Retaining Wall Design Example

Retaining Wall Design Example

CE 437/537, Spring 2011

Design a reinforced concrete retaining wall for the following conditions.

1/8

surcharge = qs = 400

f'c = 3000 psi fy = 60 ksi

psf

Fill: φ = 32o Unit wt = 100 pcf HT = 18 ft

tf wtoe tstem wheel Natural Soil: φ = 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 = φM n

a⎞ ⎛ M n = As f y ⎜ d − ⎟ 2⎠ ⎝ a⎞ ⎛ M u = φAs f y ⎜ d − ⎟ [ Eqn 1] 2⎠ ⎝ C = T , 0.85 f c' a b = As f y As = 0.85

f c' ab fy

strain compatibility :

[ Eqn 2]

0.003 ε s + 0.003 a 0.003 = , = β1 a / β1 d d ε s + 0.003

Assuming β1 = 0.85, εs 0.005 0.00785 0.010

a/d 0.319 0.235 0.196

and choosing a value for εs in about the middle of the practical design range, a = 0.235, a = 0.235 d d

[ Eqn 3]

CE 437/537, Spring 2011

Retaining Wall Design Example

2/8

Substituting Eqn. 2 into Eqn. 1: ⎛ f' ⎞ ⎛ a⎞ M u = φ ⎜ 0.85 c ab ⎟ f y ⎜ d − ⎟ ⎜ ⎟ f 2⎠ y ⎝ ⎠ ⎝

And substituting Eqn. 3 into the above: M u = φ 0.85

f c' 0.235 d ⎞ ⎛ 0.235d b f y ⎜ d − ⎟ fy 2 ⎠ ⎝ 0.883d

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 = 0.90(0.85) 3ksi (12 in )(0.235)(0.883)d 2

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 = φ 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"

Retaining Wall Design Example

CE 437/537, Spring 2011

3/8

3. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on by the horizontal earth pressure.

ft

in

h = 8 – 16 /12 ft h =16.67

in/ft

h

tf = 16in

ka γ h

ka qs

wtoe tstem wheel

calc. d: Pfill =

Pfill =

1 ( k a γ h ) h (1 ft out of page) 2 1 − sin φ 1 − sin( 32 o ) ka = = = 0.31 1 + sin φ 1 + sin( 32 o ) 1 (0.31)(100 pcf )(16.67 ft ) 2 (1 ft ) = 4310 lb 2

Psur = k a q sur h (1 ft ) = 0.31 ( 400 psf )(16.67 ft )(1 ft ) = 2070 lb h h M u = ( Earth Pressure LoadFactor)( Pfill )( ) + ( Live LoadFactor)(Psur )( ) 3 2 ft ft 16 . 67 16 . 67 M u = (1.6)(4310 lb )( ) + (1.6)(2070 lb )( ) = 65.9 k − ft 3 2

Mu =

k in 5.71

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

use t stem = 15in

CE 437/537, Spring 2011

Retaining Wall Design Example

4/8

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). 12in

Neglect soil resistance in front of the wall. Fresist = Fsliding FS FS = Factor of Safety = 1.5 for sliding set

18

ft

Fresist = (Vertical Force)(coefficient of friction) Fresist = WT (tan φnatural soil )

tf = 16in

tan φnatural soil = tan(32 o ) = 0.62 WT = W fill + Wstem + W found

W fill = (100 pcf )(16.67 ft )( wheel )(1 ft ) = 1670

15in

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

Wstem = (150 pcf )(16.67 ft )( W found

Fsliding = 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

Pfill = Psur

Fsliding = 5020 lb + 2230 lb = 7250 lb

CE 437/537, Spring 2011

Retaining Wall Design Example

5/8

⎡ 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

use wheel = 7.5 ft

5. Check Overturning.

12in

18 ft 18 ft ) + Psur ( ) 3 2 = 5.02 k (6 ft ) + 2.23k (9 ft ) = 50.2 k − ft

M over = P fill ( M over

18

ft

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 )

M resist = W fill (

M resist

12.53k

2.35k

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. WT M L + , equation is valid only if e < bL bL2 6 6 + W found

σ v at end of toe = WT = W fill + Wstem

WT = 12.45 k + 2.81k + 2.35 k = 17.69 k 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

M = M over − W fill (5.875 ft − M = 50.2 k − ft

Check that e < L/6: e=

m 29.9 k − ft L 11.75 ft L ft = = 1 . 68 , = = 1.96 ft , ∴ e < , OK k WT 6 6 6 17.69

tf = 16in

Retaining Wall Design Example

CE 437/537, Spring 2011

σv =

17.69 k ft

ft

(1 )(11.75 )

+

29.9 k − ft 1 ft (1 )(11.75 ft ) 2 6

6/8

= 2.80 ksf < 5.0 ksf = allowable bearing capacity, OK

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 16 ft )(1 ft ) 12 + 1.2(100 pcf )(16.67 ft )(1 ft )

W = 1.2(150 pcf )( + 1.6( 400 plf ) W = 2.88 Mu =

Vu

wu 16

Mu 7.5

in

ft

klf

wu L2 2.88klf (7.5 ft ) 2 = = 81.0 k − ft 2 2

k 2 d in in k 81.0 k − ft (12 ) = 5.71 d 2 , d = 13.0in for flexure ft in M u = 5.71

Shear: Vu = wu (7.5 ft ) = 2.88 klf (7.5 ft ) = 21.6 k

φVc = (0.75) 2 f c' bw d = (0.75) 2 3000 psi (12 in ) d setVu = φVc ,

21,600lb = (0.75) 2 3000 psi (12 in ) d , d = 21.9 in for shear, controls

Shear controls the thickness of the heel. 1 t heel = 21.9 in + 2 in cover + in = 24.4 in (assume #8 bar ), 2

use t heel = 21.5 in

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"

Retaining Wall Design Example

CE 437/537, Spring 2011

7/8

8. Toe Design. Earth Pressure at Tip of Toe: Wu Mu ± bL 1 2 bL 6 Wu = 1.2(W fill + Wstem + W found ) + 1.6 (Wsur )

σv =

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 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

σv = σ vA

σ vC = 2.78ksf σ vB = 0.82 ksf +

−

1.96ksf

= 0.82 ksf

A

B

4.74 ksf − 0.82 ksf (8.75 ft ) = 3.74 ksf 11.75 ft

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

k 2 d in in k 19.8k − ft (12 ) = 5.71 d 2 , d = 6.5in for flexure ft in M u = 5.71

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" σ vcritical sec tion = 0.82 ksf +

Vu =

4.74 ksf − 0.82 ksf 18 (8.75 ft + ft ) = 4.24 ksf ft 12 11.75

1 18 ( 4.74 ksf + 4.24 ksf )(3 ft − ft )(1 ft ) = 6.74 k 2 12

φVc = (.75)2 3000 psi (12in )(18in ) = 17,750lb > Vu , OK , d for flexure controls

C

CE 437/537, Spring 2011

Retaining Wall Design Example

8/8

Reinforcement in toe: M u = 47.7 ksi As d 19.8k − ft (12

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

0.79

in 2 bar (12 in ) = 8.6in ft in 2 0.28 ft

0.20

use #4 @ 8"