RCC design B.C.Punmia RETAINING WALL 18.2 TYPE OF RETAINING WALLS 1 Gravity walls 2 Cantilever retaining walls a. T- shaped b. L- shaped 3 Counterfort retainig walls. 4 Buttresssed walls. The cantilever retaining wall resist the horizontal earth pressure as wel A retaining wall or retaining structure is used for mai the ground surfgaces at defrent elevations on either side of it. embankments are involed in construction ,retaining wall are usually ne In the construction of buildins having basements, retaining walls are ma Similsrly in bridge work, the wing walls and abutments etc. are des retaining walls , to resist earth pressure along with superimposed lo material retained or supported by a retaining wall is called backfill lyi the horizontal plane at the elevation of the top of a wall is ca In the design of retaining walls or other retaining str it is necessary to compute the lateral earth pressure exerted bythe r mass of soil. The equation of finding out the lateral earth pressure retaining wall is one of the oldest in Civil Engineering field. The pla strees, when the failure is imminent, was invetigated by Rankine in1860. theoretical experiment work has been done in this field and many th Retaining walls may be classified according to their resisting the earth pressure,and according to their shape. Following are commen types of retaining walls (Fig) A gravity retaining wall shown in fig 1 is the one in w earth pressure exrted by the back fill is resisted by dead weight which is either made of masonry or of mass concrete . The stress devlo wall is very low ,These walls are no proportioned that no tension is
Transcript
RCC design B.C.Punmia
RETAINING WALL
18.2 TYPE OF RETAINING WALLS
1 Gravity walls2 Cantilever retaining walls a. T- shaped b. L- shaped3 Counterfort retainig walls.4 Buttresssed walls.
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
A retaining wall or retaining structure is used for maintaining the ground surfgaces at
defrent elevations on either side of it. Whenever embankments are involed in construction ,retaining
wall are usually necessary. In the construction of buildins having basements, retaining walls are
mandatory. Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining
walls , to resist earth pressure along with superimposed loads. The material retained or supported by
a retaining wall is called backfill lying above the horizontal plane at the elevation of the top of a wall
is called the surcharge, and its inclination to horizontal is called the surcharge angle b
In the design of retaining walls or other retaining structures, it is necessary to compute the lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical experiment work has been done in this field and many theory and hypothesis heve benn proposed.
Retaining walls may be classified according to their mode of resisting the earth pressure,and according to their shape. Following are some of commen types of retaining walls (Fig)
A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any where, and the resultant of forces remain withen the middle third of the base.
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
Pressure p' at the junction of stem with Heel slab is
p' = 67.55 -67.55 - 32.99
x 0.90 = 54.592.40
5 Design of toe slab:-
(1) Up ward soil pressure (2) Down ward weight of slab Down ward weight of slab per unit area = 0.30 x 1 x 1.00 x 25 = 7.50
Hence net pressure intensities will be = 67.55 - 7.50 = 60.05
= 50.27 - 7.50 = 42.77Total force = S.F. at E = 0.50 x( 60.05 + 42.77 ) x 1.20 = 61.69 kN
=42.77 2.00 x 60.05 x 1.20
= 0.63 m42.77 60.05 3
\ = 61.69 x 0.63 = 39.09 kN-m
Effective depth required =BM
=39.09 x
= 207 mmRxb 0.913 x 1000
Keep effective depth d = 210 mm and total thickness = 210 + 60 = 270Reduce the total thickness to = 200 mm or 0.20 m at edge say = 0.27
=61.69 x 1000
= 0.228 <1000 x 270
Ast = =39.09 x
= 896230 x 0.904 x 210
The reinforcement has to be provided at bottom face .If alternate bars of stem reiforcerment are
are bent and continued in toe slab, area available = 1/2 x 1256 = 628 (see step 7)
using 12 mm bars A == 3.14 x 12 x 12 = 113
4 4\ Spacing A x1000 / Ast = 113 x 1000 / 896 = 120 mm
Hence Provided 12 120 mm c/c
45 x 12 = 540 mm
Providing 30 mm clear side cover actual length available = 1200 - 30 = 11701170 > 540 Hence safe
Distribution steel =0.12
x 1000 x270 + 200
= 282100 2
Using 8 = =3.14 x ( 8
= 504 4
\ Spacing =1000 x 50
= 178 mm say = 170 mm c/c282
6 Design of heel slab :-Three force act on it
1. down ward weight of soil 2 weight of heel slab 3 Down ward earth pressure 4 upward soil pressure
= 0.90 x3.70 + 3.96
x 18 = 62 KN say = 63.00 kN2
Acting at =3.70 + 2 x 3.96
x0.90
= 0.455 m from B3.70 + 3.96 3
Total weight of heel slab = 0.90 x 0.27 x 1 x 25 = 6.08 kN
Acting at 0.45
\
….(I)
….(II)
Hence total force due to vertical component of earth pressure is
= Ka.y x
kN-m2
kN-m2
The upward pressure distribution on the toe slab is shown in fig 1b .The weight of soil above the toe slab is neglicted . Thus two forces are acting on it
kN-m2
kN-m2 under D
And at under E kN-m2
x from E+
+
B.M. at E 10 6
tv N/mm2 tc even at mimum steel
BM x 106 10 6
mm2
sst x j x D
mm2
3.14xdia2
mm F bar, @
Let us check this reinforcement for development length Ld=45 F =
mm2
mm F bars, Area P D2 )'2
mm2
Total weight of soil over Heel
(H1+2xH2)xb
(H1+H2)x3
m from B.
Earth pressure intencity at b = Ka.y.H1 per unit inclined area, at b to horizontal,
Earth pressure at B, on horizontal unitarea = Ka.y.H1.tan bVertical component of this, at B = Ka.y.H1 .tan b.sin b
Similarly, Vertical component of earth pressure intencity at C =Ka.y.H2 tan b. sin b.
Bend these bars into toe slab, to serve as reiforcement there. Sufficient devlopment length ia available.Between A and B some of bars can be curtailed. Cosider a section at depth below the top of stem
140+ 250 - 140
x h (where h In meter)H
d' = 140 +250 - 140
x h =( 140 + 29.49 x h )3.73
Now AsH3
or H =(d'
Hence h
=
where Ast' = reinforcement at depth h Ast =
d' = effective depthat depth h d =
= than=
1\
h=
1x
d'
2 H1 2 d
Subsituting d = 255 mm and d' =( 140 + 29.5 x h ) we get
h = x140 + 29.5 x h
2 x 255
h = 3.73 x140 x 29.5 x h
2 x 255h = 0.467 x ( 140 + 29.5 x h
h = 2.83 m 0.467 x ( 140 + 29.5 x 0.01Howerver, the bars should be extented by a distance of = 12 x 12 = 144 mm
Or d = 250 mm whichever is more beyond the point.\ h = 2.83 - 0.25 = 2.58 m. Hence curtailed half bars at at height of
2.58 m below the top . If we wish to curtailed half of the remaining bars so that remaining
=1
Hence from ….(2)4
\h
=1 x d'
= x140 + 29.5 x h
H1 4 d 4 x 255
h = 3.73x 140 x 29.5 x h
4 x 255h = 0.371 x ( 140 + 29.5 x hh = 2.19 m 0.371 x ( 140 + 29.5 x 0.00
This can be solved bytrial and error, Noting that if the effective thickness of stem w=are constant,.Howerver, the bars should be extented by a distance of = 12 x 12 = 144 mm
Or d = 250 mm whichever is more beyond the point.\ h = 2.19 - 0.25 = 1.94 m. Hence stop half bars the remaining barsby 1.94 m below the top of the stem . Continue rest of the bars to the top of the stem
Check for shear:-Shear force =
p = = 0.38 x18
x 3.73 47.52 kN2 2
\ =47.52 x 1000
= 0.19 < (see table 3.1)1000 x 250
Nomber of Bars = Ast/A = 1137 / 113 = 10.06 say = 11 No.
Hence Provided 11 bars of 12
% of steel provided =11 x 113
x 100 = 0.50 %1000 x 250
Permissible shear stress for 0.50 % = 0.3 (See Table 3.1)If tc > tvhence safe here 0.30 > 0.19 Hence Safe
Distribution and temprechure reinforcement:-Average thickness of stem = 310 + 200
= 255 mm2
\ Distribution reinforcement =0.12
x 1000 x 255 = 306
Actual AS provided mm2
The effective depth d' at section is =
Ast Ast d )1/3
Ast' d' 1/3
H1 Ast d
reinforcement at depth H1
effective depthat depth H1
if Ast 1/2 Ast
Ast' 1/3
Ast
H1 1/3
1/3
)1/3
h) 1/3- h =12 F
remaining reinforcement is one forth of that provided ar B, we have Ast'Ast
The wall is in unsafe in sliding, and hence shear key will be provided below the stem as shown in fig.Let u sprovide ashear key 300 x 310 Let Pp be the intensity of passive pressure devloped
in front of key this intencity Pp depend upon the soil pressure P in front of the key= = 1/ 0.38 = 2.64 x 50.27 = 132.46
Pp x a = 132.46 x 0.30 = 39.74 kN
Sliding force at level GJ = 0.38 x18
x 4.53 x`
2.00
= 3.42 x( 4.53 0.961 = 67.22 kN ….(2)Weight of the soil between bottom of the base and GJ = 2.40 x 18 x 0.30 = 12.96
