Design the size and length of Fillet weld for the lap joint shown below, Use SMAW E70XX process, plates are A-36 steel? 90k LL 40k DL R-X7 Gusset P 90k LL 40k DL
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- a. Use LRFD and design a welded connection for the bracket shown in Figure P8.4-19. All structural steel is A36. The horizontal 10-inch dimension is a maximum. b. State why you think your weld size and configuration are best.Use an elastic analysis and determine the maximum load per inch of weld.What size of fillet welds are required to attach a 16 mm flat plate hanger to the bottom flange of a UB 457 x 152 x 74 so that the full tensile capacity of the hanger may be developed ? Calculate the effective width (beff) of the weld. Assume the use of S355 steel for UB and the plate and the width of the plate is the same as the flange width of UB .
- The connection shows a PL10x200 loaded in tension and welded to a gusset plate. Calculate the ff, assuming A992 steel is used. Use U=0.85. The weld thickness is 6mm and the electrode is E70 (485 Mpa). What is the allowable tensile strength (neglecting block shear) in kN?4.) A 150 x 90 x 12 angular section is welded to a gusset plate as shown in the figure. The angle is A36 steel with FY=248 MPa. The weld is E 80 electrode with Fu = 550 MPa. The allowable tensile stress for the angle is 0.6Fy and the allowable shear stress for the weld is 0.3Fu. The Area of the angular section is 2751 sq.mm. with y=51mm. CS 56 Which of the following most nearly gives the design force P? a.) 400,365.9 N b.) 409,348.8 N c.) 412,793.5 N d.) 420,366.6 N 5.) A W350 x 90 steel is used as a simply supported beam 8m long. The beam carries three equal concentrated loads at every quarter points. It also carries a uniform dead load o 5 kn/m (including its own weight) and a uniform live load of 7.20 kn/m. Properties of W 350 x 90 steel: bf = 250 mm The allowable bending stress is 0.66Fy. The allowable shear stress is 0.40Fy. The allowable deflection is L/360. Use Fx=248 Mpa and E=200 Gpa. tf = 16.40 mm V d = 350 mm Determine the maximum value of each concentrated load based on…A 4 x 4 x ½ - in. angle of A36 steel is to be welded to a plate with E70XX electrodes to develop the full tensile strength of the angle. Using 3/8-in. fillet welds, compute the design lengths for the welds on the two sides of angle, assuming development of tension on the full cross section of the angles. GIVEN: A = 3.75 in2 L1 = 10.94” L2 = 5.09” IF only connected leg, L = 7.7”; 4.75 each side
- A channel C250x37 mm section is welded to a 9 mm gusset plate. Welding is not permitted on the back of the channel. All steel is A36 with Fy=250 MPa and Fu=400 MPa. Use E70electrodes having and Fu=485 MPa (SMAW) process. The maximum length of lap is 250mm. The size of fillet weld is 8mm. Assume the width of slot weld is 22 mm. Size of slot weld is 13mm Properties of C250x37 A = 4750 mm2 tw = 13.0 mm2 d = 254 mm a. Determine the force resisted by the slot weld in kN, when the full tensile capacity is 712.5 KN (from the gross yielding capacity using ASD) Hint: Full tensile Capacity = Force Resisted by Fillet and Slot Weld Round your answer to 3 decimal places.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.Material Strengths: Fy= 248 MPa, Fu= 400 MPa Method: Use ASD. A single angle L203X152X19 is welded (longitudinal and transverse welds) on one leg (shorter leg) to a gusset plate; thus, there are no holes. The length of the longitudinal welds is 150 mm. Which of the following most nearly gives the allowable tensile load (N) based on fracture on net area? CHOICES: 474 549 719 478 1424 1439 1099 957 719 949 712 732 366 1439
- 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 The angle 150x100x10mm is connected to a 8.5 mm gusset plate. The size of weld is 8 mm and using E70 Electrodes. Fu = 438 MPa. Use LRFD. Questions: a) Find the length of the weld required at the toe of the angle. b) Find the length of the weld required at the heel of the angle. c) Find the block shear strength of the gusset plate.A lap joint is shown in the figure is welded using a fillet weld with an electrode E70 (Fu - 485MPa) Fy - 250MPa. Use SAW process. a. Design the size of the fillet weld. b. Compute the value of "a". c Compute the length of end returns "b" T + 190mm mm T + ►TQ: Design Fillet welded connection using E65X electrodes to connect three plate sections (W=20in & t=0.75”) for transmitting dead load of 10k & live load of 140k.Consider minimum weld thickness (S=5/8”) with “A36” member steel (fy=36 ksi & fu=58 ksi).