Introduction
Basement wall is
constructed to retain the earth and to prevent moisture from seeping into the
building. Since the basement wall is supported by the mat foundation, the
stability is ensured and the design of the basement wall is limited to the safe
design of vertical stem.
Basement walls are
exterior walls of underground structures (tunnels and other earth sheltered
buildings), or retaining walls, which must resist lateral earth pressure as
well as additional pressure due to other type of loading. Basement walls carry
lateral earth pressure generally as vertical slabs supported by floor framing
at the basement level and upper floor level. The axial forces in the floor
structures are, in turn, either resisted by shear walls or balanced by the
lateral earth pressure coming from the opposite side of the building.
Although basement
walls act as vertical slabs supported by the horizontal floor framing, keep in
mind that during the early construction stage when the upper floor has not yet
been built, the wall may have to be designed as a cantilever.
Design of Vertical
System
The basement wall is designed as the
cantilever wall with the fixity provided by the mat foundation.
Design of Basement
Wall
Concrete
Grade =M20
Steel Grade = Fe415 (TOR)
Reference
|
Step
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Calculation
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Output
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IS 456:2000,
Cl.32.5.1
IS 456:2000,
Cl.26.5.2.1
IS 456:2000,
26.5.2.2
IS 456:2000,
Cl.32.5.b
IS 456:2000,
Cl.31.6.2.1
IS 456:2000,
Table-19
IS456:2000,
Cl.23.2.a
IS 456:2000,
Cl.32.5.d
IS 456:2000,
Cl.26.2.1
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1.
2.
3.
4.
5.
6.
7.
8.
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Design
Constants
Floor to floor heights,
Basement, h = 2.844 m
Unit weight of soil,
![]()
Angle of internal friction of the
soil,
![]()
Surcharge produced due to vehicular
movement,
Ws = 10 kN/m2
Safe bearing capacity of the soil, qs
= 125 kN/m2
Height of soil from base to ground
level (h)= 2.48 m
Moment
calculation
Here, Coefficient of Earth Pressure,
![]()
Lateral load due to soil pressure,
Pa =
![]()
=
![]()
= 22.92 kN/m
Lateral load due to surcharge load,
Ps
=
Ka
![]() ![]()
= 0.333
![]() ![]()
= 9.48 kN/m
Characteristic bending moment at the
base of wall is calculated below. Since the weight of the wall gives
insignificant moment, so this can be neglected in the design.
Mc =
![]()
=
![]()
= 35.21 kN-m/m
Design moment, M = 1.5Mc = 1.5
![]()
= 52.81 kN-m/m
Approximate
design of section
Let effective depth of wall = d
BM =0.138 ƒck bd2
(For Fe415)
or, 52.81
![]() ![]() ![]() ![]()
or, d = 138.36 mm
Let clear cover is 40mm & bar size
is 20mm-Ф
Overall depth of wall, D = 138.36 + 40 + 10
= 188.36 mm
Take D = 200 mm
So, d
= 200-40-10 = 150 mm
D = 200mm
![]()
So, two curtains of reinforcement need
to be provided.
Calculation
of Main Steel Reinforcement
![]() ![]()
or, Ast = 1161.83 mm2
Minimum Ast = 0.12%
![]() ![]() ![]() ![]()
= 240 mm2<
Ast
Maximum Diameter of bar =
![]()
Providing 16mm-Ф bar,
Spacing of bars (s) on the front as well as back face:
![]()
or, s =
![]()
Provide 16 mm-Ф bar @275 mm c/c
So, Provided
![]()
= 1462.3 mm2 > Ast min
![]()
Max. Spacing = 3d = 3x300 = 450 mm or
450 mm
OK.
Check
for Shear
The critical section for shear
strength is taken at a distance, ‘d’ from the face of the support .Thus, the
critical section is at d =(200-40-8)= 0.152 m from the top of mat foundation
i.e. at (2.844- 0.152) = 2.692 m below the top edge of wall.
Shear force at critical section is,
Vu =
![]() ![]()
= 44.242 KN
Nominal shear stress ,
![]()
= 0.29 N/mm2
Permissible shear stress ,
![]()
Here,
![]() ![]()
Hence, it is safe in shear.
Check
for Deflection
![]()
Allowable deflection =
![]()
=11.68 mm
Actual Deflection =
![]()
=
![]()
= 6.85 mm < 11.68
mm
which is less than the allowable
deflection, hence it is safe.
Calculation
of Horizontal Reinforcement Steel bar
Area of Hz. Reinforcement = 0.002Dh
= 0.002
![]() ![]()
As the temperature change occurs at
front face of basement wall, 2/3 of horizontal reinforcement is provided at
front face and 1/3 of horizontal reinforcement is provided in inner face.
Front face Horizontal Reinforcement
steel,
=
![]()
Providing 10mm-Ф bars,
No. of bars required,
N =
![]()
Spacing =
![]()
=
![]()
= 275.4 mm
Provide 10mm-Ф bar @ 250 mm c/c
Inner
face Horizontal Reinforcement steel,
=
![]()
Providing 10mm-Ф bars,
No.
of bars required,
![]()
Spacing =
![]()
=
![]()
Max. spacing = 3d = 3x152 = 456 mm or
450 mm
So, provide 10mm-Ф bar @ 450 mm c/c
Curtailment
of Reinforcement
No bars can be curtailed in less than
Ld distance from the bottom of stem.
Ld =
![]() ![]() ![]()
M1
= 0.87 fy * Ast * d * (1
-
![]()
=
0.87 * 415 * 1462.3 * 152 * (1 -
![]()
= 64.230* 106 N-mm
= 64.230 KN-m
Vu =44.242 KN
1.3
M1/Vu
+ Lo = 1.3*64.230*106/(44.242*103) + 12*16
= 2081.38 mm > Ld (ok)
The curtailment of bars can be done in
two layers, at 1/3 and 2/3 heights of the stem above the base.
Let us curtail bars at 1/3 distance
i.e. 948 mm from the base.
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Pa
= 22.92
kN/m
Ps
= 9.48 kN/m
M = 52.81
kN-m/m
D
= 200 mm
d = 150 mm
Ast
= 1161.83
mm2
s
= 275 mm on front and back face
![]()
Vu = 44.242 kN
![]()
N/mm2
![]()
N/mm2
Front
face:
Provide
10mm-Ф bar @ 250 mm c/c
Inner
face: Provide 10mm-Ф bar @ 450 mm c/c
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