Our previous article, Retaining Wall: A Design Approach discusses the principle and concept behind and when and where to consider a retaining wall in our design. We have learned the different checks against the mode of failures in the retaining wall should be considered in the design. To further understand the designed approach, here is a worked example of the design of retaining wall.
This example is intended to be readily calculated by hand through a lot of structural spreadsheets and software such as Prokon are available. The purpose of this article is for the reader to fully understand the principle behind it.
Worked Example:
Figure A.1Retaining Wall Cross Section
Consider the cantilever retaining wall with the cross section shown in the above Figure A.1, which retain a 2m depth of soil having the groundwater table at 1.0m level.
Design Parameters:
 Soil Bearing Capacity, q_{all }: 100 kPa
 Coefficient of Soil Friction, ф: 30°
 Unit Weight of Soil, ɣ_{s}: 18 kN/m^{3}
 Unit Weight of Water, ɣ_{w}: 10 kN/m^{3}
 Unit Weight of Concrete, ɣ_{c}: 25 kN/m^{3}
 Surcharge, ω: 12 kN/m^{2}
 Ground Water Level: 1m from 0.00 level
 Height of Surcharge: 0.8m
 Height of Wall: 2.0m
 f’c: 32 Mpa
 fy: 460 Mpa
 concrete cover: 75mm
1. Analytical Geometry and Variables
Before we proceed with the design, it is important for the designer to know the geometric variable and parameters of the retaining wall. Refer to Figure A.2 below.
Figure A.2Retaining Wall Geometric Variables
where:
 H: Height of the retaining wall
 L: Width of the base
 D: Thickness of the base
 B: Width of toe
 C: Stem thickness at the bottom
 T: Stem thickness at the top
2. Approximate Proportions of a Cantilever Retaining Wall
The next thing to consider is the assumptions that we can make in terms of the geometry of the retaining wall that we are designing. Given the height, H of the retaining wall, we can assume or counter check our initial design considerations should at least according to the following geometric proportions:
 Base width: L= 0.5H to 2/3H
 Thickness of base: D= 0.10H
 Stem thickness at the bottom: C=0.10H
 Width of the toe: B= 0.25L to 0.33L
 Stem thickness at the top: t=250mm (minimum)
Based on the above approximate geometric proportions, let us assumed the following parameters to be used in our design:
 Base width: L= 1.5m
 Base thickness: D= 0.25m
 Stem thickness: C=t =0.25m
 Width of Toe: B= 0.625m
3. Analytical Model
Sketches of the retaining wall forces should be considered to properly distinguish the different forces acting on our retaining wall as tackled in the previous article, Retaining Wall: A Design Approach. Based on our example in Figure A.1, we have the forces due to soil pressure, due to water and surcharge load to consider. Figure A.3 below is most likely our analytical model.
Figure A.3Retaining Wall Forces Diagram
Considering the Figure A.3, we can derive the following equation for the active pressures, Pa and passive pressure Pp. Notice that the pressures acting on the wall are equivalent to the area of the pressure distribution diagram. Hence,
 Pa_{1}=1/2 ɣkaH^{2 }→eq. 1
 Pa_{2}=1/2 ɣH_{w}^{2 }→eq.2
 Pa_{3}=ωkaH →eq.3
The passive pressure, Pp would be:
 Pp=1/2 ɣkpH_{p}^{2 }→eq.4
Values of Coefficient of Pressure, ka and kp
According to Rankine and Coulomb Formula, the following are the equation in calculating the coefficient of pressure:
Ka= (1sin ф)/(1+sin ф)
Ka= 0.33
Kp= (1+sin ф)/(1sin ф)
Kp= 3
Substituting the values, we have the following results:

 Pa_{1}=1/2 ɣkaH^{2 }= 11.88kN
 Pa_{2}=1/2 ɣH_{w}^{2}= 5kN
 Pa_{3}=ωkaH= 3.17 kN
 Pp=1/2 ɣkpH_{p}^{2}= 19.44kN
3. Stability Check:
There are two checks to consider the stability of the retaining wall. One is the check for an overturning moment and the other one is the check for sliding. The weight of the retaining wall including the gravity loads within it plays a vital role in performing the stability check. Refer to Figure A.4 for the mass or weight calculations.
Figure A.4Retaining Wall Weight Components
The selfweight component of the retaining wall should be factored down or to be multiplied by weight reduction factor (0.9) to account for uncertainty because they are “stabilizing” in this context. Hence,

 Weight due to soil: W_{1}= 18kN/m^{3 }x 0.6m x 0.625m x 1.0m = 6.75kN
 Weight due to footing: W_{2}= 0.9 x 25kN/m^{3 }x 0.25m x 1.5m x 1.0m = 8.44kN
 Weight due to wall: W_{3}= 0.9 x 25kN/m^{3 }x 0.25m x 2.0m x 1.0m = 11.25kN
 Weight due to soil: W_{4}= 18kN/m^{3 }x 0.625m x 2.0m x 1.0m = 22.5kN
 Weight due to water: W_{5}= 10kN/m^{3 }x 0.625m x 1.0m x 1.0m = 6.25kN
 Weight due to surcharge: W_{s}= 18kN/m^{3 }x 0.625m x 0.8m x 1.0m = 9kN
 Total Weight, W_{T }= 64.19kN
3.1 Check for Overturning Moment:
To satisfy the Overturning Moment Stability, the following equation should follow:
where:

 RM: Righting Moment due to the weight of the retaining wall
 OM: Overturning Moment due to lateral earth pressure
With reference to the Figure A.4 diagram and taking moment at the point, P conservatively neglecting the effect of passive pressure hence:
 RM= W_{1 }(0.313) + W_{2 }(0.75) + W_{3}(0.75) + W_{4}(1.19) + W_{5}(1.19) + W_{s}(1.19) = 61.80 kNm
 OM= Pa_{1} (0.67) +Pa_{2} (0.33) +Pa_{3} (0.4) = 10.88kNm
RM/OM = 5.68 > 2.0, hence SAFE in Overturning Moment!
3.2 Check for Sliding
To satisfy the stability against sliding, the following equation should govern:
where:

 RF: Resisting Force
 SF: Sliding Force
The sliding check should be carried out with reference to the Figure A.4 diagram and considering the summation of vertical forces for resisting force and horizontal forces for sliding force conservatively neglecting the passive pressure, hence:
 RF= W_{1}+W_{2}+W_{3}+W_{4}+W_{5}+ W_{s} = 64.19kN
 SF= Pa1+Pa_{2}+Pa_{3} = 16.70kN
RF/SF = 3.84 > 1.5, hence SAFE for Sliding!
4. Check the Wall Thickness for Shear
The nominal shear is equal to the lateral forces on the retaining wall, neglecting the effect of passive pressure which will give us:
 Nominal Shear, V_{n} = 20.05kN
 Ultimate Shear, V_{u} = 1.6Vn = 32.08kN
For the thickness of the wall to be safe in shear, the ultimate shear, V_{u} should less than the allowable shear, V_{allow }as recommended by the ACI 318 code.
_{}
V_{c =} 0.17√fc’b_{w}d
where: ф=0.75
b_{w}=1000mm
d= 250mm75mm6mm = 169mm
V_{c =} 0.17√fc’b_{w}d = 162.52kN
V_{allow}= 121.89kN
Since V_{u} < V_{allow}, hence SAFE in Shear!
5. Design the Wall Stem for Flexure
 Nominal Moment, M_{n} = 10.88kNm
 Ultimate Moment, M_{u} = 1.6Mn = 17.40kNm
Mu =φ fc’ bd^{2}ω (1 0.59 ω)
17.40×10^{6 }= 0.90 x 32 x 1000 x 169^{2} ω (10.59 ω)
ω = 0.0216
ρ = ω fc’/fy= 0.00150
As= ρbd = 0.00150x1000x169 = 254mm^{2}
As_{min}= ρ_{min}bt = 0.002 x 1000 x 250 = 500mm^{2}
Required Vertical Bar: Try T10200; As_{ act}= 392mm^{2 }x 2 sides = 785.4mm^{2}
Required Horizontal Bar: Try T10250; As_{ act}= 392mm^{2 }x 2 sides= 628.32mm^{2}
Hence: use T10200 for vertical bar and T10250 for horizontal bar.
6. Check for Bearing Pressure under Footing
The foundation bearing capacity usually governs the design of the wall. The soil, particularly under the toe of the foundation, is working very hard to resist the vertical bearing loads, sliding shear, and to provide passive resistance to sliding. The bearing capacity of the soil should be calculated taking into account the effect of simultaneous horizontal loads applied to the foundation from the soil pressure.
For the footing to be safe in soil pressure, the maximum soil pressure under working load shall be less than the allowable soil bearing capacity. The maximum soil bearing pressure under the footing considering 1m strip is:
_{}
where:

 P= 64.19kN
 A= (1×1.5) m^{2}
 M=10.88 kNm
 b= 1m
 d=1.5m
Substituting the values above will give us:
q_{max}= 71.81kPa < q_{all=}100 kPa, hence, oK!
Solving for Ultimate bearing pressure:
^{ }
where:

 P= 1.6x 6.75 + 1.4×8.44 +1.4×11.25 +1.6x 22.5 +1.6×6.25 +1.6×9= 98.76kN
 A= (1×1.5) m^{2}
 M=17.40kNm
 b= 1m
 d=1.5m
_{ }Substituting the values above will give us:
 q_{umax}= 112.24kN
 q_{umin}= 19.44kN
7. Check the Required Length of the Base
If q_{umin} is in tension check the required length otherwise ignore if it is in compression. Since our q_{umin }is tension (+), the value of L must be computed as follows:
Figure A.5Pressure Diagram under Tension
From Figure A.5:
Solve for Eccentricity:
e=M/P = 0.176
where:

 a=length of pressure
 qe= qu_{max}
 b=1 meter strip

 a= 1.75m
L= 2(e+a/3) = 1.52 say 1.6m
8. Check the Adequacy of Footing Thickness for Wide Beam Shear
Fig A.6Pressure Diagram under Compression
8.1 When qu_{min} is in Compression
Solving for y by similar triangle: referring to Figure A.6 above
y/1.044 = (112.2419.44)/1.5; y = 64.59 kPa
q_{c}= 19.44 + 64.59 = 84.03 kPa
 L’= (1.5m1.044m) = 0.456m
 B= 1m strip
 qu_{max}=112.24kPa
Vu= 44.75kN
8.2 When qu_{min} is in Tension
qc=y
Solving for y by similar triangle: (referring to Figure A.6 above, L=a=1.75)
y/1.244= 112.24/1.75; y = 79.79 kPa
qc=79.79kPa
Vu= 1/2 (q_{c }+ qu_{max}) L’b
 L’= (1.6m1.244m) = 0.356m
 B= 1m strip
 qu_{max}=112.24 kPa
Vu=34.18kN
Hence, use: Vu=44.75kN
V_{allow}= фV_{c}
V_{c =} 0.17√fc’b_{w}d
where:

 ф=0.75
 b_{w}=1000mm
 d= 250mm75mm6mm = 169mm
V_{c =} 0.17√fc’b_{w}d = 162.52kN
V_{allow}= 121.89kN
Since V_{u} < V_{allow}, hence SAFE in Shear!
9. Check the Wall Thickness for Flexure
Figure A.7Pressure Diagram for Flexure Check
9.1 When qu_{min} is in Compression
Solving for y by similar triangle:(referring to Figure A.7 above)
y/0.875 = (112.2419.44)/1.5; y = 54.13 kPa
q_{c}= 19.44 + 54.13 = 73.57 kPa
Mu= (73.57×0.625) x (0.625/2) + (38.67×0.625) (1/2) x (2/3) (0.625) → (area of trapezoid x lever arm)
Mu=19.40 kNm
9.2 When qu_{min} is in Tension
q_{c}= qu_{min} + y
Solving for y by similar triangle: (referring to Figure A.7 above. L=a=1.75)
y/1.075 = 112.24/1.75; y = 68.95 kPa
qc=19.44 + 68.95= 88.39 kPa
Mu = (88.39×0.75) x (0.75/2) + (23.85×0.75) (1/2) x (2/3) (0.75) → (area of trapezoid x lever arm)
Mu= 19.40kNm
Hence, use Mu=29.33kNm
Mu =φ fc’ bd^{2}ω (1 0.59 ω)
29.33×10^{6 }= 0.90 x 32 x 1000 x 169^{2} ω (10.59 ω)
ω = 0.0364
ρ = ω fc/fy= 0.002532
As= ρbd = 0.002532x1000x169 = 428mm^{2}
As_{min}= ρ_{min}bt = 0.002 x 1000 x 250 = 500mm^{2}
Required Vertical Bar: Try T10200; As_{ act}= 392mm^{2 }x 2 sides = 785.4mm^{2}
Required Horizontal Bar: Try T10250; As_{ act}= 392mm^{2 }x 2 sides= 628.32mm^{2}
10. Reinforcement Details of Retaining Wall
The presented calculations above are actually too tiring to perform manually especially if you are doing a trial and error design. Thanks to structural design soft wares and spreadsheets, available nowadays, our design life will be easier.
Our team developed a userfriendly spreadsheet for the design of cantilever retaining wall based on the above calculation. Grab your copy here!
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