A 16 mm-diameter steel rod AB is fitted to a round hole near end Cof the wooden member CD. For the loading shown, determine, (a) the distance b for which the average shearing stress is 7MPA on the surfaces indicated by the dashed lines, (provide FBD and answer in mm)(b) the average bearing stress on the wood (provide FBD and answer in MPa) 25 тm. 8000N 4000N 100 mm D A 4000N
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- 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.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)A16mm–diameter steel rod AB is fitted to a round hole near end Cof the wooden member CD. For the loading shown, determine, (a) the distance bfor which the average shearing stress is 7Mpa a on the surfaces indicated by the dashed lines, (provide FBD and answer in mm)(b)the average bearing stress on the wood (provide FBD and answer in Mpa)
- 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)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 answersll 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
- For the clevis connection shown, the shear stress in the 0.287-in.-diameter bolt must be limited to 29 ksi. Determine the maximum load P (in kips rounded to the nearest hundredths) that may be applied to the connection.A bolted connection shown consists of two plates 300mm x12mm connected by 4 - 22 mm diameter bolts. Edge distances = 75mm dhole for tensile and rupture = db + 3 mm dhole for bearing strength for Lc = db + 1.5 mm Fy = 248 Mpa Fu = 400 Mpa Fn = 330 Mpa Use LRFD design method. Determine the design strength due to the gross yielding of plates. (kN)Determine 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
- 3. A plate with width of 300mm and thickness of 20mm is to be connected to two plates of the same width with half the thickness by 24mm diameter bolts, as shown. The rivet holes have a diameter of 2mm larger than the rivet diameter. The plate is A36 steel with yield strength F,-248MPa and ultimate strength F,-400MPa. a. Determine the design strength of the section. b. Determine the allowable strength of the section 24mm 30mmThe 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 rivetsA plate 400 x 12 mm is to be connected to a plate of the same width and thickness by 34 mm diameter bolts, as shown. The holes are 2 mm larger than the bolt diameter. The plate is A36 steel. Assume allowable tensile stress on net area is .60Fy. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equal to the net width along bolts 1-2-4. a. calculate the value 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. d. Calculate the nominal block shear strength based on possible failure paths