Determine the maximum actored load that can be pplied to the Tension nember shown. The angle is STM A36. The diameter of he bolts is 1-in. - ENGINEERING. TEEL DESIGN W12 x 16 in 1,-0.220 -3.99 in 一十十十問 3 in 3 in L
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- Question 3 Determine the capacity of plate as well as below connection based on Balts (4) in the below Following reactions :- 7 11 A325, $4 þ Bolts use! Use: AISC manual U Moment = 10 kip-67 2 kip (2) Shear (3) Axial 2 kip رب Ź Plate (A 36) 10 " x 10" x ≤ 11 8 HSS. их их nxuxt 4 H² 72 hEe c W@ B & - For the beam shown find the value of Force (F) Ib and maximum stress due to flexure ( oy ) psi if the shear force in bolts is 300Ib and the shear stress in beam 120psi. Spacing between bolts (S) =5in F T —————————————+% 4ft 8ft 1in 1in -~ - 10in 1n § 8in= 410 N/mm? 8.20 A column section ISHB 300 @ 618 N/m (f, mand f, = 250 N/mm") is to be spliced. The factored design loads are: %3D Axial load over the column 450 kN Shear force 100 kN Bending moment (a) Design the splice plates and bolted connections using bolts of grade 4.6. (b) Design HSFG bolts and revise the splice plate size if possible. 30 kNm
- (2) The connection shown below is bolted with 3/4 inch diameter, Group A slip-critical bolts in single shear. Assume that the bearing strength is adequate and use an elastic analysis to determine the maximum factored load that can be applied. 5.5" Column Flange o o O o o Gusset Plate 4sp. @3" O O 4" 30⁰ P₁ = ?Determine the LRFD design tensile strength of the bolted connection shown. The angles are connected through the longer legs to a % x 11 in plate. Holes are made for % in bolts. Use steel F,-50 ksi F-65 ksi. L5x3x% (4, = 4.92in, x%30.947in, y=1.69in). Consider block shear in your calculations. %3D 3/4" plate 2L5 x 3 % x 5/8 3" Pr 3" 3 @ 3" 12" - 3" P,A double-channel shape, 2C8 x 18.75 is used as a tension member. The channels are bolted to a ¾-inch gusset plate with % - 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. →2" popoy |-|-|-|-|-| n оо- 0 0 0 0 x + = ³/8" 288×18.15
- Determine the controlling (lowest) net area of the member. X = 2.69 Y = (6-2.69)/2 Z=0.7 Y" -X Y" 5 @ 1/2" OD O a Do -OHD -(5)-z" bolts -L6 X 4 X 7/16The following image shows a front and side view of the column connection. The bolts have a diameter of ¾”. The W12X65 profile has a thickness of 0.605”. Determine: a. The shear stress in the bolts. Answ. 11.81 ksi b. The axial stress in the bolts. Answ. 23.62 ksi c. The support stress in the W12X65 profile. 11.50 ksiDetermine the LRFD design strength and the ASD allowable strength. Neglect block shear. A36 steel and 7/8-in Ø bolts N VA N SZ 2-MC 12 x 40 00 K All 2 in 2 in 3-in 3/in 2/in
- Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts t=12 mm t=12 mm P. P- P 300 75 75 75 Edge distances -75 mm dhole for tensile and rupture d, +3 mm dnole for bearing strength for L- de +1.5 mm Fy 248 MPa F.- 400 MPaQuestion 12. Determine the net area (An) and effective net area (A) for the channel tension member shown below. Assume bolt diameter is 5/8 in. The section is C6x10.5 (A36) (- 1½" lapttp 4@2" le ole ole ole o o ob O Ос2-PL10x1/2 A-36 Bolts are 7/8 in A-325, threads are not excluded. All edge spacings are minimum | 3" | 3" | о PL 10x3/4 A-572Gr50 a) Determine the nominal bolt capacity. b) Determine the nominal bearing capacity for each plate. c) Determine the nominal rupture capacity of each plate. d) What is the nominal capacity of the connection? e) Using LRFD, is this connection capable of supporting a dead load of 150 kip and a live load of 100 kip?