Rcc R-w Examples

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CE 437/537, Spring 2011 Retaining Wall Design Example

Design a reinforced concrete retaining wall for thefollowing conditions.

f'c = 3000 psi

fy = 60 ksi

Development of Structural Design Equations. In this example, the structural three retaining wall components is performed by hand. Two equations are deve

section for determining the thickness & reinforcement required to resist the bendin the retaining wall components (stem, toe and heel).

Equation to calculate effective depth, d: Three basic equations will be used to d

equation for d.

11

'

'

003.0

003.0,

003.0

/

003.0:

]2[85.0

85.0,

]1[2

2

! "

"

!

#

#

+

=+

=

=

==

$ % &

'( ) *=

$

%

& '

(

) *=

=

s

s

y

c s

y sc

y su

y sn

nu

d

a

d aitycompatibil strain

Eqnab f

f A

f Aba f T C

Eqna

d f A M

ad f A M

M M

Assuming !1 = 0.85,

"s a/d

Fill: # = 32o Unit wt =

Natural Soil: # = 32o allowable bearing pre

wtoe tstem wheel

HT = 18 ft

tf

surcharge = qs = 4


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Substituting Eqn. 2 into Eqn. 1:

! " #

$% & '!

! "

# $$%

& =

285.0

'a

d f ab f

f M y

y

cu (

And substituting Eqn. 3 into the above:

d

d d f bd

f

f M y

y

cu

883.0

2

235.0235.085.0

'

! " # $

% & '= (

Inserting the material properties: f'c = 3 ksi and fy = 60 ksi, and b = 12in

(1

strip of wall, in the direction out of the paper).

2)883.0)(235.0)(12(3)85.0(90.0 d M inksiu

=

Equation for area of reinforcement, As. The area of reinforcement required is ca

from Eqn. 1:

d Ad f A M ksi s y su 883.06090.0883.0 ==!

Design Procedure (after Phil Ferguson, Univ. Texas)

1. Determine HT. Usually, the top-of-wall elevation is determined by the clien bottom-of-wall elevation is determined by foundation conditions.

2. Estimate thickness of base. tf !7% to 10% HT (12" minimum)

Tf = 0.07 (18' x 12"/') = 15.1"


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3. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on

horizontal earth pressure.

calc. d:

lb ft ft ft sur a sur

lb ft ft fill

o

o

a

ft a fill

psf hqk P

pcf P

k

pageof out hhk P

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

4310)1()67.16)(100)(31.0(2

1

31.0)32sin(1

)32sin(1

sin1

sin1

)1()(2

1

2

===

==

=+

!

=+

!

=

=

"

"

#

inininin

inininin stem

inin

k

ft k

in

k

u

ft k ft

lb ft

lbu

fill u

d

barsassumet

d d

ft

in

d M

M

hh

P M

5.125.0215

)8#(,3.14)0.1(2

1cover 28.11

8.11,71.5)12(9.65

71.5

9.65)2

67.16)(2070)(6.1()

3

67.16)(4310)(6.1(

)2)()(PLoadFactor Live()3)()(LoadFactor PressureEarth(

2

2

sur

=!!=

=++=

==

=

=+=

+=

!

!

wtoe tstem wheel

h = 8 – 16n/12

n

h =16.67ft

tf = 16in

ka $

h


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calc. As:

bar

in

ft

in

wall of ft in

bar

in

inbar oneof A

in A A ft

in

d A M

s

sin

sksi ft k

sksiu

,13.712

33.1

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 forc

failure to occur in the soil (soil-to-soil has higher friction angle than soil-to-conc

Neglect soil resistance in front of thewall.

found stem fill T

o soil natural

soil natural T

sliding resist

W W W W

W

F FS

F set

++=

==

=

=

==

=

62.0)32tan(tan

)(tanF

friction)of fficientForce)(coe(VerticalF

slidingfor1.5Safetyof FactorFS

resist

resist

!

!

850200)1)(312

15)(

12

16)(150(

2810)1()12

1

2

1512)(67.16)(150(

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

+=++=

=+

=

==

heel ft ft

heel found

lb ft

in

ft inin ft

stem

heel ft

heel ft

fill

w plf ft w ft pcf W

pcf W

w ft

lbw pcf W

lb ft ft

lb ft ft fill

sur fill sliding

psfP

pcf P

P P F

2230)1)(18)(40031.0(

5020)1()18)(10031.0(2

1 2

=!=

=!=

+=

18t

tf = 16n

15 n

12n


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ft heel heel

lblb

lbheel

lbheel

lb

uww ft

lb

w ft

lbw

ft

lb

,42.7,1870366062.0

5.17250

5.1

)62.0(85020028101670

7250

=+=

!"

#$%

&+++

=

5. Check Overturning.

)2

75.11(

)2

25.13(

3),312

15

2

5.7(

2.50)9(23.2)6(02.5

)2

18()3

18(

ft

found

ft ft

stem

ft toe

ft ft

fill resist

ft k ft k ft k over

ft

sur

ft

fill over

W

W

wassume ft W M

M

P P M

+

++

=++=

=+=

+=

!

OK FS M

M

M

M

over ft k

ft k

over

resist

ft k resist

ft k ft klf ft k ft ft klf resist

,0.247.22.50

2.124

2.124

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

=>==

=

+!++!=

"

"

"

6. Check Bearing.

found ft

ft ft

stem

ft ft

fillover

k k k k T

found stem fill T

T v

W W W M M

W

W W W W

bL

M

bL

W toeof end at

+!++!!=

=++=

++=


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OKcapacity, bearingallowable0.580.2

)75.11)(1(6

1

9.29

)75.11)(1(

69.17

2

=


7/109


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

foundat recalcnot(did,7.32)1)(18)(4.0(6.1)35.281.253.12(2.1

)(6.1)(2.1

6

1 2

k ft ft ksf k k k u

sur found stem fill u

uuv

W

W W W W W

bL

M

bL

W

=+++=

+++=

±=!

[ ]

ksf ft

ft

ksf ksf ksf

v

ksf ksf ksf v

ksf ksf ksf

v

ft ft

ft k

ft ft

k

v

ft k ft k ft k ft k u

ft stem

ft soil over u

B

C

A

M

W W M M

74.3)75.8(75.11

82.074.482.0

82.096.178.2

74.496.178.2

)75.11)(1(6

1

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

=!

+=

=!=

=+=

+=

=+!=

"+"!=

!

!!

#

#

#

#

d for flexure:

flexure for d d in

k

ft

in

d in

k M

M

in ft k

u

ft k ft ft ft ksf ft

ft ft ksf u

5.6,71.5)12(8.19

71.5

8.19)33

2)(1)(3)(00.1(

2

1)

2

3)(1)(3)(74.3(

2

2

==

=

=+=

!

!

d for shear:Assume theel = ttoe = 21.5

in

Critical section for shear occurs at "d" from face of stem, d = 21.5" – 3"cove

ft

kftftksfksf

ksf ft

ft

ksf ksf ksf

v tioncritical

181

24.4)12

1875.8(

75.11

82.074.482.0

sec=+

!

+=#

A B

3' 1.25'


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Reinforcement in toe:

#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

use ft

in

ft

inbar

in

saybars smaller try ft

in

ft

in

bar

in

in A A ft in

d A M

in

in

sin

sksi ft k

sksi

u

=

=

==

=

!


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Rcc R-w Examples

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