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A reinforced concrete cantilever retaining wall is supporting a backfill of height 4.5 m above ground.

Take density of soil =18 kn/m3. Angle of repose=30. SBC of soil=200 kn/m3 and coefficient of friction between concrete and soil=0.45. Grade of concrete is M20 and steel is Fe415.

a) Design retaining wall and show all stability checks.

b) Draw reinforcement details of toe and stem with curtailment of reinforcements.

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Step 1. Design Constants

Height of backfill above the ground = $4.5 \ m$

Density of soil = $\gamma = 18 KN/m^3$

Angle of repose = $\phi = 30˚$

SBC of soil = $q = 200 KN/m^3$

Coefficient of friction = $\mu = 0.45$

Grade of conc. M20 and steal Fe 415

$K_a = \frac{1 - sin \phi }{1 + sin \phi} = \frac{1- sin (30)}{1 + sin (30)} = \frac{1}{3}$

$K_p = 3$

Moment = $0.36 fck \ x b \ (d – 0.42x)$

$X_u$ limit = $0.48 d$ for Fe 415

$M_u$ limit = $0.138 fck \ bd^2$

Step 2. Preliminary design

Since in the question height above the G.L is given we need to calculate the Depth of Foundation.

$\therefore D.O.F = \frac{SBC}{\gamma} \times ka^2 = \frac{200}{18} \times (\frac{1}{3})^2 = 1.23 \simeq 1.3m$

$\therefore$ Total height of the retaining wall is 5.8m

$\therefore$ He = 5.8 m

1. Stem wall = $\frac{He}{15} \text{ to } \frac{He}{10} = \frac{5.8}{15} \text{ to } \frac{5.8}{10}$

$\therefore$ Provide the stem having depth 0.5m at the bottom which is reduced to 0.2m at the top.

2. Base width (b) = 0.55 He to 0.75 He = 3.19 to 4.35m

$\therefore$ provide Base width of 4.3m

3. Toe projection = $\frac{b}{3} = \frac{4.3}{3} = 1.43m$

4. Base slab thickness = $\frac{He}{15} \text{ to } \frac{He}{10} = 0.387 \text{ to } 0.58$

$\therefore$ Provide 0.5m

5. Heal projection = 4.3 – 1.43 – 0.5 = 2.37 m Step 3. Lateral forces and overturning moment

Lateral force $(P_1)$ = Area of pressure distribution dia as shown.

$\therefore L.F = \frac{1}{2} \times (K_a \ rH) \times H$

$= \frac{1}{2} \times \frac{1}{3} \times 18 \times 5.8 \times 5.8 = 100.92 KN$ ---------(1)

This force i.e. $P_1$ will act at height $\frac{H}{3}$ from bottom.

$\therefore \text{Lever arm distance} = \frac{5.8}{3} = 1.933$

Overturning moments = $100.92 \times \frac{5.8}{3} = 195.112 KNm$ --------(2)

Step 4. Gravity forces and Resisting moment

Assume the unit weight of concrete no. be $25 KN/m^3$ and 1m width.

Symbol Force (KN) Distance (m) from toe Moment (KNm)
W1 $25 \times 1 \times 0.2 \times 5.3 = 26.5$ $1.43 + 0.3 + \frac{0.2}{2} = 1.83$ $48.5$
W2 $\frac{1}{2} \times 0.3 \times 5.3 \times 1 \times 25 = 19.875$ $1.43 + \frac{2}{3} \times 0.3 = 1.63$ $32.4$
W3 $25 \times 4.3 \times 0.5 \times 1 = 53.75$ $\frac{4.3}{2} = 2.15$ $115.56$
W4 $18 \times 1 \times 5.3 \times 2.37 = 226.1$ $4.3 - \frac{2.37}{2} = 3.115$ $704.3$

Total gravity forces = 326.225 KN ---------(3)

Total resisting moment = 900.76 KNm ------ (4)

Step 5. Check for stability

A] check for over turning -

$\frac{\text{Resisting moment}}{\text{overturning moment}} = \frac{\text{From 4}}{\text{From 2}} = \frac{900.76}{195.112} = 4.62 \gt 1.55 \therefore$ safe

B] check for sliding -

$\frac{u R_v}{R_H} = \frac{u \times \text{from 3}}{\text{from 1}} = \frac{0.45 \times 326.225}{100.92} = 1.45$

1.45 < 1.55 $\therefore$ UNSAFE

Since the sliding check has failed we will or we have to provide shear key.

C] Check against soil pressure -

1. Check for eccentricity -

Let $\mathbf{x}$ be the distance of point of application of the resultant.

$x = \frac{\text{Net moment}}{\text{Gravity forces}} = \frac{4 – 2}{3} = \frac{900.76 – 195.112}{326.225} = 2.163 m$

Eccentricity (e) = $\frac{b}{2} = x = \frac{4.3}{2} - 2.163 = - 0.013$

Negative sign indicates that the resultant lies on the other side of center.

As per the cause

$e \gt \frac{b}{6} = \frac{4.3}{6} 0.72 \therefore \text{Safe}$

2. Check for maximum and minimum pressure at toe sheer.

$P_{max} = \frac{\sum W}{b} [ 1 + \frac{6e}{b} ] = \frac{326.225}{4.3} [ 1 + \frac{6 \times 0.013}{4.3}]$

$77.24 KN/m^3 \gt SBC = Q = 200 KN / m^3 \therefore \text{Safe}$

$P_{min} = \frac{\sum W }{b} [ 1 - \frac{6e}{b}] = 74.5 \gt 0 \therefore \text{Safe}$ Step 6. Design of Toe Slab A] Calculation of forces and moment on toe –

1. Upward pressure acting on the

$toe = (\frac{76.33 + 77.24}{2}) \times 1.43 = 109.8$

This pressure force will act at the C.G of pressure distribution dia.

$\therefore$ point of action = $(\frac{a+2b}{1+b}) \frac{1}{3} = (\frac{76.33 + (2 \times 77.24)}{76.33 + 77.24}) \frac{1.43}{3} = 0.7164m$

2. Self-weight of Toe slab

$\therefore force = 25 \times 1 \times 0.5 \times 1.43 = 17.875 KN$

Acting at = $\frac{1.43}{2} = 0.715$

$\therefore$ take moment of these forces towards the junction of toe and stem

$\therefore moment = (109.8 \times 0.7164) – (17.875 \times 0.715) = 65.88 KNm$

Ultimate moment = $98.82 KNm (65.88 \times 1.5)$

B] Check for depth -

Depth provided = 500 - 60 (cover = 60mm) = 440mm.

Depth required $d = \sqrt \frac{M}{0.138 fck \ b} = \sqrt \frac{98.82 \times 10^6}{0.138 \times 20 \times 1000}$ = 189.22 mm

$D_q \lt d$ provided $\therefore$ Safe.

C] Calculation of area of steel required -

$Ast = \frac{0.5 fck}{fy} \bigg[ 1 - \sqrt{1 – (\frac{4.6 M}{fck \ bd^2})} \bigg] bd$

$= \frac{0.5 \times 20}{415} \bigg[ 1 - \sqrt{(1 - \frac{4.6 \times 98.82 \times 10^6}{20 \times 1000 \times 440^2})} \bigg] 1000 \times 440$

$= 641.8 mm^2$

Let’s provide bar of 12 mm $\phi$

$\therefore spacing = \frac{\frac{\pi}{4} \times 12^2}{641.8} \times 1000 = 176mm$

$\therefore \text{provide # 16mm @ 175mm c/c}$

Ast provided $= 646.27 mm^2$

D] Check for shear -

1. Shear at critical section which is at distance d from the junction is

$= (\frac{76.61 + 77.24}{2}) (1.43 – 0.44) - [25 \times 1 \times 0.5 \times (1.43 - 0.44)]$

$\therefore$ shear at c.s = 63.78 KN

Ultimate shear = 95.67 KN

2. Shear taken by concrete

$V_c = \tau c \ bd$

% of steel = $\frac{646.27}{1000 \times 440} \times 100 0.146$

From IS 4562000 page 73 for $\tau c = 0.28$

$\therefore V_c = 0.28 \times 1000 \times 440 = 123.2 \times 10^3 N = 123.2 KN$

$\therefore$ since ultimate shear < shear taken by conc. $\therefore$ safe.

E] Distribution steal -

Generally 0.12% of gross section is provided

$\therefore$ D.S Ast = $\frac{0.12}{100} \times 1000 \times 500 = 600 mm^2$

$\therefore \text{provide # 10mm @ 130mm c/c}$

Step 7. Design of heal slab A] Calculation of forces and moment

1. Upward pressure force = $(\frac{76.01 + 74.5}{2}) \times 2.37 = 178.35 KN$

Distance = $(\frac{76.01 + (2 \times 74.5)}{76.01 + 74.5}) \times \frac{2.37}{3} = 1.181 m$

2. self-weight of slab force = 25 x 1 x 0.5 x 2.37 = 29.625 KN

Acting at 1.185 m $(\frac{2.37}{2})$

3. Force or weight of soil above the heel

= 18 x 1 x 5.3 x 2.37 = 226.098 KN

Acting at 1.185 m

$\therefore$ Total vertical shear force = 77.4 KN

Ultimate shear = 116.1 KN

Total moment = 92.4 KN m

Total ultimate moment = 138.6 KN m

B] Check for depth -

d provided = 440 mm

$d = \sqrt\frac{138.6 \times 10^6}{0.138 \times 20 \times 1000} = 224.1 mm$

$D_{req} \lt d$ provided $\therefore$ safe.

C] Calculation of Ast -

$Ast = (\frac{0.5 \times 20}{415}) \times \bigg[ 1 - \sqrt{ 1 - \frac{4.6 \times 138.6 \times 10^6}{20 \times 1000 \times 440^2}} \bigg] 1000 \times 440 = 912.13 mm^2$

$\therefore$ provide 12mm dia ban

Spacing $= \frac{\pi / 4 \times 12^2}{92.13} \times 1000 = 123.99$

$\text{# 12mm @ 120mm c/c}$

$Ast \ prov = 942.5mm^2$

D] Check for shear -

1. Shear at critical section i.e. the junction of heel and stem = 116.1 KN

2. Shear taken by conctere = $\tau cbd$

$\% = \frac{942.5}{1000 \times 440} \times 100 = 0.2142$

From IS Code $\tau c = 0.28$

$V_c = 123.2 KN$

Since shear at junction is less than shear taken by concrete $\therefore$ safe.

E] Distribution steel-

(0.12% of gross section) Same as toe

$\text{# 10mm @ 130mm c/c}$

Step 8. Design for stem A] Force and moment

Lateral force = $\frac{1}{2} \times K_a \ rH \times H$

$= \frac{1}{2} \times \frac{1}{3} \times 18 \times 5.3^2 = 84.27 KN$

Acting at $= 1.77 (\frac{5.3}{3})$

$\therefore$ sheer = 84.27 KN

Ultimate sheer = 126.4 KN ultimate moment = 223.74 KNm

Moment = 149.16 KNm

B] Check for depth -

$d = \sqrt \frac{m_v}{0.138 fck \ b} = 429.2 \lt 440 \therefore \text{safe}$

C] calculation of Ast

$Ast = (\frac{0.5 \times 20}{415}) \times \bigg( 1 - \sqrt{ \frac{4.6 \times 223.74 \times 10^6}{20 \times 1000 \times 440^2}} \bigg) 1000 \times 440$

Ast $1517.73 mm^2$

$\therefore$ provide 16mm $\phi$

$Spacing = \frac{\pi /4 \times 16^2}{1517.73} \times 1000 = 132.5mm$

$\text{# 16mm @ 130mm c/c}$

Ast provided = 1546.63

D] Check for shear

1. Shear at junction = 126.4 KN

2. Shear resist by concrete

$\% = \frac{1546.63}{1000 \times 440} \times 100 = 0.35$

$C_c = 0.36$

$V_c = 0.36 \times 1000 \times 440 = 158.4 KN$

$\therefore V_u \lt Vc \therefore safe$

E] Distribution steel -

Same as Toe $\text{# 10mm @ 130mm c/c}$

Step 9. Curtailment of ban in stem

$\frac{ASH}{Ast2} = \frac{1}{2} = \frac{y^2}{5.3^2}$

Y = 3.75m

But as per IS code bar has to be extended for 12 $\phi$ or d whichever is greater.

$\therefore$ y = 3.31 m $\therefore$ curtail bar from 3.3m from top

Step 10 - Design of shear key

Provide shear key of $0.5 \times 0.5m$ 1. Lateral force = $\frac{1}{2} \times \frac{1}{3} \times 8 \times 6.3 \times 6.3 = 119.07 KN$

2. Additional weight of soil = $18 \times 4.3 \times 0.5 = 38.7 KN$

Total vertical force = 326.225 + 38.7 = 364.925 KN

Passive earth pressure $PP = k_p \times \text{depth of shear key} \times \text{upward pressure at junction of toe}$.

$= 3 \times 0.5 \times 76.33 = 114.5 KN$

Check for sliding

$\frac{\mu \sum W + P_p}{lateral force} = \frac{0.45 \times 364.92 + 114.5}{119.07}$

= 2.34 > 1.55 $\therefore$ Safe

Step 11. Reinforcement Diagram 0