1. Determine the allowable load Pa based on bearing at bolt holes. (ASD) 2. Determine the ultimate load Py based on block shear.
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- Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem The butt connection shows 8-22 mm dia. A325 bolts spaced as follows: S1 = 40 mm S3 = 50 mm t1 = 16 mm S2 = 80 mm S4 = 100 mm t2 = 12 mm Steel strength and stresses are: Fy = 248 MPa Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Bolt Hole diameter = 25 mm Questions: Calculate the allowable tensile load T, under the following conditions. a) Based on the gross area of the plate b) Based on the net area of the plate c) Based on block shear strengthDetermine 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 +40Considering 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.
- 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 5" Gusset P 8 90k LL 40k DLSTAGGERED 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.4. Determine the design strength of the connection shown in the Figure below. Thebolts are 25 mm diameter A490 bolts with the threads not in the plane of shear.A36 steel is used (Fy = 250 MPa, Fu = 400 MPa).a. Compute the shear strength for all bolts.b. Compute the bearing strength for the tension member on all bolts.c. Compute the bearing strength for the gusset plate on all bolts.d. Compute the tensile strength of the tension member.e. Compute the design strength of the connection.A bearing type connection is shown in Figure 3.19. The diameter of A 325 bolts is 22 mm and the A572 Grade 50 plate material has a width of 150 mm and thickness of 16 mm. Assume diameter hole to be 24 mm. Bolt threads are excluded from the shear plane. Allowable stress of A 325 bolts: Fv = 207 MPa Fp = 1.5 Fu (to prevent excessive hole deformation) Allowable stresses of A572 Grade 50 plate material: Fy = 345 MPa Fu = 450 MPa a. Compute the tensile capacity due to the failure of the plates. b. Compute the tensile capacity due to the failure of the bolts.
- The tension member is a PL 1⁄2 × 6. It is connected to a 3⁄8-inch-thick gusset plate with 7⁄8-inch-diameter bolts. Both components are of A242 steel. Note: A242 Fu = 70ksi dh = db + 1/16’’ Use: Consider deformation at the bolt hole what is the: minimum spacing as per AISC code provisions maximum spacing as per AISC code provisions minimum edge distance as per AISC code provisions maximum edge distance as per AISC code provisionsTopic:Bolted Steel Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A 12.5 mm x150 mm plate is connected to a gusset plate having a thickness of 9.5 mm.Diameter of bolt = 20 mm Both the tension member and the gusset plate are of A 36 steel. Fy = 248 MPa, FU = 400 MPa. Shear stress of bolt Fnv = 300MPa. Questions: a.Determine the shear strength of the connection. b.Determine the bearing strength of the connection. c. Determine the block shear strength of the connection.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 @ 4umm