9. Determine the service load capacity assuming the load is 70% live load of the tee connection as shown in Figure 6.28 and details the proper double-bevel-groove weld for the SMAW process. Assume the flange of the tee does not control design. R-3×8 (A36 Steel) 128 kips Problem 164 164 kips Solution E60
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- 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.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.
- Design a double angle tension member that is fillet welded to a gusset plate. Assume that the angles will be made from A36 steel, and the gusset plate from A572 Gr. 50 steel. Use 70ksi electrode. Design the system to resist a factored load of 200 kips. Check all applicable limit states that are possible. Provide your results on a sketch with standard weld symbols and details.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 service load capacity P for the welded bracket of the accompanying figure. The load is 70% live load and 30% dead load. Neglect the returns at the outer ends of the C-shaped weld configuration. The weld size is 3/8" (9.53 mm) and E70 electrodes are used in the shielded metal arc process. • Use strength analysis ● Use elastic method 203.50 -12.70 THK -285.75 -63.50
- PROBLEM 9.9 9.10 in the bolted connection of Figure 9.12, assume that the bolts are in. in diameter and that the allowable shear stress for the bolts is 14,500 psi. Based on bolt shear only, compute the safe allowable load P that may be applied. 9.2 CcAssume 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 longProblem 7: The load that will be applied to the connection shown has a live load - to - dead load ratio of 3.0. Investigate all limit states. All structural steel is A36, and the weld is a 1/4-inch fillet weld with E70 electrodes. Note that the tension member is a double-angle shape, and both of the angles are welded as shown. Use ASD. Determine the following. 5" 5" 2L5 X 32 X 5/16 LLBB -t = ³/8" Maximum service load that can be applied without exceeding the allowable capacity on yielding on gross area of the tension member (double angle). Maximum service load that can be applied without exceeding the allowable fracture on the net area of the tension member (double angle). Maximum service load that can be applied without exceeding the allowable block shear strength. Considering the weld metal and base metal strength, calculate the maximum service load that can be applied.
- Q#4: 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).The connection is shown in Fig. 3. Two cover plates are applied to connect the steel plates by fillet welds. The dimensions of connection are presented in Fig. 3. The leg size of fillet weld hr is 10 mm. Q235 steel, f-215 N/mm2, E43 electrode, fav=160N/mm?. Calculate the load-carrying capacity of the fillet weld connection wwww. (Note: the strength of steel plate should be calcualted. (Unit: mm) 220 ス20 091Compute 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