Design of Toe Wall PDF
Design of Toe Wall PDF
Design of Toe Wall PDF
This report contains the load calculations and structural design of the
3m height Toe Wall at 76+46 RHS for Central Expressway Project
(CEP) (Section-02).
According to the site condition 3m height toe wall is proposed against the
1:1.5 slope embankment.
The structural designs were carried out for 3m wall with given dimensions
Material Characteristics
Loading
Dead loads
In this case, self-weight of the structure was considered as a dead load.
Density of concrete was considered as 25 kN/m3. Grade 30 concrete to be
used for the construction of the toe wall.
5.7E-05 0.300 m
b c
Embankment Slope
4.500 m = 1: ###
d 1.200 0.500 m
X 0.500 m
1.100 m 0.500 m
2.200 m 𝑇𝑎𝑛
0.300 m
Where β is slope angle of backfill surface, β = 0.00
Internal friction angle(∅') is taken as 36 0
Reinforced
Design
Structures
B.C.Punmia KA = 0.260
2
Surcharge Load = 10 kN/m
3
Soil Density = 20 kN/m
3
Concrete Density = 25 kN/m
Embankment Height = 0.0 m
Embankment Slope = 1: ###
2
Surcharge Load Due to embankment fill = kN/m
2
Soil Cohesion = 2 kN/m
Demension of Wall
Top width of Stem = 0.300 m
Bottom width of Stem = 0.500 m
Wall Height = 4.500 m
Heel Length = 1.200 m
Toe Length = 0.500 m
Foundation Thickness = 0.500 m
Base Width = 2.200 m
Shear Key height = 1.100 m
Shear Key thickness = 0.300 m
Vertical Forces
Soil 117.0
Concrete Wall 45.0
Concrete Base 27.5
Surcharge Load 14.0
Total 203.5
Moment SLS (kNm)
Stability
Soil 181.2
Soil (embankment) 0.0
Concrete Wall 31.7
Concrete Base 30.3
Surcharge Load 21.0
Total 264.1
Overturning
Soil Pressure 78.9
Surcharge Pressure 26.3
Total 105.1
Checking for overturning
Stability moment = 264.1
Over turning = 105.1
F.O.S = Stability moment / Over turning
OK
= 264.1 / 105.1
Reinforced Concrete = 2.5
Designer's Hand Book
Reynolds C.E. F.O.S > 1.5 ,Hence OK
Table 187
μ = tan (24)
= 0.44
e' = Ms – Mo
V
e' = 264.1 - 105.1
146.5
= 1.085 m
e= b/2 - e'
= 0.015 m
Bending moment, M = V× e
= 2.1975 kNm
σmax = σd + σb
= 69.3 kN/m2
σmin = σd - σb
= 63.9 kN/m2
BS 5400-4
= 2
1990 0.15x 25 x 1000 x 447
Equ. 2 = 749.3 kNm > 157.7
Hence section is Singly Reinforcement
5.057𝑀
𝑧 = 0.5𝑑 1 + 1 −
𝑓 𝑏𝑑
6
z = 0.5x 447 1+ 1 -5.057 x 157.7 x10^
2
25 x 1000 x 447
z = 428.4 mm
0.95d = 0.95 x 447
= 424.7 < 428.4
Therefore z = 424.7 mm
𝑀
Designed
Design of Retaining Wall Checked
Reference Calculation
𝑀 Output
Required reinforcement area 𝐴 = 0.87𝑓 𝑧
= 157.7 x 10^6
0.87 x 500 x 424.7
2
= 854 mm
𝐴 2
BS 5400-4 Hence , = 854 mm
2
1990 Provide T 16 @ 150 , 𝐴 , = 1340 mm T 16 @
Cl 5.8.9 150 mm
1/3 1/3
1990 𝑉 = 0.27 100 x 1340 25
Table 8 1.25 1000 x 447.0
2
= 0.423 N/mm Alow shear
stress
Allowable Shear Stress = 𝑉 x ξs = 0.435
2
= 0.423 x 1 N/mm
2
= 0.435 > 0.261 N/mm Hence OK
As*fy = f 'ct* Ac
-6
Fig. 14 kL = 275 x 10 (for 70% relative humidity)
Fig. 11 kC = 0.90 (for W/C ratio-0.5, cement content - 300)
Fig. 12 ke = 0.60 (for effective thickness 250)
kj = 1.00 (for long term)
-6
єsh = 275 x 10 x 0.90 x 0.60 x 1.00
-6
= 148.5 x 10
For distribution, φ = 12 mm
2
As = 564.02 mm < As,prov = 754
Hence satisfies the Shinkage & Temparature Reinforcement requirement
= 7.51 mm < 18 mm
Hence Deflection is OK
Eccentricity of the resultant force from the edge of the section (e')
e' = Ms – Mo
V
e' = 312.8 - 157.7
171.5
= 0.904 m
e= b/2 - e'
= 0.196 m
Designed
Design of Retaining Wall Checked
Reference Calculation Output
Bending moment, M = V× e
= 33.614 m
σmax = σd + σb
= 119.6 kN/m2
σmin = σd - σb
= 36.3 kN/m2
= 14.2 kNm
Dimensions
Width of Base, BB = 2200 mm
Thickness of Base, TB = 500 mm
BS 5400-4
= 2
1990 0.15x 25 x 1000 x 447
Equ. 2 = 749 kNm > 14.16
Hence section is Singly Reinforcement
Designed
Design of Retaining Wall Checked
Reference Calculation Output
5.057𝑀
𝑧 = 0.5𝑑 1 + 1 −
𝑓 𝑏𝑑
6
z = 0.5x 447 1+ 1 -5.057 x 14.164 x10^
2
25 x 1000 x 447
z = 445.4 mm
0.95d = 0.95 x 447
= 424.7 < 445.4
Therefore z = 424.7 mm
𝑀
Required reinforcement area 𝐴 = 0.87𝑓 𝑧
= 14.164 x 10^6
0.87 x 500 x 424.7
2
= 77 mm
BS 5400-4 Minimum reinforcement required = 0.15% bd
1990 = 0.15% x 1000 x 447
2
Cl 5.8.4.1 = 671 mm
𝐴 2
BS 5400-4 Hence , = 671 mm T 16 @
2
1990 Provide T 16 @ 150 , 𝐴, = 1340 mm 150 mm
BS 5400-4 Secondary Reinforcement
1990 Minimum secondary reinforcement requirement,
Cl 5.8.4.2 𝐴 = 0.12% bd
= 0.12% x 1000 x 447
2
BS 5400-4 = 536 mm
2
1990 Provide T 12 @ 150 , 𝐴, = 754 mm T 12 @
Cl 5.8.9 To prevent shinkage & temperature cracking 150mm spacing is provided 150 mm
BS 5400-4
1990 Check for Shear
2 2
Cl 5.4.4 Maximum Shear Stress at toe = 0.101 N/mm < 4.8 N/mm
= < 0.75 𝑓
2
Hence OK 3.75 N/mm
.
𝑉= / 𝑓 /
1/3 1/3
1990 𝑉 = 0.27 100 x 1340 25
Table 8 1.25 1000 x 447.0
2
= 0.423 N/mm
Dimensions
Width of Base, BB = 2200 mm
Thickness of Base, TB = 500 mm
BS 5400-4
= 2
1990 0.15x 25 x 1000 x 447
Equ. 2 = 749 kNm > 16.86
Hence section is Singly Reinforcement
5.057𝑀
𝑧 = 0.5𝑑 1 + 1 −
𝑓 𝑏𝑑
6
z = 0.5x 447 1+ 1 -5.057 x 16.86 x10^
2
25 x 1000 x 447
z = 445.1 mm
0.95d = 0.95 x 447
Designed
Design of Retaining Wall Checked
Reference Calculation Output
= 424.7 < 445.1
Therefore z = 424.7 mm
𝑀
Required reinforcement area 𝐴 = 0.87𝑓 𝑧
= 16.86 x 10^6
0.87 x 500 x 424.7
2
= 91 mm
BS 5400-4 Minimum reinforcement required = 0.15% bd
1990 = 0.15% x 1000 x 447
2
Cl 5.8.4.1 = 671 mm
𝐴 2
BS 5400-4 Hence , = 671 mm T 16 @
2
1990 Provide T 16 @ 150 , , = 𝐴 1340 mm 150 mm
BS 5400-4 Secondary Reinforcement
1990 Minimum secondary reinforcement requirement,
Cl 5.8.4.2 𝐴 = 0.12% bd
= 0.12% x 1000 x 447
2
BS 5400-4 = 536 mm
2
1990 Provide T 12 @ 150 , 𝐴, = 754 mm T 12 @
Cl 5.8.9 To prevent shinkage & temperature cracking 150mm spacing is provided 150 mm
BS 5400-4
1990 Check for Shear
2 2
Cl 5.4.4 Maximum Shear Stress at heal = 0.049 N/mm < 4.8 N/mm
= < 0.75 𝑓
2
Hence OK 3.75 N/mm
.
𝑉= / 𝑓 /
1/3 1/3
1990 𝑉 = 0.27 100 x 1340 25
Table 8 1.25 1000 x 447.0
2
= 0.423 N/mm
-6
Fig. 14 kL = 275 x 10 (for 70% relative humidity)
Fig. 11 kC = 0.90 (for W/C ratio-0.5, cement content - 300)
Fig. 12 ke = 0.60 (for effective thickness 250)
kj = 1.00 (for long term)
-6
єsh = 275 x 10 x 0.90 x 0.60 x 1.00
-6
= 148.5 x 10
2
Single face reinforcement for thermal cracking As = 78.34 φ mm
For main reinforcement, φ = 16 mm
2
As = 1253.4 mm < As,prov = 1340
Hence satisfies the Shinkage & Temparature Reinforcement requirement
For distribution, φ = 12 mm
2
As = 940.04 mm > As,prov = 754
Hence satisfies the Shinkage & Temparature Reinforcement requirement
Summary of reinforcement
Toe Wall