Concrete Design Retaining Wall
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Transcript of Concrete Design Retaining Wall
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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.)
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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( )
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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
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!"#$ &$'()* +,-. /)
*$$& &$'()* 0 /)
)1$ &$'()* 2-. /)
!1))13 #)$3 4 /)
)15 #)$3 + /)
Base Length and position of stem
Pfill =12500lbs
Psur = 2083lbs
stem ! location=x
W = x( ) hsur +h( ) !( )
W = x( ) 250
120+25
!
"#
$
%&120
W = 3250 x( )
Ma! = 0
" 12500( ) 8.33( )! 2083( ) 12.5( )+ 3250x( ) x
2
!
"#
$
%&
x = 8.94
Base ' length =3
2
!
"#
$
%& 8.94( )
Base ! length =13.42
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6178$ 913$') :73 913$') 913$');/)> ? @-,,, +>2A+BB 4@A,.>Psur (H2) 4>@,-,, ? +4-. 4BA>,@ @AC+,
D1)"& EFG +2A.@, +,>A4>2 4>A0>>
+C,A4>>
+0A.>>
EF G ,>A4C. 9 G 4B@ABB,
2>.>&!# ? ;0=
D1)"&
H+G;4=;+,-.=;+.>=H4GI;4,=;+=J4K;+.>=
H,G;4,=;+=;+.>=
H2G;0=;4,=;+4>=
H.G;0=;4.>J+4>=+4>
LF$7 DM7'N'( 913$') EN(*)N'( 913$')
!"#$%& (")%*+ #*+ ,-$+%.+/0/1
!"#$%& (")%*+ "1"0/2% 23040/1
5$201/ #**%0/1 6$& %* 078+*-$ (! #*+ 23040/1
913$') :736178$
+0.>&!# ? ;++=
+C,4> ? ;+>=
,2.>&!# ? ;B=
+04.&!# ? ;.-+B0=
Safety ! factor !overturning=268, 663
130, 204= 2.06 > 2
OK
Safety ! factor !against! sliding =( ) Rv( )
force !causig ! sliding=
0.50( )30,29514,583
=1.04 > 2
NG
Pp = 0.5Cp!hp2
Cp =1+sin!
1! sin!=
1+ sin30
1! sin30= 3
hp = h+ l( )sin30 = 4+ 4.5sin30( ) = 6.25
Pp = 0.5 3( ) 120( ) 6.252( ) = 7, 031lb
Safety !factor !against! sliding =( ) Rv( )
Pa!
Pp=
0.50( )30, 295
14, 583!7, 031
= 2.01> 2
OK
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O$"7N'( P"5"8N)Q 6"8)17 1/ R"/$)Q
Bearing !Capacity !of! soil
qu = c 'Nc +!DfNq +0.5!BN!
qu = 0( ) 37.16( )+ 120( ) 10( ) 22.46( )+ 0.5( ) 120( ) 13.5( ) 19.13( )
qu = 28,501psf
Footing ! soil !pressure
Rv = 30, 295lb
x =268, 663"130, 204
30, 295= 4.57ft
Soil !pressure ="Rv
A
Mc
I
A = 1'( ) 13.5'( ) =13.5ft
I =1
12
!
"#
$
%& 1( ) 13.5( )
3= 205.03ft4
ftoe = !30, 295
13.5!
30, 295 6.75! 4.57( ) 6.75( )205.03
ftoe = !2, 244.07! 2,174.27 = !4, 418psf
fheel = !2, 244.07+ 2,174.27 = !70psf
FS= quqmax
qmax
= fmax
= ftoe
FS=28, 501
4, 418= 6.45> 3
OK
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S$#N(' 1/ T$$&
&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
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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
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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