The document provides details to design the reinforcement for a basement retaining wall. It includes calculating the required wall thickness, loads on the wall, bending moments, shear forces, and reinforcement requirements. The summary is as follows: 1. The thickness of the basement retaining wall is determined to be 200mm based on the given height and material properties. 2. The loads on the wall, including soil pressure, water pressure, and surcharge loads are calculated. 3. The bending moment and shear force diagrams are drawn, with the maximum bending moment found to be 33.12 kNm and maximum shear force 65.76kN. 4. The required vertical and horizontal reinforcement is calculated for different sections based on

1. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 243
SECTION 18: BASEMENT RETAINING WALL DESIGN
1. Determine the thickness and necessary reinforcement for the basement retaining wall
shown in Figure below.
GIVEN:
• oncrete Compressive Strength: '
cf 25MPa=
• Steel Yield Strength: yf 390MPa=
• Unit Weight of Reinforced Concrete: c 3
kN
24
m
 =
• Height of Basement Wall: wh 3m=
• Soil Density (Backfill): s 3
kN
18
m
 =
• Water Density: w 3
kN
10
m
 =
• Traffic/Parking Load Surcharge: s 2
kN
w 2.4
m
=
• Internal Friction Angle of Soil: 30 = 
• Concrete/Clear Cover: cv 40mm=
• Vertical Rebar Diameter: s 12mm =
• Horizontal Rebar Diameter: h 12mm =
A
B
Ha
Ps
hw/2
SURCHARGE
Hs
hw/3
Pa
hw

2. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 244
SOLUTION
❖ Step 1: Given Data
• Concrete compressive strength: '
cf 25MPa=
• Steel Yield Strength: yf 390MPa=
• Unit Weight of Reinforced Concrete: c 3
kN
24
m
 =
• Height of Basement Wall: wh 3m=
• Soil Density (Backfill): s 3
kN
18
m
 =
• Water Density: w 3
kN
10
m
 =
• Traffic/Parking Load Surcharge: s 2
kN
w 2.4
m
=
• Internal Friction Angle of Soil: 30 = 
• Concrete/Clear Cover: cv 40mm=
❖ Step 2: Determine The Thickness of the Wall
• Thickness of Wall:
w
w
h 3000mm
t max ,100mm max ,190mm 190mm 200mm
25 25
   
= = = =  
  

3. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 245
❖ Step 3: Load on Basement Wall
• Earth Pressure & Concentrated Load:
a a soil w
a
a 3
P C h b
1 Sin 1 Sin30
C 0.333333
1 Sin 1 Sin30
kN kN
P 0.3333 18 3m 1m 18
mm
= 
−  − 
= = =
+  + 
=    =
a w
a
kN
18 3mP h mH 27kN
2 2

= = =
• Water Pressure & Concentrated Load (In Case Soil is wet: 50%):
w w w
w 3
w w
w
P 50% h b
kN kN
P 50% 10 3m 1m 15
mm
kN
15 3mP h mH 22.5kN
2 2
= 
=    =

= = =
• Effect of Surcharge Load & Concentrated Load:
2
s
s
s
3
s a s s 3
s s w
kN
2.4
w mh 0.13333m
kN
18
m
kN kN
P C h b 0.333333 18 0.133333m 1m 0.8
mm
kN
H P h 0.8 3m 2.4kN
m
= = =

=  =    =
=  =  =
A A
B B
A
B
RA
RB
M
u.positive
M
u.negative
Pw
Hw
Pa
Ha
Ps
hw
Hs
B
A
x

4. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 246
❖ Step 4: Calculate Bending Moment & Shear Force
( )
( )
w w
u.neg a w s
u.neg
h h
M 1.6 H H 1.6H
7.5 8
3m 3m
M 1.6 27kN 22.5kN 1.6 2.4kN 33.12kN.m
7.5 8
= + +
= + +   =
( )
( )
( ) ( )
( ) ( )
a w w s w
u.neg
B
w
B
A a w s B
u A B
H H h H h
1.6 M
3 2
R
h
27kN 22.5kN 3m 2.4kN 3m
1.6 33.12kN.m
3 2
R 17.28kN
3m
R 1.6 H H H R 1.6 27kN 22.5kN 2.4kN 17.28kN 65.76kN
V max R ,R max 17.28kN,65.76kN 65.76kN
 +
+ − 
 =
 +  
+ − 
 = =
= + + − = + + − =
= = =
( )
( )
( ) ( )
2a w
s B
w
2
3
2s a w
u.pos B
w
2
u.pos
P P
1.6 x 1.6P x R 0
2h
18kN / m 15kN / m
1.6 x 1.6 0.8kN / m x 17.28kN 0
2 3m
x 1.33045m
P P P x
M R x 1.6 x 1.6
2 2h 3
0.8kN / m
17.18kN 1.33045m 1.6 1.33045m
2
M
18kN / m 15k
1.6
 +
+ − = 
 
+ 
+  − =  
=
  +
= − − + +  
  
−  +
= −
+
+
( )3 14.94kN.m
1.33045mN / m
2 3m 3
 
 
  =
  
    

5. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 247
➢ Using CSI Etabs 2018, we get the results as following
Bending Moment Diagram Shear Force Diagram
➢ Using Robot Structural Analysis Professional 2020, we get the results as following
Surcharge Load: (Ps)
Water Pressure: (Pw)
Soil Pressure(Backfill): (Pa)

6. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 248
Bending Moment Diagram
Shear Force Diagram

7. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 249
❖ Step 5: Calculate Required Reinforcement
➢ Case: Negative Bending Moment u.negM 33.12kNm=
• Calculate b max min, , ,   
( )'
1 c
'
c
b 1
y y
y
5
s
max b
0.85 17MPa f 25MPa 28MPa
f 600 25MPa 600MPa
0.85 0.85 0.85 0.028
f 600 f 390MPa 600MPa 390MPa
f 3900.003 0.003
E 2 10 0.028 0.
0.008 0.008
 =  = 
     
 =  =   =       + +    
   + +     =  = = 
        
( )min s y
2
v.min min w
0174
0.0015 10mm 16mm and f 390MPa 420MPa
A bt 0.0015 1000mm 200mm 300mm
 =  =  = 
=  =   =
( )
s
6
u.neg
u 2 2
'
c u
'
y c
12mm
d h cv 200mm 40mm 154mm
2 2
M 33.12 10 Nmm
R 1.3965MPa
bd 1000mm 154mm
f 2R 25MPa 2 1.3965MPa
0.85 1 1 0.85 1 1 0.0041355
f 390MPa 0.85 25MPa 0.90.85f
0.0041

= − − = − − =

= = =

     
 = − − = − − =               
 =
( ) ( )
max
2
s
2 2 2
s.use s v.min
355 0.0174
A bd 0.0041355 1000mm 154mm 636.867mm
A max A ,A max 636.867mm ,300mm 636.867mm
  =
=  =   =
= = =

8. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 250
( )
( ) ( )
( )
2
s
s 2
s1
s
max w
use max
A 636.867mm
n 5.631 6
A 12mm
4
b 1000mm
s 166.66mm 160mm
n 6
s min 3t ,450mm min 3 200mm,450mm 450mm
s min s,s 160mm
= = = 

= = = =
= =  =
= =
➢ Case: Positive Bending Moment u.posM 14.94kN.m=
• Calculate b max min, , ,   
( )
max
min s y
2
v.min min w
0.0174
0.0015 10mm 16mm and f 390MPa 420MPa
A bt 0.0015 1000mm 200mm 300mm
 =
 =  =  = 
=  =   =
( )
s
6
u.neg
u 2 2
'
c u
'
y c
12mm
d h cv 200mm 20mm 174mm
2 2
M 14.94 10 Nmm
R 0.4934MPa
bd 1000mm 174mm
f 2R 25MPa 2 0.4934MPa
0.85 1 1 0.85 1 1 0.0014243
f 390MPa 0.85 25MPa 0.90.85f

= − − = − − =

= = =

     
 = − − = − − =               
( ) ( )
( )
( )
max
2
s
2 2 2
s.use s v.min
2
s.use
s 2
s1
s
max w
0.0014243 0.0174
A bd 0.0014243 1000mm 174mm 247.83mm
A max A ,A max 247.83mm ,300mm 300mm
A 300mm
n 2.6525 3
A 12mm
4
b 1000mm
s 333.33mm 330mm
n 3
s min 3t ,450mm min 3 200mm,
 =   =
=  =   =
= = =
= = = 

= = = =
= = ( )
( )use max
450mm 450mm
s min s,s 330mm
=
= =

9. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 251
❖ Step6: Determine Shear Force & Check Shear Capacity (Section Adequacy)
• Shear Force
uV 65.76kN=
• Check Shear Capacity
' 3
c c
c u
1 1
V f bd 0.75 1 25MPa 1000mm 155mm 10 96.875kN
6 6
V 96.875kN V 65.76kN (OK)
−
 =   =       =
 =  =
Wall Thickness is Sufficient to Resist Shear Force. Shear Reinforcement is not required.
❖ Step7: Determine the Minimum Distributed Horizontal Reinforcement
( )
( )
( ) ( )
( )
min s y
2
h.min min w
2
h.min
s 2
s1
s
max w
use max
0.0025 10mm 16mm and f 390MPa 420MPa
A bt 0.0025 1000mm 200mm 500mm
A 500mm
n 4.42 5
A 10mm
4
b 1000mm
s 200mm
n 5
s min 3t ,450mm min 3 200mm,450mm 450mm
s min s,s 200mm
 =  =  = 
=  =   =
= = = 

= = =
= =  =
= =

10. CIVIL ENGINEERING TRAINING CENTER (BIM-CETC) RC DESIGN
Prepared By: Mr. SENG PHEARAK (M.ENG, S.E) PAGE: 252
Summary Result
Direction Location Bending Moment Diameter of
Rebar
Spacing of
Rebar
Vertical
Exterior Face u.negM 33.12kNm= 12mm 160mm
Interior Face u.posM 14.94kN.m= 12mm 330mm
Horizontal Both Side - 12mm 200mm
1000
DB12@160
BASEMENT WALL REINFORCEMENT DETAIL
SCALE: 1/50
3000
DB12@200
DB12@160
DB12@330