2. If P = 200 kN, determine the maximum live load that the connection could carry based on tensile rupture using LRFD.
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- Determine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40STAGGERED CONNECTIONS: A PLATE WITH WIDTH OF 400 mm AND THICKNESS OF 12 mm IS TO BE CONNECTED TO A PLATE OF THE SAME WIDTH AND THICKNESS BY 34 mm DIAMETER BOLTS, AS SHOWN IN THE FIGURE. THE HOLES ARE 2 mm LARGER THAN THE BOLT DIAMETER. THE PLATE IS A36 STEEL WITH YIELD STRENGTH Fy = 248 MPa. ASSUME ALLOWABLE TENSILE STRESS ON NET AREA IS 0.60Fy. IT IS REQUIRED TO DETERMINE THE VALUE OF b SUCH THAT THE NET WIDTH ALONG BOLTS 1-2-3-4 IS EQUAL TO THE NET WIDTH ALONG BOLTS 1-2-4. a. CALCULATE THE VALUE OF b IN MILLIMETERS. b. CALCULATE THE VALUE OF THE NET AREA FOR TENSION IN PLATES IN SQUARE MILLIMETERS. c. CALCULATE THE VALUE OF P SO THAT THE ALLOWABLE TENSILE STRESS ON NET AREA WILL NOT BE EXCEEDED.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 answer
- Design a welded connection for an MC9x23.9 of A572 Grade 50 steel connected to a 3/8-inch-thick gusset plate (Figure 6). The gusset plate is A36 steel. Show your results on a sketch, complete with dimensions. = 3/8" Figure 6 D = 48 k L = 120 k MC9 x 23.9 a. Use LRFD. b. Use ASD.1. A riveted lap joint is shown in the figure below. The rivets used are A502, Grade 2, hot-driven with diameter of 16mm. The steel is ASTM A501, 200mm wide and 10mm thick. Determine the safe value of P in all possible modes of failure and the allowable and ultimate value of P that the connection can resist. Use NSCP 2015 200 mam 110 mm P' 10 mmConsidering 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.
- Agui - 4550/2 =2276 two plates each with thickness t = 16 mm are bolted together with 6-22 mndia. bolts forming alap Joint - Bolt spacing are follows St=40mm S2 - 80mm, 53= 100mm effective bolt hole dia is 25mm A36 is used. Compute the allowable tensile strength t = 16 mm ob 27 40 60 80 40 +=16mm S1-40 S2-100 52-40 P 7 @ 4ummQ21 In the connection shown in Figure below, the bolts are %in diameter, and A36 steel is used for all components. Compute the design Strength of the connection. 12" 12" 22"Determine the maximum service load P that can be resisted safely by the welded connection shown. Use ASD. E80 10 150 300 200
- Bolted and Riveted Connections From the bolted connection shown, the diameter of bolts is 18mm0 with the hole diameter equal to 3mm bigger than the bolt, the angular section is 100 x 75 x 8 mm, with an area of 1340 mm2. Thickness of the gusset plate is 9mm. The gusset plate and angle arc A36 steel with Fy = 250 MPa and Fu = 400 MPa. Determine the tensile capacity of the connection based on gross area. Determine the tensile capacity of the connection based on effective net area using a reduction factor of 0.85. Determine the tensile capacity of the connection based on gross area.6) The connection shown in the figure uses 20mm diameter A325 bolts. The tension member is A36 steel (Fy = 248 MPa, Fu = 400 MPa) and the gusset plate is A572 steel (Fy=290 MPa, Fu = 415 MPa). Determine the design strength of the connection (in kN) considering block shear rupture. 37.5 mm 75 mm 37.5 mm 9 mm gusset P 12 mm x 150 mm 37.5 mm 75 mm 37.5 mmProblem 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.