Rigid Retaining Walls

22.4 Rigid Retaining Walls

Design lateral active earth pressures for low retaining walls (i.e., height <6 m) are often estimated using conservative design charts [see Department of the Navy, 1982] or using equivalent fluid pressures. The

equivalent fluid unit weight, g eq , equals the product of the minimum active earth pressure coefficient

and the backfill material’s unit weight (i.e., g eq =K a · g), and and the backfill material’s unit weight (i.e., g eq =K a · g), and

90.0 b /f = +1

80.0 b /f = +.4 FOR VARIOUS RATIOS OF

10 .978 .962 .946 .929 .912 .898 .881 .864 b /f = +.2 15 .961 .934 .907 .881 .854 .830 .803 .775

30 .878 .811 .746 .686 .627 .574 .520 .467 b /f = 0 35 .836 .752 .674 .603 .536 .475 .417 .362

+b FAILURE

H −δ

90 b /f = 10.0 °− f LOGARITHMIC −. P N

ASSIVE PRESSURE, K 7.0 σ P =K P γ H b /f = −. 6

PASSIVE PRESSURE

5.0 P P =K P γ H 2 /2; P N =P P COS δ ;P V =P P SIN δ

4.0 NOTE:CURVES SHOWN ARE FOR δ /f = − 1

EXAMPLE:f = 25 ; b/f = .2

COEFFICIENT OF P d/f = − .3

b /f = −. 8 K P = R(K P FOR δ /f = − 1)

R = .711 2.0 (K P FOR δ /f = −

K P = .711x3.62= b /f = −. 9 2.58

1.0 A .9

P ASSIVE ZONE

b /f = − 1 .6

d = f, b/f =+ 1 .5 TIVE +b

AC ZONE

90 °− f FAILURE

d = f, b/f = +. 8 CTIVE PRESSURE, K

.3 H + δ P N

d = f, b/f = +. 6 P

SURFACE LOGARITHMIC

d = f, b/f = 0 ACTIVE PRESSURE

d = f, b/f =−. 4 P A =K A γ H 2 /2 P N =P A COS δ

d = f, b/f =−. 1 COEFFICIENT OF A

.1 P V =P A SIN δ

ANGLE OF INTERNAL FRICTION, f, DEGREES

FIGURE 22.4 Minimum active and maximum passive lateral earth pressure coefficients developed from log-spiral solution techniques. (After Department of the Navy. 1982. Foundations and Earth Structures: Design Manual 7.2. NAVFAC DM-7.2, May.)

(22.6) Conservative estimates of g eq for a variety of backfill materials are listed in Table 22.1 . All earth retention

p a = g eq ◊ z

structures should be designed to sustain potential surcharge loadings, and typically a minimum surcharge

3m SAND

19.6 kPa

c ′ = 0, f ′ = 30 ° 6m

ρ t = 2.0 Mg/m 3 P w

58.8 kPa

69.7 kPa CLAY

29.4 kPa

175.6 kPa (a) 12m

c = 24kPa, f =0

ρ t = 1.8 Mg/m

(b)

FIGURE 22.5 Example 22.1.

TABLE 22.1 Equivalent Fluid Unit Weights (kN/m3) for Design of Low Retaining Walls

Level Backfill 2H:IV Backfill Soil At-Rest Active Active

Clean sand or gravel 7.5 5 6 Silty sand 8.5 6 8 Clayey sand 9.5 7 9 Sandy clay 11 10 11 Fat clay

0.3 m to H 12 0.25 m to 0.4 m

Gravity Retaining Wall Cantilever Retaining Wall FIGURE 22.6 Tentative gravity and cantilever wall dimensions. load equivalent to an additional 0.6-m thickness of backfill is specified. Retaining walls should be

constructed with free-draining backfill materials and with effective drainage systems, because if water can pond behind the wall the additional water pressure will dramatically increase the load on the wall. If ponding cannot be precluded, the wall should be designed to resist the higher total pressures.

The general design procedure for gravity and cantilever retaining walls follows:

1. Characterize project site and subsurface conditions. Pay particular attention to groundwater and surface water, site geology, availability of free-draining backfill soils, and potentially weak seams.

2. Select tentative wall dimensions (see Fig. 22.6 ).

3. Estimate the forces acting on the retaining wall (i.e., active earth pressure, weight, surcharge, and resultant).

4. Check overturning stability; the resultant force should act within the middle third of the base of the wall.

5. Check bearing capacity; maximum earth pressure on the wall base should be less than the allowable earth pressure regarding bearing capacity or permissible settlement.

6. Check sliding; horizontal frictional resisting force on the base of the wall should be at least 1.5 times the horizontal driving force.

7. Check for excessive settlement from deeper soil deposits.

8. Check the overall stability of the earth mass that contains the retaining structure.

9. Apply load factors and compute reactions, shears, and moments in the wall.

10. Compute the ultimate strength of the structural components.

11. Check adequacy of structural components against applied factored forces and moments. Design procedures differ for retaining systems built of reinforced soil [see Mitchell and Villet, 1987].

Example 22.2 — Cantilever Retaining Wall

Check the adequacy of the cantilever retaining wall shown in Fig. 22.7 regarding overturning, bearing capacity, and sliding. The allowable bearing pressure is 360 kPa.

Solution. (a) The overall dimensions of the wall appear to be appropriate (see Fig. 22.6 ). (b) Estimate the forces acting on wall:

W 1 =r c gA = (2.4 Mg/m 3 )(9.81 m/s 2 )(0.25 m · 8 m) = 47.1 kN/m W 2 = (2.4 Mg/m 3 )(9.81 m/s 2 )(0.5 · 0.45 m · 8 m) = 42.4 kN/m W 3 = (2.4 Mg/m 3 )(9.81 m/s 2 )(0.6 m · 4.5 m) = 63.6 kN/m

W 4 =r t gA ª (1.9 Mg/m 3 )(9.81 m/s 2 )(8.5 m · 2.8 m) = 444 kN/m W 5 = (1.9 Mg/m 3 ) (9.81 m/s 2 ) (0.9 m · 1.0 m) = 16.8 kN/m

W T = Â W i = 614 kN/m Add 0.6 m of soil behind the wall to account for surcharge; hence, H = 0.6 m + 8 m + 1 m + 0.6 m =

10.2 m. Conservatively assume P p ª 0. Use the log-spiral solution for the sloping backfill to estimate K a

(see Fig. 22.4 ).

1m 0.7m

SAND c ′ =0 f ′ = 35 °

ρ t = 1.9 Mg/m 3

W 3 0.6m

FIGURE 22.7 Example 22.2.

b/f = 24°/35° @ 0.7 and f¢ = 35° with d = f¢ Æ K a = 0.38 P a =K a g H 2 /2 = (0.38)(1.9 Mg/m 3 )(9.81 m/s 2 )(10.2 m) 2 /2 = 368 kN/m

P ah =P a cos d = (368 kN/m) cos 35° = 301 kN/m P av =P a sin d = (368 kN/m) sin 35° = 211 kN/m

N =W T +P av = 614 kN/m + 211 kN/m = 825 kN/m T = N tan d b = (825 kN/m) tan 35° = 578 kN/m

Location of resultant, N:

 M A = 0 = ( 47.1 kN m § ) 1.575 m ( ) + ( 42.4 kN m § ) 1.3 m ( ) + 63.6 kN m ( § ) 2.25 m ( ) + ( 444 kN m § ) 3.1 m ( )

+ 16.8 kN m ( § ) 0.5 m ( ) + ( 211 kN m § ) 4.5 m ( ) – ( 301 kN m § ) 3.2 m ( ) 825 kN m – ( § ) () x

2.0 m

(c) Check overturning: B3 § =

4.5 m 3 § =

1.5 m 2B 3 § = ◊ 2 1.5 m = 3m

ce N acts within middle third of base (d) Check bearing capacity:

1.5 m <

2.0 m < 3 m OK, sin

e = B --- x – = ------------- 4.5 m –

2.0 m =

0.25 m

= N ---- 1 Ê + ----- 6e ˆ

= -------------------------- 1 § Ê + ------------------------ ◊ 6 0.25 m ˆ =

(e) Check sliding:

FS = ------- T = --------------------------- 578 kN m § =

P ah 301 kN m §

1.9 > 1.5 OK