Research ArticleAnalysis of Stiffened Penstock External Pressure StabilityBased on Immune Algorithm and Neural Network

Wensheng Dong1 Xuemei Liu1 and Yunhua Li12

1 North China University of Water Resources and Electric Power Zhengzhou 450011 China2 School of Automation Science and Electrical Engineering Beihang University Beijing 100191 China

Correspondence should be addressed to Wensheng Dong dongwensh163com

Received 5 November 2013 Accepted 4 January 2014 Published 19 February 2014

Academic Editor Her-Terng Yau

Copyright copy 2014 Wensheng Dong et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

The critical external pressure stability calculation of stiffened penstock in the hydroelectric power station is very important workfor penstock design At present different assumptions and boundary simplification are adopted by different calculation methodswhich sometimes cause huge differences too In this paper we present an immune based artificial neural network model via themodel and stability theory of elastic ring we study effects of some factors (such as pipe diameter pipe wall thickness sectionalsize of stiffening ring and spacing between stiffening rings) on penstock critical external pressure during huge thin-wall procedureof penstock The results reveal that the variation of diameter and wall thickness can lead to sharp variation of penstock externalpressure bearing capacity and then give the change interval of it This paper presents an optimizing design method to optimizesectional size and spacing of stiffening rings and to determine penstock bearing capacity coordinate with the bearing capacity ofstiffening rings and penstock external pressure stability coordinate with its strength safety As a practical example the simulationresults illustrate that the method presented in this paper is available and can efficiently overcome inherent defects of BP neuralnetwork

1 Introduction

Penstock is one of the important compositions in the hydro-electric power station building It is arranged between reser-voir and underground power station house [1] In recentyears along with the construction of the large-capacitypumped storage power station and the application of high-strength materials the structure of the penstocks is turningto huge thin-walled structure For this structure its stabil-ity problem under external pressure has been particularlyprominent At home and abroad there are a lot of casesdue to external pressure caused penstock buckling failureStability problem of hydroelectric power station penstockunder external pressure has become one of the main controlconditions of penstock design

Stability analysis of stiffened penstock under externalpressure includes computing of tube shell and the criticalexternal pressure of stiffening ring At present the calculatedmethod of tube shell critical load uses mainly Mises formula

[1] Mises considered that when the instability failure tubeshell between stiffen rings takes place there will be morewave-numbers but the amplitude is relatively small Sincethere are many initial cracks between stiffened penstock andits outside concrete the outside concrete has a smaller con-straint for tube shell The calculating of critical external loadof embedded stiffened penstock can adopt the computationalformula of exposed penstock and the safety coefficient can beappropriately reduced

Actually due to penstock exists initial defects and asym-metrical cracks buckling penstock does not meet Misesassumption in some ways Reference [2] proposed a calcula-tion formula about critical load of penstock under externalpressure In the procedure of formula derivation Lai andFang adopted some basic assumptions such as elastic theoryknown wave numbers and stiffener ring stiffness infinityThe formula does have a unique novelty but due to thoseassumptions the application of formula is limited (when

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 823653 11 pageshttpdxdoiorg1011552014823653

2 Mathematical Problems in Engineering

the ring spacing is relatively large the calculated value is169 smaller than the measured value)

In some literatures [3ndash6] the derivation of penstock crit-ical external pressure formula did not consider the influenceof the external stiffened ring equivalent flange width on thecritical pressure which resulted in the computation accuracyLiu andMa presented a semianalytical finite element methodto analyze the instability problem of the stiffened penstock[7] it is a more objective computational method and it canbetter meet the actual situation

In this paper a nonlinear relationship between tubeshell critical external pressure and its influence factors isestablished by artificial immune neural network model andengages the elastic ring theory we have studied the effects ofsome factors on the critical external pressure of huge thin-walled penstock (such as the pipe diameter the pipe wallthickness the sectional size of stiffening ring and stiffenedring spacing) and have revealed the bearing capacity of hugethin-walled penstock plummeting reason and drasticallyreducing interval By optimizing sectional size of stiffenedring and spacing among stiffened rings we have presentedan optimal design method of the bearing capacity of stiffen-ing rings penstock bearing capacity coordination penstockexternal pressure stability and its strength safety coordinated

The rest of the paper is arranged as follows In Section 2we briefly introduce how to solve the critical load of thepenstock using semianalytical element method Section 3discusses the simulating of the critical pressure of penstockSection 4 studies the computation of the critical externalpressure of stiffened ring Section 5 provides one case studyof one practical project Finally the main conclusions of thepaper are inducted

2 Semianalytical Finite Element Method forStability Analysis of Penstock

In 1990 Liu and Ma proposed a semianalytic finite elementcomputation method (SA-FEM) for the stability analysis ofthe penstock under external pressure [7]Thismethod adoptsthe analytical method along circumferential direction andthe discrete finite element method along the axial directionrespectivelyThe penstock is divided into the finite cylindricalshell elements which are connected using the node circle Atypical cylindrical shell element is shown in Figure 1 where119905 119877 and 119861 denote the shell thickness node circle radiusand element axial direction length respectively To facilitatestudy we select the axial direction of shell middle curvedsurface as the 119909 coordinate (the dimensionless coordinateis 120585) the circumferential direction as the 119910 coordinate (thedimensionless coordinate 120593) and the normal directions asthe 119911 coordinate and providing positive 119911 is pointing to thedirection of the curvature centerThe displacement of a pointin middle curved surface is119882 The displacements of nodal 119894and 119895 are represented by119882

119894 120579119894119882119895 and 120579

119895 respectively

Defining the node displacement vector is as

120575 = [119882119894 120579119894 119882119895 120579119895]119879

(1)

120579i

y(120593)

Wi

z

R120593

B

ji

120579jWj

x(120585)

Figure 1 Cylindrical shell element

Introducing the Hermite interpolation polynomial vector[119867] is as

[119867] = [1198671 1198672 1198673 1198674] (2)

where 1198671= 1 minus 3120585

2+ 21205853 1198672= 119861(minus120585 + 2120585

2minus 1205853) 1198673=

31205852minus21205853 and 119867

4= 119861(120585

2minus1205853) we can express the cylindrical

shell element radial displacement function119883(120585) along the119883-axis as follows

119883(120585) = [119867] 120575 (3)

Accordingly the radial displacement function119882(120585 120593) ofany point on cylindrical shell element middle surface can bewritten as

119882(120585 120601) = 119883 (120585) cos (119899120601) (4)

where 120593 is central angle (radians) and 119899 is the unstable wavenumber along circumferential direction

According to constraint conditions we can see that thereis neither the tensile deformation along the axial nor theshear deformation on the cylindrical shell middle surfaceTherefore from the deflection function 119882 we can derivedisplacement 119880 and 119881 of any point on the middle surfaceso that the displacement function 119891 can be expressed asfollows

119891 = 119880 119881 119882119879

=

[[[[[[

[

119877

1198992119861cos (119899120601) 0 0

01

119899sin (119899120601) 0

0 0 cos (119899120601)

]]]]]]

]

[

[

[1198671015840]

[119867]

[119867]

]

]

120575

(5)

where [1198671015840] is the first order derivative of polynomial vector[119867]

Mathematical Problems in Engineering 3

According to the strain displacement relations of cylin-drical shell with large deflection we can obtain geometricequation of cylindrical shell element

120576 = 120576119871 + 120576

119873

=

1

119861

120597119880

120597120585minus119911

1198612

1205972119882

1205971205852

1

119877

120597119881

120597120601+2119911

1198772

120597119881

120597120601minus119882

119877minus119911

1198612

1205972119882

1205971206012

1

119877

120597119880

120597120601+1

119861

120597119881

120597120585+2119911

119877119861

120597119881

120597120585minus2119911

119877119861

1205972119882

120597120585120597120601

+

1

21198612(120597119882

120597120585)

2

1

21198772(120597119882

120597120601)

2

1

119877119861

120597119882

120597120585

120597119882

120597120601

(6)

where 120576119871 denotes the linear strain item and 120576119873 denotes thenonlinear strain item resulted by large deflection

According to the constitutive relationships of materialsand equilibriumequations of cylindrical shell element we canset up the matrix equation of the cylindrical shell elementThe stiffnessmatrix is composed of the elastic stiffnessmatrixand the equivalent geometric stiffness matrix The elasticstiffness matrix formula of cylindrical shell element is asfollows

[119896119864] = [119896

1198641] + [119896

1198642] + [119896

1198643] + [119896

1198644] (7)

[1198961198641] =

120587119864119877119905 (121198772+ 11989941199052)

611989941198613(1 minus 120583

2)

[[[

[

6 minus3119861 minus6 minus3119861

minus3119861 211986123119861 119861

2

minus6 3119861 6 3119861

minus3119861 11986123119861 2119861

2

]]]

]

(8a)

[1198961198642] =

1205871198641205831199053(1198992minus 1)

180119877119861 (1 minus 1205832)

[[[

[

36 minus18119861 minus36 minus3119861

minus18119861 41198612

3119861 minus1198612

minus36 3119861 36 18119861

minus3119861 minus119861218119861 4119861

2

]]]

]

(8b)

[1198961198643] =

1205871198641198611199053(1198992minus 1)2

50401198773(1 minus 120583

2)

[[[

[

156 minus22119861 54 13119861

minus22119861 4119861213119861 minus3119861

2

54 13119861 156 22119861

13119861 minus3119861222119861 4119861

2

]]]

]

(8c)

[1198961198644] =

1205871198641199053(1198992minus 1)2

1801198771198611198992(1 + 120583)

[[[

[

36 minus3119861 minus36 minus3119861

minus3119861 411986123119861 minus119861

2

minus36 3119861 36 3119861

minus3119861 minus11986123119861 4119861

2

]]]

]

(8d)

where 119864 is the modulus of elasticity (Nmm2) 120583 is Poissonratio and the other symbols are the same as previous Theequivalent geometric stiffness matrix formula of cylindricalshell element is represented as follows

[119896119866] = [119896

119866119909] + [119896

119866119910] + [119896

119866119909119910] (9)

After integration operation we found that [119896119866119909] and [119896

119866119909119910]

are zero matrix and the computing formula of [119896119866119910] is as

follows

[119896119866119910] = minus

119902119890120587119861 (119899

2+ 1)

420

[[[

[

156 minus22119861 54 13119861

minus22119861 4119861213119861 minus3119861

2

54 13119861 156 22119861

13119861 minus3119861222119861 4119861

2

]]]

]

(10)

where 119902119890 is the radial external pressure of the elementAdopting stiffness integration methods penstock we

can obtain respectively the elastic stiffness matrix andthe equivalent geometrical stiffness matrix of the overallstructure In the structural stiffness equation introducingboundary constraint conditions in the structural stiffnessmatrix crossing out the rows and columns associated withthe displacement constraints we can get the force-balancingequation of the overall penstock as follows

([119870119864] + 119902 [119870

lowast

119866]) Δ = 119875 (11)

where [119870119864] and [119870lowast

119866] are respectively elastic stiffness matrix

and equivalent geometrical stiffness matrix [119870119864] = sum

119890[119870]119890

[119870lowast

119866] = sum119890(119902119890119902)[119870

lowast

119866]119890 and 119890 is element number 119902 is external

pressure of structure Δ and 119875 are respectively vectors ofthe nodal displacement and load of the structure

It is known from structure stability theory that char-acteristic equation to describe structure stability is thatdeterminant of the overall stiffness matrix is equal to zerothat is det([119870

119864] + 119902[119870

lowast

119866]) = 0 Thus stability problem is

transformed to solve the largest eigenvalue problem of realmatrix (minus[119870

119864]minus1[119870lowast

119866]) The reciprocal of the largest eigen-

value is critical stable loadFrom the above discussions we can see that using SA-

FEM to solve the critical pressure of the penstock is verycomplicated Actually in practical engineering design andstructure analysis an analytical explicit formula to describethe relationship between critical pressure and structureparameters is more welcome In order to meet this requestwe provide a realization method based on neural networkFirstly we acquire a group of the samples that adopt SA-FEM to calculate critical pressure of different penstocksThen using nonlinear mapping ability of neural networkto get nonlinear relationship between critical pressure andrelated parameters namely penstockmaterial pipe diameterthickness of the penstock wall the spacing among stiffenerrings and so forth

3 Penstock Critical Pressure CalculatingBased on Neural Network

It is well known that neural network can approach com-plicated nonlinear map with very high accuracy Immunealgorithm is an evolutionary method that reflects immunesystem characteristic of living organisms [8 9] and it canavoid the drawbacks of traditional neural network learningalgorithms The main idea of adopting immune algorithmto design neural network is neural network structures andconnection among neurons as an antibody of biological

4 Mathematical Problems in Engineering

Table 1 Connection relationship table of neural nodes

To From1 2 sdot sdot sdot 8 9

1 times times sdot sdot sdot times times

2 times times sdot sdot sdot times times

3 times times sdot sdot sdot times times

4 11990841

11990842

sdot sdot sdot times times

5 11990851

11990852

sdot sdot sdot times times

6 0 0 sdot sdot sdot times times

7 11990871

0 sdot sdot sdot times times

8 0 0 sdot sdot sdot 11990888

times

9 0 0 sdot sdot sdot 11990898

11990899

1

2

3

5

46

79

8

Input layer

Output signal

Output layer

Input signal

Hidden layer

Figure 2 The schematic diagram of neural network

immune system Selection based on antibodies concentra-tion and self-adaptive mutation operator makes antibodypopulation continuously optimized and finally finds the bestantibodies Immune algorithm is characterized by diversitydistribution of solution group and it can better overcomethe shortcomings of that network structure and cryptic layernumbers defined difficultly

31 Neural Network Design In the paper the neural networkstructure adopted is shown in Figure 2The network neuronshave no significant hierarchical relationship In addition toinput neurons there are no restriction connections amongneurons each network node is assigned a serial number theserial number of node is only used to distinguish beginningand end of directed link [10]

Deletion of Connection Edges If the connection weights valueof a connection edge is less than specified threshold range[minus0001 0001] its weights value is set to zero that is in thesame neuron numbers circumstances different connectionform composes different network structure

Neurons Removed If all weights values of connection with aneuron are less than specified threshold range deleting thisneuron The network structure and connection weights canbe expressed as an equivalent matrix as Table 1 Concate-nating each element of matrix constitutes an antibody Anantibody expresses a neural network structure

32 Design Steps of Neural Network Based onImmune Algorithm

(a) Fitness FunctionThe antibodyw119894 constitutes the objectivefunction of network

119864 (w119894) =119899

sum

119895=1

10038161003816100381610038161003816119905119895minus 119910119895

10038161003816100381610038161003816 (12)

where 119905119895is objective output of network 119910

119895is actual output

and 119899 is sample number in training setsFitness function can be expressed as

119865 (w119894) = 1

119864 (w119894) + 119862 (w119894) (13)

where 119862(w119894) reflect impact of network complexity it isthe sum of network nodes and connections among nodes119862(w119894) = 119904(w119894) + 119897(w119894) 119897(w119894) and 119904(w119894) represent respectivelynetwork connections and network nodes

(b) Immune Selection Algorithm Based on Similarity andVector DistanceAssuming that in a population each antibodycan be represented by a one-dimensional array of119898 elementsantibodies similarity is calculated as follows assuming thatw1 = 119908

1

1 1199081

2 119908

1

119898 and w2 = 119908

2

1 1199082

2 119908

2

119898 are any

two antibodies of an antibody population with size 119899 thesimilarity of w1 and w2 is 119889(w1w2)

119889 (w1w2) = radic119898

sum

119894=1

10038161003816100381610038161199081

119894minus 1199082

119894

1003816100381610038161003816 (14)

where 119899 antibodies constitute a nonempty immune setW thedistance of two antibodies is defined as

119863(w119894) =119899

sum

119895=1119895 = 119894

119889 (w119894w119895) (15)

The concentration of antibody can be expressed asDensity(w

119894)

Density (w119894) = 1

119863 (w119894) (16)

From formula (16) we can see that the more the similarityantibodies the greater the antibody concentration and on thecontrary the smaller the antibodies concentration

The population Update Based on Antibody ConcentrationAfter the parents generated offspring through mutationaccording to the selection probability random selection ofindividuals from the population and offspring constitutes anew population The probability selective function is definedas follows

119875119904(w119894) = 120572 times density (w119894)(1 minus

119863 (w119894)sum119899

119894=1119863(w119894)

)

+ 120573

119865 (w119894)sum119899

119894=1119865 (w119894)

(17)

Mathematical Problems in Engineering 5

where 120572 120573 is adjustable parameter in (0 1) interval its valuedetermined based on experience In this paper alpha andbeta value is set to 05 meaning that antibody concentrationand fitness have equal status during the update process ofantibody population 119865(w119894) is fitness of the antibody 119894

As known from (17) the first part of right side of theequation is based on antibody concentration selection itemsthe higher concentration antibody has little selected chancebut the lower concentration antibody has bigger selectedchance the second part of right side of the equation isbased on antibody fitness selection items the higher fitnessantibody has bigger selected chance

(c) Generate the NewAntibodies Because the network param-eters and the network structure is many to one relationshiptherefore this paper only adopts mutation operation to carryout antibodies update

Defining mutation operator 120590119894= radic1 minus 119865(119908

119894) and using it

mutate all parameters of network as follows

1199081198941015840

119895= 119908119894

119895+ 120590119894times 119873119895 (0 1)

(18)

where 119908119894119895and 119908119894

1015840

119895are respectively antibodies of gene before

and aftermutation and119873119895(0 1) indicates random variable for

each subscript 119895 re-sampling

33 Simulating the Critical External Pressure of Penstock Thispaper uses [7 11 12] proposed calculationmethod to computecritical pressure and regards the computed results as thetraining sample of the network The calculation model isdescribed as follows Penstock material is 16Mn (modulusof elasticity is 119864 = 21 times 10

5MPa Poisson ratio 120583 = 03and 120590

119904= 340MPa) The ratio of penstock radius 119903 and shell

thickness 119905 (relative tube radius 119903119905) is from 20 to 400 steplength is 20 the ratio of rings spacing and tube radius (relativering spacing 119871119903) is [01 02 03 05 08 14 20 30 40]The total calculation models are 180

The transfers function among neurons uses s-function inthe Matlab the transfers function of output layer uses linearfunction Population size is 119873 = 50 The simulating resultsare shown in Table 2

34 Analysis of Calculated Results (1) Figure 3 shows thatwith the increase of 119903119905 the losing stability capability ofpenstock under external pressure is decreased acutely Thecritical external pressure decreases with increasing of 119903119905Within 119903119905 = 20 sim 260 the critical external pressure isacutely decreased beyond the range the change is less Forexample within 119871119903 = 01 sim 30 119903119905 from 20 to 260 119875cr willdecrease to 013sim016 of initial value (119903119905 = 20 119875cr) whenthe diameter of penstock is increased to a certain value thestability power of the penstock under external pressure willchange very small For example if 119871119903 = 30 and 119903119905 = 400then the 119875cr value is only 002MPa

(2)The calculated result shows that the critical pressure119875cr of penstock decreases with relative distance of reinforcingring 119871119903 increasingbut the influence of decreasing velocityis less than relative radius 119903119905 For example 119903119905 = 300

8

6

4

2

0

minus2

minus4

50 100 150 200 250 300 350 400

Mill fnish steel tube Lr = 50

01

02

040610

3050

rt

log(P

cr)

The real line is result of Mises theoryThe point is result of simulation

Figure 3 Simulating results

Table 2 Simulating results of the critical external pressure 119875cr

Critical pressure(simulation)

119871119903

01 02 03 05 08119903119905

20 202426(20242)

90187(9019)

55456(5545)

29914(2991)

17036(1704)

60 11916(11917)

5205(5214)

3197(3189)

1743(1756)

1012(1018)

100 3171(3157)

1379(1371)

850(843)

467(471)

274(277)

140 1323(1338)

575(586)

355(347)

197(201)

116(114)

180 688(692)

299(287)

186(179)

103(103)

061(067)

260 264(257)

116(123)

072(078)

040037)

024(028)

300 182(181)

079(081)

049(051)

028(025)

016(016)

360 113(114)

049(045)

031(029)

017(019)

011(015)

400 086(081)

037(036)

024(023)

013(011)

008(007)

119871119903 = 01 sim 30 and the critical external pressure 119875cr willdecrease to 219sim4341 of initial value

(3) The most effective reinforcing rings spacing and thecurve of Figure 3 shows that reducing rings spacing caneffectively improve the carrying capacity of critical externalpressure of penstock and while 119871119903 decrease 119875cr value andits increase ratio increase For example 119903119905 = 260 119871119903from 30 decrease to 08 and continue to decrease to 01 andthe increment of 119875cr is respectively 018MPa and 240MPaIn other words the reinforcing rings spacing reduces to01119903 and the average increment of 119875cr value is respectively00082MPa and 034MPa The latter is 41 multiples of theformer So the most effective reinforcing ring spacing should

6 Mathematical Problems in Engineering

meet 119871 lt 08119903The reinforcing ring spacing of Chinarsquos SanXiahydropower station is 119871 = 032119903

(4)Noneffective reinforcing rings spacing Figure 3 showsthat along with the stiffener ring spacing increases the role ofstiffening ring is gradually reduced and while 119871119903 = 40 thelog(119875cr) sim 119903119905 curves of reinforcing penstock and mill finishsteel tube (the calculated formula of mill finish steel tube119901 = 2119864(119905119863)

3 119863 is diameter of tube) are almost equal Thisshows that stability against external pressure of the two tubesis roughly equal then the reinforcing ring does not possessany sustaining effect on the stiffness

(5)The losing stability of wave numbers 119899 is a synthesisembodiment for longitudinal and circular stiffness of thepenstock With an increasing of the penstock diameter thestiffness of circular decreases and the losing stability ofwave numbers increases But in 119883-axial direction with anincreasing of the reinforcing ring spacing the stiffness of 119883-axial direction of penstock decreases and also reduces theinstability wave numbers 119899 For fine pitch large diameterstiffened penstock the losing stability of wave shape showsmultiwave form but for the sparse space and small diametershows less wave form

(6) The design of penstock in Figure 3 shows that thecurve cluster of ldquolog(119875cr) sim 119899 119903119905 119871119903rdquo is divided into twoupper and lower districts by the plastic losing stabilitycurve If the penstock diameter and relative reinforcing ringsspacing are bigger the penstock appears elastic losing stabilityunder smaller external pressure log(119875cr) value is locatedin the below district of the elastic losing stability curveWhen the penstock diameter and relative reinforcing ringsspacing are smaller the stability of penstock under externalpressure is powerful and can bear great pressure the log(119875cr)value is located in upon district of the elastic losing stabilitycurve When we design the penstock if 119903119905 value has beendetermined by use construction and so forth we can selectappropriate 119871119903 value according to Figure 3 making thecarrying capacity of penstock critical external pressure meetsnot only the requirements resistance to external pressurestability but also the instability curve as close as possible inorder to achieve full use of the material strength to ensureexternal pressure stability and strength safety coordinatedpurposes

4 Computation of the Stiffening RingrsquosCritical External Pressure

41The Structure Form of Stiffening Ring Thestructure formsof stiffening ring include the cross-section form of ringand the connected mode between ring and tube shell asshown in Figure 4 As for the huge thin-wall penstock itis advisable to adopt the structure of that both stiffeningring and penstock are rolled together as a whole which caneffectively avoid penstock initial defects that caused by weldbead and uneven weld quality The reasonable cross-sectionstructure and dimensions of stiffening ring not only shouldbe able to bear large external load in smaller cross-section sizebut also enable the critical load of tube shell close to or equalto the critical load of stiffening ring effective control range so

aL

b

t

Tube axis

(a) Rectangle ring

b

2a

t

Radius of steel tuber

Tube axis

b2

(b) T-shape ring

Figure 4 The structure form of stiffening ring

as to effectively improve external loads condition of structureand to facilitate construction

On the structure that penstock and stiffening ring arerolled together as a whole the effective control range ofstiffening ring is 078radic119903119905 Computing model is as followspenstock radius r is 6200mm penstock shell thickness 119905is 40mm ring thickness 119886 is respectively 20 40 60 and80 100mm and the variation range of relative ring spacing119871119903 is [01 30] 119887119886 isin [05 50] Calculating separately criticalpressure of rectangle ring and T-shape ring that having thesame cross-section areaThe computing formula of stiffeningring critical external pressure is as follows (calculation resultsare shown in Table 3)

119875cr =3119864119869119896

1198773

119896119871 (19)

where 119877119896is radius that is located in the gravity axis of

stiffening ring effective section (mm) and 119869119896is moment of

inertia that is located in the gravity axis of the stiffened ringeffective section (mm4)

For the rectangle ring

119877119896=(1198862) (119887 + 119905)

2+ 078 119905

2radic119903119905

119886 (119887 + 119905) + 156 119905radic119903119905

119869119896=119886

12(119887 + 119905)

3+ 119886 (119887 + 119905) (119877119896 minus

119887 + 119905

2)

2

+ 013 1199053radic119903119905 + 156 119905radic119903119905(119877

119896minus119905

2)

2

(20)

For the T-shape ring

119877119896=(1198861198874) (119887 minus 119886 + 2119905) + (1198862) (119886 + 119905 + 1198872)

2+ 078 119905

2radic119903119905

(1198861198872) + 119886 (119886 + 119905 + 1198872) + 156 119905radic119903119905

119869119896=1198863119887

24+119886119887

2(119887 minus 119886

2minus 119877119896+ 119905)

2

+119886

12(119887

2+ 119905)

3

+ 119886(119887

2+ 119905) times [

1

2(119887

2+ 119905) minus 119877

119896]

2

+ 013 1199053radic119903119905

+ 156 119905radic119903119905(119877119896minus119905

2)

2

(21)

The calculated results can be plotted as shown in Figure 5

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

the ring spacing is relatively large the calculated value is169 smaller than the measured value)

In some literatures [3ndash6] the derivation of penstock crit-ical external pressure formula did not consider the influenceof the external stiffened ring equivalent flange width on thecritical pressure which resulted in the computation accuracyLiu andMa presented a semianalytical finite element methodto analyze the instability problem of the stiffened penstock[7] it is a more objective computational method and it canbetter meet the actual situation

In this paper a nonlinear relationship between tubeshell critical external pressure and its influence factors isestablished by artificial immune neural network model andengages the elastic ring theory we have studied the effects ofsome factors on the critical external pressure of huge thin-walled penstock (such as the pipe diameter the pipe wallthickness the sectional size of stiffening ring and stiffenedring spacing) and have revealed the bearing capacity of hugethin-walled penstock plummeting reason and drasticallyreducing interval By optimizing sectional size of stiffenedring and spacing among stiffened rings we have presentedan optimal design method of the bearing capacity of stiffen-ing rings penstock bearing capacity coordination penstockexternal pressure stability and its strength safety coordinated

The rest of the paper is arranged as follows In Section 2we briefly introduce how to solve the critical load of thepenstock using semianalytical element method Section 3discusses the simulating of the critical pressure of penstockSection 4 studies the computation of the critical externalpressure of stiffened ring Section 5 provides one case studyof one practical project Finally the main conclusions of thepaper are inducted

2 Semianalytical Finite Element Method forStability Analysis of Penstock

In 1990 Liu and Ma proposed a semianalytic finite elementcomputation method (SA-FEM) for the stability analysis ofthe penstock under external pressure [7]Thismethod adoptsthe analytical method along circumferential direction andthe discrete finite element method along the axial directionrespectivelyThe penstock is divided into the finite cylindricalshell elements which are connected using the node circle Atypical cylindrical shell element is shown in Figure 1 where119905 119877 and 119861 denote the shell thickness node circle radiusand element axial direction length respectively To facilitatestudy we select the axial direction of shell middle curvedsurface as the 119909 coordinate (the dimensionless coordinateis 120585) the circumferential direction as the 119910 coordinate (thedimensionless coordinate 120593) and the normal directions asthe 119911 coordinate and providing positive 119911 is pointing to thedirection of the curvature centerThe displacement of a pointin middle curved surface is119882 The displacements of nodal 119894and 119895 are represented by119882

119894 120579119894119882119895 and 120579

119895 respectively

Defining the node displacement vector is as

120575 = [119882119894 120579119894 119882119895 120579119895]119879

(1)

120579i

y(120593)

Wi

z

R120593

B

ji

120579jWj

x(120585)

Figure 1 Cylindrical shell element

Introducing the Hermite interpolation polynomial vector[119867] is as

[119867] = [1198671 1198672 1198673 1198674] (2)

where 1198671= 1 minus 3120585

2+ 21205853 1198672= 119861(minus120585 + 2120585

2minus 1205853) 1198673=

31205852minus21205853 and 119867

4= 119861(120585

2minus1205853) we can express the cylindrical

shell element radial displacement function119883(120585) along the119883-axis as follows

119883(120585) = [119867] 120575 (3)

Accordingly the radial displacement function119882(120585 120593) ofany point on cylindrical shell element middle surface can bewritten as

119882(120585 120601) = 119883 (120585) cos (119899120601) (4)

where 120593 is central angle (radians) and 119899 is the unstable wavenumber along circumferential direction

According to constraint conditions we can see that thereis neither the tensile deformation along the axial nor theshear deformation on the cylindrical shell middle surfaceTherefore from the deflection function 119882 we can derivedisplacement 119880 and 119881 of any point on the middle surfaceso that the displacement function 119891 can be expressed asfollows

119891 = 119880 119881 119882119879

=

[[[[[[

[

119877

1198992119861cos (119899120601) 0 0

01

119899sin (119899120601) 0

0 0 cos (119899120601)

]]]]]]

]

[

[

[1198671015840]

[119867]

[119867]

]

]

120575

(5)

where [1198671015840] is the first order derivative of polynomial vector[119867]

Mathematical Problems in Engineering 3

According to the strain displacement relations of cylin-drical shell with large deflection we can obtain geometricequation of cylindrical shell element

120576 = 120576119871 + 120576

119873

=

1

119861

120597119880

120597120585minus119911

1198612

1205972119882

1205971205852

1

119877

120597119881

120597120601+2119911

1198772

120597119881

120597120601minus119882

119877minus119911

1198612

1205972119882

1205971206012

1

119877

120597119880

120597120601+1

119861

120597119881

120597120585+2119911

119877119861

120597119881

120597120585minus2119911

119877119861

1205972119882

120597120585120597120601

+

1

21198612(120597119882

120597120585)

2

1

21198772(120597119882

120597120601)

2

1

119877119861

120597119882

120597120585

120597119882

120597120601

(6)

where 120576119871 denotes the linear strain item and 120576119873 denotes thenonlinear strain item resulted by large deflection

According to the constitutive relationships of materialsand equilibriumequations of cylindrical shell element we canset up the matrix equation of the cylindrical shell elementThe stiffnessmatrix is composed of the elastic stiffnessmatrixand the equivalent geometric stiffness matrix The elasticstiffness matrix formula of cylindrical shell element is asfollows

[119896119864] = [119896

1198641] + [119896

1198642] + [119896

1198643] + [119896

1198644] (7)

[1198961198641] =

120587119864119877119905 (121198772+ 11989941199052)

611989941198613(1 minus 120583

2)

[[[

[

6 minus3119861 minus6 minus3119861

minus3119861 211986123119861 119861

2

minus6 3119861 6 3119861

minus3119861 11986123119861 2119861

2

]]]

]

(8a)

[1198961198642] =

1205871198641205831199053(1198992minus 1)

180119877119861 (1 minus 1205832)

[[[

[

36 minus18119861 minus36 minus3119861

minus18119861 41198612

3119861 minus1198612

minus36 3119861 36 18119861

minus3119861 minus119861218119861 4119861

2

]]]

]

(8b)

[1198961198643] =

1205871198641198611199053(1198992minus 1)2

50401198773(1 minus 120583

2)

[[[

[

156 minus22119861 54 13119861

minus22119861 4119861213119861 minus3119861

2

54 13119861 156 22119861

13119861 minus3119861222119861 4119861

2

]]]

]

(8c)

[1198961198644] =

1205871198641199053(1198992minus 1)2

1801198771198611198992(1 + 120583)

[[[

[

36 minus3119861 minus36 minus3119861

minus3119861 411986123119861 minus119861

2

minus36 3119861 36 3119861

minus3119861 minus11986123119861 4119861

2

]]]

]

(8d)

where 119864 is the modulus of elasticity (Nmm2) 120583 is Poissonratio and the other symbols are the same as previous Theequivalent geometric stiffness matrix formula of cylindricalshell element is represented as follows

[119896119866] = [119896

119866119909] + [119896

119866119910] + [119896

119866119909119910] (9)

After integration operation we found that [119896119866119909] and [119896

119866119909119910]

are zero matrix and the computing formula of [119896119866119910] is as

follows

[119896119866119910] = minus

119902119890120587119861 (119899

2+ 1)

420

[[[

[

156 minus22119861 54 13119861

minus22119861 4119861213119861 minus3119861

2

54 13119861 156 22119861

13119861 minus3119861222119861 4119861

2

]]]

]

(10)

where 119902119890 is the radial external pressure of the elementAdopting stiffness integration methods penstock we

can obtain respectively the elastic stiffness matrix andthe equivalent geometrical stiffness matrix of the overallstructure In the structural stiffness equation introducingboundary constraint conditions in the structural stiffnessmatrix crossing out the rows and columns associated withthe displacement constraints we can get the force-balancingequation of the overall penstock as follows

([119870119864] + 119902 [119870

lowast

119866]) Δ = 119875 (11)

where [119870119864] and [119870lowast

119866] are respectively elastic stiffness matrix

and equivalent geometrical stiffness matrix [119870119864] = sum

119890[119870]119890

[119870lowast

119866] = sum119890(119902119890119902)[119870

lowast

119866]119890 and 119890 is element number 119902 is external

pressure of structure Δ and 119875 are respectively vectors ofthe nodal displacement and load of the structure

It is known from structure stability theory that char-acteristic equation to describe structure stability is thatdeterminant of the overall stiffness matrix is equal to zerothat is det([119870

119864] + 119902[119870

lowast

119866]) = 0 Thus stability problem is

transformed to solve the largest eigenvalue problem of realmatrix (minus[119870

119864]minus1[119870lowast

119866]) The reciprocal of the largest eigen-

value is critical stable loadFrom the above discussions we can see that using SA-

FEM to solve the critical pressure of the penstock is verycomplicated Actually in practical engineering design andstructure analysis an analytical explicit formula to describethe relationship between critical pressure and structureparameters is more welcome In order to meet this requestwe provide a realization method based on neural networkFirstly we acquire a group of the samples that adopt SA-FEM to calculate critical pressure of different penstocksThen using nonlinear mapping ability of neural networkto get nonlinear relationship between critical pressure andrelated parameters namely penstockmaterial pipe diameterthickness of the penstock wall the spacing among stiffenerrings and so forth

3 Penstock Critical Pressure CalculatingBased on Neural Network

It is well known that neural network can approach com-plicated nonlinear map with very high accuracy Immunealgorithm is an evolutionary method that reflects immunesystem characteristic of living organisms [8 9] and it canavoid the drawbacks of traditional neural network learningalgorithms The main idea of adopting immune algorithmto design neural network is neural network structures andconnection among neurons as an antibody of biological

4 Mathematical Problems in Engineering

Table 1 Connection relationship table of neural nodes

To From1 2 sdot sdot sdot 8 9

1 times times sdot sdot sdot times times

2 times times sdot sdot sdot times times

3 times times sdot sdot sdot times times

4 11990841

11990842

sdot sdot sdot times times

5 11990851

11990852

sdot sdot sdot times times

6 0 0 sdot sdot sdot times times

7 11990871

0 sdot sdot sdot times times

8 0 0 sdot sdot sdot 11990888

times

9 0 0 sdot sdot sdot 11990898

11990899

1

2

3

5

46

79

8

Input layer

Output signal

Output layer

Input signal

Hidden layer

Figure 2 The schematic diagram of neural network

immune system Selection based on antibodies concentra-tion and self-adaptive mutation operator makes antibodypopulation continuously optimized and finally finds the bestantibodies Immune algorithm is characterized by diversitydistribution of solution group and it can better overcomethe shortcomings of that network structure and cryptic layernumbers defined difficultly

31 Neural Network Design In the paper the neural networkstructure adopted is shown in Figure 2The network neuronshave no significant hierarchical relationship In addition toinput neurons there are no restriction connections amongneurons each network node is assigned a serial number theserial number of node is only used to distinguish beginningand end of directed link [10]

Deletion of Connection Edges If the connection weights valueof a connection edge is less than specified threshold range[minus0001 0001] its weights value is set to zero that is in thesame neuron numbers circumstances different connectionform composes different network structure

Neurons Removed If all weights values of connection with aneuron are less than specified threshold range deleting thisneuron The network structure and connection weights canbe expressed as an equivalent matrix as Table 1 Concate-nating each element of matrix constitutes an antibody Anantibody expresses a neural network structure

32 Design Steps of Neural Network Based onImmune Algorithm

(a) Fitness FunctionThe antibodyw119894 constitutes the objectivefunction of network

119864 (w119894) =119899

sum

119895=1

10038161003816100381610038161003816119905119895minus 119910119895

10038161003816100381610038161003816 (12)

where 119905119895is objective output of network 119910

119895is actual output

and 119899 is sample number in training setsFitness function can be expressed as

119865 (w119894) = 1

119864 (w119894) + 119862 (w119894) (13)

where 119862(w119894) reflect impact of network complexity it isthe sum of network nodes and connections among nodes119862(w119894) = 119904(w119894) + 119897(w119894) 119897(w119894) and 119904(w119894) represent respectivelynetwork connections and network nodes

(b) Immune Selection Algorithm Based on Similarity andVector DistanceAssuming that in a population each antibodycan be represented by a one-dimensional array of119898 elementsantibodies similarity is calculated as follows assuming thatw1 = 119908

1

1 1199081

2 119908

1

119898 and w2 = 119908

2

1 1199082

2 119908

2

119898 are any

two antibodies of an antibody population with size 119899 thesimilarity of w1 and w2 is 119889(w1w2)

119889 (w1w2) = radic119898

sum

119894=1

10038161003816100381610038161199081

119894minus 1199082

119894

1003816100381610038161003816 (14)

where 119899 antibodies constitute a nonempty immune setW thedistance of two antibodies is defined as

119863(w119894) =119899

sum

119895=1119895 = 119894

119889 (w119894w119895) (15)

The concentration of antibody can be expressed asDensity(w

119894)

Density (w119894) = 1

119863 (w119894) (16)

From formula (16) we can see that the more the similarityantibodies the greater the antibody concentration and on thecontrary the smaller the antibodies concentration

The population Update Based on Antibody ConcentrationAfter the parents generated offspring through mutationaccording to the selection probability random selection ofindividuals from the population and offspring constitutes anew population The probability selective function is definedas follows

119875119904(w119894) = 120572 times density (w119894)(1 minus

119863 (w119894)sum119899

119894=1119863(w119894)

)

+ 120573

119865 (w119894)sum119899

119894=1119865 (w119894)

(17)

Mathematical Problems in Engineering 5

where 120572 120573 is adjustable parameter in (0 1) interval its valuedetermined based on experience In this paper alpha andbeta value is set to 05 meaning that antibody concentrationand fitness have equal status during the update process ofantibody population 119865(w119894) is fitness of the antibody 119894

As known from (17) the first part of right side of theequation is based on antibody concentration selection itemsthe higher concentration antibody has little selected chancebut the lower concentration antibody has bigger selectedchance the second part of right side of the equation isbased on antibody fitness selection items the higher fitnessantibody has bigger selected chance

(c) Generate the NewAntibodies Because the network param-eters and the network structure is many to one relationshiptherefore this paper only adopts mutation operation to carryout antibodies update

Defining mutation operator 120590119894= radic1 minus 119865(119908

119894) and using it

mutate all parameters of network as follows

1199081198941015840

119895= 119908119894

119895+ 120590119894times 119873119895 (0 1)

(18)

where 119908119894119895and 119908119894

1015840

119895are respectively antibodies of gene before

and aftermutation and119873119895(0 1) indicates random variable for

each subscript 119895 re-sampling

33 Simulating the Critical External Pressure of Penstock Thispaper uses [7 11 12] proposed calculationmethod to computecritical pressure and regards the computed results as thetraining sample of the network The calculation model isdescribed as follows Penstock material is 16Mn (modulusof elasticity is 119864 = 21 times 10

5MPa Poisson ratio 120583 = 03and 120590

119904= 340MPa) The ratio of penstock radius 119903 and shell

thickness 119905 (relative tube radius 119903119905) is from 20 to 400 steplength is 20 the ratio of rings spacing and tube radius (relativering spacing 119871119903) is [01 02 03 05 08 14 20 30 40]The total calculation models are 180

The transfers function among neurons uses s-function inthe Matlab the transfers function of output layer uses linearfunction Population size is 119873 = 50 The simulating resultsare shown in Table 2

34 Analysis of Calculated Results (1) Figure 3 shows thatwith the increase of 119903119905 the losing stability capability ofpenstock under external pressure is decreased acutely Thecritical external pressure decreases with increasing of 119903119905Within 119903119905 = 20 sim 260 the critical external pressure isacutely decreased beyond the range the change is less Forexample within 119871119903 = 01 sim 30 119903119905 from 20 to 260 119875cr willdecrease to 013sim016 of initial value (119903119905 = 20 119875cr) whenthe diameter of penstock is increased to a certain value thestability power of the penstock under external pressure willchange very small For example if 119871119903 = 30 and 119903119905 = 400then the 119875cr value is only 002MPa

(2)The calculated result shows that the critical pressure119875cr of penstock decreases with relative distance of reinforcingring 119871119903 increasingbut the influence of decreasing velocityis less than relative radius 119903119905 For example 119903119905 = 300

8

6

4

2

0

minus2

minus4

50 100 150 200 250 300 350 400

Mill fnish steel tube Lr = 50

01

02

040610

3050

rt

log(P

cr)

The real line is result of Mises theoryThe point is result of simulation

Figure 3 Simulating results

Table 2 Simulating results of the critical external pressure 119875cr

Critical pressure(simulation)

119871119903

01 02 03 05 08119903119905

20 202426(20242)

90187(9019)

55456(5545)

29914(2991)

17036(1704)

60 11916(11917)

5205(5214)

3197(3189)

1743(1756)

1012(1018)

100 3171(3157)

1379(1371)

850(843)

467(471)

274(277)

140 1323(1338)

575(586)

355(347)

197(201)

116(114)

180 688(692)

299(287)

186(179)

103(103)

061(067)

260 264(257)

116(123)

072(078)

040037)

024(028)

300 182(181)

079(081)

049(051)

028(025)

016(016)

360 113(114)

049(045)

031(029)

017(019)

011(015)

400 086(081)

037(036)

024(023)

013(011)

008(007)

119871119903 = 01 sim 30 and the critical external pressure 119875cr willdecrease to 219sim4341 of initial value

(3) The most effective reinforcing rings spacing and thecurve of Figure 3 shows that reducing rings spacing caneffectively improve the carrying capacity of critical externalpressure of penstock and while 119871119903 decrease 119875cr value andits increase ratio increase For example 119903119905 = 260 119871119903from 30 decrease to 08 and continue to decrease to 01 andthe increment of 119875cr is respectively 018MPa and 240MPaIn other words the reinforcing rings spacing reduces to01119903 and the average increment of 119875cr value is respectively00082MPa and 034MPa The latter is 41 multiples of theformer So the most effective reinforcing ring spacing should

6 Mathematical Problems in Engineering

meet 119871 lt 08119903The reinforcing ring spacing of Chinarsquos SanXiahydropower station is 119871 = 032119903

(4)Noneffective reinforcing rings spacing Figure 3 showsthat along with the stiffener ring spacing increases the role ofstiffening ring is gradually reduced and while 119871119903 = 40 thelog(119875cr) sim 119903119905 curves of reinforcing penstock and mill finishsteel tube (the calculated formula of mill finish steel tube119901 = 2119864(119905119863)

3 119863 is diameter of tube) are almost equal Thisshows that stability against external pressure of the two tubesis roughly equal then the reinforcing ring does not possessany sustaining effect on the stiffness

(5)The losing stability of wave numbers 119899 is a synthesisembodiment for longitudinal and circular stiffness of thepenstock With an increasing of the penstock diameter thestiffness of circular decreases and the losing stability ofwave numbers increases But in 119883-axial direction with anincreasing of the reinforcing ring spacing the stiffness of 119883-axial direction of penstock decreases and also reduces theinstability wave numbers 119899 For fine pitch large diameterstiffened penstock the losing stability of wave shape showsmultiwave form but for the sparse space and small diametershows less wave form

(6) The design of penstock in Figure 3 shows that thecurve cluster of ldquolog(119875cr) sim 119899 119903119905 119871119903rdquo is divided into twoupper and lower districts by the plastic losing stabilitycurve If the penstock diameter and relative reinforcing ringsspacing are bigger the penstock appears elastic losing stabilityunder smaller external pressure log(119875cr) value is locatedin the below district of the elastic losing stability curveWhen the penstock diameter and relative reinforcing ringsspacing are smaller the stability of penstock under externalpressure is powerful and can bear great pressure the log(119875cr)value is located in upon district of the elastic losing stabilitycurve When we design the penstock if 119903119905 value has beendetermined by use construction and so forth we can selectappropriate 119871119903 value according to Figure 3 making thecarrying capacity of penstock critical external pressure meetsnot only the requirements resistance to external pressurestability but also the instability curve as close as possible inorder to achieve full use of the material strength to ensureexternal pressure stability and strength safety coordinatedpurposes

4 Computation of the Stiffening RingrsquosCritical External Pressure

41The Structure Form of Stiffening Ring Thestructure formsof stiffening ring include the cross-section form of ringand the connected mode between ring and tube shell asshown in Figure 4 As for the huge thin-wall penstock itis advisable to adopt the structure of that both stiffeningring and penstock are rolled together as a whole which caneffectively avoid penstock initial defects that caused by weldbead and uneven weld quality The reasonable cross-sectionstructure and dimensions of stiffening ring not only shouldbe able to bear large external load in smaller cross-section sizebut also enable the critical load of tube shell close to or equalto the critical load of stiffening ring effective control range so

aL

b

t

Tube axis

(a) Rectangle ring

b

2a

t

Radius of steel tuber

Tube axis

b2

(b) T-shape ring

Figure 4 The structure form of stiffening ring

as to effectively improve external loads condition of structureand to facilitate construction

On the structure that penstock and stiffening ring arerolled together as a whole the effective control range ofstiffening ring is 078radic119903119905 Computing model is as followspenstock radius r is 6200mm penstock shell thickness 119905is 40mm ring thickness 119886 is respectively 20 40 60 and80 100mm and the variation range of relative ring spacing119871119903 is [01 30] 119887119886 isin [05 50] Calculating separately criticalpressure of rectangle ring and T-shape ring that having thesame cross-section areaThe computing formula of stiffeningring critical external pressure is as follows (calculation resultsare shown in Table 3)

119875cr =3119864119869119896

1198773

119896119871 (19)

where 119877119896is radius that is located in the gravity axis of

stiffening ring effective section (mm) and 119869119896is moment of

inertia that is located in the gravity axis of the stiffened ringeffective section (mm4)

For the rectangle ring

119877119896=(1198862) (119887 + 119905)

2+ 078 119905

2radic119903119905

119886 (119887 + 119905) + 156 119905radic119903119905

119869119896=119886

12(119887 + 119905)

3+ 119886 (119887 + 119905) (119877119896 minus

119887 + 119905

2)

2

+ 013 1199053radic119903119905 + 156 119905radic119903119905(119877

119896minus119905

2)

2

(20)

For the T-shape ring

119877119896=(1198861198874) (119887 minus 119886 + 2119905) + (1198862) (119886 + 119905 + 1198872)

2+ 078 119905

2radic119903119905

(1198861198872) + 119886 (119886 + 119905 + 1198872) + 156 119905radic119903119905

119869119896=1198863119887

24+119886119887

2(119887 minus 119886

2minus 119877119896+ 119905)

2

+119886

12(119887

2+ 119905)

3

+ 119886(119887

2+ 119905) times [

1

2(119887

2+ 119905) minus 119877

119896]

2

+ 013 1199053radic119903119905

+ 156 119905radic119903119905(119877119896minus119905

2)

2

(21)

The calculated results can be plotted as shown in Figure 5

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

According to the strain displacement relations of cylin-drical shell with large deflection we can obtain geometricequation of cylindrical shell element

120576 = 120576119871 + 120576

119873

=

1

119861

120597119880

120597120585minus119911

1198612

1205972119882

1205971205852

1

119877

120597119881

120597120601+2119911

1198772

120597119881

120597120601minus119882

119877minus119911

1198612

1205972119882

1205971206012

1

119877

120597119880

120597120601+1

119861

120597119881

120597120585+2119911

119877119861

120597119881

120597120585minus2119911

119877119861

1205972119882

120597120585120597120601

+

1

21198612(120597119882

120597120585)

2

1

21198772(120597119882

120597120601)

2

1

119877119861

120597119882

120597120585

120597119882

120597120601

(6)

where 120576119871 denotes the linear strain item and 120576119873 denotes thenonlinear strain item resulted by large deflection

According to the constitutive relationships of materialsand equilibriumequations of cylindrical shell element we canset up the matrix equation of the cylindrical shell elementThe stiffnessmatrix is composed of the elastic stiffnessmatrixand the equivalent geometric stiffness matrix The elasticstiffness matrix formula of cylindrical shell element is asfollows

[119896119864] = [119896

1198641] + [119896

1198642] + [119896

1198643] + [119896

1198644] (7)

[1198961198641] =

120587119864119877119905 (121198772+ 11989941199052)

611989941198613(1 minus 120583

2)

[[[

[

6 minus3119861 minus6 minus3119861

minus3119861 211986123119861 119861

2

minus6 3119861 6 3119861

minus3119861 11986123119861 2119861

2

]]]

]

(8a)

[1198961198642] =

1205871198641205831199053(1198992minus 1)

180119877119861 (1 minus 1205832)

[[[

[

36 minus18119861 minus36 minus3119861

minus18119861 41198612

3119861 minus1198612

minus36 3119861 36 18119861

minus3119861 minus119861218119861 4119861

2

]]]

]

(8b)

[1198961198643] =

1205871198641198611199053(1198992minus 1)2

50401198773(1 minus 120583

2)

[[[

[

156 minus22119861 54 13119861

minus22119861 4119861213119861 minus3119861

2

54 13119861 156 22119861

13119861 minus3119861222119861 4119861

2

]]]

]

(8c)

[1198961198644] =

1205871198641199053(1198992minus 1)2

1801198771198611198992(1 + 120583)

[[[

[

36 minus3119861 minus36 minus3119861

minus3119861 411986123119861 minus119861

2

minus36 3119861 36 3119861

minus3119861 minus11986123119861 4119861

2

]]]

]

(8d)

where 119864 is the modulus of elasticity (Nmm2) 120583 is Poissonratio and the other symbols are the same as previous Theequivalent geometric stiffness matrix formula of cylindricalshell element is represented as follows

[119896119866] = [119896

119866119909] + [119896

119866119910] + [119896

119866119909119910] (9)

After integration operation we found that [119896119866119909] and [119896

119866119909119910]

are zero matrix and the computing formula of [119896119866119910] is as

follows

[119896119866119910] = minus

119902119890120587119861 (119899

2+ 1)

420

[[[

[

156 minus22119861 54 13119861

minus22119861 4119861213119861 minus3119861

2

54 13119861 156 22119861

13119861 minus3119861222119861 4119861

2

]]]

]

(10)

where 119902119890 is the radial external pressure of the elementAdopting stiffness integration methods penstock we

can obtain respectively the elastic stiffness matrix andthe equivalent geometrical stiffness matrix of the overallstructure In the structural stiffness equation introducingboundary constraint conditions in the structural stiffnessmatrix crossing out the rows and columns associated withthe displacement constraints we can get the force-balancingequation of the overall penstock as follows

([119870119864] + 119902 [119870

lowast

119866]) Δ = 119875 (11)

where [119870119864] and [119870lowast

119866] are respectively elastic stiffness matrix

and equivalent geometrical stiffness matrix [119870119864] = sum

119890[119870]119890

[119870lowast

119866] = sum119890(119902119890119902)[119870

lowast

119866]119890 and 119890 is element number 119902 is external

pressure of structure Δ and 119875 are respectively vectors ofthe nodal displacement and load of the structure

It is known from structure stability theory that char-acteristic equation to describe structure stability is thatdeterminant of the overall stiffness matrix is equal to zerothat is det([119870

119864] + 119902[119870

lowast

119866]) = 0 Thus stability problem is

transformed to solve the largest eigenvalue problem of realmatrix (minus[119870

119864]minus1[119870lowast

119866]) The reciprocal of the largest eigen-

value is critical stable loadFrom the above discussions we can see that using SA-

FEM to solve the critical pressure of the penstock is verycomplicated Actually in practical engineering design andstructure analysis an analytical explicit formula to describethe relationship between critical pressure and structureparameters is more welcome In order to meet this requestwe provide a realization method based on neural networkFirstly we acquire a group of the samples that adopt SA-FEM to calculate critical pressure of different penstocksThen using nonlinear mapping ability of neural networkto get nonlinear relationship between critical pressure andrelated parameters namely penstockmaterial pipe diameterthickness of the penstock wall the spacing among stiffenerrings and so forth

3 Penstock Critical Pressure CalculatingBased on Neural Network

It is well known that neural network can approach com-plicated nonlinear map with very high accuracy Immunealgorithm is an evolutionary method that reflects immunesystem characteristic of living organisms [8 9] and it canavoid the drawbacks of traditional neural network learningalgorithms The main idea of adopting immune algorithmto design neural network is neural network structures andconnection among neurons as an antibody of biological

4 Mathematical Problems in Engineering

Table 1 Connection relationship table of neural nodes

To From1 2 sdot sdot sdot 8 9

1 times times sdot sdot sdot times times

2 times times sdot sdot sdot times times

3 times times sdot sdot sdot times times

4 11990841

11990842

sdot sdot sdot times times

5 11990851

11990852

sdot sdot sdot times times

6 0 0 sdot sdot sdot times times

7 11990871

0 sdot sdot sdot times times

8 0 0 sdot sdot sdot 11990888

times

9 0 0 sdot sdot sdot 11990898

11990899

1

2

3

5

46

79

8

Input layer

Output signal

Output layer

Input signal

Hidden layer

Figure 2 The schematic diagram of neural network

immune system Selection based on antibodies concentra-tion and self-adaptive mutation operator makes antibodypopulation continuously optimized and finally finds the bestantibodies Immune algorithm is characterized by diversitydistribution of solution group and it can better overcomethe shortcomings of that network structure and cryptic layernumbers defined difficultly

31 Neural Network Design In the paper the neural networkstructure adopted is shown in Figure 2The network neuronshave no significant hierarchical relationship In addition toinput neurons there are no restriction connections amongneurons each network node is assigned a serial number theserial number of node is only used to distinguish beginningand end of directed link [10]

Deletion of Connection Edges If the connection weights valueof a connection edge is less than specified threshold range[minus0001 0001] its weights value is set to zero that is in thesame neuron numbers circumstances different connectionform composes different network structure

Neurons Removed If all weights values of connection with aneuron are less than specified threshold range deleting thisneuron The network structure and connection weights canbe expressed as an equivalent matrix as Table 1 Concate-nating each element of matrix constitutes an antibody Anantibody expresses a neural network structure

32 Design Steps of Neural Network Based onImmune Algorithm

(a) Fitness FunctionThe antibodyw119894 constitutes the objectivefunction of network

119864 (w119894) =119899

sum

119895=1

10038161003816100381610038161003816119905119895minus 119910119895

10038161003816100381610038161003816 (12)

where 119905119895is objective output of network 119910

119895is actual output

and 119899 is sample number in training setsFitness function can be expressed as

119865 (w119894) = 1

119864 (w119894) + 119862 (w119894) (13)

where 119862(w119894) reflect impact of network complexity it isthe sum of network nodes and connections among nodes119862(w119894) = 119904(w119894) + 119897(w119894) 119897(w119894) and 119904(w119894) represent respectivelynetwork connections and network nodes

(b) Immune Selection Algorithm Based on Similarity andVector DistanceAssuming that in a population each antibodycan be represented by a one-dimensional array of119898 elementsantibodies similarity is calculated as follows assuming thatw1 = 119908

1

1 1199081

2 119908

1

119898 and w2 = 119908

2

1 1199082

2 119908

2

119898 are any

two antibodies of an antibody population with size 119899 thesimilarity of w1 and w2 is 119889(w1w2)

119889 (w1w2) = radic119898

sum

119894=1

10038161003816100381610038161199081

119894minus 1199082

119894

1003816100381610038161003816 (14)

where 119899 antibodies constitute a nonempty immune setW thedistance of two antibodies is defined as

119863(w119894) =119899

sum

119895=1119895 = 119894

119889 (w119894w119895) (15)

The concentration of antibody can be expressed asDensity(w

119894)

Density (w119894) = 1

119863 (w119894) (16)

From formula (16) we can see that the more the similarityantibodies the greater the antibody concentration and on thecontrary the smaller the antibodies concentration

The population Update Based on Antibody ConcentrationAfter the parents generated offspring through mutationaccording to the selection probability random selection ofindividuals from the population and offspring constitutes anew population The probability selective function is definedas follows

119875119904(w119894) = 120572 times density (w119894)(1 minus

119863 (w119894)sum119899

119894=1119863(w119894)

)

+ 120573

119865 (w119894)sum119899

119894=1119865 (w119894)

(17)

Mathematical Problems in Engineering 5

where 120572 120573 is adjustable parameter in (0 1) interval its valuedetermined based on experience In this paper alpha andbeta value is set to 05 meaning that antibody concentrationand fitness have equal status during the update process ofantibody population 119865(w119894) is fitness of the antibody 119894

As known from (17) the first part of right side of theequation is based on antibody concentration selection itemsthe higher concentration antibody has little selected chancebut the lower concentration antibody has bigger selectedchance the second part of right side of the equation isbased on antibody fitness selection items the higher fitnessantibody has bigger selected chance

(c) Generate the NewAntibodies Because the network param-eters and the network structure is many to one relationshiptherefore this paper only adopts mutation operation to carryout antibodies update

Defining mutation operator 120590119894= radic1 minus 119865(119908

119894) and using it

mutate all parameters of network as follows

1199081198941015840

119895= 119908119894

119895+ 120590119894times 119873119895 (0 1)

(18)

where 119908119894119895and 119908119894

1015840

119895are respectively antibodies of gene before

and aftermutation and119873119895(0 1) indicates random variable for

each subscript 119895 re-sampling

33 Simulating the Critical External Pressure of Penstock Thispaper uses [7 11 12] proposed calculationmethod to computecritical pressure and regards the computed results as thetraining sample of the network The calculation model isdescribed as follows Penstock material is 16Mn (modulusof elasticity is 119864 = 21 times 10

5MPa Poisson ratio 120583 = 03and 120590

119904= 340MPa) The ratio of penstock radius 119903 and shell

thickness 119905 (relative tube radius 119903119905) is from 20 to 400 steplength is 20 the ratio of rings spacing and tube radius (relativering spacing 119871119903) is [01 02 03 05 08 14 20 30 40]The total calculation models are 180

The transfers function among neurons uses s-function inthe Matlab the transfers function of output layer uses linearfunction Population size is 119873 = 50 The simulating resultsare shown in Table 2

34 Analysis of Calculated Results (1) Figure 3 shows thatwith the increase of 119903119905 the losing stability capability ofpenstock under external pressure is decreased acutely Thecritical external pressure decreases with increasing of 119903119905Within 119903119905 = 20 sim 260 the critical external pressure isacutely decreased beyond the range the change is less Forexample within 119871119903 = 01 sim 30 119903119905 from 20 to 260 119875cr willdecrease to 013sim016 of initial value (119903119905 = 20 119875cr) whenthe diameter of penstock is increased to a certain value thestability power of the penstock under external pressure willchange very small For example if 119871119903 = 30 and 119903119905 = 400then the 119875cr value is only 002MPa

(2)The calculated result shows that the critical pressure119875cr of penstock decreases with relative distance of reinforcingring 119871119903 increasingbut the influence of decreasing velocityis less than relative radius 119903119905 For example 119903119905 = 300

8

6

4

2

0

minus2

minus4

50 100 150 200 250 300 350 400

Mill fnish steel tube Lr = 50

01

02

040610

3050

rt

log(P

cr)

The real line is result of Mises theoryThe point is result of simulation

Figure 3 Simulating results

Table 2 Simulating results of the critical external pressure 119875cr

Critical pressure(simulation)

119871119903

01 02 03 05 08119903119905

20 202426(20242)

90187(9019)

55456(5545)

29914(2991)

17036(1704)

60 11916(11917)

5205(5214)

3197(3189)

1743(1756)

1012(1018)

100 3171(3157)

1379(1371)

850(843)

467(471)

274(277)

140 1323(1338)

575(586)

355(347)

197(201)

116(114)

180 688(692)

299(287)

186(179)

103(103)

061(067)

260 264(257)

116(123)

072(078)

040037)

024(028)

300 182(181)

079(081)

049(051)

028(025)

016(016)

360 113(114)

049(045)

031(029)

017(019)

011(015)

400 086(081)

037(036)

024(023)

013(011)

008(007)

119871119903 = 01 sim 30 and the critical external pressure 119875cr willdecrease to 219sim4341 of initial value

(3) The most effective reinforcing rings spacing and thecurve of Figure 3 shows that reducing rings spacing caneffectively improve the carrying capacity of critical externalpressure of penstock and while 119871119903 decrease 119875cr value andits increase ratio increase For example 119903119905 = 260 119871119903from 30 decrease to 08 and continue to decrease to 01 andthe increment of 119875cr is respectively 018MPa and 240MPaIn other words the reinforcing rings spacing reduces to01119903 and the average increment of 119875cr value is respectively00082MPa and 034MPa The latter is 41 multiples of theformer So the most effective reinforcing ring spacing should

6 Mathematical Problems in Engineering

meet 119871 lt 08119903The reinforcing ring spacing of Chinarsquos SanXiahydropower station is 119871 = 032119903

(4)Noneffective reinforcing rings spacing Figure 3 showsthat along with the stiffener ring spacing increases the role ofstiffening ring is gradually reduced and while 119871119903 = 40 thelog(119875cr) sim 119903119905 curves of reinforcing penstock and mill finishsteel tube (the calculated formula of mill finish steel tube119901 = 2119864(119905119863)

3 119863 is diameter of tube) are almost equal Thisshows that stability against external pressure of the two tubesis roughly equal then the reinforcing ring does not possessany sustaining effect on the stiffness

(5)The losing stability of wave numbers 119899 is a synthesisembodiment for longitudinal and circular stiffness of thepenstock With an increasing of the penstock diameter thestiffness of circular decreases and the losing stability ofwave numbers increases But in 119883-axial direction with anincreasing of the reinforcing ring spacing the stiffness of 119883-axial direction of penstock decreases and also reduces theinstability wave numbers 119899 For fine pitch large diameterstiffened penstock the losing stability of wave shape showsmultiwave form but for the sparse space and small diametershows less wave form

(6) The design of penstock in Figure 3 shows that thecurve cluster of ldquolog(119875cr) sim 119899 119903119905 119871119903rdquo is divided into twoupper and lower districts by the plastic losing stabilitycurve If the penstock diameter and relative reinforcing ringsspacing are bigger the penstock appears elastic losing stabilityunder smaller external pressure log(119875cr) value is locatedin the below district of the elastic losing stability curveWhen the penstock diameter and relative reinforcing ringsspacing are smaller the stability of penstock under externalpressure is powerful and can bear great pressure the log(119875cr)value is located in upon district of the elastic losing stabilitycurve When we design the penstock if 119903119905 value has beendetermined by use construction and so forth we can selectappropriate 119871119903 value according to Figure 3 making thecarrying capacity of penstock critical external pressure meetsnot only the requirements resistance to external pressurestability but also the instability curve as close as possible inorder to achieve full use of the material strength to ensureexternal pressure stability and strength safety coordinatedpurposes

4 Computation of the Stiffening RingrsquosCritical External Pressure

41The Structure Form of Stiffening Ring Thestructure formsof stiffening ring include the cross-section form of ringand the connected mode between ring and tube shell asshown in Figure 4 As for the huge thin-wall penstock itis advisable to adopt the structure of that both stiffeningring and penstock are rolled together as a whole which caneffectively avoid penstock initial defects that caused by weldbead and uneven weld quality The reasonable cross-sectionstructure and dimensions of stiffening ring not only shouldbe able to bear large external load in smaller cross-section sizebut also enable the critical load of tube shell close to or equalto the critical load of stiffening ring effective control range so

aL

b

t

Tube axis

(a) Rectangle ring

b

2a

t

Radius of steel tuber

Tube axis

b2

(b) T-shape ring

Figure 4 The structure form of stiffening ring

as to effectively improve external loads condition of structureand to facilitate construction

On the structure that penstock and stiffening ring arerolled together as a whole the effective control range ofstiffening ring is 078radic119903119905 Computing model is as followspenstock radius r is 6200mm penstock shell thickness 119905is 40mm ring thickness 119886 is respectively 20 40 60 and80 100mm and the variation range of relative ring spacing119871119903 is [01 30] 119887119886 isin [05 50] Calculating separately criticalpressure of rectangle ring and T-shape ring that having thesame cross-section areaThe computing formula of stiffeningring critical external pressure is as follows (calculation resultsare shown in Table 3)

119875cr =3119864119869119896

1198773

119896119871 (19)

where 119877119896is radius that is located in the gravity axis of

stiffening ring effective section (mm) and 119869119896is moment of

inertia that is located in the gravity axis of the stiffened ringeffective section (mm4)

For the rectangle ring

119877119896=(1198862) (119887 + 119905)

2+ 078 119905

2radic119903119905

119886 (119887 + 119905) + 156 119905radic119903119905

119869119896=119886

12(119887 + 119905)

3+ 119886 (119887 + 119905) (119877119896 minus

119887 + 119905

2)

2

+ 013 1199053radic119903119905 + 156 119905radic119903119905(119877

119896minus119905

2)

2

(20)

For the T-shape ring

119877119896=(1198861198874) (119887 minus 119886 + 2119905) + (1198862) (119886 + 119905 + 1198872)

2+ 078 119905

2radic119903119905

(1198861198872) + 119886 (119886 + 119905 + 1198872) + 156 119905radic119903119905

119869119896=1198863119887

24+119886119887

2(119887 minus 119886

2minus 119877119896+ 119905)

2

+119886

12(119887

2+ 119905)

3

+ 119886(119887

2+ 119905) times [

1

2(119887

2+ 119905) minus 119877

119896]

2

+ 013 1199053radic119903119905

+ 156 119905radic119903119905(119877119896minus119905

2)

2

(21)

The calculated results can be plotted as shown in Figure 5

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

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Mathematical Problems in Engineering

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4 Mathematical Problems in Engineering

Table 1 Connection relationship table of neural nodes

To From1 2 sdot sdot sdot 8 9

1 times times sdot sdot sdot times times

2 times times sdot sdot sdot times times

3 times times sdot sdot sdot times times

4 11990841

11990842

sdot sdot sdot times times

5 11990851

11990852

sdot sdot sdot times times

6 0 0 sdot sdot sdot times times

7 11990871

0 sdot sdot sdot times times

8 0 0 sdot sdot sdot 11990888

times

9 0 0 sdot sdot sdot 11990898

11990899

1

2

3

5

46

79

8

Input layer

Output signal

Output layer

Input signal

Hidden layer

Figure 2 The schematic diagram of neural network

immune system Selection based on antibodies concentra-tion and self-adaptive mutation operator makes antibodypopulation continuously optimized and finally finds the bestantibodies Immune algorithm is characterized by diversitydistribution of solution group and it can better overcomethe shortcomings of that network structure and cryptic layernumbers defined difficultly

31 Neural Network Design In the paper the neural networkstructure adopted is shown in Figure 2The network neuronshave no significant hierarchical relationship In addition toinput neurons there are no restriction connections amongneurons each network node is assigned a serial number theserial number of node is only used to distinguish beginningand end of directed link [10]

Deletion of Connection Edges If the connection weights valueof a connection edge is less than specified threshold range[minus0001 0001] its weights value is set to zero that is in thesame neuron numbers circumstances different connectionform composes different network structure

Neurons Removed If all weights values of connection with aneuron are less than specified threshold range deleting thisneuron The network structure and connection weights canbe expressed as an equivalent matrix as Table 1 Concate-nating each element of matrix constitutes an antibody Anantibody expresses a neural network structure

32 Design Steps of Neural Network Based onImmune Algorithm

(a) Fitness FunctionThe antibodyw119894 constitutes the objectivefunction of network

119864 (w119894) =119899

sum

119895=1

10038161003816100381610038161003816119905119895minus 119910119895

10038161003816100381610038161003816 (12)

where 119905119895is objective output of network 119910

119895is actual output

and 119899 is sample number in training setsFitness function can be expressed as

119865 (w119894) = 1

119864 (w119894) + 119862 (w119894) (13)

where 119862(w119894) reflect impact of network complexity it isthe sum of network nodes and connections among nodes119862(w119894) = 119904(w119894) + 119897(w119894) 119897(w119894) and 119904(w119894) represent respectivelynetwork connections and network nodes

(b) Immune Selection Algorithm Based on Similarity andVector DistanceAssuming that in a population each antibodycan be represented by a one-dimensional array of119898 elementsantibodies similarity is calculated as follows assuming thatw1 = 119908

1

1 1199081

2 119908

1

119898 and w2 = 119908

2

1 1199082

2 119908

2

119898 are any

two antibodies of an antibody population with size 119899 thesimilarity of w1 and w2 is 119889(w1w2)

119889 (w1w2) = radic119898

sum

119894=1

10038161003816100381610038161199081

119894minus 1199082

119894

1003816100381610038161003816 (14)

where 119899 antibodies constitute a nonempty immune setW thedistance of two antibodies is defined as

119863(w119894) =119899

sum

119895=1119895 = 119894

119889 (w119894w119895) (15)

The concentration of antibody can be expressed asDensity(w

119894)

Density (w119894) = 1

119863 (w119894) (16)

From formula (16) we can see that the more the similarityantibodies the greater the antibody concentration and on thecontrary the smaller the antibodies concentration

The population Update Based on Antibody ConcentrationAfter the parents generated offspring through mutationaccording to the selection probability random selection ofindividuals from the population and offspring constitutes anew population The probability selective function is definedas follows

119875119904(w119894) = 120572 times density (w119894)(1 minus

119863 (w119894)sum119899

119894=1119863(w119894)

)

+ 120573

119865 (w119894)sum119899

119894=1119865 (w119894)

(17)

Mathematical Problems in Engineering 5

where 120572 120573 is adjustable parameter in (0 1) interval its valuedetermined based on experience In this paper alpha andbeta value is set to 05 meaning that antibody concentrationand fitness have equal status during the update process ofantibody population 119865(w119894) is fitness of the antibody 119894

As known from (17) the first part of right side of theequation is based on antibody concentration selection itemsthe higher concentration antibody has little selected chancebut the lower concentration antibody has bigger selectedchance the second part of right side of the equation isbased on antibody fitness selection items the higher fitnessantibody has bigger selected chance

(c) Generate the NewAntibodies Because the network param-eters and the network structure is many to one relationshiptherefore this paper only adopts mutation operation to carryout antibodies update

Defining mutation operator 120590119894= radic1 minus 119865(119908

119894) and using it

mutate all parameters of network as follows

1199081198941015840

119895= 119908119894

119895+ 120590119894times 119873119895 (0 1)

(18)

where 119908119894119895and 119908119894

1015840

119895are respectively antibodies of gene before

and aftermutation and119873119895(0 1) indicates random variable for

each subscript 119895 re-sampling

33 Simulating the Critical External Pressure of Penstock Thispaper uses [7 11 12] proposed calculationmethod to computecritical pressure and regards the computed results as thetraining sample of the network The calculation model isdescribed as follows Penstock material is 16Mn (modulusof elasticity is 119864 = 21 times 10

5MPa Poisson ratio 120583 = 03and 120590

119904= 340MPa) The ratio of penstock radius 119903 and shell

thickness 119905 (relative tube radius 119903119905) is from 20 to 400 steplength is 20 the ratio of rings spacing and tube radius (relativering spacing 119871119903) is [01 02 03 05 08 14 20 30 40]The total calculation models are 180

The transfers function among neurons uses s-function inthe Matlab the transfers function of output layer uses linearfunction Population size is 119873 = 50 The simulating resultsare shown in Table 2

34 Analysis of Calculated Results (1) Figure 3 shows thatwith the increase of 119903119905 the losing stability capability ofpenstock under external pressure is decreased acutely Thecritical external pressure decreases with increasing of 119903119905Within 119903119905 = 20 sim 260 the critical external pressure isacutely decreased beyond the range the change is less Forexample within 119871119903 = 01 sim 30 119903119905 from 20 to 260 119875cr willdecrease to 013sim016 of initial value (119903119905 = 20 119875cr) whenthe diameter of penstock is increased to a certain value thestability power of the penstock under external pressure willchange very small For example if 119871119903 = 30 and 119903119905 = 400then the 119875cr value is only 002MPa

(2)The calculated result shows that the critical pressure119875cr of penstock decreases with relative distance of reinforcingring 119871119903 increasingbut the influence of decreasing velocityis less than relative radius 119903119905 For example 119903119905 = 300

8

6

4

2

0

minus2

minus4

50 100 150 200 250 300 350 400

Mill fnish steel tube Lr = 50

01

02

040610

3050

rt

log(P

cr)

The real line is result of Mises theoryThe point is result of simulation

Figure 3 Simulating results

Table 2 Simulating results of the critical external pressure 119875cr

Critical pressure(simulation)

119871119903

01 02 03 05 08119903119905

20 202426(20242)

90187(9019)

55456(5545)

29914(2991)

17036(1704)

60 11916(11917)

5205(5214)

3197(3189)

1743(1756)

1012(1018)

100 3171(3157)

1379(1371)

850(843)

467(471)

274(277)

140 1323(1338)

575(586)

355(347)

197(201)

116(114)

180 688(692)

299(287)

186(179)

103(103)

061(067)

260 264(257)

116(123)

072(078)

040037)

024(028)

300 182(181)

079(081)

049(051)

028(025)

016(016)

360 113(114)

049(045)

031(029)

017(019)

011(015)

400 086(081)

037(036)

024(023)

013(011)

008(007)

119871119903 = 01 sim 30 and the critical external pressure 119875cr willdecrease to 219sim4341 of initial value

(3) The most effective reinforcing rings spacing and thecurve of Figure 3 shows that reducing rings spacing caneffectively improve the carrying capacity of critical externalpressure of penstock and while 119871119903 decrease 119875cr value andits increase ratio increase For example 119903119905 = 260 119871119903from 30 decrease to 08 and continue to decrease to 01 andthe increment of 119875cr is respectively 018MPa and 240MPaIn other words the reinforcing rings spacing reduces to01119903 and the average increment of 119875cr value is respectively00082MPa and 034MPa The latter is 41 multiples of theformer So the most effective reinforcing ring spacing should

6 Mathematical Problems in Engineering

meet 119871 lt 08119903The reinforcing ring spacing of Chinarsquos SanXiahydropower station is 119871 = 032119903

(4)Noneffective reinforcing rings spacing Figure 3 showsthat along with the stiffener ring spacing increases the role ofstiffening ring is gradually reduced and while 119871119903 = 40 thelog(119875cr) sim 119903119905 curves of reinforcing penstock and mill finishsteel tube (the calculated formula of mill finish steel tube119901 = 2119864(119905119863)

3 119863 is diameter of tube) are almost equal Thisshows that stability against external pressure of the two tubesis roughly equal then the reinforcing ring does not possessany sustaining effect on the stiffness

(5)The losing stability of wave numbers 119899 is a synthesisembodiment for longitudinal and circular stiffness of thepenstock With an increasing of the penstock diameter thestiffness of circular decreases and the losing stability ofwave numbers increases But in 119883-axial direction with anincreasing of the reinforcing ring spacing the stiffness of 119883-axial direction of penstock decreases and also reduces theinstability wave numbers 119899 For fine pitch large diameterstiffened penstock the losing stability of wave shape showsmultiwave form but for the sparse space and small diametershows less wave form

(6) The design of penstock in Figure 3 shows that thecurve cluster of ldquolog(119875cr) sim 119899 119903119905 119871119903rdquo is divided into twoupper and lower districts by the plastic losing stabilitycurve If the penstock diameter and relative reinforcing ringsspacing are bigger the penstock appears elastic losing stabilityunder smaller external pressure log(119875cr) value is locatedin the below district of the elastic losing stability curveWhen the penstock diameter and relative reinforcing ringsspacing are smaller the stability of penstock under externalpressure is powerful and can bear great pressure the log(119875cr)value is located in upon district of the elastic losing stabilitycurve When we design the penstock if 119903119905 value has beendetermined by use construction and so forth we can selectappropriate 119871119903 value according to Figure 3 making thecarrying capacity of penstock critical external pressure meetsnot only the requirements resistance to external pressurestability but also the instability curve as close as possible inorder to achieve full use of the material strength to ensureexternal pressure stability and strength safety coordinatedpurposes

4 Computation of the Stiffening RingrsquosCritical External Pressure

41The Structure Form of Stiffening Ring Thestructure formsof stiffening ring include the cross-section form of ringand the connected mode between ring and tube shell asshown in Figure 4 As for the huge thin-wall penstock itis advisable to adopt the structure of that both stiffeningring and penstock are rolled together as a whole which caneffectively avoid penstock initial defects that caused by weldbead and uneven weld quality The reasonable cross-sectionstructure and dimensions of stiffening ring not only shouldbe able to bear large external load in smaller cross-section sizebut also enable the critical load of tube shell close to or equalto the critical load of stiffening ring effective control range so

aL

b

t

Tube axis

(a) Rectangle ring

b

2a

t

Radius of steel tuber

Tube axis

b2

(b) T-shape ring

Figure 4 The structure form of stiffening ring

as to effectively improve external loads condition of structureand to facilitate construction

On the structure that penstock and stiffening ring arerolled together as a whole the effective control range ofstiffening ring is 078radic119903119905 Computing model is as followspenstock radius r is 6200mm penstock shell thickness 119905is 40mm ring thickness 119886 is respectively 20 40 60 and80 100mm and the variation range of relative ring spacing119871119903 is [01 30] 119887119886 isin [05 50] Calculating separately criticalpressure of rectangle ring and T-shape ring that having thesame cross-section areaThe computing formula of stiffeningring critical external pressure is as follows (calculation resultsare shown in Table 3)

119875cr =3119864119869119896

1198773

119896119871 (19)

where 119877119896is radius that is located in the gravity axis of

stiffening ring effective section (mm) and 119869119896is moment of

inertia that is located in the gravity axis of the stiffened ringeffective section (mm4)

For the rectangle ring

119877119896=(1198862) (119887 + 119905)

2+ 078 119905

2radic119903119905

119886 (119887 + 119905) + 156 119905radic119903119905

119869119896=119886

12(119887 + 119905)

3+ 119886 (119887 + 119905) (119877119896 minus

119887 + 119905

2)

2

+ 013 1199053radic119903119905 + 156 119905radic119903119905(119877

119896minus119905

2)

2

(20)

For the T-shape ring

119877119896=(1198861198874) (119887 minus 119886 + 2119905) + (1198862) (119886 + 119905 + 1198872)

2+ 078 119905

2radic119903119905

(1198861198872) + 119886 (119886 + 119905 + 1198872) + 156 119905radic119903119905

119869119896=1198863119887

24+119886119887

2(119887 minus 119886

2minus 119877119896+ 119905)

2

+119886

12(119887

2+ 119905)

3

+ 119886(119887

2+ 119905) times [

1

2(119887

2+ 119905) minus 119877

119896]

2

+ 013 1199053radic119903119905

+ 156 119905radic119903119905(119877119896minus119905

2)

2

(21)

The calculated results can be plotted as shown in Figure 5

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

where 120572 120573 is adjustable parameter in (0 1) interval its valuedetermined based on experience In this paper alpha andbeta value is set to 05 meaning that antibody concentrationand fitness have equal status during the update process ofantibody population 119865(w119894) is fitness of the antibody 119894

As known from (17) the first part of right side of theequation is based on antibody concentration selection itemsthe higher concentration antibody has little selected chancebut the lower concentration antibody has bigger selectedchance the second part of right side of the equation isbased on antibody fitness selection items the higher fitnessantibody has bigger selected chance

(c) Generate the NewAntibodies Because the network param-eters and the network structure is many to one relationshiptherefore this paper only adopts mutation operation to carryout antibodies update

Defining mutation operator 120590119894= radic1 minus 119865(119908

119894) and using it

mutate all parameters of network as follows

1199081198941015840

119895= 119908119894

119895+ 120590119894times 119873119895 (0 1)

(18)

where 119908119894119895and 119908119894

1015840

119895are respectively antibodies of gene before

and aftermutation and119873119895(0 1) indicates random variable for

each subscript 119895 re-sampling

33 Simulating the Critical External Pressure of Penstock Thispaper uses [7 11 12] proposed calculationmethod to computecritical pressure and regards the computed results as thetraining sample of the network The calculation model isdescribed as follows Penstock material is 16Mn (modulusof elasticity is 119864 = 21 times 10

5MPa Poisson ratio 120583 = 03and 120590

119904= 340MPa) The ratio of penstock radius 119903 and shell

thickness 119905 (relative tube radius 119903119905) is from 20 to 400 steplength is 20 the ratio of rings spacing and tube radius (relativering spacing 119871119903) is [01 02 03 05 08 14 20 30 40]The total calculation models are 180

The transfers function among neurons uses s-function inthe Matlab the transfers function of output layer uses linearfunction Population size is 119873 = 50 The simulating resultsare shown in Table 2

34 Analysis of Calculated Results (1) Figure 3 shows thatwith the increase of 119903119905 the losing stability capability ofpenstock under external pressure is decreased acutely Thecritical external pressure decreases with increasing of 119903119905Within 119903119905 = 20 sim 260 the critical external pressure isacutely decreased beyond the range the change is less Forexample within 119871119903 = 01 sim 30 119903119905 from 20 to 260 119875cr willdecrease to 013sim016 of initial value (119903119905 = 20 119875cr) whenthe diameter of penstock is increased to a certain value thestability power of the penstock under external pressure willchange very small For example if 119871119903 = 30 and 119903119905 = 400then the 119875cr value is only 002MPa

(2)The calculated result shows that the critical pressure119875cr of penstock decreases with relative distance of reinforcingring 119871119903 increasingbut the influence of decreasing velocityis less than relative radius 119903119905 For example 119903119905 = 300

8

6

4

2

0

minus2

minus4

50 100 150 200 250 300 350 400

Mill fnish steel tube Lr = 50

01

02

040610

3050

rt

log(P

cr)

The real line is result of Mises theoryThe point is result of simulation

Figure 3 Simulating results

Table 2 Simulating results of the critical external pressure 119875cr

Critical pressure(simulation)

119871119903

01 02 03 05 08119903119905

20 202426(20242)

90187(9019)

55456(5545)

29914(2991)

17036(1704)

60 11916(11917)

5205(5214)

3197(3189)

1743(1756)

1012(1018)

100 3171(3157)

1379(1371)

850(843)

467(471)

274(277)

140 1323(1338)

575(586)

355(347)

197(201)

116(114)

180 688(692)

299(287)

186(179)

103(103)

061(067)

260 264(257)

116(123)

072(078)

040037)

024(028)

300 182(181)

079(081)

049(051)

028(025)

016(016)

360 113(114)

049(045)

031(029)

017(019)

011(015)

400 086(081)

037(036)

024(023)

013(011)

008(007)

119871119903 = 01 sim 30 and the critical external pressure 119875cr willdecrease to 219sim4341 of initial value

(3) The most effective reinforcing rings spacing and thecurve of Figure 3 shows that reducing rings spacing caneffectively improve the carrying capacity of critical externalpressure of penstock and while 119871119903 decrease 119875cr value andits increase ratio increase For example 119903119905 = 260 119871119903from 30 decrease to 08 and continue to decrease to 01 andthe increment of 119875cr is respectively 018MPa and 240MPaIn other words the reinforcing rings spacing reduces to01119903 and the average increment of 119875cr value is respectively00082MPa and 034MPa The latter is 41 multiples of theformer So the most effective reinforcing ring spacing should

6 Mathematical Problems in Engineering

meet 119871 lt 08119903The reinforcing ring spacing of Chinarsquos SanXiahydropower station is 119871 = 032119903

(4)Noneffective reinforcing rings spacing Figure 3 showsthat along with the stiffener ring spacing increases the role ofstiffening ring is gradually reduced and while 119871119903 = 40 thelog(119875cr) sim 119903119905 curves of reinforcing penstock and mill finishsteel tube (the calculated formula of mill finish steel tube119901 = 2119864(119905119863)

3 119863 is diameter of tube) are almost equal Thisshows that stability against external pressure of the two tubesis roughly equal then the reinforcing ring does not possessany sustaining effect on the stiffness

(5)The losing stability of wave numbers 119899 is a synthesisembodiment for longitudinal and circular stiffness of thepenstock With an increasing of the penstock diameter thestiffness of circular decreases and the losing stability ofwave numbers increases But in 119883-axial direction with anincreasing of the reinforcing ring spacing the stiffness of 119883-axial direction of penstock decreases and also reduces theinstability wave numbers 119899 For fine pitch large diameterstiffened penstock the losing stability of wave shape showsmultiwave form but for the sparse space and small diametershows less wave form

(6) The design of penstock in Figure 3 shows that thecurve cluster of ldquolog(119875cr) sim 119899 119903119905 119871119903rdquo is divided into twoupper and lower districts by the plastic losing stabilitycurve If the penstock diameter and relative reinforcing ringsspacing are bigger the penstock appears elastic losing stabilityunder smaller external pressure log(119875cr) value is locatedin the below district of the elastic losing stability curveWhen the penstock diameter and relative reinforcing ringsspacing are smaller the stability of penstock under externalpressure is powerful and can bear great pressure the log(119875cr)value is located in upon district of the elastic losing stabilitycurve When we design the penstock if 119903119905 value has beendetermined by use construction and so forth we can selectappropriate 119871119903 value according to Figure 3 making thecarrying capacity of penstock critical external pressure meetsnot only the requirements resistance to external pressurestability but also the instability curve as close as possible inorder to achieve full use of the material strength to ensureexternal pressure stability and strength safety coordinatedpurposes

4 Computation of the Stiffening RingrsquosCritical External Pressure

41The Structure Form of Stiffening Ring Thestructure formsof stiffening ring include the cross-section form of ringand the connected mode between ring and tube shell asshown in Figure 4 As for the huge thin-wall penstock itis advisable to adopt the structure of that both stiffeningring and penstock are rolled together as a whole which caneffectively avoid penstock initial defects that caused by weldbead and uneven weld quality The reasonable cross-sectionstructure and dimensions of stiffening ring not only shouldbe able to bear large external load in smaller cross-section sizebut also enable the critical load of tube shell close to or equalto the critical load of stiffening ring effective control range so

aL

b

t

Tube axis

(a) Rectangle ring

b

2a

t

Radius of steel tuber

Tube axis

b2

(b) T-shape ring

Figure 4 The structure form of stiffening ring

as to effectively improve external loads condition of structureand to facilitate construction

On the structure that penstock and stiffening ring arerolled together as a whole the effective control range ofstiffening ring is 078radic119903119905 Computing model is as followspenstock radius r is 6200mm penstock shell thickness 119905is 40mm ring thickness 119886 is respectively 20 40 60 and80 100mm and the variation range of relative ring spacing119871119903 is [01 30] 119887119886 isin [05 50] Calculating separately criticalpressure of rectangle ring and T-shape ring that having thesame cross-section areaThe computing formula of stiffeningring critical external pressure is as follows (calculation resultsare shown in Table 3)

119875cr =3119864119869119896

1198773

119896119871 (19)

where 119877119896is radius that is located in the gravity axis of

stiffening ring effective section (mm) and 119869119896is moment of

inertia that is located in the gravity axis of the stiffened ringeffective section (mm4)

For the rectangle ring

119877119896=(1198862) (119887 + 119905)

2+ 078 119905

2radic119903119905

119886 (119887 + 119905) + 156 119905radic119903119905

119869119896=119886

12(119887 + 119905)

3+ 119886 (119887 + 119905) (119877119896 minus

119887 + 119905

2)

2

+ 013 1199053radic119903119905 + 156 119905radic119903119905(119877

119896minus119905

2)

2

(20)

For the T-shape ring

119877119896=(1198861198874) (119887 minus 119886 + 2119905) + (1198862) (119886 + 119905 + 1198872)

2+ 078 119905

2radic119903119905

(1198861198872) + 119886 (119886 + 119905 + 1198872) + 156 119905radic119903119905

119869119896=1198863119887

24+119886119887

2(119887 minus 119886

2minus 119877119896+ 119905)

2

+119886

12(119887

2+ 119905)

3

+ 119886(119887

2+ 119905) times [

1

2(119887

2+ 119905) minus 119877

119896]

2

+ 013 1199053radic119903119905

+ 156 119905radic119903119905(119877119896minus119905

2)

2

(21)

The calculated results can be plotted as shown in Figure 5

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

meet 119871 lt 08119903The reinforcing ring spacing of Chinarsquos SanXiahydropower station is 119871 = 032119903

(4)Noneffective reinforcing rings spacing Figure 3 showsthat along with the stiffener ring spacing increases the role ofstiffening ring is gradually reduced and while 119871119903 = 40 thelog(119875cr) sim 119903119905 curves of reinforcing penstock and mill finishsteel tube (the calculated formula of mill finish steel tube119901 = 2119864(119905119863)

3 119863 is diameter of tube) are almost equal Thisshows that stability against external pressure of the two tubesis roughly equal then the reinforcing ring does not possessany sustaining effect on the stiffness

(5)The losing stability of wave numbers 119899 is a synthesisembodiment for longitudinal and circular stiffness of thepenstock With an increasing of the penstock diameter thestiffness of circular decreases and the losing stability ofwave numbers increases But in 119883-axial direction with anincreasing of the reinforcing ring spacing the stiffness of 119883-axial direction of penstock decreases and also reduces theinstability wave numbers 119899 For fine pitch large diameterstiffened penstock the losing stability of wave shape showsmultiwave form but for the sparse space and small diametershows less wave form

(6) The design of penstock in Figure 3 shows that thecurve cluster of ldquolog(119875cr) sim 119899 119903119905 119871119903rdquo is divided into twoupper and lower districts by the plastic losing stabilitycurve If the penstock diameter and relative reinforcing ringsspacing are bigger the penstock appears elastic losing stabilityunder smaller external pressure log(119875cr) value is locatedin the below district of the elastic losing stability curveWhen the penstock diameter and relative reinforcing ringsspacing are smaller the stability of penstock under externalpressure is powerful and can bear great pressure the log(119875cr)value is located in upon district of the elastic losing stabilitycurve When we design the penstock if 119903119905 value has beendetermined by use construction and so forth we can selectappropriate 119871119903 value according to Figure 3 making thecarrying capacity of penstock critical external pressure meetsnot only the requirements resistance to external pressurestability but also the instability curve as close as possible inorder to achieve full use of the material strength to ensureexternal pressure stability and strength safety coordinatedpurposes

4 Computation of the Stiffening RingrsquosCritical External Pressure

41The Structure Form of Stiffening Ring Thestructure formsof stiffening ring include the cross-section form of ringand the connected mode between ring and tube shell asshown in Figure 4 As for the huge thin-wall penstock itis advisable to adopt the structure of that both stiffeningring and penstock are rolled together as a whole which caneffectively avoid penstock initial defects that caused by weldbead and uneven weld quality The reasonable cross-sectionstructure and dimensions of stiffening ring not only shouldbe able to bear large external load in smaller cross-section sizebut also enable the critical load of tube shell close to or equalto the critical load of stiffening ring effective control range so

aL

b

t

Tube axis

(a) Rectangle ring

b

2a

t

Radius of steel tuber

Tube axis

b2

(b) T-shape ring

Figure 4 The structure form of stiffening ring

as to effectively improve external loads condition of structureand to facilitate construction

On the structure that penstock and stiffening ring arerolled together as a whole the effective control range ofstiffening ring is 078radic119903119905 Computing model is as followspenstock radius r is 6200mm penstock shell thickness 119905is 40mm ring thickness 119886 is respectively 20 40 60 and80 100mm and the variation range of relative ring spacing119871119903 is [01 30] 119887119886 isin [05 50] Calculating separately criticalpressure of rectangle ring and T-shape ring that having thesame cross-section areaThe computing formula of stiffeningring critical external pressure is as follows (calculation resultsare shown in Table 3)

119875cr =3119864119869119896

1198773

119896119871 (19)

where 119877119896is radius that is located in the gravity axis of

stiffening ring effective section (mm) and 119869119896is moment of

inertia that is located in the gravity axis of the stiffened ringeffective section (mm4)

For the rectangle ring

119877119896=(1198862) (119887 + 119905)

2+ 078 119905

2radic119903119905

119886 (119887 + 119905) + 156 119905radic119903119905

119869119896=119886

12(119887 + 119905)

3+ 119886 (119887 + 119905) (119877119896 minus

119887 + 119905

2)

2

+ 013 1199053radic119903119905 + 156 119905radic119903119905(119877

119896minus119905

2)

2

(20)

For the T-shape ring

119877119896=(1198861198874) (119887 minus 119886 + 2119905) + (1198862) (119886 + 119905 + 1198872)

2+ 078 119905

2radic119903119905

(1198861198872) + 119886 (119886 + 119905 + 1198872) + 156 119905radic119903119905

119869119896=1198863119887

24+119886119887

2(119887 minus 119886

2minus 119877119896+ 119905)

2

+119886

12(119887

2+ 119905)

3

+ 119886(119887

2+ 119905) times [

1

2(119887

2+ 119905) minus 119877

119896]

2

+ 013 1199053radic119903119905

+ 156 119905radic119903119905(119877119896minus119905

2)

2

(21)

The calculated results can be plotted as shown in Figure 5

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50

log(P

cr)

(a)

15

14

13

12

11

10

9

8

ab

02

03

05

070912162030

Lr = 01

0 10 20 30 40 50lo

g(P

cr)

(b)

Figure 5 log(119875cr) sim 119887119886 of the rectangular ring and T-shape ring

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

2

1

0

Pcr

(MPa

)

0 20 40 60

ba

a = 20mm

a = 40mm

a = 80mma = 100mm

a = 60mm

a = 30mm

Figure 6 119875cr minus 119887119886 curve with the different 119897119903 and 119886

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

Table 3 The part of calculation results of rectangle ring and T-shape ring (119886 = 60mm)

119875cr (MPa)119887119886

Rectangle shaped ring T-shaped ring05 11 65 165 05 11 65 165

119871119903

01 6575 9591 7468 2509 3882 4530 5741 248403 2192 3197 2489 837 1294 1510 1914 82820 328 479 373 125 194 226 287 12430 219 319 249 84 129 151 191 83

Table 4 Appropriate range of 119887119886 value in the conditions ofdifferent 119886

119886 (mm) With maximum 119875crcorresponding 119887119886 Appropriate range of 119887119886

20 127 10ndash1430 69 5ndash840 45 35ndash5560 25 15ndash3580 17 15ndash25

42 Calculated Result Analysis

(a) Stiffening Ring Reasonable Cross-Section Form By com-paring log(119875cr) sim 119887119886 curve of rectangular ring and T-shapering in Figure 5 we can see the ring with the same ringsspacing and cross-section area the rectangular ring possessesbigger 119875cr The smaller the 119871119903 is the greater this effect isFor example for 119886 = 60mm 119887 = 990mm 119871119903 = 01 and119871119903 = 30 the critical external pressure 119875cr of rectangularring and T-shape ring are respectively (2509 836)MPa and(2484 828)MPa While 119871119903 = 01 119875cr difference betweenrectangular ring and T-shape ring is 257MPa but 119871119903 =

30119875cr difference between rectangular ring and T-shape ringis 0085MPa It is thus clear that adopting small spacingrectangular ring ismore reasonable and themanufacture andbuilding construction are more convenient

(b) Stiffening Ring Appropriate Size The computed resultsshow the variation trend of critical external pressures ofstiffening ring with 119886 and 119887119886 For the different 119886 the risinginterval of 119875cr with 119887119886 is different Tables 4 and 5 showthe appropriate range of 119887119886 and the critical pressure in theconditions of different 119886

Figure 6 illustrates that under the same stiffening ringthickness the upper limit of 119875cr rising interval is unchangedand it has no relation with the relative ring spacing Forexample when 119886 is 40mm and 119871119903 isin [01 30] thecorresponding 119887119886 with the maximum 119875cr is 45

(c) Coupling Rule and Its Application It can be seen fromcalculated results that stiffening rings with different cross sec-tion sizeslayout spacingtheir bucking curves have couplingphenomenon Therefore by adjusting cross section sizes andlayout spacing of stiffening rings the antibuckling capacity

of penstock and stiffening ring can be coordinated to theoptimal state

5 Case Study

51 Setting of Computation Conditions The computationconditions of external pressure stability of embedded stiff-ened penstock in the Yachi river hydropower station includemaintenance working conditions and constructing condi-tions In the maintenance working conditions normal exter-nal pressurewater head119867

1is 50m In the checking condition

external pressure water head1198672is 80m In the constructing

condition grouting pressure of concrete 119875 is 03MPa In theabove computing conditions themaximumexternal pressurewater head is 080MPa (119867

2= 80m) design external load

is 119870 times 080MPa 119870 is safety coefficient and 119870 is 18 In thiscase the design external pressure of penstock is 18 times 080 =144MPa

52 Stability Design of the Penstock The penstock stabilityanalysis and design was respectively carried out by Mises [1]Lai and Fang [2] and Liu and Ma [7] Among them thestiffening ring stability design method adopts formula (19) tocompute The calculate results are shown in Table 6

Adopting semianalytical finite element method to designpenstock separately considers two situations of that simplesupported role of stiffening ring and clamped role of stiffen-ing ring

The inside radius of penstock is 25m stiffening ringspacing 119897 is 20m the penstock material is 16Mn (elasticmodulus 119864 is 210GPa Poisson ratio 120583 is 03 and yieldstrength120590 is 325MPa) and the initial crack between penstockshell and its outside concrete Δ is 05mm Using the aboveseveral calculation methods obtain the calculation results(shown in Table 5) of external pressure stability of embeddedpenstock on Chinarsquos YACIHE hydropower station

The computed results show that Mises method compu-tational results are basically situated between two computa-tional results that calculated by semianalytical finite elementmethod (two support forms of stiffened ring) Simulatedresults are close to the computational results of semianalyticalfinite element method with clamped stiffening ring

Therefore Mises calculation method can be used as mainmethod of stiffening penstock stability design under theexternal pressure Reference [2] method computed results

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

Table 5 Part of computing results of 119875cr in the case of the different parameters 119886

119875cr (MPa) 119887119886

119871119903 119886 (mm) 05 15 19 21 23 29 31 55

01

20 54301 57915 60817 62609 64629 72015 74892 1195930 55804 68273 77208 82319 87752 10512 11099 1619640 5812 8435 9944 10699 11427 13296 13786 1480960 6575 11529 12803 13166 13364 13206 12979 886480 7661 12345 12230 11927 11534 10135 9663 5629100 8816 11397 10364 97814 9206 7664 7225 4126

05

20 10860 11583 12163 12521 12926 14403 14978 2391930 11160 13664 5441 16464 17550 21025 22198 3239240 1162 16871 1989 2139 2285 2659 2757 296260 1315 2306 2560 2633 2673 2641 2596 1772880 1532 2469 2446 2385 2307 2027 19326 1126100 1763 2279 2073 1956 1841 1533 1445 825

09

20 6033 6435 6757 6756 7181 8002 8321 1328830 6200 7586 8579 946 9750 1680 12332 1799540 646 937 1105 1189 1269 1477 1532 164560 731 1281 1423 1463 1485 1467 1442 98580 851 1372 1359 1325 1282 1126 1074 625100 979 1266 1152 1087 1023 852 803 458

30

20 1801 1931 2027 2087 2154 2400 2496 398630 1860 2276 2574 2744 2925 3504 3699 539940 194 281 331 357 381 443 459 49460 219 384 427 439 445 440 433 29580 255 412 408 398 384 338 322 188100 294 379 345 326 306 255 241 138

Table 6 Computational results of various calculation methods

Shell thickness

Mises method Semianalytical finiteelement method Lai-Fan method Proposed method

in this paperStiffening ring assimple supported

Stiffening ring asfixed supported

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

119875crMPa 119870

20 1529 191 0859 107 1595 199 2716 340 1579 19722 1965 246 1144 143 2123 265 3462 433 2273 28425 2690 336 1678 210 3115 389 4795 599 2916 36427 3266 408 2114 264 3924 491 5846 731 3784 47329 3926 491 2619 327 4862 608 7031 879 4592 574

Table 7 The calculation results of stiffening ring stability

Shell-thickness (mm) Ring-thickness (mm) Ring plate-high (mm) Stiffening ring instabilityInstability pressure (MPa) 119870

20 20 300 2380 29822 22 300 2685 33625 25 300 3159 39527 27 300 3468 43629 27 300 3823 478

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

have the great deviation Meanwhile this paper method isalso validated

The computed results also illustrate that when pen-stock shell thickness respectively is 20mm and 22mm thesafe coefficient calculated by semianalytical finite elementmethod (stiffening ring played simple-supported function)is less than 18 and cannot meet the requirements Howeverthis method considers penstock resistant external pressurecapability in the case of the stiffening ring bucking butactually penstock resistant external pressure capability shouldbe higher than this value The safety coefficient of othermethod computational results is greater than 18 and meetthe requirements For security purposes penstock shell thick-ness respectively is 20mmand 22mmand the stiffening ringspacing can be adjusted to 15m at this time the calculatedresults by semianalytical finite element method (stiffeningring played simple-supported function) respectively are1557MPa (safety coefficient 195) and 2072MPa (safetycoefficient 259) and meet the requirement

53 The Buckling Analysis of Stiffening Ring Set the sizeof stiffening ring using formula (19) to calculate criticalexternal pressure of the stiffened ringThe calculation resultsare shown in Table 7

From Table 7 we can see that when the penstock shellradius is 25m stiffening rings spacing is 20m penstockthickness is 002m and the stiffening ring height is 03m thecomputational result of 119875cr is 2380MPa 119875cr = 2380MPa gt144MPa (119870 times 08) therefore the stiffening ring is stablewhen penstock shell thickness is 0022m the critical externalpressure of stiffening ring is 2685MPa and the stiffening ringstability is more reliable

6 Conclusions

This paper analyzed characteristics and drawbacks of dif-ferent calculation methods of penstock external pressurestability problem and proposed a simulation calculationmethod based on immune network Caculation exampledemonstrates the feasibility of the method The methodprovides a newdesign approach for embedded stiffening pen-stock external pressure stability problem in the hydropowerstation building engineering The main conclusions are asfollows

(1) By analyzing the shortcomings of various calculationmethods of that stiffening penstock external pressure stabilityproblem in the current design of hydropower penstockthis paper presented simulating model of the problem Insimulation solving process this paper adopts the immuneevolutionary programming designed neural network andeffectively overcomes shortcomings of hidden layer neuronsand network structure which is difficult to determine in thetraditional BP network increasing convergence speed andimproving global convergence capacity of the network Bycomparing the results calculated by this paper calculationmethod and Mises calculation method (see Table 2 andFigure 3) we verify the calculation accuracy of this paperpresented algorithm

(2) The results reveal that different section sizes anddifferent layout spacing stiffening rings their critical pressurebucking curves have coupling phenomena Therefore inexternal pressure stability design of stiffening penstock wecan appropriately adjust stiffening rings sectional size param-eters stiffening rings spacing and stiffening ring sectionstructure to make the bearing capacity of stiffening ringspenstock bearing capacity coordination penstock externalpressure stability and its strength safety coordinated

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by Research Programs forInnovative Research Team (in Science andTechnology) of theHenan University (Grant no 13IRTSTHN023) InnovationScientists and Technicians Troop Construction Projects ofZhengzhou City (Grant no 131PCXTD595) and Public Wel-fare industry Special Funding Research Project ofMinistry ofWater Resources (Grant no 201101009)

References

[1] The Ministry of Water Resources of the Peoplersquos Republic ofChina ldquoHydroelectric power station penstock design specifica-tionrdquo 2003

[2] H J Lai and C R Fang ldquoThe investigate of losing stabilityand breach for embedding reinforcing rings penstock underexternal pressurerdquo Journal of Hydraulic Engineering vol 12 pp30ndash36 1990

[3] E Amstutz ldquoBuckling of pressure-shaft and tunnel liningsrdquoInternational Water Power and Dam Construction vol 22 no11 pp 391ndash399 1970

[4] S Jacobsen ldquoBuckling of pressure tunnel steel linings with shearconnectorsrdquo International Water Power and Dam Constructionvol 20 no 6 pp 58ndash62 1968

[5] S Jacobsen ldquoPressure distribution in steel lined rock tunnelsand shaftsrdquo International Water Power and Dam Constructionvol 29 no 12 pp 47ndash51 1977

[6] F M Svoisky and A R Freishist ldquoExternal pressure analysis forembedded steel penstocksrdquo InternationalWater Power andDamConstruction vol 44 no 1 pp 37ndash43 1992

[7] D C Liu and W Y Ma ldquoResearch on the semi-analyticalfinite element methods analyzing stability of cylindrical shellsexternal pressurerdquo inProceedings of the International Conferenceon structural Engineering and Computer pp 413ndash417 1990

[8] M Srinivas and L M Patnaik ldquoAdaptive probabilities ofcrossover and mutation in genetic algorithmsrdquo IEEE Transac-tions on Systems Man and Cybernetics vol 24 no 4 pp 656ndash667 1994

[9] X-B Cao K-S Liu and X-F Wang ldquoDesign multilayer feed-forward networks based on immune evolutionary program-mingrdquo Journal of Software vol 10 no 11 pp 1180ndash1184 1999

[10] M THagan andM BMenhaj ldquoTraining feedforward networkswith the Marquardt algorithmrdquo IEEE Transactions on NeuralNetworks vol 5 no 6 pp 989ndash993 1994

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

[11] W-S Dong C-H Dang Z-C Deng and D-C Liu ldquoStabilityanalysis of penstock under external pressure based on GA-NNalgorithmsrdquoChinese Journal of AppliedMechanics vol 23 no 2pp 304ndash307 2006

[12] W-S Dong Z-C Deng D-C Liu and Y Liang ldquoA relativelysimple and fast method for calculating critical uniform externalpressure of steel cylindrical shell structurerdquo Journal of North-western Polytechnical University vol 24 no 1 pp 119ndash123 2006

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of