Retaining wall design_example

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Page 1: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 1 / 8

Design a reinforced concrete retaining wall for the following conditions. f'c = 3000 psi fy = 60 ksi Development of Structural Design Equations. In this example, the structural design of the three retaining wall components is performed by hand. Two equations are developed in this section for determining the thickness & reinforcement required to resist the bending moment in the retaining wall components (stem, toe and heel). Equation to calculate effective depth, d: Three basic equations will be used to develop an equation for d.

11

'

'

003.0003.0,003.0

/003.0:

]2[85.0

85.0,

]1[2

2

βε

εβ

φ

φ

+=

+=

=

==

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞⎜

⎝⎛ −=

=

s

s

y

cs

ysc

ysu

ysn

nu

da

daitycompatibilstrain

EqnabffA

fAbafTC

EqnadfAM

adfAM

MM

Assuming β1 = 0.85,

εs a/d 0.005 0.319

0.00785 0.235 0.010 0.196

and choosing a value for εs in about the middle of the practical design range,

]3[235.0,235.0 Eqndada ==

Fill: φ = 32o Unit wt = 100 pcf

Natural Soil: φ = 32o allowable bearing pressure = 5000psf

wtoe tstem wheel

HT = 18 ft

tf

surcharge = qs = 400psf

Page 2: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 2 / 8

Substituting Eqn. 2 into Eqn. 1:

⎟⎠⎞⎜

⎝⎛ −⎟

⎟⎠

⎞⎜⎜⎝

⎛=

285.0

' adfabffM yy

cu φ

And substituting Eqn. 3 into the above:

d

ddfbdffM yy

cu

883.0

2235.0235.085.0

'

⎟⎠⎞⎜

⎝⎛ −= φ

Inserting the material properties: f'c = 3 ksi and fy = 60 ksi, and b = 12in (1-foot-wide strip of wall, in the direction out of the paper).

22 71.5)883.0)(235.0)(12(3)85.0(90.0 dMdM ink

uinksi

u ==

Equation for area of reinforcement, As. The area of reinforcement required is calculated from Eqn. 1:

dAMdAdfAM s

ksiu

ksisysu 7.47883.06090.0883.0 === φ

Design Procedure (after Phil Ferguson, Univ. Texas) 1. Determine HT. Usually, the top-of-wall elevation is determined by the client. The bottom-of-wall elevation is determined by foundation conditions. HT = 18 feet. 2. Estimate thickness of base. tf ≈7% to 10% HT (12" minimum) Tf = 0.07 (18' x 12"/') = 15.1" use tf = 16"

Page 3: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 3 / 8

3. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on by the horizontal earth pressure. calc. d:

lbftftftsurasur

lbftftfill

o

o

a

ftafill

psfhqkP

pcfP

k

pageofouthhkP

2070)1)(67.16)(400(31.0)1(

4310)1()67.16)(100)(31.0(21

31.0)32sin(1)32sin(1

sin1sin1

)1()(21

2

===

==

=+−=

+−=

=

φφ

γ

inininin

instem

ininininstem

inink

ftk

ink

u

ftkft

lbft

lbu

fillu

d

tusebarsassumet

ddftin

dM

M

hhPM

5.125.0215

15)8#(,3.14)0.1(21cover28.11

8.11,71.5)12(9.65

71.5

9.65)267.16)(2070)(6.1()

367.16)(4310)(6.1(

)2

)()(PLoadFactorLive()3

)()(LoadFactorPressureEarth(

2

2

sur

=−−=

==++=

==

=

=+=

+=

wtoe tstem wheel

h = 8ft – 16in/12in/ft

h =16.67ft

tf = 16in

ka γ h ka qs

h

Page 4: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 4 / 8

calc. As:

inusebarin

ftin

wallofftinbarin

inbaroneofA

inAAftin

dAM

s

sin

sksiftk

sksi

u

6@8#,13.71233.1

79.0

79.08#

33.1),5.12(7.47)12(9.65

7.47

2

2

2

2

=

=

==

=

4. Choose Heel Width, wheel Select wheel to prevent sliding. Use a key to force sliding failure to occur in the soil (soil-to-soil has higher friction angle than soil-to-concrete). Neglect soil resistance in front of the wall.

foundstemfillT

osoilnatural

soilnaturalT

slidingresist

WWWW

W

FFSF

set

++=

==

==

==

=

62.0)32tan(tan

)(tanFfriction)offficientForce)(coe(VerticalF

slidingfor 1.5 Safety ofFactor FS

resist

resist

φ

φ

850200)1)(31215)(

1216)(150(

2810)1()12

121512)(67.16)(150(

1670)1)()(67.16)(100(

+=++=

=+=

==

heelftft

heelfound

lbftin

ftininft

stem

heelft

heelft

fill

wplfftwftpcfW

pcfW

wftlbwpcfW

lblblbsliding

lbftftsur

lbftftfill

surfillsliding

F

psfP

pcfP

PPF

725022305020

2230)1)(18)(40031.0(

5020)1()18)(10031.0(21 2

=+=

=×=

=×=

+=

18ft

tf = 16in

15in

12in

Page 5: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 5 / 8

ftheel

ftheelheel

lblb

lbheel

lbheel

lb

wusewwftlb

wftlbw

ftlb

5.7,42.7,1870366062.05.17250

5.1

)62.0(850200281016707250

==+=

⎥⎦

⎤⎢⎣

⎡+++

=

5. Check Overturning.

)275.11(

)225.13(

3),31215

25.7(

2.50)9(23.2)6(02.5

)218()

318(

ft

found

ftft

stem

fttoe

ftft

fillresist

ftkftkftkover

ft

sur

ft

fillover

W

W

wassumeftWM

M

PPM

+

++

=++=

=+=

+=

OKFSMM

M

M

overftk

ftk

over

resist

ftkresist

ftkftklfftkftftklfresist

,0.247.22.502.124

2.124

)875.5)(85.05.720.0()625.3)(81.2()8)(5.767.1(

=>==

=

+×++×=

6. Check Bearing.

ftkftkftkftk

foundft

ftft

stem

ftft

fillover

kkkkT

foundstemfillT

Tv

M

WWWMM

W

WWWW

bLM

bLWtoeofendat

−− =+−=

+−++−−=

=++=

++=

<+=

9.29)25.2(81.2)125.2(53.122.50

)0()875.52

25.15.7()25.7875.5(

69.1735.281.245.12

6L e ifonly validis equation,

6

Check that e < L/6:

OKLeLWme ft

ftft

k

ftk

T,6

,96.1675.11

6,68.1

69.179.29 <∴=====

12.53k 2.35k

18ft

tf = 16in

12in

3' 15" 7.5'

Page 6: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 6 / 8

OK capacity,bearingallowable0.580.2)75.11)(1(

61

9.29)75.11)(1(

69.172

=<=+=−

ksfksf

ftft

ftk

ftft

k

7. Heel Design. Max. load on heel is due to the weight of heel + fill + surcharge as the wall tries to tip over. Flexure:

surfillheel WWWW ++=

ftkftklf

uu

klf

ftft

ft

LwM

W

plfpcf

ftpcfW

−===

=

++

=

0.812

)5.7(88.22

88.2

)400(6.1)1)(67.16)(100(2.1

)1)(1216)(150(2.1

22

flexureforddink

ftin

dinkM

inftk

u

0.13,71.5)12(0.81

71.5

2

2

==

=

Shear:

controlsshearforddpsiVsetV

dpsidbfV

wV

ininlbcu

inwcc

kftklfftuu

,9.21,)12(30002)75.0(600,21,

)12(30002)75.0(2)75.0(

6.21)5.7(88.2)5.7(

'

===

==

===

φ

φ

Shear controls the thickness of the heel.

inheel

inininheel tusebarassumeint 5.21),8#(4.24

21cover29.21 ==++=

Reinforcement in heel:

"8@8#,83.8)12(07.1

79.0

07.1),9.21(7.47)12(0.81

7.47

2

2

2

useftin

ftinbarin

inAAftin

dAM

in

sin

sksiftk

sksi

u

=

==

=

7.5ft

16in Mu

wu

Vu

Page 7: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 7 / 8

8. Toe Design. Earth Pressure at Tip of Toe:

change) neglible b.c. wt foundation recalcnot (did,7.32)1)(18)(4.0(6.1)35.281.253.12(2.1

)(6.1)(2.161 2

kftftksfkkku

surfoundstemfillu

uuv

W

WWWWW

bL

MbLW

=+++=

+++=

±=σ

[ ]

ksfftft

ksfksfksf

v

ksfksfksfv

ksfksfksfv

ftft

ftk

ftft

k

v

ftkftkftkftku

ftstem

ftsoiloveru

B

C

A

M

WWMM

74.3)75.8(75.1182.074.482.0

82.096.178.2

74.496.178.2

)75.11)(1(61

0.45)75.11)(1(

7.32

0.45)1(81.2)125.2(53.122.1)2.50(6.1

)0.1125.2(2.16.1

2

=−+=

=−=

=+=

+=

=+−=

×+×−=

−−

σ

σ

σ

σ

d for flexure:

flexureforddink

ftin

dinkM

M

inftk

u

ftkftftftksfft

ftftksfu

5.6,71.5)12(8.19

71.5

8.19)332)(1)(3)(00.1(

21)

23)(1)(3)(74.3(

2

2

==

=

=+=

d for shear:

Assume theel = ttoe = 21.5in Critical section for shear occurs at "d" from face of stem, d = 21.5" – 3"cover-1/2"=18"

controlsflexurefordOKVpsiV

ftV

ft

ulbinin

c

kftftksfksfu

ksfftft

ksfksfksf

v tioncritical

,,750,17)18)(12(30002)75(.

74.6)1)(12183)(24.474.4(

21

24.4)121875.8(

75.1182.074.482.0

sec

>==

=−+=

=+−+=

φ

σ

A B C

3' 7.5' 1.25'

Page 8: Retaining wall design_example

CE 437/537, Spring 2011 Retaining Wall Design Example 8 / 8

Reinforcement in toe:

"8@4#6.8)12(28.0

20.0

4#,,33)12(28.0

79.0

28.0),18(7.47)12(8.19

7.47

2

2

2

2

2

useftin

ftinbarin

saybarssmallertryftin

ftinbarin

inAAftin

dAM

in

in

sin

sksiftk

sksi

u

=

=

==

=