e single shear connection consists of six 26-mm-diame Its. If the ultimage strength of the bolts is 720 MPa, cermine the factor of safety for the connection at an ap ad of P=570 kN.
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- • Compute the available block shear strength of the connection based on the KN. (Diameter of hole is 20mm, ST37 steel is used, and thickness of plate is 7.5mm) 2@75 mm 3@75 mmProblem 5: The lap connection shown consists of a PL1/2"X3" plate connected to a ½ in gusset plate with ¼" longitudinal fillet welds as indicated. The plates are A36 steel and the electrodes are E70XX. Calculate the length L (in) of the connection to develop the full capacity of the plate. Hint: The full capacity of the plate will depend on the governing strength of the tension member. Neglect block shear. Use ASD 6" x 1/2" P [ T L 1/4 1/4 E70 - 13²x P 3" x 1/2"If the 18 mm plate is connected to two cover plate of 10 mm on top and bottom, what is the capacity of the connection in double shear? Assume that the shear planes pass though the threaded portion. State the condition necessary for these strengths to be available. (Draw the connect first and calculate the capacity)
- Problem 17. The welded connection shown in the figure uses electrodes E70XX. The nominal weld strength of the connection is: a 184.40 kN b 245.87 kN c 430.27 kN d 485.59 kN 18. For the connection shown in problem 17, the LRFD design strength is: a 322.70 kN b 364.19 kN c 387.24 kN d 437.03 kN 19. For the connection shown in problem 17, the ASD al strength is: a 215.14 kN b 242.80 kN c 322.70 kN d 387.24 kNCalculate and check the design strength of the connection shown below. Is the connection adequate for carrying the factored load of 65 kips. -3/8 in. А36 5 x % 125 АЗ6 2.50 65 k 1.25 % in, bolts 1.25 2.50 1.25Select 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.
- M3 HW 4 4. Six rivets are used in the connection shown in the REVISED PROBLEM: figure. If both P= 50 kN, what is the minimum diameter P- 50 rivet is necessary so as not to exceed the 70MPA allow shearing stress? What thickness of plate is required so as not to exceed a bearing stress of 140 MPa? 80 mm 80 mm Pe so 100 mm7. The yoke-and-rod connection shown below is subjected to a tensile force of 5kN. Determine the minimum thickness of yoke if the bolt has a diameter of 25mm and the bearing stress is limited to 12MPA. 40 mm 5 kN. 30 mm 25 mm 5 kNa rivet lap joint is shown. The rivets used have allowable shear strength equal to 533 MPA with diameter of 16mm. The steel plate has a yield strength equal to 275 MPa and ultimate strength equal to 340 mpa; 300 mm wide and 16 mm thick. Determine the safest value of P that the connection can resist. Use NSCP, assume additional 1.6mm for the rivet hole. Neglect block shear
- 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 criticalA connection using 3 bolts with the eccentricity is shown in Figure Q3. Determine: (a) Average shear stress for the bolts, (b) Average normal stress between bolt and plate A, and (c) Maximum normal stress in plate B. plat B 12 mm P/2 plat A 25 men. P/2 12 mm 100 mm -P D- 10 mm D- 20 mm Figure Q4The five-bolt connection must support an applied load of P = 2600 lb. If the average shear stress in the bolts must be limited to 27 ksi, determine the minimum bolt diameter that may be used for this connection. cocco Answer: dmin - PI in.