Numerical modelling of chain-link steel wire nets with discrete ...

Draft

Numerical modelling of chain-link steel wire nets with

discrete elements

Journal: Canadian Geotechnical Journal

Manuscript ID cgj-2017-0540.R3

Manuscript Type: Article

Date Submitted by the Author: 22-May-2018

Complete List of Authors: von Boetticher, Albrecht; Eidgenössische Technische Hoschschule Zürich, Department of Environmental Systems Science Volkwein, Axel; WSL Swiss Federal Research Institute

Keyword: chain-link mesh, flexible barrier, discrete element simulation, natural hazard protection, rockfall

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Canadian Geotechnical Journal

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1

Numerical modelling of chain-link steel

wire nets with discrete elements

Albrecht von Boetticher and Axel Volkwein1

Abstract:2

The chain-link mesh is one of several net types used as protection against rockfall,3

shallow landslides and debris flows. The dynamic impact and the corresponding non-linear4

barrier response require numerical models. Chain-link meshes show a non-linear anisotropic5

behaviour caused by the geometry of the wire. Resolving this geometry and its deformation6

results in a bottleneck of numerical costs. We present a discrete element model which covers7

the non-linear and anisotropic behaviour of the chain-link mesh, using results from either8

small-scale quasi-static tension tests or from a detailed mechanical model as material-law9

input. The mesh stiffness, resistance and failure depend on the inner mesh opening angle10

and thus on the direction of deformation. This information enters the model through the11

transformation of the non-linear three dimensional deformation processes into a non-linear12

material-law, with an interpolated dependency on the inner mesh angle. The model maps13

the resistance of the mesh against impacting masses and covers the energy absorption and14

it is capable of predicting the dynamic behaviour of different protection barriers with high15

accuracy, optimized calculation time and minimized calibration efforts. This is illustrated16

by high impact energy tests which follow the ETAG027 standard, and also with a rockfall17

attenuating system.18

Key words: flexible barrier, wire mesh, simulation, chain-link element, rockfall protection.19

Albrecht von Boetticher1,2 and Axel Volkwein. WSL Swiss Federal Institute for Forest, Snow and LandscapeResearch, 8903 Birmensdorf, Switzerland1 Present Address: ETH Zurich, 8092 Zurich, Switzerland2 Corresponding author (e-mail: [email protected]).

Can. Geotech. J. 99: 1–43 (2018) DOI: 10.1139/Zxx-xxx Published by NRC Research Press

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2 Can. Geotech. J. Vol. 99, 2018

1. Introduction20

Among the natural hazard group of frequently occurring rapid moving mass events, rockfall events21

mobilize volumes and kinetic energies that protection barriers can still stop with reasonable construc-22

tion costs. Flexible barriers use steel nets with support cables, which guide the impact energy towards23

energy absorbing devices where the energy gets dissipated by plastic deformation and frictional pro-24

cesses. Aside from the reduced impact on the environment of a filigree steel net barrier, compared to25

rigid walls with deep foundations or massive dams, the flexible barriers distribute the deceleration of26

impacting material over time due to their dynamic response. As a consequence, peak forces within27

the structure and in the anchorage are controlled and stay below a certain level. The design of such28

barriers requires a numerical model which considers the impact process according to highly non-linear29

dynamic interactions as the impact wave travels through the barrier system, causing large deformations30

of structural components with non-linear elastic-plastic behaviour.31

Numerical modelling of flexible net systems has been performed since first mentioned by Mustoe32

(1993). The level of numerical details and performance changed a lot over the time mostly due to33

increasing computational possibilities. Numerous approaches have been published accordingly that are34

summarized in Albaba et al. (2017) or Effeindzourou et al. (2017).35

Simulation models of protection nets were implemented into different existing simulation codes36

such as YADE (Thoeni et al. , 2013), ABAQUS (Cazzani et al. , 2002), LS-DYNA (Dhakal et al. ,37

2011) or specially developed software to model rockfall protection systems such as FARO (Volkwein ,38

2005). Usually, the simulation is performed based on explicit time step algorithms due to the complex39

dynamic behaviour of the structures which incorporate structural changes, large deformations, fric-40

tional processes and non-linear material behaviour. The main task of such simulations is to map the41

dynamic behaviour of the steel net as for example done by Escallon et al. (2015) or Effeindzourou et42

al. (2017) for chain-link meshes.43

The simulation model described in this article enables the code FARO to handle chain-link meshes.44

The FARO code incorporates discrete elements, where every element corresponds to a single com-45

ponent of the barrier (e.g. posts, ropes or special energy dissipating devices). The steel nets consist46

of several discrete elements, for example single net rings or chain-link-strands as presented in this47

article. All elements are connected through common nodes. The sliding processes between pairs of48

Published by NRC Research Press

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von Boetticher and Volkwein, subm. 3

components are considered within each element, as such special contact algorithms are not necessary.49

In simple words, a chain-link mesh is formed out of zig-zag steel wires (Fig. 1). The contact points50

between the zig-zag spirals form the mesh nodes, as a result of the non-linear processes and geometric51

effects at these nodes this mesh type has superior performance under load. These non-linear processes52

and geometric effects also result in high complexity when modelling this mesh type, making it a subject53

of challenging research. A simplified model for the chain-link elements is introduced. It is fed by54

the results of both a fully detailed mechanical model (Section 3) and small-scale quasi-static tension55

tests to correctly map failure load, mesh elongation and the increase of stiffness upon reloading after56

plastic deformations. The approach is illustrated in Section 4 and has been used in von Boetticher57

(2012). Simulation results could be validated by comparison with results from field tests as described58

in Section 5. The idea of the model is to describe the full behaviour of the chain-link mesh according59

to the actual mesh opening angle α at the individual mesh nodes (Fig. 1c). The considered mesh types60

were Tecco G65-3, Tecco G65-4 and G80-4, Rombo G80-3, Spider 130-4 and Spider 230-4, which are61

all manufactured by the company Geobrugg AG. The last two mesh types use segments which consist62

of multiple strand wire threads rather than single strand wire meshes (Fig. 1b).63

2. Mechanical behaviour and determination of the mesh characteristics64

The chain-link meshes described in this article are formed of segments of straight wires which65

lead to rounded parts at the nodes (see Fig. 1). The wire between two nodes is hereafter referred to as66

segment.67

The deformation of such a mesh under load depends on the steel wire properties and geometric68

details among other ingredients. Bending deformations within the rounded part at the mesh node con-69

tribute to the mesh deformations, not only by changing the mesh opening angle α: By straightening70

the rounded wire parts, bending deformations at the nodes contribute to the increase of node-to-node71

distance of a segment under load. This increase of straight segment length by a decrease of the rounded72

part makes the chain-link meshes a challenge for approaches with classical Finite Element beams,73

because such elements need to resolve the geometry within the rounded part to capture the plastic de-74

formations close to the nodes and the corresponding effect on mesh stiffness. The overall stiffness and75

resistance of the mesh types we address is dominated by the interaction of the three dimensional ge-76

ometry of the rounded parts with the plastic deformations and the interaction of normal force, bending77


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Oh

i 0

d0

12m

w 12m

h

d0

S0

(c) (d)α

0

(b)(a)

Sr,0

Sx,0

Ltot,0

Transversal direction

Lon

git

ud

ina

l d

ire

ctio

n

Fig. 1. Top: typical undeformed chain-link mesh consisting of (a) a single wire strand (b) a multiple strand wirethread. Bottom: the geometric properties of an undeformed zig-zag spiral with view (c) orthogonal to the meshand (d) along the spiral in mesh plane.

and shear at the nodes. The failure modes observed during all plane tension test, vertical impact and78

full scale barrier tests always lead to a rupture in the rounded parts.79

Mesh directions are specified as transversal, in the direction of the chain-link spiral axis and lon-80

gitudinal, in the direction normal to the chain-link spiral axis (Fig. 1a). Figure 2 illustrates a chain-link81

net in a standard test set-up of the manufacturer, which is used to determine the net performance under82

(a) lateral and (b) transversal loading. If the mesh is pulled orthogonally to the chain-spirals it is called83

the longitudinal tension test or parallel to them as transversal tension test. Although the segments in84

the net experience approximately an equally distributed load F (Fig. 3a), the stiffness and resistance of85

a segment in a longitudinal or transversal tension test differ in a way that a longitudinal test has a 10 %86

smaller failure load at about 20 % smaller deformation (Fig. 4). The difference arises mainly from the87

effect of an increasing (transversal tension test) or decreasing (longitudinal tension test) mesh-opening88


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√©√

√©√

√©√

©√

√

√©√

√©√ √

©√

√©√

√©√

(a)(b)

(c)

Fig. 2. Sketch of (a) the longitudinal plane tension tests, (b) the transversal plane tension test and (c) a detailedillustration showing the load cell position of a selected test series. The load cells positioned on lateral bearingsmeasured the load A in the mesh bearing from which the segment load F was derived as A/(2 cos α/2) for thetransversal tension test and A/(2 sin α/2) in case of the longitudinal tension test.

angle α on the 3D geometry of the rounded segment part. At the node, the contact point between two89

meshes has a position with an eccentricity exy to the normal force F acting in the straight segment90

parts (Fig. 5). Apparently, this in-plane eccentricity depends on the curvature of the rounded part and91

on the mesh-opening angle α. In addition, an eccentricity ez normal to the mesh plane is present, due92

to the mesh clearance hi.93

The aim is now to reconstruct the mechanical behaviour solely based on the distance between94

individual nodes and the steel wire properties, without additional free parameters.95

The scope of the model presented in this section is to reproduce the quasi-static tension tests of96

chain-link meshes in general, based on geometry and steel wire properties. The subscript 0 used for97

the mesh opening angle α0 (Fig. 1c) and for the initial clearance hi,0 (Fig. 1d) denotes the undeformed98

geometry, while the same symbols without a subscripted zero (0) represent specific geometric prop-99

erties corresponding to an actual load situation. The segment reaction forces are calculated based on100

the position of the two segment ends, the mesh nodes. Due to the rounded parts, the segment’s posi-101

tion and orientation is initially different to that of a straight connection line between the two segment102

nodes with length Lnodes,0. Due to the symmetry identified in the quasi-static plane tension tests the103


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Sl,0

Lnodes,0

Sl = Sl,0 + Sn,0 – Sn

Lnodes

½Sn,0½Sn

Sseg = Lnodes – (Lnodes,0 – Sl,0)

+ 0.7(Sn,0 – Sn )

½Sn

½Sn,0

Sseg,0 = Sl,0

( )a ( )b ( )c ( )d

Fig. 3. (a) A planar quasi-static tension test that equally distributes the load causing the segment loads F by a rela-tive segment node displacement δ. Node displacements can be substituted in a static system by forced deformationδ as illustrated in (b) to represent the segment deformation under load. Figure (c) illustrates the initial straightsegment length Sseg,0 based on the initial node distance Lnodes,0 and the length of the undeformed straight seg-ment part Sl,0 and initial rounded parts Sn,0. (d) shows a segment after deformation, accounting for the new nodedistance and a reduced rounded part. Sseg approaches the node-to-node distance.

10

12

14

16 Segment load [kN]

Tecco G65-4 (hi = 7mm) segment load with increasing and decreasing mesh opening angle

0

2

4

6

8

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

Transversal Tension Test

Longitudinal Tension Test

Segment strain [-]

Fig. 4. Curves showing the segment load F (as measured according to Figure 2c) over segment strain for alongitudinal (green) and a transversal (blue) plane tension test of Tecco G65-4 with 4 mm wire (hi = 7 mm).

reaction forces acting on the mesh nodes can still be described by the tension load F acting along the104

centre-line of the straight segment (Fig. 3). In the following it is the aim to model the true geometry105


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α

2

2��2��

α

2

(a) (b)

x

y

z

Fig. 5. The rounded geometry at the node causes an eccentricity exy of the segment load F within the mesh plane(x-y) which depends on the mesh opening angle α. Figure (a) illustrates the situation in a longitudinal tension testwhile figure (b) represents a transversal tension test. In addition, the eccentricity depends on the mesh clearancein z-direction.

and bending processes, which take place in the rounded parts, depending on the mesh nodes’ positions106

and the segment tension F .107

2.1. Mesh stiffness and bending deformations at the nodes108

If the segment was modelled as a straight steel wire (or multiple wire strand thread) connecting the109

segment nodes, without considering the three-dimensional curvature/geometry of the steel wire in the110

vicinity of the mesh nodes, the segment stiffness would be overestimated by an order of magnitude.111

Furthermore, the true segment stiffness, which would describe the segment tension F depending on112

the distance between segment nodes Lnodes (Fig. 3), has to change with the mesh opening angle α113

(Fig. 1). The key aspects of the varying stiffness result from the three dimensional wire geometry and114

its subsequent deformations under load, which are described in the following subsections.115

2.1.1. Three dimensional wire geometry116

A force F applied to the straight part of the wire has an eccentricity to the contact point at the node117

which lies within the rounded section S0 of the wire. This eccentricity drives the bending deformation118


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x

z

y

x.

F F

FFβ

F

F

hi

π

2S

ki

ka

dπ

2

α

2

ki

ka

hi + d

2

hi + d

4π

exy

β

(a) (b) (c) (d)

ih

2

1

Fig. 6. Determination of the initial length of the part of the wire forming the half-helix S by using the contactlines between the rounded wire part (a) and a virtual inner and outer cylindrical envelope (b). The geometricdetermination is apparent when looking at the unreeled half of the cylindrical envelopes and their correspondingcontact lines (d) together with the geometric measures (c). With increasing load, β transforms into half of the meshopening angle α leading to a simple calculation of exy (the subscript ’xy’ refers to the mesh plane) with Equation10.

of the wire and varies with changing load. To describe the eccentricity and - as a consequence - the full119

three-dimensional geometry through node positions, we link geometry of the steel wire to the mesh120

opening angle α (Fig. 1c) and the height hi, which denotes the clear distance vertical to the mesh plane121

between two adjacent segments (Fig. 1d). In a first step, the total segment wire length of an undeformed122

segment Ltot,0 is calculated. This length is split into the rounded part Sn,0 and the straight part Sl,0123

(Fig. 3c). The original total segment length is calculated by the additional orthogonal clearance between124

two segments following a cylindrical shape (Fig. 6), which is generated in the production process of125

the chain-link where the wire is wrapped around a steel cylinder. The length of the wire centre line126

in the rounded part Sn,0 can be derived from the contact lines between two virtual cylinders and the127

wire by addressing their geometry in a planar unreeled representation of the virtual cylinder’s mantle128

(Fig. 6d).129

Due to the curvature of the wire the tension force F in the straight part of the segment has an130

eccentricity (hi + d)/2 orthogonal to the mesh plane and an eccentricity exy parallel to the mesh131

plane (Fig. 6c). To begin with, both eccentricities are calculated based on the initial mesh width mw,132

mesh height mh, the wire diameter d0, the initial clearance hi,0 and the initial mesh opening angle α0133

(Fig. 1), as this set defines the three-dimensional geometry. The rounded part ofmh is roughly equal to134


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the diameter of the cylinder the wire was wrapped 180 degrees around during production, which equals135

hi,0. Thus the length of the straight segment component in the direction ofmh is lh−straight = mh/2−136

hi,0. The straight segment component in the direction of mw therefore is lh−straight tan(α0/2) =137

(mh/2−hi,0) tan(α0/2). The length of the virtual cylinders at the rounded part is therefore calculated138

as:139

Sx,0 = 2(mw

2− (

mh

2− hi,0) · tan

α0

2).[1]140

The wire centre line lies on a cylinder of diameter hi,0 + d0, thus half of the circumference length141

is:142

Sr,0 = (hi,0 + d0) · π2.[2]143

The average length of the undeformed steel wire within the rounded part can be obtained from the144

unwrapped geometry (Fig. 6 d) and is then given as:145

Sn,0 =√S2x,0 + S2

r,0 = (hi,0 + d0) · π

2 cos β0,[3]146

with β0 as the initial angle of the contact line between the centre line of the wire and Sr,0 (Fig. 6c147

and d). It is given as148

β0 = arctan(Sx,0/Sr,0).[4]149

The straight wire part is derived from the straight segment component in direction of mh and half150

of the initial mesh angle α0/2 as151

Sl,0 =mh

2 − hi,0cos α0

2

[5]152

with the total length of an undeformed wire per segment as:153

Ltot,0 = Sl,0 + Sn,0.[6]154

2.1.2. Distance between segment nodes155

A single chain segment in a plane tension test can be idealised, due to the symmetry of load156

(Fig. 3a), as an over-determined static system loaded by a forced deformation (Fig. 3b). An increasing157

distance between two neighbouring nodes is mainly driven by the interaction of bending and normal158


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forces in the rounded part. The elastic elongation δ could be calculated depending on F by superposing159

the deformations resulting from force, bending and shearing, using the concept of virtual work with a160

virtual force F ∗ = 1 (von Boetticher , 2012).161

Due to large deformations, second order terms become important and limit the applicability of162

classic beam theory. A pragmatic solution as reported in Betschon (1977) neglects the elastic bending163

deformations (see Section 2.1.4), and only accounts for irreversible bending deformations as they begin164

when the elastic bending resistance of the wire is reached. The plastic bending deformations proceed165

until the reduction in eccentricity e causes a decline in the bending moment F × e to a level just below166

the limit of the bending resistance, Mr.167

2.1.3. Plastic deformations168

The segment normal force and strain are the result of comparing the actual length of straight wire,169

derived from the segment node positions, with the corresponding actual length of unloaded steel wire170

available in the straight part of the segment. This unstressed straight segment length Sl is calculated by171

subtracting the actual rounded part length Sn from the unstressed total length.172

Sl = Ltot,0 − Sn.[7]173

The actual length of the rounded part can be expressed in analogy to Equation 3 depending on the174

actual clearance hi175

Sn =πhi + d02 cos β

,[8]176

where β is equal to β0 initially and transforms to β = α/2 with ongoing deformation. The distance177

hi + d0 is equal to double the out-of-plane eccentricity ez . The total eccentricity then becomes e =178 √e2z + e2xy with exy being the eccentricity parallel to the mesh plane. This total eccentricity is the lever-179

arm of the acting segment load F and is restricted related to the maximum elastic bending resistance180

Mr (see Section 2.1.4). hi + d0 is therefore181

hi + d0 = 2ez = 2√e2 − e2xy = 2

√(Mr/F )2 − e2xy.[9]182

Once the angle β becomes α/2, the in-plane eccentricity exy can be expressed according to the183

actual mesh angle α and the actual clearance hi (see Figure 6d).184

exy =hi + d0

4(π − 2) sin α/2[10]185


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Based on a segment load F causing plastic bending at a current clearance hi and a corresponding186

in-plane eccentricity using Equation 10, the clearance is adjusted according to Equation 9. This in187

turn changes exy and – again – hi. So, this step is repeated until a sufficiently precise hi is found188

which results in the bending moment being just below the bending resistance Mr. Thus, for a given189

segment load F , the corresponding clearance hi and the straight length of the steel wire Sl can be190

calculated based on plastic bending deformations. This elongation from Sl,0 to Sl caused by bending191

in the rounded parts dominates above any elastic elongation of Sl. Any increase of F that causes the192

bending moment in the rounded part to exceed the bending resistance leads to further transfer of steel193

wire from the rounded to the straight part, increasing Sl.194

The force F is derived as the reaction force of the straight steel wire due to its strain,195

F = EA(SsegSl− 1),[11]196

below the elastic tension limit of the wire, where E is the elastic modulus and A the cross-section197

of the segment steel wire.198

An increase in node distance causes an increase in Sseg and thus in F . However, an increase in F199

leads to plastic bending deformations and an increase in Sl, as described above, and based on Equa-200

tion 11 this again relaxes the actual segment tension F , which is why the steps from Equation 9 to201

Equation 11 have to be solved iteratively to find the reaction of a segment to a given node distance.202

This approach is suitable to reproduce the mechanical behaviour of the mesh in detail as shown in the203

following, but the geometry of the rounded part has to be treated accurately.204

Initially, the straight part of the segment does not align with the node-to-node connection due to205

the eccentricity e, but with ongoing reduction in the clearance hi, the segment becomes almost parallel206

to a node-to-node connection Lnodes (Figure 3d). In transversal direction, the length of the rounded207

part at the node can be approximated as hi π/2 tan(β) (see Fig. 6d), and according to von Boetticher208

(2012) a more precise representation is:209

2a∗ = πhi2

tan(β) + 2d√

3(1− sin(β)) .[12]210

If we imagine ∆ to be the difference between half of the initial node-to-node connection Lnodes211

and the length of half of the straight part of the segment Sl, ∆ can be derived if the node length 2a∗212

is known (see Fig. 7). Initially the straight segment length Sseg can be formulated based on the length213


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Fig. 7. Representation of the discrepancy ∆ between the straight segment part and a node-to-node connection. Asz1 = hi/2 and z2 = a

tan β− z1, it is possible to derive that ∆ = a

sin β− z2 cos β with β determined from the

3D representation of the segments (see Fig. 6)

Lnodes derived from node positions by accounting for the difference between the initial node-to-node214

connection Lnodes,0 and the initial length of the straight part of the segment Sl,0,215

Sseg = Lnodes − Lnodes,0 + Sl,0.[13]216

However, as the eccentricity decreases at the node under plastic deformations and the straight seg-217

ment part increases accordingly, the change in ∆ needs to be account for. As ∆ was formulated based218

on β and a∗, ∆ can also be formulated based on the rounded segment part Sn of Equation 8. The219

change in ∆ can be expressed as the change in the rounded part Sn,0 − Sn using Equation 3. Thus we220

modified Equation 13 by reducing the difference between Lnodes,0 and Sl,0 with decreasing length of221

the rounded part Sn:222

Sseg = Lnodes − Lnodes,0 + Sl,0 + 0.7(Sn,0 − Sn) .[14]223

Here the factor 0.7 represents a ratio between (∆0−∆min) and (Sn,0−Sn,min), where ∆min and224

Sn,min are the corresponding remaining lengths at hi = 0. More precisely, the average value for all225

mesh types is 0.737, with a standard deviation of ±0.006.226


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2.1.4. Change of mesh stiffness227

The static system of a segment loaded by an in-plane tension test is comparable to that of a sym-228

metric s-hook with clamped and partially movable nodes (Fig. 3b). As deformations increase in size,229

second order effects appear with the reduction of the eccentricity (Betschon , 1977). The interaction230

between the normal force and the bending moment in the rounded part must then be considered for231

the following reason: the bending deformation in the node reduces the rounded section and therefore232

the lever arm of F , and thereby again the bending load. Consequently the mesh becomes stiffer with233

increasing deformation, because a greater load increase is necessary to achieve the same transfer of234

wire from the rounded part to the straight part of the segment. At the same time, the bending resistance235

of the wire, Mr, which denotes the limit below which no further plastic bending deformations occur,236

is not constant. Two counter-acting processes therefore change Mr: one increases the resistance of the237

wire against further plastic bending deformations while the other process reduces Mr. The increase238

in Mr is due to the spreading of the region within the cross section of the steel wire that reaches the239

yield stress, and furthermore, the increase of the yield stress itself affects Mr due to steel hardening240

with ongoing deformations. The counter-acting reduction of Mr starts at higher loads and is a conse-241

quence of the interaction of bending and normal load. We show our approaches to both processes in242

the following.243

When the elastic bending resistance becomes the plastic bending resistance with ongoing plastic244

bending deformations,Mr evolves to a higher bending resistance than in the beginning, with a plasticity245

shape factor of about 1.7 for a circular cross-section (Knodel , 2010). Furthermore, the high tension246

steel shows a pronounced steel hardening under plastic deformation as the failure stress is much higher247

than the yield stress. We account for this effect with an internal calibration factor By that increases248

the bending resistance, which we calibrated to By = 4 based on the longitudinal tension test of Tecco249

G65-4 (Torres-Vila et al. , 2001) to fit the measured segment stress-strain curve.250

Before the failure limit is reached, the effect of interaction between bending and the normal force251

counteracts the increasing bending resistance, which again modifies Mr. An example for such N-M-252

interaction in chain-link meshes is shown in Escallon et al. (2015). For our approach we consider the253

section close to the node, where the normal force equals F sin α/2. Very generally, we consider the254

cross section of the segment to form a generalized plastic hinge (Jirasek and Bazant , 2002) and we255

receive the remaining wire section area available for a remaining plastic bending resistance Mr by256


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excluding a central part of the wire section AN . This core area AN is considered to carry the normal257

load, thus AN is defined by the area necessary to carry the normal force load under the yield stress fy258

AN =F sin α/2

fy.[15]259

The left-over ’upper and lower’ cross-section parts not occupied by AN form the plastic bending resis-260

tance, and it is assumed that one part is under tension yield stress and the opposite part under pressure261

yield stress:262

Mr = ysegAsegfyBy,[16]263

where Aseg is the remaining cross-sectional area (the cross-section of the segment minus AN ) and264

yseg the lever arm between the regions of Aseg under tension and the region under pressure at the265

opposite side of the cross-section. The influence of α on the combined interaction of bending, normal266

force and shear and its influence on the in-plane eccentricity exy make the mesh stiffness depending267

on the opening angle. Figure 4 shows the segment load F in the mesh of two in-plane tension tests.268

The load in the single segments is reconstructed from the bearing forces measured at a lateral sliding269

fixation (Fig. 2c). The difference in segment stiffness between transversal and longitudinal tension270

tests occurs due to the increasing opening angle in the transversal tension test and its decrease in the271

longitudinal tension test. The initial stiffness of the segment is the same in both tests because the initial272

opening angle is the same. Between strains of 0.02 and 0.03, the increase in bending resistance and273

the reduction of mesh eccentricity leads to a stiffer segment, and this process is more pronounced for274

smaller mesh angles. Above a strain of 0.03, the momentum- and normal load interaction at the node275

starts to dominate and reduces the segment stiffness. Finally, the mesh fails under bending with normal276

force and shear interaction, and in case of the longitudinal tension test, the shear resistance becomes277

the limiting factor.278

2.2. Mesh resistance279

Deduced from scratches and deformations (Fig. 8) of experimentally tested chain-link nets, an280

elliptical contact area can be assumed at the nodes between two chains. With the additional assumption281

of an elastic, perfectly plastic behaviour of the steel wire, the stresses in the contact area cannot exceed282

the yield stress. By integrating the yield stress over the ellipse width, the elliptic area may be substituted283

by an equivalent line load. A minimal contact area under yield pressure can be defined. Because yield284


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Fig. 8. Plastic deformations at the contact area (orange circle) of a failed segment, such deformations justified theassumption of an elliptical contact area.

pressure multiplied by the contact area is in balance with the two adjacent segment loads F , this yield285

pressure has to oppose the parts of F that act in longitudinal direction. Thus, this simplified contact286

area has to be oriented according to Figure 9b, normal to the longitudinal direction. The length of the287

contact area 2a∗1 immediately before failure can be estimated based on the opening angle, by assuming288

that all out-of-plane eccentricity has vanished and the wires at nodes are bent by α/2 with a radius of289

half the wire diameter d. The width of the contact area can be derived to 2b∗1 from the load F because290

the contact area multiplied by the yield strength must be equal to the contact force. This in turn defines291

the elliptic shape of the contact area. Integrating over the wire length leads to the resultant normal (N),292

shear (S) and bending moment (M) in the wire (Fig. 9a). To perform this integration, it is necessary293

to approximate the contact force distribution at the connections immediately before failure with a line294

load of variable magnitude c according to the local width of the contact area times the yield stress295

(Fig. 9a). The pressed contact area is then replaced by the static equivalent of a line load q(x) over296

the distance 2a∗1, were the magnitude of q in dependency to x is defined through the width at x of the297

given elliptic shape of the contact area. The line load q(x) allows a detailed reconstruction of the wire298

stresses in the node at the moment of failure (Fig. 9c). To register the moment of failure, we select the299

section of maximal normal force-shear interaction and calculate the reduced wire bending resistance300

due to the normal force-bending interaction combined with shear.301

During loading, F is calculated for each segment based on Equation 11 with Sseg based on the new302

node distances. If the elastic tension limit of the wire is reached, F is recalculated adding to the elastic303

tension resistance NR,el the plastic deformation times a plastic modulus Epl (a hardening parameter304

equivalent to the linearized slope of the stress-strain curve of a straight pulled wire after yielding) that305


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e

N V MF F

c

a)

FF

2a*

2b*

F F

x

y

)(xq

b)

c)

x

y

x

z

Fig. 9. (a) The three dimensional contact area with a corresponding contact line load c along the centre lineof the contact area is in equilibrium with the two segment loads F , leading to the corresponding section-forcedistributions of normal load N , Shear V and bending moment M . (b) The contact area is simplified to a planeelliptic area under yield pressure with face normal in the ’longitudinal mesh direction’ y and replaced by its staticequivalence, the line load q(x) derived as the width-integrated elliptical contact area under yield stress (c).


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was derived from single wire tension tests:306

F = NR,el + EplA(SsegSl− 1− εy) ,[17]307

where εy is the segment strain at the onset of yielding, when F = NR,el. Subsequently, the normal-308

force shear interaction is calculated along the rounded part at the node, followed by a test whether the309

resistance fulfills the constraint:310

(Q

QR)2 + (

N

NR,pl)2 < 1,[18]311

where Q is the shearing force and QR is the shear resistance of the wire and N and NR,pl denote the312

normal force and the tension resistance of the wire. In case of sufficient wire resistance to normal force313

and shear interaction, the wire resistance in the section with the highest value for the left-hand side of314

Equation 18 is tested for a combined interaction between normal force, shear and bending.315

Based on Schwarzlos (2005) the reduced normal force resistance NR,red due to shear becomes316

NR,red = (1− ρw)NR,pl = fu(1− ρw)πd2

4.[19]317

where fu is the failure stress of the wire, d is the wire diameter and318

ρw = (2Q

QR− 1)2.[20]319

The remaining bending resistance MR,red at this section of normal-force shear interaction is then320

calculated by linear superposition as:321

MR,red = BR ∗MR(1− N

NR,red),[21]322

whereMR is the plastic bending resistance of the wire. The factorBR compensates the neglected bend-323

ing deformation in the straight part of a segment according to Betschon (1977). The model increases324

the bending resistance by the additional factor BR which was calibrated to BR = 2.4 for all mesh325

types to achieve a good agreement of the modelled failure loads with planar tension tests performed by326

Torres-Vila et al. (2001).327

If the bending moment at this section exceeds this bending resistanceMR,red, the combined failure328

of the segment is assumed.329


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3. Detailed numerical implementation of mesh mechanics330

Above details were implemented into a detailed chain-link discrete element, which reproduces331

the failure load of longitudinal and transversal tension tests for single wire meshes, with wires of 3332

and 4mm diameter. The model uses the wire diameter, the coating against corrosion (which has an333

influence on the steel diameter), the tension yield load of the wire and the failure load as input, as well334

as the ratio of shear strength to tension strength in the wire. Furthermore, the elastic and plastic moduli335

are required along with details of the mesh geometry in terms of mesh width, mesh height, clearance336

and the mesh opening angle. All data except the shear strength and elastic- and plastic moduli are337

available from the standard data sheets of the manufacturer. The shear strength to tension strength338

ratio was approximately determined through plane tension tests using different wire types with a mesh339

opening angle close to zero, ranging from 0.38 to 0.44. In these tests the set-up was identical to the340

usual longitudinal tension test with 1 × 1m specimen mesh, but without lateral fixation, resulting in341

a shear-failure in the segment nodes at mesh opening angles close to zero (Fig. 10). The elastic and342

plastic limits of the wire and the elastic and plastic moduli were taken from single wire tests performed343

by Torres-Vila et al. (2001), Zund (2006), Kastli (2007), Stradtner and Steidl (2009), and Stradtner344

and Steidl (2009a). The plane tension tests (transversal and longitudinal) used for validation were in345

principle set-up according to Figure 2 but in most cases the transversal test had a larger mesh size of 26346

segments in load direction. Additionally, small changes in mesh height have occurred with changes in347

the mesh clearance hi,0, e.g. from 8 mm to a minimal value of 3.5 mm applied for Tecco G65-3-FLAT,348

6 mm for Tecco G65-4, and 5 mm for Tecco G65-3. Other clearances were used for the mesh types349

Tecco G80-4 and Rombo G80-3.350

Figure 11 shows example curves of simulated and measured (Zund , 2006; Stradtner and Steidl351

, 2009,a) loads and deformations until failure. The simulation has been done quasi-statically, i.e. the352

same explicit time integration scheme has been applied as for full dynamic simulations, only the speeds353

of the moving nodes are magnitudes lower. All model input was derived from the steel wire properties354

and the mesh geometry without calibration apart from BR and By which are constant for all mesh355

types. Table 1 lists the set of input parameters to model the corresponding plane tension tests shown in356

Figure 11.357

Small offsets and imperfections in the mesh geometry at the beginning of the tests, combined with358


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Fig. 10. (left) The initial set-up of longitudinal plane tension tests without lateral bearing (top: Rombo G80-3,bottom: Spider S230-4). (right) The moment after failure under high shear load.

Table 1. Example mesh type parameter to be used for modelling with FARO – detailed modelMechanical property Unit G80/4 G80/3 G65/4 G65/3d mm 4 3 4 3hi,0 mm 7 6.5 7 4Fwire,tension−failure kN 24.200 14.055 24.200 14.055Fwire,tension−yielding kN 17.869 10.526 17.869 10.526Fwire,tension−failure

Fwire,shear−failure– 0.44 0.44 0.44 0.44

E kN/mm2 224’060 267’617 224’060 267’617Epl kN/mm2 42’276 34’411 42’276 34’411mh mm 177 180 138.6 143mw mm 102 103 83.8 83

a model representation which neglects elastic bending deformations, cause a horizontal offset between359

the measured and simulated curves. We have chosen to only compare simulated and measured loads360

and deformations above a threshold of 1.5 % of the experimentally measured failure load, as we expect361

that most imperfections of wire geometry become negligible at that level. This load level is close to the362

beginning of plastic bending deformations in most cases.363

During the simulations we observed two cases of numerical oscillations:364

(i) One was during the elastic deformation of the mesh. As the segment load increases, the elastic365

bending moment (which proceeds the bending deformations) increases which at the same time366

reduces eccentricity and thus reduces the bending moment. A reduction in eccentricity and bend-367


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0

10

20

30

40

50

60

0 0.05 0.1 0.15 0.2

Mesh Load [kN]

Mesh Elongation [-]

Rombo G80-3 (hi = 7mm) under transversal tension

Tension Test

Simulation (uncalibrated)

0

50

100

150

200

250

300

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Mesh Load [kN]

Mesh Elongation [-]

Tecco G65-4 (hi = 7mm) under longitudinal tension

Tension Test


0

20

40

60

80

100

120

140

0 0.05 0.1 0.15 0.2

Mesh Load [kN]

Mesh Elongation [-]

Tecco G65-4 (hi = 7mm) under transversal tension

Tension Test


0

50

100

150

200

250

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Mesh Load [kN]

Mesh Elongation [-]


Tension Test Simulation (uncalibrated)

0

20

40

60

80

100

120

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Mesh Load [kN]

Mesh Elongation [-]



0

20

40

60

80

100

120

140

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Mesh Load [kN]

Mesh Elongation [-]

Rombo G80-3 (hi = 7mm) under longitudinal tension

Tension Test


0

20

40

60

80

100

120

140

160

180

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Mesh Load [kN]

Mesh Elongation [-]


Tension Test


0

10

20

30

40

50

60

70

0 0.05 0.1 0.15 0.2 0.25

Mesh Load [kN]

Mesh Elongation [-]



Fig. 11. Simulated and experimental curves of mesh load (in terms of tension load per meter mesh width) overmesh deformation (elongation per meter mesh length) for different mesh types. The left column shows longitudinaland the right column transversal tension tests.

ing moment would reverse the elastic bending deformations, and for larger chain-link systems368

the model would fall into uncontrolled numerical dynamic oscillations. However, elastic bending369

deformation would have a negligible contribution to the failure deformation because the elastic370

bending contribution reduces with decreasing hi. Therefore, we neglected elastic bending defor-371


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mations.372

(ii) Also, the second order effect of plastic bending deformation may lead to numerical oscillations373

with node deformations in the order of the quasi-statically moving nodes due to the coupling374

between segment load, bending moment, bending deformation and eccentricity as visible in some375

of the diagrams in Figure 11.376

At the beginning of the plane tension tests, geometrical imperfections contribute to the measured377

load. With ongoing mesh deformations, most irregularities of mesh positions are averaged out.378

4. Simplified numerical implementation of mesh mechanics379

The numerical costs of the approach in Section 3 are too high to apply the model to full-scale mesh380

barriers. However, the characteristic mesh behaviour and its mechanical background were captured,381

providing the basis for a simplified chain-link model. We developed a numerical chain-link element that382

uses the output of modelled or measured plane tension tests as described in Section 2 to characterize383

the anisotropic mesh behaviour. The detailed approach served herein as a proof of concept that the384

anisotropic behaviour of chain-link meshes can be formulated in dependency to the mesh angle α.385

We present this efficient and robust model in the following and illustrate its application to full-scale386

rockfall barriers.387

Like in the previous section, the numerical chain-link element of the simplified numerical model is388

designed to represent one zig-zag part of the net and consists of a number of segments that represent389

the wire between neighbouring nodes (Fig. 1). For simplification, possible slip of wire through the390

node is neglected. The length of an undeformed steel wire per node, Ltot,0, can as well be seen as the391

length of an undeformed wire between two contact points / nodes (Fig. 1c).392

Failure loads and the corresponding deformations differ depending on whether the mesh is pulled in393

a longitudinal- or in transversal direction (Fig. 4). As illustrated in Section 2, the mesh-opening angle394

α, and how it evolves during loading, plays a key role as it defines the three-dimensional structure of395

the contact in the nodes. The longitudinal and transversal tension tests deliver the mesh resistance and396

stiffness for two different opening angles. At the scale of a segment, two different maximum segment397

tension forces can be defined accordingly for two different mesh angles as α declines, in the case of the398

longitudinal deformation, or increases in the case of transversal tension test. Without lateral fixation399


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the mesh angle would decline to 0◦ in a longitudinal test or increase to almost 180◦ in a transversal400

tension test. In the transversal case, the tension strength of the steel wire and its failure deformation401

give a first approximation of the maximum mesh resistance at α = 180◦. The shear resistance of the402

wire approximately corresponds to the failure load for α = 0◦ where failure is expected to happen due403

to the shear caused by the contact force between the single chain wires. The simplified model, which404

we define in the following, takes these four mesh-angle dependant failure loads and deformations from405

plane tension tests as input (longitudinal and transversal tension tests as described in EOTA (2016),406

with and without lateral fixation). With this set of failure loads and failure deformations a coarse407

interpolation of segment stiffness and maximum segment tension for all opening angles is derived.408

The angle dependency identified in the experimental plane tension tests is linearly interpolated,409

providing failure loads Fu and failure strains εu for any given angle α. The failure load is the tension410

force F in the segment at which the wire breaks at the node. The chain-link model calculates the strain411

ε in every segment based on the actual node distance Sseg in relation to a node distance Sl referring to412

the unloaded and undeformed segment:413

ε =SsegSl− 1.[22]414

The actual angles α between the segments are calculated from node positions. The corresponding415

segment load F is then derived from the angle-dependant local failure load Fu(α) and failure strain416

εu(α) as417

F = εFu(α)

εu(α).[23]418

Figure 12 shows a comparison between the resulting modelled and measured plane tension tests.419

The simulation initially has the same segment stiffness in both test cases, but with ongoing deformation420

and corresponding change in mesh angle and thus stiffness, the modelled curves deviate from each421

other. As the measured failure load and failure deformations of the longitudinal and transversal test422

cases are model input, the model reaches the same maximal values as the experiment. The simplified423

model captures the dependence of stiffness on mesh angle. However, the counter-acting non-linear424

processes of eccentricity reduction versus the interaction of bending with normal load are represented425

with a linear material law for a given angle. This simplification is justified for the net types we address426

by their almost linear stress-strain behaviour which occurred when the deformation exceeded about one427


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0

1000

2000

3000

4000

5000

6000

7000

8000

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

Segm

ent

load

F [N

]

Segment strain [-]

Tecco G65-3 (hi = 8mm)

Longitudinal test, simulated

Transversal test, simulated

Longitudinal test, measured

Transversal test, measured

Fig. 12. Strain-load diagram for the segment load in a longitudinal and a transversal tension test, and the cor-responding curves derived from the simplified model of Tecco G65-3 (hi = 8 mm). In contradiction to the testsshown in Figure 4 the bearings on the sliding beam and its opposite side were constrained and not slidable.

third of the failure deformation (Fig. 11). The simplified model overestimates the stiffness at lower load,428

but the corresponding overestimation in energy dissipation is small compared to the total dissipation429

until failure. If a net type would show a different behaviour with a load-deformation curve far from430

linear in the quasi-static tension tests, the simplified model might not be able to represent the overall431

mesh behaviour.432

To also consider plasticity, un- or reloading, the load history becomes relevant. If the segment strain433

reaches maximum in the segment’s time history, the tension force F in the segment is calculated using434

the actual linear-plastic stiffness and strain according to Eq. 23. If the maximum strain was exceeded435

before, an unloading ∆F is calculated based on the difference of actual strain and previous maximum436

strain εmax applying the linear-elastic steel wire stiffness EA (Fig. 13). However, to account for the437

angle-dependant elastic bending deformations at the nodes, we use loading and unloading sequences438

from tests, and cover the relaxation behaviour of the mesh by the ratios between the loading stiff-439

ness and the stiffness during unloading, λlong and λtrans. We may replace EA by the corresponding440


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ε εmax1 εmax2

1

EA

ε1 ε2

F

Fmax1

Fmax2

F2

F1

ΔF1

ΔF2

Fig. 13. Strain-load diagram for a modelled segment pulled with a constant opening angle with two relaxations.The decrease in deformation to ε1 after ε1max leads to a linear-elastic load decrease ∆F1 and reloading followsthe elastic path until ε1max is again exceeded. As long as the actual strain exceeds all previous strains, the plasticstiffness is used to calculate the load. In the example shown here a second reduction in load takes place at ε2maxuntil ε2 is reached. The load F2 is calculated by subtracting ∆F2 derived from the reduction in strain ε2max − ε2multiplied by the elastic stiffness EA. The segment in the model is defined as the connection of two chain-linkelement nodes. The opening angle is kept constant in this case by a prescribed displacement of the segment nodesin direction of the segment axis.

stiffness E∗A derived from loading and unloading sequences in plane tension tests as:441

E∗A = λ(α)Fu(α)

εu(α).[24]442

The relaxation ∆F then becomes:443

∆F = λ(α)Fu(α)

εu(α)(εmax − ε).[25]444

The actual tension force F is then equal to the previously reached maximum tension force Fmax445

(known from Eq. 23 with ε = εmax), minus the relaxation ∆F .446

F = Fmax −∆F.[26]447

The discontinuity in stiffness may lead to oscillations if the static equilibrium is close to the max-448

imum load experienced. In this case, numerical damping measures might be advisable. However, in449

applications with dynamic loading as presented here this effect was not observed.450

No bending stiffness is modelled as the dominant effects of bending deformations in the rounded451

parts of the segments are included in the angle-dependant material-law. The modelled segments do452

not bear pressure in analogy to the reality, where node connections loose contact in such a case. The453

reacting forces F for each segment and their directions are therefore known from node positions. Thus,454

for the numerical time integration of a DEM as e.g. described in Volkwein (2004), the assembly of455


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Table 2. Example mesh type parameters to be used for modelling with FARO – simplified model (calibrated)Mechanical property Unit S4-230 S4-130 G80/4 G80/3 G65/4 G65/3Rmax,long kN/mm 268’560 465’480 273’120 157’560 334’200 200’400Rmax,trans kN/mm 136’320 205’440 119’520 64’920 148’560 69’230Fwire,tension kN 68.85 68.85 22.95 15.64 22.95 15.64Fwire,shear kN 26.37 26.37 11.00 5.94 11.00 5.94εmax,long – 0.0980 0.12 0.1033 0.0948 0.156 0.121εmax,trans – 0.3556 0.566 0.2433 0.2280 0.410 0.410λlong – 3.3 3.6 3.3 2.7 3.6 2.9λtrans – 3.3 3.6 3.3 2.7 3.6 2.9mh mm 500 315 177 180 138 143mw mm 292 175 102 103 83 83

the nodal forces can take place. The single mesh parameters are listed in Table 2. The parameters can456

be retrieved directly from tension tests on mesh specimen: mh and mw describe the geometrical size457

of one mesh part in width and length. Rmax,long and Rmax,trans contain the maximum loads from458

longitudinal and transversal tension tests. εmax,long and εmax,trans contain the lengthening of the459

tested meshes related to their original dimensions. From maximum loads and corresponding straining460

a kind of elastic stiffness of the mesh can be determined. Fwire,tension and Fwire,shear are the failure461

loads of single mesh wires.462

4.1. Model calibration463

To calibrate above model to the specific meshes three main characteristics have to be addressed:464

failure loads, failure deformation respective net stiffness and relaxation behaviour due to unloading.465

The corresponding calibration procedures are listed in the following resulting in the model parameters466

listed in Table 2.467

4.1.1. Failure loads468

The chain-link mesh parameters for the failure limits are defined both in longitudinal and transver-469

sal directions based on the results derived from small-scale quasi-static in-plane tension tests (Figs. 2470

& 4). From these tests estimates are derived for the failure loads in longitudinal and transversal mesh471

directions, Rmax,long and Rmax,trans , as-well as estimates for the wire failure loads at 0 or 180◦472

opening angles which are named Fwire,shear for 0◦ opening angle, and Fwire,tension for 180◦ opening473

angle. The failure loads at 0 or 180◦ opening angles are either based on quasi-static in-plane tension474

tests without lateral fixation or are calculated from the tension and shear resistance of the single wire.475


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Fig. 14. Example of a longitudinal plane tension test with Tecco G65-4 (hi = 8 mm), showing the measured loadat each lateral bearing except for the lateral mesh bearing 4 in the middle (no data available). The white arrowsindicate the main direction of the spreading of the load at mesh 5 and 3 throughout the mesh.

As described above there is a short-coming of the quasi-static in-plane tension tests because the476

model does not account for (i) the slip in the node that allows the transfer of some length of wire477

between the two adjacent segments and (ii) for load concentrations at the force inlets, see Figure 14.478

The model therefore resulted in underestimated failure loads when using the model for the simulation479

of vertical drop tests as shown in Figure 16. The values Rmax,long and Rmax,trans given in Table 2480

therefore already contain an increase compared to the in-plane tension test measurements by a factor481

of 1.2 for all mesh types.482

The drop tests (Fig. 16) applied a concrete sphere of 820 kg containing accelerometers, which was483

dropped vertically into a mesh specimen positioned horizontally in a steel frame. The frame forms484

an inside square of 4.35 x 4.35 m placed about 4 m above the ground (Grassl et al. , 2003). A rope485

was attached to the inside of the frame forming a horizontal square shape. It served as a bearing for486

the mesh specimen that were attached to the rope by shackles for each mesh on all sides (Fig. 16),487

resulting in a mesh specimen with a size of about 3.5 x 3.5 m. Load cells in every frame post measured488

the vertical impact forces over time in addition to the acceleration measurements inside the concrete489

sphere. For mesh types G80, S180 and S230 the measured accelerations were derived from the load490

cells placed in each frame post, resulting in a 16% uncertainty in the measured accelerations. High-491

speed video cameras tracked the displacement path of the sphere as it dropped into the mesh, rebounded492

and fell back into the mesh a second time. Video tracking and acceleration measurements were both493

used to calculate the velocity time series of the concrete sphere. The falling height was defined after494

initial quasi-static loading, where the sphere is placed in the middle of the mesh. From this reference495


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point, the sphere is lifted until it looses contact with the net and the corresponding height difference is496

considered as elastic deformation helastic. The sphere is then lifted further to the falling height h, with497

an accuracy of ±5 cm.498

4.1.2. Failure deformation499

The initial mesh stiffness is defined through the corresponding longitudinal and transversal failure500

deformation, termed εmax,long for longitudinal in-plane tension tests and εmax,trans for transversal501

in-plane tension tests, as they define the angle-dependency of the failure deformation in Eq. 23. Both502

failure deformations were increased in analogy to the Section 4.1.1 such that the model could capture503

the impact trajectory and mesh deformation at a vertical impact test close to the failure limit (Fig. 17)504

as well as the maximum acceleration (Fig. 16c). The quasi-static failure deformations were increased505

by a factor of 1.4 – 2.0 depending on the mesh type.506

4.1.3. Un- and reloading behaviour507

In case of unloading, the stress-strain-curves of the steel wire and meshes, respectively, follow a508

different path which is covered in the model by converting the original mesh stiffness using the factors509

λlong and λtrans (see Equation 24). These relaxation factors account for the plastic energy dissipation510

during an impact and the corresponding consumption of the plastic reserves which lead to a differed511

mesh performance under follow-up impacts. For each chain-link mesh type in our research it was512

possible to simplify the relaxation factors to λlong = λtrans = λ. To calibrate this mesh-type specific513

relaxation factor also above mentioned drop tests were used. λ was calibrated to match the simulated514

and observed time delay between the first and second maximum impact caused by the rebounding515

concrete sphere falling back into the mesh (Fig. 16c).516

The second impact of the concrete sphere is a perfect indicator of kinetic energy absorbed by the517

mesh during the first impact through (i) the time duration between first impact and rebounce and (ii)518

magnitude of the second impact. When calibrating λ some deviation between simulated and measured519

second impact time is accepted in favour of a correct prediction of the second impact peak decelera-520

tions.521


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-50

0

50

100

150

200

250

300

350

400

450

-20

-15

-10

-5

0

5

500 700 900 1100 1300 1500 1700

R003S1 1.75 m

Fallen distance, measured

Fallen distance, simulated

Velocity, measured

Velocity, simulated

Acceleration, measured

Acceleration, simulated

Fig. 15. Simulation of a mesh failure in the steel-frame impact test. Here the measured and simulated time seriesof a steel-frame impact test of a Tecco G65-3 mesh are shown. The concrete sphere is released from 1.75 m abovethe mesh at time t=730 ms.

4.2. Model accuracy and verification522

Figure 16c shows a case where the simulated and measured accelerations of the first and second523

impact match in both time and amplitude. Figure 17 shows a high resolution compilation of modelled524

and measured sphere trajectories for all mesh types, with the experiment set-up and model geometry525

similar to Figures 16a, b. The contact between sphere and a mesh node is modelled as fully plastic,526

i.e. after impact there is no rebound of the colliding masses. The tangential movement of a mesh node527

along the sphere is modelled using a Coulomb friction coefficient of 0.3. The maximum deformations,528

rebound heights and accelerations for tests closest to the failure limit are provided in Table 3. In the529

test for Spider S230 a natural stone (size = 0.9 x 0.48 x 0.3 m3) of mass 350 kg was applied instead of530

the concrete sphere.531

Nine tests lead to a mesh failure at first impact. The corresponding comparison of the modelled and532

the predicted failure load is provided in Table 4. The two tests with Spider S230-4 with 4.0 m falling533

height illustrate the 16% uncertainty of measured accelerations. The minor deviations in modelled534

failure load between two tests of identical set-up are due to varying initial elastic sag of the mesh.535


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(a) (b)

(c)-50

50

150

250

350

450

550

650

750

850

950

-20

-15

-10

-5

0

5

0 500 1000 1500 2000 2500

R008S1 4.00 m

rock displacement [m] velocity [m/s] rock acceleration [m/s2]

[m/s2] [m] [m/s] [kJ]

Time [ms]

[m/s2]

Fig. 16. (a) Experimental set-up with Tecco G65-3 in the steel-frame. (b) Simulation after the concrete sphere wasdropped from the crane into the centre of the mesh. The simulation screen-shot illustrates the force distribution atthe moment of impact with the impacting body removed from the view from a Tecco G80-4 mesh simulation witha drop height of 2.5 m. (c) Measured (continuous lines) and simulated (dashed lines) time series of a 4 m drop intoa Tecco G65-4 mesh.

-7

-6

-5

-4

-3

-2

-1

0

0 200 400 600 800 1000 1200 1400

z [m]

Time [ms]

Modeled (dashed) and measured impact trajectories

Tecco G65-3, h = 1.75 mTecco G65-4, h = 4.5 mSpider 130-4, h = 6.5 mSpider 230-4, h = 4.5 m, rock 350 kg Tecco G80-4, h = 2.5 mRombo G80-3, h = 0.8 m

Fig. 17. Comparison of modelled and measured trajectories of the concrete sphere at the point of maximum meshdeformation for different mesh types.


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Table 3. Comparison of simulated and measured maximum values from tests closest below the failure limit. sdenotes the mesh deformation, ∆ t is the time delay between first and second impact and a stands for acceleration.Spider S230 was tested with a 350 kg block.

Mesh type drop height modelled measured deviationmaximum deflection s

Tecco G65-3 1.75 m 0.88 m 0.93 m -5%Tecco G65-4 4.5 m 1.13 m 1.18 m -4%Rombo G80-3 0.8 m 0.48 m 0.26 m 85%Tecco G80-4 2.5 m 0.49 m 0.32 m 53%Spider S130-4 6.5 m 0.55 m 0.51 m 8%Spider S230-4 4.5 m 0.48 m 0.19 m 253%

re-bounce time ∆t

Tecco G65-3 1.75 m 900 ms 835 ms 8%Tecco G65-4 4.5 m 1130 ms 1114 ms 1%Rombo G80-3 0.8 m 685 ms 769 ms -11%Tecco G80-4 2.5 m 945 ms 942 ms 0%Spider S130-4 6.5 m 1868 ms 1376 ms -2%Spider S230-4 4.5 m 1195 ms 1126 ms 6%

max. acceleration aTecco G65-3 1.75 m 174 m/s2 166 m/s2 5%Tecco G65-4 4.5 m 299 m/s2 315 m/s2 -5%Rombo G80-3 0.8 m 129 m/s2 136 m/s2 -5%Tecco G80-4 2.5 m 232 m/s2 248 m/s2 -6%Spider S130-4 6.5 m 456 m/s2 537 m/s2 -15%Spider S230-4 4.5 m 363 m/s2 420 m/s2 -14%

Table 4. Comparison of simulated and measured failure load from tests above the failure limit. afail stands forthe acceleration of the sphere in the moment of failure.

Mesh type Drop height Modelled Measured Deviationfailure acceleration afail

Tecco G65-3 1.75 m 159 m/s2 163 m/s2 -2%Rombo G80-3 1.0 m 138 m/s2 148 m/s2 -7%Rombo G80-3 1.0 m 134 m/s2 150 m/s2 -11%Tecco G80-4 4.0 m 127 m/s2 82 m/s2 56%Tecco G80-4 3.0 m 128 m/s2 151 m/s2 -15%Tecco G80-4 2.75 m 123 m/s2 139 m/s2 -11%Spider S230-4 4.5 m 194 m/s2 198 m/s2 -2%Spider S230-4 4.0 m 176 m/s2 182 m/s2 -3%Spider S230-4 4.0 m 177 m/s2 167 m/s2 6%

A parameter study was carried out to identify to which uncertainties the model is sensitive to. The536

relaxation factors (λlong and λtrans) are not part of the standard mesh specifications and thus may be537

a source of uncertainty. Besides this, there are several parameters like time step size, viscous damping,538

or properties of the modelled concrete sphere which are properties of the FARO software and have539

no dependency on the implemented chain-link element, although they affect the results. For a detailed540


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R006S1 2.75 m

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thick lines: measured

Velocity [m/s]

Trajectory [m]

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Acceleration [m/s2] Velocity [m/s]

Trajectory [m]

Fig. 18. Sensitivity of the modelled impact to the mesh geometry uncertainty of a Tecco G65-4 mesh, for 2.75 mrelease height.

parameter study we refer to von Boetticher (2012). Here we simply want to highlight that a variation541

of the two relaxation factors (λlong and λtrans) by ±15% has the same effect as a ±3% variation in542

the initial unloaded mesh geometry (Fig. 18).543

In general, the modelled meshes with a higher node density, for example G65 or S130, perform544

better than those with less nodes per mesh area, e.g. the maximum deflection of meshes with more545

nodes per square meter are captured more precisely than those of identical segment cross section but546

larger segment lengths. A possible explanation might be that the definition of the maximum deflection547

is linked to the mesh shape prior to the impact, which is influenced by elastic deformations and mesh548

imperfections. The contributions of modelled processes in the mesh nodes to the overall deflection549

become larger with increasing number of nodes, reducing the influence of the unloaded mesh shape.550

5. Validation with fully featured rockfall protection systems551

The model was validated with large scale rockfall tests. We selected the structural protection sys-552

tems GBE-500A (TSUS , 2009) and GBE-1000A (TSUS , 2010) which contain a Tecco G80-4 mesh553

held by an upper and lower support rope. The test results with these systems were used to assess554

the model’s capability to predict maximum forces and deformations (Section 5.1). In Section 5.2, a555

so-called attenuating drapery system (Glover , 2010) compares the field test with the corresponding556


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simulation.557

A Coulomb friction coefficient of 0.3 was used between the net and a block made of concrete with558

a smooth surface. This friction coefficient was calculated from calibration procedures performed for559

the software FARO by Volkwein (2004). For the simulation of drapery net systems with an impacting560

sphere made of steel we used a friction coefficient of 0.15 instead. For comparison, Thoeni et al. (2014)561

used tan 30◦ = 0.58 for blocks with edges and corners.562

5.1. Model validation with vertical impact tests on GBE rockfall protection barriers563

The upper and lower support ropes of the GBE rockfall protection barriers were held in position564

by four posts that subdivided the barrier into three fields. The support ropes could slide perpendicular565

to the posts which allowed the transfer of the impact energy wave away from the impact location to566

anchors with energy dissipating devices. The bases of the posts were hinge supported to the slope and567

the top of each post was connected to two up-slope ropes. The beginning of the mesh at the first post568

and the end of the mesh at the last post were shackled to a rope that was spanned from the base of the569

post to its top. Two lateral ropes connected the first and last post to anchors positioned aside. All ropes570

were equipped with load cells which recorded the forces exerted on the ropes throughout the test. The571

up-slope and lateral ropes were not in contact with the mesh.572

The barriers were placed on an almost vertical wall at the test site Lochezen in Walenstadt, Switzer-573

land (Gerber , 2001), and retained a concrete block dropped vertically from about 32m above the bar-574

rier into the central area between the inner posts according to the testing standard ETAG 027 (EOTA575

, 2013). For the GBE-500A barrier the dropped block had a mass of 1590 kg resulting in an impact576

energy greater than 500 kJ . The GBE-1000A system retained a block of mass 3200 kg resulting in an577

impact energy of about 1000 kJ .578

For the corresponding simulations, a time step of 5 ∗ 10−6 s was used. The energy dissipating579

devices were modelled as truss elements which undergo plastic elongation. Their modelled material580

behaviour was based on the force-deflection diagrams in which the onset of elongation was taken from581

quasi-static tension tests. The final elongation measured at the MEL-test together with the measured582

peak load at the dissipating devices (taken from test reports for the corresponding MEL-test) formed a583

second point in the force-deflection diagram of the modelled truss element.584

The simulation predicted the measured maximum values well (Table 5, Fig. 19) considering that585


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Table 5. Comparison of simulated and measured maximum valuesGBE-500A Simulation Measurement DeviationStopping distance 4.99 m 4.95 m 1 %Upper support rope (left anchor) 102.6 kN 100.4 kN 2 %Upper support rope (right anchor) 103.1 kN 103.9 kN -1 %Bottom support rope 110.1 kN 95.3 kN 15 %GBE-1000AStopping distance 7.59 m 8.60 m -15 %Upper support rope (left anchor) 119.4 kN 108.1 kN 10 %Upper support rope (right anchor) 118.7 kN 98.2 kN 20 %Bottom support rope 95.3 kN 109.4 kN 11 %

(a)

Time [s](b)

Upper left support rope, measuredUpper right support rope, measuredUpper left support rope, simulatedUpper right support rope, simulated

Load [N]

Load [N]

Time [s]

Bottom support rope, simulatedBottom support rope, measured

Fig. 19. Comparison of simulated and measured rope forces in (a) the bottom support rope and (b) the uppersupport ropes of barrier type GBE-500A, MEL test 63.2

the model applies a mesh type once calibrated based on small-scale tests. Failed segments of the steel586

mesh appear at the same locations both in the simulation (red segments in Figure 20) and in the test.587

The same comparison of a modelled and measured time series was performed for the GBE-1000A588

MEL test (Table 5). A temporal average of the modelled rope forces increase continuously over time,589


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Fig. 20. Front view perspective of the modelled barrier GBE-500A at the moment of maximum deformationwith colours corresponding to the degree of utilization of the structure. The rock begins to move upward andsome segments at the upper support rope failed, both in simulation and experiment, when the sliding mesh hit theconnection between the upper support rope and the central posts.

with a clear signal from the onset of elongation at the dissipating devices right before 0.1 s (Fig. 21).590

The evolution of the measured rope forces indicate that the material law of the modelled dissipating591

devices after the onset of elongation was simplified to a point were it cannot reproduce the dynamic592

response in detail. However, the overall energy dissipation was captured well, as the decrease in mea-593

sured rope forces towards the end is represented adequately by the model for both time and amplitude.594

5.2. Wave propagation595

So-called attenuating systems Glover et al. (2012) present a different form of protection, which596

guides the rock downhill between the mesh and the slope. Here, a main visible effect of rock impact is597

the wave propagation along the mesh drape, without any influence from energy dissipating elements,598

resulting in a test case of pure chain-link dynamics. The reference system was tested using a 2700 kg599

steel sphere which impacted horizontally at 10 m/s into an inclined mesh of type Spider 4-130 (Glover600

, 2010) connected to a 40 m long and 17.5 m wide drape mesh of type Spider 4-250. The mesh was601

held on its upper edge by a support rope and laid loosely on the 50◦ inclined hill-slope (Fig. 22). The602

forces were introduced orthogonally and tangentially to the mesh, the latter by friction between the603

steel-sphere and the mesh. The impact wave propagation was dominated by the mesh stiffness under604

low load. Aside from the use of segments formed from multiple wire strand threads for different mesh605


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MEL Test 66.2 Upper Support Ropes

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upper right support rope measured

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upper right rope simulated

(a)

(b)

Upper left support rope, measured Upper right support rope, measured Upper left support rope, simulated Upper right support rope, simulated

Load [N]

MELTest 63.2 Upper Support Ropes

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MELTest 63.2 Upper Support Ropes

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0 0.1 0.2 0.3 0.4 0.5 Time [s]

0 0.1 0.2 0.3 0.4 0.5 Time [s]

Fig. 21. Comparison of simulated and measured rope forces in (a) the bottom support rope and (b) the uppersupport ropes of barrier type GBE-1000A, MEL-test 66.2

types, the test covered the stiffness model under a small load and thereby introduced a new perspective606

to the model validation.607

The overall shape of the dynamic drape deformation is captured by the model (Fig. 22). The wave608

propagation is represented in the load signal registered at the support rope and up-slope rope. The dy-609

namic oscillation of the modelled and measured rope forces is directly related to the wave propagation610

throughout the mesh and thereby represents the mesh stiffness and the contact between steel sphere611

and mesh. The dynamic signal of the simulated rope forces differs from the measured one in terms612

of amplitude, because the contact between mesh drape and slope was modelled without any friction613

(Fig. 23). Due to the neglection of friction between mesh and slope, the modelled impact causes an614

initial decrease in the retaining load because the sphere lifts the drape hanging on the support rope. In615


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Retaining rope Posts Support rope

Shackles

Fig. 22. (a) Experimental impact in the attenuating system and (b) the corresponding model perspective. Theimpact wave propagation results in the same angle, in both the experiment and simulation, between the mesh infront and the mesh behind the steel sphere which is a key aspect for the friction between the steel sphere and themesh. The dashed lines indicate shape structures that appeared in the experiment.

the experiment, the impact area is under less static load due to the friction between drape and slope,616

which results in reduced local stiffness and amplitudes of force in the rope compared to the simulation.617

However, the modelled and measured dynamic signals evolve with a comparable swinging which in-618

dicates accuracy in the stiffness model. Never-the-less, the non-linear change in stiffness with ongoing619

deformation as it is especially apparent under little load, is not represented by the simplified model.620

Modelled attenuating systems with other net types with a more pronounced curvature in the quasi-static621

plane tension tests may not perform well.622

5.3. Execution time623

The time taken to run the simulation on a single processor with 2.67 GHz (Pentium i7) was 52624

seconds for each second of simulated impact for a steel-frame impact test with 587 chain-link nodes625

(24 chain-link elements with 12 segments each) and 134 other mesh nodes for the frame truss, shackles626

and cable elements. The model of the GBE-500A barrier included 360 chain-link elements with 32627

segments each, resulting in 11732 chain-link mesh nodes. Together with 181 additional nodes for 170628

other elements of the barrier, the simulation of the GBE-500A barrier took 17 minutes per second of629

impact with the processor mentioned above. For comparison, the simulation of plane tension tests with630


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JG, JS StLeonardFaroAuswertung.xls

WSL Birmensdorf 29.08.2011 Page 1

AT4 - 27.05.10 Test Nr. 2 Static load removed

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Retaining rope force

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Retaining rope simulated

Support rope simulated

Fig. 23. Comparison of modelled and measured dynamic rope forces of attenuator-test AT4-2.

the detailed mesh mechanics implementation of Section 3 took about 30 minutes on the same machine.631

6. Conclusions632

We have developed a detailed analysis showing how the three-dimensional geometry of chain-link633

meshes can be determined in dependency of the node-to-node distance and the mesh opening angle634

α. It was possible to investigate two dominant processes that take place in the rounded wire parts at635

the nodes when the chain-link mesh is under load: (i) a decrease of the eccentricity of the straight636

wires with a corresponding increase in mesh stiffness and (ii) a decrease in wire bending resistance637

due to M-N interaction. Both effects show a clear dependency on the mesh opening angle α. The638

findings were implemented in a detailed numerical model that can reproduce the behaviour of different639

chain-link net types and thus allows to investigate the effect of changes in initial mesh geometry and640

steel wire properties. However, high numerical costs and possible numerical oscillations under certain641

loads make the detailed model inappropriate to address full-scale rockfall protection barriers. Instead,642

the mesh behaviour as captured by the detailed model was inherited into an α-dependent material643

law implemented into a simplified model. The counter-acting processes of changing mesh stiffness644


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mentioned above were simplified to a linear stress-strain relation but with an α-dependent gradient645

and an α-dependent failure limit. The simplified discrete element model can simulate the chain-link646

mesh based on either the results of small-scale quasi-static tension tests or corresponding results of647

simulations using the detailed chain-link model, both serving as material-law input.648

In this study, all necessary information was extracted from the small-scale quasi-static tension tests649

and introduced to the model by transforming the complex non-linear three dimensional deformation650

processes within the mesh nodes into the material-law, with an interpolated dependency on the inner651

mesh opening angle. The α-dependent interpolation of the simplified model allows to account for the652

dynamic dependency of the mesh stiffness, resistance and failure on the inner mesh opening angle. The653

mesh stiffness within an unloading and reloading cycle was captured using a ratio between the elastic654

stiffness of a second loading and the stiffness of the first loading, such that the simplified model can655

capture the energy dissipation of a mesh impact. Under low load, this simplification does not represent656

the true mesh stiffness.657

Some calibration of the overall mesh stiffness is necessary to account for the local impact as loading658

of the mesh is not equally distributed, unlike quasi-static plane tension tests. This mesh stiffness is659

defined by the failure deformation. The failure deformations derived from plane tension tests needed660

to be increased by a factor of 1.4 – 2.0 depending on the mesh type and based on results from medium-661

scaled tests (mesh specimen size about 3.5 x 3.5 m) with a vertical impact. It was further possible to662

calibrate the elastic stiffness of the simplified chain-link mesh, such that the energy loss of a rebounding663

impact body was covered well by the model. The model accurately predicted all first impact mesh664

failures observed at the medium-scale experiments. Furthermore, the change in mesh stiffness by the665

first impact was predicted by the model, such that relatively high loads caused by second impacts were666

accurately captured: Figure 16 shows that a 25% rebound height (1 m) for the second impact can still667

reach 70% of the first impact’s maximum acceleration. Considering rockfall barrier design, this result668

highlights the importance of considering rockfall events with more than one impact.669

Regarding the full-scale tests the simplified model accurately predicts the maximum load as well670

as the maximum deformation of the GBE-500A barrier under an MEL-test (Table 5). Single failed671

segments were predicted by the model at the correct locations. Considering the GBE-1000A rockfall672

barrier, the simulation covered the duration of the impact (Fig. 21), but rope forces were overestimated673

and the maximum deformation was underestimated (Table 5). A possible source for the deviation be-674


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tween modelled and measured rope forces in the GBE-1000A rockfall barrier test may be the modelled675

energy dissipating devices with a simplified material law and by underestimating the energy dissipa-676

tion in the modelled mesh and the friction between the mesh and the ropes. Also, the underestimated677

maximum deformation suggests that the plastic deformation of the segments for large opening angles678

was underestimated. It therefore might be useful to perform additional transversal tension tests without679

lateral fixation to determine the mesh stiffness for large mesh angles more accurately.680

Plane tension tests can serve as a first estimate for the chain-link properties because the limitations681

incurred by not considering an orthogonal impact appear to be systematic. It is possible to account682

for the difference between plane tension tests and orthogonal impacts with standard corrections that683

increase the longitudinal and transversal failure loads and deformations by a predefined factor. In case684

of the mesh types considered here, this resulted in an increase of failure loads by a factor of 1.2 for all685

mesh types. However, as soon as the chain-link design development reaches a state of large scale test686

application, orthogonal impact tests will be necessary to prove if the standard correction also applies687

for the modified mesh.688

Plastic energy dissipation depends on the elastic stiffness, which can be determined by a temporal689

relaxation during a plane tension test. During a vertical impact test orthogonal to the mesh plane the690

dissipation can be measured with high accuracy by considering the time delay between the first and691

second impact when the rebounding sphere falls back into the net, or by the difference in release and692

rebound height. von Boetticher (2012) carried out about 60 experiments with the frame with mesh693

specimen of type Tecco G65-3, Tecco G65-4 and G80-4, Rombo G80-3, Spider 130-4 and Spider 230-694

4 to calibrate the elastic and plastic stiffness for each mesh type. The parameter study indicates that a695

source of uncertainty, considering the rebound height and plastic energy dissipation in the mesh, may696

arise from small variations in the initial mesh geometry. In cases where these variations are distributed697

randomly across the mesh, segment deformations will balance out under high load showing little effect.698

However, if meshes are produced with in general shorter or longer segments or different hi,0 due to an699

update in production, the mesh performance and especially the plastic dissipation is affected.700

The model is capable of predicting the dynamic behaviour and resistance of different protection701

barriers with high accuracy, optimized calculation time and minimized calibration efforts. The detailed702

comparison between simulation and experiment for different scales and types of impact have led to the703

following conclusions:704


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1. The new chain-link discrete element model was able to simulate the behaviour of all different705

chain-link mesh types tested in von Boetticher (2012).706

2. The model can be applied to different barrier types without recalibration. However, different707

mesh types will require additional testing.708

3. The model can accurately capture the dynamic wave propagation and the corresponding dynamic709

loading throughout the mesh for high energy vertical impact as well as for low energy tangential710

impact.711

4. The model can reproduce the energy dissipation due to plastic deformation and the correspond-712

ing change in mesh stiffness while some uncertainty remains concerning the sensitivity of this713

process to small deviations in the manufactured mesh geometry.714

5. Situations where the sphere broke the net and fell through were captured by the simulation in all715

cases. The reduction of the remaining mesh stiffness due to such holes can be accounted for by716

an unravelling approach (Hendrich , 2010).717

Acknowledgements718

The project that lead to this article has been co-financed by Swiss Commission for Technology and719

Innovation (CTI 9266.1 PFIW-IW) in collaboration with the Company Geobrugg AG. The authors are720

grateful for the possibility to develop the numerical models based on valuable experimental data. The721

conduction of the corresponding tests was enabled by Geobrugg. The authors further thank the two722

reviewers who significantly helped to improve the article.723

References724

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impacted by a dry granular flow. Rock Mechanics and Rock Engineering, 50(11), 3029-3048, 2017.726

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Cazzani, A., Mongiov, L., Frenez, T. Dynamic finite element analysis of interceptive devices for falling729

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