A 3/4" x 10" X 5' long plate is fabricated from A36 steel and has standard holes for 3/4" bolts at each end for attachment to other structural members. The figure below shows a face view of the plate. The service loads that the member will be subject to are 140 kips of dead load and 30 kips of live load. Calculate the factored load, tensile yield strength and rupture strength. Compare the results to the factored loads and comment. Ignore block shear. +3" 3" + 3" **- - 1 1/2" 10" 5' SYM 5
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- 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 CSTopic: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”Topic: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.
- 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.The Aluminum plates below are connected by two 3/4" diameter bolts. What is the shear stress in the bolts given the load applied?
- A steel plate is to be attached to a support with three bolts. The cross-sectional area of the plate is 800 mm² and the yield strength of the steel is 260 MPa. The ultimate shear strength of the bolts is 570 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. Note: consider only the gross cross-sectional area of the plate-not the net area. Support Plate ↓ PTopic: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.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 80
- 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.SITUATION 14: A PL 300 x 20 mm is to be connected to two plates of the same material with half the thickness by 25 mm Ø rivets as shown in the figure. The rivet holes have a diameter 2 mm larger than the rivet diameter. The plate is A36 steel with Fy= 250 MPa, allowable tensile stress of 0.60Fy and allowable bearing stress of 1.35Fy. The rivets are A502, Grade 2, hotdriven rivets with allowable shear stress of 150 MPa. 25 mm Ø rivets + + PL 300x20 P/2 P P/2 PL 300x10 43. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable shear stress in the rivets? a. 675 KN b. 490 KN c. 598 KN d. 790.5 kN 44. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable bearing stress between the plates and the rivets? a. 490 KN b. 675 KN c. 598 KN d. 790.5 kN 45. Which of the following most nearly gives the max. load in kN that can be applied to the…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