The ultimate shear strength of the bolts is 250 MPa, and a factor of safety of 2 is required. Determine the minimum required bolt diameter to resist an applied load of P= 1,300 kN. Support Bar
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A: Given data dead load = 60 kN Live load=40kN fy=248N/mm2 fu=400N/mm2
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- Determine 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 IThe butt connection shows 8-22 mm diameter bolts spaced as shown below. P- 50 100 50 50 100 50 16 mm +HHHH 40 80 40 12 mm 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 Based on the gross area of the plate. Based on the net area of the plate. Based on block shear strength. Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions:A flanged bolt coupling consists of 6 - 12 mm - diameter bolts on a bolt circle of 320 mm in diameter. Compute the number of additional 10 mm - diameter bolts on a bolt circle 200 mm in diameter to increase the torque capacity to 7.65 kNm. The shearing stress of the bolts should not exceed 60 MPa. Select one: O a. 6 bolts O b. 4 bolts O c. 3 bolts O d. 2 bolts NEXT PAGE
- 5. a bolted lap joint shown in the figure below. The bolts are 20m in diameter in 23 mm holes. The plates are 12m thick X350 Allowable stress of Plates: Tenaion in gross area0.60Fy Tension in net area0.50 Fu Shear on Net Area 0.30Fu Yield Strength of plate, Fy 248 Npa Ultinate tensile strength of Plate. Fu400 Mpa x, Find the safe load P based on a- Gross area yielding b. net area rapture c. block shear 2 of 3The bolted connection shown must support an applied load of P = 5300 kN. The average shear stress in the bolts must be limited to 250 MPa. Determine the minimum bolt diameter that may be used in the connection. Answer: Minimum bolt diameter = i mmSITUATION II: Calculate the permissible tensile load P 350 under the following conditions: Fy-248 MPa; Fu=600 MPa. The diameter of the bolts is 16 mm. 716 100 50 50 100 100 50 50 Allowable stresses: Tensile stresses on gross area: 0.6Fy Shear stress of the bolts: Fv=130MPA Bearing stress of the bolts: Fp=1.2Fu 7. Based on shear capacity of bolts. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 k 8. Based on the bearing capacity of bolts. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 kN 9. Based on the yielding of plates. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 kN
- 4. As shown below, the splicing joint of a beam was made with bearing type high-strength bolts. The loads transferred through the joint is V=3000KN, M=90KN.m. The bolts are of grade 10.9, nominal diameter 22mm, effective area Ae=2.45cm2;f=310N/ mm?; f-500N/mm²; End plates made of Q235B steel, thickness 22mm, fe-470N/mm?. Check the strength of this bolt connection. -24mm 35 130 35 80 V 1 og1 081 1 081The plate shown is attached by using three M12x2 grade 8.8 bolts. Which screw is subjected to the largest stress? What safe load ( static) F can be supported by the screw for n=1.5.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.
- 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|YA 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.Determine the minimum bolt diameter (mm) required if the allowable shear stress is 140 MPa. 40 kN 40 kN 30 mm 80 KN 30 mm