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Design of Masonary Retaining Wall with Back Face in Batter Design parameters : River/Nalla Discharge Q =
144.70
Ground water Table LVL =
368.50 m 371.20 m
LVL of Retaining wall top head = LVL of Retaining wall foundation LVL = Lacey's Silt factor = Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Width of Ret Wall top Head (a)= Height of Retaining wall top head (b) = Height of batter above foundation block (c) = Bottom width of batter at foundation block (d) = Width of foundation block (e) = Height of foundation block (f) = Friction coeff betn base of footing and soil Coefficient of active earth pressure Ka = Coefficient of passive earth pressure Kp =
cumecs
367.50
4
1 16.5 2 1.75 0.6 0 3.7 2.6 3.2 0
2
t/m2 t/m3 t/m3 m m m m m
1
3
B
m
0.5 0.49 2.04
e
Evaluation of forces acting on the Retaining wall at El
367.50
(A) Vertical Forces
(i) Self weight of Retaining wall (a) Weight of Triangular section (1)= acting at a distance of (b) Weight of Rectangular section (2)= acting at a distance of (c) Weight of Rectangular section (3)= acting at a distance of (d) Weight of Rectangular soil backfill (4) = acting at a distance of (e) Weight of Triangular soil backfill (5) = acting at a distance of (v) Up lift Force Uplift force = acting at a distance of (B) Horizontal Forces
12.03 1.47 5.55 0.3 0.00 1.60 0.00 1.90 8.42 2.33 1.60 2.13
tonnes m from tonnes m from tonnes m from tonnes m from tonnes m from
368.50 m
Intensity of pressure at HFL = Horizontal force = acting at height = (b) Wet earth pressure effect below FSL due to wet soil
2.32 3.13 1.90 above 2.32 2.32
B B B B B
tonnes m from B
(i) Wet earth pressure (a) Wet earth pressure above HFL of
Intensity of pressure at FSL = Horizontal force =
a
t/m2 tonnes m above B t/m2 tonnes
acting at height = (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom
0.50
m above B
0.49
Hor. force due to submerge earth pressure =
0.25 0.33
t/m2 tonnes m above B
acting at height (b) Water pressure
Intensity of pressure at bottom = 1.00 t/m2 Horizontal force due to water pressure = 0.50 tonnes acting at height = 0.33 m above B (iii) Live Load Surcharge Intesity of loading on backfill = 0.578 tonnes per sqm Horizontal force = 1.05 tonnes acting at height = 1.85 m above B (vi) Passive Earth pressure of soil retained in front of Abutment Int of Passive pressure at found block top 0.0 t/m2 Int Passive pressure at found block bottom LVL = 0.00 t/m2 Total Passive earth pressure on found block 0.00 t Acting at height above found LVL 0.00 m
Consider stability of section at Level
367.50
(A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (d) Weight of Rectangular soil backfill (4) (d) Weight of Triangular soil backfill (5) (ii) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge SUM Base width at foundation block bottom Level (B)
m about X-X Axis Vertical Force (Tons)
Particular of Forces
8
Horizontal Force Lever (Tons) arm (m)
12.03
1.47 0.3 1.60 1.90 2.33 2.13
5.55 0.00 0.00 8.42 -1.60
24.39 367.50
3.2
m
3.13 2.32
1.90 0.50
0.25 0.50 1.05 7.23
0.33 0.33 1.85
Net stabilising moment =
M M
26.25
M
The resultant will strike base at distance (x) =
ton.m
M
V
The eccentricity (e) about C.G. of Section at foundation block LVL =B/2 - x = Check eccentricity e as compared to B/6 as (e 2.0 Hence Safe
1.69
Safe in Sliding
m 1
tonnes/m2= Safe
m from A
0.00
ton/m2 Safe
e
in Batter
4
b 5 c
1
d
f
M+ (t.m)
M-(t.m)
17.64 1.67 0.00 0.00 19.64 3.41
5.94 1.16
38.94
0.08 0.17 1.94 12.69
N/mm2
0.15
N/mm2
Design of Masonary Retaining Wall with Back Face in Batter Design parameters : River/Nalla Discharge Q =
144.70
Ground water Table LVL =
366.50 m 369.20 m
LVL of Retaining wall top head = LVL of Retaining wall foundation LVL = Lacey's Silt factor = Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Width of Ret Wall top Head (a)= Height of Retaining wall top head (b) = Height of batter above foundation block (c) = Bottom width of batter at foundation block (d) = Width of foundation block (e) = Height of foundation block (f) = Friction coeff betn base of footing and soil Coefficient of active earth pressure Ka = Coefficient of passive earth pressure Kp =
cumecs
365.50
4
1 16.5 2 1.75 0.6 0 3.7 2.6 3.2 0
2
t/m2 t/m3 t/m3 m m m m m
1
3
B
m
0.5 0.49 2.04
e
Evaluation of forces acting on the Retaining wall at El
365.50
(A) Vertical Forces
(i) Self weight of Retaining wall (a) Weight of Triangular section (1)= acting at a distance of (b) Weight of Rectangular section (2)= acting at a distance of (c) Weight of Rectangular section (3)= acting at a distance of (d) Weight of Rectangular soil backfill (4) = acting at a distance of (e) Weight of Triangular soil backfill (5) = acting at a distance of (v) Up lift Force Uplift force = acting at a distance of (B) Horizontal Forces
12.03 1.47 5.55 0.3 0.00 1.60 0.00 1.90 8.42 2.33 1.60 2.13
tonnes m from tonnes m from tonnes m from tonnes m from tonnes m from
366.50 m
Intensity of pressure at HFL = Horizontal force = acting at height = (b) Wet earth pressure effect below FSL due to wet soil
2.32 3.13 1.90 above 2.32 2.32
B B B B B
tonnes m from B
(i) Wet earth pressure (a) Wet earth pressure above HFL of
Intensity of pressure at FSL = Horizontal force =
a
t/m2 tonnes m above B t/m2 tonnes
acting at height = (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom
0.50
m above B
0.49
Hor. force due to submerge earth pressure =
0.25 0.33
t/m2 tonnes m above B
acting at height (b) Water pressure
Intensity of pressure at bottom = 1.00 t/m2 Horizontal force due to water pressure = 0.50 tonnes acting at height = 0.33 m above B (iii) Live Load Surcharge Intesity of loading on backfill = 0.578 tonnes per sqm Horizontal force = 1.05 tonnes acting at height = 1.85 m above B (vi) Passive Earth pressure of soil retained in front of Abutment Int of Passive pressure at found block top 0.0 t/m2 Int Passive pressure at found block bottom LVL = 0.00 t/m2 Total Passive earth pressure on found block 0.00 t Acting at height above found LVL 0.00 m
Consider stability of section at Level
365.50
(A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (d) Weight of Rectangular soil backfill (4) (d) Weight of Triangular soil backfill (5) (ii) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge SUM Base width at foundation block bottom Level (B)
m about X-X Axis Vertical Force (Tons)
Particular of Forces
8
Horizontal Force Lever (Tons) arm (m)
12.03
1.47 0.3 1.60 1.90 2.33 2.13
5.55 0.00 0.00 8.42 -1.60
24.39 365.50
3.2
m
3.13 2.32
1.90 0.50
0.25 0.50 1.05 7.23
0.33 0.33 1.85
Net stabilising moment =
M M
26.25
M
The resultant will strike base at distance (x) =
ton.m
M
V
The eccentricity (e) about C.G. of Section at foundation block LVL =B/2 - x = Check eccentricity e as compared to B/6 as (e 2.0 Hence Safe
1.69
Safe in Sliding
m 1
tonnes/m2= Safe
m from A
0.00
ton/m2 Safe
e in Batter
e
4
b 5 c
1
d
f
M+ (t.m)
M-(t.m)
17.64 1.67 0.00 0.00 19.64 3.41
5.94 1.16
38.94
0.08 0.17 1.94 12.69
N/mm2
0.15
N/mm2
Design of Masonary Retaining Wall with Back Face in Batter Design parameters : River/Nalla Discharge Q =
144.70
Ground water Table LVL =
362.50 m 366.00 m
LVL of Retaining wall top head = LVL of Retaining wall foundation LVL = Lacey's Silt factor = Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Width of Ret Wall top Head (a)= Height of Retaining wall top head (b) = Height of batter above foundation block (c) = Bottom width of batter at foundation block (d) = Width of foundation block (e) = Height of foundation block (f) = Friction coeff betn base of footing and soil Coefficient of active earth pressure Ka = Coefficient of passive earth pressure Kp =
cumecs
361.50
4
1 16.5 2 1.75 0.6 0 4.5 3.15 3.75 0
2
t/m2 t/m3 t/m3 m m m m m
1
3
B
m
0.5 0.49 2.04
e
Evaluation of forces acting on the Retaining wall at El
361.50
(A) Vertical Forces
(i) Self weight of Retaining wall (a) Weight of Triangular section (1)= acting at a distance of (b) Weight of Rectangular section (2)= acting at a distance of (c) Weight of Rectangular section (3)= acting at a distance of (d) Weight of Rectangular soil backfill (4) = acting at a distance of (e) Weight of Triangular soil backfill (5) = acting at a distance of (v) Up lift Force Uplift force = acting at a distance of (B) Horizontal Forces
17.72 1.65 6.75 0.3 0.00 1.88 0.00 2.18 12.40 2.70 1.88 2.50
tonnes m from tonnes m from tonnes m from tonnes m from tonnes m from
362.50 m
Intensity of pressure at HFL = Horizontal force = acting at height = (b) Wet earth pressure effect below FSL due to wet soil
3.00 5.25 2.17 above 3.00 3.00
B B B B B
tonnes m from B
(i) Wet earth pressure (a) Wet earth pressure above HFL of
Intensity of pressure at FSL = Horizontal force =
a
t/m2 tonnes m above B t/m2 tonnes
acting at height = (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom
0.50
m above B
0.49
Hor. force due to submerge earth pressure =
0.25 0.33
t/m2 tonnes m above B
acting at height (b) Water pressure
Intensity of pressure at bottom = 1.00 t/m2 Horizontal force due to water pressure = 0.50 tonnes acting at height = 0.33 m above B (iii) Live Load Surcharge Intesity of loading on backfill = 0.578 tonnes per sqm Horizontal force = 1.27 tonnes acting at height = 2.25 m above B (vi) Passive Earth pressure of soil retained in front of Abutment Int of Passive pressure at found block top 0.0 t/m2 Int Passive pressure at found block bottom LVL = 0.00 t/m2 Total Passive earth pressure on found block 0.00 t Acting at height above found LVL 0.00 m
Consider stability of section at Level
361.50
(A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (d) Weight of Rectangular soil backfill (4) (d) Weight of Triangular soil backfill (5) (ii) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge SUM Base width at foundation block bottom Level (B)
m about X-X Axis Vertical Force (Tons)
Particular of Forces
8
Horizontal Force Lever (Tons) arm (m)
17.72
1.65 0.3 1.88 2.18 2.70 2.50
6.75 0.00 0.00 12.40 -1.88
35.00 361.50
3.75
m
5.25 3.00
2.17 0.50
0.25 0.50 1.27 10.27
0.33 0.33 2.25
Net stabilising moment =
M M
44.07
M
The resultant will strike base at distance (x) =
ton.m
M
V
The eccentricity (e) about C.G. of Section at foundation block LVL =B/2 - x = Check eccentricity e as compared to B/6 as (e 2.0 Hence Safe
1.70
Safe in Sliding
m 1
tonnes/m2= Safe
m from A
0.00
ton/m2 Safe
e in Batter
e
4
b 5 c
1
d
f
M+ (t.m)
M-(t.m)
29.24 2.03 0.00 0.00 33.49 4.69
11.38 1.50
64.75
0.08 0.17 2.87 20.68
N/mm2
0.19
N/mm2
Design of Concrete Retaining Wall at KSB of CAD Chambal Project Design parameters : River/Nalla Discharge Q =
144.70
Ground water Table LVL =
101.70 m 102.30 m
LVL of Retaining wall top head = LVL of Retaining wall foundation LVL = Lacey's Silt factor = Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Width of Ret Wall top Head (a)= Height of Retaining wall top head (b) = Height of batter above foundation block (c) = Bottom width of batter at foundation block (d) = Width of foundation block (e) = Height of foundation block (f) =
cumecs
a
100.00
4
1 10.5 2.00 1.75 0.45 0 2.3 2.00 2.45 0
2
t/m2 t/m3 t/m3 m m m m m
1
3
B
m
Friction coeff betn base of footing and soil 0.5 Coefficient of active earth pressure Ka = 0.33 Coefficient of passive earth pressure Kp = 3.03 Concrete Grade M-15 to be used as bulk gravity Material
e
Evaluation of forces acting on the Retaining wall at El
100.00
(A) Vertical Forces
(i) Self weight of Retaining wall (a) Weight of Triangular section (1)= acting at a distance of (b) Weight of Rectangular section (2)= acting at a distance of (c) Weight of Rectangular section (3)= acting at a distance of (d) Weight of Rectangular soil backfill (4) = acting at a distance of (e) Weight of Triangular soil backfill (5) = acting at a distance of (v) Up lift Force Uplift force = acting at a distance of (B) Horizontal Forces
5.75 1.12 2.59 0.225 0.00 1.23 0.00 1.45 4.03 1.78 2.08 1.63
tonnes m from tonnes m from tonnes m from tonnes m from tonnes m from
101.70 m
Intensity of pressure at HFL = Horizontal force = acting at height = (b) Wet earth pressure effect below FSL due to wet soil
0.35 0.10 1.90 above 0.35
B B B B
tonnes m from B
(i) Wet earth pressure (a) Wet earth pressure above HFL of
Intensity of pressure at FSL =
B
t/m2 tonnes m above B t/m2
Horizontal force = acting at height = (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom
0.59 0.85
tonnes m above B
0.56
Hor. force due to submerge earth pressure =
0.48 0.57
t/m2 tonnes m above B
acting at height (b) Water pressure
Intensity of pressure at bottom = 1.70 t/m2 Horizontal force due to water pressure = 1.45 tonnes acting at height = 0.57 m above B (iii) Live Load Surcharge Intesity of loading on backfill = 0.578 tonnes per sqm Horizontal force = 0.44 tonnes acting at height = 1.15 m above B (vi) Passive Earth pressure of soil retained in front of Abutment Int of Passive pressure at found block top 0.0 t/m2 Int Passive pressure at found block bottom LVL = 0.00 t/m2 Total Passive earth pressure on found block 0.00 t Acting at height above found LVL 0.00 m
Consider stability of section at Level
Particular of Forces (A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (d) Weight of Rectangular soil backfill (4) (d) Weight of Triangular soil backfill (5) (ii) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge SUM
100.00
8
m about X-X Axis Vertical Force (Tons)
Horizontal Force Lever (Tons) arm (m)
5.75
1.12 0.225 1.23 1.45 1.78 1.63
2.59 0.00 0.00 4.03 -2.08
10.28
0.10 0.59
1.90 0.85
0.48 1.45 0.44 3.05
0.57 0.57 1.15
Base width at foundation block bottom Level (B) Net stabilising moment =
100.00
M M
2.45
8.49
M
The resultant will strike base at distance (x) =
m
ton.m
M
V
The eccentricity (e) about C.G. of Section at foundation block LVL =B/2 - x =
0.83
m from A
0.40
m
1
Check eccentricity e as compared to B/6 as (e 2.0 Hence Safe
1.68
Safe in Sliding
0.00
ton/m2 Safe
As per IS-456 2000 clause 34.5.2 provide nominal reinforcement @ 140 mm spa cing c/c in both directions
bal Project Kota
e
4
b 5 c
1
d
f
M+ (t.m)
M-(t.m)
6.42 0.58 0.00 0.00 7.18 3.40
0.20 0.50
14.18
0.27 0.82 0.50 5.69
N/mm2
directions
0.08
N/mm2
Design of Masonary Retaining Wall with Back Face in Batter Design parameters : River/Nalla Discharge Q =
144.70
Ground water Table LVL =
364.50 m 369.50 m
LVL of Retaining wall top head = LVL of Retaining wall foundation LVL = Lacey's Silt factor = Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Width of Ret Wall top Head (a)= Height of Retaining wall top head (b) = Height of batter above foundation block (c) = Bottom width of batter at foundation block (d) = Width of foundation block (e) = Height of foundation block (f) = Friction coeff betn base of footing and soil Coefficient of active earth pressure Ka = Coefficient of passive earth pressure Kp =
cumecs
363.50
4
1 16.5 2 1.75 0.6 0 6 4.25 4.85 0
2
t/m2 t/m3 t/m3 m m m m m
1
3
B
m
0.5 0.49 2.04
e
Evaluation of forces acting on the Retaining wall at El
363.50
(A) Vertical Forces
(i) Self weight of Retaining wall (a) Weight of Triangular section (1)= acting at a distance of (b) Weight of Rectangular section (2)= acting at a distance of (c) Weight of Rectangular section (3)= acting at a distance of (d) Weight of Rectangular soil backfill (4) = acting at a distance of (e) Weight of Triangular soil backfill (5) = acting at a distance of (v) Up lift Force Uplift force = acting at a distance of (B) Horizontal Forces
31.88 2.02 9.00 0.3 0.00 2.43 0.00 2.73 22.31 3.43 2.43 3.23
tonnes m from tonnes m from tonnes m from tonnes m from tonnes m from
364.50 m
Intensity of pressure at HFL = Horizontal force = acting at height = (b) Wet earth pressure effect below FSL due to wet soil
4.29 10.72 2.67 above 4.29 4.29
B B B B B
tonnes m from B
(i) Wet earth pressure (a) Wet earth pressure above HFL of
Intensity of pressure at FSL = Horizontal force =
a
t/m2 tonnes m above B t/m2 tonnes
acting at height = (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom
0.50
m above B
0.49
Hor. force due to submerge earth pressure =
0.25 0.33
t/m2 tonnes m above B
acting at height (b) Water pressure
Intensity of pressure at bottom = 1.00 t/m2 Horizontal force due to water pressure = 0.50 tonnes acting at height = 0.33 m above B (iii) Live Load Surcharge Intesity of loading on backfill = 0.578 tonnes per sqm Horizontal force = 1.70 tonnes acting at height = 3.00 m above B (vi) Passive Earth pressure of soil retained in front of Abutment Int of Passive pressure at found block top 0.0 t/m2 Int Passive pressure at found block bottom LVL = 0.00 t/m2 Total Passive earth pressure on found block 0.00 t Acting at height above found LVL 0.00 m
Consider stability of section at Level
363.50
(A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (d) Weight of Rectangular soil backfill (4) (d) Weight of Triangular soil backfill (5) (ii) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge SUM Base width at foundation block bottom Level (B)
m about X-X Axis Vertical Force (Tons)
Particular of Forces
8
Horizontal Force Lever (Tons) arm (m)
31.88
2.02 0.3 2.43 2.73 3.43 3.23
9.00 0.00 0.00 22.31 -2.43
60.76 363.50
4.85
m
10.72 4.29
2.67 0.50
0.25 0.50 1.70 17.45
0.33 0.33 3.00
Net stabilising moment =
M M
99.68
M
The resultant will strike base at distance (x) =
ton.m
M
V
The eccentricity (e) about C.G. of Section at foundation block LVL =B/2 - x = Check eccentricity e as compared to B/6 as (e 2.0 Hence Safe
1.74
Safe in Sliding
m 1
tonnes/m2= Safe
m from A
0.00
ton/m2 Safe
e in Batter
e
4
b 5 c
1
d
f
M+ (t.m)
M-(t.m)
64.28 2.70 0.00 0.00 76.61 7.84
28.58 2.14
143.59
0.08 0.17 5.09 43.91
N/mm2
0.25
N/mm2
Design of R.C.C. Bridge Piers Design parameters : River / Nalla discharge Q =
144.70
Clear span between piers = Thickness of road way deck slab = F.S.L. of canal U/S of Cross regulator = F.S.L. of canal D/S of Cross regulator = Canal bed Level D/S of Cross Regulator = Lacey's Silt factor (f) =
6.35 0.64 258.97 258.97 254.30 1.00
m m
Bearing Capacity of Soil = Width of roadway slab of bridge = Weight of Railing per m run of bridge = Weight of Gates per m length of bridge = Eccen of C/L road deck slab from pier center= Height of Pier upto Bed block LVL from CBL = Thickness of Bed block = Area of bridge Hoisting platform = Top Level of Hoisting Arrangement = Eccen of hoisting platform from pier center= Width of pier Length of pier at top bed block level = Length of pier at CBL = LVL of wearing top of road way deck slab = LVL of deck slab bearing LVL of top of foundation block of Pier No of bays of canal Length of Pier at FSL D/S of CR Gates Length of Pier at CBL D/S of CR Gates Length of foundation block at foundation LVL Width of foundation block at foundation LVL Height of foundation block at Pier-footing junct Height of foundation block at edge of footing (i) Foundation Level of Piers Normal depth of scour (R) =0.47(Q/f)1/3
75 6.55 0.7 4.00 2.97 5.67 0.6 19.34 263.55 -3.02 1.15 14 14.55 261.27 260.57 254.3 9 7.75 8.02 18.55 5 1.6 1
t/m2 m ton/m ton/m m m m m2
Anticipated depth of scour around piers = 2.0 ( R ) =
cumecs
m m
m m m m m m m Nos m m m m m m
2.47
m
4.93
m
Level of foundation of Piers should not be less than 1 m below anticipated depth of scour Level of foundation = 253.04 m say 252.70 m
Evaluation of forces acting on the Pier (a) Dead weight of super structure (i) Weight of slab with wearing coat = (ii) Weight of Kerbs = (iii) Weight of Railings = (v) Weight of Gates hoisting structures =
78.60 9.00 10.50 55.60
tonnes tonnes tonnes tonnes
(vi) Weight of Gates and movement assembly Total
30.00 183.70
tonnes tonnes
(b) Live Loads Maximum live load reaction on pier is obtained when fourth axle load 11.4 t at (i) Reaction from right span = 14.48 tonnes (ii) Reaction from left span = 11.41 tonnes Total Live Load reactio on the Pier from live loads = 25.89 Impact Factor = 0.34 So Maximum live load reaction with impact = 34.60 tonnes Eccentricity of live load eccentricity of L.L. from center of pier about Y-Y axis = 3.75 eccentricity of L.L from center of pier about X-X axis = 0.00
center of pier
tonnes
m m
(c) Braking Force It is equal to 20% of total live load present on the bridge = Reduced level of application of braking force = Height of appication force above bearing level = Increase in reaction due to braking force = 2.12
8.36 262.47 1.900 tonnes
m m
(d) Temperature force due to sliding friction Reaction on sliding end when loads are placed so as to produc maximum reaction at sliding end Live load Reaction at sliding end = 12.24 tonnes Impact factor = 0.34 Live load with impact = 16.36 tonnes Dead Load reaction = 183.70 tonnes Increase due to braking force = 2.12 tonnes Total : 202.51 tonnes Friction in sliding bearing ( with coeff. of sliding friction as 0.03) = 6.08 R.L. of point of application = 260.57 m
(e) Wind force
Since intensity of wind pressure depends upon the height of the point above mean retarding surface, so two cases for calculation of wind forces have been considered (i) When level of water in the canal is at F.S.L. (ii) When depth of water is zero (i) When level of water in the canal is at F.S.L. Average height of elevation area =
4.58
Intensity of wind pressure =
46.3
Exposed Elevation area = Total Wind force on structure = R.L. of point of application =
34.35 1.59 261.26
Wind force on live load Intensity of wind force = Length of live load = length of one span = Total wind force on live load =
300.00 7.50 2.25
m kg/m
2
m2 tonnes m
kg/m m tonnes
R.L. of point of application =
262.47
m
(ii) When depth of water is zero Average height of area of elevation above bed level = Intensity of wind pressure = Exposed Elevation area = Total wind force on structure = R.L. of point of application =
85.0 66.8 5.68 259.10
8.91 kg/sqm sqm tonnes m
Wind force on live load Total wind force on live load = R.L. of point of application =
2.25 262.47
tonnes m
m
(f) Dead Loads of Sub Structure (i) Dead Load of bed block Plan area = Thickness = Weight of bed block =
21.35 0.6
Sqm m 32.03
tonnes
89.33
tonnes
(ii) Dead Load of pier Dead load of pier =
(iii) Force due to buoyancy Plan area at F.S.L. = 15.76 Weight of Water displaced = 68.22 Force due to buoyancy allowing 15% upward force =
Sqm tonnes 10.23
(g) Force on pier due to current of water Maximum Velocity of flow (V) = Q/PR =
m/s
0.87 V= 2
Maximum value of depth of scour = 2.0 R = Assuming variation of V2 to be linear than its value at bed level = Pressure due to water current = K V2 Pressure at water level =
1.05 4.98 0.05 3.03
m 0.04
where K = 19.62
Pressure at bed level = Area on which pressure acts = Total force due to water current = Height of center of pressure from bed level =
0.75 4.93
tonnes
26 kg/m
for pointed Nose 45 deg
2
kg/m2 Sqm tonnes m
(h) Force on pier due to current of water perpendicular to length of pier Maximum variation in direction of current = 200 Maximum Velocity at top = V sin 20 =
0.30 V2=
Value of V2 at bed level = Assuming K =
80
0.09
0.00 for rectangular shape
Pressure at water level = Pressure at bed level = Area on which pressure acts = Total force due to water current = Height of center of pressure from bed level =
7.06
kg/m2
0.16 66.66 0.24 3.08
kg/m2 Sqm tonnes m
(i) Force on pier due to pressure of water perpendicular to length of pier when water flows i Lateral Forces on pier when one gate is opened for flow and adjacent one are closed F = h A h= 2.35 m Area under influence for water pressure = 34.14 sqm Total pressure force F = h A = 80.17 tonnes Center of pressure below water surface yp =( I0/A h ) + h I0= h3(a2+4ab+b2)/36 a b= yp= 2.56 Level of center of pressure above base =
16.98 m 256.410
(j) Force on Pier due to prestressed anchorage system Force due to prestressed anchoring bars in grout = Eccentricity of anchorage about X-X axis = 0.400 Eccentricity of anchorage about Y-Y axis = 3.220
m3 m
35.0
tonnes
m m
Check for stresses in the pier at Bed Level with R.L. = Area (A) = Ixx= Iyy=
16.07
254.30
Sqm
1.73
m4
262.03
m4
Case 1 : When Live load on the bridge and canal is running at F.S.L. (i) Vertical dead loads (a) Dead Load of super structure = (b) Dead load of bed block = (c) Dead Load due to self weight of Pier = (d) Increase in reaction due to braking = (ii) Live load with impact factor = (iii) Buoyancy force acting upwards = (iv) Vertical reaction from anchorage system =
So Net vertical downward force (P) =
183.70 32.03 89.33 2.12 34.60 -10.23 35.00
tonnes tonnes tonnes tonnes tonnes tonnes tonnes
366.55 tonnes
Moment about x-x axis Due to braking force = 68.30 Due to sliding friction = 38.09 Due to water current in perpendicular direction = Due to water pressure when alternate gates open = Due to vertical reaction from ground anchorage system = Total Moment (Mx) = 262.28 say 263.00
t.m t.m 0.74 169.14 -14.00 t.m t.m
t.m t.m t.m
Moment about y-y axis Due to eccentricity of dead Load of super structure (slab+Kerbs+Railing) = Due to eccentricity of dead Load of super structure (Gate hoisting platform) = Due to eccentricity of Live Load = 129.59 t.m Due to wind force on super structure = 11.07 t.m Due to wind force on live load = 18.38 t.m Due to water current force in long. Dir = 0.16 t.m Due to vertical reaction from ground anchor = -112.70 t.m Total Moment (My) = 79.34 t.m say 80.00 t.m Now stress =
My P Mx ( xm ) ( ym ) A I xx I yy
Maximum compressive stress σ Minimum compressive stress σ
com,max com,min
=
112.03 -66.41
=
t/m2 t/m2 =
< 3000 t/m2 for M30 gra -0.664
< 0.67 N/mm2 for M30 concrete grade permissible tensile stress casted in full height at a ti
Case 2 : When Live load on the bridge and No water in canal (i) Vertical dead loads (a) Dead Load of super structure = (b) Dead load of bed block = (c) Dead Load due to self weight of Pier = (d) Increase in reaction due to braking = (ii) Live load with impact factor =
183.70 32.03 89.33 2.12 34.60
So Net vertical downward force (P) =
tonnes tonnes tonnes tonnes tonnes
341.78 tonnes
Moment about x-x axis Due to braking force = Due to sliding friction = Total Moment (Mx) = say
68.30 38.09 106.39 107.00
t.m t.m t.m t.m
Moment about y-y axis Due to eccentricity of dead Load of super structure (slab+Beam+Kerbs+Railing) = Due to eccentricity of dead Load of super structure (Gate hoisting platform) = Due to eccentricity of Live Load = 129.59 t.m Due to wind force on super structure = 11.07 t.m Due to wind force on live load = 18.38 t.m Total Moment (My) = 191.88 t.m say 192.00 t.m Now stress =
My P Mx ( xm ) ( ym ) A I xx I yy
Maximum compressive stress σ Minimum compressive stress σ
com,max
com,min
=
=
61.64 -19.11
t/m2 (Tensile) t/m2 =
< 3000 t/m2 for M30 gra -0.191
< 0.67 N/mm2 for M30 concrete grade p
Check for stresses in the pier at foundation Level with R.L. = Area (A) = Ixx= Iyy=
92.75
252.70
Sqm
193.23
m4
2659.63
m4
Case 1 : When Live load on the bridge and canal is running at F.S.L. (i) Vertical dead loads (a) Dead Load of super structure = (b) Dead load of bed block = (c) Dead Load due to self weight of Pier = (d) Increase in reaction due to braking = (e) Live load with impact factor = (f) Buoyancy force acting upwards = (g) Dead load of foundation block =
So Net vertical downward force (P) = Moment about x-x axis Due to braking force = Due to sliding friction = Due to water current in perpendicular direction = Due to water pressure when alternate gates open = Total Moment (Mx) = say
183.70 32.03 89.33 2.12 34.60 -10.23 313.49
tonnes tonnes tonnes tonnes tonnes tonnes tonnes
645.04 tonnes
81.68 47.81
428.11 429.00
t.m t.m 1.21 297.41 t.m t.m
t.m t.m
Moment about y-y axis Due to eccentricity of dead Load of super structure (slab+Beam+Kerbs+Railing) = Due to eccentricity of dead Load of super structure (Gate hoisting platform) = Due to eccentricity of Live Load = 129.59 t.m Due to wind force on super structure = 13.61 t.m Due to wind force on live load = 21.98 t.m Due to water current force in long. Dir = 0.26 t.m Total Moment (My) = 198.28 t.m say 199.00 t.m Now stress =
My P Mx ( xm ) ( ym ) A I xx I yy
Maximum compressive stress σ Minimum compressive stress σ
com,max com,min
=
=
13.20
(Compressive) t/m2
0.71
t/m2 =
Case 2 : When Live load on the bridge and No water in canal (i) Vertical dead loads (a) Dead Load of super structure = (b) Dead load of bed block = (c) Dead Load due to self weight of Pier = (d) Increase in reaction due to braking = (e) Live load with impact factor =
183.70 32.03 89.33 2.12 34.60
tonnes tonnes tonnes tonnes tonnes
(f) Dead Load of foundation block =
371.00
So Net vertical downward force (P) =
tonnes
712.78 tonnes
Moment about x-x axis Due to braking force = Due to sliding friction = Total Moment (Mx) = say
81.68 47.81 129.49 130.00
t.m t.m t.m t.m
Moment about y-y axis Due to eccentricity of dead Load of super structure (slab+Beam+Kerbs+Railing) = Due to eccentricity of dead Load of super structure (Gate hoisting platform) = Due to eccentricity of Live Load = 129.59 t.m Due to wind force on super structure = 13.61 t.m Due to wind force on live load = 21.98 t.m Total Moment (My) = 198.02 t.m say 199.00 t.m Now stress =
My P Mx ( xm ) ( ym ) A I xx I yy
Maximum compressive stress σ Minimum compressive stress σ
com,max com,min
=
=
10.06 5.31
t/m2 t/m2 =
< 75 t/m2 (Compressive 0.053
Area of steel 0.3% of gross sectional area as per IRC 21 Provide 20 mm dia main reinforcement at spacing of 190 mm with ties bar 8 mm dia 190 mm.
Design of Prestressed anchors : Design load for anchorage system : 35 tonnes Taking load factor as 1.75 capacity of anchorage system : 61.25 tonnes Minimum breaking load for 12.7 mm dia HT strands : 18737 kg Factor of safety for tension in strand wire : 1.5 Ultimate tensile strength of tension wire : 12491.333 kg No of strands required for anchorage system : 4.9033997 Nos say 5 Nos Hence use 12K13 anchorage system i.e 5 Nos HT wire strands for each anchor. FIXED LENGTH : Fixed length of anchorage bar : Diameter of hole for each anchorage system = 110 mm Permissible Bond stress between strata and grout = 2.9 kg/cm2 Fixed length required for each tendon = 3.84 m say 3.90 m Free length of strand as per IS 10270:1982 5.00 m Total length of anchor for passsing design load : 9.90 m
12.49
Load 11.4 6.8 6.8 6.8 6.8
tonnes
tonnes
surface, so
Nose 45 degree angled
ter flows in alternate bays
m
291.36 -258.52
t.m t.m
for M30 grade concrete N/mm2
eight at a time
291.36 -258.52
t.m t.m
for M30 grade concrete N/mm2
rete grade permissible tensile stress
m
291.36 -258.52
t.m t.m
< 75 t/m2 (Compressive strength of foundation rock) 0.007 N/mm2 O.K.
291.36 -258.52
t.m t.m
ompressive strength of foundation rock) N/mm2 O.K.
tonnes
position wrt heaviest load for L.L
Span 5.6 6.60
position wrt front wheel load for sliding friction cal
0 4.3 7.3
-13.3 -9 -6
10.3 13.3
-3 0 Thickness of slab as per MOST 0.6 0.65
Design of Reinforcement for foundation Block : Bearing capacity of Soil/Rock : Angle of internal friction of soil mass :
75 30.0
kN/m2 degree
Unit Weight of soil : 18.0 kN/m3 Coefficient of friction between soil and concrete : 0.5 Concrete Grade M 30 with σcbc : 10.0 N/mm2 Steel of Grade Fe 415 with σst 230.0 N/mm2 Design constants : For M 30 concrete and Fe 415 steel reinforcement we have the following : Modular ratio m =280/3σcbc= 9.33 Kbal=1/(1+(σst/mσcbc))= 0.289 5 jbal=1-kbal/3 0.904 Rc=0.5σcbckbaljbal= 1.304 Ka= (1-sin)/(1+sin)= +X 0.333 C Consider dimensions of footing to be checked : Length of column in x-x dir = 14.55 m Width of column in y-y dir = 1.15 m Length of footing base in x-x dir = 18.55 m Width of footing base in y-y dir = 5 m -Y 18.55 Moment about x-x direction = 4290.00 kN.m Moment about y-y direction = 1990.00 kN.m Vertical load on column = 6450.40 kN Maximum soil pressure at toe = 131.99 kN/m2 Minimum soil pressure at heel = 7.10 kN/m2 Pressure intensity under column axis = 69.55 kN/m2 Intensity of soil pressure below the column face = 118.525 kN/m2 Cantilever length for bending about critical face = 2.00 m Total force under cantilever length = 1252.57416 kN D Distance of centroid about critical face = 0.51 m 1.15 Bending moment about critical face = 637.51 kN.m Width of section at crtical face = 1.631 m Depth of footing base as required from bending consideration = 547.36 mm Provide depth of footing base as 1540 mm AS PER SCOURING DEPTH CRITERION AND IRC 21 DESIG Effective cover of reinforcement= 60 mm Total depth of footing base at column face = 1600 mm Provide total depth of footing base at end = 1000 mm Check for Shear Force (a) Punching Shear force : For punching shear stress point of view the crritical section occurs at distance d/2 from coloumn face. Position of critical face = 770 mm from coloumn face Eff. Length of critical plane for punching shear stress = 15.32 m Eff. Width of critical plane for punching shear stress = 1.92 m Punching shear force (F) = 4404.74 kN Depth of section at punching shear line of influence = 1.3 m Punching shear stress at section for critical punching shear force= 98.267 kN/m2
Permissible shear stress in punching shear = ks x 0.16 fck =
876.36
kN/m2
The section is safe from punching shear point of view O.K. (a) One way Shear force : For one way shear stress point of view the critical section occurs at Position of critical face for oneway shear = 1540 Cantilever length to right of critical plane = 0.46 Intensity of pressure at critical plane = 128.89296 The section at critical plane will be trapezoidal with Top width of trapezoid = 4.1145 m Effective depth of critical section= 1078 mm Depth of Neutral Axis = kd' = 311.18 mm Width of section at neutral axis level = 6.111 Shear force at critical section = 300.02 kN Shear stress at crtical section = 0.189 N/mm2 Permissible shear stress at section due to direct shear=
distance d from column face. mm from column face m kN/m2
m
0.28
N/mm2
Section is safe in d
Design for bending Tension : Area of steel required for bending (Ast) = Area of steel wire 20 mm dia = 314 No of bars required for bending = 7 Provide 20 mm dia HYSD bars at spacing @ However provide 20 mm dia bars at spacing Provide same spacing of bars in other direction also.
1991.47221 mm2 mm2 Nos 714.3 mm c/c 200.0 mm c/c at bottom of footing
5
+X
A
+Y 14.55
B 1.15
-X
ON AND IRC 21 DESIGN CRITERION
ction is safe in direct shear O.K.
Design of Reinforcement for foundation Block : Bearing capacity of Soil/Rock : Angle of internal friction of soil mass :
75 30.0
kN/m2 degree
Unit Weight of soil : 18.0 kN/m3 Coefficient of friction between soil and concrete : 0.5 Concrete Grade M 30 with σcbc : 10.0 N/mm2 Steel of Grade Fe 415 with σst 230.0 N/mm2 Design constants : For M 30 concrete and Fe 415 steel reinforcement we have the following : Modular ratio m =280/3σcbc= 9.33 Kbal=1/(1+(σst/mσcbc))= 0.289 jbal=1-kbal/3 0.904 Rc=0.5σcbckbaljbal= 1.304 Ka= (1-sin)/(1+sin)= +X 0.333 C Consider dimensions of footing to be checked : Length of column in x-x dir = 6.55 m Width of column in y-y dir = 7 m Length of footing base in x-x dir = 6.55 m Width of footing base in y-y dir = 8.35 m -Y Moment about x-x direction = 27188.98 kN.m Moment about y-y direction = 0.00 kN.m Vertical load on column = 6672.80 kN Maximum soil pressure at toe = 856.79 kN/m2 Minimum soil pressure at heel = 741.48 kN/m2 Pressure intensity under column axis = 791.89 kN/m2 Intensity of soil pressure below the column face = 840.220 kN/m2 Cantilever length for bending about critical face = 1.20 m Total force under cantilever length = 6669.25626 kN D Distance of centroid about critical face = 0.50 m Bending moment about critical face = 3345.48 kN.m Width of section at crtical face = 6.550 m Depth of footing base as required from bending consideration = 625.75 mm Provide depth of footing base as 1540 mm AS PER SCOURING DEPTH CRITERION AND IRC 21 DESIG Effective cover of reinforcement= 60 mm Total depth of footing base at column face = 1600 mm Provide total depth of footing base at end = 1000 mm Check for Shear Force (a) Punching Shear force : For punching shear stress point of view the crritical section occurs at distance d/2 from coloumn face. Position of critical face = 770 mm from coloumn face Eff. Length of critical plane for punching shear stress = 6.55 m Punching shear force (F) = 4442.45 kN Depth of section at punching shear line of influence = 1.2 m Punching shear stress at section for critical punching shear force= 587.218 kN/m2 Permissible shear stress in punching shear = ks x 0.16 fck = 876.36 kN/m2
The section is safe from punching shear point of view O.K.
Design for bending Tension : Area of steel required for bending (Ast) = Area of steel wire 20 mm dia = No of bars required for bending = Provide 20 mm dia HYSD bars at spacing @ However provide 20 mm dia bars at spacing
314 34
10450.7489 mm2 mm2 Nos 245.6 mm c/c 150.0 mm c/c at bottom of footing
8.35
+X
A
+Y 6.55
B 7
-X
ON AND IRC 21 DESIGN CRITERION
Design of R.C.C. Bridge Abutments Design parameters : Canal discharge Q =
207.30
Clear span between piers = Thickness of road way deck slab = FSL of canal D/S of Cross regulator = Width of Pier = Canal foundation block top Level = Lacey's Silt factor (f) =
6.35 0.68 258.97 1.15 254.3 1
Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Thickness of Dirt Wall = Height of Dirt Wall above bed block LVL = Thickness of Bed block = Width of Bed block = Height of Bed Block above foundation block = Height of batter above foundation block= Bottom width of batter at foundation block = Width of foundation block = Height of foundation block = Length of abutment wall = Friction coeff betn base of footing and rock (i) Foundation Level of Piers Normal depth of scour (R) =0.47(Q/f)1/3 Anticipated depth of scour around Abut = 2.0 ( R ) =
75 2 1.75 0.45 0.7 0.6 0.95 5.67 6.37 5.75 8.35 1.6 6.55
cumecs m m m m m t/m2 t/m3 t/m3 m m m m m m m m m m
0.5 2.78
m
5.56
m
Level of foundation of abutment should not be less than 1 m below anticipated depth of scour Level of foundation = 252.41 m say 252.70 m Evaluation of forces acting on the Abutment at El 254.30 m about X-X Axis (A) Vertical Forces
(i) Self weight of Abutment (a) Weight of Triangular section (1)= 299.89 tonnes acting at a distance of 3.17 m from A (b) Weight of Rectangular section (2)= 51.36 tonnes acting at a distance of 1.025 m from A (c) Weight of Rectangular section (3)= 82.14 tonnes acting at a distance of 0.40 m from A (ii) Vertical Reaction due to Live Loads and dead load of super structure (a) Reaction due to dead load of super structure = 98.10 tonnes acting at a distance of 0.4 m from A (b) Maximum live load reaction on abutment is obtained when third axle load 11.4 t at center of bearing of abutm Reaction from right span = 14.48 tonnes
Total Live Load reaction on the abutment from L.L = 14.48 tonnes Impact Factor = 0.34 So Maximum live load reaction with impact = 19.35 tonnes acting at a distance of 0.4 m from A (iii) Vertical force due to wt of earth retained as backfill The soil upto FSL will be saturated while above that it will be a wet soil (a) Wet earth of section (4) = 39.55 tonnes acting at a distance of 4.13 m from A (b) Wet earth of section (5) = 14.95 tonnes acting at a distance of 2.27 m from A (c) Wet earth of section (6) = 82.144 tonnes acting at a distance of 4.892 m from A (d) Saturated earth of section (7) = 128.94 tonnes acting at a distance of 5.59 m from A (iv) Vertical load due to equivalent live load surcharge of 1m Intensity of loading = 0.578 tonnes per sqm Vertical load due to surcharge = 21.75 tonnes acting at a distance of 4.13 m from A (v) Up lift Force Uplift force = 107.06 tonnes acting at a distance of 4.67 m from A (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL 258.97 m Intensity of pressure at FSL = 8.70 t/m Horizontal force = 1/2 x 9.19 x 2.43 = 10.01 tonnes acting at height = 5.44 m above A (b) Wet earth pressure effect below FSL due to wet soil above Intensity of pressure at FSL = 8.70 t/m Horizontal force = 40.63 tonnes acting at height = 2.34 m above A (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom 10.09 t/m Hor. force due to submerge earth pressure = acting at height (b) Water pressure Intensity of pressure at bottom = Horizontal force due to water pressure = acting at height = (iii) Live Load Surcharge Horizontal force = acting at height = (iv) Braking Force Equal to 20% of total live load on the bridge =
23.57 1.56
tonnes m above A
30.59 71.42 1.56
t/m tonnes m above A
26.36 3.49
tonnes m above A
8.36
tonnes
4 5 6
7 1
R.L. of application of braking force = 262.47 Height of application force above bearing level = 1.900 m Increase in reaction due to braking force = 2.12 tonnes (v) Temperature force or due to sliding friction Reaction on sliding end when loads are placed so as to produc maximum reaction at sliding end Live load Reaction at sliding end = 12.24 tonnes Impact factor = 0.34 Live load with impact = 16.36 tonnes Dead Load reaction = 183.70 tonnes Increase due to braking force = 2.12 tonnes Total : 202.18 tonnes Friction in sliding bearing ( with coeff. of sliding friction as 0.03) = 6.07 R.L. of point of application = 260.57 m So acting at height = 6.27 m above A Consider stability of section at Level
Particular of Forces (A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (ii) Vertical Reaction due to L.L and D.L (a) Reaction due to dead load of super structure (b) Reaction due to L.L (iii) Vertical force due to wt of earth (a) Wet earth of section (4) (b) Wet earth of section (5) (c) Wet earth of section (6) (d) Saturated earth of section (7) (iv) Vertical load due to L.L. surcharge (v) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge (v) Temp or due to sliding friction force SUM
254.3 m about X-X Axis
Vertical Force (Tons)
Horizontal Force (Tons)
299.89 51.36 82.14 98.10 19.35 39.55 14.95 82.14 128.94 21.75 -107.06
10.01 40.63
731.11
23.57 71.42 26.36 6.07 178.06
Base width at foundation block top Level (B) Net stabilising moment =
M
254.3
M
7.0
1565.59 ton.m M M
The resultant will strike base at distance (x) =
m
2.14
V
Eccentricity (e) about C.G. of Section at foundation block LVL =B/2 - x = Check eccentricity e as compared to B/6 as (eB/6, so tension will develop at heel Maximum tension at heel =
V
B
1
6e B
1.36 0
-17.19
< 0.67 N/mm
tonnes/m2= Safe
2
As Permissible tensile stress for M30 grade concrete= 0.67 N/mm however steel provided for temperature and shrinkage will take tensile stresses in the concrete mass at bed level 2
V
Maximum compression at toe = Sliding Factor =
H V
Factor of safety in overturning = Factor of safety in sliding =
B
M M
1
6e B
0.24
< N/mm2 < 0.65 Hence Safe
2.69
> 1.50 Hence Safe
2.05
Safe in Sliding
226.07
of bearing of abutment
260.57
F S 2L
258.97
3
A 254.3 8
tonnes
Lever M+ (t.m) arm (m)
3.17 1.025 0.40
949.64 52.64 32.85
0.4 0.4
39.24 7.74
4.13 2.27 4.892268 5.59 4.13 4.67
163.13 33.98 401.87 721.43 89.72
M-(t.m)
499.61
5.44 2.34
54.39 94.87
1.56 1.56 3.49 6.27
36.69 111.18 91.88 38.03 926.66
2492.25
m from A m
-0.17
N/mm2
temperature
ton/m2 Safe
2.26
N/mm2
Design of R.C.C. Bridge Abutments Design parameters : Canal discharge Q =
207.30
Clear span between piers = Thickness of road way deck slab = FSL of canal D/S of Cross regulator = Width of Pier = Canal foundation block top Level = Foundation Level of Abutment = Lacey's Silt factor (f) =
6.4 0.68 258.97 1.15 254.3 252.70 1
Bearing Capacity of Soil = Density of saturated soil= Density of wet soil = Thickness of Dirt Wall = Height of Dirt Wall above bed block LVL = Thickness of Bed block = Width of Bed block = Height of Bed Block above foundation block = Height of batter above foundation block= Bottom width of batter at foundation block = Width of foundation block = Height of foundation block = Length of abutment wall =
75 2 1.75 0.45 0.7 0.6 0.95 5.67 6.37 5.75 8.35 1.6 6.55
Friction coeff betn base of footing and rock (i) Foundation Level of Piers Normal depth of scour (R) =0.47(Q/f)1/3
0.65
Anticipated depth of scour around Abut = 2.0 ( R ) =
cumecs m m m m m m t/m2 t/m3 t/m3 m m m m m m m m m m
2.78
m
5.56
m
Level of foundation of abutment should not be less than 1 m below anticipated depth of scour Level of foundation = 252.41 m say 252.70 m Evaluation of forces acting on the Abutment at El 252.70 m about X-X Axis (A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1)= 299.89 tonnes acting at a distance of 4.37 m from B (b) Weight of Rectangular section (2)= 51.36 tonnes acting at a distance of 2.225 m from B (c) Weight of Rectangular section (3)= 82.14 tonnes acting at a distance of 1.60 m from B (d) Weight of foundation block (8) = 218.77 tonnes acting at a distance of 4.18 m from B (ii) Vertical Reaction due to Live Loads and dead load of super structure (a) Reaction due to dead load of super structure = 98.10 tonnes
acting at a distance of 1.6 m from B (b) Maximum live load reaction on abutment is obtained when third axle load 11.4 t at center of bearing of abutm Reaction from right span = 14.48 tonnes Total Live Load reaction on the abutment from L.L = 14.48 tonnes Impact Factor = 0.34 So Maximum live load reaction with impact = 19.35 tonnes acting at a distance of 1.6 m from B (iii) Vertical force due to wt of earth retained as backfill The soil upto FSL will be saturated while above that it will be a wet soil (a) Wet earth of section (4) = 39.55 tonnes acting at a distance of 5.33 m from B (b) Wet earth of section (5) = 14.95 tonnes acting at a distance of 3.47 m from B (c) Wet earth of section (6) = 82.144 tonnes 4 acting at a distance of 6.092 m from B (d) Saturated earth of section (7) = 128.94 tonnes acting at a distance of 6.79 m from B (iv) Vertical load due to equivalent live load surcharge of 1m 6 Intensity of loading = 0.578 tonnes per sqm Vertical load due to surcharge = 22.32 tonnes 5 acting at a distance of 5.55 m from B (v) Up lift Force Uplift force = 171.46 tonnes 7 acting at a distance of 5.57 m from B 1 (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL of 258.97 m Intensity of pressure at FSL = 8.70 t/m Horizontal force = 10.01 tonnes acting at height = 7.04 m above B (b) Wet earth pressure effect below FSL due to wet soil above Intensity of pressure at FSL = 8.70 t/m Horizontal force = 54.55 tonnes acting at height = 3.14 m above B (ii) Saturated earth pressure (a) Submerged earth pressure Intensity of horizontal pressure at bottom 13.55 t/m Hor. force due to submerge earth pressure = acting at height (b) Water pressure Intensity of pressure at bottom = Horizontal force due to water pressure = acting at height = (iii) Live Load Surcharge Horizontal force =
42.49 2.09
tonnes m above B
41.07 t/m 128.75 tonnes 2.09 m above B 32.42
tonnes
8
8.35
4
acting at height = 4.29 m above B (iv) Braking Force Equal to 20% of total live load on the bridge = 8.36 tonnes R.L. of application of braking force = 262.47 Height of application force above bearing level = 1.900 m Increase in reaction due to braking force = 2.12 tonnes (v) Temperature force or due to sliding friction Reaction on sliding end when loads are placed so as to produc maximum reaction at sliding end Live load Reaction at sliding end = 12.24 tonnes Impact factor = 0.34 Live load with impact = 16.36 tonnes Dead Load reaction = 183.70 tonnes Increase due to braking force = 2.12 tonnes Total : 202.18 tonnes Friction in sliding bearing ( with coeff. of sliding friction as 0.03) = 6.07 R.L. of point of application = 260.57 m So acting at height = 7.87 m above B (vi) Passive Earth pressure of soil retained in front of Abutment Int of Passive pressure at found block top LVL 254.30 0.0 Int Passive pressure at found block bottom LVL 252.70 = 77.01 Total Passive earth pressure on found block 61.61 t Acting at height above found block bottom LVL (252.70) 0.65
Consider stability of section at Level
Particular of Forces (A) Vertical Forces (i) Self weight of Abutment (a) Weight of Triangular section (1) (b) Weight of Rectangular section (2) (c) Weight of Rectangular section (3) (d) Weight of Rectangular section (4) (ii) Vertical Reaction due to L.L and D.L (a) Reaction due to dead load of super structure (b) Reaction due to L.L (iii) Vertical force due to wt of earth (a) Wet earth of section (4) (b) Wet earth of section (5) (c) Wet earth of section (6) (d) Saturated earth of section (7) (iv) Vertical load due to L.L. surcharge
252.70
t/m t/m m
m about X-X Axis Vertical Force (Tons)
299.89 51.36 82.14 218.77 98.10 19.35 39.55 14.95 82.14 128.94 22.32
Horizontal Force (Tons)
(v) Up lift Force (B) Horizontal Forces (i) Wet earth pressure (a) Wet earth pressure above FSL (b) Wet earth pressure below FSL (ii) Saturated earth pressure (a) Submerged earth pressure (b) Water pressure (iii) Live Load Surcharge (v) Temp or due to sliding friction force (vi) Passive Earth Pressure of soil SUM
-171.46
10.01 54.55
667.28
Base width at foundation block bottom Level (B) Net stabilising moment =
M
M
42.49 128.75 32.42 6.07 -61.61 274.27
The resultant will strike base at distance (x) =
254.3
2718.90 M
8.35
m
ton.m
M
V
4.07
The eccentricity (e) about C.G. of Section at foundation block LVL =B/2 Check eccentricity e as compared to B/6 as (e 1.50 Hence Safe
1.58
Safe in Sliding
85.68 2
nts
er of bearing of abutment
260.57
F S L2
258.97
3
A 8
254.3
4
B
tonnes
Lever arm (m)
M+ (t.m)
4.37 2.225 1.60 4.18
1309.51 114.28 131.42 913.36
1.6 1.6
156.96 30.97
5.33 3.47 6.092268 6.79 5.55
210.58 51.93 500.44 876.16 123.86
M-(t.m)
5.57
954.47
7.04 3.14
70.40 171.01
2.09 2.09 4.29 7.87 0.65
88.80 269.09 138.91 47.73 39.8409374 4459.30
1740.41
m from A m
0.74
ton/m2 Safe
N/mm2
0.86
N/mm2
Design Data : Height of counterfort retaining wall above NSL : Depth of foundation of Retaining Wall :
6.78 2.5
m m
Bearing capacity of Soil/Rock : Angle of internal friction of soil mass :
165 20.0
kN/m2 degree
Unit Weight of soil : Coefficient of friction between soil and concrete : Concrete Grade M 20 with σcbc : 7.0 Steel of Grade Fe 415 with σst 230.0
17.5 0.5
kN/m3
N/mm2 N/mm2
Solution : 1. Design constants : For M 20 concrete and Fe 415 steel reinforcement we have the following : Modular ratio m =280/3σcbc= 13.33 Kbal=1/(1+(σst/mσcbc))= 0.289 jbal=1-kbal/3
0.904
Rc=0.5σcbckbaljbal=
0.913
Ka= (1-sin)/(1+sin)=
0.490
2. Dimension of various parts. The ratio of length of toe slab (DE) to the base of width b may be found by the expression α : 1-q α= 0.536 Base width b = 0.95 H SQRT(Ka/(1-α)(1+3α)) b= 5.579 Normal range of b is between 0.5 H to 0.6 H. However, keep minimum b = 0.5 H to 0.6H. However keep minimum length of base slab = 0.5 H = 4.62 m say Length of base slab as = 4.10 m As length of toe slab 2.20 m keep the length of toe slab = 1.6 m Taking thickness of stem = 260 mm length of heel slab BC = 2.24 m Let the thickness of base slab as : 500 mm 1/4 Clear spacing of counterforts l= 3.5 (H/) l= 2.98 m However keep counterforts at 3.00 m apart Let us also provide counterforts over toe slab, upto ground level, at 3 m clear distance.
3. Stability of wall. The preliminary dimensions of the wall are marked as per fig
The calculation are arranged in Table below Force(kN) S.No.
Desig-nation Horizontal Force (H)
1
W1 weight of stem, per metre length.
L.A.(m)
Vertical Force (F)
Moments about toe (k
56.745
1.73
98.17
51.25
2.05
105.06
342.216
2.98
1019.80
3.08
-1123.82
Sum M =
99.21
W2 2
weight of base slab.
3
W3 weight of soil on heel slab.
4
PH Horizontal Earth Pressure Sum F =
365.27 365.27
x= eccentricity e =b/2-x =
0.22 1.83
450.211
Revise Base Width
Pressure P1 under the toe is given by (Sum F/b)(1+6e/b)=
403.82
kN/m2
Pressure P2 under the heel is given by (Sum F/b)(1-6e/b)=
-184.20
kN/m2
Pressure P'1 under point E is =
174.35
kN/m2
Pressure P'2 under point B is =
137.06
kN/m2
4. Design of Heel Slab : Clear spacing between counterforts of Ret wall = Consider a strip of 1m wide near outer edge C.
3.00
m
Upward minimum pressure intensity at outer edge
-184.20
kN/m2
Downward load due to weight of earth is calc as :
152.775
kN/m2
12.5
kN/m2
Downward load due to self weight of base slab :
Net downward intensity of load p is given by : 349.48 kN/m2 As maximum negative bending moment in heel slab, at counterforts is M1 = Pl2/12 M1 =
262.11 Depth of heel slab d = Shear force V is given as :
kN.m 535.77 524.22
Area of steel for balanced section is given as = Percentage of steel % Ast = 0.4393 Hence c= 0.33 N/mm2
mm kN 2353.47
mm2
Depth required from shear point of view d=V/t c b =
1588.54 mm However, keep Overall depth of heel slab as D= 500 mm Effective cover of slab reinforcement provided = 60 mm Effective depth of heel slab d= 440 mm 2 v= N/mm > c hence shear reinforcement will be necessary 1.191 Area of steel at supports is given as Ast = Provide 12 mm dia bars with area of bar Spacing of 12 mm dia bars = 39.45 However Provide 12 mm dia bars at spacing of
2865.75 113.04 mm 130
Actual Area of steel provided=
mm
869.5384615385
Maximum positive B.M.= pl2/16= Area of bottom steel Ast2=
196.58
mm
2
kN.m
652.15 Provide 12 mm dia bars with area of bar Spacing of 12 mm dia bars = 173.33 However Provide 12 mm dia bars at spacing of
mm2 113.04 mm 150
Actual Area of steel provided= Reinforcement Near B
mm
753.60
mm2 mm
mm mm
2
Net downward load near B : 28.22 kN/m2 It is about 0.081 of Load intensity at C Hence spacing of 12 mm dia steel bars near supports at top face is given by 1610.1 However provide steel bars 12 mm dia at supports on top face with spac 300 Spacing of 12 mm dia steel bars at mid span at bottom face is given by = 1857.77 Provide steel bars 12 mm dia at mid span on bottom face with spacing = 300
Distribution steel : Area of distribution reinforcement for temp and shrinkage= 0.12 bD/100 = Using 12 mm bars with area of bar as = Spacing of distribution steel = 188.4 However provide distribution steel at spacing of
113.04 mm 180
600
mm
2
mm
Shear Reinforcement : Shear stress at C is given = Per Steel at C as 100As/bd= Hence c= 0.22
1.191 0.198 N/mm2
N/mm2 % Shear Reinforcement is necessary
Vc=c bd =
96.8 kN Consider a section distance x1 from face of counter fort where shearforce is Check distance x1 = 0.49 m Shear force at distance x1 = 352.44
96.8
Hence provide shear reinforcement upto a distance 0.5 m on either side of counterfort Consider a point at distance y1 form end of heel where SF is valued= 96.8 kN Check distance y1 = 1.25 m Upward Pressure intensity at y1= Net downward pressure at y1=
-4.42
kN/m2
169.69
kN/m2
Shear force at face of counterforts at distance y1 from heel =
254.54
kN
However provide shear reinforcement upto a distance of y 1 =
1.3 m from heel of base slab So provide in rectangular portion of size 1.3 m x 0.5 m on either side of counterfort in heel slab portion Let us provide 4 legged stirrups of 8 mm dia bars having area of401.92 reinf= mm2 So spacing of shear reinforcement is gven as Sv= 95.16 mm However provide shear reinforcement at spacing of600 mm c/c on either side of counterfort
Design of Toe Slab : As toe slab is also lengthy so front counterforts will be necessary Provide front counterforts at toe slab spacing o 3.00 m c/c Depth of toe slab =
500
mm
Upward maximum pressure intensity at edge D
403.82
kN/m2
Downward load due to self weight of base slab :
12.5
kN/m2
391.32
kN/m2
Net upward intensity at edge D is given by :
Net upward intensity at edge E is given by : 161.85 kN/m2 Consider a strip of unit width at D. Maximum Negative moment is given by M1 = wl2/12 M1 =
293.49 Depth of toe slab d = Shear force V is given as :
kN.m 566.94 586.98
Area of steel for balanced section is given as = Percentage of steel % Ast = 0.5648 Hence c= 0.33 N/mm2
mm kN 2823.78
mm2
Depth required from shear point of view d=V/t c b =
1778.72 mm However, keep Overall depth of heel slab as D= 500 mm Effective cover of slab reinforcement provided = 60 mm Effective depth of heel slab d= 440 mm v= N/mm2 > c hence shear reinforcement will be necessary 1.334 Area of steel at supports is given as Ast = Provide 12 mm dia bars with area of bar Spacing of 12 mm dia bars = 35.23 However Provide 12 mm dia bars at spacing of
3208.84 113.04 mm 90
Actual Area of steel provided=
mm
Maximum positive B.M.= wl /16= Area of bottom steel Ast2= 2
1256 220.12
mm2 mm mm
2
kN.m
942.00 Provide 12 mm dia bars with area of bar Spacing of 12 mm dia bars = 120.00 However Provide 12 mm dia bars at spacing of
mm2 113.04 mm 120
Actual Area of steel provided= Reinforcement Near E
942.00
mm
Net upward load at 1 m from E :
305.27
kN/m2
2
mm mm
It is about 0.780 of Load intensity at C Hence spacing of 12 mm dia steel bars near supports at bottom face is given by However provide steel bars 12 mm dia at supports on bottom face with Spacing of 12 mm dia steel bars at mid span at top face is given by = Provide steel bars 12 mm dia at mid span on top face with spacing =
115.4 110 153.83 140
Distribution steel : Area of distribution reinforcement for temp and shrinkage= 0.12 bD/100 = Using 12 mm bars with area of bar as = Spacing of distribution steel = 188.4 However provide distribution steel at spacing of
113.04 mm 180
600
mm
2
mm
Shear Reinforcement : Shear stress at D is given = Per Steel at D as 100As/bd= Hence c= 0.24
1.334 0.2855 N/mm2
N/mm2 % Shear Reinforcement is necessary
Vc=c bd =
105.6 kN Consider a section distance x1 from face of counter fort where shearforce is Check distance x1 = 0.68 m Shear force at distance x1 = 320.97 Hence provide shear reinforcement upto a distance 0.7 Consider a point at distance y1 from end of toe where SF is valued= Check distance y1 = Upward Pressure intensity at y1=
2.8 6.00
105.6
m on either side of counterfort 105.6
kN
m kN/m2
Net upward pressure at y1=
4.75 kN/m2 Shear force at face of counterforts at distance y1 from toe =
7.13 kN It is more than length of toe slab 1.6 m hence S.F. at E is more than value Actual S.F. at E is given by 259.64 kN Consider a section distance z from face of counter fort where shearforce is 105.6 Check z = 0.44 m SF at section distant z from face of front counterfort= 183.06 kN So provide in trapezoidal portion at distant 0.45 m and 0.7 m from face of counterfort in toe slab port Let us provide 4 legged stirrups of 8 mm dia bars having area of reinf= 401.92 mm2 So spacing of shear reinforcement is given as Sv= 84.50 mm However provide shear reinforcement at spacing of400 mm c/c on either side of counterfort Design of Stem (Vetical Slab) The stem acts as a continuous slab. Consider 1 m strip at B. Intensity of earth pressure at B, ph=KaH1= 74.9 2 Negative B.M. in Vertical slab near back Counterforts M 1=ph.l /12= So depth of vertical slab (d) = 247.97 mm 200 However provide effective depth of vertical slab as Effective Cover of slab main reinforcement 60
kN/m2 56.15 mm mm
kN.m
Total Depth of vertical slab 260 mm Shear Force V at supports = 112.29 kN v= 0.56 N/mm2 Area of steel near counterforts= 1350.51 mm2 Area of steel @ 0.5% of reinforcement = 1000 mm2 > 1350.51 Checkc at 0.5% of steel reinforcement is greater than v= 0.56 N/mm2 O.K. Spacing of 12 mm dia bars = 113.04 mm However provide 12 mm dia main reinforcement bars at spacing = 90.0 mm Actual Area of steel provided = 1256 mm2 Percent of steel area (%) = 0.63 Max. Positive B.M. = 3/4 M1= 42.11 kN.m Area of steel for positive mid span moment in slab = 942 mm2 Spacing of 12 mm dia bars = 120 mm However provide 12 mm dia bars at spacing of 120 mm Actual Area of steel provided = 942 mm2 The spacing of reinforcement may be increased gradually upwards as earth pressure decreases with Ht Spacing of 12 mm dia bars at top level of Stem = 300 mm Area of Distribution Steel = 0.12 b D /100 = 312 mm2 Let us choose Steel 10 mm dia bars for dist reinf area 78.5 mm2 Spacing of distribution steel = 251.60 mm Hence provide distribution steel at spacing of 230 mm c/c
Design of Main Counterforts
Let us assume thickness of main counterforts as 500 mm Spacing of counterforts center to center 3.50 m At any section below depth h from top A, active earth pressure on Counterfort : 30.01 Simillarly net downward pressure on heel at C = 349.48 kN/m2 Simillarly net downward pressure on heel slab at B = 28.22 kN/m2 Hence reaction transferred to each counterfort at C = 1223.18 kN Hence reaction transferred to each counterfort at B = 98.76 kN The critical section for counterfort will be at F since below this enormous depth is available by front counte Pressure intensity at section F is given as, h= 6.78 m = 203.49 Shear Force at F is given by 689.83 kN B.M. at Section F is given by 1559.01 kN.m Counterfort acts like a T-Beam, however even as a rectangular Beam depth required (d)= Total depth required from bending moment point of view = 1908.00 mm However keep overall depth D 1720 mm Effective depth at Section F = 1660 mm Angle of face AC is given by tan 0.26 14.39 degree Sin = 0.248 Cos 0.969 Depth F1G1 of counterfort = 1684.23 mm Depth FG of Counterfort =
1944.2
mm
Which is more than required O.K.
Assuming reinf with 20 mm dia bars is provided in 2 layers with 20 mm space betn bars and nominal cover Effective depth at Section F = 1854.2 mm So Area of steel to be provided = 4044.78 Using 20 mm dia bars As= 314.00 No of bars to be provided = 14 Provide these in two layers. Effective shear force =Q-(M/d') tanwhere d'=d/cos S.F. = 480857.5 N Nominal shear stress at section = 0.52 100 As/bd at section is given as : 0.474 Hence c= 0.236 N/mm2
mm2 mm2 Nos
N/mm2
Thus shear stress at section is more than v . However the horizontal and vertical ties will take care of exce Height where half of the reinforcement can be curtailed is giv AI1= 3.14 m AG1= 7.00 m I1G1= 3.86 m O.K.
3.04
m below A at point I1
Design of horizontal Ties
The vertical stem has a tendency to separate out from the counterfort so it should be tied by horizontal ties Active Earth Pressure at height 6.78 m below A is 203.49 kN/m Steel area required for horizontal ties at level above :
884.74
mm2 per m height of vertical stem
Using 10 mm dia bars with 2 legged ties, A(10)= 157 mm2 Spacing of 2 legged bars horiz along counterfort : 177 mm Provide 2 legged horizontal stirrups at spacing of 250-300 mm c/c along main counterforts, tow
Design of Vertical Ties The heel slab has a tendency to separate out from the counterfort so it should be tied by vertical ties. Net downward force at C 1223.2 kN Steel area required for horizontal ties at level above : 5318.15 mm2 per m length of heel slab at C Using 12 mm dia bars with 2 legged ties, A(12)= 226.08 mm2 Spacing of 2 legged bars vertically along counterfort : 42 mm Provide 2 legged vertical stirrups at spacing of 130 mm c/c along main counterforts C. Net downward force at B 98.8 kN Steel area required for horizontal ties at level above : 429.40 mm2 per m length of heel slab at C Using 12 mm dia bars with 2 legged ties, A(12)= 226.08 mm2 Spacing of 2 legged bars vertically along counterfort : 526 mm Provide 2 legged vertical stirrups at spacing of 300 mm c/c along main counterforts at B. So spacing can be gradually increased from 130 mm at C to 300 mm at B
Design of front counterforts :
Consider width of front counterfort to be provided : 500 Upward pressure intensity at D : 403.82 kN/m2 Upward pressure intensity at E : 174.35 kN/m2 Considering weight of toe slab net upward intensity of pressure At D : 391.32 kN/m2 At E : 161.85 kN/m2 C/C spacing of front counterfort as proposed : 3.50 Upward force transmitted to counterforts at D per m = 1369.62 Upward force transmitted to counterforts at E per m = 566.46 Total upward force on front counterfort transmitted through toe slab distance of center of pressure from E is given as : 0.91 B.M. at E = 1410.43 kN.m Depth of front counterfort required to resist bending: 1757.65 However provide total depth of 1100 mm Effective depth available with 80 mm effective cover : 1020 Area of steel required Ast = 6652.13 mm2 Using 20 mm dia bars A(25)= 490.63 mm2 No of bars required for bending : 14 Nos These bars should be continued to a distance 45 beyond E = tan= 0.375 Net shear force at E is given as : F-(M/d) tan = 1030.32 Nominal shear stress at section at E is given as : 2.02 100As/bd= 1.347 %
mm
m kN/m length along toe slab kN/m length along toe slab : 1548.86 kN m mm mm
1125
mm beyond E
kN N/mm2
So c=
0.285 N/mm2 Shear reinorcement is necessary Using 12 mm dia 2 legged stirrups with Asv = 226.08 mm2 Vc= 145350 N Vs=V-Vc= 884972.64 N 6.8 Spacing of shear reinforcement = 59.93 mm However provide shear reinforcement at spacing of 200 mm c/c NSL 2.5
E 20.6
D 0.5
1.6
Fixing effects in stem, toe and heel slab Fixing moments are induced at junctions of stem, toe and heel slabs. (i) In stem @ 0.24% of cross section to be provided at the inner face of stem for a length of 45 As= 624 mm2 Use 10 mm dia bars area A(10)= 78.5 mm2
Spacing = 125.80 mm However provide 10 mm dia fixing bars at spacing o 100 mm (ii) In Toe slab @ 0.12% of cross section to be provided at the lower face of slab for a length of 45 As= 600 mm2 Use 10 mm dia bars area A(10)= 78.5 mm2 Spacing = 130.83 mm However provide 10 mm dia fixing bars at spacing o 120 mm (iii) In Heel slab @ 0.12% of cross section to be provided at the upper face of slab for a length of 45 As= 600 mm2 Use 10 mm dia bars area A(10)= 78.5 mm2 Spacing = 130.83 mm However provide 10 mm dia fixing bars at spacing o 120 mm
Design of Shear Key The wall is unsafe in sliding. Therfore provide shear key. Let the depth of shear key to be provided is "a" Check "a": 0.3 m Passive earth pressure in front of shear key is : 83.91 kN Weight of soil in between base slab and shear key bottom level = 21.525 Total downward force acting at shear key bottom level, for friction = 471.736 Total frictional and passive earth pressure resistance developed = 319.78 Net Horizontal force acting for sliding of Retaining wall upto S.K. LVL= 389.40 Factor of safety for sliding = 0.82 However provide shear key with Depth = 450 mm Width = 500 mm Actual horizontal force resisted by shear key is : 348.24 kN Shear stress at bottom face of base slab at shear key = 0.696 N/mm2 Bending stress at bottom face = 1.880 N/mm2 safe
kN kN kN kN
safe
wing :
α : 1-q 0/2.2H
per fig
ments about toe (kN-m)
98.17
105.06
1019.80 -1123.82 99.21
mm mm mm mm mm2
kN
ounterfort
heel of base slab slab portion
nterfort
mm mm mm mm mm2
kN
ounterfort
105.6
kN
rt in toe slab portion
nterfort
mm2
eases with Ht
h
kN
e by front counterfort kN/m
1847.91
equired O.K.
mm
nd nominal cover of 50 mm
take care of excess stress. A at point I1
by horizontal ties. vertical stem
ounterforts, towards top.
vertical ties. heel slab at C
orts C. heel slab at C
orts at B.
oe slab oe slab
ond E 0.26
A All dim in metre 14.39
sary
f 45
B
4.10
2.24
C
h of 45
gth of 45
