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

Civil Eng > Sem 8 > Design and Drawing of Reinforced Concrete structure

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modified 6 months ago by gravatar for Yashbeer Yashbeer170 written 8 months ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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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

enter image description here


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}$

    enter image description here


Step 6. Design of Toe Slab

enter image description here

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

enter image description here

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

enter image description here

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 $0.5 \times 0.5m$

enter image description here

  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

enter image description here

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modified 6 months ago  • written 6 months ago by gravatar for Yashbeer Yashbeer170
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