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 effective net area of the L7 × 4 × 1/2 shown in the figure. Assume the holes are for 1-in Ø bolts. Use the U values given in Table 3.2. (Ans. 3.97 in^2)Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.Determine the nominal tensile strength of the single angle. Consider 7/8" diameter bolt and A242 steel with Fy = 50 ksi and Fu = 70 ksi. L6x6x3/4 d = 6 in. b = 6 in. t = 0.75 in. k = 1.25 in. wt./ft. = 28.8 plf. A = 8.46 in.^2 x = 1.77 in.
- Instructions:-Write down each step clearly in calculation suggestion what you are doing.( you need to tell why you selected particular value ofø (phi) and U). What and why you are calculating certain step Consider R= 433.5 PROBLEM A welded tension member is consisting of two channels placed 400mm back to back with flanges turned out. Select a channel for factored tensile of 4RKN using A36 steel and AISC specifications.The member is to be 15m long. PLASE USE THE EXAMPLE AS GUIDE OF ALL STEPS AND FORMULAS TO BE USEA 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.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 13-mm thick plates of the lap joint shown is made of A36 steel whose yield strength and tensile strength is 248 MPa and 400 MPa, respectively. The bolts are A325-M22. The holes are standard sizes and the threads are excluded from the shear plane. Assume that deformation at bolt holes is a design consideration. Determine the design (LRFD) strength of the lap joint based on shear strength (kN) of the bolts. [Group of Answers: 424, 521, 390, 318] Determine the nominal strength of the lap joint based on shear strength (kN) of the bolts [Group of Answers: 424, 566, 521, 695]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 shearSubject : Steel design 1. A tension number is an angular section 150mm x 110mm x 12.5 mm having a cross sectional area of 3064mm^2. The section is connected with three 18mm diameter bolt spaced at 100mm on centers as shown on the figure.Minimum tensile strength Fu=400Mpa and Fy=248.8Mpa. Diameter of Hole is 21mm. a. Compute the block shearing strength of the member.
- The joint shown in Figure 2 is made from steel Grade 350 (AS/NZS 3678). Four bolts are used. The diameters of all holes are 5 mm. Note that the middle hole on both Plates A is just a hole with no bolt. The design shear capacity (ϕVf) of the bolts in single shear is 10 kN. Dimensions given below are in mm.The bolts in the lap joint shown are 22mm in diameter and 25mm diameter holes. The plates are A36 steel with Fy = 250MPa and Fu = 400Mpa. For this problem, x1 = 500mm, x2 = 160mm, x3 = 60mm and t = 12mm. The allowable stresses are: Bearing Stress on projected area of the plates Fp = 1.5Fu Tension on net area of plates, Ft = 0.5Fu Shear on plates, Fy= 0.3Fu Shear strength of bolt, Fv= 210 Mpa a.) Determine the value of P based on bolt shear b.) Determine the value of load P based on bearing on projected area on plate (area in contact with the plate and bolts) c.) Determine the value of P based on block shear (use net area for block shear and use the diameter of hole)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.