Retaining Wall Design Calculation   Example Posted on April 29, 2012by 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 => OKPosted in Uncategorized | Tagged retaining wall design calculation example | Leave a comment

Check for Bearing Capacity of the retaining   wall 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     whereMnet-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)  andqmin= 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=y2DB’=B-2e

Fcd=1+0.4D/B’

FØd=1+2tanØ’2(1-sinØ’2)2D/B’Fyd=1Fqi=(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.Aside | Posted on April 28, 2012 by retainingwalldesign| Tagged check for bearing capacity of the retaining

wall | Leave a comment

Check for Sliding along   base 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’2and  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.

Aside | Posted on April 27, 2012 by retainingwalldesign| Tagged check for the retaining's wall sliding along

base | Leave a comment

Check for   Overturning 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

1A1 W1=y1xA1 X1 M1

2A2 W2=y2xA2 X2 M2

3A3 W3=ycxA3 X3 M3

4A4 W4=ycxA4 X4 M4

5A5 W5=ycxA5 X5 M5

6A6 W6=ycxA6 X6 M6

PvB

Mv

ƩVƩMR

where:  yi-unit weight of backfillYc-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.

Aside | Posted on April 27, 2012 by retainingwalldesign| Tagged check for overturning | Leave a comment

Stability of retaining   walls 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.

Aside | Posted on April 27, 2012 by retainingwalldesign| Tagged stability of retaining walls | Leave a

comment

 

Recent Posts Retaining Wall Design Calculation   Example

Check for Bearing Capacity of the retaining   wall

Check for Sliding along   base

Check for   Overturning

Stability of retaining   walls

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