Question

This system consists of two of the same same flange couplers that are full penetration welded to the drive and driven shafts on the weld faces shown in Fig 1. The two flange couplers are bolted with 2" diameter SAE grade 8 bolts. The coupler design is required to transmit a torque of 500,000 lbf-ft and is to be made of a High Strength Low Alloy (HSLA) steel. The application is for a drill-bore head drive shaft in an off-shore oil well. The HSLA steel grade chosen for this application due to its corrosion resistance has a yield strength of 82,100 psi and the flange couplers require a minimum factor of safety to yield of 1.5. SAE grade 8 bolts have a proof strength of 120,000 psi and require a minimum factor of safety to proof of 2.0. 2 SAE GRADE 8 BOLT WELD FACE T=500,000 LBF-FT FLANGE COUPLER what is the maximum shear stress in the flange coupler ?

Answer 1

solution in we have gwen in the question Meld étrength) Syt = 82,160 pudi - 560 Mpa Fos = 1.5 [ Torque. 9 Td = 500, 600 167+ft = 677908.974 Nm we need to determine the shear stress on flange Where Tf = shear stress of flange Та - Чоrары I= IC (d-dy) T = Tar I= polar moment of mention 32 du= diameter of hub d = diametio of shaft du= ad do we can find the diameter of shaft wing Porque equation of shaft Ts &st Where Tg = shaft shear strees 16 Td सवर To = Syt Fos 560 1.5 To = 373. MPa be howe To = 16Td राव? 373 16 x 677908.974 to da

d 0.029 d = 0.17m dy= 2X0.17 = 0.34m dy 0.17 0.085m Clay-d 4) = (0.344-0.174) 32 32 I= 1.23x10 my Substitutung the values an first equation T = 677908.974% 0.085 1.23x10 3 T = 466 47367/m2 4 = = 46.85 MPa Answer The maximum shear stress on flange will be 46.85 MPa