Jtam-A4.dvi


JOURNAL OF THEORETICAL

AND APPLIED MECHANICS

51, 3, pp. 697-710, Warsaw 2013

ON AUTOFRETTAGE OF CYLINDERS BY LIMITING CIRCUMFERENTIAL

RESIDUAL STRESS BASED ON MISES YIELD CRITERION

Ruilin Zhu

Polytechnic College, Hunan Normal University, Changsha, China

e-mail: zrl200701@sina.com

Guolin Zhu

Basic Courses Department, Jiangxi Police College, Nanchang, China

The autofrettage technique is an effective and important measure to improve load-bearing
capacity and safety and to even distributions of stresses for pressure vessels. Based on the
classical fundamental theory on autofrettage and by theoretical analysis of residual stresses,
the total stresses, the overstrainand the load-bearing capacity, etc., in viewof the cylindrical
pressure vessels with outside-to-inside radius ratio larger than the critical ratio, the laws
contained in the autofrettage theory are revealed, the essential cause and reason for the
obtained laws are analyzed, the inherent andmeaning relations between various parameters
in the autofrettage theory are brought to light, and the safe depth of the plastic zone
(overstrain) as well as the conditions of loading or optimum operation conditions are found
out. It is shown that under the optimum operation conditions, pressure vessels are not
only safe but also economic, and the equations of the autofrettage theory are simplified
greatly and are quite terse, as a result, the essential relations between various parameters
in the autofrettage theory are distinct, and these equations are convenient for application
in engineering practice.

Key words: pressure vessel, autofrettage, load-bearing capacity, strength theory

Nomenclature

ri,rj,ro – Inside radius, radius of elastic-plastic juncture, outside radius, respectively
k – outside to inside radius ratio (k= ro/ri)
kj – depth of plastic zone (kj = rj/ri)
kc – critical radius ratio
p,pa,py – internal, autofrettage and entire yield pressure, respectively
pe – maximumelastic load-bearing capability of unautofrettaged pressure vessel

or initial yield pressure
σr,σθ,σz,σe – radial, circumferential, axial and equivalent stress, respectively
σy – yield strength

Superscripts: ′,T,p – residual, total and stress caused by p, respectively.

1. Introduction

Cylindrical pressure vessels are important mechanical equipment and are widely applied in the
industry. To raise load-bearing capacity of a pressure vessel and even distribution of stresses
in the wall of a pressure vessel, the “autofrettage” technique for thick pressure vessels was
created, which is an interesting and absorbing subject. Parker (2001a,b) extended a procedure
to model the addition of pressure or material to a tube, providing an associated residual stress



698 R. Zhu, G. Zhu

profile. Hameed et al. (2004) studied the residual stress distribution in an autofrettaged thick-
-walled cylinder. Maleki et al. (2010) evaluated residual stress distributions in autofrettaged
homogenous spherical pressure vessels subjected to different autofrettage pressures. Park et al.
(2008) presented an analytical and numerical study of a machined compound cylinder by using
the finite element code ANSYS10.0. Lee et al. (2008), conducted residual stress analysis of an
autofrettaged compound cylinder based on elastic-perfectly plastic and strain hardeningmodels.
Farrahi et al. (2009) presented an analytical method for reconstructing residual stresses within
axially symmetric cylinders.The optimumautofrettage pressure of a thick-walled cylinder under
thermo-mechanical loadings was analyzed by Zheng and Xuan (2010).
The autofrettage damage mechanics model was setup from the ultra-high pressure vessel

autofrettage damage mechanism (Lin et al., 2009) and a brief introduction of the autofrettage
method, such as mechanical extrusion, direct pressure, expansion of the explosion pressure and
solid autofrettage was given byGao et al. (2008). In terms of the point of view of avoiding com-
pressive yield for cylinders experiencing autofrettage handling and raising load-bearing capacity
as far as possible simultaneously, Zhu (2008a,b) studied autofrettage of cylinders based on the
3rd and 4th strength theory, respectively. By studying the results published by Zhu (2008a,b),
it is found that there are still some problems needed to be studied further, especially when the
strength theory (Mises yield criterion) is applied to study the autofrettage for pressure vessels.
It is also necessary to pay attention to some problems, such as overstrain to avoid compressive
yield, load-bearing capacity, safe conditions for operation, and so on. Based on thework by Zhu
(2008a,b), this paper discusses the stresses and depth of the plastic zone (overstrain), which is
done in Section 2 and followed by conclusions in Section 3.

2. Discussion about stresses and safe conditions for operation

Figure 1 shows a cross section of a cylinder wall with partially plastic and partially elasticity.

Fig. 1. Cross section of a cylinder wall with marked regions of partial plasticity and partial elasticity

The residual stress at a general radial location within the plastic zone (Zhu, 2008a,b)

σ′z
σy
=
1√
3

[k2j
k2
+ln
(r/ri)

2

k2j
−
(

1−
k2j
k2
+lnk2j

) 1

k2−1
]

σ′r
σy
=
1√
3

[k2j
k2
−1+ln (r/ri)

2

k2j
−
(

1−
k2j
k2
+lnk2j

) 1

k2−1
(

1− k
2

(r/ri)2

)]

σ′θ
σy
=
1√
3

[k2j
k2
+1+ln

(r/ri)
2

k2j
−
(

1−
k2j
k2
+lnk2j

) 1

k2−1
(

1+
k2

(r/ri)
2

)]

(2.1)

Therefore, the equivalent residual stress at a general radial location within the plastic zone
is (Zhu, 2008a,b)



On autofrettage of cylinders by limiting circumferential residual stress ... 699

σ′e
σy
=

√
3

2

(σ′θ
σy
−
σ′r
σy

)

=1−
k2−k2j +k2 lnk2j
(k2−1)(r/ri)2

(2.2)

which is the same as that based on the 3rd strength theory (Tresca yield criterion). The reason
is as follows.

For a cylinder with closed ends σz =(σr+σθ)/2 (Yu, 1990), then, under the identical stress
state and according to the 3rd and 4th strength theories, there is: σIVe =

√
3σIIIe /2, while each

component of the residual stresses based on the 4th strength theory is 2/
√
3 times of that based

on the 3rd strength theory (Zhu, 2008a,b) for the same k and kj. This is the reason why the
equivalent residual stress based on the 3rd strength theory equals that based on the 4th strength
theory.

The residual stress at a general radial location within the elastic zone is (Zhu, 2008a,b)

σ′z
σy
=
1√
3

[k2j
k2
−
(

1−
k2j
k2
+lnk2j

) 1

k2−1
]

σ′r
σy
=
1√
3

(

1−
k2

(r/ri)2

)[k2j
k2
−
(

1−
k2j
k2
+lnk2j

) 1

k2−1
]

=
(

1−
k2

(r/ri)2

)σ′z
σy

σ′θ
σy
=
1√
3

(

1+
k2

(r/ri)2

)[k2j
k2
−
(

1−
k2j
k2
+lnk2j

) 1

k2−1
]

=
(

1+
k2

(r/ri)2

)σ′z
σy

(2.3)

Therefore, the equivalent residual stress at a general radial location within the elastic zone
is

σ′e
σy
=

√
3

2

(σ′θ
σy
− σ

′
r

σy

)

=
k2(k2j −1− lnk2j)
(k2−1)(r/ri)2

(2.4)

which is also the sameas thatbasedonthe3rd strength theory.Thereason is as abovementioned.

In the whole elastic zone σ′e/σy > 0 for k
2
j − 1 > lnk2j [14], but in the plastic zone σ′e/σy

may be positive or negative. For example, on the inside surface of the cylinder where r/ri =1,
from Eq. (2.2), the equivalent residual stress is

σ′ei
σy
=
k2j −1−k2 lnk2j
k2−1

(2.5)

Since lnk2j = 2(k
2
j − 1)/(k2j + 1) + . . . (Zhu, 2008a,b), lnk2j > 2(k2j − 1)/(k2j + 1), hence,

k2 lnk2j > 2k
2(k2j −1)/(k2j +1)>k2j −1.

This means that on the inside surface of the cylinder σ′e/σy is always negative. It is not
difficult to know that dσ′e/dr > 0 within the plastic zone and dσ

′
e/dr < 0 within the elastic

zone, and, at the elastic-plastic juncture where r/ri = kj, both Eq. (2.2) and (2.4) become

0<
σ′e
σy
=
σ′ej
σy
=
k2(k2j −1− lnk2j)
(k2−1)k2j

< 1 (2.6)

Therefore,within theplastic zone σ′e/σy goes up fromthenegative value on the inside surface
till a positive value at the elastic-plastic juncture, and goes downwithin the elastic zone fromthe
positive value at the elastic-plastic juncture to a smaller positive value on the outside surface.

In Eq. (2.2), letting

1−
k2−k2j +k2 lnk2j
(k2−1)(r/ri)2

=0



700 R. Zhu, G. Zhu

one obtains

r

ri
=

√

k2−k2j +k2 lnk2j
k2−1

=

√

σ′ei
σy
+1<kj (2.7)

On the other hand, the solution for σ′z = σ
′
r, σ

′
r = σ

′
θ and σ

′
θ = σ

′
z within the plastic zone

(see Eqs. (2.1)) is also Eq. (2.7). This means that the three curves for the residual stress at
a general radial location collect at one point within the plastic zone, and the abscissa of the
intersection is just Eq. (2.7), where σ′e/σy =0.
Since the depth of the plastic zone (kj) affects the residual stresses and the greater the kj,

the greater the equivalent residual stress, so it is inadvisable to raise kj, otherwise, compressive
yieldmay occur after removing the autofrettage pressure pa. However, raising kj can raise load-
-bearing capacity of the cylinder thereupon, Zhu (2008a,b) put forward an equation for kj∗, the
maximum and optimum plastic depth kj for a certain k to avoid the compressive yield

k2 lnk2j∗ −k2−k2j∗ +2=0 or k=

√

√

√

√

k2j∗−2
lnk2j∗−1

(kj∗ ­
√
e) (2.8)

where
√
e ¬ kj∗ ¬ kc = 2.2184574899167 . . . and k ­ 2.2184574899167 . . ., when

k¬ 2.2184574899167 . . .= kc, |σ′ei/σy| ­ 1 never occurs irrespective of kj.
When kj is determined byEq. (2.8) – kj = kj∗, Eqs. (2.1)-(2.4) can be rearranged as follows

within the plastic zone

σ′z
σy
=
ln(r/ri)

2

√
3
− 1√
3
=
lnx2√
3
− 1√
3

σ′r
σy
=
lnx2√
3
+ 2√

3x2
− 2√

3

σ′θ
σy
= lnx

2

√
3
− 2√

3x2

σ′e
σy
=1− 2

(r/ri)2

(2.9)

within 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′e
σy
=
k2j −2
(r/ri)2

σ′θ
σy
=
(

1+
k2

(r/ri)2

)σ′z
σy
=
(

1+
k2

x2

)σ′z
σy

(2.10)

where x represents r/ri, the same below.
By taking k=3, kj =1.5,1.748442 and 2, respectively, the residual stresses are illustrated

as shown in Fig. 2.
In Fig. 2b, kj = 1.748442 = kj∗, as is just determined by Eq. (2.8). kj = 1.5 6= kj∗ in

Fig. 2a, and kj = 2 6= kj∗ in Fig. 2c or none of them is determined by Eq. (2.8). The former
residual stresses are less than those when kj =1.748442 = kj∗, and the latter residual stresses
are greater than those when kj =1.748442= kj∗.
Substituting Eq. (2.7) into Eq. (2.9)1−3 one obtains

σ′z
σy
=
σ′r
σy
=
σ′θ
σy
=
ln2−1√
3

When kj = kj∗, Eq. (2.7) becomes r/ri =
√
2. This means that when kj = kj∗, for any k,

the three curves for the residual stress at a general radial location collect at the same point
within the plastic zone, and the intersection is

(√
2,
ln2−1√
3

)

=(1.414214 . . . ,−0.17716 . . .)



On autofrettage of cylinders by limiting circumferential residual stress ... 701

Fig. 2. Distributions of the residual stresses

The residual stresses at the inside surface (r= ri) can be obtained by letting r= ri in Eqs.
(2.1) or (2.3)

σ′zi
σy
=−

1√
3(k2−1)

(1−k2j +k2 lnk2j)< 0
σ′ri
σy
=0

σ′θi
σy
=− 2√

3(k2−1)
(1−k2j +k2 lnk2j)= 2

σzi
σy

(2.11)

The equivalent residual stress at ri is

σ′ei
σy
=

√
3

2

(σ′θi
σy
−
σ′ri
σy

)

=
√
3
σ′zi
σy
=−

1

k2−1
(1−k2j +2k2 lnkj) (2.12)

If kj = kj∗ then, Eqs. (2.11) and (2.12) become

σ′zi
σy
=−

1√
3

σ′ri
σy
=0

σ′θi
σy
=−

2√
3
<−1

σ′ei
σy
=
σ′ej
σy
=−1 (2.13)

Therefore, if kj = kj∗, the residual stresses and their equivalent stress at the inside surface
are always compressive and constant irrespective of k and kj or the thickness and overstrain
of the cylinder. The equivalent residual stress at ri just reaches −σy. However, |σ′θi| exce-
eds the compressive yield strength σy. This is an inexorable law, for σ

′
ei =

√
3(σ′θi −σ′ri)/2,

at the inside surface σ′ri = 0, then σ
′
ei =

√
3σ′θi/2, when σ

′
ei/σy = −1, there must be

σ′θi =−(2/
√
3)σy <−σy.

The residual stresses at elastic-plastic juncture (r = rj) can be obtained by letting r = rj
in Eqs. (2.1) or (2.3)



702 R. Zhu, G. Zhu

σ′zj
σy
=
1√
3

[r2j
r20
−
(

1−
r2j
r20
+2ln

rj
ri

) 1

k2−1
]

=
1√
3

[k2j
k2
−
(

1−
k2j
k2
+2lnkj

) 1

k2−1
]

=
k2j −1− lnk2j√
3(k2−1)

σ′rj
σy
=
(

1− k
2

k2j

)σ′zj
σy

σ′θj
σy
=
(

1+
k2

k2j

)σ′zj
σy

(2.14)

The equivalent residual stress at rj (σ
′
ej/σy) is

σ′ej
σy
=
2k2

k2j

σ′zj
σy
=
k2(k2j −1− lnk2j)
k2j(k

2−1)
= 1−

1

k2−1
(k2

k2j
−1+

2k2

k2j
lnkj
)

=
k2j −1
k2j
−
σ′ei/σy
k2j

(2.15)

InEq. (2.14)1, when kj =1, the numerator=0, and d(k
2
j−1−lnk2j)/dkj =2(k2j−1)/kj > 0,

so, σ′zj > 0 or tension. Then, σ
′
rj < 0, σ

′
θj > 0.

If kj = kj∗, then Eqs. (2.14) and (2.15) become

σ′zj
σy
=
lnk2j√
3
−
1√
3

σ′rj
σy
=
(

1−
k2

k2j

)σ′zj
σy

σ′θj
σy
=
(

1+
k2

k2j

)σ′zj
σy

σ′ej
σy
=
k2

k2j
(lnk2j −1)=

k2j −2
k2j
> 0

(2.16)

It is clear fromEq. (2.15) that σ′ej/σy < 1, as a result σ
′
ei/σy < 1 in Eq. (2.15) or nomatter

whether kj = kj∗, the equivalent residual stress at the elastic-plastic juncture is certainly tension.
So, it is quite necessary to pay close attention to the residual stresses at the inside surface instead
of those at the elastic-plastic juncture.
When k = 3, kj∗ = 1.748442, at the inside surface |σ′ei| and |σ′θi| just reach σy according

to the 3rd strength theory, but |σ′θi| exceeds σy according to the 4th strength theory. This may
lower the safety of a pressure vessel. So, it is necessary to limit |σ′θi| to find the maximum kj
for |σ′θi| to be below σy, written as kjθ. Then, letting

σ′θi
σy
=− 2√

3(k2−1)
(1−k2j +k2 lnk2j)=−1

obtains

2k2 lnk2jθ−2k2jθ −
√
3k2+2+

√
3=0 or k=

√

√

√

√

2(k2jθ −1)−
√
3

2lnk2jθ−
√
3

(2.17)

From Eq. (2.17), k → ∞, kjθ = e
√
3/4 = 1.541896 . . . . Letting kjθ = k in Eq. (2.17), one

obtains

k2 lnk2

k2−1
=
2+
√
3

2
(2.18)

The solution to Eq. (2.18) is k = 2.024678965 . . . = kcθ, which means that when k ¬ kcθ, no
matter howdeep is the depth of the plastic zone, evenwhen thewhole cylinder is yielded kj = k,
|σ′θi| can not exceed σy after removing pa from the cylinder. When k > kcθ, if kj is greater
than what is determined by Eq. (2.17) (kj >kjθ), |σ′θi|will exceed σy after removing pa. This
paper is in the light of the cylindrical pressure vessels with k ­ kcθ. The relation between kjθ
and k is shown in Fig. 3, the solution to Eq. (2.8) is shown in this figure too.



On autofrettage of cylinders by limiting circumferential residual stress ... 703

Fig. 3. The depth of plastic zone

The pressure which a cylinder with kj can contain is (Yu, 1990)

p

σy
=
2√
3
lnkj +

k2−k2j√
3k2

(2.19)

Combining Eq. (2.17) with Eq. (2.19), one obtains the ultimate allowable loading of the
cylinderwith kj determinedbyEq. (2.17) (kj = kjθ) or byFig. 3 under the condition σ

′
θi =−σy

(|σ′ei|<σy) based on the 4th strength theory

p

σy
=

√
3+2

2
√
3

k2−1
k2
=

√
3+2

2

pe
σy

(2.20)

Therefore, when k > kcθ and under the condition σ
′
θi =−σy (|σ′ei|<σy), the load-bearing

capacity of the autofrettaged cylinder shouldbedetermined byEq. (2.20). Thevalue determined
by Eq. (2.20) is less than that determined by another equation proposed by Zhu (2008b) based
on the 4th strength theory and when kj = kj∗ : p4/σy = (2/

√
3)(k2 − 1)/k2) but greater

than that proposed by Zhu (2008a) based on the 3rd strength theory and when kj = kj∗ :
p3/σy = (k

2 − 1)/k2, and greater than the maximum elastic load-bearing capability of the
unautofrettaged cylinder: pe/σy = (k

2 −1)/(
√
3k2) or 1+

√
3/2(= 1.866025) times the initial

yield pressure of the unautofrettaged cylinder. It is seen that p = (p3 +p4)/2, and this is too
big to be coincidental – the authors think. Another interesting thing is that letting Eq. (2.20)
equal the entire yield pressure py/σy or p/σy =(2

√
3+3)/6(k2 −1)/k2 = py/σy =(2/

√
3)lnk

one obtains Eq. (2.18). p, p3, p4 and pe are plotted in Fig. 4.

Fig. 4. Comparison between load-bearing capacity



704 R. Zhu, G. Zhu

When kj = kjθ, uniting Eq. (2.17) and Eq. (2.1)-(2.4), one obtains

σ′z
σy
=
1√
3
ln
( r

ri

)2

− 1
2
=
1√
3
lnx2− 1

2

σ′r
σy
=
1√
3

(

lnx2+
2+
√
3

2x2
− 2+

√
3

2

)

σ′θ
σy
=
1√
3

(

lnx2−
2+
√
3

2x2
+
2−
√
3

2

) σ′e
σy
=1−

√
3+2

2(r/ri)2
=1−

√
3+2

2x2

(2.21)

It is seen that Eqs. (2.21) are unconcerned with kj and k, meaning that when σ
′
θi is con-

trolled, or kj = kjθ, the residual stresses and their equivalent stress are determined only by
the radial relative location (r/ri) within the plastic wall of the cylinder and independent of kj
and k or at a certain r/ri, the residual stresses and their equivalent stress within the plastic
zone are identical for various kj and k. From Eq. (2.21)4, there are:

when x¬
√

(
√
3+2)/2, σ′e ¬ 0, at the inside surface where x= 1, σ′e =−

√
3/2 reaching the

minimum for dσ′e/dr > 0.When x>
√

(
√
3+2)/2,σ′e > 0.Within thewholewall of a cylinder,

|σ′e|<σy. At the elastic-plastic juncture where x= kj, 0<σ′e/σy =1−(
√
3+2)/(2k2j)< 1, for

k2j > 1+
√
3/2, meaning that σ′e at the elastic-plastic juncture is tension and can not reach σy.

At r/ri =
√

(
√
3+2)/2 = 1.366025 . . ., σ′e = 0, and this is just the location where

σ′θ = σ
′
r = σ

′
z, that is to say, when kj = kjθ, no matter how great k is, the three curves

for the residual stress at a general radial location collect at the same point within the plastic
zone, and the intersection is

(

√√
3+2

2
,
1√
3
ln

√
3+2

2
− 1
2

)

=(1.366025 . . . ,−0.13984 . . .)

where r/ri =
√

(
√
3+2)/2 is just Eq. (2.7) when k and kj meet the relation expressed by Eq.

(2.17)

σ′z
σy
=
k2j − (

√
3+2)/2
√
3k2

σ′r
σy
=
(

1−
k2

(r/ri)2

)σ′z
σy
=
(

1−
k2

x2

)σ′z
σy

σ′θ
σy
=
(

1+
k2

(r/ri)
2

)σ′z
σy
=
(

1+
k2

x2

)σ′z
σy

σ′e
σy
=
k2j − (

√
3+2)/2

x2

(2.22)

Thedistribution of equivalent residual stresswithin thewholewall is demonstrated inFig. 5,
the related parameters are shown in the figure as well.
kj(= 1.748442) for k = 3 in Fig. 2b is determined by Eq. (2.8) or kj = kj∗ = 1.748442 . . ..

It is seen from the figure that |σ′ei/σy|=1, but |σ′θi/σy|> 1. For |σ′θi/σy|=1, letting k=3 in
Eq. (2.17), one obtains kj = kjθ =1.603502 . . . . Similarly, letting k=4 in Eq. (2.17) one finds
kjθ =1.571211 . . ., letting k =4 in Eq. (2.8) kj∗ =1.694172 . . .. For k =3, kj∗ =1.748442 . . .
and k=3, kjθ =1.603502 . . ., the residual stresses in three directions and their equivalent stress
are plotted in Fig. 6a. For k=4, kj∗ =1.694172 . . . and k=4, kjθ =1.571211 . . ., the residual
stresses in the three directions and their equivalent stress are plotted in Fig. 6b.
The stresses caused by the internal pressure p at a general radius location are

σpz
σy
=

1

k2−1
p

σy

σpr
σy
=
(

1−
k2

(r/ri)2

)σpz
σy

σ
p
θ

σy
=
(

1+
k2

(r/ri)2

)σpz
σy

(2.23)

The equivalent stress of the stresses caused by the internal pressure p is

σpe
σy
=

√
3

2

(σ
p
θ

σy
−
σpr
σy

)

=

√
3k2

k2−1
p

σy

( r

ri

)−2
(2.24)



On autofrettage of cylinders by limiting circumferential residual stress ... 705

Fig. 5. The distribution of equivalent residual stress within the whole wall

Fig. 6. Residual stresses and their equivalent stress; (a) k=3, kj∗ =1.748442 . . . and
k=3, kjθ =1.603502 . . ., (b) k=4, kj∗ =1.694172 . . . and k=3, kjθ =1.571211 . . .

The equivalent stress of the total stress σT/σy (ETS) is

σTe
σy
=

√
3

2

σTθ −σTr
σy

=

√
3

2

(σ′θ+σ
p
θ

σy
−
σ′r+σ

p
r

σy

)

=

√
3

2

(σ′θ −σ′r
σy

+
σ
p
θ −σpr
σy

)

=
σ′e
σy
+
σpe
σy
(2.25)

Then, within the plastic zone

σTe
σy
=1−

k2−k2j +k2 lnk2j
(k2−1)(r/ri)2

+

√
3k2

k2−1
p

σy

( r

ri

)−2
(2.26)

within the elastic zone

σTe
σy
=
k2(k2j −1− lnk2j)
(k2−1)(r/ri)2

+

√
3k2

k2−1
p

σy

( r

ri

)−2
(2.27)

At the elastic-plastic juncture (r/ri = kj), Eqs. (2.26) and (2.27) both become

σTe
σy
=
k2(k2j −1− lnk2j)
(k2−1)k2j

+

√
3k2p/σy
(k2−1)k2j

(2.28)

If kj = kjθ, Eqs. (2.26) and (2.27) become respectively

σTe
σy
=1−

√
3+2

2x2
+

√
3k2

k2−1
p/σy
x2

σTe
σy
=
k2j − (

√
3+2)/2

x2
+

√
3k2

k2−1
p/σy
x2

(2.29)



706 R. Zhu, G. Zhu

From Eq. (2.29)1, it is known that:

(1) provided p/σy > −[4x2 − (
√
3+2)]/(2

√
3)(k2 − 1)/k2 (negative), σTe /σy > −1, this is

definitely feasible for p/σy > 0 in engineering;

(2) as long as p/σy > [(
√
3 + 2) − 2x2]/(2

√
3)(k2 − 1)/k2, σTe > 0, while

[(
√
3+2)−2x2]/(2

√
3)(k2−1)/k2 <pe/σy, so when p>pe, σTe > 0;

(3) so long as p/σy ¬ (
√
3+2)/(2

√
3)(k2−1)/k2 =(

√
3+2)/2pe/σy, σ

T
e ¬σy. When x=1,

if p/σy =(k
2−1)/2k2 =(

√
3/2)pe/σy, then σ

T
ei/σy =0.

From Eq. (2.29)2:

(1) provided p/σy >−[k2j−(
√
3+2)/2]/

√
3(k2−1)/k2 (negative), σTe /σy > 0, this is definitely

feasible for p/σy > 0 in engineering, so the equivalent residual stress within the elastic
zone is always tension;

(2) as long as p/σy < [x
2 − k2j + (

√
3 + 2)/2]/

√
3(k2 − 1)/k2, σTe < σy, so when

p/σy ¬ (
√
3+2)/(2

√
3)(k2−1)/k2 =(

√
3+2)/2pe/σy, σ

T
e ¬σy.

At the inside surface, x= r/ri =1, then, from Eq. (2.29)1

σTei
σy
=

√
3k2

k2−1
p

σy
−
√
3

2
(2.30)

Unless p/σy <−(2−
√
3)/(2
√
3)(k2−1)/k2 (negative), σTei can not be lower than −σy. Unless

p/σy > 2pe/σy, σ
T
ei/σy can not be higher than σy. So, when 0 ¬ p/σy ¬ (

√
3+ 2)/2pe/σy,

−1<σTei/σy ¬ 1. Especially, when p/σy =(
√
3+2)/2pe/σy,σ

T
e /σy ≡ 1within thewhole plastic

zone.
At the elastic-plastic juncture, x= r/ri = kj, from (2.29)1 or (2.29)2

σTej
σy
=
k2j − (

√
3+2)/2

k2j
+

√
3k2

k2−1
p/σy
k2j

(2.31)

Clearly, σTej > 0 within the whole elastic zone. If p/σy ¬ (
√
3+2)/2pe/σy, σ

T
e can not be

higher than σy. So, when 0 ¬ p/σy ¬ (
√
3+ 2)/2pe/σy, 0 < σ

T
ej/σy ¬ 1. Especially, when

p/σy = (
√
3+2)/2pe/σy, σ

T
ej/σy = 1 at the elastic-plastic juncture and σ

T
e /σy = k

2
j/x
2 at a

general radial location within the elastic zone.
When p/σy = (

√
3+2)/2pe/σy or the load-bearing capacity is determined by Eq. (2.20),

Eqs. (2.29) become respectively

σTe
σy
≡ 1 σ

T
e

σy
=
k2j
x2

(2.32)

Figure 7 shows the distribution of the equivalent stress of the total stress.
The supplemental explaination for Fig. 7 is as follows.

(1) Horizontal line baa: k = 2.024678965 . . . = kcθ, kj = k = kjθ = 2.024678965 . . ..
Within the plastic zone, σT/σy is a horizontal line: σ

T/σy = 1, x varies from 1 to
kjθ(= kcθ = 2.024678965 . . .) (from point b to a), and then from 2.024678965 . . . (kjθ)
to 2.024678965 . . . (k) (from point a to a) within the elastic zone (no elastic zone, the
“curve” of the equivalent stress of the total stress is actually a point within the elastic
zone).

(2) Curve bcd: k=2.1 . . ., kjθ =1.806908 . . .. Within the plastic zone, σ
T/σy is a horizontal

line: σT/σy = 1, x varies from 1 to kjθ(1.806908 . . .) (from point b to c), and then from
1.806908 . . . to 2.1 . . . (k) (from point c to d) within the elastic zone.



On autofrettage of cylinders by limiting circumferential residual stress ... 707

Fig. 7. Distribution of the equivalent stress of the total stress for p/σy =(
√
3+2)/2pe/σy and kj = kjθ

(3) Curve bef: k=2.5 . . ., kjθ =1.6522121 . . ..Within the plastic zone, σ
T/σy is a horizontal

line: σT/σy =1, x varies from 1 to kjθ(1.6522121 . . .) (from point b to e), and then from
1.6522121 . . . to 2.5 . . .(k) (from point e to f) within the elastic zone.

(4) Curve bgh: k = 3, kjθ = 1.60350225 . . .. Within the plastic zone, σ
T/σy is a horizontal

line: σT/σy = 1, x varies from 1 to kjθ(1.844363 . . .) (from point b to g), and then from
1.844363 . . . to 3(k) (from point g to h) within the elastic zone.

(5) Curve bkl: k = 4, kjθ = 1.57121054 . . .. Within the plastic zone, σ
T/σy is a horizontal

line: σT/σy =1, x varies from 1 to kjθ(1.57121054 . . .) (from point b to k), and then from
1.57121054 . . . to 4(k) (from point k to l) within the elastic zone.

(6) Curve bmn: k → ∞, kjθ = e
√
3/4. Within the plastic zone, σT/σy is a horizontal line:

σT/σy = 1, x varies from 1 to kjθ = e
√
3/4 = 1.541896 . . . (from point b to m), and then

from kjθ to ∞(k) (from pointm to n) within the elastic zone.

The prerequisite to the above arguments is that k and kj meet Eq. (2.17) or kj = kjθ and
p/σy = (

√
3+2)/(2

√
3)(k2−1)/(k2) = (

√
3+2)/2pe/σy. Grasping these laws is helpful to the

design of high and ultrahigh pressure vessels. If kj 6= kjθ or p/σy 6=(
√
3+2)/(2

√
3)(k2−1)/k2,

the above facts are untenable, and |σTe |may exceed σy.
When p/σy =(

√
3+2)/2pe/σy, Eqs. (2.23) and (2.24) become

σpz
σy
=

√
3+2

2
√
3k2

σpr
σy
=

√
3+2

2
√
3k2
−
√
3+2

2
√
3x2

σ
p
θ

σy
=

√
3+2

2
√
3k2
+

√
3+2

2
√
3x2

σpe
σy
=

√
3+2

2x2

(2.33)

When p/σy =(
√
3+2)/2pe/σy and kj = kjθ, the components of the total stresses are:

—within the plastic zone

σTz
σy
=
lnx2√
3
− 1
2
+

√
3+2

2
√
3k2

σTr
σy
=
lnx2√
3
+

√
3+2

2
√
3k2
−
√
3+2

2
√
3

σTθ
σy
=
lnx2√
3
+

√
3+2

2
√
3k2
+
2−
√
3

2
√
3

(2.34)

From Eqs. (2.34)2 and (2.34)3, σ
T
e /σy =(

√
3/2)[σTθ /σy−σTr /σy]≡ 1, the same as fromEq.

(2.32)1. This proves the above arguments are correct and reasonable.



708 R. Zhu, G. Zhu

Within the elastic zone

σTz
σy
=
σ′z
σy
+
σpz
σy
=
k2j√
3k2

σTr
σy
=
σ′r
σy
+
σpr
σy
=−
k2j(x

−2−k−2)
√
3

σTθ
σy
=
σ′θ
σy
+
σ
p
θ

σy
=
k2j(x

−2+k−2)
√
3

(2.35)

From Eqs. (2.35)2 and (2.35)3, σ
T
e /σy = k

2
j/x
2, the same as from Eq. (2.32)2. This as well

proves that the above arguments are correct and reasonable.
Figure 8 shows a comparison between the equivalent stresses of the total stresses un-

der different internal pressures and kj = kjθ, from which it is known that only when
p/σy = (

√
3+ 2)/(2

√
3)(k2 − 1)/k2 = (

√
3+2)/2pe/σy and kj = kjθ, is the operation state

optimum, otherwise, or p/σy 6= (
√
3+2)/2pe/σy and/or kj 6= kjθ, either σTe >σy or the load-

-bearing capacity is lowered or compressive yield occurs. In Fig. 8, curve 1 is just curve bgh in
Fig. 7, in this case, p/σy = (

√
3+2)/2pe/σy, σ

T
e ≡ σy within the whole plastic zone. Curve 2:

p/σy =(
√
3+2.1)/2pe/σy,σ

T
ei >σy. Curve 3: p/σy =(

√
3+1.9)/2pe/σy, 0<σ

T
ei <σy. Curve 4:

p = pe, 0 < σ
T
ei ≪ σy. Curve 5: p/σy = (k2 − 1)/2k2 = (

√
3/2)pe/σy, σ

T
ei/σy = 0. Curve 6:

p=0.8pe, σ
T
ei/σy < 0.

Fig. 8. Comparison between the equivalent stresses of the total stresses under different internal
pressures and kj = kj∗

Besides, for certain k, when kj < kjθ, though the residual stresses are less than when
kj = kjθ, the load-bearing capacity is dropped. For example, for k = 3, if kj = kjθ
(= 1.60350225 . . .), from Eq. (2.19) or (2.20), p/σy = 0.9576 . . .; while if kj = 1.5, from Eq.
(2.19), p/σy =0.9012 . . . < 0.9576 . . ..

3. Conclusions

• For |σ′ei| ¬ σy, the depth of the plastic zone is k2 lnk2j∗ − k2 − k2j∗ + 2 = 0, where√
e ¬ kj∗ ¬ kc = 2.2184574899167 . . ., when k ­ 2.2184574899167 . . ., the load-bearing
capacity is

p4
σy
=
2√
3

k2−1
k2
=2
pe
σy

• For |σ′θi| ¬ σy, the depth of the plastic zone is 2k2 lnk2jθ −2k2jθ −
√
3k2 +(2+

√
3) = 0,

where e
√
3/4 ¬ kjθ ¬ kcθ = 2.024678965 . . ., when k ­ 2.024678965 . . ., the load-bearing

capacity is

p

σy
=

√
3+2

2
√
3

k2−1
k2
=

√
3+2

2

pe
σy



On autofrettage of cylinders by limiting circumferential residual stress ... 709

The data 2.024678965 . . . is the solution to (k2 lnk2)/(k2−1)= (2+
√
3)/2.

• For an autofrettaged cylinder, kj = kjθ with

p

σy
=

√
3+2

2
√
3

k2−1
k2
=

√
3+2

2

pe
σy

is the optimum operation state – not only safe but also highly capable, under the state
σTe ≡σy within the plastic zone and σTe /σy = k2j/x2 < 1 within the elastic zone.

• For the same k, kjθ < kj∗, as means that the depth of plastic zone necessary to ensure
|σ′θi| ¬σy is less than that to ensure |σ′ei| ¬σy.

• Irrespective of k, the three curves for the residual stress at a general radial location collect
at the same point within the plastic zone, and the intersection is

(

√√
3+2

2
,
1√
3
ln

√
3+2

2
−
1

2

)

=(1.366025 . . . ,−0.13984 . . .) when kj = kjθ

or

(√
2,
ln2−1√
3

)

=(1.414214 . . . ,−0.17716 . . .) when kj = kj∗

• When σ′θi is controlled, or kj = kjθ, the residual stresses and their equivalent stress
are determined only by the radial relative location (r/ri) within the plastic zone and
independent of kj and k, or once the location r/ri is determined, the residual stresses
and their equivalent stress within the plastic zone are identical for any kj and k.

Acknowledgement

This project is supported by Scientific Research Fund of Hunan Provincial Education Department

(Grant No. 12A087).

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Manuscript received August 17, 2012; accepted for print November 30, 2012