What is the minimum bolt diameter required in mm.
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- e. Calculate "P" in KN considering failure on bearing on block shear Two A36 16mm Thick Steel Plates Are Connected By Four Rivets With Fv=152 MPa As Shown. Given: Rivet Dia 20 mm x=100 mmWhat is the minimum required bolt diameter (mm) at C ifthe allowable bolt shear stress is 68 Mpa.a. 9 mm c. 11 mmb. 10 mm d. 12 mm Determine the stress (Mpa) in strut BC.a. 5.6 MPa c. 3.2 MPab. 2.8 MPa d. 6.4 MPa Calculate the bolt stress (Mpa) at A if its diameter is 20mm.a. 7.64 MPa c.11.46 MPab. 3.82 MPa d.14.92 MPa4. The plate shown in the figure is fastened to thefixed member by seven 10-mm-diameter rivets.Compute the value of the loads P so that theaverage shearing stress in any rivet does notexceed 70 MPa.
- A steel bracket of solid circular crosssection is subjected to two loads, each of which isP = 4.5 kN at D (see figure). Let the dimension variablebe b = 240 mm.(a) Find the minimum permissible diameter dminof the bracket if the allowable normal stress is110 MPa.(b) Repeat part (a), including the weight ofthe bracket. The weight density of steel is77.0 kN/m3.The angle bracket shown, 75 mm x 75 mm x 15 mm thick, is 200 mm long. It is connected to the 16 mm thick flange of the column support with 2 of 16 mm diameter bolts. The bracket is subjected to a load of 42 kN from beam A. Assume that the load is applied concentrically at the connection. 1. Determine the resulting bolt shear stress in MPa 2. Determine the resulting bolt shear stress in MPa 3. Determine the critical bolt shearing stress in MPaIf the allowable bearing stress in the bolt is 18.855 ksi. what is the minimumrequired diameter of the bolt at B? b= .75” h= 7.874” Distances are in feet.
- Given the following allowable stresses:• 50 MPa for shear in rivets• 100 MPa for bearing between a plate and a rivet• 80 MPa for tension in the platesNOTE: Draw FBD(a) If we increase the diameter of each bar to 20 mm, determine the maximum load Pthat can be applied to the lap joint.(b) Using the value computed for maximum load P from (a) compute for the over-allfactor of safety given that the actual P= 20kN.b. Calculate "P" in KN considering tension failure on net area Two A36 16mm Thick Steel Plates Are Connected By Four Rivets With Fv=152 MPa As Shown. Given: Rivet Dia 20 mm x=100 mmDetermine the smallest diameter (mm) that can be used in the clevis shown in the figure below if P = 451 kN. The shearing stress of the bolt is 320 MPa. Final answer should have 2 decimal places.
- pls help! Write the complete solutions and legibly. Answer in 2 decimal places. UPVOTE WILL BE GIVEN! MECHANICS OF DEFORMABLE BODIES Let Q = 0 and T = 8 (Example: 12Q°C = 120°C). Just substitute the number in letter. Thank you! For the frame shown, a 4QQT-N load is acting on member ABD at D. If the allowable material shear stress for is 4Q MPa, determine the required diameter (rounded off to the nearest 2.5 mm) of the pins at C and D. Pin C and pin D are subjected to double shear and single shear, respectively. If the thickness of member BC is 12 mm and that of member DE is 16 mm, determine the maximum bearing stress at C.Prob 01: An 18” depth beam is bolted to a 24” depth girder is connected similar to that in figure shown below. Thediameter of the rivet is 20mm and the angles are each 100mm x 100mm x 12mm thick. For each bolt, assume thatthe allowable bearing stress is 220 MPa. Determine the allowable load P in KN on the connection. *show all theload P in kN for bearing stressBeam web thickness = 5mmGirder web thickness =1. Two A36 16mm Thick Steel Plates Are Connected By Four Rivets With Fv=152 MPa As Shown. Given: Rivet DIa 20 mm x=100 mm a. Calculate "P" in KN considering tension failure on gross area b. Calculate "P" in KN considering tension failure on net area c. Calculate "P" in KN considering failure on bearing on projected area. d. Calculate "P" in KN considering failure on shear on rivets e. Calculate "P" in KN considering failure on bearing on block shear f. Calculate the maximum safe load "P" in KN