Jtam-A4.dvi JOURNAL OF THEORETICAL AND APPLIED MECHANICS 55, 1, pp. 17-27, Warsaw 2017 DOI: 10.15632/jtam-pl.55.1.17 MECHANICAL ANALYSIS OF AUTOFRETTAGED HIGH PRESSURE APPARATUS Ruilin Zhu College of Engineering and Design, Hunan Normal University, Changsha, China e-mail: zrl200701@sina.com Guolin Zhu Basic Courses Department, Jiangxi Police College, Nanchang, China Aifeng Mao Moshanjie School, Nanchang, China High pressure apparatus is widely used in industries, the design of them depends on stress distributions in their walls.Most of high pressure apparatuses aremade in formof cylinders. To raise load-bearing capacity and extend operation life for high pressure apparatus, the autofrettage technology is often used. To design autofrettaged high pressure apparatus, it is necessary to study characteristics of stresses in the wall of thick-wall cylinders, including residual stressesandtotal stresses, etc. In this study, through investigating the characteristics of stresses of cylinders subjected to internal pressure according to the maximum distortion strain energy theory, a set of simplified equations for residual stresses and total stresses are obtained, the safe and optimum load-bearing conditions for autofrettaged cylinders are found out, which are the basis for design of autofrettaged high pressure apparatus. Keywords: thick-wall cylinder, autofrettage, total stress, residual stress, load-bearing capa- city Nomenclature ri,rj,ro – inside radius, radius of elastic-plastic juncture, outside radius, respectively k – ratio of outside to inside radius, k= ro/ri kj – depth of plastic zone or plastic depth, kj = rj/ri kj∗ – optimum kj kc – critical radius ratio, kc =2.2184574899167 . . . x – relative location, x= r/ri p,py – internal and entire yield pressure pa – autofrettage pressure pe – maximum elastic load-bearing capability of unautofrettaged cylinder or initial yield pressure; σy – yield strength σe – equivalent stress Superscripts p,t, ′ – quantity related with internal pressure, total and residual stress, respectively. Subscripts z,r,θ – axial, radial and circumferential direction, respectively. 18 R. Zhu et al. 1. Introduction Cylindersarewidelyused inmanufacturinghighandultra-pressurevessels, high-pressurepumps, battleship and tank cannon barrels as well as fuel injection systems for diesel engines, etc. The autofrettage technique is an effectivemethod to raise load-bearing capacity and extendoperation life of cylinders. Usually, in the most commonly employed autofrettage process, a cylinder is pressurized to a quite high internal hydraulic pressure, as a result, the portion of the cylinder from inner radius to some intermediate radius becomes plastic while the remaining portion remains elastic. After releasing the pressure, the residual stresses are set up in the wall of the cylinder. Studies on autofrettage about specific engineering problems have been done widely. Finite element simulations and experiments, the interaction betweenmanufacturing processes with re- spect to residual stresses and deformationswas studied byBrünnet andBähre (2014). Farrahi et al. (2012) investigated the residual stress distribution at the wall of a thick-walled tube affected by the re-autofrettage process. The effects of thermal autofrettage on the residual stresses in a titanium-copper brazed joint were studied by Hamilton et al. (2015). Lin et al. (2009) built the autofrettage damage mechanics model from an ultra-high pressure vessel autofrettage damage mechanism. By using continuumdamagemechanics approach, Lvov andKostromitskaya (2014) analyzed the autofrettage process and derived general set of government equations of elastic- plastic bodies byusing the effective stress concept.Afinite elementmodel of the swaging process was developed in ANSYS and systematically refined to investigate the mechanism of deforma- tion and subsequent development of residual stresses by Gibson et al. (2014). Noraziah et al. (2011a,b) set an analytical autofrettae procedure to predict the required autofrettage pressure for various levels of allowable pressure and to achieve maximum fatigue life. By using Huang’s model for modeling reverse yielding due to Bauschinger effect, Bhatnagar (2013) presented an original concept of an autofrettage compounded tube which was modeled for the autofrettage process.Byusing theKendallmodel,whichwas adopted byASMECode, Shim et al. (2010) pre- dicted the accurate residual stress of SNCM8high strength steel. Zheng andXuan (2010, 2011) analyzed the optimum autofrettage pressure of a thick wall cylinder under thermo-mechanical loadings and investigated theoretically and validated by the finite element method (FEM) the closed form solution of the limit thermal load of autofrettage and the optimum autofrettage pressure under plane strain and open-ended conditions. Zhu (2008) investigated the optimum plastic depth and load-bearing capacity of an autofrettaged cylinder in terms of the point of view of avoiding compressive yield after removing autofrettage pressure and raising load-bearing capacity as far as possible simultaneously. Zhu and Zhu (2013) studied autofrettage of cylinders by limiting circumferential residual stress and according to Mises Yield criterion. Zhu and Li (2014) presented equations of optimum overstrain (ελ) and depth of the plastic zone (kjλ) for a certain load-bearing capacity and radius ratio (k). This paper is intended to investigate the varying tendency and distribution laws of stresses in autofrettaged cylinders so as to provide the theoretic basis for the design of high pressure apparatus. Engineering conditions are in endless variety. This paper is based on ideal conditions including (1) the material of a cylinder is perfectly elastic-plastic and the Bauschinger effect is neglected, the compressive yield limit is equal to the tensile one; (2) strain hardening is ignored; (3) there is not any defect in the material. 2. General residual stresses After removing autofrettage pressure, residual stresses remain in the wall of a cylinder. Yu (1980) put forward the residual stresses at a general radius location which has been re-arranged as follows: Mechanical analysis of autofrettaged high pressure apparatus 19 — in the plastic zone σ′z σy = 1√ 3 [k2j k2 +ln x2 k2j − ( 1− k2j k2 +lnk2j ) 1 k2−1 ] σ′r σy = 1√ 3 [k2j k2 −1+ln x2 k2j − ( 1− k2j k2 +lnk2j ) 1 k2−1 ( 1− k2 x2 )] σ′θ σy = 1√ 3 [k2j k2 +1+ln x2 k2j − ( 1− k2j k2 +lnk2j ) 1 k2−1 ( 1+ k2 x2 )] (2.1) Accordiong to theMises criterion, the equivalent residual stress is σ′e σy = √ 3 2 (σ′θ σy − σ′r σy ) =1− k2−k2j +k2 lnk2j (k2−1)x2 (2.2) — in the elastic zone σ′z σy = 1√ 3 [k2j k2 − ( 1− k2j k2 +lnk2j ) 1 k2−1 ] σ′r σy = 1√ 3 ( 1− k 2 x2 )[k2j k2 − ( 1− k2j k2 +lnk2j ) 1 k2−1 ] = ( 1− k 2 x2 )σ′z σy σ′θ σy = 1√ 3 ( 1+ k2 x2 )[k2j k2 − ( 1− k2j k2 +lnk2j ) 1 k2−1 ] = ( 1+ k2 x2 )σ′z σy (2.3) The equivalent residual stress at a general radius location based on theMises criterion is σ′e σy = √ 3 2 (σ′θ σy − σ′r σy ) = k2(k2j −1− lnk2j) (k2−1)x2 (2.4) Since σ′e = σ ′ θ − σ′r based on the maximum shear stress theory (Tresca criterion) and σ′e =( √ 3/2)(σ′θ −σ′r) based on theMises criterion, while the components of the residual stress based onMises criterion are 2/ √ 3 times those based onTresca criterion, the equivalent residual stresses based on both criterions must be the same. At the inner surface, x=1. For Eq. (2.2), letting x=1 and σ′e =−σy obtains an equation for kj∗, themaximumand optimumplastic depth (kj) for a certain k to avoid compressive yield after removing pa k2 lnk2j∗ −k2−k2j∗ +2=0 kj∗ ­ √ e (2.5) where √ e¬ kj∗ ¬ kc and k­ kc. When k¬ kc, |σ′e/σy|< 1, irrespective of kj, Eq. (2.5) is just the equation proposed by Zhu (2008) in another method. 3. Residual stresses and total stresses when p= pa The total stresses σt include the residual stresses plus the stresses caused by p, or σt =σ′+σp. To produce plastic depth kj, the pressure subjected to a cylinder is the autofrettage pressu- re pa p σy = 2√ 3 lnkj + k2−k2j√ 3k2 = pa σy (3.1) 20 R. Zhu et al. Letting k=∞ in Eq. (3.1), one obtains p∞ σy = 2√ 3 lnkj + 1√ 3 (3.2) Inappropriate kj causes compressive yield or reduces load-bearing capacity. To avoid com- pressive yield, theplastic depthkj for a certain kmustbe smaller than or equal to themagnitude calculated byEq. (2.5).Then, to raise load-bearing capacity fully, combiningEqs. (2.5) and (3.1), one obtains p σy = 2(k2−1)√ 3k2 = 2pe σy (3.3) Equation (3.3) is the optimum load-bearing capacity of an autofrettaged cylinder, it is just two times the initial yield load. The limit of Eq. (3.3) with k→∞ is p/σy =2/ √ 3, which can be obtained by letting kj = √ e in Eq. (3.2) as well. The stresses caused by p at a general radius location are σpz σy = 1 k2−1 p σy σpr σy = ( 1− k2 x2 )σpz σy σ p θ σy = ( 1+ k2 x2 )σpz σy (3.4) The equivalent stress of Eq. (3.4) based on theMises criterion is σpe σy = √ 3 2 (σ p θ σy − σpr σy ) = √ 3k2 k2−1 p σy 1 x2 (3.5) When p= pa, Eq. (3.5) becomes σpe σy = 2k2 k2−1 p σy 1 x2 = k2−k2j +k2 lnk2j (k2−1)x2 (3.6) The equivalent total stress σte is σte = √ 3 2 (σtθ−σtr)= √ 3 2 [(σ′θ+σ p θ )−(σ′r+σpr)] = √ 3 2 [(σ′θ−σ′r)+(σ p θ −σpr)] =σ′e+σpe (3.7) Then, generally, in the plastic zone σte σy =1− k2−k2j +k2 lnk2j (k2−1)x2 + √ 3k2 k2−1 p σy 1 x2 (3.8) in the elastic zone σte σy = k2(k2j −1− lnk2j) (k2−1)x2 + √ 3k2 k2−1 p σy 1 x2 (3.9) At the elastic-plastic juncture (x= kj), Eqs (3.8) and (3.9) both become σte σy = k2(k2j −1− lnk2j) (k2−1)k2j + √ 3k2p σy(k2−1)k2j (3.10) When p= pa, the first one of Eq. (3.4) becomes σpz σy = k2 lnk2j +k 2−k2j√ 3k2(k2−1) (3.11) Mechanical analysis of autofrettaged high pressure apparatus 21 Using Eqs. (3.11) and (3.4), the general residual stress, Eqs. (2.1)-(2.4), can be rewritten as follows: — in the plastic zone σ′z σy = 1√ 3 (k2j k2 +lnx2− lnk2j ) − σ p z σy σ′r σy = 1√ 3 (k2j k2 −1+lnx2− lnk2j ) − σ p r σy σ′θ σy = 1√ 3 (k2j k2 +1+lnx2− lnk2j ) − σ p θ σy σ′e σy = √ 3 2 (σ′θ σy − σ′r σy ) =1− (σ p θ σy − σpr σy ) =1− σpe σy (3.12) — in the elastic zone σ′z σy = 1√ 3 k2j k2 − σ p z σy σ′r σy = 1√ 3 (k2j k2 − k2j x2 ) − σ p r σy σ′θ σy = 1√ 3 (k2j k2 + k2j x2 ) − σ p θ σy σ′e σy = √ 3 2 σ′θ σy − σ ′ r σy = k2j x2 − √ 3 2 (σ p θ σy − σ p r σy ) = k2j x2 − σ p e σy (3.13) Therefore, when p= pa, irrespective of kj, the total stresses are: — in the plastic zone σtz σy = σ′z σy + σpz σy = 1√ 3 (k2j k2 +lnx2− lnk2j ) σtr σy = σ′r σy + σpr σy = 1√ 3 (k2j k2 −1+lnx2− lnk2j ) σtθ σy = σ′θ σy + σ p θ σy = 1√ 3 (k2j k2 +1+lnx2− lnk2j ) σte σy ≡ 1 (3.14) — in the elastic zone σtz σy = σ′z σy + σpz σy = 1√ 3 k2j k2 σtr σy = σ′r σy + σpr σy = 1√ 3 (k2j k2 − k2j x2 ) σtθ σy = σ′θ σy + σ p θ σy = 1√ 3 (k2j k2 + k2j x2 ) σte σy = k2j x2 (3.15) The components of total stresses based on the Mises criterion are 2/ √ 3 times those based on the Tresca criterion, but the equivalent total stresses based on both theories in the plastic and elastic zone are the same, respectively. The reason is that the equivalent total stress based on the Tresca criterion is 2/ √ 3 times that based on the Mises criterion since σz = (σr +σθ)/2 for cylinders. 4. Residual stresses and total stresses when kj = kj∗ and p= pa If kj is determined by Eq. (2.5), or kj = kj∗, Eqs. (3.8) and (3.9) become respectively σte σy =1− 2 x2 + √ 3k2 k2−1 p/σy x2 σte σy = k2j −2 x2 + √ 3k2 k2−1 p/σy x2 (4.1) From Eq. (4.1)1, it is seen that: (1) provided p/σy >−2(k2−1)(x2−1)/ √ 3k2 (negative), σte >−σy, this is definitely feasible for p> 0 in engineering; 22 R. Zhu et al. (2) as long as p/σy > (k 2−1)(2−x2)/ √ 3k2, σte > 0, while (k 2−1)(2−x2)/ √ 3k2 pe, σ t e > 0; (3) so long as p< 2pe, σ t e <σy. Thus, when pe −(k2 −1)(k2j − 2)/ √ 3k2 (negative), σte > 0, this is certain for p > 0 in engineering, so the equivalent residual stress in the elastic zone is always tensile; (2) so long as p/σy < (k 2−1)(x2−k2j +2)/ √ 3k2, σte <σy, so when p< 2pe, σ t e <σy. At the inside surface, x=1, then, from Eq. (4.1)1 σte σy = √ 3k2 k2−1 p σy −1 (4.2) Unless p< 0, σte can not be lower than −σy. Unless p> 2pe, σte can not be higher than σy. So, when 0 0 in the elastic zone. If p < 2pe, σ t e can not be higher than σy. So, when 0 < p < 2pe, 0 < σ t e/σy < 1. Especially, when p = 2pe, σ t e = σy at the elastic-plastic junc- ture and σte/σy = k 2 j/x 2 at a general radius location in the elastic zone. When kj = kj∗, by the aid of Eq. (2.5), Eqs (2.1)-(2.4) can be simplified as follows: — in the plastic zone σ′z σy = lnx2√ 3 − 1√ 3 σ′r σy = lnx2√ 3 + 2√ 3x2 − 2√ 3 σ′θ σy = lnx2√ 3 − 2√ 3x2 σ′e σy =1− 2 x2 (4.4) — in the elastic zone σ′z σy = k2j −2√ 3k2 σ′r σy = ( 1− k2 (r/ri)2 )σ′z σy = ( 1− k2 x2 )σ′z σy σ′θ σy = ( 1+ k2 x2 )σ′z σy σ′e σy = k2j −2 x2 (4.5) The equations of residual stresses are greatly simplified, and cylinders are safe after remo- ving pa. When k = ∞, kj = √ e, from Eq. (4.5)4, the equivalent residual stress at a general radius location in the elastic zone is σ′e σy = e−2 x2 σ′e σy =1− 2 e → 0 when x= √ e→∞ (4.6) When kj = kj∗, the distributions of equivalent residual stresses in the plastic and elastic zones – which are the same as those based on Tresca criterion – are illustrated in Fig. 1. In Fig. 1: (1) Curve BAA: k = kj = kc, x varies from 1 to kj in the plastic zone (from point B to A), and from 2.2184574899167 . . . (kj) to 2.2184574899167 . . . (k) (from point A to A) in the elastic zone (no elastic zone). Mechanical analysis of autofrettaged high pressure apparatus 23 (2) Curve BCD: k = 2.25, kj = 2.046308 . . ., x varies from 1 to 2.046308 . . . in the plastic zone (frompointB toC), and from 2.046308 . . . to 2.25 (from pointC toD) in the elastic zone. (3) Curve BEF: k = 3, kj = 1.748442 . . ., x varies from 1 to 1.748442 . . . in the plastic zone (from pointB toE), and from 1.748442 . . . to 3 (from pointE to F) in elastic zone. (4) Curve BMN: k =∞, kj = √ e, x varies from 1 to √ e= 1.648721 . . . in the plastic zone (from pointB toM), and from √ e to k=∞ (from pointM toN) in the elastic zone. The above results and Fig. 1 are fit for both the Tresca andMises criterion. Fig. 1. The distributions of equivalent residual stresses in the whole wall Figure 1 and Eq. (4.4)4 show that all curves of equivalent residual stresses for any k and kj in the plastic zone are located on the identical curve AB and pass through the same point ( √ 2,0), except that a different curve for different k and kj is located in a different section of the curveAB, i.e. curvesBA,BC,BE andBM are all on the curveBA or they coincide with each other. However, if kj 6= kj∗ or the relation between kj and k does not satisfy Eq. (2.5), the above argument is untenable, or even |σ′e| > σy. This case is illustrated in Fig. 2, where the curves BEF andBKL coincide with each other in the plastic zone and both pass through point ( √ 2,0) because kj = kj∗, but curves HSI and GQJ neither coincide with each other in the plastic zone nor pass through the point ( √ 2,0), and they do not coincide with the curves BEF andBKL in the plastic zone for kj 6= kj∗. When p=2pe, Eq. (3.4) and (3.5) become Eq. (4.7) and Eq. (4.8), respectively σpz σy = 2√ 3k2 σpr σy = 2√ 3k2 − 2√ 3x2 σ p θ σy = 2√ 3k2 + 2√ 3x2 (4.7) and σpe σy = 2 x2 (4.8) When kj = kj∗, p = pa = 2pe, thus, when p = pa = 2pe, Eqs. (3.14) and (3.15) for the total stresses become: — in the plastic zone σtz σy = lnx2√ 3 − 1√ 3 + 2√ 3k2 σtr σy = lnx2√ 3 − 2√ 3 + 2√ 3k2 σtθ σy = lnx2√ 3 + 2√ 3k2 σte σy = √ 3 2 (σtθ σy − σtr σy ) ≡ 1 (4.9) 24 R. Zhu et al. Fig. 2. Comparison between equivalent residual stresses for different k and kj — in the elastic zone σtz σy = σ′z σy + σpz σy = k2j√ 3k2 σtr σy = σ′r σy + σpr σy = k2j√ 3 ( 1 k2 − 1 x2 ) σtθ σy = σ′θ σy + σ p θ σy = k2j√ 3 ( 1 k2 + 1 x2 ) σte σy = k2j x2 σte σy = e x2 (4.10) where k=∞, x∈ ( √ e,∞). The equations of total stresses are greatly simplified, and cylinders are safe after removing pa and in operation. Figure 3 shows the distribution of equivalent stress of the total stress when p=2pe and kj = kj∗. In Fig. 3: (1) Horizontal line baa: k= kj = kc. In the plastic zone, σ t e/σy is a horizontal line: σ t e/σy =1, x varies from 1 to kj (frompoint b to a) and from kj to k (frompoint a to a) in the plastic zone (no elastic zone). (2) Curve bcd: k = 2.25, kj = 2.046308 . . .. In the plastic zone, σ t e/σy is a horizontal line, bc: σte/σy = 1, x varies from 1 to kj (from point b to c) and from kj to k (from point c to d) in the elastic zone. (3) Curve bef: k = 3, kj = 1.748442 . . .. In the plastic zone, σ t e/σy is a horizontal line, be: σte/σy =1, x varies from 1 to kj (from point b to e) and from kj to k (from point e to f) in the elastic zone. (4) Curve bmn: k = ∞, kj = √ e. In the plastic zone, σte/σy is a horizontal line: σ t e/σy = 1, bm: x varies from 1 to kj (from point b to m) and from kj to k(∞) (from point m to n) in the elastic zone. If kj 6= kj∗ or p 6= 2pe, the above traits can not arise. Figure 4 is comparison between the equivalent total stresses under different internal pressure and kj = kj∗ from which it is known that only when p = 2pe and kj = kj∗, the operation state is optimum, otherwise, or p 6= 2pe and/or kj 6= kj∗, either σte >σy or load-bearing capacity is lowered or compressive yield occurs. In Fig. 4, curve 1 is just curve bef in Fig. 3. Besides, for a certain k, when kj 2pe and kj = kj∗, the equivalent total stress is greater than σy and uneven.When p < 2pe and kj = kj∗, the equivalent total stress is lower than σy but load-bearing capacity is reduced, and equivalent total stress is uneven. When p < pe, σ t e at the inside surface is lower 26 R. Zhu et al. than 0, σte can not be lower than −σy in the whole plastic zone, the load-bearing capacity is reduced greatly and the equivalent total stress is uneven. As long as an autofrettaged cylinder contains the autofrettage pressure pa, theremust be an inexorable law irrespective of kj and k:σe/σy ≡ 1 in the plastic zone and 0<σe/σy = k2j/x2 < 1 in the elastic zone. Nevertheless, too great kj causes compressive yield after removing pa, too small kj reduces the load-bearing capacity of a cylinder. The optimumplastic depth is kj = kj∗, and when kj = kj∗, p= pa =2pe. 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