1. The tension member shown in Figure 3.4-2 is a PL 58 x 10, and the steel is A36. The bolts are 3/4-inch in diameter. a. Draw the different potential failure lines. b. Compute the effective net area.
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Q: 1. The tension member shown in Figure 3.4-2 is a PL 58 x 10, and the steel is A36. The 3/4-inch in…
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- A L 3 x 2 x ¼ is connected to a gusset plate via six bolts. The nominal diameter of the bolt is 0.25 inches, the pitch spacing is 1.75 inches, the gage spacing is 2 inches, and the thickness of the connection is (¼) inch. The yield stress is 50 Ksi and the ultimate stress is 60 Ksi. Consider section line (a-a') for the analysis. . a. What is the effective net area (Ae) of the angle section in inches2 [a-a']?(The shear lag factor is "0.80") b. What is the design tensile yielding strength in Kips for the steel member? c. What is the design tensile rupture strength in Kips for the steel member? d. What is the minimum Factor of Safety if a tensile load of 23 Kips is applied to the angle section? please make sure the answer is correct 100% I only need the final answersDetermine 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 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0.a) Determine the design tensile strength of the section based on yielding of the gross area.b) Determine the critical net area of the connection shown.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 30mmTwo 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 shear
- A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. Determine the critical net area of the connection shown. Determine the design tensile strength of the section based on yielding of the gross area.The given angle bar L125x75x12 with Ag = 2,269 sq.mm. is connected to a gusset plate using 20 mm diameter bolts as shown in the figure. Using A36 steel with Fy = 248 MPa and Fu = 400 MPa, determine the following: 2. Determine the nominal tensile strength of the 12 mm thick, A36 angle bar shown based on: a. Gross yielding b. Tensile rupture Bolts used for the connection are 20 mm in diameter. O O O O O O O Effective net area of the tension member if the shear lag factor is 0.80. Select the correct response: 1,516.1 1,354.4 1,431.2 1,221.6WT12 x 38 Longitudinal welds Given: Properties of WT12 x 38: A, = 11.2 in? y = 3.0 in. b = 8.99in. Use A992 Steel: F, = 50 ksi F = 65 ksi LL = 3 DL %3D %3D tw = centroidal distance bf A. Governing Ultimate Tensile Capacity based on Yielding of gross section Round your answer to 0 decimal places.
- 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.A bolted connection shown consists of two plates 300mm x12mm connected by 4 - 22 mm diameter bolts. Edge distances = 75mm dhole for tensile and rupture = db + 3 mm dhole for bearing strength for Lc = db + 1.5 mm Fy = 248 Mpa Fu = 400 Mpa Fn = 330 Mpa Use LRFD design method. Determine the design strength due to the gross yielding of plates. (kN)Considering 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.