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22.314/1.56/2.084/13.14 Fall 2006 Problem Set II Solution 1. Stress intensity = max{|σr − σθ |, |σθ − σz |, σz − σr |} For thin wall approximation: Pi + Po 2 Pi R2 − Po (R + t)2 σz = − (R + t)2 − R2 Pi − P o t σθ = (R + ) t 2 σr = − (1) (2) (3) Therefore: Sthin = σθ − σr = Pi − P o t Pi + Po (R + ) + t 2 2 (4) Thick wall solution: Equilibrium in radial direction gives: dσr σr − σθ + =0 dr r (5) Hook’s law: 1 (σr − νσθ − νσz ) E 1 �θ = (σθ − νσr − νσz ) E 1 �z = (σz − νσr − νσθ ) E �r = Since �θ = u/r, �r = du , dr (6) (7) (8) we get: d�θ 1 = (�r − �θ ) dr r (9) For this close end cylinder far from the end, plane stress condition is assumed, i.e., σz is const. Plug Eq 6 and Eq 5into Eq 9, we get d (σθ + σr ) = 0 dr (10) Plug Eq 10 into Eq 5, we get d 1 d 2 (r σr ) = 0 dr r dr (11) 1 With B.C. σr (r = R) = −Pi and σr (r = R + t) = −Po , we get: R R −Po (R + t)2 + Pi R2 σr = −Pi ( )2 + (1 − ( )2 ) (R + t)2 − R2 r r R R −Po (R + t)2 + Pi R2 σθ = Pi ( )2 + (1 + ( )2 ) (R + t)2 − R2 r r R (Pi − Po )(R + t)2 σθ − σr = 2( )2 (R + t)2 − R2 r (12) (13) (14) Maximum stress intensity is at the location of inner radius: Sthick = 2 (Pi − Po )(R + t)2 (R + t)2 − R2 The error in thin wall approximation is: |1 − Sthin | Sthick Results are tabulated below: t/R Pi = 2Po Pi = 20Po 0.03 1.41% 1.31% 0.10 0.15 4.13% 5.67% 4.09% 5.88% 0.30 8.88% 10.46% 2. We use thin shell approximation to solve this problem. For a region of cylinder far from a junction, stresses are: PR t σr = −P/2 PR σx = 2t σθ = (15) (16) (17) Radial displacement: uc = P R2 νt (2 − ν + ) R 2Et (18) For the sphere, stresses are: PR 2t σr = −P/2 σθ = σφ = (19) (20) Radial displacement: us = P R2 νt (1 − ν + ) 2Et R (21) 2 At the junction, Note C Eq 30–33 give that at the edge of cylinder: u0 = u c + φ0 = − V0 M0 + 2β 3 D 2β 2 D (22) V0 M0 − 2β 2 D βD (23) at the edge of hemisphere: u0 = u s − φ0 = − 2Rλ 2λ2 V0 + M0 Et Et (24) 2λ2 4λ3 V0 + M0 Et REt (25) 2 where, λ = βs R, βs = ( 3(1−ν )1/4 , u0 is the radial displacement at the junction, φ0 is the R2 t2 1 slope at the junction. Due to the continuity of the displacement and slope, the four unknowns u0 , φ0 , M0 , and V0 1 Note that the direction of the shear force Q in Note L4 is different from that in Note C. If you assume the direction of Q0S is same as QOC , the continuity of shear force should give that: Q0S = −Q0C . Vo Mo Mo z Vo x Figure 1: Junction of hemisphere and cylinder 3 can be solved by above equations. It can be found that M0 = 0 V0 = −P R2 /(4Rλ + Et/β 3 D) V0 u = uc + 2Dβ 3 P R σx = 2t PR EV0 σθ = + t 2DRβ 3 σr = −P/2 (26) (27) (28) (29) (30) (31) With mean radius R = inner radius + t/2 = 1.155 m, t = 0.11 m, E = 200 GPa, ν = 0.3, we get (a) At the junction, from Eq 29–Eq 31: σx = 81.38 MPa σθ = 122.06 MPa σr = -7.75 MPa The maximum stress is the hoop stress: 122.06 MPa. (b) Radial displacement as a function of radial postion z is: (R + z)�θ . Thus, from Eq 18, the radial displacement of cylinder is: PR νt (2 − ν + )(R + z) 2Et R From Eq 21, the radial displacement of hemishpere is: PR νt (1 − ν + )(R + z) 2Et R From Eq 28, the radial displacement of junction is: ( PR νt V0 (2 − ν + ) + )(R + z) 2Et R 2DRβ 3 Therefore: Max. displacement (m) Cylinder Sphere 0.00089 0.00037 4 Junction 0.00063