Assume ASTM A992 Gr. 50 steel for beams and columns. Check that the design satisfies the strong-column/weakbeam requirements, as well as all other applicable detailing requirements. If one or many limits of applicability are found to be violated for a specific connection type, just highlight the violations and continue calculations as if the connection was permitted. Answer: (a) Welded unreinforced flange-welded web (WUF-W) connections Material Properties: Beam Column



Fy = Fu = Fy = Fu =



344737.9 448159.3 344737.9 448159.3



kN/m2 kN/m2 kN/m2 kN/m2



Ry = Rt = Lbeam = Lcolumn =



1.1 1.1 8000 mm 3500 mm



Geometric Properties:



Shape W30X173 W14X311



Beam: W30X173



Depth



Flange width



mm 772.16 434.34



mm 381 411.48



Flange Web thickness thickness mm 27.178 57.404



mm 16.637 35.814



Area



Sx



Sy



mm2 32903.16 58967.62



mm3 8.87E+06 8.29E+06



mm3 1.31E+06 3.26E+06



Column: W14X311



2



Design Procedure Step 1: Calculate probable maximum moment at plastic hinge Mpr



Step 2: Determine the beam plastic hinge location Step 3: Compute shear force Vh at the plastic hinge location at each end of the beam



Step 4: Check column-beam relationship limitations



Formula



Result and Calculation



𝑀𝑝𝑟 = 𝐶𝑝𝑟 𝑅𝑦 𝐹𝑦𝑏 𝑍𝑒 Cpr = 1.4, connection overstrength [AISC C.7] Ry = 1.1, ratio of the expected yield stress to the specified minimum yield Fyb = Specified minimum yield stress of the beam (kN/m2) Ze = Effective plastic modulus of section (or connection) at location of a plastic hinge (mm2)



Ze = The value of Ze shall be taken as equal to Zx of the beam section [AISC Ch.7] = 8865421.1 mm3



𝑀𝑝𝑟 = (1.4)(1.1)(344737.9)(8865421.1𝑥10−9 ) = 4706.619 𝑘𝑁 − 𝑚



𝑆ℎ = The plastic hinge location shall be taken to be at the face of the column



𝑆ℎ = 0



2𝑀𝑝𝑟 𝑉ℎ = ( ′ ) 𝐿



𝐿′ = (8000) − 2 (0) = 8000 𝑚𝑚



where: Mpr = maximum moment at the plastic (kN-m) L’ = determine distance between plastic hinges, L’ = Lo – 2Sh (mm) Lo = Length of beam (mm)



Calculate the moment expected at the face of the column flange



𝑉ℎ = (



2(4706619.708) ) = 1176.654 𝑘𝑁 8000𝑥10−3



𝑀𝑓 = 4706.619 + 1176.654(0) = 4706.619



𝑀𝑓 = 𝑀𝑝𝑟 + 𝑉ℎ 𝑆ℎ where: Mpr = maximum moment at the plastic hinge (kN-m) Vh = shear force at the beam plastic hinge location (kN) Sh = beam plastic hinge location (mm)



Step 5: Check beam design shear strength, Vu



𝑉𝑢 = 𝑉ℎ ± 𝑉𝑔𝑟𝑎𝑣𝑖𝑡𝑦 where: Vh = shear force at the plastic hinge location at each end of the beam (kN) Vgravity = gravity shear (kN) =



𝑤𝑢 = 1.2 (𝑤𝐷𝐿 ) + 0.5(𝑤𝐿𝐿 ) = 1.2(15) + 0.5(10) = 23 𝑘𝑁/𝑚 𝑉𝑢 = 1176.654 ±



𝑤𝑢𝐿′ 2



(23)(8000𝑥10−3 ) 2



𝑉𝑢 = 1199.655 𝑘𝑁 (𝑙𝑒𝑓𝑡) 𝑉𝑢 = 1153.655 𝑘𝑁 (𝑟𝑖𝑔ℎ𝑡)



Step 6: Check column continuity plate



𝑅𝑢 ≤ 1.8bbf 𝑡𝑏𝑓 𝐹𝑦𝑏 𝑅𝑦𝑏 where:



𝑅𝑢 =



𝑀𝑓 4706.619 = 𝑑𝑏 − 𝑡𝑏𝑓 (772.2 − 27.2)𝑥10−3 = 6317.609 𝑘𝑁



3



Design Procedure



Formula Ru = Ultimate strength of fillet weld (kN), 𝑅𝑢 =



Result and Calculation



𝑀𝑓 𝑑𝑏 −𝑡𝑏𝑓



𝑅𝑢 ≤ 1.8bbf 𝑡𝑏𝑓 𝐹𝑦𝑏 𝑅𝑦𝑏



bbf = width beam flange (mm) tbf = Thickness of beam flange (mm) Fyb = Specified minimum yield stress of the beam (kN/m) Ryb = Ratio of expected yield stress to specified minimum yield stress of beam



𝑅𝑢 ≤ 1.8(381)(27.2)(344737.9)(1.1) 𝑅𝑢 ≤ 1.8(381𝑥10−3 )(27.2𝑥10−3 )(344737.9)(1.1) 6317.609 𝑘𝑁 ≤ 7073.724



(e) Bolted flange plate (BFP) connections Material Properties: Beam Plate (assumption) Column



Fy = Fu = Tp = bfp = bp = Fy = Fu =



344737.9 448159.3 35 575 700 344737.9 448159.3



kN/m2 kN/m2 mm mm mm kN/m2 kN/m2



Ry = Rt = Lbeam = Lcolumn =



1.1 1.1 8000 mm 3500 mm



Geometric Properties:



Shape W30X173 W14X311



Beam: W30X173



Depth



Flange width



mm 772.16 434.34



mm 381 411.48



Flange Web thickness thickness mm 27.178 57.404



mm 16.637 35.814



Area



Sx



Sy



mm2 32903.16 58967.62



mm3 8.87E+06 8.29E+06



mm3 1.31E+06 3.26E+06



Column: W14X311



4



Design Procedure Step 1: Calculate probable maximum moment at plastic hinge, Mpr



Step 2: Calculate maximum bolt diameter db



Formula 𝑀𝑝𝑟 = 𝐶𝑝𝑟 𝑅𝑦 𝐹𝑦𝑏 𝑍𝑒



𝐶𝑝𝑟 =



𝑑𝑏 ≤



where: Fnv = nominal shear strength of bolt from the AISC Specification (kN/m2) Fub = specified minimum tensile strength of beam material (kN/m2) Fup = specified minimum tensile strength of plate material (kN/m2) Ab = nominal unthreaded body area of bolt (mm2) db = nominal bolt diameter (mm) tf = beam flange thickness (mm) tp = flange plate thickness (mm)



2 𝐹𝑦



=



344737.9+448159.3 2 (448159.3)



= 1.1 < 1.2 OK



𝑀𝑝𝑟 = (1.1)(1.1)(344737.9)(9946900) = 4337.78 𝑘𝑁 − 𝑚



𝑏𝑓 𝑅𝑦 𝐹𝑦 (1 − ) − 3𝑚𝑚 2 𝑅𝑡 𝐹𝑢



𝑟𝑛 = 1.0 𝐹𝑛𝑣 𝐴𝑏 𝑟𝑛 = 2.4 𝐹𝑢𝑏 𝑑𝑏 𝑡𝑓 𝑟𝑛 = 2.4 𝐹𝑢𝑝 𝑑𝑏 𝑡𝑝



𝐹𝑦 +𝐹𝑢



𝑅𝑦 = 1.1, 𝐴𝐼𝑆𝐶 341, 2005 𝐹𝑦𝑏 = 344737.9 𝑘𝑁/𝑚2 𝑍𝑒 = 9946900 𝑚𝑚3



where: Cpr = Factor to account for peak connection strength, Cpr < 1.2 Ry = Ratio of the expected yield stress to the specified minimum yield Fyb = Specified minimum yield stress of the beam (kN/m2) Ze = Effective plastic modulus of section (or connection) at location of a plastic hinge (mm2)



where: db = depth of beam (mm) Fy = Specified minimum yield stress of the yielding element (kN-m) Fu = Specified minimum tensile strength of yielding element (kN-m) Ry = Ratio of the expected yield stress to the specified minimum yield Rt = Ratio of expected tensile strength to specified minimum tensile



Step 3: Calculate nominal shear strength per bolt rn



Result and Calculation



=



381 (1.1) (344737.9) (1 − )−3 2 (1.1) (448159.3) = 187.5 𝑚𝑚



From Table J3.4M, db = 36 mm Min. edge distance = 46 mm Check, 36 𝑚𝑚 ≤ 187.5 𝑚𝑚



𝑂𝐾



𝑟𝑛 = 2.4 𝐹𝑛𝑣 𝐴𝑏 = 1.0() = 𝑘𝑁/𝑏𝑜𝑙𝑡 𝑟𝑛 = 2.4 𝐹𝑢𝑏 𝑑𝑏 𝑡𝑓 = 2.4(448159.3)(36𝑥10−3 )(27.178𝑥10−3 ) = 1052.36 𝑘𝑁/𝑏𝑜𝑙𝑡



𝑟𝑛 = 2.4 𝐹𝑢𝑝 𝑑𝑏 𝑡𝑝 = 2.4(448159.3)(36𝑥10−3 )(35𝑥10−3 ) = 1355.23𝑘𝑁/𝑏𝑜𝑙𝑡



5



Design Procedure Step 4: Select a trial number of bolts



Formula 𝑛≥



1.25 𝑀𝑝𝑟 ∅𝑛 𝑟𝑛 (𝑑 + 𝑡𝑝 )



Result and Calculation =



1.25 (4337.78) (1)(1052.36)(772.16 + 35)𝑥10−3 = 6.38 = 8 𝑏𝑜𝑙𝑡𝑠



where: n = number of bolts rounded to next higher even number increment Mpr = maximum moment at the plastic hinge (kN-m) ∅n = 1 for nonductile limit state [AISC



Ch. 2.4] rn = nominal shear strength per bolt d = beam depth (mm) tp = flange plate thickness (mm)



Step 5: Determine the beam plastic hinge location



Step 6: Compute the shear force at the beam plastic hinge location at each end



Step 7: Calculate the moment expected at the face of the column flange



𝑛 𝑆ℎ = 𝑆1 + 𝑠 ( − 1) 2



𝑆1 = 1.7 𝑑𝑜 = 1.7(3.9) = 66.3 𝑚𝑚 𝑠 = 2 𝑑𝑏 = 2(36) = 72



where: S1 = distance from face of column to nearest row of bolts, 1.7do (mm) do = dia. bolt hole (mm) s = spacing of bolt row, 2 db – 3 db (mm) db = nominal bolt diameter (mm) n = number of bolts



8 𝑆ℎ = 66.3 + 72 ( − 1) 2 𝑆ℎ = 282.3 𝑚𝑚



1.1𝑅𝑦 𝑀𝑝 𝑤𝑢 𝐿′ 𝑉ℎ = 2 ( ) ± 𝐿′ 2



𝑀𝑝 = 𝐹𝑦 𝑍𝑏 = 3.06 𝑥103 𝑘𝑁 − 𝑚



where: Ry = Ratio of the expected yield stress to the specified minimum yield Mp = Fy.Zb L’ = determine distance between plastic hinges, L’ = Lo – 2Sh Lo = length of beam (mm) Sh = beam plastic hinge location wu = Distributed load on beam (kN/m) wDL = dead load per meter (kN/m) wLL = live load per meter (kN/m)



𝑀𝑓 = 𝑀𝑝𝑟 + 𝑉ℎ 𝑆ℎ



𝐿′ = (8000𝑥10−3 ) − 2 (282.3) = 7435.4 𝑚𝑚 𝑤𝑢 = 1.2 (𝑤𝐷𝐿 ) + 0.5(𝑤𝐿𝐿 ) = 1.2(15) + 0.5(10) = 23 𝑘𝑁/𝑚 𝑉ℎ = 2 (



1.1(1.1)(3.06 𝑥103 ) 23(7435.4𝑥10−3 ) )+ 7435.4 2 = 1080.224 𝑘𝑁 (𝑙𝑒𝑓𝑡)



𝑉ℎ = 2 (



1.1(1.1)(3.06 𝑥103 ) 23(7435.4𝑥10−3 ) )− 7435.4 2 = 909.210 𝑘𝑁 (𝑟𝑖𝑔ℎ𝑡)



𝑀𝑓 = 4337.78 + (1080.224)(282.3𝑥10−3 ) = 4.64𝑥103 𝑘𝑁 − 𝑚 (𝑙𝑒𝑓𝑡)



where: Mpr = maximum moment at the plastic hinge (kN-m) Vh = shear force at the beam plastic hinge location (kN) Sh = beam plastic hinge location (mm)



𝑀𝑓 = 4337.78 + (909.210 )(282.3𝑥10−3 ) = 4.59𝑥103 𝑘𝑁 − 𝑚 (𝑟𝑖𝑔ℎ𝑡)



6



Design Procedure Step 8: Compute Fpr, the force in the flange plate due to Mf



Formula Fpr =



Mf 𝑑 + 𝑡𝑝



where: Mf = moment expected at the face of the column flange (kN-m) d = beam depth (mm) tp = flange plate thickness (mm)



Step 9: Confirm that the number of bolts



𝑛 ≥



Result and Calculation Fpr =



4.64𝑥103 = 5751.9269 𝑘𝑁 (𝑙𝑒𝑓𝑡) (772.16 + 35)𝑥10−3



Fpr =



4.59𝑥103 = 5692.115 𝑘𝑁 (𝑟𝑖𝑔ℎ𝑡) (772.16 + 35)𝑥10−3



Fpr ∅𝑛 𝑟𝑛



𝐶ℎ𝑒𝑐𝑘, 𝑛 ≥



where: n = number of bolts Fpr = force in the flange plate (kN) ∅n = 1 for nonductile limit state [AISC Ch. 2.4] rn = nominal shear strength per bolt (kN/bolt)



Step 10: Check that the thickness of the flange plate



𝑡𝑝 ≥



Step 11: Check the flange plate for the limit state of tensile rupture



5751.9269 (1)(1052.36) 8 ≥ 5.465 𝑂𝐾 (𝑙𝑒𝑓𝑡) 𝑛 ≥



5692.115 (1)(1052.36) 8 ≥ 5.408 𝑂𝐾 (𝑟𝑖𝑔ℎ𝑡) 𝑛 ≥



Fpr ∅𝑑 𝐹𝑦 𝑏𝑓𝑝



where: Fpr = force in the flange plate (kN) ∅d = 1 [AISC Ch. 2.4] Fy= specified minimum yield stress of material (kN/m2) bfp = width of flange plate (mm)



Fpr ∅𝑛 𝑟𝑛



𝐶ℎ𝑒𝑐𝑘, 𝑡𝑝 ≥



Fpr ∅𝑑 𝐹𝑦 𝑏𝑓𝑝



𝑡𝑝 ≥



5751.9269 = 0.0290 𝑚 (𝑙𝑒𝑓𝑡) (1)(344737.9)(575𝑥10−3 ) 35 𝑚𝑚 ≥ 29 𝑚𝑚 𝑂𝐾



𝑡𝑝 ≥



5692.115 = 0.0287 𝑚 (𝑟𝑖𝑔ℎ𝑡) (1)(344737.9)(575𝑥10−3 ) 35 𝑚𝑚 ≥ 28.7 𝑚𝑚 𝑂𝐾



𝐹𝑝𝑟 ≤ ∅𝑛 𝑅𝑛



𝐴𝑛 = (𝑏 − 𝑛𝑑𝑜 )𝑡𝑝 = (700 − (8)(39))35 = 13580 𝑚𝑚2



∅n = 1 for nonductile limit state [AISC Ch. 2.4] Rn = block shear rupture, 𝑅𝑛 = 𝐹𝑢 𝐴𝑒 Fu = Specified minimum tensile strength of yielding element (kN-m) Ae = Effective net area, Ae = An.U (mm2) U = shear lag factor = 1 [Steel code, Table D3.1] An = Net effective area at critical section, An = (b-ndo)tp (mm2) b = width of plate (mm) n = number of bolts do = dia. bolt hole (mm) tp = flange plate thickness (mm)



𝐴𝑒 = 𝐴𝑛 𝑈 = (13580)(1) = 13580 𝑚𝑚2



where: 𝑅𝑛 = 𝐹𝑢 𝐴𝑒 = (448159.3)(13580𝑥10−6 ) = 6086.003 𝑘𝑁



𝐶ℎ𝑒𝑐𝑘, 𝐹𝑝𝑟 ≤ ∅𝑛 𝑅𝑛 𝐹𝑝𝑟 ≤ (1)(6086.003) 5751.9269 𝑘𝑁 ≤ 6086.003 𝑘𝑁 𝑂𝐾 (𝑙𝑒𝑓𝑡) 5692.115 𝑘𝑁 ≤ 6086.003 𝑘𝑁



𝑂𝐾 (𝑟𝑖𝑔ℎ𝑡)



7



Design Procedure Step 12: Determine the requires shear strength, Vu



Formula 𝑉𝑢 =



Result and Calculation



2𝑀𝑝𝑟 + 𝑉𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝐿′



where: Mpr = maximum moment at the plastic hinge (kN-m) L’ = distance between plastic hinges (mm) Vgravity = gravity shear (kN) =



𝑤𝑢 = 1.2 (𝑤𝐷𝐿 ) + 0.5(𝑤𝐿𝐿 ) = 1.2(15) + 0.5(10) = 23 𝑘𝑁/𝑚 𝑉𝑢 =



2 (4337.78) (23)(8000𝑥10−3 ) + (7435.4𝑥10−3 ) 2 𝑉𝑢 = 1258.79 𝑘𝑁



𝑤𝑢𝐿′ 2



∅𝑛 𝑅𝑣 < 𝑅𝑢 ,



𝑑𝑝 = 𝑑𝑏 − 𝑡𝑏𝑓 = 772.16 − 27.178 = 744.982 𝑚𝑚



𝑑𝑜𝑢𝑏𝑙𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 2 3𝑏𝑐𝑓 𝑡𝑐𝑓 𝑅𝑣 = ∅𝑛 0.6 𝐹𝑦𝑐 𝑑𝑐 𝑡𝑤 (1 + ) 𝑑𝑏 𝑑𝑐 𝑡𝑤



Step 13: Check the column panel zone



where: ∅n = 1 for nonductile limit state [AISC Ch. 2.4] Rv = panel zone shear strength (kN-m) Fyc = specified minimum yield stress of column web material (kN/m2) dc = depth of column (mm) db = depth of beam (mm) tcf = flange thickness of column (mm) bcf = flange width of column (mm) tw = web thickness of flange (mm) Ru = Ultimate strength of fillet weld, Ru = ∑𝑀𝑓 𝑑𝑝



𝑅𝑢 =



∑𝑀𝑓 (4.64𝑥103 + 4.59𝑥103 ) = 𝑑𝑝 (744.982𝑥10−3 ) = 12399.2 𝑘𝑁 − 𝑚



𝑅𝑣 = (1)( 0.6)(344737.9)(434.34𝑥10−3 ) (35.814𝑥10−3 ) 3(411.48𝑥10−3 )(57.404𝑥10−3 )2 (1 + ) (772.16𝑥10−3 )(434.34𝑥10−3 )(35.814𝑥10−3 ) 𝑅𝑣 = 3217.53 𝑘𝑁 − 𝑚 𝐶ℎ𝑒𝑐𝑘, ∅𝑛 𝑅𝑣 < 𝑅𝑢 (1) (3217.53) < 𝑅𝑢 3217.53 𝑘𝑁 − 𝑚 < 12399.2 𝑘𝑁 − 𝑚 𝑂𝐾 𝑑𝑜𝑢𝑏𝑙𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑



(kip-in)



Mf = moment expected at the face of the column flange (kN-m) dp = diameter of panel zone = db - tbf (mm)



8



Problem 8.3 Design a prequalified RBS connection for the beam of the SMRF structure shown. More specifically: (1) Select an appropriate geometry for the RBS and location of the RBS along the beam length. (2) Check whether the selected beams and columns meet the specified limitations and details of the prequalified connection. (3) Check whether moment at face of column is acceptable. (4) Check whether column panel zone strength is acceptable.



Answer: Material properties: Beam Column



Fy = Fu = Fy = Fu =



50 65 50 65



ksi ksi ksi ksi



Ry = Rt = Lbeam = Lcolumn =



1.1 1.1 30 15



ft ft



Geometric properties:



Shape W30X173 W14X311



Beam: W30X173



Depth



Flange width



Flange Web thickness thickness



(in)



(in)



(in)



30.7 17.1



10.5 16.2



1.18 2.26



Area



Sx



Sy



(in)



(in2)



(in3)



(in3)



0.65 1.41



43.5 91.4



436 506



43.3 199



Column: W14X311



9



Design Procedure Step 1:



Formula



-



Trial values for RBS dimensions a, b and c



Choose plastic hinge configuration and location



𝑎 = 0.5 𝑏𝑏𝑓 ≤ 0.75 𝑏𝑏𝑓 𝑏 = 0.65𝑑𝑏 ≤ 0.85 𝑑𝑏 𝑏𝑏𝑓 𝑐 = 0.45 ( ) 2



Result and Calculation 0.5 𝑏𝑏𝑓 ≤ 0.75 𝑏𝑏𝑓 where, bbf = 10.5 in 5.25 ≤ 7.875 , 𝑐ℎ𝑜𝑜𝑠𝑒 𝑎 = 6 𝑖𝑛 0.65𝑑𝑏 ≤ 0.85 𝑑𝑏 where, db = 30.7 in 19.955 ≤ 26.095 , 𝑐ℎ𝑜𝑜𝑠𝑒 𝑏 = 23 𝑖𝑛 𝑏𝑏𝑓 ) 2 where, bbf = 10.5 in 𝑐 = 2.362 𝑖𝑛 𝑐 = 0.45 (



-



Determine the radius of the flange cut, R 𝑅=



-



𝑅=



4𝑐 2 + 𝑏 2 8𝑐



4(2.362)2 + (23)2 8(2.362) 𝑅 = 29.17 𝑖𝑛



Determine distance to RBS 𝑆ℎ =



𝑑𝑐 𝑏 +𝑎+ 2 2



𝑆ℎ =



17.1 23 +6+ 2 2



𝑆ℎ = 26.05 𝑖𝑛



-



Determine distance between plastic hinges



𝐿𝑜 = 30 𝑓𝑡 26.05 𝑆ℎ = 𝑖𝑛 = 2.17 𝑓𝑡 12



′



𝐿 = 𝐿𝑜 − 2(𝑆ℎ )



𝐿′ = 30 − 2(2.17) 𝐿′ = 25.658 𝑓𝑡



Step 2: Determine plastic section modulus at the RBS



𝑍𝑒 = 𝑍𝑥𝑏 − 2 𝑐 𝑡𝑏𝑓 (𝑑𝑏 − 𝑡𝑏𝑓 )



Step 3: 𝑀𝑝𝑟 = 𝐶𝑝𝑟 𝑅𝑦 𝐹𝑦𝑏 𝑍𝑒 Determine probable maximum moment at where: the RBS C = Factor to account for peak connection pr



strength, Cpr < 1.2 Ry = Ratio of the expected yield stress to the specified minimum yield Fyb = Specified minimum yield stress of the beam (kN/m2) Ze = Effective plastic modulus of section (or connection) at location of a plastic hinge (mm2)



𝑍𝑒 = 436 − 2 (2.362)1.18(30.7 − 1.18) 𝑍𝑒 = 271.411 𝑖𝑛3 = 𝑍𝑅𝐵𝑆



𝐶𝑝𝑟 =



𝐹𝑦 +𝐹𝑢 2 𝐹𝑦



=



50+65 2 (50)



= 1.15 < 1.2 OK



𝑅𝑦 = 1.1, 𝐴𝐼𝑆𝐶 341, 2005 𝐹𝑦𝑏 = 50 𝑘𝑖𝑝𝑠 𝑍𝑒 = 271.411 𝑚𝑚3 𝑀𝑝𝑟 = (1.15)(1.1)(50)(271.411) = 17166.76 𝑘𝑖𝑝 − 𝑖𝑛



10



Design Procedure



Step 4: Compute the share force at the center of each RBS



Formula



Result and Calculation



𝑉 𝑅𝐵𝑆 = 𝑉ℎ ± 𝑉𝑔𝑟𝑎𝑣𝑖𝑡𝑦



2(17166.76) 𝑉ℎ = ( ) = 111.5087 𝑘𝑖𝑝𝑠 25.658



where: Vh = shear force at the plastic hinge location at each end of the beam (kip) = 2 𝑀𝑝𝑟



𝑉 𝑅𝐵𝑆 = 111.5087 ± 0 𝑉 𝑅𝐵𝑆 = 111.5087 𝑘𝑖𝑝𝑠 (𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑟𝑖𝑔ℎ𝑡)



𝐿′



Mpr = maximum moment at the plastic (kN-m) L’ = determine distance between plastic hinges, L’ = Lo – 2Sh (mm) Lo = Length of beam (mm) Vgravity = gravity shear (kip) =



Step 5: Compute the probable maximum moment at the face of the column



Step 8: Determine the required shear strength



2



=0 𝑀𝑓 = 17166.76 + (1111.5087)(17.5) = 19118.16 𝑘𝑖𝑝𝑠 − 𝑖𝑛 (𝑙𝑒𝑓𝑡 𝑎𝑛𝑑 𝑟𝑖𝑔ℎ𝑡)



𝑀𝑓 = 𝑀𝑝𝑟 + 𝑉𝑅𝐵𝑆 𝑆ℎ where: Mpr = maximum moment at the plastic hinge (kip-in) Vh = shear force at the beam plastic hinge location (kip) Sh = beam plastic hinge location (in), Sh = a+b/2)



Step 6: Compute the expected plastic moment of the beam Step 7: Check that Mf does not exceed ∅dMpe



𝑤𝑢𝐿′



𝑀𝑝𝑒 = (436)(1.1)(50) = 23980 𝑘𝑖𝑝 − 𝑖𝑛



𝑀𝑝𝑒 = 𝑍𝑥𝑏 𝑅𝑦 𝐹𝑦𝑏



𝑀𝑓 ∅𝑑 𝑀𝑝𝑒 where: ∅d = 1 Mf = Mpe = plastic moment of the beam (kip-in)



The demand at the RBS was determined in Step 4: 𝑉𝑢 =



𝑀𝑓 19118.16 = = 0.797 < 1.0 𝑂𝐾 ∅𝑑 𝑀𝑝𝑒 (1)(23980)



𝑉𝑢 =



2𝑀𝑝𝑟 + 𝑉𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑉𝑅𝐵𝑆 𝐿′



2(17166.76) + 0 = 111.5087 𝑘𝑖𝑝𝑠 25.658



where: Mpr = maximum moment at the plastic hinge (kip-in)) L’ = distance between plastic hinges (in) Vgravity = gravity shear (kN) =



𝑤𝑢𝐿′ 2



=0



The results of Steps 2 through 7 indicate that no change is required for the dimensions of the reduced



11



Design Procedure



Formula



Step 9: Check column panel zone



Result and Calculation



∅𝑛 𝑅𝑣 < 𝑅𝑢 , 𝑑𝑜𝑢𝑏𝑙𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑



𝑅𝑣 = ∅𝑛 0.6 𝐹𝑦𝑐 𝑑𝑐 𝑡𝑤 (1 +



2 3𝑏𝑐𝑓 𝑡𝑐𝑓 ) 𝑑𝑏 𝑑𝑐 𝑡𝑤



where: ∅n = 1 for nonductile limit state [AISC Ch. 2.4] Rv = panel zone shear strength (kip-in) Fyc = specified minimum yield stress of column web material (ksi) dc = depth of column (in) db = depth of beam (in) tcf = flange thickness of column (in) bcf = flange width of column (in) tw = web thickness of flange (in) Ru = Ultimate strength of fillet weld, Ru = ∑𝑀𝑓 𝑑𝑝



𝑑𝑝 = 𝑑𝑏 − 𝑡𝑏𝑓 = 30.7 − 1.18 = 29.52 𝑖𝑛 𝑅𝑢 =



∑𝑀𝑓 (19118.16 + 19118.16 ) = 𝑑𝑝 (29.52) = 11295.268 𝑘𝑖𝑝𝑠 3(16.2)(2.26)2 ) (30.7)(17.1)(1.41) 𝑅𝑣 = 965.8994 𝑘𝑁 − 𝑚



𝑅𝑣 = (1)( 0.6)(50)(30.7) (1.41) (1 +



𝐶ℎ𝑒𝑐𝑘, ∅𝑛 𝑅𝑣 < 𝑅𝑢 (1) (965.8994) < 𝑅𝑢 965.8994 𝑘𝑖𝑝𝑠 < 11295.268 𝑘𝑖𝑝𝑠 𝑂𝐾 𝑑𝑜𝑢𝑏𝑙𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑



(kip-in)



Mf = moment expected at the face of the column flange (kip-in)



12



Problem 8.7 Which of the prequalified types of welded connections can be used to connect a W36 × 256 to a W14 column? Answer: Section property of W36x256:



Shape



(in)



Flange width (in)



Flange thickness (in)



Web thickness (in)



37.4



12.2



1.73



0.96



Depth



Area



Sy



Sy



(in2)



(in3)



(in3)



75.4



895



86.5



W36X256 Berdasarkan properti dari baja diatas, tipe prakulaifikasi sambungan las yang cocok untuk sambungan Balok profil W36X256 dengan Kolom profil W14 adalah Tipe Reduced Beam Section (RBS) untuk frame OMF (Ordinary Moment Frame) maupun SMF (Special Moment Frame). Berikut merupakan Batasan geometri untuk tipe sambungan RBS berdasarkan FEMA 350. Geometric Limits of FEMA 350 Prequalified Connection Critical Beam Parameters Type



Frame Maximum Beam Size



RBS



OMF SMF



W36 W36



Min. Span (I) to Depth (db) Ration (I/db) 5 7



Max. Beam Flange Thickness (tbf) (in) 1.75 1.75



Critical Column Parameter Max Column Size No Limit W12, W14



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RANGKUMAN BAB 8. Design of Ductile Moment-Resisting Frames Respon Dasar dari Frame Momen-Tahan Daktil terhadap Beban Lateral



dimana:, tfc = tebal sayap profil kolom (m), twc = tebal badan profil kolom (m), Fyf = kuat leleh sayap baja profil (kN), Fyw = kuat leleh badan baja profil (kN), k = parameter kelangsingan, tfb = tebal sayap profil balok (m), dan twc = tebal badan profil kolom (m), d = lebar badan kolom (m) Gaya Geser pada Panel Zone



Gaya dalam saat respons seismik gempa: momen lentur, gaya geser, dan gaya aksial dalam bingkai momen di bawah beban lateral. - Kebutuh Rotasi Plastis Diperoleh dari analisis inelastic response-histrory. Cara perkiraan untuk memperkirakan permintaan rotasi plastik dalam frame adalah dengan memeriksa mekanisme keruntuhan plastik pada titik drift maksimum - Lateral Bracing dan Local Buckling Tekuk lokal flange dan web serta tekuk lateral-puntir tidak akan terhindarkan dari rotasi plastik yang sangat besar (setidaknya dalam bentuk struktural yang umum digunakan), tetapi tekuk dapat diperlambat dengan beberapa syarat yang dapat dilakukan. Panel Zone Panel zone direncanakan untuk menahan gaya tarik horisontal yang terjadi pada sambungan balokkolom.



Gaya geser horisontal pada pada panel zone: 𝑀1 𝑀2 𝑀1 + 𝑀2 𝑉𝑤 = + − 𝑑𝑏1 𝑑𝑏2 ℎ dimana: M1 = momen balok kiri (kN-m), M2 = momen balok kanan (kN-m), db1 = tinggi balok kiri (m), db2 = tinggi balok kanan (m) Panel Zone Desain Perencanaan panel zone ini diperkuat dengan pengaku diagonal (double plate) yang dipasang tegak lurus dengan sayap kolom dan sejajar dengan sayap balok. 𝑃𝑢 ≤ 0.75 𝑃𝑦 (𝐴𝐼𝑆𝐶 360) 3𝑏𝑐𝑓 𝑡𝑐𝑓 2 𝑉𝑢 = 0.60 𝐹𝑦 𝑑𝑐 𝑡𝑤 (1 + ) 𝑑𝑏 𝑑𝑐 𝑡𝑤 dimana, Pu = gaya aksial kolom pada joint, 𝑃𝑦 = 𝐴𝑠 . 𝐹𝑦 As = luas penampang bruto profil kolom (m2), dan Fy = kuat leleh baja



Flange Distortion and Column Web Yielding/ Crippling Prevention Ketika pada gaya yang terjadi pada sayap balok 𝑃𝑏𝑓 >∅𝑅𝑛 , maka pengaku pada kolom dibutuhkan, dimana nilai ∅𝑅𝑛 , Pelelehan sayap lokal, ∅𝑅𝑛 = ∅6.25 𝑡𝑓𝑐 2 𝐹𝑦𝑓 ( ∅ = 0.90) Pelelehan badan lokal, ∅𝑅𝑛 = ∅(5𝑘 + 𝑡𝑓𝑏 )𝐹𝑦𝑤 𝑡𝑤𝑐 ( ∅ = 0.10) Pelipatan pelat badan,



𝐸𝐹𝑦𝑤 𝑡𝑓𝑐 𝑁 𝑡𝑤𝑐 ( ∅ = 0.75) ∅𝑅𝑛 = ∅0.8𝑡𝑤𝑐 2 (1 + 3 ( ) ( ) 1.5 )√ 𝑑 𝑡𝑓𝑐 𝑡𝑤𝑐



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