A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 215 x 3 x %6 LLBB Determine the allowable tensile strength for ASD. Select one: a. 52.1 b. 43.9 c. 87.8 d. 104.2
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- Determine the design tensile strength of the 12 in. x 1/2 in. steel plate shown in the figure. The bolts are 3/4 in. diameter. The steel is A572 Gr. 50. Check yielding and fracture. Check Block Shear. T 3in. 73im 13in 1 3in tDetermine 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 +40A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. Ae 1 2 Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN AL Section 2L5 x 3 x 516 LLBB
- A plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. 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 equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue 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.PROBLEM NO. 2: Two plates each with thickness t = 16mm are bolted together with 6-22mm diameter bolts forming a lap connection. Bolt Spacing are as follows: S₁ = 40mm, S₂ = 80mm, S3 = 100mm. Bolt diameter = 22mm. Use A36 steel. Calculate the allowable strength due to block shear only. I= 16 mm S₁ S₂ S₂ S1 40 80 80 40 18-16mm S-40 S₁-100 S-40Q1:A: The Ix6 in. plate shown in Figure below is connected to a lx10 in. plate with longitudinal fillet welds to transfer a tensile load. Determine the LRFD design tensile strength of the member if F, = 50 ksi and Fu = 65 ksi. PLI X 10 in PL1 x 6 in P P w= 6 in Longitudinal fillet welds L=8 in
- Two plates each with thickness t=16mm are bolted together with g-22 mm dia•bolts forming a lap connection.bolts spacing are as follows S1=40mm, S2=80mm,S3=100. Bolt hole dia=25 mm Fu=483Mpa Fy=345Mpa Solve the allowable strength and the ultimate strength in: 1. Yielding 2.rupture 3.shear 4.block shearWhat is the safe load "P" that can be transmitted by the fillet-weld joint shown in figure below if allowable shear stress in the fillet weld is 108 MPa and F.O.S. for load is 1.5? P4 www 60mm 777 50mm 8mm PThe tension member is a PL 12 x 6. It is connected to a 38-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 3" PL ... PL ½x 6 12" 24" 24" 12" a. What is the minimum spacing as per AISC code provisions Round your answer to 3 decimal places.
- 3. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mmQUESTION 8 3. check the capacity of the shear connection. A992: Fy=50ksi, Fu=65ksi A36: Fy=36ksi, Fu=58ksi effective bolt hole diameter = bolt hole + 1/16" beam properties: tw=0.295 in. Fexx=70ksi what is the shear yielding capacity of shear plate in kips? (2 decimal places)A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 2L5 x 3 x16 LLBB Determine the design tensile strength for LRFD. Select one: а. 78 b. 66 С. 132 d. 156