Retaining Wall Design Calculation Example

Posted on April 29, 2012 by retainingwalldesign

The cross section of a cantilever retaining wall is shown in figure 8.10.Calculate the factors of safety in regards to overturning,sliding and bearing capacity.

Solution:

From the figure ,

H’=H1+H2+H3=6tan10⁰+18+2.75=21.81ft   ,                             Pa=1/2y1KaH’2   ;

For Ø1’=34⁰ and α=10⁰, the value of Ka=0.294,so                Pa=1/2(117)(0.294)(21.81)2/1000=8.18kip/ft;

Pv=Pasin10⁰=1.42kip/ft; Ph=Pacos10⁰=8.06kip/ft;

Factor of safety against Overturning

The following table should be prepared now for determination of the resisting moment.

Section

Weight(kip/ft)

Moment arm from C(ft)

Moment about C(kipft/ft)
1 (1.5)(18)(0.15)=4.05 5.75 23.29
2 ½(1)(18)(0.15)=1.35 4+2/3(1)=4.67 6.3
3 (12.5)(2.75)(0.15)=5.15 6.25 32.23
4 (18+19.06)/2(6)(0.117)=13.01 ~4+2.5+6/2=9.5 123.6
Pv=1.42 12.5 17.75
ƩV=24.986 ƩMR=203.17

Note:yconcrete=150lb/ft3

The overturning moment ,MO, is:

MO=PhH’/3=(8.06)(21.81/3)=58.6 kip/ft

FS(overturning)=ƩMR/MO=203.17/58.6=3.47   >2 => OK

Factor of Safety against Sliding

FS(sliding)=((ƩV)tan(k1Ø’2)+Bk2c’2+Pp)/(Pvcosα);    where we consider k1=k2=2/3 and Pp=0. So

FS(sliding)=((ƩV)tan(2/3Ø’2)+B(2/3c’2))/(Pvcosα)=((24.98)tan(2/3(18))+(12.5)2/3(0.9))/8.06=

=1.59  >1.3   =>OK

Factor of Safety against Bearing Capacity Failure

e=B/2-(ƩMR-ƩM0)/ƩV=6.25-(203.17-58.6)24.98=0.464   <  B/6=12.5/6=2.08ft;

qtoe=ƩV/B(1+6e/B)=24.98/12.5(1+6(0.464)/12.5)=2.44 kip/ft2 ;

The ultimate bearing capacity of the soil can be determinate from the following equation:

qu=c’2NcFcdFel+qNqFqdFqi+1/2B’NyFydFyi          where:

q=y2D=(4)(0.107)=0.428 kip/ft2;

B’=B-2e=12.5-(2)(0.464)=11.572ft;

Fcd=1+0.4(D/B’)=1+0.4(4/(11.572)=1.138;

Fqd=1+2tanØ’2(1-sinØ’2)2(D/B’)=1+0.31(4/11.57)=1.107;

Fyd=1;

Fel=Fqi=(1-ψ⁰/90⁰)2

Ψ=tan-1(Pacosδ/ƩV)=(8.06/24.98)=17.88⁰ so

Fel=Fqi=(1-17.88/90)2=0.642;

Fyi=(1-ψ/Ø’2)=(1-17.88/18)2=0   ;      thus :

qu=(0.9)(13.1)(1.138)(0.642)+(0.428)(5.26)(1.107)(0.642)+1/2(0.107)(11.572)(4.07)(1)(0)=10.21 kip/ft2

FS(bearing capacity)=qu/qtoe=10.21/2.44=4.18    >2 => OK

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The vertical pressures transmitted to the soil by the base slab of the retaining wall should be checked against the ultimate bearing capacity of the soil.The nature of variation of the vertical pressure transmitted by the base slab into the soil is shown in fig.8.9. Note that qtoe and qheel are the maximum and the minimum pressures occurring at the ends of the toe and heel sections ,respectively .The magnitudes of qtoe and qheel can be determined in the following manner:

-the sum of the vertical forces acting on the base slab is ƩV,and the horizontal force Ph=Pacosα

R=ƩV+PA

-the net moment of these forces about point C  is     Mnet=ƩMR-ƩMO

-let the line of action of the resultant R intersect the base slab at E,then the distance  CE=X=Mnet/ƩV

Hence, the eccentricity of the resultant R may be expressed as:                   e=B/2-CE .

The pressure distribution under the base slab may be determined by using simple principles from the mechanics of materials.First we have:

q=ƩV/A+/-Mnety/I     where

Mnet-moment=(ƩV)e;

I-moment of inertia per unit length of the base section=1/12(1)B2.

For maximum and minimum pressures ,the value of y in the main equation equals B/2 resulting:

qmax=ƩV/B(1+6e/B)  and

qmin= ƩV/B(1-6e/B)  .

Note that ƩV includes the weight of the soil ,and that when the value of the eccentricity e becomes greater than B/6,qmin becomes negative thus will result some tensile strength at the end of the heel section.This stress is not desirable,because the tensile strength of the soil is very small.If the analysis of a  design shows e>B/6,the design should be reproportioned and calculations redone.

The ultimate bearing capacity of a shallow foundation is defined by the following equation:

qu=c’2NcFcdFØl+qNqFØdFqi+1/2y2B’NyFydFyi    where:

q=y2D

B’=B-2e

Fcd=1+0.4D/B’

FØd=1+2tanØ’2(1-sinØ’2)2D/B’

Fyd=1

Fqi=(1-ψ⁰/90⁰)2

ψ⁰=tan-1(Pacosα/ƩV)

Fyl=(1- ψ⁰/Ø’2⁰)2

Once the ultimate bearing capacity of the soil has been calculated ,the factor of safety can be determinated (generally a factor of 2..3) is required   FS(bearing capacity)=qu/qmax.

Posted on April 28, 2012 by retainingwalldesign | Leave a comment

The factor of safety against sliding may be expressed by the equation:

FS(sliding)=ƩFR’/ƩFsl

where: ƩFR’-sum of the horizontal resisiting forces ƩFsl-sum of the horizontal driving forces Figure 8.2 indicates that the shear strength of the soil immediately below the base slab may be expressed as:

s=σ’tanδ’+c’a

where: δ’-angle of the friction between the soil and the base slab; c’a-adhesion between the soil and the base slab Thus,the maximum resisiting force that can be derived from the soil per unit length of the wall along the bottom of the base slab is:

R’=s(area of cross section)=s(Bx1)=B σ’ tanδ’+Bc’a

however     B σ’=sum of the vertical forces=ƩV   , so

R’=(ƩV) tanδ’+Bc’a

Figure 8.7 shows that the passive force Pp is also a horizontal resistance force,hence

ƩFR’-=(ƩV)tanδ’ +Bc’a+Pp .

The only horizontal force that will tend to cause the wall to slide (a driving force)is the horinzontal component of the active force Pa,so:

ƩFsl=Pa cosα  .

 From the equations above results:

FS(sliding)=((ƩV)tanδ’+bc’a+Pp)/(Pacosα)

A minimum factor of safety of 1.3 against sliding is generally required.In many cases, the passive force Pp is ignored in calculating the factor of safety with respect to sliding.In general , we can write δ’=k1Ø’2 and  c’a=k2c’2 .In most cases k1 and k2 are in the range of 1/2to 2/3.Thus

FS(sliding)=((ƩV)tanδ’+bc’a+Pp)/(Pacosα+Pp)

If the desired value of FS(sliding)is not achieved ,several alternatives may be investigated: -increasing the width of the base slab(the heel of the footing); -use a key to the base slab ,if a key is included,the passive force per unit lengthof the wall becomes:

Pp=1/2y2Df2Kp+2c’2Df√Kp

-use a deadmen anchor at the stem of the retaining wall.

Posted on April 27, 2012 by retainingwalldesign | Leave a comment

Figure 8.6 shows the forces acting on a cantilever and a gravity retaining wall,based on the assumption that the Rankine active pressure is acting along a vertical plane AB drawn through the heel of the structure.Pp is the Rankine passive pressure ,recall that its magnitude is from the following equation:

Pp=1/2Kpy2D2+2c’√KpD                       where:

Y2-unit weight of the soil in front of the heel and under the base slab(kN/m3)

Kp-the Rankine passive earth pressure coefficient=tan2(45+Ø’2/2)

C2’-cohesion (kPa=kN/m2)

Ø’2-effective soil friction angle .

The overturning moment is:                     ƩMO=Pa(H’/3)                     where:

Ph=Pacosα.

Table 8.1-Procedure for Calculating ƩMR

Section

Area

Weight/unit length of wall

Momentum arm measured from C

Momentum about C

1

A1

W1=y1xA1

X1

M1

2

A2

W2=y2xA2

X2

M2

3

A3

W3=ycxA3

X3

M3

4

A4

W4=ycxA4

X4

M4

5

A5

W5=ycxA5

X5

M5

6

A6

W6=ycxA6

X6

M6

Pv

B

Mv

ƩV

ƩMR

where:  yi-unit weight of backfill

Yc-unit weight of concrete.

To calculate the resisting moment ,ƩMR(neglecting Pp),a table such as Table 8.1 should be prepared.The weight of the soil above the heel and the weight of the concrete (or masonry)are both forces that contribute to the resisting moment.Pv is the vertical component of the active force Pa ,or Pv=Pa sinα.

The moment of the force Pv,about c is

Mv=PvB=PasinαB  where B=width of the base slab.

Once ƩMR is known ,the factor of safety can be calculated as:

FS(overturning)=(M1+M2+M3+M4+M5+M6+Mv)/(Pacosα(H’/3)

The usual minimum desirable vealue of the factor of safety with respect to overturning is 2 to 3.

Posted on April 27, 2012 by retainingwalldesign | Leave a comment

Stability of Retaining Walls

A retaining wall could fail to ensure stability in any of the following ways:

-it may overturn about its toe(fig.8.4a)

-it may slide along its base(fig.8.4b)

-it may fail do to loss of the bearing capacity of the soil supporting the base(fig.8.4c)

-it may undergo deep- seated shear failure(fig.8.4d)

-it may go to excessive settlement (when a weak  soil layer is located at a shallow depth ,that is within a depth of 1.5 times the width of the base slab of the retaining wall the possibility of excessive settlement should be considered problem that is usually solved by using a lightweight backfill material behind the retaining wall.

 Deep shear failure can occur along a cylindrical surface ,such as abc shown in fig.8.5, as a result of the existanece of a weak layer of soil underneath the wall at a depth of about 1.5 times the width of the base slab .In such cases the cylindrical failure surface abc has to be determined by trial and error ,using various centers such as O.The failure surface along wich the minimum factor of safety is obtained is the critical surface of sliding.For the backfill slope width with α less than about 10⁰, the critical failure circle apparently passes through the edge of the heel slab (such as def curve defined in the drawing ).In this situation ,the minimum factor of safety also has to be determined by trial and error by changing the center of the trial circle.

Posted on April 27, 2012 by retainingwalldesign | Leave a comment