Problem 2 The unbraced column length of Round HSS 16 x0.625 in A500 Grade B (Fy = 46 ksi) is 27 ft. Determine the design compressive strength of the column using the equations in Chapter E in Part 16 of the AISC Manual. Confirm your answer using the AISC Manual Table 4-5 (pages 4-85). The column has pin supports at both ends (i.e Kx = Ky = 1.0) columns.
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- Topic:Bolted Steel Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A channel shown is attached to a 12 mm gusset plate with 9-22 mm diameter A 325 bolts as shown. Use LRFD. Fnv = 300 MPa Fu = 400 MPa Fy = 248 MPa Ag = 3354 mm2 Question: Determine the capacity of the channel based on the bearing strength of the connection.The 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.b. The axial stress in the bolts.c. The support stress in the W12X65 profile.ll zain IQ 37 MPa no one 10/10 find the tensile stress in upper plate in section 3-3 5 1 4 80 kN 80 kN B-bolt DIA. 10 MM 80 kN 80 kN B-bolt DIA. 10 MM 10 MPa 12.5 MPa 15 MPa 8.5 MPa no one 10/ find the bearing stress between bolts and plate
- 4. The tension member shown below is a C 310 x 30.8 of A572M gr 345 steel materials. Will it safely support a service dead load of 265 kN and a service live load of 555 kN. L 50 102.5 102.5 50 65 65 65 O O 22 veny diameter bolts C310 x 30.8 Iso A36M stool matorial and select a double anglo tonsion member to rosistThe given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. Use a501 plate and a502 gr2 rivet a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivetsFor the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WI
- Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|YA RA * 21 0.5 kip 6⁰ (J Re 4' 0.75 kips F 2 Kips 4 5' Steel 10" Calculate the strass in each member and identify what the factor of safety to failure is?Topic:Bolted Steel Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A 12.5 mm x150 mm plate is connected to a gusset plate having a thickness of 9.5 mm.Diameter of bolt = 20 mm Both the tension member and the gusset plate are of A 36 steel. Fy = 248 MPa, FU = 400 MPa. Shear stress of bolt Fnv = 300MPa. Questions: a.Determine the shear strength of the connection. b.Determine the bearing strength of the connection. c. Determine the block shear strength of the connection.
- Situation 2. Two plates each with thicknessF16mm are bolted together with6 -22mm dimater bolts forming a lap conne ction. Bolt spacing are as follows: S1 = 40mm, S2 = 80mm, Sa = 100mm. Bolt hole diameter=25 mm 50 250mm 30 30 60 75 Allowable stress: Tensile stess on gross area of the plate=0.60 Fy Tensile stress on net area of the plate=0.5Fy Shear Stress of the bolt Fv=120MPA Bearing Stress of the bot Fp=1.2 Fu Calculate the permiss ible tensile load P under the following Conditions: 4. Based on shear capacity of bolts 5. Based on bearing capacity of bolts 6. Based on block shear strength (taking into consideration the failure path given in the figure below) 40 80 16 mm 40 180 mm 40 Bearing Failure Path #1 140 209 Bearing Failure Path #2= 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 kNmThe built-up section shown below is fastened together by passing two 10 mm diameter rivets through the top and bottom plates into the flanges of the beam. Each rivet will withstand 11.8 kN in shear. Determine the required spacing of the rivets along the length of the beam if it carries a shearing force of 199 kN. 12 mm x 180 mm plates (2) I IPE I 400x180x642.2 steel 10 mm rivets