Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4- 22 mm diameter bolts. 12 mm bolts t12 mm t12 mm P, P- P. 300 75 75 75 Edge distances 75 mm dhole for tensile and rupture - de +3 mm dhole for bearing strength for Le d + 1.5 mm Fy 248 MPa F.-400 MPa
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- A bolted connection shown consists of two plates 300mm x12mm connected by 4 - 22 mm diameter bolts. Edge distances = 75mm dhole for tensile and rupture = db + 3 mm dhole for bearing strength for Lc = db + 1.5 mm Fy = 248 Mpa Fu = 400 Mpa Fn = 330 Mpa Use LRFD design method. Determine the design strength due to the gross yielding of plates. (kN)Considering the following steel connection. The plates in Pink are 9mm steel plates. The middle plate (Yellow) is 18mm thick. The width of the plate is 100mm. The maximum allowable tension stresses on any of the plates is 100Mpa in Gross Area Yielding and 150 Mpa for Net Area or Tension Rupture. The bolts used are 8mm in diameter, the holes are 10mm in diameter, no need to add 1.6mm. The bolts allow a maximum of 280 Mpa of shear. Determine the maximum allowable "P" of the connection in kN.Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts T12 mm te12 mm P 300 75 75 75 Edge distances = 75 mm %3D dnole for tensile and rupture = dp +3 mm %3D dnole for bearing strength for Lec = d + 1.5 mm Fy = 248 MPa Fu = 400 MPa %3D En 330 MPa %3D Use LRFD design method. Determine the design strength due to tensile rupture strength of plates. (kN)
- The connection shown is subjected to a tensile force of P = 100 kN. The allowable shear stress for the bolts is 100 MPa. Assume each bolt supports an equal portion of the load. Determine the required diameter of the bolts. a. 20.6 mm b. 25.2 mm c.35.7 mm d.17.8 mm choose letter of correct answerBolted and Riveted Connections From the bolted connection shown, the diameter of bolts is 18mm0 with the hole diameter equal to 3mm bigger than the bolt, the angular section is 100 x 75 x 8 mm, with an area of 1340 mm2. Thickness of the gusset plate is 9mm. The gusset plate and angle arc A36 steel with Fy = 250 MPa and Fu = 400 MPa. Determine the tensile capacity of the connection based on gross area. Determine the tensile capacity of the connection based on effective net area using a reduction factor of 0.85. Determine the tensile capacity of the connection based on gross area.A76 x 76 x6 mm angular section shown is welded to an 8 mm thick gusset plate. The length L1 is 65 mm, L2 is 125 mm, and the cross sectional area of the angle is 929 sq.mm. Fy-248MPA and Fu -400MPA. Gusset Plate Alowable Sresse Alowable tensle stress igross ara)0.00 Alowable tereile stres net area)0SOF. Allowable shear stress (net ares)0.30F. Determine the value of P based on net area using a strength reduction coefficient of 85% O 138.24 KN O 140.14 kN O 157.93 kN O 304.00 N Next
- Check the bolted section of the given gusset plate connection assuming that the connection is done by M12 bolts with a staggered arrangement. The maximum design tensile axial force to be sustained is Ned = 168 kN. Assume a grade of S235 steel. P. Gusset plate BỘ do=13mm Gusset plate O DO t=16mm t-16mm1 Ni.Ed FO NtEd do=13mm do3D13mm G. 40mm 40mm 140 mm 240mm 30mmThe given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. Use a501 plate and a502 gr2 rivet a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivetsThe diagonal at the left to the connection is a double angle 90 mm x 90 mm x 8 mm, with area of 2700 mm?, bolted to the 8 mm thick gusset plate. Bolt diameter = 16 mm Bolt hole diameter = 18 mm Bolt bearing capacity, Fp = 480 MPa Bolt shear strength, Fv = 68 MPa Steel plate strength and stresses are as follows: Yield strength, Fy 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 0.60 Fy Allowable tensile stress on the net area = %3D 0.50 Fu Allowable shear stress on the net area = 0.30 Fu Bolt bearing capacity, Fp = 1.2 Fy Calculate the allowable tensile load, P(kN) under the following conditions: %3D
- 4. Determine the design strength of the connection shown in the Figure below. Thebolts are 25 mm diameter A490 bolts with the threads not in the plane of shear.A36 steel is used (Fy = 250 MPa, Fu = 400 MPa).a. Compute the shear strength for all bolts.b. Compute the bearing strength for the tension member on all bolts.c. Compute the bearing strength for the gusset plate on all bolts.d. Compute the tensile strength of the tension member.e. Compute the design strength of the connection.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 channel shown is attached to a 12 mm gusset plate with 9-22 mm diameter A 325 bolts as shown. Use LRFD. Fnv = 300 MPa Fu = 400 MPa Fy = 248 MPa Ag = 3354 mm2 Question: Determine the capacity of the channel based on the bearing strength of the connection.Determine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40