Retainning Wall

7/30/2019 Retainning Wall

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Retaining Wall

Retaining walls are structures used to retain earth which would not be able to stand verticallyunsupported.

Different types of retaining wall

are

Free standing (gravity wall)

Reinforced cantilever wall

Counterforts

Which of the three types of wall is

appropriate in a given case dependson variety of conditions, such as

local availability and price of

construction materials and propertyrights.

In general gravity walls areeconomical only relatively low

walls. Possibly up to 10 ft.

Cantilever wall are economicalfor heights from 10 to 20 ft.

Counterforts are used for graterheights.

Figure - 1 Types of retaining walls and

back drains: (a) gravity walls; (b)

Cantilever walls; (c) Counterfort

3-d view of Counterfort


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Main use:

At bridge abutment (figure-2)

Retaining to retain vertical cut surface(figure -3)

At water channel or swimming poll (figure-4)

At basement (figure-5)

Figure -3

Figure -2

Figure -4

Figure -5


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Earth Pressure for common load conditions

Figure-6 Earth pressure for (a) horizontal surface; (b) Sloping Surface; (c)

horizontal surface with surcharge s.

If the ground water level is above the base of the wall, additional water pressure is

to be considered.

Possible Failure mode

Failure mode Remedy

Individual parts may fail becausethey are not so strong to resist theacting forces

By providing necessary dimensions,thickness, and reinforcement toresist the moments and shares

The wall as a whole may be bodilydisplaced by the earth pressure,without breaking up initially

By ensuring the external stability

KahwhKahwh

Kahw(h+h)

(a) (b) (c)


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Design of cantilever retaining wall

Different parts of cantilever retaining wall

External force acting on the wall

Heel

ToeBase

Steam (sometime

may be tapered)

Figure -7; Different parts of cantilever retaining wall

Heel

ToeBase

Steam (sometime

may be tapered)

Figure -7; Different parts of cantilever retaining wall

Figure -8; External force acting on the wall

1

23

4

5

6

1. Earth pressure on the steam2. Weight of soil above heel3. Weight of base and steam4. Friction between base and soil5. Passive earth pressure on the face of toe6. Soil pressure on the base


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For external stability

Sliding must be prevented by the friction between the soil and footing i.e.

the summation of4 and 5 must be greater than force 1

Compression must act through out the base i.e. resultant of1, 2, and 3

must be pass through the center third of the base

Pressure under the footing does not exceed the permissible bearing pressure

For giving the required strength

Tension

Tension

Tension

Figure 9; Tension zone and placement of rebar

Rebar


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Dimensions Estimations

Example 1 [ McCormac 13.2]

Using the approximate rules presented in this section,estimate the sizes of the parts of the retaining wall

shown in Figure 11. The soil weighs 100 lb/ft3, and a

surcharge of 300 psf is present. Assume Kah= 0.32. (Formany practical soils such as clays or silts, kahwill be twoor more times this large.)

(8" absolute minimum)

(12" preferable minimum)Minimum batter

0.25 in/ft

Greater than frost penetration

and depth for which there is

a seasonal change in volume

0.07h to 0.1h

(10 in. to 12 in. min.)

b/3

0.07h to

0.12h

b = 0.4h to 0.6h

h

Figure-10, Economical Dimensions

Figure-11


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Solution:

Stem ThicknessAssume 12-in. thickness at top.

Assume bottom thickness = 0.07h = (0.07)(21) = 1.47'

Base Thickness

Assume base t = 7% to 10% of overall wall height:

t = (0.07)(21) = 1.47'

Height of stem = 21 0" minus 16" = 19'6"

Base Length and Position of StemCalculating horizontal forces without load factors, as shown in Figure 12.

Pa = kahwh = (0.32)(100)(21) = 672 lb

H1 = (0.5)(21)(672) = 7056 lb

H2 = (21)(96) = 2016 lb

W = (x)(24)(l00) = 2400x

Ma = 0-(7056)(7.00) - (2016)(10.5) + (2400x)(x/2) = 0

x = 7.67'

b = (3/2)(7.67) = 11.505'

Say 1-6

Say 1-6

Say 11-6

Figure-12


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Example 2[Mccormac 13.3]

Complete the design of the cantilever retaining wall whose dimensions were estimated inExample 1 and are shown in Figure 12, if = 3000 psi, = 60,000 psi, = 4000 psf,and the coefficient of sliding friction equals 0.50 for concrete on soil. Use p approximatelyequal to 0.18/ to maintain reasonable deflection control.

Solution:The safety factors against overturning and sliding and the soil pressures under the heel and

toe are computed using the actual unfactored loads.

Safety Factor against Overturning (with Reference to Figure 13)

Overturning Moment

Force Moment arm Moment

H1 = (

)(21)(672) = 7056 lb X 7.00 =49,392 ft-lb

H2 = (21)(96) = 2016lb X 10.50 =21,168 ft-lb

Total 70,560 ft-lb

Figure-13


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Rightining Moment

Force Moment arm Moment

W1 = (1.5)(11.5)(150) = 2588lb X 5.75 =14,881 ft-lb

W2 = ()(l9.5)( )(150) = 731lb X 4.08 =2982 ft-lb

W3 = (19.5)()(l50)

= 2925 lb X 4.75 =13,894 ft-lb

W4 = (22.5)(6.25)(100) = 14,062 lb X 8.37 =117,699 ft-Ib*

Rv = 20,306 lb M = 149,456 ft-Ib

*including surcharge

Safety factor against overturning = = 2.12 > 2Factor of Safety against Sliding

Force causing sliding = H1 + H2 + 9072 lb

Resisting force = Rv = (0.50)(20,306) = 10,153 lb

Safety factor = = 1.12 < 1.5little wider footing on the heel side will easily take care of the situation. In addition to or

instead of this solution a key, perhaps 1 ft-6 in. X 1 ft-6 in.

Footing Soil Pressures

Rv = 20,306 lb and is located a distance from the toe of the footing = = = 3.89

Soil pressure = - A= (1)(11.5) = 11.5 ft2

I = (1)(11.5)3 = 126.74 ft4ftoe =

= - 1766 - 1714 = -3480 psf

fheel = - 1766 + 1714 = -52 psf

Design of Stem

The lateral forces applied to the stem are calculated using a load factor of 1.6 as shown in

Figure 14.

OK

N.G.

Just inside the middle third


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Design of Stem for Moment

Mu = (H1)(6.50) + (H2)(9.75) = (9734)(6.50) + (2995)(9.75)Mu = 92,472 ft-lb

= approximately 0.18/ = = 0.009

= 482.6 psi [from Graph A.1b]

h = 14.59 + 2" + = 17.09"

Figure-14

Say 18(d= 15.50)


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= 0.00786 [from Graph A.1b]As = (0.00786)(12)(15.5) = 1.46 in.2

Minimum vertical by ACI Section 14.3 = 0.0015Minimum horizontal Ag = (0.0025)(12)(average stem t)

( ) (say one-third inside face and two-thirds outside face).

Checking Shear Stress in Stem

Actually, Vu at a distance d from the top of the footing can be used, but for simplicity.

Vu= H1+ H2= 9734 + 2995 = 12,729 lb

= 15,281 lb > 12,729 lb

Design of HeelThe upward soil pressure is conservatively neglected, and a load factor of 1.2 is used for

calculating the shear and moment because soil and concrete make up the load.

Vu= (22.5)(6.25)(100)(1.2) + (1.5)(6.25)(150)(1.2) = 18,563 lb

Try 24-in. Depth (d = 20.5 in.)

Neglecting slight change in Vu with different depth

Use # 8 @ 6 (1.57 in.2)

OK

Use # 4 @ 7.5out side face and 15 in side

OK

No Good

OK


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Muat face of stem = lb-ft

Using 0.00333,

Ax= (0.00333)(12)(20.5) = 0.82 in.2/ft

ldrequired calculated with ACI Equation 12-1 for #8 top bars with c = 2.50 in. and Ktr= 0is 43 in. < 72 in. available.

Note: Temperature and shrinkage steel is normally considered unnecessary in the heel and toe.

However, the author has placed #4 bars @ 8 in. on center in the long direction, as shown in Figures15 and 16 to serve as spacers for the flexural steel and to form mats out of the reinforcing.

Design of Toe

Vu= 10,440 + 7086 = 17,526 lb

Muat face of stem =

lb-ft

Therefore, use

As= (0.00333)(12)(20.5) = 0.82 in. 2/ft

ldrequired calculated with ACI Equation 12-1 for #8 bottom bars with c = 2.50 in. and Ktr=

0 equals 33 in. < 42 in. available.

Use =0.0033

Use # 8 @ 11 c/c

OK

OK

Use # 8 @ 11 c/c


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Figure-16Figure-15 Heel reinforcement

Figure-16 Toe reinforcement


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Details of Vertical reinforcement

Distance

from the top

of stem

MU (lb-ft)

Effective

Stem depth d

(in.)

As required

(in.

2

/f)

Bars

needed5 2987 11.04 Use min = 0.0033 0.44 #8 @ 18"10 16,213 12.28 Use min = 0.0033 0.50 #8 @ 18"15 46,080 14.12 0.00452 0.77 #8 @ 12

19.5 92,472 15.50 0.00786 1.46 #8 @ 6"

Contraction and Expansion joint

To handling shrinkage problems and differential settlements contraction joints are useful.They need to be spaced at intervals about 25 ft on center (the AASHTO says not greater

than 30 ft).

Expansion joints ate vertical joints that completely separate the different parts of a wall.

They are placed approximately 50 to 100 ft on centers (the AASHTO says maximum spacing

should not be greater than 90 ft).

Drainage

Weep Holes consisting of 6 or 8 in. pipe embedded in the wall, as shown as infigure 1 (c), are usually spaced horizontally about 5 to 10 ft. In addition to bottomrow, additional rows should be provided in walls of substantial height. To facilitate

drainage and prevent clogging, 1 ft3 or more of crushed stone is placed at the rear

end of weeper. Longitudinal drains figure 1 (b)

Continuous back drain figure 1 (a)

Why back sands and gravels are superior to all other soils as a backfill material

Sands and gravels are fee-draining

Not susceptible to frost action

Do not become less stable with the passage of time

Retainning Wall

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