5. Find the allowable strength of the welded connection with a minimum throat size shown below. Base metal: $275 F₂ = 275N/mm² Fu = 430N/mm² Weld metal tensile strength: F = 480N/mm² [all dimensions are in mm] A -10mm thick gusset plate 150 125 125 -10mm thick gusset plate 125 VIEW A-A a 125 PL 150x12 PL 150x12 LPL 150x12 A
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- 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.Calculate the maximum tensile load, Ta or Tu of the connection shown in Figure 7. Theconnection consists of two 3/8" plates fastened to each side of one 13/16" plate.Assume 7/8" diameter bolts and standard punched holes. Use A36 steel. The plate is 13” wide. Note: Be sure to check all failure modes; bolt failure modes, and member failure modes(including block shear)
- 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 answerShown in Figure 1 is a centrically loaded bolted connection. The diameter of the bolts is 19 mm while holes have 21 mm diameter. Use depth 4y = 200 mm. 1. Determine the critical net width (Net Width) for the staggered bolted connection as shown in the figure. 2. Determine the value of "x" so that the net width along bolts A-B-C-D-E is equal to the net width along bolts A-B-D-E.Topic: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 tension member consists of a double angle section with long legs back to back. The angles are attached to a 9.5 mm thick gusset plate. Fu = 400 MPa Fy = 248 MPa for angular section. Fw = 480 MPa for 8 mm fillet weld. Reduction factor U = 0.80 Prop. of One Angle L 125m x 75m x 12.7 m A= 2419 mm2 y=44.45 mm Questions: a) Compute the design strength capacity of one angle. b) Compute the base metal shear strength (gusset plate) per unit length. c) Compute the length L1 and L2.
- 10. determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksij) what is the bolt shear capacity of the flange connection in kips? (2 decimal places)W 10x45 is connected to two plates with two lines of ¾” bolts as shown below. Determine the tensile strength of the system if Fy = 50 ksi and Fu = 65 ksi.Design a welded connection. The given loads are service loads. Use Fy =50 ksi for the angle tension member and Fy=36 ksi for the gusset plate. Show your results on a sketch, complete with dimensions. a. Use LRFD. b. Use ASD.
- 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.Two steel plate tension members have been connected using 0.72 inch diameter bolts arranged in an equally-spaced 9 by 3 rectangular formation. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 51.2 ksi, clear distance for each edge bolt as 0.12 in, and the clear distance for the other bolts as 1.072 in. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load and live load assuming live load is half as much as dead load. Consider ASD.Determine the critical net width for the staggered bolted connection as shown in the figure. The diameter of the bolts is 19mm. Use depth 4y = 200 mm. Determine also the value of that will make the Path A-B-C- D-E critical