Question

A pressurized pipe with an outside diameter of 11.35 in and a wall thickness of 0.375 in. is subjected to an axial force of P - 25700 lb and a torque of T = 10000 lb-ft, as shown. If the internal pressure in the pipe is 290 psi, determine the principal stresses (op 1>op 2 ), the maximum in-plane shear stresst max, and the absolute maximum shear stress tabs max on the outside surface of the pipe. Answers: Op 1 - 4684204 psi. op 2 - psi Tmax psi Tabs max psi

Answer 1

Solution : 11.35 in 288.29 mm Pipe diameter di 0: 375 in = 9.525 mm a waul thickness t = P = 25700 16 114.32 KN axial force P= a torque 10000 lb-ft = 1.3558 km 290 psi internal pressure in pipe ? P = 1999.48 kPa T P 88 D. ot 뷰

A = ū (di-dž) = ( 288.292-269.24%) A=8341.67 mm? उ T 32 ( 288.29 294 269.20 69.249) 6 = 162.246 X10 mn ming X103 Cascial = Р A 114.32 8341.67 0.0137 MPa = 13.1 MPa TC 288.29 (1.3558x106) 키 2 162.246 X 106 T = 1.204 1.204 MPa Shear stress value is tve q 1:204 MPa

stresses acting on the shaft due to internal press wre P. = ed ut (1999.48)(269.24) 4 L 9.525) = 14129.65 kPa a=14.13 MPa pa hoop (1999.48)(288.29) 2(9.525) 20 = 30253.549 kPa = 30.253 MPa hoop is a tensile stress or=o-o arial 14813 - 13.7 는 0.43 MPa by= "hoop 30.25.0 MPa 1.204 MPa = P = Tay Principal stresses at stresses at element (Pipe) satry Geory) ². 2 Spiropa Tay 2 y

0.43-30,25 SPP2= 0.43 + 30,25 sy +61:204) 2 2 oploop2 = 15.34 $14.958 6p, = 30.298 MPa op2 -0.382 MPa Therefore, principal stresses are 30.29 MPa and 0.382 MPa Maximiam plane shear stress 2 2 T mar = 2) Pay 14.958 MPa z Mipa Therefore, Maximum plane shear stress is 14.95 MPa 2 Tag tan 20p 2 og 21-1.204) 20p = tan 0.43 -30.25 Op = 0.040 rad Op= 2.319 Therefore, 1319 principle angle is ?.

op3 = 82=0 que abs mare = omase-6min 2 30,298-0 2 15.149 MPa Therefore, The absolutte maximum . shear stress is 15.149 MPa.