10-Retaining Wall and Piles

8/17/2019 10-Retaining Wall and Piles

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(10-RET. WALL & RET. PILES ) ENG. BAHI ABU ZAID MO:- 01111073437Page

DESIGN

10 – RETAINING WALL AND RETAINING

PILES


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(10-RET. WALL & RET. PILES ) ENG. BAHI ABU ZAID MO:- 01111073437Page2

Retaining wall

-. اد

ص

وخدم

ار

واب

د

خدم

وا ط

و

TYPES OF RETAINING WALL:-

- GRAVITY WALL:- like masonry wall , R.c. wall and

counterfort wall

Masonry wall


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R.c. wall

Counterfort wall


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Stability of rigid walls

Rigid wall failures due to

- Sliding failure

- Over turning failure

- Bearing capacity failure

States of equilibrium

- Earth pressure at rest

- Active earth pressure

- Passive earth pressure


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• Active earth pressure coefficient (Ka):

It is the ratio of horizontal and vertical principal effective

stresses when a retaining wall moves away (by a small amount)

from the retained soil.

• Passive earth pressure coefficient (Kp):


stresses when a retaining wall is forced against a soil mass.

• Coefficient of earth pressure at rest (K0):


stresses when the retaining wall does not move at all, i.e. it is

“at rest”.


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Stability analysis:-

- Sliding failure

- Over turning failure

- Bearing capacity failure


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Check of sliding:-

F.O.S = £V * ( tan α ) / £h

F.O.S . > OR = 2


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Check of overturning:-

F.O.S . > OR = 2


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Check of Bearing capacity

When £ moment = net moment

= resisting moment – disturbed

Moment

Pressure on foundation soil < B/C


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Types of retaining wall

ا ون ا ط اد وح رر واد ن اين او ون-

وودة

ون

د

وادة

ررا

غير

وح

غير

ا ط

ارض

را ط ي با ط


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(10-RET. WALL & RET. PILES ) ENG. BAHI ABU ZAID MO:- 01111073437Page

design retaining wall-Ex:

-س

اد

ط

ا

وادة

يدا

ادة

ك

رض

وا

ةدووا داوا خ

- مينيو ر ودا ل1.ا1ض رر دة

- Check of Sliding failure

- Check of Overturning failure

- Check Bearing capacity failure


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Stability forces

W1 = 0.25 *3*2.5 = 1.875 t/m

W2 = 1.5*0.6*2.5= 2.25 t/m

W3 = 1.9*0.4*2.2 = 1.672 t/m

W4 = 1.25*3*1.8 = 6.75 t/m

W5 = 0.4*3.6*1.8 = 2.6 t/m

Total forces N = 15.15 t/m


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Earth pressure forces

Ka = ( 1-sin ϕ ) / ( 1+sin ϕ )

Φ from soil report ( 30 to 45 )Ka = ( 1-sin 30 ) / ( 1+sin 30 ) = 0.5 / 1.5 = 0.33333

e 1 = q * Ka = 0.3 * 0.33333 = 0.1 t/m2

e2 = ɣ * H * ka = 1.8 * 4 * 0.3333 = 2.4 t/m2

then get F1 and F2

F1 = 0.1*4 = 0.4 t/m

F2 = 0.5 * 4 * 2.4 = 4.8 t/m

horizontal forces = 0.4+4.8 = 5.2 t/mTotal

Check of slidingN = 15.15 t/m

H = 5.2 t/m

F.O.S = £V * ( tan α ) / £h

( tan α ) = 0.5 to 0.7

F.O.S = 0.7 * 15.15 / 5.2 = 2.04 > 2

OK SAFE


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of over turningCheck

Stability forces

MOMENT ) O1ARMN

3.321.775W1 = 1.875

2.61.15W2 = 2.25

1.60.95W3 = 1.672

6.91.025W4 = 6.75

0.520.2W5 = 2.6

14.94TOTAL

forcesOver turning

MOMENT ) O1ARMH

0.82F1 = 0.4

6.41.33F2 = 4.8

7.2TOTAL

F.O.S = 14.94 / 7.2 = 2.08 ( OK SAFE )


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of stressCheck

M net = 14.94- 7.2 = 7.74 t.m.

X = Mnet / N = 7.74 / 15.15 = 0.51 m

e = ( B/2 ) – X = ( 1.9/2) – 0.51 = 0.44 m

q1 = (-N / B )*( 1+6e/B ) = ( -15.15/1.9)*(1+6*0.44/1.9)

q1 = - 19 t/m2 ( comp.)

q2 = (-N / B )*( 1-6e/B ) = ( -15.15/1.9)*(1-6*0.44/1.9)

q2 = + 3.1 t/m2 ( tension )

q2/q1 = 3.1 / 19 = 0.16 < 0.3333 ( ok )

if q2/q1 > 0.33333 ( unsafe )

q* = 19 + 3.1 = 22.1 t/m2 < B/C ( ok safe )


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Design of R.c. footing

Stability forces

MOMENT ) O2ARMN

2.581.375W1 = 1.875

1.70.75W2 = 2.25

4.20.625W4 = 6.75

8.5TOTAL

forcesOver turning

MOMENT ) O1ARMH

0.651.8F1 = 0.1*3.6=0.36

4.71.2F2 =

(1.8*3.6*0.33)*0.5*3.6

=3.9

5.35TOTAL

Mnet = 8.5 -5.35 = 3.15 t.m.

N = 10.9 ton

X = 3.15/10.9 = 0.29 m

e = ( 1.5/2 ) – 0.29 = 0.46 m


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q1 = (-N / B )*( 1+6e/B ) = ( -10.9/1.5)*(1+6*0.46/1.5)

q1 = 20.65 t/m2 ( comp.)

q2 = (-N / B )*( 1+6e/B ) = ( -10.9/1.5)*(1-6*0.46/1.5)

q2 = 6.1 t/m2 ( tension )

q* = 26.75 t/m2 ,,,, arm = 1.25 m

M = 26.75 * 1.25*1.25/2 = 20.9 t.m.

Mu = 20.9*1.5 = 31.35 t.m , d = 55cm

As = 31.35*100000/3600/55/0.8 = 20 cm2

Use 8 bars 18 mm / m (bot. reinforcement )

of R.c. wallDesign

Reinforcement from h.l. forces

e 1 = 0.1 t/m2

e 2 = 1.8*3*0.333 = 1.8 t/m2

F1 = 0.1 *3 = 0.3 t/m

F2 = 0.5*1.8*3 = 2.7 t/m

M = 0.3 * 1.5 + 2.7 * 1 = 3.15 t.m.

Mu = 3.15*1.5 = 4.725 t.m.

For t = 25 cm ,,, d = 20 cm

As = 4.725 *100000/3600/0.8/20 = 8.2 t.m

Use 5bars dim 16mm /m


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Notes

-د

ا ط

وود

نوي

ر

واد

live loadsيت

33

/ م /م1.ر ار ث ونb/cو ي2م او2م

1 / و ذه ا ج ا ط رر يرة وي ار 2طن

-يوا هي دوو يا موزا ريو يا ىوا ريي

ار

يو

يرة

رر

ا

يا

وج


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e1 = q * Ka = 0.3 * 0.3333 = 0.1 t /m2

e2 = ɣ * h * Ka = 1.8 *2 * 0.333 = 1.2 t/m2

e3 = ɣsub * h * Ka = 0.8 * 2 * 0.3333 = 0.5333 t/m2

e water = ɣ water * h = 1 *2 = 2 t/m2

F1 = 0.1 * 4 = 0.4 t/m

F2 = 0.5 * 2 * 1.2 = 1.2 t/m

F3 = 2 * 1.2 = 2.4 t/m

F4 = 0.5 * 2 * 0.5333 = 0.5333 t/m

F5 = 0.5 *2 *2 = 2 t/m


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Retaining piles

خدم د واب ار ل ي ار ويم صو خوازيق-

ي ار ثا وا لخاد ا بارا ليي وا

) ) ان ودت وا لخاد برا ن يوا هيا

ل ظ اشوارع اي ن اير-

ل ظ ات ار ن اير-

- اريد

داخل

اخزوق

ارول

ارع

وصف

رة

يون اطول ا خزوق يوى Hاذا ن ارع ار-

L = H + 1.5 H = 2.5 H

ي اخزوق ي ن ال اي ا ن د اراب-

-ايويت

ن

واخر

رخ

ادا

او

رخ

اخوازيق

ون

يو هي دوو نوي ريخاو


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Ex : design retaining pile for 4m height of

excavation

Solution

L = 4 + 1.5 * 4 = 4 +6 = 10 m

Assume d=30 cm


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Earth pressure forces

Ka = ( 1-sin ϕ ) / ( 1+sin ϕ )

Φ from soil report ( 30 to 45 )Ka = ( 1-sin 30 ) / ( 1+sin 30 ) = 0.5 / 1.5 = 0.33333

e 1 = q * Ka = 0.3 * 0.33333 = 0.1 t/m2

e2 = ɣ * H * ka = 1.8 * 4 * 0.3333 = 2.4 t/m2

then get F1 and F2

F1 = 0.1*4 = 0.4 t/m

F2 = 0.5 * 4 * 2.4 = 4.8 t/m

M = 0.4 * 2 + 4.8 * 4/3 = 7.2t.m. / m

For spacing 0.3 m

M = 7.2 * 0.3 = 2.16 t.m.

Mu = 2.16 *1.5 = 3.24 t.m.

T = 30 cm d = 25 cm

As = 3.24*100000/3600/0.8/25= 4.5 cm2

Use 3 bars dim. = 16 mm at tension

bars dim. = 16 mmUse total reinforcement = 6


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Notes

33يت live loadsوود ا ط د واد ر يون-

/ م /م1.ر ار ث ونb/cو ي2م او2م

1 / اخزوق ا طر ار ويجو ذه ا2 طن

ار

-يا موزا ريو يا ىوا ريي يوا هي دوو

) ال ايا طر ار وي اراياخزووج

يو

م

retaining wall

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10-Retaining Wall and Piles