A ½ x 5 plate of A36 steel is used as a tension member. It is connected to a gus- set plate with four 5/8-inch-diameter bolts as shown in Figure. Assume that the effective net area A, equals the actual net area A,, (we cover computation of effec- tive net area in Section 3.3). a. What is the design strength for LRFD? b. What is the allowable strength for ASD? DI 1/2 in.
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- A 1 ⁄2 × 5-inch plate of A588 steel is used as a tension member. It is connected to a gusset plate with four 5 ⁄8-inch-diameter bolts as shown in Figure 1. Assume that the effective net area Ae equals the actual net area An. (a) Calculate the design strength for LRFD? (b) Calculate the allowable strength for ASD?Determine the Net Area for the following Plate Section consider all possible Cases. The Plate thickness is 1-in. and the bolts' diameter is 7/8-in. s = 6.0 " g = 2.0 "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)
- 5. a bolted lap joint shown in the figure below. The bolts are 20m in diameter in 23 mm holes. The plates are 12m thick X350 Allowable stress of Plates: Tenaion in gross area0.60Fy Tension in net area0.50 Fu Shear on Net Area 0.30Fu Yield Strength of plate, Fy 248 Npa Ultinate tensile strength of Plate. Fu400 Mpa x, Find the safe load P based on a- Gross area yielding b. net area rapture c. block shear 2 of 3an angle bar 100x100x11 mm tension member is connected with 20-mm-dia bolts as shown in the fig. both legs of the angles are connected. use fy=248 MPa and Fu=400 MPa. 1. determine the nominal strength for tensile yielding in the gross section. a. 515.59 kN b. 429.67 kN c. 545.60 kN d. 390.10 kN 2. determine the effective net area of member a. 1573 mm^2 b. 1632 mm^2 c. 1605 mm^2 d. 1815 mm^2 3. determine the nominal strength for tensile rupture in the net area. a. 642 kN b. 730 kN c. 554 kN d. 650 kNIf the allowable bearing stress in the bolt is 18.855 ksi. what is the minimumrequired diameter of the bolt at B? b= .75” h= 7.874” Distances are in feet.
- The Plate shown in the fIgure has thickness of 12 mm. Diameter od bolts is 16 mm. A 36 steel is used with Fy = 248 MPa and Fu = 400 MPa. Standard nominal hole diameter of 16 mm bolts = 17 mm. Compute the minimum pitch ‘S” for which only two and one half bolts need be subtracted at any one section in calculating the net area. Compute the effective net area of the section if the reduction factor u = 0.75 Compute the allowable tensile strength of this sectionThe 1 × 8 plate is shown in the figure. The holes are for 3/4-in Ø bolts. Compute the net area of each of the given members. (Ans. 6.44 in^2)A channel shape, C8 × 18.75, is used as a tension member. The channels are boltedto a 10mm-thick gusset plate with 22mm- diameter bolts. The tension member is A572 Grade 50 steel (Fy = 345 MPa, Fu = 450 MPa) and the gusset plate is A36 (Fy = 250 MPa, Fu = 410MPa). What is the ult. capacity of the channel in kips, if yielding on the gross area governs? What is the ult. capacity of the channel in kips, if fracture on the net area governs?
- 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.3. A single angle tension member L89x89x9.5 with an area of 1,800mm2 connected to a gusset plate with 22 mm diameter bolts as shown in the figure. A36 steel is used with Fy=248 MPa and Fu= 400 MPa. The service loads are 150 KN dead load and 65 KN live load. Investigate the adequacy of the member. Diameter of hole is 3 mm bigger than bolt diameter and use net area equal to 85% of the computed net area. Neglect block shear. a. Check using ASD Load Combination b. Check using LRFD load combinationThe tension member shown in Figure is an L6 × 3 ½× 5/16. The bolts are ¾ inch ameter. If A36 steel is used, is the member adequate for a total load of 100 kips?