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Analysis of the mechanical properties of a fibreglass reinforced flexible pipe (FGRFP)

Gao, Yifan; Bai, Yong; Cheng, Peng; Fang, Pan; Xu, Yuxin

DOI10.1080/17445302.2020.1857568Publication date2020Document VersionFinal published versionPublished inShips and Offshore Structures

Citation (APA)Gao, Y., Bai, Y., Cheng, P., Fang, P., & Xu, Y. (2020). Analysis of the mechanical properties of a fibreglassreinforced flexible pipe (FGRFP). Ships and Offshore Structures.https://doi.org/10.1080/17445302.2020.1857568

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Analysis of the mechanical properties of afibreglass reinforced flexible pipe (FGRFP)

Yifan Gao , Yong Bai , Peng Cheng , Pan Fang & Yuxin Xu

To cite this article: Yifan Gao , Yong Bai , Peng Cheng , Pan Fang & Yuxin Xu (2020): Analysisof the mechanical properties of a fibreglass reinforced flexible pipe (FGRFP), Ships and OffshoreStructures, DOI: 10.1080/17445302.2020.1857568

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Analysis of the mechanical properties of a fibreglass reinforced flexible pipe (FGRFP)Yifan Gaoa, Yong Baia, Peng Chenga, Pan Fangb and Yuxin Xua

aCollege of Civil Engineering and Architecture, Zhejiang University, Hangzhou, People’s Republic of China; bDepartment of Maritime and TransportTechnology, Delft University of Technology, Delft, The Netherlands

ABSTRACTFibreglass reinforced flexible pipe (FGRFP) is a kind of composite thermoplastic pipe serving as apreferred application in the field of oil transportation. This paper studies the mechanical behaviour ofFGRFPs under pure bending by experimental, numerical and theoretical methods. Full-scale four-pointbending tests are conducted and the curvature-bending moment relations of specimens are recorded.In the numerical simulation method (NSM), a detailed finite element model considering both materialand geometric nonlinear behaviour is established, and the composite is defined as an orthotropicelastic-plastic material. Based on the Euler–Bernoulli beam theory, a simplified theoretical method(STM) is proposed to predict the ultimate bending moment. In the parametric study, a simple formulais introduced to modify STM to make it more accurate. Good agreements proves the reasonability ofthe proposed NSM and STM. Additionally, STM could make a contribution to engineers in terms of aconcise and relatively accurate way in ultimate status analysis.

ARTICLE HISTORYReceived 6 May 2020Accepted 8 November 2020

KEYWORDSBending; fiberglassreinforced flexible pipes;numerical simulationmethod; simplifiedtheoretical method; full-scaleexperiment

1. Introduction

Zero pollution is the goal especially in the remote offshoreareas where equipment for pollution control response is eitherlimited or challenging to mobilise. As a result, the improve-ments of bonded flexible pipe are primarily driven by environ-mental safety, at the same time, applied to offshoredevelopment where mobile offshore production unit(MOPU) are used (Northcutt 2000). Until 1989, there hadbeen scarcely new development in bonded flexible pipe, but,this situation changed after the MOPS were introduced(Northcutt 2000). In 1959, the first use of bonded flexible mar-ine hoses for offshore loading was in offshore Miri, Sarawak(Gibson 1989). After that, in the British sector of the NorthSea, the first bonded flexible flowlines and risers were installedin 1988. From then on, flexible pipes were extensively appliedin various engineering practices, such as oil transportation(Gibson 1989). As a kind of bonded flexible pipe, fibrereinforced flexible pipe (FRFP) is composed of two kinds ofmaterials, fibre and resin or polyethylene (PE). Fibres usuallyincluding carbon, aramid, Kevlar and glass are used inreinforced layers due to their excellent tensile strength andmodulus. Resins (used as matrix in the reinforced layers), onthe other hand, are capable of transferring stress amongfibres, hence, enables the fibres inside the reinforced layerswork together. However, Kevlar and carbon fibre are notusually used in deep-sea pipelines because of high cost andelectrochemical corrosion (Xu et al. 2019). Recently, FGRFPbecomes a favourable choice for its high corrosion resistance,light-weight characteristic, and relatively low fabrication andfacility cost.

The cross section of the FGRFP studied in this paper,shown in Figure 1, consists of a polyethylene liner, eight layers

of reinforcement and an outer polyethylene coating. The innerliner pipe is made of ultra-high molecular weight polyethylene(UHMWPE), and the outer coating pipe is high-density poly-ethylene (HDPE). The reinforcement considered here is pro-duced by the helical tape wrapping method, using prepregtapes (shown in Figure 2) in which impregnated twistedglass fibres are embedded in the HDPE matrix. Since thetwisted glass fibres are impregnated firmly, there is enoughbonding between the fibres and the matrix. This is the maindifference between FGRFP and other reinforced thermoplasticpipe (RTP) (Kruijer et al. 2005).

PE, the crucial part of FGRFP, exhibits a complicatedcharacteristic, which comprises elasticity, plasticity, and vis-cosity, and its behaviours are also strongly dependent on temp-erature, time, and loading conditions. Many works have beendone to investigate the non-linear behaviour of PE by usingexperimental and theoretical methods (Bodner 1987; Zhangand Moore 1997a, 1997b; Colak and Dusunceli 2006). Never-theless, some simple models are widely used among the analy-sis of bonded flexible pipe. In the work of Kruijer et al. (2005),PE was treated as a linear elastic material in the analysis ofRTP’s reinforced layers. By using a linear elastic materialmodel, Dhar and Moore (2006) made an investigation onthe evaluation of local bending in profile-wall PE pipes.Zheng et al. (2006) and Li et al. (2009) also consideredHDPE as a linear elastic material in analysing the propertiesof PSP (plastic pipe reinforced by cross-winding steel wire).In the investigation, made by Fang et al. (2018), on mechanicalbehaviour of FGRFPs under torsion, matrix made of HDPE inreinforced layers was treated as a linear elastic material both intheoretical model and finite element model. Fredriksson et al.(2007), who had made some improvements on simplification

© 2020 Informa UK Limited, trading as Taylor & Francis Group

CONTACT Peng Cheng chengpeng812@163.com Room A812, An-zhong Building, 866 Yuhangtang Road, Hangzhou, Zhejiang, PR People’s Republic of ChinaThis article has been republished with minor changes. These changes do not impact the academic content of the article.

SHIPS AND OFFSHORE STRUCTUREShttps://doi.org/10.1080/17445302.2020.1857568

of HDPE material model, modelled the HDPE as an elastic-perfectly-plastic material to analysis the HDPE plastic net pens.

Regarding the greater complication of the reinforced layers,the classical laminated-plate theory has been adopted in moststudies. In recent ten years, the classical laminated-plate theoryhas been adopted in most studies on the mechanical propertiesof reinforced layers of bonded flexible pipe (Xia et al. 2001;Zhu 2007; Menshykova and Guz 2014; Hu et al. 2015; Xinget al. 2015; Fang et al. 2018). It is remarkable that Menshykovaand Guz (2014) proposed a theoretical method in analysing themulti-layer thick-wall composite pipe under bending, in whichcomposite tube bending stiffness calculation equations pro-posed are of reference to this paper. In the classical lami-nated-plate theory, the reinforced layers are considered as anintegral part which is merged by matrix and fibres or steelwires. Each reinforced layer is regarded as a 3D orthotropiccylinder, and the stiffness of the reinforced layer is determinedby the rule of mixture. Nevertheless, rather than consideringthe elastic–plastic properties, the classical laminated-plate

theory assume that the material of reinforced layers is elastic.Therefore, the elastic–plastic mechanic behaviour of thebonded flexible pipe is not captured among those works. Onthe other hand, the embedded elements technique, which isused to account for the fact that the reinforcement plies andthe coil are embedded into the matrix, is employed to investi-gate the non-linear behaviour of bonded flexible pipes in thecommercial software Abaqus™ (Bai et al. 2013; Tonattoet al. 2016; Tonatto et al. 2017, 2018). After that, Fang et al.(2018) proposed a finite element model in which the glassfibre reinforcement was considered as an integral part, and thematerial orientation assignment technique was used to analyzethe linear behaviour of FGRFP under torsion in the Abaqus™.Recently, Edmans et al. (2019) provided a multi-scale method-ology in analysing the flexible pipe under bending load, whichis innovative in the area of numerical simulation of risers.

In this paper, a uniaxial tensile test of both HDPE andUHMWPE is conducted to obtain the non-linear behaviourof PE used in the FGRFP specimens. Then, an experimentalstudy of the FGRFP in a typical four-point bending test is pre-sented. Curvature-bending moment relations were recordedduring the test. In the numerical simulation method (NSM), adetailed finite element model (FEM) considering both materialand geometric nonlinear behaviour is established to investigatethe non-linear behaviour of the FGRFP by using a modifiedmaterial orientation assignment technique. The composite inthe reinforced layer is defined as an orthotropic elastic–plasticmaterial, meanwhile, the inner and outer layer is regarded as anisotropic elastic–plastic material. The result of NSM shows excel-lent agreement with the experiments. Besides, a simplified theor-eticalmethod (STM)based on the Euler–Bernoulli beam theory isproposed to predict the ultimate moment of the FGRFP underpure bending. By modifying the classical laminated-plate theory,the STM takes the non-linear behaviour of both reinforced layersand inner and outer layers into consideration. This simplifiedmethod only takes into consideration the axial tangent module’scontribution to the ultimate bending moment. The double inte-gral method is applied to solve the equilibrium equations in theMatlab™. Then, a factor is introduced to modify STM, as STMdoes not consider the effect of geometric non-linearity in thecross section, unlike NSM. After that, a simple formula is usedto calculate the factor which is obtained by the linear fittingmethod. The modified STM shows great agreement with NSMand the experiment, which proves the applicability of STM. Fur-thermore, a detailed parametric study on the structural responseis conducted by using STM and NSM. Meanwhile, based on theultimate curvature calculated by NSM, a simple formula of theultimate curvature is proposed by using the surface fittingmethod. Finally, a profound understanding of the function ofthe FGRFP can be achieved, which can provide valuable advicefor this kind pipe’s design and application.

2. Experiment

2.1. Material experiments

Aiming at evaluating the mechanical characteristics of theHDPE and UHMWPE used in this pipe, uniaxial tensile testsare conducted by an electronic universal testing machine.

Figure 1. Structure of the FGRFP.

Figure 2. Prepreg tape of the FGRFP.

2 Y. GAO ET AL.

According to the stander ISO527-2012 (ISO 2012), theHDPE and UHMWPE are made into the dumb-bell shape,and the elastic modulus of them is calculated as the secantmodulus when the true strain is between 0.05% and 0.25%.

After the material test, the engineering stress–strain datacan be downloaded from the computer of the electronic uni-versal testing machine. Then, the engineering stress–straindata is reshaped into the true stress–strain data, and theobtained curves are shown in Figure 3. From Figure 3, it canbe observed that the stress of the HDPE begins to level offwhen the strain is over 7%∼10%. It means that the HDPEbegins to lose its axial tensile strength for axial strains higherthan these values.

The mechanical characteristics data of fibre-glass shown inTable 1 are provided by the manufacturer of the pipe.

2.2. Experiments of the pipe

The bending experiment is usually known as the three-pointbending test and the four-point bending test. Troina et al.(2003) used a bending test, in which the flexible pipe wasregarded as a cantilever beam, and the bending stiffness ofthe specimen can be measured by applying a displacementon the end of the beam. Kagoura et al. (2003) employed thethree-points bending test to investigate the bending stiffness

of metallic flexible pipe. After that, Lu et al. (2010) used thesame experimental method to get the bending stiffness ofsteel wires reinforced flexible pipeline. However, the test sec-tion is not under pure bending load in the three-point bendingtest, as there is a stress concentration on the test section. Toavoid this flaw, Bai et al. (2015) investigated the bendingbehaviour of reinforced thermoplastic pipe (RTP) by using afour-point bending test, and the results of experiment andnumerical simulation fit each other very well. Hence, the bend-ing moment-curvature relationship of the FGRFP in this paperis also obtained by the four-point bending test.

2.3. Experimental facility

A typical four-point bending test is carried out on an exper-imental facility, whose diagrammatic sketch is shown inFigure 5. The dimensions of the experimental facility areshown in Table 2. As shown in Figure 4, the four-point bend-ing test is carried out in the horizontal plane to eliminate theinfluence of the weight of the specimen. A loading beampushed by a jack moves on the slider. The displacement ofthe beam is recorded by a displacement gauge. The load gen-erated by the jack is recorded by a force sensor. Accordingto the load and the displacement recorded by the force sensorand the displacement gauge, a moment-curvature relationshipcan be figured out.

The straight specimen, initially, is in the horizontal pos-ition, which is indicated by the solid line in Figure 5. Afterthat, the loading beam and the loading roller move downwardstogether to exert a bending moment on the specimen. Conse-quently, the specimen, indicated by the dash line in Figure 5,tends to be bent and the rigid region of it begins to rotate anangle of a.

2.4. Specimen

The FGRFP specimen is formed by an inner UHMWPElayer, an outer HDPE layer and eight reinforced layers.HDPE (matrix) and fibreglass (embedded in the matrix)constitute the eight reinforced layers in which the windingangle of fibreglass in odd layers is +54.7°, while that ineven layers is −54.7° (the opposite direction against+54.7°). It should be noted that the fibre volume ratio is

Figure 3. Stress-strain curves from tensile test.

Table 1. Material properties of the testing specimens.

Materials Young’s modulus E (MPa) Poisson’s Ratio n

Fibreglass 45100 0.3HDPE 961 0.4UHMWPE 1080 0.4

Table 2. Dimensions of the test facility.

SymbolValue(mm) Note

L 600 Length of the test sectionL1 800 Distance between loading pointsL2 1400 Distance between support pointsl 300 Horizontal distance between

loading point and support point

SHIPS AND OFFSHORE STRUCTURES 3

50%. The nominal manufacturer dimensions of the speci-mens are shown in Table 3.

2.5. Experiment process

The thickness, outer diameter and length of the test specimenswere measured before the test. The detailed valid dimensionsof three specimens are shown in Appendix (Tables A1–A3).The general measured dimensions of the specimens areshown in Table 4.

The relative standard deviation (RSD) between the maxi-mum and the minimum diameter is 0.17%, while the averageof the outer diameter is 0.38% deviated from the nominal man-ufacturer size. It also can be found that the deviation of themeasured thickness is 1.53%, which is also within the errorlimit. It presents that the experiment results of them are com-parable. Furthermore, initial ovality of the specimens is allpretty small and no bigger than 1%. Thus, it can be concludedthat the initial imperfection of the three specimens is quitesmall.

As shown in Figure 6. During the test, the loading beam waspushed by a jack, generating an increasing thrust (2F), and itslid forward at a constant speed. The displacement of the load-ing beam was recorded as D by the displacement gauge. Whenthe loading beam was sliding forward, the FGRFP tended tobend under the bending moment generated by the four rollers.

The rigid region was formed by inserting a cylindrical rigid rodinto the specimen. It should be noted that the length of therigid rod is around 600 mm, which means that the length ofrigid region is also around 600 mm. According to the loadingcondition of the specimens, the test section can be regarded asa pure bending region. In order to make sure that the pipe issubjected to a static load, the loading process must be slow,stable and constant by keeping the speed of the loadingbeam at about 0.2∼0.4 mm/s. After the test, the displacementof the loading beam (D) and the load (2F) was convertedinto the curvature k and the moment M from the followingexpression:

D− [l · tana+ (2r + D0)− 2r + D0

cosa] = 0 (1)

k = 2aL

(2)

M = Fcosa

· [ lcosa

− (2r + D0)tan a] (3)

Where a is the inclination angle of the rigid region withrespect to the horizontal line, r is the diameter of the roller,D0 is the outer diameter of the rigid region, l is the horizontaldistance between the loading roller and the support roller, L isthe length of the test section, and F is half the thrust generatedby the jack.

Table 3. Geometric parameters of testing specimens.

Parameter Value

Inner radius (mm) 25Outer radius (mm) 38Thickness of inner PE layer (mm) 4Thickness of outer PE layer (mm) 3Number of reinforced layers 8Winding angle of the fibreglass (°) ±54.7Thickness of reinforced layers (mm) 6

Table 4. Valid length and diameter of specimens.

Specimen Outer Diameter (mm)Wall thickness

(mm)Initial ovalization

(%)

#1 76.44 12.96 0.51#2 76.58 13.35 0.10#3 76.71 12.90 0.36Figure 4. The four-point facility.

Figure 5. Diagrammatic sketch of facility.

Figure 6. Geometric relationship between a and D.

4 Y. GAO ET AL.

According to the geometric relationship between a and Dshown in Figure 6, Equation 1 can be obtained. Equation 3are proved in Appendix (shown in Figures A1 and A2).

2.6. Experimental results

At the end of the test, the photographs of the samples weretaken, and the curvature-moment relationships were obtained.From Figures 7(a–c), it can be observed that the samples bentexcessively under bending load.

Figure 8 shows that the moment gradually rises as the curva-ture increases, and the three test curves are close to each other.Although the test curves have a slight fluctuation, they maintaina relatively steady rise. This means that the test result is reason-able. The summary of the bending test data is shown in Table 5.

3. Numerical simulation method (NSM)

In this part, a finite element model (FEM) is established tostudy the mechanical behaviour of the FGRFP by using theAbaqus™/Standard nonlinear finite element analysis tool.The geometrical dimensions of the FEM are consistent withthose from manufacture.

3.1. Parts and properties

As shown in Figure 9, a 300 millimetres long numerical model(semi-structure) of the FGRFP consists of ten layers: an outerlayer, an inner layer and eight reinforced layers. During themanufactory process, each layer of the pipe is bonded firmly.The reinforcement considered here is produced by the helicaltape wrapping method, using prepreg tapes (shown in Figure 2)in which impregnated twisted glass fibres are embedded in theHDPE matrix. Since the twisted glass fibres are impregnatedfirmly, the reinforced layers are considered as an integral part.Considering how the pipes are made, extrusion and partitioncommands were used to separate the 300 millimetres longpart into ten cells in its radial direction, and each one of themrepresents one layer. The inner and outer layers, which aremade of ordinary HDPE and UHMWPE respectively, are iso-tropic, while the reinforced layers made of fibreglass reinforcedHDPE are orthotropic. Therefore, the three principal orien-tations need to be assigned to the different reinforced layersby using the material orientation assignment technique.

As shown in Figure 10, the local material coordinate systemof the reinforced layers is designated as (1,2,3), where 1 is thedirection of the fibre, 2 is the direction perpendicular to theglass-fibre strand in the plane, and 3 is the normal directionin the cylindrical coordinate system. Before the orientationassignment, the coordinate axis 1, 2 and 3 are directed at theradial, hoop and axial direction respectively. It means thedirection of the fibre is along the axial direction of the pipe.For instance, one reinforced layer of the model before theorientation is shown in Figure 10(a). Then the coordinate sys-tem (1,2,3) rotates about the axis 3 by an additional rotation(−54.7° for even reinforced layers or 54.7° for odd reinforcedlayers). As shown in Figure 10(b), the coordinate system(1,2,3) of the same reinforced layer is rotated by 54.7° afterthe orientation assignment.

Figure 7. Bending deformation of the three specimens.

Figure 8. Curvature-moment curves of three test specimens.

SHIPS AND OFFSHORE STRUCTURES 5

3.2. Simplified model of reinforced layer

3.2.1. Simplification in linear elasticity stageAs described in the introduction part, the reinforced layer con-sists of HDPE matrix and fibreglass, hence the mechanicalcharacteristics of fibreglass and HDPE matrix shown inTable 1 can be transformed into the nine local direction’s elas-tic constants of each reinforced layer by using the simplifica-tion model of the reinforced layer. Then, nine elasticconstants (E1, E2, E3, G12, G13, G23, n12, n13, n23) can be usedto define the engineering elastic constants of the numericalmodel’s reinforced layers in the Abaqus™.

In this part, the glass fibres, which are in each of thereinforced layers or prepreg tapes, are simplified as equidistantfrom each other. Hence, it can be considered that each of pre-preg tapes consists of a large quantity of representative units asshown in Figure 11. The length of the representative unit isequal to the space between fibres. Then, according to thevolume ratio of the glass-fibre in the prepreg tape, the spacebetween fibres can be obtained. For a convenient calculationof the nine elastic constants of the reinforcement, the represen-tative unit is taken as an equivalent unit. Compared to therepresentative unit, the equivalent unit have the same sizeand volume ratio of the glass-fibre, but, a different cross-sec-tion of the fibre (one is circular and another is a square).

As shown in Figure 11, a global cylindrical coordinate systemis established. The coordinate axis r, u and z denote the radial,hoop and axial direction respectively. The local material coordi-nate system of the reinforced layers is designated as (1,2,3),where 1 is the direction of the fibre, 2 is the direction perpendicu-lar to the glass-fibre strand in the plane, and 3 is the normal direc-tion in the cylindrical coordinate system. w is the winding angle(54.7°) of fibres in reinforced layers. dG is the diameter of glass-fibre, and h is the thickness of each reinforced layer or prepreg

tape. According to the dimension of prepreg tape provided bymanufacturer, dG = h. b is the space between fibres. This spacecan be calculated by following formula (Zhu 2007).

b = 2pRi

2pRihVFB/SFB= ph

2≈ 1.18mm (4)

Where Ri is diameter of i th reinforced layer; VFB is the volume

Table 5. Summary of the bending test data.

SpecimenDisplacement

(mm)Curvature(m−1)

Moment(×106N ·mm)

FGRFP1 182.20 2.036 1.552FGRFP2 182.12 2.035 1.637FGRFP3 180.91 2.023 1.581

Figure 9. Break out the section of FEM.

Figure 10. Discrete field of one layer before and after orientation.

Figure 11. Simplification of the reinforced layer.

6 Y. GAO ET AL.

ratio of the glass-fibre VFB = 50%; SFB is the cross-section area of

glass-fibre SFB = ph2

4.

As shown in Figure 11, the cross-section of fibre is a square inthe equivalent unit, while the actual shape is circular. In order tomake sure that the volume ratio of the matrix and fibre is notchanged after the simplification, the area of the square in theequivalent unit should be equal to the actual area of the cross-sec-tion of the fibre, which is SFB. Therefore, the length of each side of

the square in the equivalent unit is����SFB

√ =��p

√h

2.

Based on simplification of reinforced layer shown in Figure 11,nine elastic constants of the reinforced layersE1, E2, E3, G12, G13, G23, n12, n13, n23 can be determined (Zhu2007).

E1 = EFBVFB + EPEVPE (5)

E2 = EFBEPEVI

EPEVFB

VI+ EFB(1− VFB

VI)+ EPE(1− VI) (6)

E3 =EFBEPE

VFB

VI

EPEVI + EFB(1− VI)+ EPE(1− VFB

VI) (7)

G12 = G13 =GFBGPE

VFB

VI

GPEVI + GFB(1− VI)+ GPE(1− VFB

VI) (8)

G23 = GFBGPEVI

GPEVFB

VI+ GFB(1− VFB

VI)+ GPE(1− VI) (9)

n12 = n13 = nFBVFB + nPE(1− VFB) (10)

n23 =nFBVFB + nPE

EFBEPE

(VI − VFB)

VFB

VI+ EFB

EPE(1− VFB

VI)

+ nPE(1− VI) (11)

Where EPE, nPE, GPE is Young’s modulus, Poisson’s ratioand shear modulus of the HDPE matrix, EFB, nFB, GFB isYoung’s modulus, Poisson’s ratio and shear modulus of thefibre-glass, VPE, VFB is the volume ratio of the matrix andfibre. VI is the volume ratio of area iii and area iv, as shownin Figure 11. VI can be obtained by:

VI =��p

√h/4

h/2=

��p

√2

(12)

In specific, the volume ratio of the representative or equiv-alent unit equals one. Therefore, VPE = 1− VFB.

The calculation result of the reinforced layer’s elastic con-stants is used to define the engineering elastic constants ofthe numerical model’s reinforced layers in the Abaqus™.

The calculation result of the reinforced layer’s elastic con-stants is shown in Table 6.

3.2.2. Simplification in elastic-plastic stageIn this part, the reinforced layer’s true stress–strain curves ofthree global cylindrical directions (r, u, z) can be calculated

by using Equations 4–17. For a convenient formulation, vectorCi in this section represents any of the nine engineering con-stants vector E1, E2, E3, G12, G13, G23, n12, n13, n23. Firstly,the tangent modulus vector EPE of the HDPE can be trans-formed intoCi. Then, nine vectors Ci are transformed intothree reinforced layer’s modulus vectors Ez, Er , Eu. In theend, the reinforced layer’s true stress–strain curves of threeglobal cylindrical directions (r, u, z) can be calculated byusing the numerical integration method. All this procedureis constructed in the Matlab™. The detailed equations of Ci,E1, E2, E3, G12, G13, G23, n12, n13, n23, Ez, Er, Eu, EPE are illus-trated in Appendix (Equations A1–A14).

By employing first-order interpolation in the Matlab™, aset of stress–strain discrete data point is obtained based onthe stress–strain curve of HDPE (shown in Figure 3). Everydata point can be used to calculate a corresponding secantmodulus EkPE by using numerical differentiation method asshown in Equation 13. It should be noted that when the num-ber of data points is large enough, the secant modulus EkPEapproaches to the tangent modulus of the correspondingdata point.

EkPE = sk+1PE − sk

PE

1k+1PE − 1kPE

, (k = 1, 2 . . . , n) (13)

Where skPE is the HDPE’s true stress of the kth data point; 1kPE

is the HDPE’s true strain of the kth data point n is the numberof data points.

Then, the EPE of the HDPE can be obtained. By substitutingEPE into Equations 5–11, the nine vectors Ci are constructed.

In the local cylindrical coordinate system, the flexibilitymatrix of each reinforced layer can be expressed as (Zhu 2007):

Sk =

1

Ek1− nk12

Ek2− nk13

Ek30 0 0

− nk12Ek1

1

Ek2− nk23

Ek30 0 0

− nk13Ek1

− nk23Ek2

1

Ek30 0 0

0 0 01

Gk23

0 0

0 0 0 01

Gk13

0

0 0 0 0 01

Gk12

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

k

(14)

In the global cylindrical coordinate system, the flexibility

Table 6. Elastic constants of reinforced layer.

Elastic constants Value

E1 23151.80 MPaE2 2502.73 MPaE3 5465.85 MPaG12 1971.57 MPaG13 1971.57 MPaG23 896.00 MPan12 0.35n13 0.35n23 0.40

SHIPS AND OFFSHORE STRUCTURES 7

matrix of each reinforced layer can be expressed as:

�Sk= TSkTT (15)

Where:

T=

cos2w sin2w 0 0 0 coswsinwsin2w cos2w 0 0 0 −coswsinw0 0 1 0 0 00 0 0 cosw sinw 00 0 0 −sinw cosw 0

−2coswsinw 2coswsinw 0 0 0 cos2w−sin2w

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

(16)

�Sk =

1Ekz

−nkrzEkr

−nkuzEku

0 0 0

−nkzrEkz

1Ekr

−nkurEku

0 0 0

−nkzuEkz

−nkruEkr

1

Eku0 0 0

0 0 01

Gkru

0 0

0 0 0 01

Gkzu

0

0 0 0 0 01Gkzr

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

k

(17)

Where k= 1, 2 . . . .., n.After that, the nine vectors Ci are substituted into Equation

15, and the global modulus vectors Ez, Er, Eu of reinforcedlayers shown in Figure 12 is obtained by solving thoseequation.

Finally, the reinforced layer’s true stress–strain discretecurves of the three directions (r, u, z) can be calculatedbased on the global modulus vectors Ez, Er, Eu of reinforcedlayers by using the numerical integration method in theMatlab™. The three true stress–strain curves of the reinforcedlayer are shown in Figure 13.

According to the knowledge of material mechanics, theaxial stress is a type of primary stress compared with theradial and the circumferential stress when the pipe isunder pure bending. When the radial stress and the cir-cumferential stress are small, the stress–strain curve isapproximately linear, as shown in Figure 13. Additionally,the results of the numerical simulation show that the cir-cumferential and the radial deformation are pretty smallcompared to the axial deformation. Furthermore, thestress–strain curve in axial direction represents the effectof stress–strain curve in direction (1,2,3) to direction z.Hence, the stress–strain curve in the circumferential andthe radial direction can be regarded as approximately linear,and only the stress–strain in direction z is used to definethe yield stress-plastic strain curve of the reinforced layersin NSM. This is also the reason why STM only considersthe axial tangent module’s (Ez) contribution to the ultimatebending moment (stated in the Section 4.1).

The true stress–strain curve of the reinforced layer inthe axial direction (direction z) is used to define theyield stress-plastic strain curve of FEM’s reinforced layersand also used to calculate the stress of axial direction inSTM.

3.3. Boundary conditions and Mesh

As shown in Figure 14, symmetric boundary condition aboutthe axial direction (z axis) is applied to the right side of thepipe (U3 = UR1 = UR2 = 0) and a kinematic coupling con-straint is established at the other end. This type of constraintcan apply a uniform rotation on the left side and freely allowthe ovalization deformation on the right side. The bendingprocess is achieved by applying a rotation about the x-axis atthe reference point of the coupling, and the nonlinear staticanalysis including non-linear geometric effects is selected forlarge displacement effects.

3D solid element (C3D8I) is selected for the inner layers,the outer layers and the reinforced layers. This type of elementis more effective to provide relatively high accuracy thansecond-order elements. As shown in Figure 15, the wholemodel contains 39000 elements and 43472 nodes. The globalsize of the elements is 4 mm.

Figure 12. Reinforced layer’s modulus (Ez , Er , Eu)-strain curves in three globaldirections.

Figure 13. Reinforced layer’s true stress-strain curves in three global directions.

8 Y. GAO ET AL.

4. Simplified theoretical method (STM)

4.1. Assumptions of STM

Based on the actual stress characteristics of the samples, thefollowing assumptions are proposed to focus on the bendingproblem with the engineering applications point of view.

(a) The FGRFP consists of the outer layers, inner layers andreinforced layers, and each reinforced layer consists ofthe HDPE matrix and filament glass fibres.

(b) The materials including HDPE and filament glass fibresare continuous, homogeneous and flawless.

(c) The inner and outer layers are isotropic, and thereinforced layers are orthotropic.

(d) STM only considers the axial tangent module’s (Ez) con-tribution to the ultimate bending moment of the FGRFP

(e) STM doesn’t take the deformation in the cross section ofthe pipe into consideration.

(f) The neutral surface does not move during the bendingprocess

4.2. Definition of STM

A long, circular pipe with radius R and wall thickness t loadedby a pure bending load is shown in Figures 16 and 17. The cur-vature and deformation of the cross section are uniform alongthe length of the pipe under the pure bending load.

According to the Euler–Bernoulli beam theory neglectingthe deformation of cross-section, the axial strain is given:

1z(r, u, k) = j(r, u)k (18)

where:

j(r, u) = r cos u (19)

u, r–the coordinates of the integral point in polar coordinates;k–the curvature of the 600 mm pipe; j–the perpendiculardistance from the integral point to the neutral surface of thepipe.

Then, the stress (sz(r, u, k)) in the axial direction can beobtained based on the stress–strain curve in the axial direction(Figure 13) by the linear interpolation method in theMatlabTM.

Finally, the moment under different curvature can beexpressed as:

M(k) =∫2p0

∫Ro

Ri

j(r, u)sz(r, u, k)drdu (20)

Where Ro–the outer diameter of the pipe; Ri–the innerdiameter of the pipe. In the STM calculation, the pipecross section is discretized into Nr and Nu elements alongradial and circumferential directions, respectively. It meanthe cross section of the pipe is divided into Nr × Nu

elements totally. The centre point of each elements rep-resents the coordinate of this element. According to the

Figure 14. Boundary condition of the FEM model.

Figure 15. Break out the section of meshed model.

Figure 16. Infinite long FGRFP under pure bending.

Figure 17. Diagrammatic sketch of the cross section.

SHIPS AND OFFSHORE STRUCTURES 9

coordinate (u, r), the axial strain can be calculated byEquation 18. Then, linear interpolation is used to calculatethe corresponding axial stress in MatlabTM.

The specific algorithm is shown in Figure 18:Where ti– the thickness of the inner layer; to– the thickness

of the outer layer; t– the thickness of one reinforced layer; rn–the number of the reinforced layer; tn– the thickness of thereinforced layer;Nr– the number of the radial integral element;Nu– the number of the circular integral element.

4.3. Results of experiment, STM and NSM

The curvature-moment curves of the experiment, STM andNSM are shown in Figure 19. This curve shows nonlinear-ity with the increasing bending moment, which is due tothe nonlinearities of material and deformation of cross sec-tion. Due to the displacement limit of the loading beam(mainly because of the displacement limit of the jack),the three test curves don’t show their ultimate bendingmoment. Nevertheless, the experiment curve shows almostthe same uptrend with the numerical simulation result.Compared to the NSM, the curve of STM fails to declinewhen the curvature is larger than 3.55 m−1. The reasonfor this phenomenon is mainly that STM does not takeinto consideration the deformation in cross section of the

pipe. It means that the cross section of the model inSTM remains the same shape instead of becoming ovaliza-tion. By comparison, the cross section of FEM can becomeovalization which causes a decline in the section stiffness ofthe model. This is also the reason that the curve of STM isa little bit higher than NSM in the up stage. However, theuptrend of the curve obtained by STM still agrees withboth NSM and experiment. In general, results from NSM,STM and experiment show consistence with each other.Furthermore, the ultimate moment and curvature obtainedfrom NSM, which are 1.79× 106N ·mm and 3.55 m−1,are marked with a black pentagram in Figure 19. The ulti-mate moment obtained from STM is very close to the onecollected from NSM, which are 1.80× 106N ·mm and1.79× 106N ·mm respectively, and the RSD between themis 0.43%. In conclusion, it is safe to say that STM andNSM all achieve their expected goals.

5. Parametric study

In this section, the effects of several important parametersare investigated based on STM and NSM. Those par-ameters’ influences on the ultimate curvature and moment,the deformation development of the cross-section and thecomparison between the two methods (STM and NSM) isinvolved into discussion. It should be noted that thelegends of the figures in this section follow a fixed format.For instance, one of the legends in Figure 22 is‘D_130_t_13 _D/t_10_A_30’. ‘D_130’ means the outerdiameter of the model is 130 mm. ‘t_13’ means the wallthickens of the model is 13 mm. ‘D/t_10’ means the diam-eter-thickness ratio of the model is 10. ‘A_30’ means thewinding angle of the fibres is 30 degrees.

As the NSM takes the deformation of cross-section intoconsideration, every moment-curvature curve obtained byNSM has a peak point due to both the non-linear of materialand the ovalization deformation of the cross-section. Then,the moment value will drop down after this point. Hence,the moment and curvature of the peak point can be regardedas the ultimate moment and ultimate curvature respectively.However, the moment-curvature curves calculated by STMdon’t have a peak point, for this method doesn’t consider thedeformation in the cross-section. In spite of it, the curve ofSTM also can reach horizontal due to the non-linear ofmaterial (as shown in Figure 13). Therefore, the moment onthe horizontal line can be considered as the ultimate moment.

5.1. Deformation of cross-section

The deformation development of the cross-section is pre-sented in the ovality-curvature curves (Figures 20–22). Theovality corresponding to the ultimate curvature is markedwith a pentagram on the curves. This ovality can be expressedas OvalityU.

The ovality of the pipe can be calculated by:

Ovality = Dmax − Dmin

Dinitial× 100% (21)

Figure 18. Algorithms of STM.

10 Y. GAO ET AL.

Where Dmax–the maximum outer diameter of the pipe; Dmin–the minimum outer diameter of the pipe; Dinitial–the initialouter diameter of the pipe.

5.1.1. Different outer diametersEight cases have the same wall thickness (13 mm) and windingangle (54.7 degrees), but different outer diameters, as shown inFigure 20. Furthermore, the pipe wall consists of a 4 mm innerlayer, 3 mm outer layer and 6 mm reinforced layer.

Obviously, the pipe with a larger outer diameter has largerovalization deformation in the cross section when the curva-ture is the same. For all curves, the ovality increases as the cur-vature gets larger, but at different rates. In other words, thecross section of the pipe with a bigger outer diameter deformsmore easily.

From the pentagrams on Figure 20, it can be observed thatthe bigger the diameter-thickness ratio of the pipe is, the largerthe OvalityU is. The detailed data in Table 7 proves this con-clusion. The mean value of the last column is 16.88%, mean-while, the RSD of them is 34%. The value of the RSD provesthat the OvalityU and the diameter-thickness ratio do have aclear correlation with each other. More specifically, there is apositive relationship between them.

5.1.2. A constant diameter-thickness ratioAs shown in Figure 21, eight cases have the same diameter-thickness ratio (D/t = 5.8). The winding angle of the cases is54.7 degrees. In addition, it should be noted that the wall thick-ness is changed just by adjusting the number of the reinforcedlayers and the wall thickness of the inner and outer layer (4 and3 mm separately) does not change.

Different from Figure 20, although each group in Figure 21has the same diameter-thickness ratio, the pipe with a largerouter diameter becomes oval at a larger rate. It means thatthe cross section of the pipe deforms more easily when thewall thickness and outer diameter get larger in the same ratio.

As shown in Figure 21, it seems like that the pentagrams onthese curves are on the same horizontal line. It means that theOvalityU of each pipe doesn’t change evidently compared toFigure 20. The concrete data in Table 8 is consistent with it.Calculated based on the detailed number of this table, themean value and the RSD of the last column are 9.90% and3% separately. It presents that the OvalityU are nearly unchan-ging when the diameter-thickness ratio is a constant. It alsomeans that there is a weak relationship between the outerdiameter and the OvalityU.

By comparing Table 8 with Table 7, it can be observed thatthe diameter-thickness ratio of Table 8 is between 5 and 6,

Figure 19. Comparison of results from experiment, STM and NSM.

Figure 20. Ovality-curvature curves of NSM with different outer diameters.

Figure 21. Ovality-curvature curves of NSM with a constant D/t ratio.

Figure 22. Ovality-curvature curves of NSM with different winding angles.

SHIPS AND OFFSHORE STRUCTURES 11

correspondingly, the mean value of the last column in Table 8is between 7.43% and 10.25%. It accords with the positive cor-relation between the OvalityU and the diameter-thicknessratio.

5.1.3. Different winding anglesAs shown in Figure 22, those nine cases have the same outerdiameter (130 mm) and wall thickness (13 mm), but withdifferent winding angles of the fibres in the reinforced layers.

It can be observed that the shape of these curves is similar toeach other. The deformation rate of pipe with different wind-ing angle differs from each other. Specifically, the pipe with asmaller winding angle deforms at a higher rate. It means thatthe cross section of the pipe deforms more easily when thewinding angle is smaller. However, compared to Figures 20and 21, the influencing of the winding angle on the defor-mation development of the cross section is smaller than theouter diameter.

As shown in Figure 22, it can be observed that the penta-grams are concentrated in an area on the figure. It meansthat the OvalityU of each pipe doesn’t change evidently com-pared to Figure 20. The data in Table 9 proves it. Calculatedbased on the concrete numbers of the table, the mean valueand the RSD of the last column are 21.30% and 4% separately.It means that the OvalityU are nearly unchanging when thediameter-thickness ratio is a constant. It also means thatthere is a weak relationship between the winding angle andthe OvalityU.

5.1.4. Simple formula of OvalityUBased on the parametric study from Section 5.1 (a)–(c), it canconclude that the diameter-thickness ratio is the main par-ameter in influencing on the OvalityU. More specifically,they have a linear correlation with each other, as shown inFigure 23.

By linear fitting inMatlabTM, the relationship between themcan be expressed as:

OvalityU = 2.45× Dt− 3.91 (22)

Where D is the outer diameter of the pipe; t is the wall thick-ness of the pipe.

The R-Square of the result is 0.95. It proves that the linearcorrelation between the OvalityU and diameter-thickness ratioexists and Equation 22 obtained by the linear fitting method isreasonable.

Finally, the ovality of the FGRFP can be obtained when thepipe is under the ultimate bending moment or curvature. Inother words, the FGRFP very likely reaches its bendingcapacity (ultimate bending moment or curvature) when theovality of the cross section is equal to the OvalityU.

5.2. Comparison between STM and NSM

In this section, moment-curvature curves are obtained by bothSTM and NSM. kNSMu is the ultimate curvature calculated byNSM.

The STM does not take into consideration the deformationin the cross section of the pipe, while, NSM dose. Hence, therewill be a difference between them, as shown in Figures 24–29.Then, a factor is introduced to modify STM. The correctionfactor can be expressed as:

KMu =MNSM

u

MSTMu

(23)

Table 7. OvalityU of the pipes with different outer diameters.

D(mm) t(mm) D/tOvalityU(%)

65 13 5 7.4378 13 6 10.2591 13 7 13.39104 13 8 15.58117 13 9 19.59130 13 10 22.24143 13 11 23.80156 13 12 22.73

Table 8. OvalityU of the pipes with a constant D/t ratio.

D(mm) t(mm) D/tOvalityU(%)

49.3 8.5 5.8 9.5058.0 10.0 5.8 10.0166.7 11.5 5.8 9.8875.4 13.0 5.8 9.6284.1 14.5 5.8 10.2992.8 16.0 5.8 10.21101.5 17.5 5.8 10.13110.2 19.0 5.8 9.54

Table 9. OvalityU of the pipes with different winding angles

D(mm) t(mm) Winding angle (°)OvalityU(%)

130 13 30 21.91130 13 35 20.25130 13 40 22.55130 13 45 21.23130 13 50 20.49130 13 55 22.15130 13 60 21.69130 13 65 20.60130 13 70 20.80

Figure 23. Linear fitting of OvalityU-D/t ratio.

12 Y. GAO ET AL.

Where: MNSMu – Ultimate moment calculated by NSM; MSTM

u –Ultimate moment calculated by STM. Based on Equation 20,MSTM

u can be expressed as:

MSTMu = max (M(k)) (24)

5.2.1. Different outer diametersEight cases have the same wall thickness (13 mm) and thedifferent outer diameter, as shown in Figure 24. Furthermore,the pipe wall consists of a 4 mm inner layer, 3 mm outer layerand 6 mm reinforced layer. The winding angle of fibres in thereinforced layers is 54.7 degrees.

As shown in Figure 24, the dash line and solid line representcurves calculated by STM and NSM separately. The pentagramis the peak point of the curve. Every curve obtained by NSMhas a peak point, then, they drop down after the peak point.Intuitively, it can be observed that the deviation between thetwo method increases as the outer diameter of the pipe getslarger.

As shown in Figure 25, the larger the outer diameter of thepipe is, the larger the ultimate moment is, meanwhile, thesmaller the kNSMu is. The MNSM

u goes up from 1.26 kN ·m to7.42 kN ·m as the outer diameter rises from 65 mm to156 mm, while the MSTM

u goes up from 1.20 kN ·m to9.57 kN ·m. Specifically, the ultimate moment increases by 6times in NSM and 8 times in STM. Meanwhile, the kNSMudecreases from 4.80 m−1 to 0.88 m−1 and decreases by 5.5times. Hence, it can conclude that there will be a markedincrease in bending capacity and decrease in bending flexibilitywhen the outer diameter of the pipe gets larger.

More detailed data is shown in Table 10. The KMu decreasesfrom 1.04 to 0.78 as the diameter-thickness ratio increasesfrom 5 to 12. The RSD of the last column is 10%. It presentsthat there is a clear relationship between the KMu and the diam-eter-thickness ratio. More specifically, there is an anti-positiverelationship between them.

5.2.2. A constant diameter-thickness ratioAs shown in Figure 26, eight cases have the same diameter-thickness ratio (D/t = 5.8), which is the same as the diam-eter-thickness ratio of the specimens. The winding angle ofthe cases is 54.7 degrees. In addition, it should be noted thatthe wall thickness is changed just by adjusting the number of

the reinforced layers. Instead, the wall thickness of the innerand outer layer (4 and 3 mm separately) doesn’t change.

Similarly, the pentagram is the peak point of the curve.Every curve obtained by NSM has a peak point, then theydrop down after the peak point, as shown in Figure 26.

As shown in Figure 27, along with the value of the outerdiameter and wall thickness increasing, the ultimate momentof both methods increases, meanwhile, the kNSMu decreases.The MNSM

u goes up from 0.37 kN ·m to 6.44 kN ·m as theouter diameter rises from 49 mm to 110 mm, while theMSTM

u goes up from 0.36 kN ·m to 6.88 kN ·m. Specifically,the ultimate moment increases by 17 times in NSM and 19times in STM. Meanwhile, the kNSMu decreases from6.10 m−1 to 2.39 m−1 and decreases by 2.5 times. It showsanother conclusion that there will be a sharp increase in bend-ing capacity and decrease in bending flexibility as the wallthickness and outer diameter get larger in the same ratio.

Concrete data in Table 11 presents that the KMu decreasesfrom 1.03 to 0.94 as the outer diameter increases from49 mm to 110 mm. The RSD of the last column is 3%. Itmeans that the KMu is nearly unchanging when the diameter-thickness ratio is a constant.

5.2.3. Different winding anglesAs shown in Figure 28, those nine cases have the same outerdiameter (130 mm) and wall thickness (13 mm), but withdifferent winding angles of the fibres in reinforced layers.

Similar to Figures 24, 26, and 28 also shows that every curveobtained by NSM has a peak point, then, they drop down afterthe peak point. It means that all the moment-curvature of thepipe has peak points and the curve will drop down after thispoint, no matter what the dimension parameters (outer diam-eter, wall thickness and winding angle) of the pipe are.

As shown in Figure 29, along with the value of the windingangle increasing, the ultimate moment of both methodsdecreases, meanwhile, the kNSMu increases. The MNSM

u goesdown from 9.60 kN ·m to 4.68 kN ·m as the winding anglerises from 30 degrees to 70 degrees, while the MSTM

u goesdown from 11.34 kN ·m to 5.21 kN ·m. Specifically, the ulti-mate moment decreases by 2 times in NSM and 2.17 times

Figure 24. Moment-curvature curves of NSM and STM with different outerdiameters.

Figure 25. The ultimate moment and curvature of the pipes with different outerdiameters.

SHIPS AND OFFSHORE STRUCTURES 13

in STM. Meanwhile, the kNSMu goes up from 1.20 m−1 to1.43 m−1 and increases by 1.2 times. It presents that therewill be a clear decrease in bending capacity and a smallincrease in bending flexibility as the winding angle increases.

Hence, compared to changing the winding angle, decreas-ing the outer diameter is a more efficient way to increase ulti-mate curvature.

Concrete data in Table 12 presents that the range of KMu

increases from 0.83 to 0.90. The RSD of the last column is3%. It means that the KMu is nearly unchanging when thewinding angle changes.

5.2.4. Simple formula of KMu

Based on the parametric study from Section 5.2 (a)–(c), it canconclude that the diameter-thickness ratio is the main par-ameter in influencing on the KMu . More specifically, theyhave a linear correlation with each other, as shown inFigure 30.

By linear fitting in the MatlabTM, the relationship can beexpressed as:

K̂Mu = −0.04Dt+ 1.21 (25)

Where K̂Mu is the predicted value of KMu ; D is the outer diam-eter of the pipe; t is the wall thickness of the pipe.

The R-Square of the result is 0.97. It proves that the linearcorrelation between the KMu and the diameter-thickness ratioexists and the Equation 25 obtained by the linear fittingmethod is reasonable.

Finally, the ultimate moment of the FGRFP can be calcu-lated by using Equations 24 and 25.

MMSTMu = K̂MuM

STMu (26)

Where MMSTMu is the ultimate moment calculated by the

modified STM.As shown in Tables A4–A6, the differences of the ultimate

moment between the modified STM and the NSM are prettysmall. Hence, it is safe to say that the ultimate moment calcu-lated by the modified STM are well consistent with the NSM.

Figure 26. Moment-curvature curves of NSM and STM with a constant D/t ratio.

Figure 27. The ultimate moment and curvature of the pipes with a constant D/tratio.

Figure 28. Moment-curvature curves of NSM and STM with different windingangles.

Figure 29. The ultimate moment and curvature of the pipes with different wind-ing angles.

Table 10. KMu of the pipes with different outer diameters.

D(mm) t(mm) D/tMSTM

u(×106N ·mm)MNSM

u(×106N ·mm) KMu

65 13 5 1.20 1.26 1.0478 13 6 1.92 1.89 0.9891 13 7 2.81 2.61 0.93104 13 8 3.85 3.42 0.89117 13 9 5.05 4.31 0.85130 13 10 6.50 5.40 0.83143 13 11 7.91 6.30 0.80156 13 12 9.57 7.42 0.78

14 Y. GAO ET AL.

5.2.5. Simple formula of ultimate curvatureAs stated in Section 5.2 (a)–(c), both the outer diameter andthe wall thickness have a clear influence on the ultimate curva-ture of the FGRFP, while, the influence of the winding angle onthe ultimate curvature is pretty small. Accordingly, the outerdiameter and the wall thickness are the key parameters ininfluencing on the ultimate curvature.

Then, based on data shown in in Tables A7 and A8 (includ-ing the ultimate curvature kNSMu , the outer diameter D and thewall thickness t) a simple formula (Equation 27) of theultimate curvature can be obtained by a surface fittingmethod. k̂u is the predicted ultimate curvature calculated byEquation 27.

k̂u = 1D(0.27D+ 2.26) · t−0.71

(27)

The R-Square of the surface fitting result is 0.99. It proves thatEquation 27 obtained by the surface fitting method isreasonable.

The concrete data shown in Tables A7 and A8 indicates thatthe differences between kNSMu and k̂u are pretty small. There-fore, it is safe to say that the predicted ultimate curvature k̂ucalculated by Equation 27 are well consistent with the ultimatecurvature obtained from the NSM.

6. Conclusions

In this paper, a simplified theoretical method (STM) is pro-posed to predict the ultimate moment of FGRFP under purebending. Then the experimental method is used to prove therationality of the numerical simulation method (NSM). Afterthat, NSM is used as a benchmark to modify STM by introdu-cing a factor. Furthermore, a simple formula used to calculatethe factor is obtained by the linear fitting method. Finally, anextensive parametric study using both NSM and STM is car-ried out to analyze the influencing mechanisms on the defor-mation development of cross section, the ultimate momentand curvature of the FGRFP. Some conclusions can bedrawn as follows:

(a) The results of the proposed NSM show great agreementwith the experiment in the up stage of the curvature-moment curve. While the curves of the test fail to climbto the peak points due to the displacement limit of theloading beam.

(b) The Simple Formula of the correction factor obtained bythe linear fitting method is reasonable. Meanwhile, theultimate moment calculated by the modified STM iswell consistent with the NSM.

(c) The Simple Formula of the ultimate curvature obtained bythe surface fitting method is reasonable. Meanwhile, theultimate curvature calculated by the Equation 27 is wellconsistent with the NSM.

(d) The diameter-thickness ratio is the main parameter ininfluencing on the OvalityU. More specifically, they havea linear correlation with each other

(e) All three of the outer diameter, the wall thickness and thewinding angle have a pretty clear influence on the ultimatemoment of the FGRFP.

(f) Both the outer diameter and the wall thickness have anobvious influence on the ultimate curvature of theFGRFP, while the influence of the winding angle on theultimate curvature is relatively small compared to theother two parameters.

The proposed methods and their results can not only pro-vide references for the factory engineers during initial designand estimation, but also guide further investigation intoother more complicated flexible pipes. Especially, the modifiedSTM and the simple formula of the ultimate curvature providea concise and relatively accurate way to predict the ultimatemoment and curvature, which may be useful for engineeringapplications.

Acknowledgments

Thanks to Zhejiang University offering the experiment equipment, a full-scale experiment can be done at Research Lab for Civil Engineering in

Table 12. KMu of the pipes with different winding angles.

D(mm) t(mm)Windingangle(°)

MSTMu(×106N ·mm)

MNSMu(×106N ·mm) KMu

130 13 30 11.34 9.60 0.85130 13 35 9.36 8.14 0.87130 13 40 8.03 7.12 0.89130 13 45 7.11 6.37 0.89130 13 50 6.46 5.81 0.90130 13 55 6.50 5.40 0.83130 13 60 5.65 5.09 0.90130 13 65 5.40 4.86 0.90130 13 70 5.21 4.68 0.90

Figure 30. The linear fitting of KMu -D/t ratio.

Table 11. KMu of the pipes with a constant D/t ratio.

D(mm) t(mm) D/tMSTM

u(×106N ·mm)MNSM

u(×106N ·mm) KMu

49.3 8.5 5.8 0.36 0.37 1.0358.0 10.0 5.8 0.70 0.70 0.9966.7 11.5 5.8 1.20 1.16 0.9775.4 13.0 5.8 1.87 1.79 0.9684.1 14.5 5.8 2.76 2.60 0.9492.8 16.0 5.8 3.86 3.63 0.94101.5 17.5 5.8 5.22 4.90 0.94110.2 19.0 5.8 6.88 6.44 0.94

SHIPS AND OFFSHORE STRUCTURES 15

China to study the mechanical behaviour of FGRFP subjected to purebending. The specimens used in the experiment are manufactured byNingbo OPR Offshore Engineering Co. Ltd.

Disclosure statement

No potential conflict of interest was reported by the author(s).

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16 Y. GAO ET AL.

Appendix

(a) Detailed valid dimensions of specimensTable A1. Detailed valid dimensions of #1 FGRFP.

#1Wall thickness(mm)

A B C D1 13.32 13.52 12.40 13.022 12.56 12.72 13.10 13.10

Mean value 12.96RSD 2.74%

Outter diameter(mm)

AC BD1 76.32 76.042 76.94 76.44

Mean value 76.44Initial Ovalization 0.51%

Table A2. Detailed valid dimensions of #2 FGRFP.

#2Wall thickness(mm)

A B C D1 13.26 13.58 13.42 13.002 13.16 13.76 13.06 13.58

Mean value 13.35RSD 1.92%

Outter diameter(mm)

AC BD1 76.62 76.102 76.62 76.98

Mean value 76.58Initial Ovalization 0.10%

Table A3. Detailed valid dimensions of #3 FGRFP.

#3Wall thickness(mm)

A B C D1 12.90 12.74 12.38 13.242 13.32 13.00 12.94 12.72

Mean value 12.90RSD 2.17%

Outter diameter(mm)

AC BD1 76.70 76.122 77.00 77.02

Mean value 76.71Initial Ovalization 0.36%

(b) The proof of Equation 3

There are some considerations needed to be noted.

(1) The pipe wall offers a supporting force (FM) to the roller and direc-tion of the force is perpendicular to the pipe wall of the rigid region.

(2) The direction of static friction force between the pipe wall and theroller is along the tangent direction of the edge of the roller. Inaddition, that force can make the roller to roll.

(3) Because the loading process is slow enough, the roller is in the equi-librium with any value of a and the roller is not rolling at a particularpoint in time. It means that there is no static friction force betweenthe roller and the pipe wall at a particular point in time. (If there isstatic friction force between the roller and the pipe wall at a particu-lar point in time, the roller will roll under the static friction force)

(4) There is no friction in the bearing of the roller.

Therefore, FM = Fcosa

.

It should be noted that the straight line BE is parallel to the rigidregion of the bent specimen indicated by the dash line in Fig. A2 andthe straight line BD is parallel to the rigid region of the initial specimenindicated by the solid line in Fig. A2. Hence, the included angle of BE andBD is equal to a.

(1) Known conditions:

AC = D0 + 2r

BD = l

(2) Solutions:

[ In theDACE, CE = AC · tana = (2r + D0)tana

In the DBDE, BE = BDcosa

= lcosa

[ BC = lcosa

− (2r + D0)tana

Bending moment exerted on the pipe can be expressed as:

M = FM · BC = Fcosa

· [ lcosa

− (2r + D0)tana].

(c) The detailed equation in section 3.2

Symbols used in Section 3.2 can be calculated by the following expression.

E1 = (E11, E21, · · · · · · , Ek1, · · · · · · , En1) (A1)

E2 = (E12, E22, · · · · · · , Ek2, · · · · · · , En2) (A2)

E3 = (E13, E23, · · · · · · , Ek3, · · · · · · , En3) (A3)

Figure A1. The force diagrams of the pipe wall and the roller.

Figure A2. The proof of Equation 3.

SHIPS AND OFFSHORE STRUCTURES 17

G12 = (G112, G

212, · · · · · · , Gk

12, · · · · · · , Gn12) (A4)

G13 = (G113, G

213, · · · · · · , Gk

13, · · · · · · , Gn13) (A5)

G23 = (G123, G

223, · · · · · · , Gk

23, · · · · · · , Gn23) (A6)

n12 = (n112, n212, · · · · · · , nk12, · · · · · · , nn12) (A7)

n13 = (n113, n213, · · · · · · , nk13, · · · · · · , nn13) (A8)

n23 = (n123, n223, · · · · · · , nk23, · · · · · · , nn23) (A9)

Ez = (E1z , E2z , · · · · · · , Ekz , · · · · · · , Enz ) (A10)

Er = (E1r , E2r , · · · · · · , Ekr , · · · · · · , Enr ) (A11)

Eu = (E1u, E2u, · · · · · · , Eku, · · · · · · , Enu) (A12)

EPE = (E1PE, E2PE, · · · · · · , EkPE, · · · · · · , EnPE) (A13)

C = (E1, E2, E3, G12, G13, G23, n12, n13, n23) (A14)

The figures for modulus of the composite material depend on thespecific value of the strain. To be more specific, Equations A1–A13demonstrate the different modulus under various strain.

(d) Comparison between MMSTMu and MNSM

uTable A4. Comparison between modified STM and NSM with different outerdiameters.

D(mm) t(mm) D/tMMSTM

u(×106N ·mm)MNSM

u(×106N ·mm)Error(%)

65 13 5 1.23 1.26 2.4478 13 6 1.89 1.89 1.0091 13 7 2.65 2.61 0.99104 13 8 3.49 3.42 0.98117 13 9 4.38 4.31 0.98130 13 10 5.41 5.40 1.00143 13 11 6.28 6.30 1.00156 13 12 7.24 7.42 1.02

Table A5. Comparison between modified STM and NSM with a constant D/t ratio.

D(mm) t(mm) D/tMMSTM

u(×106N ·mm)MNSM

u(×106N ·mm)Error(%)

49.3 8.5 5.8 0.36 0.37 4.2258.0 10.0 5.8 0.70 0.70 1.0066.7 11.5 5.8 1.18 1.16 0.9875.4 13.0 5.8 1.85 1.79 0.9784.1 14.5 5.8 2.73 2.60 0.9592.8 16.0 5.8 3.82 3.63 0.95101.5 17.5 5.8 5.17 4.90 0.95110.2 19.0 5.8 6.80 6.44 0.95

Table A6. Comparison between modified STM and NSM with different windingangles.

D(mm) t(mm)Windingangle(°)

MMSTMu(×106N ·mm)

MNSMu(×106N ·mm)

Error(%)

130 13 30 9.70 9.60 1.01130 13 35 8.00 8.14 1.66130 13 40 6.87 7.12 3.50130 13 45 6.08 6.37 4.58130 13 50 5.52 5.81 5.00130 13 55 5.56 5.40 2.99130 13 60 4.83 5.09 5.07130 13 65 4.62 4.86 4.97130 13 70 4.46 4.68 4.82

(e) Comparison between kNSMu and k̂u

Table A7. Comparison between kNSMu and k̂u with different outer diameters.

D(mm) t(mm) D/tkNSMu(m−1)

k̂u(m−1)

Error(%)

65 13 5 4.80 4.80 0.0378 13 6 3.40 3.40 0.0991 13 7 2.52 2.53 0.56104 13 8 1.92 1.96 2.17117 13 9 1.57 1.56 0.42130 13 10 1.43 1.27 10.98143 13 11 1.08 1.06 2.41156 13 12 0.88 0.89 1.03

Table A8. Comparison between kNSMu and k̂u with a constant D/t ratio.

D(mm) t(mm) D/tkNSMu(m–1)

k̂u(m–1)

Error(%)

49.3 8.5 5.8 6.10 5.95 2.4158.0 10.0 5.8 4.90 4.93 0.7066.7 11.5 5.8 4.15 4.19 0.9575.4 13.0 5.8 3.55 3.62 2.0684.1 14.5 5.8 3.20 3.18 0.6392.8 16.0 5.8 2.89 2.82 2.32101.5 17.5 5.8 2.64 2.53 4.07110.2 19.0 5.8 2.39 2.29 4.13

18 Y. GAO ET AL.