Geogrid reinforcement in harsh environments. HUESKER Synthetic GmbH, Manager AFRICA. HUESKER Synthetic GmbH, Engineering department.

Transcription

1 Geogrid reinforcement in harsh environments Peter Assinder HUESKER Synthetic GmbH, Manager AFRICA Laura Carbone HUESKER Synthetic GmbH, Engineering department Morne Breytenbach HUESKER Synthetic GmbH, Manager MINING

2 Presentation contents 1. Role of geosynthetic veneer reinforcement in barrier systems 2. Reinforcement performance in elevated temperatures

3 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 ) g/m 2 nonwoven protection geotextile kn/m (ultimate) uniaxial geogrid reinforcement - 0.5m thick gravel drainage layer Design worst case interface friction angle = 17.1 (15.4 )

4 Why do we need veneer reinforcement?

5 Veneer cover example layout t Geogrid (solution A and B2) Smooth (solution A) Textured (solution B) b =13

6 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)

7 Model properties Material Solution A Smooth gmb lining system Properties Interface friction angles Granular soil g soil = 18 kn/m f soil/gtx = 29 f soil = 32 t soil = 0.5m K c = 485 kn/m Solution B Textured gmb lining system Properties Interface friction angles g soil = 18 kn/m f soil/gtx = 29 f soil = 32 t soil = 0.5m K c = 485 kn/m Geotextile (GTX) Geomembrane (GMB) Geogrid (PET) (GR) K t,gtx = 50 kn/m f GTX/GMB = 12 K t,gtx = 50 kn/m f GTX/GMB = 31 K t,gmb = kn/m f GMBs/clay = 11 (lab test value) f GMBs/clay = 9 (design value) K t,gr = 1100 kn/m / K t,gr = 350 kn/m K t,gmbt = kn/m f GMBt/clay = 14 (lab test value) f GMBt/clay = 9 (design value) (for Solution B2) /

8 Estimation of tension load in geosynthetic layers Example Solution B1 b = 13, t = 0.5m, g = 18 kn/m 3, f sec = 9, K t = kn/m, K c = 485 kn/m L es = 30m, L eg = 0m K t /K c = 0.74, L eg /L es = 0, L t /l es = f net = 1 tanφ (tan9) tan β (tan 13) = τ gs = φ net L t/ L es = 0.15 Tgs = τ gs (γtl es sinβ) = 9.1 kn/m X

9 Estimation of tension load in geosynthetic layers Solution Tension load in geosynthetic layers (kn/m) A 12.1 B1 9.1 B2 10

10 Distribution of tensile forces 1. Smooth GMB with geogrid reinforcement % carried by GTX = K tgtx K ttot = = 3% % carried by GMB = K tgmb K ttot = = 21% % carried by GR = K tgr K ttot = = 76% 2. Textured GMB no reinforcement % carried by GTX = K tgtx K ttot = = 14% % carried by GMB = K tgmb K ttot = = 86%

11 Distribution of tensile forces continued 3. Textured GMB with geogrid reinforcement % carried by GTX = K tgtx K ttot = = 7% % carried by GMB = K tgmb K ttot = = 43% % carried by GR = K tgr K ttot = = 50% 4. Textured GMB with stiffer geogrid reinforcement (same UTS) % carried by GTX = K tgtx K ttot = = 5% % carried by GMB = K tgmb K ttot = = 33% % carried by GR = K tgr K ttot = = 62%

12 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 Guidelines on durability

13 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 Guide to the derivation of reduction factors for soil reinforcement materials = 19.5 c c Source: worldweatheronline.com

14 Long term strength as a function of conditions Characteristic Value of the Long Term Reinforcement Strength R B,k = A 1 *A 2 R *A B,k 3 0 *A 4 *A 5 [kn/m] and R B,d = R B, k g M [kn/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

15 Tensile load [%] Influence of temperature on creep rupture behaviour, A1 Creep-Rupture Behavior Design life t D [years] PET- 50 C PET- 20 C PET- 10 C PVA- 50 C PVA- 20 C PVA- 10 C

16 Influence of temperature on creep rupture behaviour, A1 Effect of temperature on reduction factor for a given design life for PET and PVA reinforcement Design life [years] Design temperature [ C] Retained strength [%] A 1 PET PVA PET PVA

17 Influence of temperature on creep rupture behaviour, A1 Temperature ( C) Time required to reach % residual strength (days) PET reinforcement 95% strength 90% strength 85% strength

18 Influence of temperature on creep rupture behaviour, A1 for HDPE Source: Kasozi et al, 2015

19 Retained strength T [%] Influence of temperature on chemical degradation, A4 Hydrolytic Degradation Curve for PET Products Design life [years] 50 C 20 C 10 C

20 Influence of temperature on chemical degradation, A4 Effect of temperature on reduction factor for a given design life for PET reinforcement Design life [years] Design temperature [ C] Retained strength [%] R 4 /RF CH , Fail

21 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 x 1.2 x 1.0 x 1.32 x 1.0 = 13.7 kn/m = 11.4 kn/m Increase in-situ temperature from 35 C to 50 C x 1.2 x 1.0 x Fail x 1.0 = 0 kn/m Note: Assumes environment for internal hydrolysis

22 What happens to the reinforcement if we change the in-situ temperature? Source ISO guidelines on durability

23 Summary PVA reinforcement with a higher stiffness modulus (but same ultimate strength) reduces the relative tensile load acting on the gmb PVA is less sensitive to temperature increase than PET in relation to creep rupture PVA appears to be less sensitive to temperature increase than PET in relation to chemical degradation (research ongoing) Simple data acquisition systems to monitor temperature and moisture conditions in Southern Africa will increase confidence in design Accommodations should be made when selecting design tensile strength to address the corresponding A1 and A4 reduction in strength in relation to an increased temperature. This would help to ensure satisfactory mechanical performance of polymeric reinforcement subjected to elevated temperature conditions

24 Thank you for your attention. Any questions? Peter Assinder HUESKER Synthetic GmbH, Manager AFRICA Laura Carbone HUESKER Synthetic GmbH, Engineering department Morne Breytenbach HUESKER Synthetic GmbH, Manager MINING