Question

The thin-walled pipe has an Inner diameter of 12 mm and a thickness of 0.6 mm. It is subjected to an internal pressure of 2 MPa and the axial tension and torsional loadings shown in (Figure 1). Part A Determine the principal stresses at a point on the surface of the pipe. Express your answers using three significant figures separated by a comma O AED Ivec ROO? MPа Submit Figure 1 of 1 > < Return to Assignment Provide Feedback

Answer 1

Soln - Given Thinwall pipe I P = 1000N - QA to T= 30 N-m p=gMpa di = 12mm - 18+ 2x0.6 = 13.2 mm t = 0.6 mm If a pressure vassel apply AL a external force P parallel to the anis of vassel, then internal pressure changes its direction to balance external force. 2 6Lx(A.-A:) Drawing FBD of a cut. & PxAi section of pipe Se (A.-Ai) internal crossection area of pipe Ai= I di2 mm2 outer crossection area of pipe Ao = Ide mm2 16. is longitudenal Stress in the wall of the pipe wall. Now by balancing force P = bx Ai + L+(A.- Ai) IMA → 1000 = 3x (19"+ 6 x[(19-272-cons] 4 → 6L = 32.6 MPa

External applied force p and torque I will not effect the hoop Stress in the pike wall as their direction of load is perpendicular to the direction of hoop stress (6H). Thus it can be calculated simply by using formula of a hoop Stress in thinpressure Vassel which is given by 6H = $Dm 2t •Mean diameter of pipe Don=dot di = 13.2 + 12 Dm = 12.6 mm therefore 6H = 2 x 12.6 = 21 MPa 2x0.6 due to applied torque T Now by calculating Shear stress using torsion equation outer radius Polar moment so = do = 6.6 mm of area in case of thin tube J = Tom't TX (12.6) 3 x 0.6_942.65 mm" 4 - 4 30x10² thus Hin - * 6.6 = 210.05 MPa 942. 65

sses Now point the principal stresses at A on the surface of pipe. 2 61,2 = (+69) + Sirem agy? + 4 Top ] 6.2 = 1 (166+64) + 161 - 6H)² +4 07] 61,2 = 32.2 + 21) + / 132.2-2152 + 4x1210.05)2] 2 - 5 62 = 236.75 MPa (Major principal stress) = – 183.525 MPa (Minor principal stress)
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