Presentation contents 1. Role of geosynthetic veneer reinforcement - PowerPoint PPT Presentation

Landfill 2015 Geogrid reinforcement in harsh environments Peter er Assinde der HUESKER Synthetic GmbH, Manager AFRICA Laura a Ca Carbone ne HUESKER Synthetic GmbH, Engineering department Morne ne Breyten ytenbach bach HUESKER Synthetic GmbH, Manager MINING

Presentation contents 1. Role of geosynthetic veneer reinforcement in barrier systems 2. Reinforcement performance in elevated temperatures Conference & Exhibition Landfill 2015 September 2015

Why do we need veneer reinforcement? 1989 test section on German landfill 100m x 20m section 1 in 4 (14 ° ) slope - Clay foundation - HDPE geomembrane (“semi - textured”) - 2000g/m 2 nonwoven protection geotextile - 110 kN/m (ultimate) uniaxial geogrid reinforcement - 0.5m thick gravel drainage layer Design worst case interface friction angle = 17.1 ° (15.4 ° ) Conference & Exhibition Landfill 2015 September 2015

Why do we need veneer reinforcement? ° Conference & Exhibition Landfill 2015 September 2015

Veneer cover example layout t Geogrid (solution A and B2) Smooth (solution A) Textured (solution B) b =13 ° Conference & Exhibition Landfill 2015 September 2015

Veneer reinforcement Liu & Gilbert methodology • Simple analytical model to estimate geosynthetic loads during the placement of cover soils and/or waste • Graphical method based on an accurate and test validated mathematical model • Considers strain compatibility between the layers • Models tensile stress distribution between individual layers • For each layer any induced load is proportional to it’s stiffness (relative to the total stiffness) Conference & Exhibition Landfill 2015 September 2015

Model properties Solution A – Solution B – Material Smooth gmb lining system Textured gmb lining system Properties Interface friction Properties Interface friction angles angles g soil = 18 kN/m f soil/GTX = 29 ° g soil = 18 kN/m f soil/GTX = 29 ° Granular soil f soil = 32 ° f soil = 32 ° t soil = 0.5m t soil = 0.5m K c = 485 kN/m K c = 485 kN/m f GTX/GMB = 12 ° f GTX/GMB = 31 ° Geotextile K t,GTX = 50 kN/m K t,GTX = 50 kN/m (GTX) f GMBs/clay = 11 ° K t,GMBt = 308.3 kN/m f GMBt/clay = 14 ° Geomembrane K t,GMB = 308.3 kN/m (GMB) (lab test value) (lab test value) f GMBs/clay = 9 ° f GMBt/clay = 9 ° (design value) (design value) Geogrid (PET) K t,GR = 1100 kN/m / K t,GR = 350 kN/m / (GR) (for Solution B2) Conference & Exhibition Landfill 2015 September 2015

Estimation of tension load in geosynthetic layers – Example Solution B1 b = 13 ° , t = 0.5m, g = 18 kN/m 3 , f sec = 9 ° , L es = 30m, L eg = 0m K t = 708.3 kN/m, K c = 485 kN/m K t /K c = 0.74, L eg /L es = 0, 𝑢𝑏𝑜𝜒 (𝑢𝑏𝑜9) f net = 1 – tan 𝛾 (tan 13) = 0.314 L t /l es = 0.475 𝑀 𝑢/ ∆𝜐 𝑕𝑡 = 𝜒 𝑜𝑓𝑢 ∆𝑈𝑕𝑡 = ∆𝜐 𝑕𝑡 (𝛿𝑢𝑀 𝑓𝑡 𝑡𝑗𝑜𝛾) = 9.1 kN/m 𝑀 𝑓𝑡 = 0.15 X Conference & Exhibition Landfill 2015 September 2015

Estimation of tension load in geosynthetic layers Solution Tension load in geosynthetic layers (kN/m) A 12.1 B1 9.1 B2 10 Conference & Exhibition Landfill 2015 September 2015

Distribution of tensile forces 1. Smooth GMB with geogrid reinforcement 𝐿 𝑢𝑕𝑢𝑦 50 𝐿 𝑢𝑢𝑝𝑢 = 1458.3 = 3% % carried by GTX = 𝐿 𝑢𝑕𝑛𝑐 308.3 1458.3 = 21% % carried by GMB = 𝐿 𝑢𝑢𝑝𝑢 = 𝐿 𝑢𝑕𝑠 350 1458.3 = 76% % carried by GR = 𝐿 𝑢𝑢𝑝𝑢 = 2. Textured GMB no reinforcement 𝐿 𝑢𝑕𝑢𝑦 50 𝐿 𝑢𝑢𝑝𝑢 = 358.3 = 14% % carried by GTX = 𝐿 𝑢𝑕𝑛𝑐 308.3 358.3 = 86% % carried by GMB = 𝐿 𝑢𝑢𝑝𝑢 = Conference & Exhibition Landfill 2015 September 2015

Distribution of tensile forces continued 3. Textured GMB with geogrid reinforcement 𝐿 𝑢𝑕𝑢𝑦 50 𝐿 𝑢𝑢𝑝𝑢 = 708.3 = 7% % carried by GTX = 𝐿 𝑢𝑕𝑛𝑐 308.3 708.3 = 43% % carried by GMB = 𝐿 𝑢𝑢𝑝𝑢 = 𝐿 𝑢𝑕𝑠 350 708.3 = 50% % carried by GR = 𝐿 𝑢𝑢𝑝𝑢 = 4. Textured GMB with stiffer geogrid reinforcement (same UTS) 𝐿 𝑢𝑕𝑢𝑦 50 𝐿 𝑢𝑢𝑝𝑢 = 941.6 = 5% % carried by GTX = 𝐿 𝑢𝑕𝑛𝑐 308.3 941.6 = 33% % carried by GMB = 𝐿 𝑢𝑢𝑝𝑢 = 𝐿 𝑢𝑕𝑠 583 941.6 = 62% % carried by GR = 𝐿 𝑢𝑢𝑝𝑢 = Conference & Exhibition Landfill 2015 September 2015

What happens to the reinforcement if we change the in-situ temperature? Temperature plays a major role in all degradation mechanisms and in mechanical behaviour (creep and rupture) The temperature of the soil is constant (to within ± 0.5ºC) only at a depth of 10 m or more. Its value is then equal to the annual average atmospheric temperature at the surface. Daily and seasonal variations occur with decreasing intensity as the distance from the surface increases. Since higher temperatures increase the rates of ageing and creep of polymers disproportionally, their effect on geotextile behaviour may need to be considered… Source: ISO 13434 – Guidelines on durability Conference & Exhibition Landfill 2015 September 2015

What happens to the reinforcement if we change the in-situ temperature? Effective design soil temperature - In the absence of other information the effective design temperature can be taken conservatively as the average of the mean annual air temperature and the mean air temperature for the hottest month of the year Source: ISO 20432 Guide to the derivation of reduction factors for soil reinforcement materials = 19.5 ° c = 23.25 ° c Conference & Exhibition Landfill 2015 September 2015 Source: worldweatheronline.com

Long term strength as a function of conditions Characteristic Value of the Long Term Reinforcement Strength R R = = B, k , B k R [kN/m] and R [kN/m] 0 g B, k B, d A * A * A * A * A 1 2 3 4 5 M R B,d Design value of the tensile strength of geosynthetic reinforcement R B,k Characteristic value of the long-term tensile strength R B,k0 Characteristic value of the short-term tensile strength A 1 Reduction factor for creep strain and creep rupture behaviour (depending on the load duration) A 2 Reduction factor for damage caused during installation, transportation and compaction A 3 Reduction factor for processing (seams, connections, joints) if applicable A 4 Reduction factor for environmental impacts (resistance to weathering, chemicals, microorganisms, animals) A 5 Reduction factor for the impact of dynamic action ɣ M Partial safety factor for the structural resistance of flexible reinforcement elements Conference & Exhibition Landfill 2015 September 2015

Influence of temperature on creep rupture behaviour, A1 Creep-Rupture Behavior 79.0 77.0 75.0 Tensile load [%] 73.0 71.0 69.0 67.0 65.0 0.00 0.01 0.10 1.00 10.00 100.00 1000.00 Design life t D [years] PET- 50 °C PET- 20 °C PET- 10 °C PVA- 50 °C PVA- 20 °C PVA- 10 °C Conference & Exhibition Landfill 2015 September 2015

Influence of temperature on creep rupture behaviour, A1 Effect of temperature on reduction factor for a given design life for PET and PVA reinforcement Retained strength A 1 Design [%] Design life temperature [years] [ ° C] PET PVA PET PVA 100 10 68.8 72.7 1.45 1.38 100 20 66.2 71.6 1.51 1.40 100 35 62.3 69.7 1.61 1.43 100 50 58.5 67.8 1.71 1.48 Conference & Exhibition Landfill 2015 September 2015

Influence of temperature on creep rupture behaviour, A1 Time required to reach % residual strength Temperature (days) – PET reinforcement ( ° C) 95% strength 90% strength 85% strength 60 273.1 546.2 819.3 70 99 198 296.9 80 33.8 67.5 101.3 90 10 20.1 30.1 Conference & Exhibition Landfill 2015 September 2015

Influence of temperature on creep rupture behaviour, A1 for HDPE Source: Kasozi et al, 2015 Conference & Exhibition Landfill 2015 September 2015

Influence of temperature on chemical degradation, A4 Hydrolytic Degradation Curve for PET Products 100.0 99.5 99.0 Retained strength T [%] 98.5 98.0 97.5 97.0 96.5 96.0 95.5 95.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 Design life [years] 50 °C 20 °C 10 °C Conference & Exhibition Landfill 2015 September 2015

Influence of temperature on chemical degradation, A4 Effect of temperature on reduction factor for a given design life for PET reinforcement Retained Design life Design temperature [ ° C] strength R 4 /RF CH [years] [%] 100 10 99.5 1.01 100 20 97,3 1.03 100 35 75.5 1.32 100 50 0 Fail Conference & Exhibition Landfill 2015 September 2015

What happens to the reinforcement design strength if we change the in-situ temperature? Example B2 (textured gmb with geogrid ) Based on a PET geogrid with UTS of 35 kN/m Assume 100 year design life at 35 ° C in-situ temperature Total load carried by geosynthetics = 10 kN/m 50% load carried by geogrid = 5 kN/m 35 13.7 1.61 𝑦 1.2 𝑦 1.0 𝑦 1.32 𝑦 1.0 = 13.7 𝑙𝑂/𝑛 1.2 = 11.4 𝑙𝑂/𝑛 Increase in-situ temperature from 35 ° C to 50 ° C Note: Assumes 35 environment for internal 1.71 𝑦 1.2 𝑦 1.0 𝑦 𝐺𝑏𝑗𝑚 𝑦 1.0 = 0 𝑙𝑂/𝑛 hydrolysis Conference & Exhibition Landfill 2015 September 2015

What happens to the reinforcement if we change the in-situ temperature? Source – ISO 13434 – guidelines on durability Conference & Exhibition Landfill 2015 September 2015