CVEEN 7330 Table of Contents Table of Contents... 1 Objectives... 2 ackground... 2 Model Setup and Analysis Input... 2 Data for Interfaces... 5 Required Outputs from FLAC... 10 Required Calculations and Discussion... 10 FLAC Solution:... 10 Discussion... 15 FISH Code for model... 15
Objectives In this exercise, the student develops a FLAC model to replicate the dynamic behavior of geofoam embankment for the case of a rectangular geofoam mass resting on a soil foundation with sliding allowed in the geofoam under horizontal sinusoidal input motion only. The lump mass (load distribution slab, untreated base course and concrete pavement) is placed atop the geofoam and bonded to the top layer, so slippage is not allowed at this interface. ackground Dynamic sliding is one mode of failure and can be explored using interface elements in FLAC and the appropriate friction angle for the interfaces. Model Setup and Analysis Input 1. Model Type: Elastic 2. Elastic material properties ρ (kg/m 3 ) γ (lb/ft 3 ) E (MPa) v K (MPa) G (MPa) Soil 2000 115.00 500 0.35 555.6 185.2 Goafoam 20 1.25 10 0.103 4.20 4.53 UTC 2241 140.00 570 0.35 633.3 211.1 LDS & PCCP 2401 150.00 30000 0.18 15625.0 12711.9 Foundation soil properties (Vs = 1000 ft/s and = 0.35) prop bulk 500e6 shear 185e6 den 2000 Geofoam properties prop bulk 4.198e6 shear 4.53e6 den 20 Concrete mass above geofoam properties prop bulk 15625e6 shear 12712 den 2305
3. Geometry (Figure 1 below) 4. oundary Conditions a. Static conditions with gravity on see Figure 2. b. Dynamic conditions with horizontal velocity wave - see Figure 3. JO TITLE :. FLAC (Version 5.00) LEGEND 1.400 24-Apr-08 7:54 step 6907-1.111E+00 <x< 2.111E+01-5.611E+00 <y< 1.661E+01 6 5 1.000 Grid plot 4 0 5E 0 interface id#'s Fixed Gridpoints -direction oth directions 3 2 1 0.600 0.200-0.200 Steven artlett University of Utah 0.200 0.600 1.000 1.400 1.800 Figure 1. Geofoam/Soil model and boundary conditions for static loading under gravity.
JO TITLE :. FLAC (Version 5.00) LEGEND 1.400 24-Apr-08 7:57 step 6907-1.111E+00 <x< 2.111E+01-5.611E+00 <y< 1.661E+01 6 5 1.000 shear_mod 4.530E+06 1.850E+08 1.270E+10 Fixed Gridpoints -direction oth directions Grid plot 0 5E 0 interface id#'s 4 3 2 1 0.600 0.200-0.200 Steven artlett University of Utah 0.200 0.600 1.000 1.400 1.800 Figure 2. Plot of shear modulus for model. JO TITLE :. FLAC (Version 5.00) LEGEND 1.400 24-Apr-08 8:11 step 102725 Dynamic Time 4.0000E+00-1.067E+00 <x< 2.116E+01-5.586E+00 <y< 1.664E+01 shear_mod 4.530E+06 1.850E+08 1.270E+10 Fixed Gridpoints -direction oth directions Grid plot 0 5E 0 interface id#'s 6 5 4 3 2 1 1.000 0.600 0.200-0.200 Steven artlett University of Utah 0.200 0.600 1.000 1.400 1.800 Figure 3. oundary conditions for dynamic sliding case.
5. Fish code for sliding interfaces within geofoam (Fig. 1) a. Generate grid b. Declare model type (note: gaps in model type are left for interfaces) c. Assign elastic properties d. Assign grid to coordinate space (with generate command) e. Create interfaces for sliding planes f. Assign interface properties sand-geofoam interface friction = 31 degrees geofoam-geofoam interface friction = 32 degress normal and shear spring constants 1. Normal Stiffness = 10(K+4/3G)/Δzmin, normal 2. Shear Stiffness = 10(K+4/3G)/Δzmin, normal Data for Interfaces int Normal Stiffness = kn = ks (MPa) Friction Cohesion Dilation z Geofoam-soil 1-2 102 35 0 6 1 Geof-Geof 3-10 102 32 0 0 1 Geofoam-lds 21 102 glued 0 0 1 lds-utbc 22-23 15247 glued 0 0 0.6 utbc-pccp 23-24 15247 glued 0 0 0.6 **Use least stiff of two adjacent layers for kn and ks FLAC code for Step 5 generation of grid grid 20 17 declare model type model elastic j 1 5 model elastic i 6 15 j 7 model elastic i 6 15 j 9 model elastic i 6 15 j 11 model elastic i 6 15 j 13 model elastic i 6 15 j 15 model elastic i 6 15 j 17
assign elastic properties prop bulk 500e6 shear 185e6 den 2000 i 1 20 j 1 5 foundation properties vs = 1000 ft/s prop bulk 4.20e6 shear 4.53e6 den 20 i 6 15 j 7 geofoam properties prop bulk 4.20e6 shear 4.53e6 den 20 i 6 15 j 9 geofoam properties prop bulk 4.20e6 shear 4.53e6 den 20 i 6 15 j 11 geofoam properties prop bulk 4.20e6 shear 4.53e6 den 20 i 6 15 j 13 geofoam properties prop bulk 4.20e6 shear 4.53e6 den 20 i 6 15 j 15 geofoam properties prop bulk 15625e6 shear 12712e6den 2305 i 6 15 j 17 mass above geofoam Assign grid to coordinate space gen 0 0 0 5 20 5 20 0 i 1 21 j 1 6 gen 5 5 5 6 15 6 15 5 i 6 16 j 7 8 gen 5 6 5 7 15 7 15 6 i 6 16 j 9 10 gen 5 7 5 8 15 8 15 7 i 6 16 j 11 12 gen 5 8 5 9 15 9 15 8 i 6 16 j 13 14 gen 5 9 5 10 15 10 15 9 i 6 16 j 15 16 gen 5 10 5 11 15 11 15 10 i 6 16 j 17 18 create interfaces int 1 Aside from 6,6 to 16,6 side from 6,7 to 16,7 int 2 Aside from 6,8 to 16,8 side from 6,9 to 16,9 int 3 Aside from 6,10 to 16,10 side from 6,11 to 16,11 int 4 Aside from 6,12 to 16,12 side from 6,13 to 16,13 int 5 Aside from 6,14 to 16,14 side from 6,15 to 16,15 int 6 Aside from 6,16 to 16,16 side from 6,17 to 16,17 assign inteface properties int 1 kn 102e6 ks 102e6 friction = 35 coh = 0 dilation = 6 tbond=0 bslip on int 2 kn 102e6 ks 102e6 friction = 32 coh = 0 dilation = 0 tbond=0 bslip on int 3 kn 102e6 ks 102e6 friction = 32 coh = 0 dilation = 0 tbond=0 bslip on int 4 kn 102e6 ks 102e6 friction = 32 coh = 0 dilation = 0 tbond=0 bslip on int 5 kn 102e6 ks 102e6 friction = 32 coh = 0 dilation = 0 tbond=0 bslip on int 6 glue kn 102e6 ks 102e6 6. Perform static analysis a. Turn gravity on b. Establish boundary conditions c. Turn off dynamic module d. Make history of unbalanced forces (his 20) e. Solve for static case by stepping 9000 steps FLAC code for Step 6 GRAVIT ON FOR STATIC FORCES set grav 9.81
OUNDAR CONDITIONS FOR STATIC CASE fix right side of soil box fix x i 1 fix left side of soil box fix x i 21 fix x i 6 j 7 18 (don't fix x in geofoam for static case) fix x i 16 j 7 18 (don't fix x in geofoam for static case) fix base fix y j 1 fix x j 1 STATIC LOADING set dyn off his 20 unbal solve 7. Turn on dynamic mode and establish large strain conditions and damping properties. (Note that no damping other than Rayleigh damping has been used for the soil.) (Note hysteretic damping has been used for geofoam using sigmoidal model.) set dyn on set large set dy_damp rayl 0.05 15 j 1 5 damping in geofoam ini dy_damp=hyst sig3 1-0.45.3 j 7 ini dy_damp=hyst sig3 1-0.45.3 j 9 ini dy_damp=hyst sig3 1-0.45.3 j 11 ini dy_damp=hyst sig3 1-0.45.3 j 13 ini dy_damp=hyst sig3 1-0.45.3 j 15 Calculations of hysteretic damping using sigmoidal model a = 1 b = -0.45 x o = 0.3 Geofoam (lab) Geofoam (sig3 model) L Ms g L Ms 0.0001-4 1 0.0001-4 0.999929 0.0003-3.52288 1 0.0003-3.52287875 0.999796 0.001-3 1 0.001-3 0.999347 0.003-2.52288 1 0.003-2.52287875 0.998117 0.01-2 1 0.01-2 0.994007 0.03-1.52288 0.98 0.03-1.52287875 0.98289 0.1-1 0.95 0.1-1 0.947294 0.3-0.52288 0.88 0.3-0.52287875 0.861597 1 0 0.69 1 0 0.660756 3 0.477121 0.41 3 0.47712125 0.40285 10 1 0.18 10 1 0.174285
G/Gmax L = log 10 Ms = G/Gmax 1.2 1 Clay (Seed and Sun) Geofoam (Lab) Geofoam (sig3 model) 0.8 0.6 0.4 0.2 0 0.0001 0.001 0.01 0.1 1 10 Shear strain (percent) Figure 4. Sigmodial 3 model fit to geofoam data
8. Establish boundary conditions for dynamic case ESTALISH OUNDARIES FOR DNAMIC CASE free x free y OUNDAR CONDITIONS FOR DNAMIC CASE for horizontal wave only fix y j 1 6 fix x j 1 9. Create horizontal wave (1 g amplitude wave at fundamental frequency of geofoam) a. Horizontal velocity wave amplitude = 0.65 m/s for 1 g acceleration. b. Horizontal wave period = 0.41 s (fundamental period of geofoam) CREATE WAVES wave 1 is horizontal wave def wave1 for free vibration wave1=amp1*cos(2*pi/period1*dytime)/(dytime+1)^10 for forced vibration wave1=amp1*cos(2*pi/period1*dytime) end set amp1 = 0.65 for 10 m/s2 accel at base set period1 = 0.41 period1 of 0.41 s is resonant frequency 10. Input waves into bottom of model FOR HOR. WAVE apply xvel 1 hist=wave1 j=1 11. Establish locations for dynamic histories (See Fig. 3) HISTORIES his 1 dytime his 2 wave1 his 3 xvel j 1 i 11 his 4 xdisp j 1 i 11 his 5 xvel j 6 i 11 his 6 xdisp j 6 i 11 his 7 xvel j 18 i 11 his 8 xdisp j 18 i 11 his 10 yvel j 1 i 11
his 11 ydisp j 1 i 11 his 12 yvel j 6 i 11 his 13 ydisp j 6 i 11 his 14 yvel j 18 i 11 his 15 ydisp j 18 i 11 his 16 xacc j 1 i 11 his 17 xacc j 6 i 11 his 18 xacc j 18 i 11 12. Reset displacements to start dynamic case ini xdisp = 0 ydisp = 0 13. Solve for dynamic case for 4 seconds solve dytime 4.0 Required Outputs from FLAC 1. Plot of the following at the resonant period for a duration t = 5 s. Make all plots at the middle (centerline) of the model: a. Unbalanced forces from static and dynamic loading b. displacement at bottom of soil c. displacement at top of soil d. displacement top of pavement e. acceleration at bottom of soil f. acceleration at top of soil g. acceleration at top of pavement h. Final vector displacement pattern at time = 4 seconds 2. Animation of displacement vectors using an increment of 500 steps for t = 4 s. Required Calculations and Discussion 1. Discuss how the change in the amplitude of the horizontal wave affects the displacement at the top of the geofoam and the sliding within the geofoam. Use graphs or figures to do this. FLAC Solution: