Foundation EngineeringFoundation Engineering

Lecture #Lecture #2424

Reinforced Concrete WallsReinforced Concrete Walls

L. Prieto-Portar 2008

This New York City reinforced concrete retaining wall unexpectedly collapsed in May, 2005.

The Stability of Retaining Walls.

The preliminary design of most retaining walls is based on the experience of previously successful projects. Figure 1 and 1a shows the most common proportions used for rigid and semi-rigid reinforced concrete walls. Other types of walls, such as slurry walls, are developed for specific tasks that require project tailored dimensions.

Figure 1. Typical proportions of rigid retaining walls.retaining walls.

Figure 1a. Typical proportions of semi-rigid retaining walls.

Wall Planes of Action.

For simplicity, the lateral earth pressure upon a wall is assumed to act upon an imaginary vertical plane of action AB as shown in Figure 2. This model is used for both the Rankine and Coulomb analyses.

Figure 2. Simplified plane

of lateral pressure loading.

Note that the soil block Ws above the heel helps stabilize the wall. The assumption that the lateral pressure is indeed exerted along the plane AB is theoretically correct if the shear zone bounded by AC is not obstructed by the wall stem, i.e. that the angle � (CAB) is found from,

Most walls are designed by ignoring the presence of a hydrostatic head on the wall. This is accomplished with drainage systems, either behind the wall or below it. On the other hand, marine walls typically experience equal, or almost equal pressures, on both sides of the wall.

Types of Stability Analysis for walls.

)( 1 - 2

- 2

+ 45 = - φαφαηsin

sinsin°

Types of Stability Analysis for walls.

There are seven basic types of stability checks required for routing walls:

1) Overturning about the wall toe;2) Sliding failure along its base;3) Bearing capacity failure of the base;4) Settlement failure;5) Deep and shallow shear failures;6) Bottom heave in clays;7) Piping during dewatering.

1. Overturning Failure.

Consider Figure 3 for conditions for overturning:

Figure 3

The factor of safety against overturning (FS)O about the toe (point O) is expressed by,

(FS)O = Sum of resisting moments = � MR

Sum of overturning moments � MO

As an example, the resisting moments MR can be summarized on the table shown on the next slide, based on Figure 3b.

Note that the contribution of the passive force Fp (in front of the wall toe) is neglected; this choice essentially acts as an additional factor of safety.

On the other hand, the overturning moment MO is,

�MO = Fa cos � (H’ / 3)

Therefore, in order to attain a minimum factor of safety of 1.5 for temporary walls, or at least 2 for permanent walls, the following equation is generally used,

cos

i

i v1

O

a

+M M(FS = )

HF ( )

3α ′

�

Table: Procedure for the calculation of �MR

Weight/Unit Moment arm

length of measured Moment

Section Area Wall from C about C

(1) (2) (3) (4) (5)

1 A1 W1 = �1 x A1 X1 M12 A2 W2 = �2 x A2 X2 M22 A2 W2 = �2 x A2 X2 M23 A3 W2 = �c x A3 X3 M34 A4 W2 = �c x A4 X4 M45 A5 W2 = �c x A5 X5 M56 A6 W2 = �c x A6 X6 M6

PV______

B MV____

�V �MR

Note �1 = unit weight of backfill�c = unit weight of concrete

Sliding Failure Along the Wall Base.

The factor of safety against sliding failure (FS)s along the base is expressed by,

(FS)s = Sum of forces resisting sliding = � FR

Sum of forces driving the slide � FD

Figure 4.

The shear strength s of the soil beneath the wall base is s = c2 + �' tan �2 , and the resistive force R' along the wall base is,

R' = (shear strength) x (base area) + Fp = s B(1) + Fp

where Fp = (1/2) �2 D2 Kp + 2 c2D √Kp

Therefore, R' = Bc2 + �’ B tan�2 + Fp

The driving force on the wall is Fh from the active earth pressure, where Fh = Fa cos �.

Therefore, the factor of safety against sliding (FS)s is,

Typically, a minimum factor of safety of 1.5 is required for temporary walls. If this factor is not readily attained, a base key may be incorporated as shown in Figure 4. It is proportioned by changing D to D,' such that Fp increases.

tancos

p2 2S

a

+ B( V) + Bc F(FS =) F

φα

�

Bearing Capacity Failure

The pressure distribution beneath a wall base may be represented as the condition shown in Figure 5.

Figure 5.

The pressure qheel is the minimum and qtoe is the maximum; both can be easily found following an analysis similar to that of a footing design.

The resultant R is made up of two components: Fh and �V, such that their vector relation is,

Taking moments about the toe (point O),

Mnet = �MR - �MO (as determined previously)

cosaR = V + ( )P α�

The point of application of R is at a distance x such thatx = Mnet/�V

so that the eccentricity, e, of R with respect to the wall base centerline is,

e = B/2 - x

The pressure distribution q along the wall base is:

_ _

_

_

net

3

V Y V V eMq = + = + AI AI B(1) I

V e ( V) B/2 = +

1B 1B12

V =

eB(1 + 6 )

B

� � �

� �

�

max

min

.

BV

. =q. eB(1 + 6 )

BV

and =qe

B(1 - 6 )B

�

�

Note that, in order for qmin not to become negative, e > B/6. The soil capacity for carrying tension under the heel is usually very small or zero (eg. in granular soils). Therefore, if R falls at a distance greater than B/6 from the base centerline, it will be necessary to re-proportion the wall.

The ultimate bearing capacity qu of the soil below the wall base is determined by a procedure much like a strip shallow foundation. This qu is then compared with the maximum pressure qmax found at the toe, where

qu = 0.5�2B�N�d�i� + c Ncdcic + qqNqdqiq

Recall that B� = B - 2e is the "reduced" wall base, andd� = 1dc = 1 + 0.4 D/B�dq = 1 + 2tan�2 (1 - sin�2)2 D/B�ic = (1 - �/90)2 = iqi� = (1 - �/�)2

The angle � is the inclination of the load with respect to the vertical and is given by

� = tan-1((Pa cos�)/�V)

Therefore, the factor of safety against bearing capacity failure is given by the relation,

(FS)bc = qu/qmax � 3.

As in shallow foundations, the ultimate bearing capacity qu occurs at an approximate settlement � = B/10. For very narrow walls, a FS = 3 is far from adequate.

Settlement Failure.Settlement Failure.

A very flexible behavior V of the wall could induce ground settlement behind it. Figures 6 and 7 show an example of such a ground loss through sidewalk cracking and large cavities beneath pavements adjacent to the World Trade Center's perimeter wall. These settlements are called ground loss.

Preliminary estimator of possible ground settlement is found from figure 8, based on Peck's (1969), and figure 9, based on field data from Mana and Clough (1981), which relates maximum lateral yield (H)max with the maximum ground settlement (V)max, where

(V)max = 0.5 to 1.0(H)max

Shear Failure

A "shallow" shear failure (Figure 11) occurs directly under the wall heel when the resisting shear strength of the soil along the surface abc is equal to the driving stresses. This failure can usually be prevented by having a factor of safety against sliding greater than 1.5, ie. (FS)S ≥≥≥≥ 1.5 .

Figure 11.

A "deep" shear failure (Figure 12) takes place when there is a weak stratum (eg., weak clay) very proximate to the wall base. Several methods are used to analyze shear failure conditions. Most are based on the use of a computer analysis that permits an iterative location of points of rotation, to determine which surface has the smallest factor of safety.

Figure 12.

The surface that yields the smallest factor of safety is called the critical surface of sliding. In Figure 12, the backfill with � < 10°°°° yield critical surfaces that touch the wall heel.

Teng (1962) proposed a method to evaluate the factor of safety against deep shear failure; which is as follows,

1. Draw wall and soil profiles to scale, such as Figure 12;

2. Establish a trial location for the point O and draw a circle abcd with radius r1;

3. Determine the driving forces by dividing the area efgh into a number of slices 3. Determine the driving forces by dividing the area efgh into a number of slices (Figure 13).

4. Determine the area of each slice, and the soil or concrete weights W in each slice;

5. Draw a vertical line through the centroid of each slice and intersect the trial circle;

6. Join point O with each intersection;

7. Determine angle w that vertical line makes with radial r;

8. Calculate W sin(w) for each slice;

9. Determine the active force Fa on face df,Fa = 1/2 �1H'2 Ka

10. The total driving force FD is FD = � [Wsin(w)] + Fax/�

where x is the perpendicular distance between the point O and line of action of Fa.

12. Draw a vertical line through the centroid of each slice and intersect the trial curve;12. Draw a vertical line through the centroid of each slice and intersect the trial curve;

13. Join the point O with intersection at angle w1.

14. For each slice obtain W1 tan�2 cos(w1)

15. Calculate c2l1 + c3l2 + c2l3where l1 = arc (ab)

l2 = arc (bi)l3 = arc (id)

16. Find the maximum resisting force Fr along the failure surface byFR = � [W1 tan(�2) cos(w1)] + c2l1 + c3l2 + c2l4

17. Determine the factor of safety against deep shear failure for this trial circle,18. Repeat for other trial circles; the smallest (FS)DS governs, and should be at least equal to 1.5.

( )rxFwW

lclclcwWFS

aDS /)sin

costan)( 42231211

++++Σ= ϕ

equal to 1.5.

Figure 13.

Backfill Drainage

Permanent retaining walls are commonly used as bridge abutments, and in steep topography for grade divided highways, residential and high-rise dwellings, parking, etc. For many of these structures, a drained backfill is essential, especially if the wall was designed to retain a dry soil. The accidental saturation of the backfill would impose a sizable increase in the horizontal loading upon the wall. A fail proof system of drainage is very important, either through weepholes or geotextile backwall drainage systems (Figure 22).

Figure 22.

Weep-holes are holes of at least 0.1 m diameter, spaced every 2-4 m horizontally. In order to prevent weep-hole clogging, or loss of backfill fines, with a consequent settlement of the backfill, a filter blanket is placed between the weep-holes and the backfill.

When soil filters are used, two factors are used: (1) the backfill fines can not wash into the filter, and (2) an excessive hydrostatic pressure head is not created behind the filtered weep-hole due to the low permeability of the filter soil. These two conditions are met when, for

(1) D15(F)/D85(B) < 5 and15 85

(2) D15(F)/D15(B) > 4

Where D15(F) is the diameter of the particles of soil that could pass 15% in the filter fabric, and D85(B) is the diameter of the particles of soil in the backfill. Where F refers to filter soil and B to backfill soil. The D15 and D85 refer to the diameter of the particles of the soil that the percentage of the soil would pass through.

Example: A wall with a backfill, within the following values.

D85 = 0.25 mm D50 = 0.13 mm D15 = 0.04 mm

therefore, D15 (F) < 5D85 (B) = 5 x 0.25 = 1.25 mmand D15 (F) > 4D15 (B) = 4 x 0.04 = 0.16 mm

The range found is from 0.16 to 1.25 mm.

Joints in Retaining Walls

Depending on the construction procedure and the type of wall, a number of joints may be built. Construction joints are present both horizontally and vertically between two successive concrete pours. Keys between these separate concrete bodies, may be required to insure shear transfer, although roughened surfaces may be adequate (Figure 24).

Figure 24.

Contraction joints are vertical grooves placed on the excavation side of the wall to allow the concrete mass to shrink without cracking (Figure 25).

Expansion joints are separation placed between two adjacent concrete masses in orderto permit expansion, due to temperature or prefabrication, etc. The joint is filled witha flexible and impermeable material (bitumen, silicone, etc.). If structurally connectedwith steel dowels, these are sheathed with epoxy paint, grease or even cardboard, toallow expansion to take place (Figure 26).

Inverted T and L WallsGravity walls are typically designed with wall thicknesses of 30 % of height. Reinforced concrete walls are much thinner, and have supplanted the gravity walls from an economical point. A common shape of these reinforced concrete walls are the inverted L or T sections. Many of these reinforced concrete walls have been standardized in design manuals. Their elements are shown in the Figure 27. These walls are analyzed using a Rankine pressure from the backfill, such as the Hairsine method. Typical dimensions are shown.

Figure 27.

Example # 1:

Design a reinforced concrete wall 20 ft high to retain a backfill of sand with � = 114.5 pcf and � = 29°

Stem design

�ah = Ka �zcos

The moment at any depth z M = (1/6) �ah z2

The ultimate design moment

Mu = 1.7 M = (1.7/6) �ah z3

= 0.28 Ka �z3cos= 0.28 Ka �z cos

Plots of Mu vs As (area of steel) would give the required reinforcing:

also Mu = �Asfy(d-(a/2)) using � = 0.9 and As = 0.04a

∴ Mu = (0.9)(0.04a)(60,000 lb/in2)(d-(a/2))

Assume that d = stem - 3" (at any depth z)

2cS

y

0.85 ab f 0.85 a(1)(3000) = = = 0.04 a ( )ftA

60,000f′

*As< As(min) ∴ use As(min) for walls (ACI 14.3.2)As = As(min) = (0.0015)(gross wall area)

= (0.0015)(1 ft)(2 ft)= 0.003 ft2 = 0.43 in2

z (ft) stem (ft) d (ft) Mu (kft/ft) a (in) A (in2)

0 1.5 1.38 0 0 0

6 1.62 1.5 3.16 0.071 0.12

12 1.82 1.7 25 0.504 0.84

20 2.02 1.9 85.4 1.57 2.64

= 0.003 ft = 0.43 in

Could use 1 #6 bar every 6 in (0.88 in2 per foot of wall) for top 15 feet of wall, and 1 #11 bar every 6 in (3.12 in2 per foot of wall) for bottom 5 feet of wall, and since we will lap #11 with #11 with #6 use ACI 12.15

Lap distance = 1.7 Ld = 1.7(0.04 Abfy/√f'c)

∴ Lap distance = (1.7)(0.04)(1.56 in2)(60,000/√3000)= 116 in = 9.7 ft.

∴ Cutoff #11's at mid wall height.

Determine the length of development of main reinforcing bars into the foundations,

Ld = 0.04(Ab)(fy/√f'c)= 0.04(1.56 in2)(60,000 psi/√3,000 psi)= 68 in

Since 68" exceeds the thickness of the base slab (~2')So std. hooks may be used, (ACI 12.5.1)basic development length ldh = 1200 db/√f'c

= 1200 (1.41 in)/√3,000 = 31"= 1200 (1.41 in)/√3,000 = 31"

Check shear strength at base of wall,

Vu = 1.7V = 1.7((1/2)�H2Kacos�)

= 1.7(0.114 k/ft3)(20 ft)2(0.35)(0.99)

= 13.5 kips/ft

A shear key is provided between wall and base, such that the nominal stress in the concrete does not exceed 0.2f'c

Vu/Akey = 13.5 kips/(4" x 12") = 0.28 ksi < 0.2�fc'

< 0.2(0.85)(3) = 0.51 ksi O.K.

∴ Use a 2" x 4" shear key at the base of the stem.

Check temperature and shrinkage steel (ACI 14.3.3)

Horizontal steelHorizontal steel

As = (0.0025)(gross area of wall)

= (0.0025)(1 ft)(2 ft)(144 in2/ft2)

= 0.72 in2/ft of wall

Provide 2 #6 bars every foot of wall (As = 0.88 in2) in both directions (see figure).

Heel design

the load qq = qsoil + qconc - qo

= q1 + q2 - q3where

q1 = �H = (0.114)(20 ft)= 2.3 ksf

q2 = �ch = (0.150)(2)= 0.3 ksf

q3 = soil reaction= q + mx= qh + mx

from static

qheel = 0.96 ksf ∴ q3 = 0.96 + 0.286x

qtoe = 3.96 ksf

∴ m = 0.286 ∴ q = 2.3 + 0.3 -0 .96 - 0.286x

= 1.64 - 0.286x

∴ Shear V = �qdx = 1.64x - (0.286/2)x2 + c1but at x = 0 V = Pv = 1.9 kips/ft

∴ V = 1.64x - (0.286/2)x2 + 1.9

the moment M = �Vdx = (1.64/2)x2 - (0.286/6)x3 + 1.9X

Critical section for shear and moment x = 8.5 ft

∴ V = 1.64(8.5) - (0.286/2)(8.5)2 + 1.9∴ V = 1.64(8.5) - (0.286/2)(8.5)2 + 1.9= 5.5 k/ft

∴ M = 0.82(8.5)2 - (0.286/6)(8.5)3 + 1.9(8.5)= 46.1 k-ft/ft

Hence,Vu = 1.7V = 1.7(5.5)

= 94 k/ft

Mu = 1.7M = 1.7(46.1)= 78.4 k-ft/ft

Check for shear (f/member subject to shear and flexure), nom. concrete shear strength

Vc = 2√f'c bd =[2√3,000)(12")(2'-3"-(1.41/2)"]= 26 k/ft of heel

and �Vc = 0.85(26) = 23 k/ft > 9.4 k/ft

Check flexural reinforcement

78.4 = Mu = �Asfy(d-(a/s)) and As = 0.04a

∴ 78.4 = 0.9(0.04a)(60,000)(3.3-(a/2) → a = 1.81"

∴ As = 0.04a = 0.07 in2 < Asmin ∴ use min. 0.60 in 2

Use 1 #6 every 6" each way

Bridge Abutments

DESIGN OF SUBSTRUCTURE• Given

• f’c = 4,000 psi• fy = 60,000 psi

• Dead Load•• DL = 1.057• DL = 82 / 2 = 43.337•• Reaction of each beam: RDL = 43.337

• Live Load HS--20--44• Live Load HS--20--44•• Reaction of beam to live load: RDL = 19.02• Taking beam at center of two piles•• (wl2/8 + pl/4)• 1.3 ( RDL + (5/3)RLL) = 97.548• P = 97.548•• Use cap of 4’ * 3’ and backwall of 5.9*15’

• End Bent• (43 + 5.9 + 1.9) * 1.5 = 3.127• 1.3 * (3.127) = 4.065• w = 4.065

• DESIGN OF SUBSTRUCTURE (cont.)

• w (11.9)2 / 8 + P (11.9) / 4 = 362.161

• M = 362.161

• Ru = M(12000) / .9(48)(31)2

• M(12000) / .9(48)(31)2 = 104.683

• .85 (4/60) ( 1 – sqrt(1-( 2Ru / (.85 * 4,000))) = 1.772 x 10 -3

• = 1.773 x 10 -3

• According to ACI, • According to ACI,

• min = 200/fy = .033

• �s = min(49)(31) = 49.6

• a = (5)(60) / (.85)(4)(48) = 1.838

• d = 31

• Mu = �s fy ( d – a/2) = �s fy ( d – a/2) * 1/12 = 7.46 x 10 5

• Mu = 742

Check Crack Moment:

• fr = 7.5�fc = 474.342

• b = 48

• (b*u2)/6 = 1.037*104

• S = (b*u2)/6

• Mcr = 1.2*fr*S

• Mcr = 1.2*fr*S*(1/12000) = 491.797

Shear Design

• VV = 110 +(11.9*4.07) = 158.433

• Vc = 2(�fc)b*d

• 2(�fc)b*d*(1/1000) = 188.219

• Use minimum: #4 @ 12

Part 2: Design of backwall

• H of Backwall:• 8 + 54 + 2 + 4 + 1 = 69 in• H = 5.75 ft• Assume: • q = 250• ø = 30• r = 115• r = 115• k = 0.5• p = k*q*H = 718.75• Ps = (1/2)(k*r*H)*H = 950.547• Pt = Ps + p = 1.669*103

• M0 = 0.95*H*(1/3) + 0.72*H*(1/2) = 3.891• Mbll = (M0*H2)/8 = 16.08

Part 3: Capacity of piles

• Total load on beam:

• 13*RDL + 3.13*82 + RLL*13 = 1.067*103

• T = 13*RDL = 3.13*82 + RLL*13

• Each Pile:

• 1.1*(T/13) = 90.31

• So we can use 30”�

References.

- http://tc17.poly.edu/mse.htm- http://www.state.in.us/dot/div/contracts/standards/rsp/sep03rev/731r202.pdf