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- Check the adequacy of an L4x4x1/2 . ASTM A36 , with one line of ( 4 ) 3/4 in . diameter bolts in standard holes . The dead load and a live load is 500 KN . Calculate at what length this tension member would cease to satisfy the recommended slenderness limit . Assume Ae = 0.8 AgSolve: Select the lightest 28 ft long W12 available to support a factored tensile load of Pu = 400 k. Assume there are 2 lines, (lines are 5” apart) of 7/8" bolts in each flange (4 lines total) and 3 bolts per line @ 4” o.c. and 4” to end. Assume A992 steel. Dimensions are from centers of bolt holes.Use load and resistance factor design and select the lightest American Standard Channel shape to resist a factored tensile load of 180 kips. The length is 15 ft, and there will be two lines of 7/8-in diameter bolts in the web. Use a shear lag factor of 0.85. Use A36 steel (Fy = 36 ksi, Fu = 58 ksi). Using the selected member, analyze the section again by computing for its ultimate strength for LRFD and comparing it to the factored load experienced by the member.
- A single angle tension member, L 4 x 4 x 3/8 in. made from A36 steel is connected to a gusset plate with 5/8 in. diameter bolts, as shown in Figure below. The service loads are 35 kips dead load and 15 kips live load. Determine the adequacy of this member using AISC specification. Assume that the effective net area is 85% of the computed net area.A C8x11.5 single channel section us bolted to a gusset plate with M22 bolts. Assume A36 steel, 5.6 mm thick web for the channel and an 8mm gusset plate. U=0.80 The channel has a gross area of 2181 mm^2. The bolts have a shear stress of 457 Mpa. What is the design tensile strength due to gross yielding (kN)?The plate shown is attached by using three M12x2 grade 8.8 bolts. Which screw is subjected to the largest stress? What safe load ( static) F can be supported by the screw for n=1.5.
- A double angle tension member L5 x 3 x 1/4, is connected to a gusset plate with 7/8 inch diameter bolts as shown in the figure. The steel used has Fy = 248.8 MPa and Fu = 400 MPa. The service loads are 35 kips for dead load and 15 kips for live load. Gross area of L5 x 3 x 1/4. Determine the effective net area, in sq.in.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 "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 combination
- A double-channel shape, 2C8 x 18.75 is used as a tension member. The channels are bolted to a ⅜ - inch gusset plate with ⅞ - inch diameter bolts. The tension member is A572 grade 50 steel and the gusset plate is A36. If LRFD is used, how much factored tensile load can be applied? Consider all limit states.Select the lightest 28 ft long W12 available to support a factored tensile load of Pu = 400 k. Assume there are 2 lines, (lines are 5” apart) of 7/8" bolts in each flange (4 lines total) and 3 bolts per line @ 4” o.c. and 4” to end. Assume A992 steel. Dimensions are from centers of bolt holes.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?