pe 22 rc retaining walls

8/3/2019 PE 22 RC Retaining Walls

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22 - Reinforced Concrete Retaining Walls

01: Design a reinforced concrete wall for a sloped backfill.

02: Design a reinforced concrete bridge abutment wall.

539


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**RC Retaining Walls01:Design a RC wall for a sloped backfill.(Revision: Oct-08)

1'

2'

15'

16'-6"

1'-6"

18

.33'

1.8

3'

PvPh

3

1

1

23

4

5

66

.11'

3' 1'-6" 5'-6"

10'

Design a reinforced concrete wall with a backfill = 125 pcf, an allowable soil bearing

capacity ofqall= 3 ksf, and a friction at the base of = 30. Design the wall and check for

its stability under working loads. (Note: All loads, shears and moments are per linear ft. of retaining wall).

560


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Solution:

( )2 2 2

Step 1: Find the lateral pressures and forces.

The active lateral earth pressure is,

30tan 45 tan 45 tan 30 0.33

2 2

The presure at the base (invert) of the wall footing is,

a

a

b a

K

K

p HK

= = = =

= ( ) ( ) ( )0.120 18.33 0.33 0.733

The forces on the wall (per unit length) are,

(12)(18.33) 2.0

(39)(18.33) 6.6

Step 2: Stability analysis for sliding and overturning.

v

h

kcf ft ksf

P kips

P kips

= =

= =

= =

Moment About A

Area Area Force(kip)

Arm(ft)

Moment(kip-ft)

1 x 5.5x 1.83 = 5.03 x 0.125 = 0.63 1.83 1.22 5.5 x 15.0 = 82.5 x 0.125 = 10.31 2.75 28.4

3 1.0 x 15.0 = 15.0 x 0.150 = 2.25 6.00 13.5

4 x 0.5 x 15.0 = 3.75 x 0.150 = 0.56 6.67 3.7

5 10.0 x 1.5 = 15.0 x 0.150 = 2.25 5.00 11.3

6 3.0 x 2.0 = 6.0 x 0.125 = 0.75 8.50 6.4

Pv 2.00

Ph H = 6.60 6.11 40.4

V=18.75 M=104.9

Location of Resultant

From point A, 104.9 = 5.6 ft

18.75

then e = 5.6 10 = 0.6 ft o.k. < B

2 6

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Soil Pressure at Toe of Base

qmax = 18.75 (1 + 6 x 0.6 ) = 1.875 (1 + 0.36) = 2.55 ksf OK < 3 ksf

10 10

Check F against SlidingShear available along base = 18.75 kips x 0.58 = 10.9 kips

Passive force at toe

Use S = 2/3 (30) = 20 , Pp = ( cos ) = 5.8 kips H

Pp = 5.8 (0.125) (3.5) = 4.7 kips2 (0.940)

Min. F = 10.9 = 1.7 kips , Max F = 10.9 + 4.7 = 15.6 = 2.4 kips OK without Key6.6 6.6 6.6

Step 2: Design parameters.

Load Factors

Stem Use 1.7 Ph

Base (toe and heel) distribute V uniformly over front B/3

Concrete and Steel Data

Capacity reduction factors: 0.90(flexure); 0.85(shear)

562

Fc = 3,000psi x 0.85 = 2,550 psi (for stress block)

Vc = 2 3,000 = 110 psi

fy = 40,000 psi ; min = 0.005 ; max = 0.0278 ; shrinkage

= 0.002

ld = 0.04 Ab (40,000) = 29.2 Ab (bottom bars) x 1.4 =

40.9 (top bars)3,000

Step 3: Design the stem of the wall.

Vertical Reinforcement

Ph = 1.7 39 (15) = 7.46

M = 7.46 x 5 x 12 = 448 kip-in

Use 6 batter on front, then t = 12 + 6 = 18

P v

P h

14"

5'

15

'

12"


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Use d = 18 4 = 14

Assume arm = d a/2 = 13

T = 448/13 = 34.5 kip

As = 34.5 / (40 x 0.90) = 0.96 in/ft

At bottom of wall Use #6 @ 5 ctrs

As = 1.06 in/ft

=1.06/(14 x 12) = 0.0063 >0.005 and


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Compute level of cutoff for #6 @ 10 ctrs. :

As = 0.53 in/ft ; = 0.53 / (11 x 12) = 0.004 > 0.002 < 0.005

Therefore, cutoff and develop bars above level where:

M = 0.53 (40) (11) (0.90) = 158 kip-in

1.33

M = 158kip-ft @ 10-6 level

Developmental length = 1-6

Level of Cutoff = 9-0

Step 4: Design the toe of the base of the wall.

Distribute V over front B/3. Assume t = 18 , d =14

V = 18.75 kips = qmax = 5.63 ksf(10 / 3)

wt. of soil over toe = 2.0 (0.125) = 0.25 ksf

wt of concrete base = 1.5 (0.150) = 0.23 ksf

Net toe pressure for design = qmax - wt. of soil over toe - wt of concrete base

= 5.63ksf - 0.25ksf - 0.23 ksf = 5.15 ksf

5.15 kip

1'-6"

3'

d =14"

front face of stem

V = 5.15 x (3.0) = 15.45 kips

M = 15.45 x (3/2) x 12 =278 kip-in

Assume arm = 13 in

T = 278/13 = 21.4 kips = 0.59 in/ft

(40 x 0.9)

Check = 0.59 = 0.0035 > 002, shrinkage OK14 x 12

< 0.005, so increase As by 1/3 , then As = 0.59 x 1.33 = 0.80 in/ftuse #6 @ 6 ctrs. As = 0.88 in/ft

Compressive stress block and shear OK by inspection after stem computations

Development length : Extend full base width, therefore ld OK

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Step 5:Design the heel of the base of the wall.

At Stem:

Weight of soil above heel at back face of stem = 15.0 (0.125) = 1.88 ksf

Weight of concrete base = 1.5 (0.150) = 0.23 ksf

Net pressure for design = 1.88 + 0.23 = 2.11 ksf

At Back:

Weight of soil above hell at back = 16.83 (0.125) = 2.10 ksf

Weight of concrete base = 0.23 ksf

Net pressure for design = 2.33 ksf

1'-6"

5'-6"

d =14"

2.33 ksf

2.11 ksf

V1 = (2.11) (5.5) = 5.80 kips

V2 = (2.33) (5.5) = 6.41 kips

Total V = 5.80 + 6.41 = 12.21 kips

M1= 5.8 (1/3) (5.5) (12) = 128 kip-in

M2 = 6.41 (2/3) (5.5) (12)=282 kip-in

Total M = 128 + 282 = 410 kip-in

Assume arm = 13

T= 410/13 = 31.5 kipAs = 31.5kip = 0.88 in/ft

(40 x 0.9)

Use #6 @ 6 ctrs. As = 0.88 in/ft > 0.002 and >0.005 and < 0.0278 OK

Compressive stress block and shear OK

Development length: Extend full base width, therefore ld OK

Horizontal shrinkage steel in stem:

Required: 0.002 (15) (12) = 0.36 in/ft of height

Use #4 @ 9 ctrs. front As = 0.27 in/ft

Use #4 @ 18 ctrs back As = 0.13 in/ft

Total As = 0.27 + 0.13 = 0.40 in/ft

Horizontal shrinkage steel in base:

Use #4 @ 12 ctrs. top and bottom As = 0.40 in/ft

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Step 6:Finished sketch of the wall.

3

1

12"

16'-6"

2'

1'-6"

3' 1'-6" 5'-6"

Ver

t.ba

rs,

Ver

t.ba

rs,

bac

kface

bac

kface

#6@

5"

#6@

10"

Provide 3" clear to

all bars except 112 "clear to bars infront face of stem

#4 @ 2'-0"

6" drains

@ 10'-0"

#6 @ 6" 2"x6" key

#6 @ 6"

#4 @ 12"

#4@ 9" front face

#4@18" back face

566


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**RC Retaining Walls02: Design a RC bridge abutment wall.

Design a bridge abutment for a backfill of = 110 lb/ft3

, an equivalent surcharge of 3

feet, an allowable soil pressure of 3.5kips/ft2, and an allowable shear between soil and

base of 45% of the vertical shear.

12'-0"

5'-6"

12"12"

3'-0"

2'-6" 12" 3'-6"

1'-6"

7'-0"

4'-0

" 6'-0"

9"

12"

A

5

72

8

4

1

3

6

3.88'

L =5.26 k

L =0.24 k

v

h

N

ote:

All loads and moments are per lineal foot of abutment.

Solution:

Phs = 30(3.0)(12) = 1.08 kips

Ph = 1/2(30)(122) = 2.16 kips

Stability Computations:

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Moments about A

Area Force Arm Moment

1 2.5 x 10.5 x 0.110 = 2.89 1.25 3.61

2 1.0 x 5.5 x 0.110 = 0.61 3.00 1.83

3 1/2 x 1.0 x 1.0 x 0.110 = 0.06 2.83 0.15

4 0.75 x 3.0 x 0.150 = 0.34 2.88 0.98

5 2.0 x 1.0 x 0.150 = 0.30 3.50 1.05

6 1/2 x 1.0 x 1.0 x 0.150 = 0.08 3.17 0.25

7 1.0 x 6.5 x 0.150 = 0.97 4.00 3.88

8 7.0 x 1.5 x 0.150 = 1.58 3.50 5.53

Lv 5.26 3.88 20.40

Lh 0.24 9.00 2.16

Phs 1.08 6.00 6.48

Ph 2.16 4.00 8.64

SH = 3.48 kips SV = 12.09 kips SMA= 54.96 ft-kips


From point A,09.12

96.54= 4.55 then e = 4.55 -

2

0.7= 1.05 1.5 ok

571


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**RC Retaining Walls02:Design a RC bridge abutment wall.

Design a bridge abutment for a backfill of_= 110 lb/ft3, an equivalent surcharge offeet, an allowable soil pressure of 3.5kips/ft2, and an allowable shear between soil

and base of 45% of the vertical shear.

All loads and moments are per lineal foot of abutment.

Solution:

Phs = 30(3.0)(12) = 1.08 kips

Ph = 1/2(30)(122) = 2.16 kips


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Stability Computations:

Moments about A

Area Force Arm Moment

1 2.5x10.5x0.11 = 2.89 1.25 3.61

2 1x5.5x0.11 = 0.61 3 1.83

3 1/2x1x1x0.11 = 0.06 2.83 0.15

4 0.75x3x0.15 = 0.34 2.88 0.98

5 2x1x0.15 = 0.3 3.5 1.05

6 .5x1x1x0.15 = 0.08 3.17 0.25

7 1x6.5x.15 = 0.97 4 3.88

8 7x1.5x0.15 = 1.58 3.5 5.53

Lv 5.26 3.88 20.4

Lh 0.24 9 2.16

Phs 1.08 6 6.48

Ph 2.16 4 8.64

H 3.48kips

V 12.09kips

MA 54.96kips


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Soil Pressure at Base:

Sliding:


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Design the stem:

Vertical Reinforcement

Ph + Phs = 8.64 +6.48 = 15.12

M = 15.12 x 12/3 x 12 = 726 kip in

Use 6 batter on front, then t = 9 + 6 = 15

Assume arm 15 4 = 11T = 726/11 = 66 kipAs = 66/ 40 x 0.9 = 1.83 in^2/ft

At bottom of wall Use #5 @ 2in =1.86in^2/ft or #9 at 6 c.c. = 2.00in^2/ft

p = 2.00/ 14 x 9 = 0.015 >.005 < 0.0278 OK

Check compressive stress block

C = T = 66kip / 2.55 x 12 x 0.9 = 2.39 >2 NOT GOOD

Check Shear

15.12 / 14x9 x .75 = 160psi > 110 psi. Need Shear Design for stirrups


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Resisting Moment

Top : 2 x 40 x .9 x 9 = 648 kip in

Bottom : 2 x 40 x.9 x 14 = 1008 kip in

Horizontal shrinkage in stem

Required = 0.002 x 14 x 9 = 0.336in^2/ft

Use #4@9 c.c front As = 0.27in^2/ft

Use #4@18 c.c back As = 0.13in^2/ft

Total As = 0.27 + 0.13 = 0.40

pe 22 rc retaining walls

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