17-retaining Wall Design 6f4r3u

15. Retaining Walls Design 15.1 Introduction Conventional retaining walls can generally be classified into four varieties: 1. Gravity retaining walls 2. Semigravity retaining walls 3. Cantilever retaining walls 4. Counterfort retaining walls

Braja M. Das (2007). Principle of Foundation Engineering, sixth edition.

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Jack C. McCormac, James K. Nelson (2006). Design of Reinforced Concrete ACI 318-05 Code Edition, seventh edition.

15.2 Proportioning Retaining Walls Page-255-

In deg retaining walls, an engineer must assume some of their dimensions. Called proportioning, such assumptions allow the engineer to check trial sections of the walls for stability. In figure below shows the general proportions of various retaining wall components that can be used for initial checks.

Bowles,J. E. (1977).

Jack C. McCormac, James K. Nelson (2006). Design of Reinforced Concrete ACI 318-05 Code Edition, seventh edition.

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Braja M. Das (2007). Principle of Foundation Engineering, sixth edition. 1.9-3 Stability of Retaining Walls

A retaining wall may fail in any of the following ways: - It may oveturn about its toe. - It may slide along its base. - It may fail due to the loss of bearing capacity of the soil ing the base. - It may undergo deep-seated shear failure. - It may go through excessive settlement.

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Failure of retaining wall: (a) by overturning; (b) by sliding; (c) by bearing capacity failure; (d) by deep-seated shear failure. Braja M. Das (2007). Principle of Foundation Engineering, sixth edition. 1. Checiking for Overturning

The Rankine active pressure Pa =

1 2 ⋅ γ1⋅ H ⋅ Ka 2

Ka =

1 − sinϕ1 1 + sinϕ1

The horizontal active force Ph = Pa⋅ cos ( α) The vertical active force Pv = Pa⋅ sin ( α) The Rankine ive pressure 1 2 Pp = ⋅ Kp⋅ γ2⋅ D + 2⋅ c2⋅ Kp⋅ D 2

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2 ϕ2 ⎞ ⎛ Kp = tan ⎜ 45deg + ⎟ 2 ⎠ ⎝

Braja M. Das (2007). Principle of Foundation Engineering, sixth edition.

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Procedure for Calculating ΣMR Section

Area

Weight/unit length of wall

Moment arm measured from C

1

A1

w1 = γ1⋅ A1

x1

M1

2

A2

w2 = γ2⋅ A2

x2

M2

3

A3

w3 = γc⋅ A3

x3

M3

4

A4

w4 = γc⋅ A4

x4

M4

5

A5

w5 = γc⋅ A5

x5

M5

6

A6

w6 = γc⋅ A6

x6

M6

Pv

B

Mv

ΣV

Moment about C

ΣMR

Note: γ1 = unit weight of backfill γc = unit weight of concrete = 24

kN 3

m

The factor of safety against overturning can be calculated as

The usaul minimum desirable value or the factor of safety with respect to overturning is 2 to 3 FSoverturing =

ΣMR

M0

=

M1 + M2 + M3 + M4 + M5 + M6 + Mv H Pa⋅ cos ( α) ⋅ ⎛⎜ ⎞⎟ ⎝3⎠

2. Checking for Sliding along the Base The factor of safety against sliding can be calculated as

FSsliding = where

ΣFR ΣF d

=

ΣV⋅ tan ( δ') + B⋅ c'a + Pp

Pa⋅ cos ( α)

ΣFR = sum of the horizontal resisting forces ΣF d

= sum of the horizontal driving forces

δ'

= angle of friction between the soil and the base slab

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c'a

= adhesion between the soil and the base slab

A minimum factor of safety of 1.5 against sliding is generally required.

Check for sliding along the Braja M. Das base (2007). Principle of Foundation Engineering, sixth edition. In many cased, the ive force Pp is ignored in calculating the factor of safety with respect to sliding. In general, we can write δ' = k1⋅ ϕ2 and c'a = k2⋅ c2. In most cases, k1a 1 2 nd k2 are in range from to . Thus, 2 3 FSsliding =

(

)

ΣV⋅ tan k1⋅ ϕ2 + B⋅ k2⋅ c2 + Pp Pa⋅ cos ( α)

If the desired value of FSsliding is not achieved, several alternatives may be investigated: 1. Increase the width of the base slab (i.e., the heel of the footing). 2. Use a key to the base slab. If a key is include, the ive force per unit length of the wall becomes Pp =

1 2 ⋅ γ2⋅ D1 ⋅ Kp + 2c2⋅ D1⋅ Kp 2

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Alternatives for increasing the factor of safety with respect to sliding Braja M. Das (2007). Principle of Foundation Engineering, sixth edition. 3. Checking for Bearing Capacity Failure

The net moment of these forces about point C Mnet = ΣMR − ΣM0 The distance from point C to E X=

Mnet ΣV

The ecectricity of the resultant R e=

B −X 2

The pressure distribution under the base slab may be determined by using simple principle from the mechanics of materials. First, we have q=

ΣV Mnet⋅ y ±⋅ A I

where Mnet = moment = ( ΣV) ⋅ e I

= moment of inertia per unit length of the base section =

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1 2 ⋅ ( 1m) ⋅ B 2

Check for bearing capacity failure Braja M. Das (2007). Principle of Foundation Engineering, sixth edition. ( ΣV) ⋅ e⋅

B 2

ΣV ±⋅ B⋅ ( 1m) 1 3 ⋅ B 12 The maximum and minimum pressure

q=

( )

6⋅ e ⎞ ΣV ⎛ qmax = qtoe = ⋅ ⎜1 + ⎟ B ⎝ B ⎠

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6⋅ e ⎞ ΣV ⎛ qmin = qheel = ⋅ ⎜1 − ⎟ B ⎝ B ⎠ The ultimate bearing capacity of a shallow foundation 1 qu = c2⋅ Nc⋅ Fcd⋅ Fci + q⋅ Nq⋅ Fqd⋅ Fqi + ⋅ γ2⋅ B'⋅ Nγ⋅ Fγd⋅ Fγi 2 where q = γ2⋅ D

B' = B − 2⋅ e

(

Fcd = 1 + 0.4⋅

D

) 2 B'

Fqd = 1 + 2⋅ tanϕ2⋅ 1 − sinϕ2 ⋅ 2 ψ⎞ ⎛ Fci = Fqi = ⎜ 1 − ⎟Fγi = ⎝ 90 ⎠

2 ⎛1 − ψ ⎞ ⎜ ϕ ⎟ 2⎠ ⎝

D B'

F γd = 1 ψ = tan

P ⋅ cosα ⎞ − 1⎛ a

⎜ ⎟ ⎝ ΣV ⎠

The factor of safety against bearing capacity failure can be calculated

FSbearing =

qu qmax

Generally, a factor of safety of 3 is required. 15.4 Effect of Surcharge

Differnt type of loads are often imposed on the surface of the backfill behind a retaining wall. If the load is uniform, an equivalent height of soil, hs may be assumed acting on the wall to for the increased pressure. For the wall shown in Figure below, the horizontal pressure due to the surcharge is constant throughout the depth of the retaining wall.

ws

H

Pp

Ph2 Ph1

D

H/3

D/3

surcharge effect under a uniform load ws hs = w where

hs

= equivalent height of soil

ws

= pressure of the surcharge

w

= unit weight of soil Page-264-

The total pressure is

⎞ ⎛ H2 Ph = Ph1 + Ph2 = Ka⋅ w⋅ ⎜ + H⋅ hs⎟ ⎝ 2 ⎠ In the case of a partial uniform load acting at a distance from the wall, only a portion of the total surcharge pressure affects the wall as in Figure below. ws 45o H

Pp

D

h’

Ph2

Ph1

H/3

D/3

surcharge effect under a partial uniform load at a distance from the wall It is common practice to assume that the effective height of pressure due to partial surcharge is h' measured from point B to the base of retaining wall. The line AB form an angle of 45deg with horizontal. 15.5 Design Requirements

The ACI Code provides methods for bearing wall design. The main requirements are as folllow 1 the ed height or lenght, 25 whichever is shorter, but not less than 100mm. 1. The minimum thickness of bearing wall is

1 tmin = ⋅h 25 wall

where hwall = the height of wall

2. The minimum area of the horzontal reinforcement in the wall is 0.0025⋅ b⋅ h, but this value may be reduced to 0.0020b⋅ h if diameter 16mm or smaller deformed bars with fy ≥ 400MPa are used. Ah.min =

0.0025⋅ b⋅ h if diameter > 16mm for fy ≥ 400MPa 0.0020⋅ b⋅ h if diameter ≤ 16mm

3. The minimum area of the vertical reinforcement is 0.0015⋅ b⋅ h, but it may be reduced to 0.0012⋅ b⋅ h if diameter 16mm or smaller deformed bars with fy ≥ 400MPa are used. Av.min =

0.0015⋅ b⋅ h if diameter > 16mm for fy ≥ 400MPa 0.0012⋅ b⋅ h if diameter ≤ 16mm

4. The maximum spacing of the vertical or the horzontal reinforcing bars is the smaller of 450mm or three times the wall thickness.

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smax = min ( 3t , 450mm)

where t = the wall thickness

5. If the wall thickness exceed 250mm, the vertical and horzontal reinforcement should be placed in two layers parallel to the exterior and interior wall surface, as follows: a/ For exterior wall surface, at leat 0.5 of the reinforcement As ( but not more than 2 A ) should have a minimum concrete cover of 50mm but not more than 1/3 of 3 s the wall thickness. 2 1 As.exterior = As and covermin = 50mm ≤ hwall 3 3 b/ For interior wall surface, the balance of the reinforcement in each direction should have a minimum concrete cover of 25mm but not more than 1/3 of the wall thickness. 1 covermin = 25mm ≤ hwall 3 c/ The minimum steel area in the wall footing (heel or toe) according to ACI Code is that required for shrinkage and temperature reinforcement, which is 0.0018⋅ b⋅ h when fy = 400MPa and 0.002⋅ b⋅ h when fy = 350MPa or fy = 275MPa.Becau se this minimum steel area is relately small, it is a common practice to increase it to that minimumn As.min required for flexure: As.shrinkage =

0.0018⋅ b⋅ h if fy = 400MPa 0.002⋅ b⋅ h if fy = 350MPa ∨ fy = 275MPa

As.min =

0.25⋅ f'c 1.4 ⋅ b⋅ d ≥ ⋅ b⋅ d fy fy

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Example15.1

b

ws

γ1 ϕ1

configuration of retaining wall

D

H1

c1

γ2 ϕ2 c2

H2 B1

B2

B3

B

Geometry of wall:

B1 := 1m

base slab width

B2 := 0.6m

B3 := 1.9m

B := B1 + B2 + B3 = 3.5 m stem width of upper wall

b := 0.3m

wall height

H1 := 5.4m

Soil properties:

backfill soil

soil in font

H2 := 0.6m

H := H1 + H2 = 6 m γ1 := 18

kN

ϕ1 := 35deg

3

m

γ2 := 19.5

kN 3

m

ϕ2 := 20deg

kN allowable soil bearing capacity qa := 190 2 m depth of soil in font

D := 0.9m Page-267-

kN c2 := 20 2 m

Material:

concrete compressive strength

f'c := 25MPa

steel yield strength

fy := 400MPa

unit weight of concrete

γc := 24

surcharge uniform load

ws := 12

kN 3

m

kN 2

m

Solution

1. Checking of wall against overturning a/ the moment resisting

⎡⎢ B3⋅ H1 ⎤⎥ ⎢ b⋅ H ⎥ 1 ⎢ ⎥ A := ⎢ ( B2 − b) ⋅ H1 ⎥ ⎢ ⎥ 2 ⎢ ⎥ ⎢ B⋅ H 2 ⎥ ⎣ ⎦

⎛ 10.26 ⎞ ⎜ ⎟ 1.62 ⎟ m2 A=⎜ ⎜ 0.81 ⎟ ⎜ ⎟ ⎝ 2.1 ⎠

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ws

A1

A2

Ph1

Ph2 A3 Pp

H 3

A4 qimn

qmax ΣR

x 0.5B

CL e

forces acting on the wall

⎛ γ1 ⎞ ⎜ ⎟ ⎜ γc ⎟ γ := ⎜ ⎟ ⎜ γc ⎟ ⎜γ ⎟ ⎝ c⎠ ORIGIN := 1

Wi := Ai⋅ γi

⎛ 18 ⎞ ⎜ ⎟ 24 kN γ = ⎜ ⎟⋅ ⎜ 24 ⎟ m3 ⎜ ⎟ ⎝ 24 ⎠ i := 1 .. rows ( γ)

⎛ 184.68 ⎞ ⎜ ⎟ 38.88 ⎟ ⋅ kN W=⎜ ⎜ 19.44 ⎟ m ⎜ ⎟ ⎝ 50.4 ⎠

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H 2

B ⎤ ⎡ ⎢ B− 3 ⎥ 2 ⎢ ⎥ ⎢ b ⎥ ⎢ B1 + B2 − ⎥ 2 ⎥ X := ⎢ ⎢ B2 − b) ⎥ ( ⎢ B1 + ⎥ 2 ⎢ ⎥ ⎢ ⎥ B ⎢ ⎥ 2 ⎣ ⎦

⎛ 2.55 ⎞ ⎜ ⎟ 1.45 ⎟m X=⎜ ⎜ 1.15 ⎟ ⎜ ⎟ ⎝ 1.75 ⎠ ⎛ 470.934 ⎞ ⎜ ⎟ 56.376 ⎟ ⋅ kN⋅ m M=⎜ ⎜ 22.356 ⎟ 1m ⎜ ⎟ ⎝ 88.2 ⎠

Mi := Wi⋅ Xi

ΣR :=

∑

ΣMR :=

W = 293.4⋅

∑

kN m

M = 637.866⋅

kN⋅ m 1m

Section No. Area (m2) Unit weight (kN/m) Moment arm (m) Moment(kNm/m) 1 2 3 4

10.260 1.620 0.810 2.100

184.680 38.880 19.440 50.400 293.400

2.550 1.450 1.150 1.750

470.934 56.376 22.356 88.200 637.866

b/ the unfactor force acting on the wall using Rankine equation Ka := Ph1 :=

( ) ( )

1 − sin ϕ1 = 0.271 1 + sin ϕ1

ws hs := = 0.667 m γ1

1 kN 2 ⋅ Ka⋅ γ1⋅ H = 87.801⋅ 2 m

kN Ph2 := Ka⋅ γ1⋅ hs⋅ H = 19.511⋅ m c/ the overturning moment H H kN⋅ m Mo := Ph1⋅ ⎛⎜ ⎞⎟ + Ph2⋅ ⎛⎜ ⎞⎟ = 234.135⋅ 1m ⎝3⎠ ⎝2⎠

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the safety of factor against overturning FSoverturning := wall :=

ΣMR

Mo

= 2.724

"is not overturning" if FSoverturning ≥ 2 "is overturning" otherwise

wall = "is not overturning" 2. Checking of wall against sliding Rankine ive force per unit length let

2 k1 := = 0.667 3

k2 := k1 = 0.667

2 ϕ2 ⎞ ⎛ Kp := tan ⎜ 45deg + ⎟ = 2.04 2 ⎠ ⎝

Pp :=

1 kN 2 ⋅ Kp⋅ γ2⋅ D + 2c2⋅ Kp⋅ D = 67.521⋅ 2 m

kN Ph := Ph1 + Ph2 = 107.312⋅ m the safety of factor against sliding

FSsliding := wall :=

(

)

ΣR⋅ tan k1⋅ ϕ2 + B⋅ k2⋅ c2 + Pp = 1.712 Ph

"is not sliding along the base" if FSoverturning ≥ 1.5 "is sliding along the base" otherwise

wall = "is not sliding along the base" 3. Checking of wall against bearing capacity failure a/ the eccentricity of the resultant x := e :=

ΣMR − Mo ΣR

= 1.376 m

B − x = 0.374 m 2 Page-271-

footing :=

"is not upward" if e ≤

B 6

"is upward" otherwise footing = "is not upward" b/ the maximum and minimum pressure ΣR ⎛ 6⋅ e ⎞ kN qmax := ⋅ ⎜1 + ⎟ = 137.569⋅ 2 B ⎝ B ⎠ m

6⋅ e ⎞ kN ΣR ⎛ qmin := ⋅ ⎜1 − ⎟ = 30.088⋅ 2 B ⎝ B ⎠ m wall :=

"is not failure" if qmax ≤ qa "is failure" otherwise

wall = "is not failure" 4. Design of stem a/ main reinforcement The lateral forces applied to the wall are calculate using a load factor of 1.6. The critical section for bending moment is at the bottom of the wall. Calculate the applied ultimate forces: kN Pu1 := 1.6⋅ Ph1 = 140.481⋅ m kN Pu2 := 1.6⋅ Ph2 = 31.218⋅ m H1 H1 ⎞ ⎛ kN⋅ m Mu1 := 1.6⋅ ⎜ Pu1⋅ + Pu2⋅ ⎟ = 539.448⋅ 1m 3 2 ⎠ ⎝ h := B2 = 0.6 m R :=

Mu1 2

d1 := h − ⎛⎜ 30mm +

⎝

20mm ⎞ ⎟ = 560⋅ mm 2 ⎠

= 1.911⋅ MPa

0.9⋅ d1

f'c ⎛ ρ := 0.85⋅ ⋅ ⎜ 1 − fy

⎝

1−

2⋅ R ⎞ = 0.00501 0.85⋅ f'c ⎟

⎠

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f'c ⎛ ⎞ ⎜ 0.25MPa⋅ ⎟ MPa 1.4MPa ⎟ ⎜ ρmin := max , = 0.0035 ⎜ fy ⎟ fy ⎝ ⎠ 2

mm As1 := max ρ , ρmin ⋅ d1 = 2.808⋅ mm

(

)

2

π⋅ ( 20mm) 2 As0 := = 314.159⋅ mm 4

⎛ As0 ⎞ s1 := Floor ⎜ , 10mm⎟ = 110⋅ mm ⎝ As1 ⎠ Because the moment decrease along the height of the wall, the reinforcement area may be reduced according to the moment requirement. It is practical to use one As1 or spacing s1 for the lower half and a second As2 or spacing s2 for the upper half of the wall. To calculate the moment at midheight of wall from the top: H1 ⎞ ⎛ Pu1 H1 kN⋅ m Mu2 := 1.6⋅ ⎜ ⋅ + Pu2⋅ ⎟ = 337.155⋅ 1m 2 ⎠ ⎝ 2 3 h := R :=

B2 + b

= 450⋅ mm d2 := h − ⎛⎜ 30mm +

⎝

2 Mu2 2

20mm ⎞ ⎟ = 410⋅ mm 2 ⎠

= 2.229⋅ MPa

0.9⋅ d2 ρ := 0.85⋅

f'c ⎛ ⋅ 1− fy ⎜

⎝

1−

2⋅ R ⎞ = 0.0059 0.85⋅ f'c ⎟

⎠

2

mm As2 := max ρ , ρmin ⋅ d2 = 2.419⋅ mm

(

)

⎛ As0 ⎞ s2 := Floor ⎜ , 10mm⎟ = 120⋅ mm ⎝ As2 ⎠

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b/ Temperature and shrinkage reinforcement the minimum horizontal reinforcement at the base of wall is: 2

cm As.min := 0.002⋅ B2 = 12⋅ 1m for the bottom third

H1 3

= 1.8⋅ m

As.min_0.33 := 0.002⋅

for the upper two-thirds

B2⋅ 2

2H1 3

3

2

= 8⋅

cm 1m

= 3.6 m

because the front face of the wall is mostly exposed to temperture changes, use half to two-thirds of the horizontal bars at the external face of wall and place the balance at the internal face. 2

cm As.tem := 0.5⋅ As.min = 6⋅ 1m

use diameter of shrinkage bar ds := 14mm

⎞ ⎛ π⋅ ds2 ⎜ ⎟ 4 ⎜ ⎟ sshrinkage := Floor , 10mm = 250⋅ mm ⎜ As.tem ⎟ ⎝ ⎠ c/ Design for shear the critical section for shear is at a distance d := d1 = 560⋅ mm from the bottom of the stem, at this section the distance for the top equal Hd := H1 − d = 4.84 m 1 kN 2 Hu := 1.6⋅ ⎛⎜ ⋅ Ka⋅ γ1⋅ Hd + Ka⋅ γ1⋅ hs⋅ Hd⎞⎟ = 116.595⋅ 1m ⎝2 ⎠ ϕVc := 0.75⋅ 0.17MPa⋅

stem_wall :=

f'c kN ⋅ d = 357⋅ MPa 1m

"no need shear reinforcement" if ϕVc ≥ Hu "need required shear reinforcement" otherwise

stem_wall = "no need shear reinforcement"

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5/ Design of the heel kN Vu := 1.2⋅ B3⋅ H1⋅ γ1 + B3⋅ H2⋅ γc + 1.6⋅ B3⋅ hs⋅ γ1 = 290.928⋅ 1m

(

)

d := h − ⎛⎜ 50mm −

h := H2 = 600⋅ mm

⎝

ϕVc := 0.75⋅ 0.17MPa⋅

heel :=

(

)

25mm ⎞ ⎟ = 562.5⋅ mm 2 ⎠

f'c kN ⋅ d = 358.594⋅ MPa 1m

"no need shear reinforcement" if ϕVc ≥ Vu "need required shear reinforcement" otherwise

heel = "no need shear reinforcement" H1 kN⋅ m Mu := Vu⋅ = 785.506⋅ 1m 2 R :=

Mu 2

= 2.758⋅ MPa

0.9⋅ d

ρ := 0.85⋅

f'c ⎛ ⋅ 1− fy ⎜

1−

⎝

2R ⎞ = 0.00741 0.85⋅ f'c ⎟

⎠

2

mm As := max ρ , ρmin ⋅ d = 4.17⋅ mm

(

⎡ ⎢ sheel := Floor ⎢ ⎣

)

π ⋅ ( 25mm) 4

As

2

⎤ ⎥ , 10mm⎥ = 110⋅ mm ⎦ 2

mm As.dis := 0.0018⋅ d = 1.012⋅ mm

⎞ ⎛ π⋅ ds2 ⎜ ⎟ ⎜ 4 , 10mm⎟ = 150⋅ mm sdis := Floor ⎜ As.dis ⎟ ⎝ ⎠

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6. Design of the toe The toe of base acts as a cantiveler beam subject to upward pressure. The critical section for the bending moment is at the front face of the stem. The critical for shear is at a distance d from the front face of stem. The toe is subject to an upward pressure from the soil and downward pressure due to self weight of the toe slab.

critical section for shear in toe

q1

qmin

q2 B1 − d

d

detail stress of soil under the base kN q1 := qmax = 137.569⋅ 2 m

q2 :=

(

)

q1⋅ B − B1 − d kN = 76.154⋅ 2 B m

⎛ q1 + q2 ⎞ kN ⎟ ⋅ ( B1 − d) − 1.2⋅ ( H1⋅ γc) ⋅ ( B1 − d) = 6.763⋅ 1m ⎝ 2 ⎠

Vu := 1.6⋅ ⎜ toe :=

"no need shear reinforcement" if ϕVc ≥ Vu "need required shear reinforcement" otherwise

toe = "no need shear reinforcement"

⎡⎢ B 2 2B1 ⎥⎤ B1 1 Mu := 1.6⋅ ⎢q2⋅ + ( q1 − q2) ⋅ B1⋅ − 1.2 H ⋅ γ ⋅ B − d ⋅ ( 1 c) ( 1 ) 2 3 ⎥⎦ 2 ⎣ kN⋅ m Mu = 92.412 1m

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R :=

Mu 2

= 0.325⋅ MPa

0.9⋅ d

ρ := 0.85⋅

f'c ⎛ ⋅ 1− fy ⎜

1−

⎝

2R ⎞ = 0.00082 0.85⋅ f'c ⎟

⎠

2

mm As := max ρ , ρmin ⋅ d = 1.969⋅ mm

(

⎡ ⎢ stoe := Floor ⎢ ⎣

)

π ⋅ ( 20mm) 4

As

2

⎤ ⎥ , 10mm⎥ = 150⋅ mm ⎦ upper stem wall steel bar

distibution steel bar

shrinkage steel bar

stem base wall steel bar heel steel bar

toe steel bar steel bar sketch for retaining wall

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