The design shear strength phi rnv of one A325 X-bolt, 1 1/8 inch in diameter (Group A, X type) in double shear is equal to: Group of answer choices 112 kips 125 kips 74.2 kips 101 kips
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The design shear strength phi rnv of one A325 X-bolt, 1 1/8 inch in diameter (Group A, X type) in double shear is equal to:
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- the design shear strength phi rnv of one A325-X bolt (Group A, X type), 7/8 inch in diameter in double shear is equal to: 46.8 kips 30.7 kips 27.9 kips or 61.3 kipsCalculate the capacity of the shear connection below. A36 Plate: Fyp= 248MPa, Fup= 400MPa. A992 Beam; Fyb= 345MPa, Fub= 450MPa. Fexx= 480MPa. twb= 10.92mm tp= 12, X= 45, nr= 3(no. of bolts) What is the bolt shear capacity in kN? What is the bearing capacity of the shear plate in kN? What us the bearing capacity of the beam in kN? What is the shear yielding capacity of shear plate in kN? What is the shear rupture capacity of the shear plate in kN? What is the block shear rupture capacity of shear plate in kN? What is the weld capacity in kN?Calculate the capacity of the shear connection below. A36 Plate: Fyp= 248MPa, Fup= 400MPa. A992 Beam; Fyb= 345MPa, Fub= 450MPa. Fexx= 480MPa. twb= 10.92mm tp= 12, X= 45, nr= 3(no. of bolts) What is the shear yielding capacity of shear plate in kN? What is the shear rupture capacity of the shear plate in kN? What is the block shear rupture capacity of shear plate in kN? What is the weld capacity in kN?
- Two steel plate tension members have been connected using 0.72 inch diameter bolts arranged in an equally-spaced 9 by 3 rectangular formation. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 51.2 ksi, clear distance for each edge bolt as 0.12 in, and the clear distance for the other bolts as 1.072 in. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load and live load assuming live load is half as much as dead load. Consider ASD.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?For two lines of M20 bolts shown in the figure, determine the pitch (s) in mm that will give it a net area DEFG equal to the one along ABC and determine the tensile strength (kN) of the plate if it is 12 mm thick Use the pitch (s) that gives it a net area DEFG equal to the one along ABC
- Determine the tensile capacity of a 200 x 100 x 12 mm angle in S275 steel assuming that the angle is connected through the longer leg by one row of M24 bolts (Fig. 3). (3 bolts in a row and p = 70 mm)A PL 10 X 140 A36 steel plate is connected to a 10mm thick gusset plate using M20 bolts. Determine the design block shear strength (kN) of PLx140 neglecting block shearDetermine the effective net area using the U values given on Table 3.2 An MC12x45 is connected through its web with 3 gages lines of 22mm diameter bolts. The gage lines are 76mm on center and the bolts are spaced 80mm on center along the gage line. If the center row of bolt is staggered with respect to the outer row, determine the effective net cross-sectional area of the channel. Assume there are four bolts in each line. Draw the figure first and label it properly. Draw the fracture path using different color or line weights.
- 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]A plate 400x12 mm is to be connected to a plate of the same width and thickness by 34 mm diameter bolts, as shown. The holes are 2 mm larger than the bolt diameter. The plate is A36 steel. Assume allowable tensile stress on net area is .60Fy. It is requireed 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 1-2-4An angle of ASTM A36 steel (with Fy=36 ksi) is connected by two-rows of 7⁄8-in. bolts, as shown. It is exposed to a dead load of 20 k, a live load of 40 k. Design a 4 in. size (4×?) member considering the strength of the angle only.