of program Written by: Ing. Veronika Vaněčková, Ph.D. Edited by: Ing. Jiří Laurin Verze: 1.0-en 1989-2009 Fine Ltd. www.finesotware.eu
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 1992-1-1 [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 http://www.finesoftware.eu/references/verification-manuals/) using our program GEO5 Gravity Wall (http://www.finesoftware.eu/geotechnical-software/gravity-wall/).
Contents 1 Example 1 1 1.1 Problem description.............................. 1 1.2 Verification of the whole wall........................ 1 1.3 Bearing capacity of the foundation soil................... 11 1.4 Dimensioning - analysis of wall section................... 11 1.5 Verification of the wall - influence of earthquake.............. 16 1.6 Bearing capacity of the foundation soil................... 23 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). 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. 2. 1.2 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, www.finesotware.eu 1 1989-2009 Fine Ltd.
MS SC bulk weight γ 18.000 18.500 [ 3 ] bulk weight of satur. soil γ sat 20.000 20.500 [ 3 ] angle of internal friction ϕ 26.500 27.000 [ ] cohesion c 12.000 8.000 [] angle of friction struc-soil δ 15.000 15.000 [ ] Poisson ratio ν 0.350 0.350 - Table 1: Soil properties - effective values MS SC angle of internal friction ϕ d 24.091 24.545 [ ] cohesion c d 8.571 5.714 [] angle of friction struc-soil δ d 13.636 13.636 [ ] Table 2: Design parameters of soil i.e. γ = 23-10 = 13 3. 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 1 0.350 0.700 2.450 23 56.350 1.950-2.550 109.883-143.693 2 0.350 0.700 1.225 23 28.175 1.367-1.967 38.506-55.411 3 0.200 2.300 0.460 23 10.580 1.150-0.700 12.167-7.406 4 0.600 2.300 1.380 13 17.940 1.150-0.300 20.631-5.382 5 0.230 2.300 0.265 13 3.439 1.533 0.077 5.272 0.264 TOTAL 116.484 186.459-211.628 Table 3: Weight of construction and centroids of its parts Centroid of the construction: x = 5 W i x i i=1 W = 186.459 116.484 = 1.601 m z = 5 W i z i i=1 W = 211.628 116.484 = 1.817 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: www.finesotware.eu 2 1989-2009 Fine Ltd.
K r = ν 1 ν = 0.35 1 0.35 = 0.538 Hydraulic gradient (h w - water tables difference, d d - seepage path downwards, d u - seepage path upwards): i = h w d d + d u = 3.7 1.5 3.03 + 0.6 = 0.606 Unit weight of soil in the area of ascending flow: γ = γ sat γ w i γ w = 20.5 10 0.606 10 = 4.439 3 Vertical normal effective stress σ z in the footing bottom: σ z = γ h = 4.439 0.6 = 2.664 Pressure at rest σ r in the footing bottom: σ r = K r σ z = 0.538 2.664 = 1.434 Resultant force of stress at rest S r : S r = 1 2 σ r h = 1 1.434 0.6 = 0.430 2 Resultant force S r is horizontally oriented, therefore: S rx = S r = 0.430 S rz = 0 Point of action of resultant force S r : x = 0 m z = h 3 = 0.6 3 = 0.200 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). www.finesotware.eu 3 1989-2009 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 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 = 5.711 K a1 = cos 13.636 [ 1 + cos 2 24.091 sin(24.091+13.636) sin(24.091 5.711) cos 13.636 cos( 5.711) ] 2 = 0.410 K ac1 = cos 24.091 cos 5.711 cos 13.636 (1 + tan 0 tan 5.711) (1 + sin(24.091 + 13.636 5.711)) cos 13.636 = 0.594 www.finesotware.eu 4 1989-2009 Fine Ltd.
β 2 = arctan γ 1 tan β 1 18 tan 5.711 = arctan γ 2 18.5 = 5.557 K a2 = cos 13.636 [ 1 + cos 2 24.545 sin(24.545+13.636) sin(24.545 5.557) cos 13.636 cos( 5.557) ] 2 = 0.402 K ac2 = cos 24.545 cos 5.557 cos 13.636 (1 + tan 0 tan 5.557) (1 + sin(24.545 + 13.636 5.557)) cos 13.636 = 0.588 Unit weight of soil SC in the area of descending flow: γ 2 = γ sat γ w + i γ w = 20.5 10 + 0.606 10 = 16.561 3 Vertical geostatic stress σ z in 2 levels: σ z1 = γ 1 h 1 = 18 1.5 = 27.000 σ z2 = σ z1 + γ 2 h 2 = 27 + 16.561 3.03 = 77.179 Active pressure σ a in 2 levels: σ a1 = K a1 σ z1 2 c d1 K ac1 = 0.410 27.000 2 8.571 0.594 = 0.886 σ a2a = K a2 σ z1 2 c d2 K ac2 = 0.402 27.000 2 5.714 0.588 = 4.121 σ a2b = K a2 σ z2 2 c d2 K ac2 = 0.402 77.179 2 5.714 0.588 = 24.274 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 = 2 8.571 0.594 18 0.410 = 1.380 m Resultant forces of the active pressure S a and horizontal, resp. vertical components: www.finesotware.eu 5 1989-2009 Fine Ltd.
S a1 = 0.5 σ a1 (h 1 h 0 ) = 0.5 0.886 (1.5 1.38) = 0.053 S ax1 = S a1 cos δ d1 = 0.053 cos 13.636 = 0.052 S az1 = S a1 sin δ d1 = 0.053 sin 13.636 = 0.013 S a2 = 0.5 (σ a2b σ a2a ) h 2 + σ a2a h 2 S a2 = 0.5 (24.274 4.121) 3.03 + 4.121 3.03 = 43.019 S ax2 = S a2 cos δ d2 = 43.019 cos 13.636 = 41.807 S az2 = S a2 sin δ d2 = 43.019 sin 13.636 = 10.142 Points of action of resultant forces S a : x 1 = 2.300 m z 1 = 0.8 2 (1.5 1.38) 3 = 2.840 m x 2 = 2.300 m 1 z 2 = 6 3.032 (24.274 4.121) + 1 2 3.032 4.121 + 0.23 = 0.927 m 1 2 3.03 (24.274 4.121) + 3.03 4.121 Total resultant force of the active pressure S a and horizontal, resp. vertical component: www.finesotware.eu 6 1989-2009 Fine Ltd.
2 S ax = S axi = 0.052 + 41.807 = 41.859 i=1 2 S az = S azi = 0.013 + 10.142 = 10.155 i=1 S a = S 2 ax + S 2 az = 41.859 2 + 10.155 2 = 43.073 Point of action of the resultant force S a : x = 2 i=1 S azi x i S az = 0.029 + 23.327 10.155 = 2.300 m z = 2 i=1 S axi z i S ax = 0.147 38.738 41.859 = 0.929 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 = 10 2.2 = 22.000 Resultant forces of water pressure S w in 2 levels: S w1 = 1 2 σ w1 h 1 = 1 22 2.2 = 24.200 2 S w2 = 1 2 σ w1 h 2 = 1 22 0.83 = 9.130 2 Points of action of resultant forces: x 1 = 2.300 m z 1 = 0.6 2.2 3 = 1.333 m www.finesotware.eu 7 1989-2009 Fine Ltd.
Figure 3: Hydrodynamic pressure σ w x 2 = 2.300 m z 2 = 0.6 + 0.83 3 = 0.323 m Total resultant force of the water pressure S w : 2 S w = S wi = 24.2 + 9.130 = 33.330 i=1 Total point of action of the resultant force S w : x = 2.300 m z = 2 i=1 S wi z i S w = 32.267 2.952 33.330 = 1.057 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: www.finesotware.eu 8 1989-2009 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 = 74.017 knm/m Result from program : M o = 74.02 knm/m Usage: V u = M o 100 = 74.017 100 = 39.2 % O.K. M r 188.833 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 = 116.484 + 10.155 = 126.638 www.finesotware.eu 9 1989-2009 Fine Ltd.
F hor = 0.430 + 41.859 + 33.330 = 74.758 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 = 126.638 cos 5.711 + 74.758 sin 5.711 = 133.449 Shear reaction in the footing bottom: T = F vert sin α b + F hor cos α b T = 126.638 sin 5.711 + 74.758 cos 5.711 = 61.786 Eccentricity of the normal force (inclined width of the footing bottom d = 2.311 m): e = d 2 M r 0.9 M o N = 2.311 2 Maximal allowable eccentricity: 188.830 74.017 0.9 133.449 = 0.138 m e alw = d 3 = 2.311 3 = 0.770 m > e = 0.138 m O.K. Resisting force H r : H r = γ s (N tan ϕ d + c d (d 2 e)) H r = 0.9 (133.449 tan 24.545 + 5.714 (2.311 2 0.138)) = 65.316 Result from program : H r = 65.38 Driving force H d : www.finesotware.eu 10 1989-2009 Fine Ltd.
H d = T = 61.786 Result from program : H d = 61.79 Usage: V u = H d 100 = 61.786 100 = 94.6 % O.K. H r 65.316 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 = 0.138 100 = 17.9 % O.K. e alw 0.770 Result from program : V u = 17.4 % O.K. Stress in the footing bottom: σ = N (d 2 e) = 133.449 (2.311 2 0.138) Result from program : σ = 65.20 Usage: V u = σ R d 100 = 65.570 100 = 65.570 100 = 65.57 % 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 = 20000 ). Verification of cross-section made from plain concrete is realized according to EN 1992 1-1 [1]. www.finesotware.eu 11 1989-2009 Fine Ltd.
Figure 5: Dimensioning of cross-section Calculation of weight and centroid of the wall. Weight of the construction: W = 23 (3.5 0.7 + 0.5 3.5 0.7) = 84.525 m Centroid of the construction: x = 23 (3.5 0.7 (0.7 + 0.35) + 0.5 3.5 0.7 2 0.7 ) 3 84.525 = 0.856 m z = 23 (3.5 0.7 3.5 + 0.5 3.5 0.7 3.5 2 3 ) = 1.556 m 84.525 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 = 27.0 + 16.561 2 = 60.121 Active pressure σ a2b at the end of second level (value at the beginning is same, i.e. σ a2a = 4.121 ): σ a2b = 0.402 60.121 2 5.714 0.588 = 17.424 www.finesotware.eu 12 1989-2009 Fine Ltd.
Resultant force of the active pressure S a2 and horizontal, resp. vertical component: S a2 = 0.5 (17.424 4.121) 2 + 4.121 2 = 21.545 S ax2 = S a2 cos δ d2 = 21.545 cos 13.636 = 20.938 S az2 = S a2 sin δ d2 = 21.545 sin 13.636 = 5.079 Points of action of resultant forces: x 1 = 1.400 m z 1 = 2.84 + 0.8 = 2.040 m x 2 = 1.400 m 0.5 4.121 2 2 + 1 (17.424 4.121) 22 z 2 = 6 4.121 2 + 0.5 (17.424 4.121) 2 = 0.794 m Total resultant force of the active pressure S a and horizontal, resp. vertical component: S ax = 20.989 S az = 5.092 S a = 21.598 Point of action of resultant force S a : x = 1.400 m z = 2.040 0.052 0.794 20.938 21.598 = 0.797 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 = 20.000 Resultant force of the water pressure S w : S w = 1 2 σ w h = 1 20 2 = 20.000 2 www.finesotware.eu 13 1989-2009 Fine Ltd.
Point of action of the resultant force: x = 2.300 m z = 2 3 = 0.667 m Verification of shear strength. Shear and normal force, bending moment and shear strength of cross-section are calculated. Design shear force: V Ed = 20.989 + 20 = 40.989 Result from program : V Ed = 40.94 Design normal force: N Ed = 84.525 + 5.092 = 89.617 Result from program : N Ed = 89.57 Area of cross-section: A cc = b h = 1 1.4 = 1.4 m 2 Stress in cross-section area: σ cp = N Ed A cc 1000 = 89.617 1.4 1000 = 0.064 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 = 0.8 20 1.5 = 10.667 MPa Lower value of characteristic strength of concrete in tension: f ctk.005 = 0.7 (0.3 f 2 3 ck ) = 0.7 (0.3 20 2 3 ) = 1.547 MPa Design strength of concrete in tension (coefficients α ct.pl = 0.8, γ c = 1.5): f ctd = α ct.pl fctk.005 = 0.8 1.547 γ c 1.5 = 0.825 MPa www.finesotware.eu 14 1989-2009 Fine Ltd.
Limit stress: σ c.lim = f cd 2 f ctd (f cd + f fctd ) Shear strength: σ c.lim = 10.667 2 0.825 (10.667 + 0.825) = 4.508 MPa ( f cvd = f 2 max(0, σcp σ c.lim ) ctd + σ cp f ctd 2 ) 2 ( ) 2 0 f cvd = 0.8252 + 0.064 0.825 = 0.857 MPa 2 Design shear strength (k = 1.5): V Rd = f cvd A cc k = 0.857 1.4 1.5 = 799.523 Result from program : V Rd = 799.37 Usage: V u = V Ed 100 = 40.989 100 = 5.1 % O.K. V Rd 799.523 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 = 84.525 (0.856 0.7) 5.092 0.7 + 20.989 0.797 + 20 0.667 = 13.355 Result from program : M Ed = 13.32 knm/m Normal force eccentricity: e = M Ed N Ed = 13.355 89.617 = 0.149 m www.finesotware.eu 15 1989-2009 Fine Ltd.
0.9 h 2 = 0.9 1.4 2 = 0.63 m > e = 0.149 m Effective height: χ = h 2 e = 1.4 2 0.149 = 1.102 m Design normal strength (η = 1.0): N Rd = χ η f cd = 1.102 1 10.667 = 11754.239 Result from program : N Rd = 11758.56 Usage: V u = N Ed 100 = 89.617 100 = 0.8 % O.K. N Rd 11754.239 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 = 116.484 (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 = 0.05 116.484 = 5.824 W eq.z = k v W = 0.04 116.484 = 4.659 www.finesotware.eu 16 1989-2009 Fine Ltd.
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 = 0.430. 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 = 27.000 and σ z2 = 77.179. Coefficients of the active pressure in both levels: K a1 = cos 15 [ 1 + cos 2 26.50 sin(26.5+15) sin(26.5 5.711) cos 15 cos( 5.711) ] 2 = 0.371 K ac1 = cos 26.5 cos 5.711 cos 15 (1 + tan 0 tan 5.711) (1 + sin(26.5 + 15 5.711)) cos 15 = 0.562 K a2 = cos 15 [ 1 + cos 2 27 sin(27+15) sin(27 5.557) cos 15 cos( 5.557) ] 2 = 0.363 K ac2 = cos 27 cos 5.557 cos 15 (1 + tan 0 tan 5.557) (1 + sin(27 + 15 5.557)) cos 15 = 0.556 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 = 2 12 0.562 18 0.371 = 2.019 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 = 0.363 27 2 8 0.556 = 0.903 σ a2b = K a2 σ z2 2 c 2 K ac2 = 0.363 77.179 2 8 0.556 = 19.126 Total resultant force of the active pressure S a and horizontal, resp. vertical component: www.finesotware.eu 17 1989-2009 Fine Ltd.
S a = 0.5 (σ a2b σ a2a ) h 2 + σ a2a h 2 S a = 0.5 (19.126 0.903) 3.03 + 0.903 3.03 = 30.344 S ax = S a cos δ 2 = 30.344 cos 15 = 29.310 S az = S a sin δ 2 = 30.344 sin 15 = 7.854 Point of action of the resultant force: x = 2.300 m 1 z = 6 3.032 (19.126 0.903) + 1 2 3.032 0.903 + 0.23 = 0.826 m 1 2 3.03 (19.126 0.903) + 3.03 0.903 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 1 + 0.04 = 2.752 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 20.5 0.05 (20.5 10) (1 + 0.04) = 5.362 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) ] 2 www.finesotware.eu 18 1989-2009 Fine Ltd.
K ae1 = cos 2.752 cos(2.752 + 15) cos 2 (26.5 2.752) [ 1 + sin(26.5+15) sin(26.5 2.752 5.711) cos(2.752+15) cos(5.711) ] 2 = 0.410 K ae2 = cos 5.362 cos(5.362 + 15) cos 2 (27 5.362) [ 1 + sin(27+15) sin(27 5.362 5.557) cos(5.362+15) cos(5.557) ] 2 = 0.443 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 ) = 16.561 3.03 (1 + 0.04) = 52.186 σ d0 = σ d1 + γ 1 h 1 (1 k v ) = 52.186 + 18 1.5 (1 + 0.04) = 80.266 Increase of the active pressure from earthquake effects in the first layer of soil MS: σ ae0 = σ d0 (K ae1 K a1 ) (0.410 0.371) = 80.266 1 k v 1 + 0.04 = 3.014 σ ae1a = σ d1 (K ae1 K a1 ) (0.410 0.371) = 52.186 1 k v 1 + 0.04 = 1.959 Increase of the active pressure from earthquake effects in the second layer of soil SC: σ ae1b = σ d1 (K ae2 K a2 ) (0.443 0.363) = 52.186 1 k v 1 + 0.04 = 4.003 σ ae2 = 0 Resultant forces of increase of the active pressure S ae and horizontal, resp. vertical components: www.finesotware.eu 19 1989-2009 Fine Ltd.
S ae1 = 0.5 (σ ae0 σ ae1a ) h 1 + σ ae1a h 1 S ae1 = 0.5 (3.014 1.959) 1.5 + 1.959 1.5 = 3.730 S aex1 = S ae1 cos δ 1 = 3.730 cos 15 = 3.603 S aez1 = S ae1 sin δ 1 = 3.730 sin 15 = 0.965 S ae2 = 0.5 σ ae1b h 2 = 0.5 4.003 3.03 = 6.064 S aex2 = S ae2 cos δ 2 = 6.064 cos 15 = 5.857 S aez2 = S ae2 sin δ 2 = 6.064 sin 15 = 1.569 Points of action of resultant forces: x 1 = 2.300 m z 1 = 2.8 1 2 1.52 1.959 + 1 3 1.52 (3.014 1.959) 1.5 1.959 + 0.5 1.5 (3.014 1.959) = 3.603 m x 2 = 2.300 m z 2 = 0.23 2 3.03 = 1.790 3 m Total resultant force of increase of the active pressure S ae and horizontal, resp. vertical component: www.finesotware.eu 20 1989-2009 Fine Ltd.
2 S aex = S aexi = 3.603 + 5.857 = 9.460 i=1 2 S aez = S aezi = 0.965 + 1.569 = 2.535 i=1 S ae = S 2 aex + S 2 aez = 9.460 2 + 2.535 2 = 9.794 Point of action of the resultant force S ae : x = 2.300 m z = 2 i=1 S aexi z i S aex = 12.981 10.485 9.460 = 2.480 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 = 33.330. 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 = (116.484 + 4.659) 1.601 + (7.854 + 2.535) 2.3 = 217.810 knm/m Result from program : M r = 217.83 knm/m Driving moment M o : M o = 0.43 0.2+29.31 0.826+9.46 2.48+33.33 1.057+5.824 1.817 = 93.377 knm/m Result from program : M o = 93.43 knm/m Usage: V u = M o 100 = 93.377 100 = 42.9 % O.K. M r 217.810 www.finesotware.eu 21 1989-2009 Fine Ltd.
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 = 116.484 + 4.659 + 7.854 + 2.535 = 131.531 F hor = 0.430 + 29.310 + 9.460 + 33.330 + 5.824 = 77.494 Normal force in the footing bottom (inclination of footing bottom α b = 5.711, see Fig. 4): N = 131.531 cos 5.711 + 77.494 sin 5.711 = 138.589 Shear reaction in the footing bottom: T = 131.531 sin 5.711 + 77.494 cos 5.711 = 64.022 Eccentricity of the normal force (inclined width of the footing bottom d = 2.311 m): e = d 2 M r M o N = 2.311 2 217.810 93.43 138.589 = 0.258 m Maximal allowable eccentricity: e alw = d 3 = 2.311 3 = 0.770 m > e = 0.258 m O.K. Resisting force H r : H r = 138.589 tan 27 + 8 (2.311 2 0.258) = 84.981 Result from program : H r = 85.07 Driving force H d : H d = T = 64.022 Result from program : H d = 64.05 www.finesotware.eu 22 1989-2009 Fine Ltd.
Usage: V u = H d 100 = 64.022 100 = 75.3 % O.K. H r 84.981 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 = 0.258 100 = 33.5 % O.K. e alw 0.770 Result from program : V u = 33.1 % O.K. Stress in the footing bottom: σ = N (d 2 e) = 138.589 (2.311 2 0.258) Result from program : σ = 76.72 Usage: V u = σ R d 100 = 77.178 100 = 77.178 100 = 77.2 % O.K. Result from program : V u = 76.7 % O.K. References [1] Czech normalization institute: EN 1992-1-1, 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, 1926 www.finesotware.eu 23 1989-2009 Fine Ltd.