Determine the cross-sectional area of the bronze rod (in mm )? 2m 3m 60 kN nsmm I5 mm ED Mm branzerod
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- Mex Force 15 KN APPlied on nail what is The Max V If spacing between Nail is 75cm 2 15cm 5cm 15cm en 10cm 5cm 9cmQ1: Define all the symbols of the scheme below. h b Ac A₂ 000 Section Neutral axis Ecu3 Strainsted 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 max
- •Calculate all two force members of the plane •Support Reaction at E -Include direction (upward, downward, N, E, W, S, or if Tensile or Compressive -Include 4 decimal points -Provide a readable solutionPermissible As in sechion to ensure tensile Compute the design strength of beam shown it 420 MPa , fc = 30 MPa . check the maximum %3D Permissible As in sechion to ensure tensile failure. 4¢25 stimpp10 q00 8432 300 mmDetermine the change in temperature required to close the 2 mm gap between the sections shown. For brass: E = 100 GPa, a =21 x 10 /°C. For steel: E = 200 GPa, a = 12 x 10 /°C ... Brass Steel 600 mm 500 mm 2 mm 108°C B) 85°C © 20°C D 55°C
- 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= sectionTopic: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 14: A PL 300 x 20 mm is to be connected to two plates of the same material with half the thickness by 25 mm Ø rivets as shown in the figure. The rivet holes have a diameter 2 mm larger than the rivet diameter. The plate is A36 steel with Fy= 250 MPa, allowable tensile stress of 0.60Fy and allowable bearing stress of 1.35Fy. The rivets are A502, Grade 2, hotdriven rivets with allowable shear stress of 150 MPa. 25 mm Ø rivets + + PL 300x20 P/2 P P/2 PL 300x10 43. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable shear stress in the rivets? a. 675 KN b. 490 KN c. 598 KN d. 790.5 kN 44. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable bearing stress between the plates and the rivets? a. 490 KN b. 675 KN c. 598 KN d. 790.5 kN 45. Which of the following most nearly gives the max. load in kN that can be applied to the…
- What is the approximate value of the tensile strength governed by yielding of the cross- section of a roof truss diagonal 100 x 75 x 6 mm (fy = 250 MPa) connected to the gusset plate by 4 mm welds as shown in figure? (Take partial safety factor as 1.10) 1 H 100 4 mm weld, 140 mm long 75 A₂ = 1010 mm² Weld 310 mm longCalculate the ultimate moment of the given triangular column cross-section section Mr? Paste your PEG or PNG Proof paper using menu above Na=150 kN 90 mm A = 3024 A2 =024 As2 600mm fot = 17 MPa fya = 365 MPa As1 1 60 mm 600 тmThe diagonal at the left to the connection is a double angle 90 mm x 90 mm x 8 mm, with area of 2700 mm?, bolted to the 8 mm thick gusset plate. Bolt diameter = 16 mm Bolt hole diameter = 18 mm Bolt bearing capacity, Fp = 480 MPa Bolt shear strength, Fv = 68 MPa Steel plate strength and stresses are as follows: Yield strength, Fy 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 0.60 Fy Allowable tensile stress on the net area = %3D 0.50 Fu Allowable shear stress on the net area = 0.30 Fu Bolt bearing capacity, Fp = 1.2 Fy Calculate the allowable tensile load, P(kN) under the following conditions: %3D