The plate with dimension 18 mm x 400 mm (thickness by width) is to be connected to a gusset plate using 6-Ø24 mm bolts in arrange in 3 row (2 each row) along the x-axis. The plate is to be design for tension. Using A36 steel and assuming Ae= An. Use NSCP 2015. a. Compute the design strength considering yielding and tensile rupture. (LRFD) b. Compute the allowable strength considering yielding and tensile rupture. (ASD)
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- The plate with dimension 18 mm x 400 mm (thickness by width) is to be connected to a gusset plate using 6-Ø24 mm bolts in arrange in 3 row (2 each row) along the x-axis. The plate is to be design for tension. Using A36 steel and assuming A= An. Use NSCP 2015. a. Compute the design strength considering yielding and tensile rupture. (LRFD} b. Compute the allowable strength considering yielding and tensile rupture. (ASD)Determine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40Problem 2 A tension plate shown below is used to support suspended load "T. Gusset Plate F, = 248 MPa Fu = 400 MPa 200 mm a) Determine the allowable tensile capacity of the plate if L= 240 mm, (Assume weld strength is satisfactory).-
- A L 3 x 2 x ¼ is connected to a gusset plate via six bolts. The nominal diameter of the bolt is 0.25 inches, the pitch spacing is 1.75 inches, the gage spacing is 2 inches, and the thickness of the connection is (¼) inch. The yield stress is 50 Ksi and the ultimate stress is 60 Ksi. Consider section line (a-a') for the analysis. . a. What is the effective net area (Ae) of the angle section in inches2 [a-a']?(The shear lag factor is "0.80") b. What is the design tensile yielding strength in Kips for the steel member? c. What is the design tensile rupture strength in Kips for the steel member? d. What is the minimum Factor of Safety if a tensile load of 23 Kips is applied to the angle section? please make sure the answer is correct 100% I only need the final answersA plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. Assume allowable tensile stress on net area is 0.60Fy. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue of b in millimeters. b. Calculate the value of the net area for tension in plates in square millimeters. c. Calculate the value of P so that the allowable tensile stress on net area will not be exceeded.4.) A 150 x 90 x 12 angular section is welded to a gusset plate as shown in the figure. The angle is A36 steel with FY=248 MPa. The weld is E 80 electrode with Fu = 550 MPa. The allowable tensile stress for the angle is 0.6Fy and the allowable shear stress for the weld is 0.3Fu. The Area of the angular section is 2751 sq.mm. with y=51mm. CS 56 Which of the following most nearly gives the design force P? a.) 400,365.9 N b.) 409,348.8 N c.) 412,793.5 N d.) 420,366.6 N 5.) A W350 x 90 steel is used as a simply supported beam 8m long. The beam carries three equal concentrated loads at every quarter points. It also carries a uniform dead load o 5 kn/m (including its own weight) and a uniform live load of 7.20 kn/m. Properties of W 350 x 90 steel: bf = 250 mm The allowable bending stress is 0.66Fy. The allowable shear stress is 0.40Fy. The allowable deflection is L/360. Use Fx=248 Mpa and E=200 Gpa. tf = 16.40 mm V d = 350 mm Determine the maximum value of each concentrated load based on…
- A structural tee bracket is attached to a column flange with six bolts as shown in Figure . All structural steel is A992. Check this connection for compliance with the AISC Specification. Assume that the bearing strength is controlled by the bearing deformation strength of 2.4dtFu. a. Use LRFD. b. Use ASD.Topic:Bolted Steel Connection - Civil Engineering *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem The bracket shown in the figure is supported by four 22 mm diameter bolts in single shear. The bracket is subject to an eccentric load of 150 kN. Use LRFD. Use A36 steel. Questions a) Determine the critical force on the most stressed bolt. b) Determine the nominal shear stress of the most stressed bolt.4 A flanged bolt coupling has ten 12-mm di ameter steel bolts on 500 mm diameter b olt circle and six 16 mm diameter aluminu m bolts on 300 mm diameter bolt circle. T he maximum shear stresses of the materi als are 60 MPa in steel and 40 MPa in al uminum. Use G = 80 GPa for steel and 3 0 GPa for aluminum. What is the maximu m required shear strength of each alumin um bolt to determine the maximum torqu e that can be applied to the system? Dra wing not included in this problem.
- Considering the following steel connection. The plates in Pink are 9mm steel plates. The middle plate (Yellow) is 18mm thick. The width of the plate is 100mm. The maximum allowable tension stresses on any of the plates is 100Mpa in Gross Area Yielding and 150 Mpa for Net Area or Tension Rupture. The bolts used are 8mm in diameter, the holes are 10mm in diameter, no need to add 1.6mm. The bolts allow a maximum of 280 Mpa of shear. Determine the maximum allowable "P" of the connection in kN.Tension Member Design Problem 1. A channel shape is under 50 kips dead and 100 kips live tensile axial load as shown in the figure. The member is connected to a gusset plate with 10 inch longitudinal welds. Find the lightest channel shape to carry the loading. Use only vielding and rupture limit states to design. Use 50 ksi steel (Fy=50 ksi, Fu=65 ksi). (a) Assume yielding limit state controls in the design process; (b) After selecting the lightest section, check the rupture limit state. Do not redesing if needed. 1 Pa=50 kips PL=100 kips 10"1. If the allowable stress of a revolving steel is 150 MPa, what is the equivalent angular velocity which will produce such stress. The mean diameter is 440mm. 2. See connection figure below. The rivets is 7/8 in. in diameter. According to the rivert supplier, the allowable stresses for the material are t = 15 ksi and Ob = 32 ksi. What is the allowable load of the connection? L4 x 3-1/2 W18 x 86 W24 x 117 Girder Beam