Determine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. <=0.7 = 3.8 Z=1/2 1½" x" bolts- 1½" Gusset plate -L3 X 3 XZ -Tu
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- A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. Ae 1 2 Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN AL Section 2L5 x 3 x 516 LLBBDetermine the maximum factored LRFD tensile force capacity in tension only (no block shear). The angle is ASTM A36 steel. X = 0.7 Y = 3.8 Z=1/2 Round your answer to 2 decimal places. Your Answer: Incorrect The answer is 77.15 ± 1%. 1/2" X" bolts 1½" Gusset plate -L3 X 3 X Z Tu IDetermine the design tensile strength of the 12 in. x 1/2 in. steel plate shown in the figure. The bolts are 3/4 in. diameter. The steel is A572 Gr. 50. Check yielding and fracture. Check Block Shear. T 3in. 73im 13in 1 3in t
- The 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 rivetsAn A36 tension member shown in the figure is connected with three 19 mm diameter bolts. Dia. Of bolt hole = 22 mm. %3D 50, 100 100 99.45 50.55 100 Properties of 150 mm x 100 mm x 12.5 mm angle A - 3065 mm- y - 50.55 mm Fy - 248 MPa Fu 400 MPa Determine the design strength due to yielding in the gross section. (kN)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. Section 215 x 3 x %6 LLBB Determine the design tensile strength for LRFD. Select one: а. 66 b. 156 c. 78 d. 132
- sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the maxTopic: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.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|Y
- Question 4 Complete the table below for the steel sections shown. All sections are bending about their horizontal axis. Section Location hep Ney Slenderness (show calculations) of element bf Flange te ld ++ tw be = 800 mm Web tf = 20 mm d = 1440 mm tw = 16 mm Grade 300, heavily 入。= Whole Asp= Asy= welded plates section br Flange tr d Web bf = 800 mm te = 20 mm d = 1420 mm tw = 16 mm Grade 300, heavily welded plates 入。= Whole Asp= Asy= sectionSituation 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 #2An A36 tension member shown in the figure is connected with three 19 mm diameter bolts. Dia. Of bolt hole = 22 mm. %3D 50, 100 100 99.45 T- 50.55 100 Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm? y = 50.55 mm Fy = 248 MPa Fu = 400 MPa Determine the design strength due to fracture in the net section. UseU 0.85. (kN)