For the splice connection shown in the figure, Compute the design strength based on shear and bearing. (Use LRFD) The plates are PL 12" x 7" steel ASTM A36 (Fy = 36 ksi, Fu = 58 ksi) and the bolts are 7/8-in. diameter A325 (Fnu = 68 ksi) in standard holes. P₁ Pa 0 O C O 3 in 3 in 3 in 3 in 3 in | -in bolts 3 in 6 in 3 in in P₁ P 12 in

For the splice connection shown in the figure, Compute the design strength based on shear and bearing. (Use LRFD) The plates are PL 12" x 7" steel ASTM A36 (Fy = 36 ksi, Fu = 58 ksi) and the bolts are 7/8-in. diameter A325 (Fnu = 68 ksi) in standard holes. P₁ Pa 0 O C O 3 in 3 in 3 in 3 in 3 in | -in bolts 3 in 6 in 3 in in P₁ P 12 in

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter3: Tension Members

Section: Chapter Questions

Problem 3.5.4P

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For the splice connection shown in the figure, a. Check all spacing and edge-distance requirements. (For 1-in diameter bolts minimum edge distance is 1.25 in. AISC Table J3.4) b. Compute the design strength based on shear and bearing. (Use LRFD) The plates are PL 1/2" x 7" steel ASTM A36 (Fy = 36 ksi, Fu = 58 ksi) and the bolts are 1-in diameter A325X (Fnv = 60 ksi) in standard holes.
A36 steel and 3/4 in bolts are used in the bolted splice shown in figure P4.48 . Consider yield, net section rupture, and block shear rupture. a) Determine design strength by LRFD and b) the allowable strength by ASD.
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.
subject: Steel Design 1. A992 steel is used for the tension member shown. The bolts are 3/4 inch in diameter. The connection is to a 3/8 in.thick gusset plate. a. Determine the nominal strength based on the gross area. b. Determine the nominal strength based on the effective net are.
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.a) Determine the allowable tensile strength for ASD.b) Determine the design tensile strength for LRFD.
410UB 59.7 (Grade 300 Plus) Primary steel beam as shown in Figure below is connected to a column using a simple cleat plate connection. Primary beam needs to transfer an ultimate shear force, V* = 200 kN through the connection. Cleat plate connection details are shown in the Figure.Cleat plate - Grade 250 fy(cleat plate)=280 MPa, fup(cleat plate)=410 MPaBolts - four 8.8/S M20 bolts, threads are excluded from the shear plane,All bolt holes diameter is 22 mm. 8 mm thick 280X90 mm cleat plate is welded to column using 6 mm fillet welds(E48XX) both sides (strength of weld = 0.978 kN/m) ∅=0.9 a) Design shear capacity of beam web against tearing is b)Calculate and verify the design shear capacity of cleat plate against tearing and bearing failure. c) Check cleat plate against bearing failure. d) Calculate stresses in the weld
A992 steel is used for the tension member shown. The bolts are ¾ inch in diameter. The connection is to a ⅜ inch thick gusset plate. A. Determine the nominal strength based on the gross area. B. Determine the nominal strength based on the effective net area.
Use load and resistance factor design and select a W shape with a nominal depth of 10 inches (a W 10) to resist a dead load of 175 kips and a live load of 175 kips. The connection will be through the flanges with two lines of 11 4 -inch-diameter bolts in each flange, as shown in Figure P3.6-6. Each line contains more than two bolts. The length of the member is 30 feet. Use A588 steal.
Use ASD to determine the width of the 5/8 inch thick A36 plate subjected to tension as shown below. The dead load effect is PD=85 kip while the live load effect is PL=80 kip. Two rows of 5/8 inch standard bolts will be used for the connection and the plate width can be manufactured in 1 inch increments. Assume that the larger connected plate does not fail. Consider the limit states of gross tensile yield and tensile rupture. Find the required tension strength, Pa, for the controlling ASD load combination. Determine the width necessary for the plate to resist gross tensile yield. Determine the width necessary for the plate to resist tensile rupture. Report the required plate width.
The 7/8 × 14 plate is shown in the figure. The holes are for 7/8-in Ø bolts. Compute the net area of each of the given members. (Ans. 10.54 in^2)
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.