Retaining Wall Design Example
Retaining Wall Design Example
Retaining Wall Design Example
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Design a reinforced concrete retaining wall for the following conditions. f'c = 3000 psi fy = 60 ksi
HT = 18 ft
surcharge = qs = 400
psf
tf wtoe tstem wheel Natural Soil: = 32o allowable bearing pressure = 5000psf
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.
M u = M n a M n = As f y d 2 a M u = As f y d [ Eqn 1] 2 C = T , 0.85 f c' a b = As f y As = 0.85 f c' ab fy [ Eqn 2]
strain compatibility :
and choosing a value for s in about the middle of the practical design range,
[ Eqn 3]
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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).
k
M u = 5.71 in d 2
Equation for area of reinforcement, As. The area of reinforcement required is calculated from Eqn. 1:
M u = As f y 0.883d = 0.90 As 60 ksi 0.883 d M u = 47.7 ksi As d
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"
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3. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on by the horizontal earth pressure.
h = 8 16 /12 ft h =16.67
ft
in
in/ft
tf = 16in
wtoe tstem wheel
ka h
ka qs
calc. d:
Pfill = 1 ( k a h ) h (1 ft out of page) 2 1 sin 1 sin( 32 o ) ka = = = 0.31 1 + sin 1 + sin( 32 o ) 1 (0.31)(100 pcf )(16.67 ft ) 2 (1 ft ) = 4310 lb 2
Pfill =
Mu =
d2
k
in ) = 5.71 in d 2 , d = 11.8in ft 1 t stem = 11.8in + 2 in cover + (1.0 in ) = 14.3in , ( assume #8 bars ) 2 in in in d = 15 2 0.5 = 12.5in 65.9 k ft (12
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calc. As:
M u = 47.7 ksi As d 65.9 k ft (12 in ) = 47.7 ksi As (12.5in ), As = 1.33 in 2 ft
As of one #8 bar = 0.79 in 2 in 2 in in bar 12 = 7.13 , 2 ft bar in 1.33 ft of wall 0.79 use #8 @ 6 in
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.
Fresist = Fsliding FS FS = Factor of Safety = 1.5 for sliding set Fresist = WT (tan natural soil ) tan natural soil = tan(32 o ) = 0.62 WT = W fill + Wstem + W found
15in 18
ft
12in
W fill = (100 pcf )(16.67 ft )( wheel )(1 ft ) = 1670 Wstem = (150 pcf )(16.67 ft )( W found
lb wheel ft
12 in + 15in 1 ft (1 ft ) = 2810 lb in 2 12 ) 16 15 = (150 pcf )( ft )( wheel + ft + 3 ft )(1 ft ) = 200 plf wheel + 850 12 12
Fsliding = Pfill + Psur Pfill = Psur 1 (0.31 100 pcf )(18 ft ) 2 (1 ft ) = 5020 lb 2 = (0.31 400 psf )(18 ft )(1 ft ) = 2230 lb
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lb lb lb lb 1670 ft wheel + 2810 + 200 ft wheel + 850 (0.62) 7250lb = 1.5 1.5 lb 7250lb = 3660lb + 1870 wheel , wheel = 7.42 ft , 0.62 ft
5. Check Overturning.
M over = P fill ( M over 18 ft 18 ft ) + Psur ( ) 3 2 = 5.02 k (6 ft ) + 2.23k (9 ft ) = 50.2 k ft
12in
18
ft
M resist = W fill (
M resist
7.5 ft 15 + ft + 3 ft ), assume wtoe = 3 ft 2 12 1.25 ft + Wstem (3 ft + ) 2 3' 15" 7.5' 11.75 ft + W found ( ) 2 = (1.67 klf 7.5 ft )(8 ft ) + ( 2.81k )(3.625 ft ) + (0.20klf 7.5 ft + 0.85k )(5.875 ft )
12.53k 2.35k
tf = 16in
M resist = 124.2 k ft M resist 124.2 k ft = = 2.47 > 2.0 = FSover , OK M over 50.2 k ft
6. Check Bearing.
v at end of toe =
WT = W fill + Wstem WT M L + , equation is valid only if e < bL bL2 6 6 + W found
WT = 12.45 k + 2.81k + 2.35 k = 17.69 k M = M over W fill (5.875 ft M = 50.2 k ft 7.5 ft 1.25 ft ) + Wstem (7.5 ft + 5.875 ft ) + W found (0) 2 2 12.53k ( 2.125 ft ) + 2.81k ( 2.25 ft ) = 29.9 k ft
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v =
17.69 k (1 )(11.75 )
ft ft
7. Heel Design. Max. load on heel is due to the weight of heel + fill + surcharge as the wall tries to tip over. Flexure:
W = Wheel + W fill + Wsur
W = 1.2(150 pcf )( + 1.6( 400 plf ) W = 2.88 Mu =
klf
Vu Mu
wu 16 7.5
ft in
Shear controls the thickness of the heel. 1 t heel = 21.9 in + 2 in cover + in = 24.4 in (assume #8 bar ), 2 Reinforcement in heel: M u = 47.7 ksi As d
in 81.0 k ft (12 ) = 47.7 ksi As (21.9 in ), As = 1.07in 2 ft in 2 0.79 bar (12 in ) = 8.83 in , ft in 2 1.07 ft use #8 @ 8"
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v =
Wu = 1.2(12.53k + 2.81k + 2.35k ) + 1.6(0.4 ksf )(18 ft )(1 ft ) = 32.7 k , (did not recalc foundation wt b.c. neglible change) M u = 1.6 M over 1.2(Wsoil 2.125 ft + Wstem 1.0 ft ) 3' 1.25' 7.5'
M u = 1.6(50.2 k ft ) 1.2 12.53k ( 2.125 ft ) + 2.81k (1 ft ) = 45.0k ft
v = vA
32.7 k 45.0k ft + (1 ft )(11.75 ft ) 1 (1 ft )(11.75 ft ) 2 6 ksf = 2.78 + 1.96ksf = 4.74 ksf 1.96ksf = 0.82 ksf
A B C
d for flexure:
M u = (3.74 ksf )(3 ft )(1 ft )( 3 ft 1 2 ) + (1.00 ksf )(3 ft )(1 ft )( 3 ft ) = 19.8k ft 2 2 3
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"
vcritical sec tion = 0.82 ksf +
4.74 ksf 0.82 ksf 18 (8.75 ft + ft ) = 4.24 ksf ft 12 11.75
Vu =
Vc = (.75)2 3000 psi (12in )(18in ) = 17,750lb > Vu , OK , d for flexure controls
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Reinforcement in toe:
M u = 47.7 ksi As d 19.8k ft (12 0.79 in ) = 47.7 ksi As (18in ), As = 0.28 in 2 ft
in 2 bar (12 in ) = 33in , try smaller bars , say #4 ft in 2 0.28 ft use #4 @ 8"
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