Design of Concrete Cantilever Retaining Wall

Design of concrete cantilever retaining wall Introduction

Common failure of retaining wall: 1.

Overturning.

2.

Sliding.

3.

Bearing capacity.

4.

Bending or shear failure of stem.

5.

Bending or shear failure of heel.

6.

Bending or shear failure of toe.

7.

Bending or shear failure of key.

All items above should be considered in designing a retaining wall. There is also a rotational stability failure that is not normally checked except when a retaining wall is located on a slope.

Design procedure for cantilever retaining wall:

Stability analysis 1.

Check factor of safety against overturning.

2.

Check soil bearing pressure.

3.

Check factor of safety against sliding.

Reinforced concrete design 1.

Check thickness of stem for shear stress.

2.

Design stem reinforcement for bending.

3.

Check thickness of heel for shear stress.

4.

Design heel reinforcement.

5.

Check shear stress for toe when the toe is long.

6.

Design toe reinforcement for bending.

7.

Check shear stress in key when key is deep and narrow.

8.

Design key reinforcement for bending.

source: www.CE-Ref.com

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Stability analysis of retaining wall Check factor of safety against overturning. Check soil bearing pressure. Check factor of safety against sliding. Example 5.1 Cantilever retaining wall with horizontal backfill, no surcharge Example 5.2 Cantilever retaining wall with horizontal backfill, surcharge, and key Example 5.3 Cantilever retaining wall with slope backfill (Free excel file for wall stabilty, e-mail [email protected] ) Important Notes: 1. Lateral earth pressure is normally calculated based on Rankine or Coulomb’s theories. Lateral earth pressure is assumed distributed triangularly. The location of resultant is at 1/3 of height. If there is surcharge, lateral earth pressure from surcharge is distributed uniformly. The resultant is at ½ of height. The lateral earth pressure is calculated at the edge of heel. 2. The rotating point for overturning is normally assumed at bottom of toe. The height of soil used to calculate lateral earth pressure should be from top of earth to the bottom of footing. 3. Elements that resisting overturning are weight of stem, weight of footing, weight of soil above footing. If there is a surcharge, the weight of surcharge can also be considered. 4. The factor of safety against overturning is resisting moment divided by overturning moment. Acceptable factor of safety is between 1.5 to 2. 5. The driving force that causes retaining wall to slide is the lateral earth pressure from soil and surcharge. The forces that resist sliding are passive pressure at toe, the friction at the base of the footing; and the passive pressure against the key if used. 6. The factor of safety against sliding is the total resisting force divided by total driving force. Acceptable factor of safety is between 1.5 to 2.

Overturning: The factor of safety against overturning is calculated as 1.

The overturning moment is calculated as

Where

is unit weight of soil, Ka is active pressure coefficient, and H is the height from top of earth

to bottom of footing, q is surcharge.

2.

The resisting moment is calculated as


2


where Ws,Wf,We,Wk,Wq are weight of stem, footing, earth, key and surcharge, Xs,Xf,Xe,Xk,Xq are distance from the center of stem, footing, earth, key, and surcharge to the rotation point at toe.

3.

The factor of safety against over turning is calculated as

Bearing pressure: The bearing pressure is calculated as follows

1.

The center of the total weight from the edge of toe is

Where W is total weight of retaining wall including stem, footing, earth and surcharge.

2.

The eccentricity, e = B/2-X

3.

If e

B/6, the maximum and minimum footing pressure is calculated as

Where, Qmax, Qmin are maximum and minimum footing pressure, B is the width of footing. 4.

If e > B/6, Qmin is zero,

5.

Qmax should be less than allowable soil bearing capacity of footing soil.

Sliding: 1.

The driving force for sliding is calculated as

2.

The friction resisting force at the base of footing is calculated as

where

is friction coefficient between concrete and soil.

is often taken as tan (2/3

).

is internal

friction of the soil.


3


3.

The passive resistance at the toe of retaining wall is calculated as

Where Kp is passive earth pressure coefficient, h is the height from top of soil to bottom of footing at toe. If a key is used to help resist sliding, h is the height from top of soil to the bottom of the key.

4.

The factor of safety is calculated as

Example 5.1 Cantilever retaining wall with horizontal backfill, no surcharge Given: Height of stem: 10 ft Thickness of stem: 1 ft Thickness of footing: 1 ft Width of footing: B = 6 ft Length of heel: 4 ft Length of toe: 1 ft Height of soil above heel: 10 ft Height of soil above toe: 1 ft Unit weight of backfill soil:

= 115 pcf,

Internal friction angle of soil: 30 degree, Friction coefficient between concrete and soil: 0.5 Allowable soil bearing capacity for footing soil: 3000 psf

Requirement: Check stability against overturning and sliding, and soil bearing capacity. 1. Check overturning stability Rankine 's active earth coefficient: Ka = tan (45- /2)2=0.333 Height from top of backfill soil to bottom of footing: H = 10 +1 = 11 ft Consider one-foot width of soil Overturning moment: Mo=

3

Ka H /6 = 8504 ft-lb

Calculate Resisting moment: Weight of stem: Ws=150x10x1=1500 lbs Distance from center of stem to edge of toe: Xs=1.5 ft Weight of footing: Wf=150x6x1=900 lbs Distance from center of footing to edge of toe: Xf=3 ft Weight of earth: We=115x10x4=4600 lbs


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Distance from center of earth to edge of toe: Xe=4 ft Resisting moment: MR = 1500x1.5+900x3+4600x4 = 23350 ft-lb Factor of safety: FS = MR/Mo = 23350/8504 = 2.75

>1.5

O.K.

2. Check soil bearing capacity: Total weight of retaining wall: W=1500+900+4600 = 7000 lbs Distance from center of resultant to toe, X = (MR-Mo)/W = (23350-8504)/7000 = 2.12 ft Eccentricity: e = 6/2 – 2.12= 0.88 ft

< 1/6 width of footing, 1 ft

Maximum and minimum footing pressure:

< 3000 psf O.K.

3. Check sliding stability Driving force: Ph = =

2

Ka H /2 = 2319 lbs

Friction resistance at bottom of footing: Fr =

W = 0.5x7000 = 3500 lbs

Rankine's passive earth coefficient: Kp = tan (45+ /2)2=3 Height from top of backfill soil to toe: h = 2 ft Passive resistance at toe: Pp = =

2

Kp H /2 = 115x3x22 =690 lbs

Factor of safety against sliding: FS= (Fr+Pp)/Ph = (3500+690)/2319 = 1.81

>1.5 O.K.

Example 5.2 Cantilever retaining wall with horizontal backfill, surcharge, and key Given: Height of stem: 10 ft Thickness of stem: 1 ft Thickness of footing: 1 ft Width of footing: B = 6 ft Length of heel: 4 ft Length of toe: 1 ft Height of soil above heel: 10 ft Height of soil above toe: 1 ft Depth of key: 1 ft Unit weight of backfill soil: Internal friction angle of soil:

=115 pcf = 28 degree

Friction coefficient between concrete and soil:

= 0.5

Allowable soil bearing capacity for footing soil: 4000 psf Surcharge above backfill soil: q = 200 psf


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Requirement: Check stability against overturning and sliding, and check soil bearing capacity. 1. Check overturning stability Active earth coefficient: Ka = tan (45- /2)2=0.361 Height from top of backfill soil to bottom of footing: H = 11 ft Consider one foot width of soil: 3

Overturning moment: Mo=

Ka H /6 = 13580 ft-lb

Calculate Resisting moment: Weight of stem: Ws=150x10x1=1500 lbs Distance from center of stem to edge of toe: Xs=1.5 ft Weight of footing: Wf=150x6x1=900 lbs Distance from center of footing to edge of toe: Xf=3 ft Weight of earth: We=115x10x4=4600 lbs Distance from center of earth to edge of toe: Xe=4 ft Weight of key: Wk = 150x1x1=150 lbs Distance from center of key to edge of toe: Xk=2 ft Weight of surcharge: Wq = 200 x 4 = 800 lbs Distance from center of surcharge to edge of toe: Xq = 4 ft Resisting moment: MR = 1500x1.5+900x3+4600x4+150x2+800x4 = 26850 ft-lb Factor of safety: FS = MR/Mo = 268500/13580 = 1.98

>1.5

O.K.

2. Check soil bearing capacity: Total weight of retaining wall: W=1500+900+4600+150+800 = 7950 lbs Distance from center of resultant to toe: X = (26850-13580)/7950=1.67 ft Eccentricity: e = 6/2-1.67 = 1.33 ft

> 1/6 width of footing, 1 ft

Maximum footing pressure:

< 4000 psf O.K.


2

Ka H /2 + q Ka H= 115x0.333x112+ 200x0.333x11 = 3306 lbs


W = 0.5x7950 = 3975 lbs

Rankine's passive earth coefficient: Kp = tan (45+ /2)2=2.77 Height from top of backfill soil to bottom of key: h = 3 ft Passive resistance at toe: Pp = =

2

Kp H /2 = 115x2.77x32 =1433 lbs



>1.5 O.K.

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Example 5.3 Cantilever retaining wall with slope backfill Given: Height of stem: 10 ft Thickness of stem: 1 ft Thickness of footing: 1 ft Width of footing: B = 6 ft Length of heel: 4 ft Length of toe: 1 ft Height of soil above heel at retaining wall: 10 ft Height of soil above toe: 1 ft Unit weight of backfill soil: Slope of backfill soil:

=115 pcf

= 10 degree

Internal friction angle of soil:

= 30 degree

Friction coefficient between concrete and soil:

= 0.5

Allowable soil bearing capacity for footing soil: 3000 psf

Requirement: Check stability against overturning and sliding, and check soil bearing capacity. 1. Check overturning stability Rankine's active earth coefficient:

Height from top of backfill soil to bottom of footing: H = 10+1+4 tan

= 11.71 ft

Consider one foot width of wall, Overturning moment: Mo=

3

Ka H /6 = 115x0.35*11.713/6 = 10740 ft-lb

Calculate Resisting moment: Weight of stem: Ws=150x10x1=1500 lbs Distance from center of stem to edge of toe: Xs=1.5 ft Weight of footing: Wf=150x6x1=900 lbs Distance from center of footing to edge of toe: Xf=3 ft Divide the earth above heel to two portions: a rectangular portion and a triangular portion Weight of rectangular portion of earth: We1=115x10x4=4600 lbs Distance from center of rectangular portion of earth to edge of toe: Xe1=4 ft Weight of triangular portion of earth: We2=115x(4x4tan /2)=162 lbs Distance from center of earth of triangular to edge of toe: Xe2=1+1+4x2/3 = 4.67 ft Resisting moment: MR = 1500x1.5+900x3+4600x4+162x4.67 = 24110 ft-lb Factor of safety: FS = MR/Mo = 24110/10740 = 2.24

>1.5

O.K.

2. Check soil bearing capacity: Total weight of retaining wall: W=1500+900+4600+162 = 7162 lbs


7


Distance from center of resultant to toe: X = (24110-10740)/7162=1.87 ft Eccentricity: e = 6/2-1.13 = 1.13 ft

> 1/6 width of footing, 1 ft

Maximum footing pressure:

< 3000 psf O.K.


2 Ka H /2 = 115x0.35x1.732/2 =2754 lbs


W = 0.5x7162 = 3581 lbs

Rankine's passive earth coefficient:

Height from top of backfill soil to toe: h = 2 ft Passive resistance at toe: Pp = =

2

Kp H /2 = 115x2.78x22 =638 lbs



>1.5 O.K.

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Reinforced concrete design of retaining wall Contents •

Design forces and critical sections

•

Design of stem

•

Design horizontal reinforcement for shrinkage and expansion

•

Design of heel

•

Design of toe

Design forces and critical sections


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Design of stem Design thickness of stem

1. Calculate factored shear force at bottom of stem 2

Vu = 1.7*(γ Ka H /2+q Ka H)

[6.1]

Where 1.7 is load factor, γ is unit weight of soil, Ka is active lateral earth coefficient, h is height of earth, q is surcharge.

2. Calculate shear strength of stem φVc=0.85*(2√fc’) b d

[6.2]

Where 0.85 is strength reduction factor, fc’ is compressive strength of concrete, b is one foot width of wall, d is effective depth of stem and is equal to thickness of stem minus 2” cover and half bar size. 3. Compare shear force with shear strength, design shear reinforcement when necessary. If φVc≥ Vu no shear reinforcement is required If φVc< Vu increase thickness of stem or design shear reinforcement Design vertical reinforcement of stem

1. Calculated factored moment at base of stem 3

2

Mu=1.7*(γ Ka H /6+q Ka H /2)


[6.3]

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2. Design flexural reinforcement for stem Reinforcement ratio: [6.4] Where 2

R=Mu/(0.9bd ), m =Fy/(0.85fc’), Fy is yield strength of steel. The required reinforcement, As = ρbd should be within maximum reinforcement. The required minimum reinforcement is the smaller of

As,min=(3√fc’/Fy) or 1.33 As. if As is less than As,min

(ACI 10.5)

The minimum total vertical reinforcement ratio for wall (both faces) is 0.0012 for deformed bars #5 or smaller or 0.0015 for other bars

(ACI 14.3.2)

Design horizontal reinforcement for shrinkage and expansion

When retaining wall gets too long, the wall will crack due to shrinkage of concrete. Vertical control joints and horizontal reinforcement are normally used to control cracks in the stems. The spacing of control joist depends on the amount of horizontal reinforcement. Larger spacing requires heavier reinforcement. The reinforcement ratio recommended by Concrete Reinforcing Steel Institute (CRSI) is shown below.

1. Design horizontal reinforcement to avoid shrinkage cracks.

Figure 1: Joint spacing related to steel for shrinkage. (Reproduced from CRSI handbook)


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The minimum total horizontal reinforcement ratio for wall (both faces) is 0.002 for deformed bars #5 or smaller or 0.0025 for others.

2. Determine minimum width of expansion joints. In some case, when temperature change is large and the retaining wall has to be water tied, expansion joist are used. The width of expansion joint depends on temperature change and the length between joints. Without consider the contribution of horizontal reinforcement, the width of expansion joints can be calculated as

D=1.5*(0.0000065*T*L)

[6.6]

Where 0.0000065 is coefficient of expansion of concrete per degree F, T is maximum range of temperature difference, L is the length of wall between expansion joints, 1.5 is factor of safety.


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Design of heel Forces that apply to the heel are weight of soil, footing, surcharge, and footing bearing pressure. Weight of soil, footing, and surcharge are downward forces. Footing bearing pressure is upward forces. Sometime, footing bearing pressure are neglected to be conservative. Otherwise, factored footing pressures are calculated as follows: Calculated factored footing pressure

1.

The center of the total weight from the edge of toe is

[6.7]

Where W is total weight of retaining wall including stem, footing, earth and surcharge. 2.

The eccentricity, eu = B/2-Xu

3.

If eu ≤ B/6, the maximum and minimum footing pressure is calculated as

[6.8]

[6.9]

Where, Qmax, Qmin are maximum and minimum factored footing pressure, B is the width of footing. The factored footing pressure at any point in the footing is calculated as Q = Qmin + (Qmax-Qmin)*(B-L)/B Where B is the width of footing, L is the distance from toe

If eu > B/6, the maximum footing pressure is calculated as


13


[6.10]

The length of bearing area is Lb = 3*Xu The footing pressure at any point in the bearing zone is

Q = Qmax*(Lb-L)/Lb

[6.11]

L is the distance from toe

Design thickness of footing


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The critical section of shear in the heel is taken at the face of stem instead of at one-effective depth from the stem because it does not produce compression to the stem according to ACI code. 1. Calculated factored shear force at face of stem Vu = 1.4*(We +Whl+Wq)-R

[6.12]

Where 1.4 is load factor, We is weight of earth, Whl is weight of heel, Wq is weight of surcharge, and R is resultant of factored bearing pressure. 2. Calculated shear strength of stem. φVc=0.85*(2√fc’) b d

[6.13]

Where 0.85 is strength reduction factor, fc’ is compressive strength of concrete, b is one foot width of wall, d is effective depth of stem and is equal to thickness of stem minus 2” cover and half bar size. 3. Compare shear force with shear strength, if φVc< Vu, increase thickness of stem. Design heel reinforcement in transverse direction

The critical section of moment is at the face of stem. The heel reinforcement is calculated as follows: 1. Calculate factored moment at face of toe Mu=1.4*(We+Whl+Wq)*C/2-R*Xr

[6.14]

Where C is the length of heel, Xr is the distance from R to face of stem. 3. Design flexural reinforcement for heel Reinforcement ratio:

Where 2


As,min=(3√fc’/Fy) or 1.33 As if As is less than As,min

(ACI 10.5)

Design longitudinal reinforcement for shrinkage and temperature

Reinforcement ratio: 0.002 for grade 40, 50 deformed bars, 0.0018 for grade 60 deformed bars.


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Design of toe

The forces that apply to the bottom of toe is footing bearing pressure. In a normal situation, the length of toe is shorter than that of heel. The maximum shear force is less than of heel. The depth of footing for heel is usually enough for toe. It is also a normal practice to bend the dowel bars at the bottom of stem for toe reinforcement. It is normally sufficient for toe reinforcement. In some situation, when toe is extra long, then, it will be necessary to check shear strength and design reinforcement for toe.

Design thickness of toe

1. Calculate factored shear at one-effective depth from face of stem If eu ≤ B/6, the factored footing pressure at one-effective depth from face of stem is Q = Qmin + (Qmax-Qmin)*(B-Lc)/B

[6.15]

If eu > B/6, the factored footing pressure at one-effective depth from face of stem is Q = Qmax*(Lb-Lc)/Lb

[6.16]

Where Lc is the distance from edge of toe to one effective depth from front face of stem. The factored shear force at the critical section is

Vu = (Q + Qmax)*Lc/2-Wc

[6.17]

Where Lc is weight of concrete and soil above toe.


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2. Calculate shear strength of toe The shear strength of the concrete is φVc=0.85*(2√fc’) b d Design reinforcement for toe

1. Calculate factored moment at the front face of stem If eu ≤ B/6, the factored footing pressure at one-effective depth from face of stem is Q = Qmin + (Qmax-Qmin)*(B-Ld)/B If eu > B/6, the factored footing pressure at one-effective depth from face of stem is Q = Qmax*(Lb-Ld)/Lb Where Ld is the distance from edge of toe to front face of stem. The factored moment at the critical section is

Mu=R*Xr-Wt*Ld/2

[6.18]

Where Xr is the distance from the resultant force to the front face of stem, Wt is weight of concrete and soil above toe.

2. Design flexural reinforcement reinforcement ratio:

Where 2


As,min=(3√fc’/Fy) or 1.33 As if As is less than As,min


(ACI 10.5)

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Design of Concrete Cantilever Retaining Wall

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