concrete design retaining wall

7/24/2019 Concrete Design Retaining Wall

1/12

INPUT DATA & DESIGN SUMMARY

CONCRETE STRENGTH fc' = 3 ksi

REBAR YIELD STRESS fy = 60 ksi

LATERAL SOIL PRESSURE Pa = 14,583

Friction angle ! = 30

SOIL BEARING CAPACITY qu = 28,501 psf

BACKFILL SPECIFIC WEIGHT gb = 120 pcf

SURCHARGE WEIGHT ws = 250 psf

FRICTION COEFFICIENT = 0.5

ALLOW SOIL PRESSURE Qa = 4.4 ksf

THICKNESS OF TOP STEM tt = 12 in

THICKNESS OF KEY & STEM tb = 24 in

TOE WIDTH LT = 4.5 ft

HEEL WIDTH LH = 7 ft

TOTAL HEIGHT OF WALL HT = 25 ft

CONCRETE SPECIFIC WEIGHT "conc. = 150 pcf

KEY DEPTH dkey = 2 ft

FOOTING THICKNESS hf = 3 in

SOIL OVER TOE hp = 2 in

STEM VERT. REINF. Exposed face (As) 8 bar @ 4 in o.c.

8 bar @ 72 in o.c.

STEM HORIZ. REINF Exposed face. (ACI 14.3.4) # 4 bar @ 6.5 in o.c.

STEM HORIZ. REINF Soil face. (ACI 14.3.4) # 4 bar @ 13 in o.c.

HEEL REINF. @ TOP OF FOOTING WITH 2.5" COVE # 8 bar @ 7 in o.c.

TOE REINF. @ BOTTOM OF FOOTING WITH 3" COV # 8 bar @ 11 in o.c.

ANALYSIS

SERVICE LOADS

Pfill = 12,500 lbs

Ps = 2,083 lbs

Pp = 7,031 lbs

W1 = 4,050 lbs

W2 = 1,725 lbs

W3 = 3,450 lbs

W4 = 19,320 lbs

W5 = 1,750 lbs

STEM VERT. REINF. Soil face (As)

(only needed to supprt horiz. Reinf.)


2/12

bar # Diameter Area

4 0.50 0.20

5 0.625 0.31

6 0.75 0.44

7 0.875 0.6

8 1 0.79

ACI Code Section 10.2.7

!= 0.9

#1 0.85

ACI Code Section 10.2.7

Concrete Strength < 4000 psi use

$ = 0.009 #1 0.85

d = 21.01

tb = 23.51 use tb = 24 inches

Calc As

use d = 21.5 inches

$ = 0.008540

As = 2.20 in^2

use #8 bar 0.79 in^2

bar span = 4.30 inches

Calculating (d) for bottom of stem

Using f'c=3, fy=60, and b=12 in

Rebar Design

use #8 bars at 4" = 2.37 in^2

! = approximately 0.18f 'c

fy

!

"

##

$

%

&&

Mu

!bd2 = 482.6! from " table "A.12

bd2 =Mu 12( )!482.6

use !b =12" 1" foot! strip !of !wall( )

solve !for !d

tb

= d+ 2in

!cover+

1

2

"

#$

%

&' bar

!diameter

( )

bar ! span =Asof !chosen !bar

Ascalculated(12)=

0.79

2.20(12)= 4.31in

Substitute ! values ! for !b !d

Mu

!bd2 =

191666.67 12( )

0.9 12( ) 21.5( )2 = 460.7

for ! 460.7

! = 0.00854

As = ! bd( )


3/12

Page

Designing Demensions for Retaining Wall

1. Determine Htgiven by site condition 25

1.75 use 2 ft

3. Design Stem (tstem, As-stem)

Calc d.

ka = 0.33333

Pfill 12500 lbs h= 23 feet

Psur 2083.33 lbs

Earth Pressure Load Factor = 1.6Live Load Factor = 1.6

Mu = 191666.67 lb-ft 191.7 k-ft

Check Shear stress in stem

2. Estimate thickness of base use tf= 7 - 10%

of Ht

pfill =1

2ka!bh( )h 1'( ) ka =

1! sin!

1+sin!

h =Ht!

tf

psur =

kaqsurh( ) 1'( )

Mu = Earth !pressure ! load! factor( ) pfill( ) h

3

"

#$

%

&'+ Live ! load! factor( ) psur( )

h

2

"

#$

%

&'

Vu =Pfill +Psur =12500 + 2083 =14583lbs

!Vc =!2 f'cbd

!Vc = 0.75 2

( ) 3000

( ) 12

( ) 21.5

( )!Vc = 21197 >14583

Shear !at! stem !OK


4/12

!"#$ &$'()* +,-. /)

*$$& &$'()* 0 /)

)1$ &$'()* 2-. /)

!1))13 #)$3 4 /)

)15 #)$3 + /)

Base Length and position of stem

Pfill =12500lbs

Psur = 2083lbs

stem ! location=x

W = x( ) hsur +h( ) !( )

W = x( ) 250

120+25

!

"#

$

%&120

W = 3250 x( )

Ma! = 0

" 12500( ) 8.33( )! 2083( ) 12.5( )+ 3250x( ) x

2

!

"#

$

%&

x = 8.94

Base ' length =3

2

!

"#

$

%& 8.94( )

Base ! length =13.42


5/12

6178$ 913$') :73 913$') 913$');/)> ? @-,,, +>2A+BB 4@A,.>Psur (H2) 4>@,-,, ? +4-. 4BA>,@ @AC+,

D1)"& EFG +2A.@, +,>A4>2 4>A0>>

+C,A4>>

+0A.>>

EF G ,>A4C. 9 G 4B@ABB,

2>.>&!# ? ;0=

D1)"&

H+G;4=;+,-.=;+.>=H4GI;4,=;+=J4K;+.>=

H,G;4,=;+=;+.>=

H2G;0=;4,=;+4>=

H.G;0=;4.>J+4>=+4>

LF$7 DM7'N'( 913$') EN(*)N'( 913$')

!"#$%& (")%*+ #*+ ,-$+%.+/0/1

!"#$%& (")%*+ "1"0/2% 23040/1

5$201/ #**%0/1 6$& %* 078+*-$ (! #*+ 23040/1

913$') :736178$

+0.>&!# ? ;++=

+C,4> ? ;+>=

,2.>&!# ? ;B=

+04.&!# ? ;.-+B0=

Safety ! factor !overturning=268, 663

130, 204= 2.06 > 2

OK

Safety ! factor !against! sliding =( ) Rv( )

force !causig ! sliding=

0.50( )30,29514,583

=1.04 > 2

NG

Pp = 0.5Cp!hp2

Cp =1+sin!

1! sin!=

1+ sin30

1! sin30= 3

hp = h+ l( )sin30 = 4+ 4.5sin30( ) = 6.25

Pp = 0.5 3( ) 120( ) 6.252( ) = 7, 031lb

Safety !factor !against! sliding =( ) Rv( )

Pa!

Pp=

0.50( )30, 295

14, 583!7, 031

= 2.01> 2

OK


6/12

O$"7N'( P"5"8N)Q 6"8)17 1/ R"/$)Q

Bearing !Capacity !of! soil

qu = c 'Nc +!DfNq +0.5!BN!

qu = 0( ) 37.16( )+ 120( ) 10( ) 22.46( )+ 0.5( ) 120( ) 13.5( ) 19.13( )

qu = 28,501psf

Footing ! soil !pressure

Rv = 30, 295lb

x =268, 663"130, 204

30, 295= 4.57ft

Soil !pressure ="Rv

A

Mc

I

A = 1'( ) 13.5'( ) =13.5ft

I =1

12

!

"#

$

%& 1( ) 13.5( )

3= 205.03ft4

ftoe = !30, 295

13.5!

30, 295 6.75! 4.57( ) 6.75( )205.03

ftoe = !2, 244.07! 2,174.27 = !4, 418psf

fheel = !2, 244.07+ 2,174.27 = !70psf

FS= quqmax

qmax

= fmax

= ftoe

FS=28, 501

4, 418= 6.45> 3

OK


7/12

S$#N(' 1/ T$$&

&U 7$VMN7$U /17 W@ )15 !"7# "'U 8G4-. N# 2, X'8*$# Y @2Z "F"N&"!&$ ;1["Q=

M#$ W @ !"7# ") 0Z #5"' 1' )15 \N)* 4-. 81F$7

Vu = hsoil + hsur( ) lheel( ) !soil( ) load!factor( )+ theel( ) lheel( ) !conc( ) load!factor( )

Vu = 25.5( ) 7( ) 120( ) 1.2( )+ 2( ) 7( ) 150( ) 1.2( )

Vu = 28, 224lb

!Vc =!2 f'cbd

!Vc = 0.75( ) 2( ) 3000( ) 12( ) 20.5( ) = 20, 211lbs < 28, 224NG

Try ! tf = 36in

Vu = 25.5( ) 7( ) 120( ) 1.2( )+ 3( ) 7( ) 150( ) 1.2( ) = 29, 484lbs

!Vc = 0.75( ) 2( ) 3000( ) 12( ) 32.5( ) = 32, 042lbs > 28, 484lbs

OK

Mu = 28, 224( ) 7

2

!

"#

$

%&= 98, 784ft' lb

Mu

!bd2 =

12( )98, 784

0.9 12( ) 32.5( )2 =103.9

!= 0.0018 < !min

= 0.00333

use ! !min

As(heel) = 0.00333 12( ) 32.5( ) =1.3in2

/ ft

bar ! span =0.79

1.312( ) = 7.29

ld =3

40

fy

! f 'c

"t"

e"

s

cb +Ktrdb

!

"#

$

%&

!

"

####

$

%

&&&&

! =1

"t =1.3

"e ="

s =1

cb +Ktr

db= 2.5

ld = 43in


8/12

S$#N(' 1/ )1$

:PX P1U$ +4-+

&U 7$VMN7$U /17 W@ !1))13 !"7# "'U 8G4-. N# ,, X'8*$# Y @2Z "F"N&"!&$ ;1["Q=

M#$ W @ !"7# ") 0Z #5"' 1' !1))13 \N)* ,Z 81F$7

Vu = load! factor( ) ftoe( )+ load! factor( ) ftoe" fheel( ) lbase" ltoe( )

lbase

#

$

%&

'

(+ fheel

Vu = 1.6( ) 4, 418 4.5( )

2

#

$%

&

'(+ 1.6( )

4, 418" 70( ) 13.5" 4.5( )13.5

#

$%

&

'(+ 70

)

*+

,

-./ 4.5( )

2

0

1

22

3

22

4

5

22

6

22

Vu =15, 905+10, 687= 26, 592lb

Mu =15, 905 4.5

3

#

$%

&

'(+10,687

2

3/4.5

#

$%

&

'(= 55, 919ft" lb

Mu

!bd2 =

12

( ) 55, 919

( )0.9( ) 12( ) 32.5( )2 = 58.8

!< !min

use !!min

= 0.00333

As(toe) = 0.00333 12( ) 32.5( ) =1.3in2

/ ft

bar ! span =0.79

1.3(12)= 7.29

ld =3

40

fy

! f 'c

"t"e"s

cb +Ktrdb

!

"#

$

%&

!

"

####

$

%

&&&&

!=1

"t ="s ="e =1

cb +Ktr

db= 2.5

ld = 33in


9/12

Rebar Design cont.

As = 0.54N']4

use #4 bars 0.2N']4

bar span = 4.44

, W 2 !"7# '$$U$U ;>-4>?,=G>-B>

Front (exposed face) = 4.44in/(2/3)=6.66

Rear face = 4.44in/(1/3)=13.32

use #4 bars at 4" = 0.60 in^2

M#$ W2 !"7 ") B-.Z #5"'

M#$ W2 !"7 ") +,Z #5"'

Minimum vertical $by ACI section 14.3

Minimum horizontal As

Exposed face of wall will have bars placed at no more than one third of total reinforceme

per ACI 14.3.4

! = 0.0015

concrete design retaining wall

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