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- 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.A bearing type connection is shown in Figure 3.19. The diameter of A 325 bolts is 22 mm and the A572 Grade 50 plate material has a width of 150 mm and thickness of 16 mm. Assume diameter hole to be 24 mm. Bolt threads are excluded from the shear plane. Allowable stress of A 325 bolts: Fv = 207 MPa Fp = 1.5 Fu (to prevent excessive hole deformation) Allowable stresses of A572 Grade 50 plate material: Fy = 345 MPa Fu = 450 MPa a. Compute the tensile capacity due to the failure of the plates. b. Compute the tensile capacity due to the failure of the bolts.Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem The butt connection shows 8-22 mm dia. A325 bolts spaced as follows: S1 = 40 mm S3 = 50 mm t1 = 16 mm S2 = 80 mm S4 = 100 mm t2 = 12 mm Steel strength and stresses are: Fy = 248 MPa Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt Hole diameter = 25 mm Questions: Calculate the allowable tensile load T, under the following conditions. a) Based on the gross area of the plate b) Based on the net area of the plate c) Based on block shear strength
- Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|YCalculate the tightening torque for an SAE 5, ASTM449, 13UNC grade bolt with a diameter of 1/2 inch. Also determine the tensile stress and torsional stress. Indicate on the Mohr circle the principal stresses in the bolt and determine the factor of safety FS.The 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 rivets
- Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.A double cover butt joint is shown in figure 2 below. If the steel is Grade S355JR and the bolts are M20 Grade 8.8 S, determine the strength of the connection. Assume the contact surface of the bolts are blast – cleaned with Class B coatings. The connection is required not to slip at serviceability. 80 120 80 80 120 80 -13 All dimensions are in mm Figure 2: Double cover butt joint. 300 O O O O o o O o oThe butt connection shows 8-22 mm diameter bolts spaced as shown below.Use NSCP2001 Steel strength and stresses are: Yield strength, Fy 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions: 1Based on the gross area of the plate. 2Based on the net area of the plate. 3Based on block shear strength. 50 100 50 50 100 50 40 80 40 12 mm 16 mm
- The 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: %3D3. Determine Rn based on bolt limit states. Slip critical connection is used for design and Grade 10.9 M22 Bolts with Class A surface. Check minimum edge distance and spacing with TSSDC 2016. All members are $355. There is no filler between plates and Du=1.0. Fy= 355 MPa; Fu=510 MPa; E=200000 MPa. All units are in mm. $R.=? 100 100 O 60 Geset Plate 75 Plate Gusset Plate $R.=? *20*201 PlateDetermine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. X = 0.7 Y = 3.8 Z=1/2 Round your answer to 2 decimal places. Your Answer: Incorrect The answer is 77.15 ± 1%. 1/2" X" bolts 1½" Gusset plate -L3 X 3 X Z Tu I