Basement wall design

Basement Wall Design Example by ACI Code

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: '
c
f 25MPa=
• Steel Yield Strength: y
f 390MPa=
• Unit Weight of Reinforced Concrete: c 3
kN
24
m
=
• Height of Basement Wall: w
h 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
P
s
h
w
/2
SURCHARGE
Hs
h
w
/3
P
a
h
w

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: '
c
f 25MPa=
• Steel Yield Strength: y
f 390MPa=
• Unit Weight of Reinforced Concrete: c 3
kN
24
m
=
• Height of Basement Wall: w
h 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
=
−  − 
= = =
+  + 
=    =
aw
a
kN
18 3m
Ph
m
H 27kN
22

= = =

• Water Pressure & Concentrated Load (In Case Soil is wet: 50%): w w w
w 3
ww
w
P 50% h b
kN kN
P 50% 10 3m 1m 15
mm
kN
15 3m
Ph
m
H 22.5kN
22
=
=    =

= = =

• Effect of Surcharge Load & Concentrated Load: 2
s
s
s
3
s a s s 3
s s w
kN
2.4
w
m
h 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
R
A
R
B
M
u.positive
M
u.negative
Pw
Hw
Pa
Ha
Ps
h
w
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
( )
( )
ww
u.neg a w s
u.neg
hh
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 sw
u.neg
B
w
B
A a w s B
u A B
H H h Hh
1.6 M
32
R
h
27kN 22.5kN 3m 2.4kN 3m
1.6 33.12kN.m
32
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
+
+−

=
 + 
+−

==
= + + − = + + − =
= = =
( )
( )
( ) ( )
2aw
sB
w
2
3
2s a w
u.pos B
w
2
u.pos
PP
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.neg
M 33.12kNm=
• Calculate b max min
, , ,    ( )
'
1c
'
c
b1
yy
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
390
0.003 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 22
'
cu
'
y c
12mm
d h cv 200mm 40mm 154mm
22
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
A636.867mm
n 5.631 6
A 12mm
4
b 1000mm
s 166.66mm 160mm
n6
s min 3t ,450mm min 3 200mm,450mm 450mm
s min s,s 160mm
= = = 

= = = =
= =  =
==

➢ Case: Positive Bending Moment u.pos
M 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 22
'
cu
'
y c
12mm
d h cv 200mm 20mm 174mm
22
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
n3
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 u
V 65.76kN=

• Check Shear Capacity '3
cc
cu
11
V f bd 0.75 1 25MPa 1000mm 155mm 10 96.875kN
66
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
n5
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.neg
M 33.12kNm= 12mm 160mm
Interior Face u.pos
M 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