Question

Figure 1(A) through to (F) shows a pressure vessel. During a test program to determine its capacity, it is subjected to a combined internal pressure, P and torsion, T. The vessel has an external diameter, D = 1 m, length 3 m, and the wall thickness is h mm. Considering the wall is thin, show by applying equilibrium conditions, that, Oy PD 0x = 4h PD and oy = T Тух Under these conditions, you are required to: - a Txy b T Ox с Evaluate the stress in the direction of cylinder axis due to P, O. (hint: use Figure 1(C), Evaluate the stress in the circumferential direction of cylinder due to P, Oy (hint: use Figure 1(D)) Evaluate the shear stress in the x-y plane of cylinder due to T, Txy (hint: use Figure 1(E)) Construct the Mohr's circle for the stress system of element of the cylinder and indicate the locations of 01, 02, and Tmax. Predict from the Mohr's circle, the value of principle stresses, and o2, and maximum shear stress, Tmax. Predict from the Mohr's circle, the locations of principle planes on the cylinder, 01 and 02 in degrees. (E) (F) d Figure 1 Pressure vessel f P (A) (B) P P (C) (D)

P = 0.44 MPa

T = 108 kN•m

h = 2.4mm

Common Data: [Young’s modulus, E=210 GN/m²]  [Poisson’s ratio, ν=0.33] [Yield stress σ₀ = 210 MN/m² ]

Answer 1

Thin Cylindrical Pressure vessel. P = 0.44 Mpa (Internal Pressure). Diameter D. im length L = 3m. thickness h = 2.4 mm Torque T= 108 kN-m of Cylinder axis stress in the direction due to P PD x = uh 3 0.44 x 10 4x 2.4 45.84 MPa 6 stress in circumferential direction of Cylinder due to P, Jy = 0.44 x103 PD 2h 2x02.4 oy 91.67 MPa N POCO X2

C Shear stress in xy plane due to Torque T, T- x R Try J= Polar moment inertia. JE 2X R3t Rm 2 - t 3 1000 2 2.4 JF 2Rx 497.6 x 2.4 497.6 mm J = 1-86 6x109 mm". T J XR Txy 108 x 106 x 500 1.86 x 10 9 Txy - 29.06 MPa A ry B С Try = 29,06 MPa x = 45.84 MPa A 91.67 MPa OCO IOT ON POCO X2

d) mohr's circle - Coordinates - (Ox, Txy) & (ty ,- txy) (45.84, 29.06) & (91.67, -24.06) Centre (ox at by 1 2 (45.84 45.84+9167 ,0) 2 (68.75510) (45.84, 29.06) H (AB, CDZF Tman. 202 o A B 20, T2 (BC, AD). G.(91.67,8-29.06). ch (Tatay) 68.755 from diagram Radius of of mohrs Circle max shear stress CH = CF = CG - AC Principal stress o OT - OE yu ON POCO X2 T2 = OA

BC = CD = он - Ту from DBCOF CF² BC ² BF2 2 (F² = (man) - + Try 2 CF Imax (1.61-45A 191-67-45.841 +29.062 Zmaxt 37 MPa now, max Principal stress J = OE = ОС + СЕ OC = Ox toy ontoy 2 & 68.755 CE = Tmax rading of mobers circle. 37 0= OC + CE 68.755 +37 105.755 MPa min Principal stress - OA = oc- AC 68.755 - 37 O2 = UN POCO X2 31.755 MPa 2 =

(6 location of Principle planes tau 20 = GD Txy tanzo, 2x 29.06 45.84-91-67 20,3 ton (1-963 20 =51.74 0,= 25.87 location of max Principle shes Plane from Plane Bc/AD is 25.81 - location of min Prinipteal plane 202 = 20.+ 180 202 = 51.74 +180 02 - 115-87 from Plaue Bc/AD.,
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