ickness (S=5/8") with "A36" member steel (fy=36 ksi & fu=58 ksi).
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- 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.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.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.
- Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. WT12 x 38 Longitudinal welds 11.2 in? y = 3.0 in. Given: Properties of WT12 × 38: Ag = Use A992 Steel: F, = 50 ksi Fu = 65 ksi bf = 8.99in. %3D %3D LL = 3 DL %3D %3D tw y = centroidal distance bf C. What is the Governing Ultimate Tensile Capacity based on Net Fracture Round your answer to 3 decimal places.The plate with dimension 18 mm x 400 mm (thickness by width) is to be connected to a gusset plate using 6-Ø24 mm bolts in arrange in 3 row (2 each row) along the x-axis. The plate is to be design for tension. Using A36 steel and assuming Ae= An. Use NSCP 2015. a. Compute the design strength considering yielding and tensile rupture. (LRFD) b. Compute the allowable strength considering yielding and tensile rupture. (ASD)Assume that the gusset plate and welds are satisfied. A gap allowance of 1/16 inch is to be considered to fit the gusset plate in the HSS. The steel used in the tension member is ASTM A500 Grade C HSS 6 x 0.5 with a length of 30 feet. a.) Find the shear lag factor b.) Find the Effective net area (Ae) c.) Solve for the tensile strenngth 16-inch-long fillet weld HSS 6 x 0.5 1/2 in. thick gusset plate 17 inch long
- Q: 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).Determine the maximum service load P that can be resisted safely by the welded connection shown. Use ASD. E80 10 150 300 200Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. Longitudinal welds bf Given: Properties of WT12 x 38: Ag = 11.2 in² y = 3.0 in. bf = 8.99in. Use A992 Steel: Fy = 50 ksi Fu = 65 ksi LL = 3 DL tw WT12 x 38 y = centroidal distance C. What is the Governing Ultimate Tensile Capacity based on Net Fracture Round your answer to 3 decimal places. T
- The plate with dimension 18 mm x 400 mm (thickness by width) is to be connected to a gusset plate using 6-Ø24 mm bolts in arrange in 3 row (2 each row) along the x-axis. The plate is to be design for tension. Using A36 steel and assuming A= An. Use NSCP 2015. a. Compute the design strength considering yielding and tensile rupture. (LRFD} b. Compute the allowable strength considering yielding and tensile rupture. (ASD)Select an American Standard Channel shape for the following tensile loads: dead load = 54 kips, live load = 80 kips, and wind load = 75 kips. The connection will be with longitudinal welds. Use an estimated shear lag factor of U = 0.85. (In a practical design, once the member was selected and the connection designed, the value of U would be computed and the member design could be revised if necessary.) The length is 17.5 ft. Use Fy=50 ksi and Fu=65 ksi. a. Use LRFD. b. Use ASD.Determine the adequacy of the hanger connection in Figure P7.8-2 Account for prying action. a. Use LRFD. b. Use ASD.