Problem 1: The single angle L8x8x1/2 steel tension member (TM) (A36, Fy = 36 ksi, Fu = 58 ksi) is welded to the Gusset Plate/Connecting Element (CE). Calculate the governing tension design strength (R₂) considering limit states (i) Gross Section Yield (GSF) of TM, (ii) Net Section Fracture (NSF) of TM and (iii) Block Shear of CE. USE EQUATIONS NOT DESIGN TABLES. CE Gusset Plate: 5/8" PI A572, Grade 65 TM oRn
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- 2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80Topic:Welded Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Use attached formula picture for block shear *Please use hand written to solve this problem A channel is used as a tension member with the web of the channel welded to a 9.5 mm thick gusset plate as shown in the figure. The tension member is subjected to the following axial loads. Use LRFD Service dead load = 200 kN Wind load = 276 kN Service live load = 260 kN For channel: Fy = 345 MPa For gusset plate: Fy = 248 MPa Fu = 400 MPa Size of E 70 electrodes = 4 mm Ultimate tensile strength of E 70 electrodes = 480; Fw = 0.6(480) Questions : a) Determine the design factored tensile force. b) Determine the length of longitudinal welds "L". c) Determine the block shear strength of the gusset plate.Topic:Welded Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A 150 mm x 90 mm x 12 mm angular section is welded to a gusset plate as shown. Area of the angle is 2736 sq.mm, allowable shear Fv is 150MPa, Allowable tensile stress Ft = 0.6Fy with Fy = 250 MPa. Questions: a) Design force P b) Total required length of weld using 12 mm fillet weld c) Value of “b”
- 2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area c. Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 90m²m 90 mm 90mm 45 45mm In my comin P Scanned with CamScanner CSProblem2. The compression member is shown in figure. Find the following: a. The Euler stress Fe. b. The buckling stress Fcr c. The design strength d. The allowable strength e. Does the member satisfactorily meet the design requirements? Why? HSS 8x 8x4 ASTM AS00, Grade B steel (Fy = 46 ksi) 15'Topic:Bolted Steel Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A 12.5 mm x150 mm plate is connected to a gusset plate having a thickness of 9.5 mm.Diameter of bolt = 20 mm Both the tension member and the gusset plate are of A 36 steel. Fy = 248 MPa, FU = 400 MPa. Shear stress of bolt Fnv = 300MPa. Questions: a.Determine the shear strength of the connection. b.Determine the bearing strength of the connection. c. Determine the block shear strength of the connection.
- The given angle bar L125x75x12 with Ag = 2,269 sq.mm. is connected to a gusset plate using 20 mm diameter bolts as shown in the figure. Using A36 steel with Fy = 248 MPa and Fu = 400 MPa, determine the following: 2. Determine the nominal tensile strength of the 12 mm thick, A36 angle bar shown based on: a. Gross yielding b. Tensile rupture Bolts used for the connection are 20 mm in diameter. O O O O O O O Effective net area of the tension member if the shear lag factor is 0.80. Select the correct response: 1,516.1 1,354.4 1,431.2 1,221.6Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem A lap joint is made up of two steel plates 225mm by 10mm. It is connected by seven rivets having a diameter of 22mm snug fit in the drilled holes. The allowable stresses in the rivets and plates are 110 MPa in shear, 220 MPa n bearing and 138 MPa in tension. Use LRFD. Questions: a) Determine the nominal strength for one bolt due to shear. b) Determine the capacity of the plate based on the gross area and effective area.For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WI
- Situation 2. Two plates each with thicknessF16mm are bolted together with6 -22mm dimater bolts forming a lap conne ction. Bolt spacing are as follows: S1 = 40mm, S2 = 80mm, Sa = 100mm. Bolt hole diameter=25 mm 50 250mm 30 30 60 75 Allowable stress: Tensile stess on gross area of the plate=0.60 Fy Tensile stress on net area of the plate=0.5Fy Shear Stress of the bolt Fv=120MPA Bearing Stress of the bot Fp=1.2 Fu Calculate the permiss ible tensile load P under the following Conditions: 4. Based on shear capacity of bolts 5. Based on bearing capacity of bolts 6. Based on block shear strength (taking into consideration the failure path given in the figure below) 40 80 16 mm 40 180 mm 40 Bearing Failure Path #1 140 209 Bearing Failure Path #2Define and describe local buckling. Enumerate and differentiate classifications of structural steel sections considering local buckling. I give upvote :)TENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)