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Architecture/math - steel structures
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- 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 answer4. 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.Bolted 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.
- 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 +40Use ASD to determine the width of the 5/8 inch thick A36 plate subjected to tension as shown below. The dead load effect is PD=85 kip while the live load effect is PL=80 kip. Two rows of 5/8 inch standard bolts will be used for the connection and the plate width can be manufactured in 1 inch increments. Assume that the larger connected plate does not fail. Consider the limit states of gross tensile yield and tensile rupture. Find the required tension strength, Pa, for the controlling ASD load combination. Determine the width necessary for the plate to resist gross tensile yield. Determine the width necessary for the plate to resist tensile rupture. Report the required plate width.4. An UPE 300 is to be used as a tension member and it is bolted to a 10mm thick gusset plate with bearing type 8-M20 8.8 bolts as shown in the figure below. a. Check all spacing and edge distance requirements. b. Compute the allowable tensile strength of the channel UPE 300. Don't consider shear strength of the bolts and bearing strength of the bolt holes. [all dimensions are in mm] Material both for gusset plate and UPE 300: S275 F-275N/mm² Fu=430N/mm² UPE 300 Ag = 5660 mm² tw 9.5 mm x = 28.9 mm 70 160 70 + 10mm thick gusset plate +50+ 80 -80-80-50 o +50+80 -80 80-50 UPE 300
- 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: %3DTopic: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 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)
- 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 strengthPrepare an excel file and to find a) area of the bolt, b) Average Shear Stress Bolt c) Bearing Area Stress and d) allowable stress. Applied force range – 500 to 1000 N (in step of 100 N) Bolt diameter range – 10 to 25 mm (in step of 5 mm) Plate thickness – 10 mm Consider ultimate tensile strength as 460 N/mm2. And consider factor of safety as 1.5.