Vm 01 Gravity Wall

Verification Manual no. 1 Update 02/2016

Verification Analysis of the Gravity Wall

Program: Gravity Wall

File: Demo_vm_en_01.gtz

In this verification manual you will find hand-made verification analysis calculations of gravity wall in permanent and seismic design situations. The results of the hand-made calculations are compared with the results from the GEO5 – Gravity Wall program.

Terms of Reference:

In Figure 1, an example of a gravity wall with inclined footing bottom in 1:10 inclination is shown. The earth body is comprised of two soil layers and the terrain is adjusted in 1:10 inclination. The top layer of the earth body (depth 1.5 m) is formed of sandy silt (MS). The lower layer of the earth body is formed of clayey sand (CS), which is at the front face of the wall too. The groundwater table is in the depth of 1.5 m behind the wall and 3.7 m in front of the wall. The properties of soils (effective values) are in Table 1. The gravity wall is made from plain concrete C20/25 with unit weight  = 23 kN/m3_{. A verification analysis of the wall is performed with the help of the theory of limit states.}

Soil Unit weight _{ [kN/m}3_] Saturated unit weight sat  [kN/m3_] Angle of internal friction ef  [°] Cohesion of soil ef c [kPa] Angle of friction struc.-soil [°] Poisson’s ratio  [-] MS 18.00 20.00 26.50 12.00 15.00 0.35 SC 18.50 20.50 27.00 8.00 15.00 0.35

Table 1 Soil properties – characteristic effective values

The angle of friction and cohesion enter the first phase of the calculation as design values and that’s why the soil parameters from Table 1 are reduced with coefficients _m 1.1 and_mc 1.4. The design values used in calculation are in Table 2.

Soil Angle of internal friction

d ef ,  [°] Cohesion of soil d ef c _, [kPa]

Angle of friction struc.-soil d

 [°]

MS 24.091 8.571 13.636

SC 24.545 5.714 13.636

Table 2 Soil properties – design values

1. The First Stage - Permanent Design Situation

Verification of the Whole Wall

Calculation of the weight force and the centroid of the wall. The wall is divided into 5 parts, which are shown in Figure 1. Parts 4 and 5 are under the groundwater table, therefore the unit weight of concrete is reduced by the unit weight of waterw= 10 kN/m

3_{. Table 3 shows the dimensions of the} parts of the wall, their weight forces and centroids.

Part height i h [m] width i b [m] Area i A [m2_] Part weight i  [kN/m3_] Weight force i W [kN/m] Point of action xi Gi Gizi i x [m] zi[m] 1 3.500 0.700 2.450 23 56.350 1.950 -2.550 109.883 -143.693 2 3.500 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 5.273 0.264 Total 116.484 - - 186.460 -211.628

 Centroid of the construction: m Wi xi Wi x_t 1.601 484 . 116 460 . 186 5 1 5 1    





m Wi zi Wi zt 1.817 484 . 116 628 . 211 5 1 5 1     





Calculation of the front face resistance. The depth of soil in front of the wall is 0.6 m. Pressure at rest is considered.

 Hydraulic gradient:

(hw- water tables difference, dd- seepage path downwards, du- seepage path upwards)

606 . 0 6 . 0 03 . 3 5 . 1 7 . 3 _      u d w d d h i

 Coefficient of earth pressure at rest:

(For cohesive soils the Terzaghi formula for computing of the coefficient of earth pressure at restKr is used) 538 . 0 35 . 0 1 35 . 0 1       r K

 Unit weight of soil in the area of ascending flow:

3 / 439 . 4 606 . 0 10 0 . 10 5 . 20 kN m i w w sat            

 Vertical normal effective stress zin the footing bottom:

kPa h

z   4.4390.62.663 

 Pressure at rest in the footing bottom: kPa K_r

z

r   2.6630.5381.433 

 Resultant force of stress at restSr:

(Resultant force

S

_r acts only in horizontal direction, thereforeSrSrx andSrz 0) m kN h S_{r r} 1.433 0.6 0.430 / 2 1 2 1      

 Point of action of the resultant forceSr: m x0.000 m h z 0.6 0.200 3 1 3 1 _{  }  

Calculation of the active pressure. There are two layers of soils in the area, where we solve the active pressure. The structure is therefore divided into two sections, in both of which the geostatic pressurez, the active earth pressurea and the resultant forces Sa1 andSa2 are calculated. The

active earth pressure is calculated using Coulomb’s theory.

Figure 2 Geostatic pressure zand active pressure a  Coefficients of active earth pressure in both sections:

(0- back face inclination of the structure, 0-inclination of the terrain; design values of soils from Table 2 are used in the calculation)

a

K - coefficient of active earth pressure

                    2 2 2 ) cos( ) cos( ) sin( ) sin( 1 ) cos( ) ( cos ) ( cos              a K

ac

K - coefficient of active earth pressure due to cohesion





) cos( 1 ) sin( 1 ) ( ) ( 1 ) cos( ) cos( ) cos(

























              tg tg K_ac

Calculation for the first section:           5.711 10 1 1



arctg



4097 . 0 ) 711 . 5 0 cos( ) 636 . 13 0 cos( ) 711 . 5 091 . 24 sin( ) 636 . 13 091 . 24 sin( 1 ) 636 . 13 0 cos( ) 0 ( cos ) 0 091 . 24 ( cos 2 2 2 1                      a K





5936 . 0 ) 0 636 . 13 cos( 1 ) 711 . 5 0 636 . 13 091 . 24 sin( 1 ) 711 . 5 ( ) 0 ( 1 ) 0 636 . 13 cos( ) 711 . 5 cos( ) 091 . 24 cos( 1 _{   }  _          tg tg K_ac

Calculation for the second section:

                  5.557 5 . 18 ) 711 . 5 ( 0 . 18 ) ( 2 tg arctg tg arctg i     4016 . 0 ) 557 . 5 0 cos( ) 636 . 13 0 cos( ) 557 . 5 545 . 24 sin( ) 636 . 13 545 . 24 sin( 1 ) 636 . 13 0 cos( ) 0 ( cos ) 0 545 . 24 ( cos 2 2 2 2                      a K





5882 . 0 ) 0 636 . 13 cos( 1 ) 557 . 5 0 636 . 13 545 . 24 sin( 1 ) 557 . 5 ( ) 0 ( 1 ) 0 636 . 13 cos( ) 557 . 5 cos( ) 545 . 24 cos( 2 _{   }  _          tg tg K_ac

 Unit weight of soil SC in the area of descending flow:

3 2 sat w wi20.510.0100.60616.561kN/m



 Vertical geostatic pressure zin two sections:

kPa h z1 1 118.01.527.000  kPa h h z2 1 12  2 18.01.516.5613.0377.180 

 Calculation of high in first layer of soil MS, where the active earth pressure is neutral: m K K c h a ac d ef 380 . 1 4097 . 0 0 . 18 5936 . 0 571 . 8 2 2 1 1 1 1 , 0 _         

 Active earth pressure ain two sections: kPa K c K_{a ef d ac} z a1 1 12 , 1 1 27.000.4097 28.5710.59360.886  kPa K c K_{a ef d ac} z a a2  1 2 2 , 2 2 27.000.401625.7140.58824.121  kPa K c K_{a ef d ac} z b a2  2  2 2 , 2  2 77.180.4016 25.7140.5882 24.274 

 Resultant forces of active earth pressure Sa1,Sa2and vertical and horizontal components:

m kN h h S_{a a} 0.886 (1.500 1.380) 0.053 / 2 1 ) ( 2 1 0 1 1 1         m kN S Sa1,x  a1cos()0.053cos(13.636)0.052 / m kN S S_a1,z  _a1sin()0.053sin(13.636)0.013 / m kN h h S_{a a b a a a a} (24.274 4.121) 3.03 4.121 3.03 43.018 / 2 1 ) ( 2 1 2 2 2 2 2 2               m kN S Sa2,x  a2cos()43.018cos(13.636)41.806 / m kN S S_a2,z  _a2 sin()43.018sin(13.636)10.142 /  Points of action of resultant forces Sa1andSa2:

m x₁2.300 m z 2.840 3 38 . 1 5 . 1 0 . 2 8 . 0 1       m x₂ 2.300 m z 0.927 2 03 . 3 ) 121 . 4 274 . 24 ( 03 . 3 121 . 4 23 . 0 3 03 . 3 2 03 . 3 ) 121 . 4 274 . 24 ( 23 . 0 2 03 . 3 03 . 3 121 . 4 2           _{ }         _{ }   

 Total resultant force of active earth pressureSa: m kN S S Sax  a1,x a2,x 0.05241.80641.858 / m kN S S Saz  a1,z  a2,z 0.01310.14210.155 /

m kN S

S

S_a  _ax2  _az2  41.8582 10.1552 43.072 /

 Point of action of total resultant force: m x_a 2.300 m S z S z z ai i z ai a 0.929 155 . 10 ) 927 . 0 ( 142 . 10 ) 840 . 2 ( 013 . 0 2 1 , 2 1 ,          

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

Calculation of the water pressure. The heel of the structure is sunken in a permeable subsoil, which allows free water flow below the structure. Therefore, the hydrodynamic pressure must be considered and its resultant force is calculated as shown in Figure 3. The area of the hydrodynamic pressure is divided into two sections.

Figure 3 Hydrodynamic pressure w

 Horizontal water pressurew at interface of section 1 and section 2 (depth 3.7 m):

kPa

w

w (3.71.5)102.222.000 

 Resultant force of water pressure Swin two sections: m kN h S_{w w w} 22.000 2.2 24.200 / 2 1 2 1 1 1        m kN h S_{w w w} 22.000 0.83 9.130 / 2 1 2 1 2 2       

 Points of action of resultant forces: m x₁2.300 m z 1.333 3 2 . 2 6 . 0 1    m x₂ 2.300 m z 0.323 3 23 . 0 6 . 0 6 . 0 2          

 Total resultant force of water pressureSw: m kN S Sw wi 24.200 9.130 33.330 / 2 1 ,    



 Point of action of resultant force Sw:

m x_w 2.300 m S z S z wi i wi w 1.056 33 . 33 ) 323 . 0 ( 13 . 9 ) 333 . 1 ( 2 . 24 2 1 2 1         





Checking for overturning stability. The moments calculated in the analysis rotate about the origin of the coordinate system (left bottom corner of the structure). Resisting momentMres and overturning momentMovr are calculated for verification.

 Calculation of resisting moment Mres and its reduction by coefficient S 1.1:

m kNm r S r W Mres   1  az  2116.4841.60110.1552.3209.847 / m kNm M S res _{190.770 /} 1 . 1 847 . 209  



 Calculation of overturning moment Movr:

m kNm M_ovr 0.4300.241.8580.92933.331.05673.997 /

Result from the GEO5 – Gravity Wall program: Movr 74.02kNm/m

 Usage: % 8 . 38 100 770 . 190 997 . 73 100     res ovr u M M V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: V_u 38.8%, SATISFACTORY

Checking for slip. Slip in the inclined footing bottom (Figure 4).

Figure 4 Forces acting in the footing bottom  Total vertical and horizontal forces Fverand Fhor:

m kN F_ver 116.48410.155126.639 / m kN Fhor 0.4341.85833.3374.758 /  Normal force N :  5.711 b  m kN coc F F

 Shear force T:

m kN F

F

T  _ver sin(_b) _horcos(_b)126.639sin(5.711)74.758cos(5.711)61.785 /

 Eccentricity of normal force:

d- inclined width of footing bottom alw

e - maximal allowable eccentricity

m d b 311 . 2 ) 711 , 5 cos( 3 . 2 ) cos( 3 . 2 _{ }   m N d N M M e res ovr 138 . 0 450 . 133 2 311 . 2 450 . 133 847 . 209 997 . 73 2 _        

In the program, eccentricity is calculated as a ratio. 060 . 0 311 . 2 138 . 0    d e e_ratio 060 . 0 333 . 0    _ratio alw e e , SATISFACTORY

 Resisting horizontal force Hres and its reduction by coefficient S 1.1:



- reduction coefficient of contact base - soil

0 . 1   (without reduction) res F - resisting force kN F_res 0       _{ }                  0 . 1 ) 138 . 0 2 311 . 2 ( 714 . 5 ) 545 . 24 ( 450 . 133 ) 2 ( tg F e d c tg N H_{res d} d _res   +0 m kN Hres 72.571 / m kN H S res / 974 . 65 1 . 1 571 . 72 _  

Result from the GEO5 – Gravity Wall program: H_res 65.98kNm/m

 Acting horizontal force Hact:

m kN T

H_act  61.785 /

 Usage: % 7 . 93 100 974 . 65 785 . 61 100     res act u H H V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: V_u 93.6%, SATISFACTORY

Bearing Capacity of the Foundation Soil

The bearing capacity of the foundation sol is set to Rd 100kPa, and is compared with the stress in the inclined footing bottom.

 Usage – eccentricity: % 0 , 18 100 333 , 0 060 , 0 100     alw u e e V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: V_u 18.0%, SATISFACTORY

 Stress in the footing bottom:

kPa e d N 577 . 65 138 . 0 2 311 . 2 450 . 133 2       

Result from the GEO5 – Gravity Wall program: 65.57kPa

 Usage: % 6 . 65 100 100 577 . 65 100     d u R V  , SATISFACTORY

Result from the GEO5 – Gravity Wall program: Vu 65.6%, SATISFACTORY

Dimensioning – Wall Stem Check

In this example, a section in the level of x-axis in Figure 5 is verified. The verified cross-section is made from plain concrete C 20/25 (characteristic cylindrical strength of concrete in compression f_ck 20000 kPa, characteristic strength of concrete in tensionf_ctm 2200 kPa) with height h1.40mand widthb1.00m. The verification of a cross-section made from plain concrete is realized in accordance with EN 1992-1-1.

Figure 5 Dimensioning – wall stem check, cross-section  Calculation of the weight force and the centroid of the wall:

m kN W 0.7 3.5) 84.525 / 2 1 5 . 3 7 . 0 ( 23       m x_t 0.856 525 . 84 3 7 . 0 2 5 . 3 7 . 0 2 1 7 . 0 2 7 . 0 5 . 3 7 . 0 23                   _     m z_t 1.556 525 . 84 3 5 . 3 5 . 3 7 . 0 2 1 2 5 . 3 5 . 3 7 . 0 23                            

Calculation of the active earth pressure. The area behind the evaluated part of the construction is divided into two sections. In the first section, the active pressure is the same as in the analysis of the whole wall. The centroids of all forces must be recalculated.

 Vertical geostatic stressz2at the end of the second section:

kPa h

z

z2  12  2 27.016.5612.060.122



 Active earth pressure a 2bat the end of the second section:

kPa

b

a2 0.401660.12225.7140.5882 17.423

 Resultant force of active earth pressureSa2 and vertical and horizontal component:

(Resultant force Sa1 at the beginning is the same)

m kN S_a (17.423 4.121) 2.0 4.121 2.0 21.544 / 2 1 2        m kN S S_a2,x  _a2cos()21.544cos(13.636)20.937 / m kN S S_a2,z  _a2sin()21.544sin(13.636)5.079 /  Calculation of points of action:

m x₁1.400 m z (1.50 1.38) 2 2.040 3 1 1     m x2 1.400 m z 0.794 2 00 , 2 ) 121 . 4 423 . 17 ( 00 . 2 121 . 4 3 00 . 2 2 00 . 2 ) 121 . 4 423 . 17 ( 2 00 . 2 00 . 2 121 . 4 2                       

 Total resultant force of active earth pressureSa and horizontal and vertical component: m kN S S Sax  a1,x  a2,x 0.05220.93720.989 / m kN S S Saz  a1,z  a2,z 0.0135.0795.092 / m kN S S S_a  _ax2  _az2  20.9892 5.0922 21.598 /

 Point of action of resultant forceSa:

m x_a 1.400 m z_a 0.797 937 . 20 052 . 0 ) 794 . 0 ( 937 . 20 ) 800 . 0 840 . 2 ( 052 . 0          

Calculation of the water pressure. Water pressure becomes higher with the increasing depth.  Horizontal water pressurew in depth of 3.5 m under the surface of the adjusted terrain:

kPa

w

w (3.51.5)102.020.000 

 Resultant force of water pressureSw: m kN h S_{w w w} 20.000 2.0 20.000 / 2 1 2 1        

 Point of action of resultant force Sw:

m x₁1.400 m z 2,0 0.667 3 1 1   

Verification of the shear strength. The design shear force, design normal force, design bending moment and shear strength of the cross-section are calculated. The design bending moment rotates about the middle of the verified cross-section.

 Design shear force VEd :

m kN S

S

VEd  w  ax 20.00020.98940.989 /

Result from the GEO5 – Gravity Wall program: V_Ed 40.94kN/m

 Design normal force NEd:

m kN W

S

N_Ed  _az 5.09284.52589.617 /

Result from the GEO5 – Gravity Wall program: N_Ed 89.57kN/m

 Design bending moment MEd:

4 3 2 1 S r S r S r r W M_Ed    _az   _ax  _w m kNm M_Ed 84.525(0.8560.700)5.0920.70020.9890.79720.0000.667 13.311 /

Result from the GEO5 – Gravity Wall program: M_Ed 13.32kNm/m

 Calculation of the area of compressed concrete Acc:

Determining the compressed area of concrete is necessary in order to determine the stresses on the front and the rear edges of the verified cross-section. The normal force NEd makes compressive stress and therefore is considered to be a negative force.

Stress from the design normal forceNEd:

kPa A N A N Ed _64.012 40 . 1 00 . 1 617 . 89 _    

Stress from the design bending momentMEd: kPa W M W M y Ed 748 . 40 40 . 1 00 . 1 6 1 311 . 13 2        

Figure 6 Course of tension on a cross-section of a wall stem

From Figure 6 it can be seen, that the whole area of the cross-section is compressed. 2 40 . 1 40 . 1 00 . 1 .h m b A_cc  _c   

 Stress in cross-section areacp:

MPa A N cc Ed cp 0.064 1000 1 400 . 1 617 . 89 1000 1 _{  }   

 Design strength of concrete in compression fcd:

MPa f f c ck pl cc cd 10.667 5 . 1 00 . 20 8 . 0 ,       

 Design strength of concrete in tension fctd: 005

,

ctk

f - lower value of characteristic strength of concrete in tension

MPa f f f c ctm pl ct c ctk pl ct ctd 0.821 5 . 1 20 . 2 7 . 0 8 . 0 7 . 0 , 005 , ,               Limit stress: MPa f f f f_{cd ctd cd ctd} c,lim 2 (  )10.6672 0.821(10.6670.821)4.525   Shear strength fcvd: 2 2 2 lim , 2 2 525 . 4 064 . 0 ; 0 max( 821 . 0 064 . 0 821 . 0 2 , 0 max(                       cp c ctd cp ctd cvd f f f   

MPa f_cvd 0.852

 Design shear strength VRd: 5 . 1  k m kN k A f V_Rd cvd cc 1000 795.200 / 5 . 1 4 . 1 852 . 0 1000       

Result from the GEO5 – Gravity Wall program: V_Rd 795.74kN/m

 Usage: % 2 . 5 100 200 . 795 989 . 40 100      Rd Ed u V V V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: V_u 5.1%, SATISFACTORY

Verification of a cross-section loaded by bending moment and normal force.  Calculation of eccentricity e :



0.149;0.047;0.02



02 . 0 ; 30 400 . 1 ; 617 . 89 311 . 13 02 . 0 ; 30

; h m Max abs Max

N M abs Max e Ed Ed _                             m e0.149

 Effective height of cross-section: m e

h2 1.40020.1491.102 



 Design normal strengthNRd:

0 . 1 200 50 50 0 . 1 200 50 ) 50 ; 20 ( 0 . 1 200 50 ) 50 ; ( 0 . 1           Max fck Max  m kN f b N_Rd (  _cd)1000 (1.01.1021.010.667)1000 11754.667 /

Result from the GEO5 – Gravity Wall program: N_Rd 11758.60kNm/m

 Usage: % 8 . 0 100 667 . 11754 617 . 89 100     Rd Ed u N N V , SATISFACTORY

2. The Second Stage – Seismic Design Situation

Verification of the Whole Wall

The second stage of calculation uses the same wall influenced by an earthquake. The calculation of earthquake effects is made according to the Mononobe-Okabe theory. The factor of horizontal acceleration is k_h 0.05(inertial force acts horizontally in an unfavourable direction) and the factor of vertical acceleration is k_v0.04(inertial force acts downwards). The coefficients of reduction of soil parameters and the coefficients of overall stability of construction are equal to one. Therefore, the design values of soil properties are the same as the characteristic values in Table 1.

Calculation of the weight force of the wall. To determine the horizontal and vertical components of a force from an earthquake, it is necessary to calculate the weight force of wall without the buoyancy exerted on the wall by the groundwater. The calculation is shown in Table 4.

Block Height i h [m] Width i b [m] Area i A [m2_] Block weight i  [kN/m3_] Weight force i W [kN/m] Point of action xi Gi Gizi i x [m] zi[m] 1 3.500 0.700 2.450 23 56.350 1.950 -2.550 109.883 -143.693 2 3.500 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 23 31.740 1.150 -0.300 36.501 -9.522 5 0.230 2.300 0.265 23 6.095 1.533 0.077 9.344 0.469 Total 132.940 - - 206.407 -215.563

Table 4 Dimensions, weight force and centroids of the individual blocks  Centroid of the structure:

m Wi xi Wi x_t 1.553 940 . 132 407 . 206 5 1 5 1    





m Wi zi Wi zt 1.622 940 . 132 563 . 215 5 1 5 1     





 Horizontal and vertical component of the force from the earthquake: m kN W k W_eq,x _h 0.05132.9406.647 / m kN W k W_eq,z  _v (0.04)132.9405.318 /

Calculation of the front face resistance. The pressure at rest on the front face of the wall is the same as in the first stage of the calculation. The resultant force of the pressure at rest is

m kN S_r 0.430 / .

Calculation of the active earth pressure. The course of the geostatic pressure is the same as in the first stage of calculation. In the calculation coefficients of active earth pressure Kaand Kacare used as characteristic values of soil properties. The active earth pressure a and the resultant force of active earth pressureSa are calculated.

 Course of geostatic stress: kPa z127.000  kPa z2 77.180 

 Coefficients of active earth pressure in both sections:

(0- back face inclination of the structure, 0-inclination of terrain; characteristic values of soils from Table 1 are used in calculation)

Calculation for the first layer:           5.711 10 1 1



arctg



3711 . 0 ) 711 . 5 0 cos( ) 0 . 15 0 cos( ) 711 . 5 5 . 26 sin( ) 0 . 15 5 . 26 sin( 1 ) 0 . 15 0 cos( ) 0 ( cos ) 0 5 . 26 ( cos 2 2 2 1                      a K





5619 . 0 ) 0 0 . 15 cos( 1 ) 711 . 5 0 0 . 15 5 . 26 sin( 1 ) 711 . 5 ( ) 0 ( 1 ) 0 0 . 15 cos( ) 711 . 5 cos( ) 5 . 26 cos( 1 _{   }  _          tg tg K_ac

Calculation for the second layer:

                  5.557 5 . 18 ) 711 . 5 ( 0 . 18 ) ( 2 tg arctg tg arctg i     3631 . 0 ) 557 . 5 0 cos( ) 0 . 15 0 cos( ) 557 . 5 0 . 27 sin( ) 0 . 15 0 . 27 sin( 1 ) 0 . 15 0 cos( ) 0 ( cos ) 0 0 . 27 ( cos 2 2 2 2                      a K





5563 . 0 ) 0 0 . 15 cos( 1 ) 557 . 5 0 0 . 15 0 . 27 sin( 1 ) 557 . 5 ( ) 0 ( 1 ) 0 0 . 15 cos( ) 557 . 5 cos( ) 0 . 27 cos( 2 _{   }  _          tg tg K_ac

 Calculation of height in the first layer of soil MS, where the active earth pressure is neutral: m K K c h a ac ef 019 . 2 3711 . 0 0 . 18 5619 . 0 12 2 2 1 1 1 1 , 0           > 1.50m

 Active earth pressurea is calculated only for the second section (in the first section it’s equal to zero): kPa K c K_{a ef d ac} z a a2  1 2 2 , 2  2 27.000.363128.00.55630.903  kPa K c K_{a ef d ac} z b a2  2 2 2 , 2 2 77.180.363128.00.556319.123 

 Resultant force of the active earth pressureSaand its horizontal and vertical components: m kN h h S_{a a b a a a a} (19.123 0.903) 3.03 0.903 3.03 30.339 / 2 1 ) ( 2 1 2 2 2 2 2                 m kN S S_ax  _a cos()30.339cos(15.0)29.306 / m kN S S_az  _a sin()30.339sin(15.0)7.852 /  Point of action of the resultant force:

m x2,300 m z 0.826 2 03 . 3 ) 903 . 0 123 . 19 ( 03 . 3 903 . 0 23 . 0 3 03 . 3 2 03 . 3 ) 903 . 0 123 . 19 ( 23 . 0 2 03 . 3 03 . 3 903 . 0            _{ }         _{ }   

Increase of the active earth pressure caused by an earthquake. An earthquake increases the effect of active earth pressure.

 Calculation of the seismic inertia angle in the first layer (without restricted water influence):                    2.752 ) 04 . 0 ( 1 05 . 0 1 1 arctg k k arctg v h 

 Calculation of the seismic inertia angle in the second layer (with restricted water influence):









                                      5.362 04 . 0 1 ) 10 5 . 20 ( 05 . 0 5 . 20 ) 1 ( ) ( ) 1 ( ,2 2 , 2 , 2 , 2 arctg k k k k arctg v w sat h sat v su h sat      

 Coefficient Kaefor the active earth pressure in both sections:                          2 2 2 ) cos( ) cos( ) sin( ) sin( 1 ) cos( ) ( cos cos ) ( cos                   ae K                          2 2 2 1 ) 711 . 5 0 cos( ) 752 . 2 0 . 15 0 cos( ) 711 . 5 752 . 2 5 . 26 sin( ) 0 . 15 5 . 26 sin( 1 ) 0 . 15 0 752 . 2 cos( ) 0 ( cos ) 752 . 2 cos( ) 0 752 . 2 5 . 26 ( cos ae K 4102 . 0 1 ae K                          2 2 2 2 ) 557 . 5 0 cos( ) 362 . 5 0 . 15 0 cos( ) 557 . 5 362 . 5 0 . 27 sin( ) 0 . 15 0 . 27 sin( 1 ) 0 . 15 0 752 . 2 cos( ) 0 ( cos ) 362 . 5 cos( ) 0 362 . 5 0 . 27 ( cos ae K 4429 . 0 2  ae K

 Calculation of normal stress dfrom the earthquake effects. The normal stress is calculated from the bottom of the wall:

kPa

d2 0.000

 - normal stress in the footing bottom





kPa k h v d12  2 (1 )16.5613.031(0.04) 52.187 





kPa k h _v d d0  1 1 1(1 )52.187 18.01.50 1(0.04) 80.267 

 Increase of the active earth pressure caused by the earthquake in both sections:

kPa K K_{ae a} d a ae,1  0 ( 1  1)80.267(0.41020.3711)3.138  kPa K K_{ae a} d b ae,1  1( 1 1)52.187(0.41020.3711)2.041  kPa K K_{ae a} d a ae,2  1( 2  2)52.187(0.44290.3631)4.165 

 Resultant forces of the increase of the active earth pressureSae in both sections:

m kN h h S_{ae ae a ae b ae b} (3.138 2.041) 1.50 2.041 1.50 3.884 / 2 1 ) ( 2 1 1 1 , 1 1 , 1 , 1              m kN S Sae1x  ae1cos()3.884cos(15.0)3.752 /

m kN S S_ae1z  _ae1sin()3.884sin(15.0)1.005 / m kN h S_{ae ae a} 4.165 3.03 6.310 / 2 1 2 1 2 2 , 2        m kN S S_ae2x  _ae2cos()6.310cos(15.0)6.095 / m kN S Sae2z  ae2sin()6.310sin(15.0)1.633 /

 Points of action of the resultant forces: m x12.300 2 50 . 1 ) 041 . 2 138 . 3 ( 50 . 1 041 . 2 ) 23 . 0 03 . 3 ( ) 50 . 1 ( 3 2 2 50 . 1 ) 041 . 2 138 . 3 ( ) 23 . 0 03 . 3 ( 2 50 . 1 50 . 1 041 . 2 1           _{   }         _{  }    z m z1 3.603 m x₂ 2.300 m z ( 3.03) 0.23 1.790 3 2 2     

 Total resultant force of the increase of active earth pressureSaeand its horizontal and vertical component: m kN S S S_ae,x  _ae1x  _ae2x 3.7526.0959.847 / m kN S S S_ae,z  _ae1z  _ae2z 1.0051.6332.638 / m kN S S S_ae  _ae,x2  _ae,z2  9.8472 2.6382 10.194 /  Point of action of the resultant forceSae:

m x_ae 2.300 m z_ae 2.481 095 . 6 752 . 3 ) 790 . 1 ( 095 . 6 ) 603 . 3 ( 752 . 3         

Calculation of water pressure. The water pressure is the same as in the verification of the whole wall in the first stage. The resultant force of the water pressure is Sw 33.330kN/m and has the same point of action as in the first stage.

Calculation of the hydrodynamic pressure acting on the front face of the wall. The action of the hydrodynamic pressure caused by the earthquake is calculated from the groundwater table to the bottom of the wall. The direction of the force is the same as the direction of the horizontal acceleration.  Calculation of the resultant force of the hydrodynamic pressure Pwdcaused by the earthquake:

m H0.60.230.83 m kN H k P_{wd h w} 0.05 10.0 0.83 0.201 / 12 7 12 7 _{  } 2 _{   } 2 _  

 Point of action of the resultant force Pwd: m x2.300 m H y z wd 0.23(0.4 )0.23(0.40.83)0.230.102

Checking for overturning stability. The moments calculated in the analysis rotate about the origin of the coordinate system (left bottom corner of the structure). Resisting momentMres and overturning momentMovr are calculated for verification.

 Calculation of the resisting moment Mres:

300 . 2 638 . 2 300 . 2 852 . 7 553 . 1 318 . 5 601 . 1 484 . 116 4 , 3 2 , 1                W r W r S r S r M_{res eqz az aez} 300 . 2 638 . 2 300 . 2 852 . 7 553 . 1 318 . 5 601 . 1 484 . 116         res M m kNm M_res 218.877 /

Result from the GEO5 – Gravity Wall program: Mres 218.86kNm/m

 Calculation of the overturning moment Movr:

6 5 4 , 3 2 , 1 W r S r S r S r P r r S M_ovr  _r   _eqz   _ax  _aex  _w  _wd  102 . 0 201 . 0 056 . 1 330 . 33 481 . 2 847 . 9 826 . 0 306 . 29 622 . 1 647 . 6 200 . 0 43 . 0              ovr M m kNm Movr 94.550 /

 Usage: % 2 . 43 100 877 . 218 550 . 94 100     res ovr u M M V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: Vu 43.2%, SATISFACTORY

Checking for slip. The slip in the inclined footing bottom in 1:10 inclination is checked (Figure 4.).

 Total vertical and horizontal forces Fvera Fhor:

m kN F_ver 116.4845.3187.8522.638132.292 / m kN Fhor 0.436.64729.6069.84733.3379.000 /  Normal force in the footing bottom N :

 5.711 b  m kN coc F F

N _ver cos(_b) _hor sin(_b)132.292 (5.711)79.000sin(5.711)139.496 /

 Shear force in the footing bottom T:

m kN F

F

T  _ver sin(_b) _horcos(_b)132.292sin(5.711)79.000cos(5.711)65.444 /

 Eccentricity of the normal force:

d- inclined width of the footing bottom alw

e - maximal allowable eccentricity

m d b 311 . 2 ) 711 . 5 cos( 3 . 2 ) cos( 3 . 2 _{ }   m N d N M M e res ovr 264 . 0 496 . 139 2 311 . 2 496 . 139 877 . 218 550 . 94 2 _        

In the program, the eccentricity is calculated as a ratio. 114 . 0 311 . 2 264 . 0    d e e_ratio 114 . 0 333 , 0    _ratio alw e e , SATISFACTORY

 Resisting horizontal force Hres and its reduction by coefficient S 1.1:



- reduction coefficient of contact base - soil

0 . 1 

res F - resisting force kN Fres 0       _{ }                  0 . 1 ) 264 . 0 2 311 . 2 ( 00 . 8 ) 0 . 27 ( 496 . 139 ) 2 ( 2 , 2 , F tg e d c tg N H res ef ef res  _ +0 m kN Hres 85.341 /

Result from the GEO5 – Gravity Wall program: H_res 85.33kNm/m

 Acting horizontal force Hact:

m kN T

Hact  65.444 /

Result from the GEO5 – Gravity Wall program: H_act 65.37kNm/m

 Usage: % 7 . 76 100 341 . 85 444 . 65 100     res act u H H V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: V_u 76.6%, SATISFACTORY

Bearing Capacity of the Foundation Soil

The bearing capacity of the foundation soil is set to Rd 100kPa, and is compared with the stress in the inclined footing bottom.

 Usage – eccentricity: % 2 . 34 100 333 . 0 114 . 0 100     alw u e e V , VYHOVUJE

Result from the GEO5 – Gravity Wall program: Vu 34.6%, SATISFACTORY

 Stress in the footing bottom:

kPa e d N 237 . 78 264 . 0 2 311 . 2 496 . 139 2       

 Usage: % 2 . 78 100 100 237 . 78 100     d u R V  , SATISFACTORY

Result from the GEO5 – Gravity Wall program: Vu 78.3%, SATISFACTORY

Dimensioning – Wall Stem Check

In this example, the section on the level of the x-axis in Figure 5 is verified. The cross-section is made from plain concrete C 20/25 (characteristic cylindrical strength of concrete in compression f_ck 20000 kPa, characteristic strength of concrete in tensionf_ctm 2200 kPa) with height h1.40mand widthb1.00m. The verification of the cross-section made from plain concrete is realized in accordance with EN 1992-1-1.

 Calculation of the weight force and the centroid is the same as in the first stage: m kN W84.525 / m x_t 0.856 m z_t 1.556

 Horizontal and vertical component of the force caused by the earthquake (the centroid is the same as the centroid of the weight force):

m kN W k W_eq,x  _h 0.0584.5254.226 / m kN W k W_eq,z  _v (0.04)84.5253.381 /

Calculation of the active earth pressure. The area behind the evaluated part of the construction is divided into two sections. The centroids of all forces must be recalculated.

 Vertical geostatic stressz1 and z2 in both sections:

kPa h z1 1 2 18.01.527.000  kPa h z z2  12  2 27.016.5612.060.122 

 Active earth pressure in the second section a 2a and a 2b(the active earth pressure in the first section is equal to zero):

kPa a a2 0.363127.00028.00.55630.903  kPa b a2 0.363160.12228.00.556312.929 

 Total resultant force of the active earth pressureSa and its horizontal and vertical components: m kN S_a (12.929 0.903) 2.0 0.903 2.0 13.832 / 2 1        m kN S S_a,x  _a2cos()13.832cos(15.0)13.360 / m kN S S_a,z  _a2sin()13.832sin(15.0)3.580 /  Point of action of the resultant force Sa

m x1.400 m z 0.710 2 00 . 2 ) 903 . 0 929 . 12 ( 00 . 2 903 . 0 3 00 . 2 2 00 . 2 ) 903 . 0 929 . 12 ( 2 00 . 2 00 . 2 903 . 0                         

Increase of the active earth pressure caused by an earthquake. An earthquake increases the effect of the active earth pressure.

 Calculation of the normal stress dfrom the earthquake effects. The vertical pressure is calculated from the lower part of the stem:

kPa

d2 0.000

 - normal stress at the level of the lower part of the stem





kPa k h _v d12 ( 2)(1 )16.561(2.00)1(0.04) 34.447 





kPa k h v d d0  11 1(1 )34.447 18.01.501(0.04) 62.527 

 Increase of the active earth pressure caused by the earthquake effects in both sections: kPa K K_{ae a} d a ae,1  0( 1  1)62.527(0.41020.3711)2.445  kPa K K_{ae a} d b ae,1  1( 1 1)34.447(0.41020.3711)1.347  kPa K K_{ae a} d a ae,2  1( 2  2)34.447(0.44290.3631)2.749 

 Resultant forces of the increase of the active earth pressureSae in both sections:

m kN h h S_{ae ae a ae b ae b} (2.445 1.347) 1.50 1.347 1.50 2.844 / 2 1 ) ( 2 1 1 1 , 1 1 , 1 , 1              m kN S S_ae1x  _ae1cos()2.844cos(15.0)2.747 /

m kN S S_ae1z  _ae1sin()2.844sin(15.0)0.736 / m kN h S_{ae ae a} 2.749 2.00 2.749 / 2 1 2 1 2 2 , 2        m kN S S_ae2x _ae2 cos()2.749cos(15.0)2.655 / m kN S S_ae2z  _ae2 sin()2.749sin(15.0)0.711 /

 Points of action of the resultant forces: m x11.400 m z 2.824 2 50 . 1 ) 347 . 1 445 . 2 ( 50 . 1 347 . 1 00 . 2 ) 50 . 1 ( 3 2 2 50 . 1 ) 347 . 1 445 . 2 ( 200 2 50 . 1 50 . 1 347 . 1 1            _{  }         _{ }    m x2 1.400 m z ( 2.0) 1.333 3 2 2    

 Total resultant force Saeand its horizontal and vertical components: m kN S S S_ae,x  _ae1x  _ae2x 2.7472.6555.402 / m kN S S S_ae,z  _ae1z  _ae2z 0.7360.7111.447 / m kN S S S_ae  _ae,x2  _ae,z2  5.4022 1.4472 5.592 /  Point of action of the resultant force Sae:

m xae 1.400 m z_ae 2.091 655 . 2 747 . 2 ) 333 . 1 ( 655 . 2 ) 824 . 2 ( 747 . 2         

Calculation of the water pressure. Water pressure becomes higher with the increasing depth.  Horizontal water pressurew in depth of 3.5 m under the surface of the adjusted terrain:  _w_w(3.51.5)102.020.000kPa

 Resultant force of the water pressureSw: m kN h S_{w w w} 20.000 2.0 20.000 / 2 1 2 1        

 Point of action of the resultant force Sw:

m x₁1.400 m z 2,0 0.667 3 1 1   

Verification of shear strength. The design shear force, design normal force, design bending moment and shear strength of the cross-section are calculated. The design bending moment rotates about the middle of the verified cross-section.

 Design shear force VEd :

m kN W S S S V_Ed  _w  _a,x  _ae,x  _eq,x 20.00013.3605.4024.22642.988 /

Result from the GEO5 – Gravity Wall program: VEd 42.95kN/m

 Design normal force NEd:

m kN W W S S NEd  a,z  ae,z   eq,z 3.5801.44784.5253.38192.933 /

Result from the GEO5 – Gravity Wall program: N_Ed 92.89kN/m

 Design bending moment MEd:

8 , 7 , 6 , 5 , 4 3 2 1 S r S r S r W r S r W r S r r W M_Ed    _az   _ax  _w  _eqz  _aez   _eqx  _aex 091 . 2 402 . 5 556 . 1 226 . 4 700 . 0 447 . 1 ) 700 . 0 856 . 0 ( 381 . 3 667 . 0 0 ., 2 710 . 0 360 . 13 700 . 0 580 . 3 ) 700 . 0 856 . 0 ( 525 . 84                     Ed M m kNm M_Ed 23.458 /

Result from the GEO5 – Gravity Wall program: MEd 23.46kNm/m

 Calculation of the area of compressed concrete Acc:

Determining the compressed area of concrete is necessary in order to determine the stresses on the front and the rear edges of the verified cross-section. The normal force NEd makes compressive stress and therefore is considered to be a negative force.

kPa A N A N Ed 381 . 66 40 . 1 00 . 1 933 . 92 _    

Stress from design bending momentMEd:

kPa W M W M y Ed 810 . 71 40 . 1 00 . 1 6 1 458 . 23 2        

Figure 7 Course of tension on the cross-section of wall stem

From Figure 7 it can be seen, that not the whole cross section is compressed, only a part, hc:

m h_c 1.347 400 . 1 429 . 5 191 . 138 191 . 138 _   2 347 . 1 347 . 1 00 . 1 .h m b A_cc  _c   

 Stress on the cross-section area cp:

MPa A N cc Ed cp 0.06899 1000 1 347 . 1 933 . 92 1000 1 _{  }   

 Design strength of concrete in compression fcd: MPa f f c ck pl cc cd 10.667 5 , 1 00 . 20 8 . 0 ,       

 Design strength of concrete in tension fctd:

005 ,

ctk

f - lower value of characteristic strength of concrete in tension

MPa f f f c ctm pl ct c ctk pl ct ctd 0.821 5 . 1 20 . 2 7 . 0 8 . 0 7 . 0 , 005 , ,             

 Limit stress: MPa f f f f_{cd ctd cd ctd} c,lim 2 (  )10.6672 0.821(10.6670.821)4.525   Shear strength fcvd : 2 2 2 lim , 2 2 525 . 4 066 . 0 ; 0 max( 821 . 0 06899 . 0 821 . 0 2 , 0 max(                       cp c ctd cp ctd cvd f f f    MPa fcvd 0.855

 Design shear strength VRd: 5 . 1  k m kN k A f V cvd cc Rd 1000 767.790 / 5 . 1 347 . 1 855 . 0 1000      

Result from the GEO5 – Gravity Wall program: VRd 767.58kN/m

 Usage: % 6 . 5 100 790 . 767 988 . 42 100     Rd Ed u V V V , SATISFACTORY

Result from the GEO5 – Gravity Wall program: Vu 5.6%, SATISFACTORY

Verification of a cross-section loaded by bending moment and normal force.  Calculation of eccentricity e :



0.252;0.047;0.02



02 . 0 ; 30 400 . 1 ; 933 . 92 458 . 23 02 . 0 ; 30

; h m Max abs Max

N M abs Max e Ed Ed _                             m e0.252

 Effective high of cross-section: m e

h2 1.40020.2520.896 



 Design normal strength NRd:

0 . 1 200 50 50 0 , 1 200 50 ) 50 ; 20 ( 0 , 1 200 50 ) 50 ; ( 0 , 1           Max fck Max  m kN f b NRd (  cd)1000 (1.00.8961.010.667)1000 9557.632 /

Result from the GEO5 – Gravity Wall program: NRd 9543.22kNm/m

% 0 . 1 100 632 . 9557 933 . 92 100     Rd Ed u N N V , SATISFACTORY