Verification Manual. of GEO5 Gravity Wall program. Written by: Ing. Veronika Vaněčková, Ph.D. Verze: 1.0-en Fine Ltd.

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1 of program Written by: Ing. Veronika Vaněčková, Ph.D. Edited by: Ing. Jiří Laurin Verze: 1.0-en Fine Ltd.

2 INTRODUCTION This Gravity Wall program Verification Manual contains hand-made calculation of following problems: calculation of earth pressures calculation of hydrodynamic pressure verification of the wall with the help of theory of limit states dimensioning of wall section according to EN [1] verification of the wall influenced by earthquake according to Mononobe-Okabe theory [2], [3] In general this Verification Manual has been made to show the ability of the GEO5 software to provide a solution consistent with the result of hand-made calculations or textbook solutions. Results of hand-made calculations are compared with the results from GEO5 Gravity Wall program. For further details please refer to our verification example file (Demo.gtz download using our program GEO5 Gravity Wall (

3 Contents 1 Example Problem description Verification of the whole wall Bearing capacity of the foundation soil Dimensioning - analysis of wall section Verification of the wall - influence of earthquake Bearing capacity of the foundation soil Reference 23

1 Example 1 1.1 Problem description In Fig. 1 is shown example of gravity wall with inclined footing bottom (inclination 1:10).

4 1 Example Problem description In Fig. 1 is shown example of gravity wall with inclined footing bottom (inclination 1:10). Earth body is compounded from two soil layers and terrain is inclined (1:10). Top layer (depth 1.5 m) is formed from sandy silt (MS), lower layer is clayey sand (SC) at front face and back face of the wall. Groundwater table is in the level of layer SC, i.e. 1.5 m behind the wall and 3.7 m in front of the wall. Soil properties (effective parameters) are mentioned in the Tab. 1. Wall is made from plain concrete (bulk weight γ = 23 3 ). Verification analysis of the wall is performed with the help of theory of limit states. Figure 1: Wall - geometry Soil parameters (angle of friction and cohesion) are reduced by coefficients γ mϕ = 1.1 a γ mc = 1.4. Design values used in calculation are in Tab Verification of the whole wall Calculation of weight and centroid of the wall. Wall is divided into 5 parts (Fig. 1). Parts 4 and 5 are under groundwater table, therefore bulk weight of concrete is reduced, Fine Ltd.

5 MS SC bulk weight γ [ 3 ] bulk weight of satur. soil γ sat [ 3 ] angle of internal friction ϕ [ ] cohesion c [] angle of friction struc-soil δ [ ] Poisson ratio ν Table 1: Soil properties - effective values MS SC angle of internal friction ϕ d [ ] cohesion c d [] angle of friction struc-soil δ d [ ] Table 2: Design parameters of soil i.e. γ = = In the Tab. 3 are shown dimensions of the parts of the wall, their weights and centroids. height width area b. weight weight force point of action block h i [m] w i [m] A i [m 2 ] γ[ 3 ] W i [] x i [m] z i [m] W i x i W i z i TOTAL Table 3: Weight of construction and centroids of its parts Centroid of the construction: x = 5 W i x i i=1 W = = m z = 5 W i z i i=1 W = = m Calculation of front face resistance. Depth of soil in front of the wall is 0.6 m. Pressure at rest is considered. For cohesive soils the Terzaghi formula is used for computing of coefficient of earth pressure at rest Kr: Fine Ltd.

6 K r = ν 1 ν = = Hydraulic gradient (h w - water tables difference, d d - seepage path downwards, d u - seepage path upwards): i = h w d d + d u = = Unit weight of soil in the area of ascending flow: γ = γ sat γ w i γ w = = Vertical normal effective stress σ z in the footing bottom: σ z = γ h = = Pressure at rest σ r in the footing bottom: σ r = K r σ z = = Resultant force of stress at rest S r : S r = 1 2 σ r h = = Resultant force S r is horizontally oriented, therefore: S rx = S r = S rz = 0 Point of action of resultant force S r : x = 0 m z = h 3 = = m Calculation of active pressure. Construction is at interface between soils MS and SC divided into two levels, in each is calculated geostatic pressure σ z, active pressure σ a and resultant force S a (Fig. 2) Fine Ltd.

Figure 2: Geostatic pressure σ z and active pressure σ a Coefficients of the active pressure in both levels (back face inclination of the structure α = 0, inclination of terrain β 0 ): K a = cos 2 α

7 Figure 2: Geostatic pressure σ z and active pressure σ a Coefficients of the active pressure in both levels (back face inclination of the structure α = 0, inclination of terrain β 0 ): K a = cos 2 α cos(α + δ d ) cos 2 (ϕ d α) [ 1 + sin(ϕ d +δ d ) sin(ϕ d β) cos(α+δ d ) cos(α β) ] 2 K ac = cos ϕ d cos β cos(δ d α) (1 + tan( α) tan β) (1 + sin(ϕ d + δ d α β)) (cos α + δ d ) β 1 = arctan 1 10 = K a1 = cos [ 1 + cos sin( ) sin( ) cos cos( 5.711) ] 2 = K ac1 = cos cos cos (1 + tan 0 tan 5.711) (1 + sin( )) cos = Fine Ltd.

8 β 2 = arctan γ 1 tan β 1 18 tan = arctan γ = K a2 = cos [ 1 + cos sin( ) sin( ) cos cos( 5.557) ] 2 = K ac2 = cos cos cos (1 + tan 0 tan 5.557) (1 + sin( )) cos = Unit weight of soil SC in the area of descending flow: γ 2 = γ sat γ w + i γ w = = Vertical geostatic stress σ z in 2 levels: σ z1 = γ 1 h 1 = = σ z2 = σ z1 + γ 2 h 2 = = Active pressure σ a in 2 levels: σ a1 = K a1 σ z1 2 c d1 K ac1 = = σ a2a = K a2 σ z1 2 c d2 K ac2 = = σ a2b = K a2 σ z2 2 c d2 K ac2 = = Depth h 0 in the first layer of soil MS, where is neutral active pressure (σ a = 0 ): h 0 = 2 c d1 K ac1 γ 1 K a1 = = m Resultant forces of the active pressure S a and horizontal, resp. vertical components: Fine Ltd.

9 S a1 = 0.5 σ a1 (h 1 h 0 ) = ( ) = S ax1 = S a1 cos δ d1 = cos = S az1 = S a1 sin δ d1 = sin = S a2 = 0.5 (σ a2b σ a2a ) h 2 + σ a2a h 2 S a2 = 0.5 ( ) = S ax2 = S a2 cos δ d2 = cos = S az2 = S a2 sin δ d2 = sin = Points of action of resultant forces S a : x 1 = m z 1 = ( ) 3 = m x 2 = m 1 z 2 = ( ) = m ( ) Total resultant force of the active pressure S a and horizontal, resp. vertical component: Fine Ltd.

10 2 S ax = S axi = = i=1 2 S az = S azi = = i=1 S a = S 2 ax + S 2 az = = Point of action of the resultant force S a : x = 2 i=1 S azi x i S az = = m z = 2 i=1 S axi z i S ax = = m Calculation of water pressure. The heel of a structure is sunk into permeable subsoil, which allows free water flow below the structure. Therefore acting of hydrodynamic pressure is considered and its resultant force is calculated (Fig. 3). Horizontal water pressure σ w at interface of level 1 and 2: σ w1 = γ w h 1 = = Resultant forces of water pressure S w in 2 levels: S w1 = 1 2 σ w1 h 1 = = S w2 = 1 2 σ w1 h 2 = = Points of action of resultant forces: x 1 = m z 1 = = m Fine Ltd.

11 Figure 3: Hydrodynamic pressure σ w x 2 = m z 2 = = m Total resultant force of the water pressure S w : 2 S w = S wi = = i=1 Total point of action of the resultant force S w : x = m z = 2 i=1 S wi z i S w = = m Checking for overturning stability. The moment rotates around the beginning of system of coordinates (left bottom corner of the footing). Resisting moment M r is reduced by coefficient of overall stability of structure γ s = 0.9: Fine Ltd.

M r = 0.9 (116.484 1.601 + 10.155 2.3) = 188.833 knm/m Result from program : M r = 188.83 knm/m Driving moment M o : M o = 0.430 0.2 + 41.859 0.929 + 33.33 1.057 = 74.

12 M r = 0.9 ( ) = knm/m Result from program : M r = knm/m Driving moment M o : M o = = knm/m Result from program : M o = knm/m Usage: V u = M o 100 = = 39.2 % O.K. M r Result from program : V u = 39.2 % O.K. Checking for slip. Slip in the inclined footing bottom is checked. (Fig. 4). Figure 4: Forces acting in the footing bottom - normal and shear soil reaction N and T, total vertical and horizontal forces acting on the structure F vert and F hor Total vertical and horizontal forces F vert and F hor : F vert = = Fine Ltd.

13 F hor = = Normal force in the footing bottom (inclination of footing bottom α b = 5.711, see Fig. 4): N = F vert cos α b + F hor sin α b N = cos sin = Shear reaction in the footing bottom: T = F vert sin α b + F hor cos α b T = sin cos = Eccentricity of the normal force (inclined width of the footing bottom d = m): e = d 2 M r 0.9 M o N = Maximal allowable eccentricity: = m e alw = d 3 = = m > e = m O.K. Resisting force H r : H r = γ s (N tan ϕ d + c d (d 2 e)) H r = 0.9 ( tan ( )) = Result from program : H r = Driving force H d : Fine Ltd.

14 H d = T = Result from program : H d = Usage: V u = H d 100 = = 94.6 % O.K. H r Result from program : V u = 94.5 % O.K. 1.3 Bearing capacity of the foundation soil Bearing capacity of the foundation soil is set as R d = 100 and is compared with stress in the inclined footing bottom. Usage - eccentricity: V u = e 100 = = 17.9 % O.K. e alw Result from program : V u = 17.4 % O.K. Stress in the footing bottom: σ = N (d 2 e) = ( ) Result from program : σ = Usage: V u = σ R d 100 = = = % O.K. Result from program : V u = 65.2 % O.K. 1.4 Dimensioning - analysis of wall section Cross-section in the level of the x-axis in Fig. 5 is analysed. Rectangular section (width b = 1 m, height h = 1.4 m) is made from plain concrete C20/25 (characteristic strength of concrete in tension f ctk = 2200, characteristic cylindrical strength of concrete in compression f ck = ). Verification of cross-section made from plain concrete is realized according to EN [1] Fine Ltd.

15 Figure 5: Dimensioning of cross-section Calculation of weight and centroid of the wall. Weight of the construction: W = 23 ( ) = m Centroid of the construction: x = 23 ( ( ) ) = m z = 23 ( ) = m Calculation of active pressure. In the first level is same active pressure as in the verification of the whole wall (part 1.2). Centroids of all forces are recalculated. Vertical geostatic stress σ z2 in the second level: σ z2 = = Active pressure σ a2b at the end of second level (value at the beginning is same, i.e. σ a2a = ): σ a2b = = Fine Ltd.

16 Resultant force of the active pressure S a2 and horizontal, resp. vertical component: S a2 = 0.5 ( ) = S ax2 = S a2 cos δ d2 = cos = S az2 = S a2 sin δ d2 = sin = Points of action of resultant forces: x 1 = m z 1 = = m x 2 = m ( ) 22 z 2 = ( ) 2 = m Total resultant force of the active pressure S a and horizontal, resp. vertical component: S ax = S az = S a = Point of action of resultant force S a : x = m z = = m Calculation of water pressure. Water pressure grows with height (h = 2 m). Horizontal water pressure σ w in the level of analysed cross-section: σ w = γ w h = 10 2 = Resultant force of the water pressure S w : S w = 1 2 σ w h = = Fine Ltd.

17 Point of action of the resultant force: x = m z = 2 3 = m Verification of shear strength. Shear and normal force, bending moment and shear strength of cross-section are calculated. Design shear force: V Ed = = Result from program : V Ed = Design normal force: N Ed = = Result from program : N Ed = Area of cross-section: A cc = b h = = 1.4 m 2 Stress in cross-section area: σ cp = N Ed A cc 1000 = = MPa Design strength of concrete in compression (coefficients α cc.pl = 0.8, γ c = 1.5): f cd = α cc.pl fck γ c = = MPa Lower value of characteristic strength of concrete in tension: f ctk.005 = 0.7 (0.3 f 2 3 ck ) = 0.7 ( ) = MPa Design strength of concrete in tension (coefficients α ct.pl = 0.8, γ c = 1.5): f ctd = α ct.pl fctk.005 = γ c 1.5 = MPa Fine Ltd.

18 Limit stress: σ c.lim = f cd 2 f ctd (f cd + f fctd ) Shear strength: σ c.lim = ( ) = MPa ( f cvd = f 2 max(0, σcp σ c.lim ) ctd + σ cp f ctd 2 ) 2 ( ) 2 0 f cvd = = MPa 2 Design shear strength (k = 1.5): V Rd = f cvd A cc k = = Result from program : V Rd = Usage: V u = V Ed 100 = = 5.1 % O.K. V Rd Result from program : V u = 5.1 % O.K. Mo- Verification of cross-section loaded by bending moment and normal force. ment turns around the cross-section centroid. Design bending moment: M Ed = ( ) = Result from program : M Ed = knm/m Normal force eccentricity: e = M Ed N Ed = = m Fine Ltd.

19 0.9 h 2 = = 0.63 m > e = m Effective height: χ = h 2 e = = m Design normal strength (η = 1.0): N Rd = χ η f cd = = Result from program : N Rd = Usage: V u = N Ed 100 = = 0.8 % O.K. N Rd Result from program : V u = 0.8 % O.K. 1.5 Verification of the wall - influence of earthquake Second stage of calculation shows the same wall with the influence of earthquake. Calculation of earthquake effects is made according to Mononobe-Okabe theory [2], [3]. Factor of horizontal acceleration is k h = 0.05 (inertia force acts horizontally in unfavorable direction) and factor of vertical acceleration k v = 0.04 (inertia force acts downwards). Coefficients of reduction of soil parameters and coefficient of overall stability of construction are equal to one (γ mϕ = γ mc = γ s = 1.0). Therefore design soil properties are same as characteristic values in Tab. 1. Calculation of weight of the wall. Weight of the wall W = (see part 1.2) is increased by weight force from earthquake. Horizontal and vertical component of weight force from earthquake: W eq.x = k h W = = W eq.z = k v W = = Fine Ltd.

20 Calculation of front face resistance. Pressure at rest on front face of the wall is considered same as in part 1.2, i.e. resultant force is S r = Calculation of active pressure. Active pressure σ a and resultant force S a are calculated similarly as in the calculation without earthquake (part 1.2), geostatic stress σ z is same, i.e. σ z1 = and σ z2 = Coefficients of the active pressure in both levels: K a1 = cos 15 [ 1 + cos sin( ) sin( ) cos 15 cos( 5.711) ] 2 = K ac1 = cos 26.5 cos cos 15 (1 + tan 0 tan 5.711) (1 + sin( )) cos 15 = K a2 = cos 15 [ 1 + cos 2 27 sin(27+15) sin( ) cos 15 cos( 5.557) ] 2 = K ac2 = cos 27 cos cos 15 (1 + tan 0 tan 5.557) (1 + sin( )) cos 15 = Depth h 0 in the first layer of soil MS, where is neutral active pressure (σ a = 0 ): h 0 = 2 c 1 K ac1 γ 1 K a1 = = m > h 1 = 1.5 m Active pressure is in the whole first level equal to zero. Active pressure σ a in the second layer of soil SC: σ a2a = K a2 σ z1 2 c 2 K ac2 = = σ a2b = K a2 σ z2 2 c 2 K ac2 = = Total resultant force of the active pressure S a and horizontal, resp. vertical component: Fine Ltd.

21 S a = 0.5 (σ a2b σ a2a ) h 2 + σ a2a h 2 S a = 0.5 ( ) = S ax = S a cos δ 2 = cos 15 = S az = S a sin δ 2 = sin 15 = Point of action of the resultant force: x = m 1 z = ( ) = m ( ) Increase of active pressure caused by earthquake. of active pressure. Seismic inertia angle in the first layer of soil MS: Earthquake increases the effect ψ 1 = arctan k h 0.05 = arctan 1 k v = Seismic inertia angle in the second layer of soil SC, where is restricted water: ψ 2 = arctan γ sat k h (γ sat γ w ) (1 k v ) = arctan ( ) ( ) = Coefficients K ae for the active pressure in both levels: K ae = cos 2 (ϕ ψ α) [ cos ψ cos 2 α cos(ψ + α + δ) 1 + sin(ϕ+δ) sin(ϕ ψ β) cos(α+ψ+δ) cos(β α) ] Fine Ltd.

22 K ae1 = cos cos( ) cos 2 ( ) [ 1 + sin( ) sin( ) cos( ) cos(5.711) ] 2 = K ae2 = cos cos( ) cos 2 ( ) [ 1 + sin(27+15) sin( ) cos( ) cos(5.557) ] 2 = Normal stress σ d for calculation of earthquake effects (stress grows from the bottom of the wall - σ d2 is in the footing bottom level and σ d0 is in the terrain level): σ d2 = 0 σ d1 = γ 2 h 2 (1 k v ) = ( ) = σ d0 = σ d1 + γ 1 h 1 (1 k v ) = ( ) = Increase of the active pressure from earthquake effects in the first layer of soil MS: σ ae0 = σ d0 (K ae1 K a1 ) ( ) = k v = σ ae1a = σ d1 (K ae1 K a1 ) ( ) = k v = Increase of the active pressure from earthquake effects in the second layer of soil SC: σ ae1b = σ d1 (K ae2 K a2 ) ( ) = k v = σ ae2 = 0 Resultant forces of increase of the active pressure S ae and horizontal, resp. vertical components: Fine Ltd.

23 S ae1 = 0.5 (σ ae0 σ ae1a ) h 1 + σ ae1a h 1 S ae1 = 0.5 ( ) = S aex1 = S ae1 cos δ 1 = cos 15 = S aez1 = S ae1 sin δ 1 = sin 15 = S ae2 = 0.5 σ ae1b h 2 = = S aex2 = S ae2 cos δ 2 = cos 15 = S aez2 = S ae2 sin δ 2 = sin 15 = Points of action of resultant forces: x 1 = m z 1 = ( ) ( ) = m x 2 = m z 2 = = m Total resultant force of increase of the active pressure S ae and horizontal, resp. vertical component: Fine Ltd.

24 2 S aex = S aexi = = i=1 2 S aez = S aezi = = i=1 S ae = S 2 aex + S 2 aez = = Point of action of the resultant force S ae : x = m z = 2 i=1 S aexi z i S aex = = m Calculation of water pressure. Water pressure is same as in the verification of the whole wall (part 1.2), i.e. resultant force of the water pressure is S w = Checking for overturning stability. The moment rotates around the beginning of system of coordinates (left bottom corner of the footing). Resisting moment M r : M r = ( ) ( ) 2.3 = knm/m Result from program : M r = knm/m Driving moment M o : M o = = knm/m Result from program : M o = knm/m Usage: V u = M o 100 = = 42.9 % O.K. M r Fine Ltd.

25 Result from program : V u = 42.9 % O.K. Checking for slip. Slip in the inclined footing bottom is checked. (Fig. 4). Total vertical and horizontal forces F vert and F hor : F vert = = F hor = = Normal force in the footing bottom (inclination of footing bottom α b = 5.711, see Fig. 4): N = cos sin = Shear reaction in the footing bottom: T = sin cos = Eccentricity of the normal force (inclined width of the footing bottom d = m): e = d 2 M r M o N = = m Maximal allowable eccentricity: e alw = d 3 = = m > e = m O.K. Resisting force H r : H r = tan ( ) = Result from program : H r = Driving force H d : H d = T = Result from program : H d = Fine Ltd.

26 Usage: V u = H d 100 = = 75.3 % O.K. H r Result from program : V u = 75.3 % O.K. 1.6 Bearing capacity of the foundation soil Bearing capacity of the foundation soil is set as R d = 100 and is compared with stress in the inclined footing bottom. Usage - eccentricity: V u = e 100 = = 33.5 % O.K. e alw Result from program : V u = 33.1 % O.K. Stress in the footing bottom: σ = N (d 2 e) = ( ) Result from program : σ = Usage: V u = σ R d 100 = = = 77.2 % O.K. Result from program : V u = 76.7 % O.K. References [1] Czech normalization institute: EN , Eurocode 2: Design of concrete structures - Part 1-1: General rules and rules for buildings. CNI, Prague, 2006 [2] Mononobe N., Matsuo, H.: On the determination of earth pressure during earthquakes. In Proc. Of the World Engineering Conf., Vol. 9, pg. 176, 1929 [3] Okabe S.: General theory of earth pressure. Journal of the Japanese Society of Civil Engineers, Tokyo, Japan, Fine Ltd.

Verification Analysis of the Gravity Wall

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