f'c = 3000 psi fy = 60 ksi
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. 1 1 ' ' 003 . 0 003 . 0 , 003 . 0 / 003 . 0 : ] 2 [ 85 . 0 85 . 0 , ] 1 [ 2 2 β ε ε β φ φ + = + = = = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = = s s y c s y s c y s u y s n n u d a d a ity compatibil strain Eqn ab f f A f A b a f T C Eqn a d f A M a d f A M M M Assuming β1 = 0.85, εs a/d 0.005 0.319 0.00785 0.235 0.010 0.196
and choosing a value for εs in about the middle of the practical design range,
] 3 [ 235 . 0 , 235 . 0 a d Eqn d a = = Fill: φ = 32o Unit wt = 100 pcf Natural Soil: φ = 32o
allowable bearing pressure = 5000psf wtoe tstem wheel
HT = 18 ft
Substituting Eqn. 2 into Eqn. 1: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 2 85 . 0 ' ab f d a f f M y y c u φ
And substituting Eqn. 3 into the above:
d d d f b d f f M y y c u 883 . 0 2 235 . 0 235 . 0 85 . 0 ' ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =φ
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).
2 2 5.71 ) 883 . 0 )( 235 . 0 )( 12 ( 3 ) 85 . 0 ( 90 . 0 d M d M in k u in ksi u = =
Equation for area of reinforcement, As. The area of reinforcement required is calculated
from Eqn. 1: d A M d A d f A Mu =φ s y0.883 =0.90 s60ksi0.883 u =47.7ksi s
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)
horizontal earth pressure. calc. d: lb ft ft ft sur a sur lb ft ft fill o o a ft a fill psf h q k P pcf P k page of out h h k P 2070 ) 1 )( 67 . 16 )( 400 ( 31 . 0 ) 1 ( 4310 ) 1 ( ) 67 . 16 )( 100 )( 31 . 0 ( 2 1 31 . 0 ) 32 sin( 1 ) 32 sin( 1 sin 1 sin 1 ) 1 ( ) ( 2 1 2 = = = = = = + − = + − = = φ φ γ in in in in in stem in in in in stem in in k ft k in k u ft k ft lb ft lb u fill u d t use bars assume t d d ft in d M M h h P M 5 . 12 5 . 0 2 15 15 ) 8 # ( , 3 . 14 ) 0 . 1 ( 2 1 cover 2 8 . 11 8 . 11 , 71 . 5 ) 12 ( 9 . 65 71 . 5 9 . 65 ) 2 67 . 16 )( 2070 )( 6 . 1 ( ) 3 67 . 16 )( 4310 )( 6 . 1 ( ) 2 )( )(P LoadFactor Live ( ) 3 )( )( LoadFactor Pressure Earth ( 2 2 sur = − − = = = + + = = = = = + = + = − −
wtoe tstem wheel
h = 8ft – 16in/12in/ft h =16.67ft
tf = 16in
ka γ h ka qs
calc. As: in use bar in ft in wall of ft in bar in in bar one of A in A A ft in d A M s s in s ksi ft k s ksi u 6 @ 8 # , 13 . 7 12 33 . 1 79 . 0 79 . 0 8 # 33 . 1 ), 5 . 12 ( 7 . 47 ) 12 ( 9 . 65 7 . 47 2 2 2 2 = = = = = −
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. found stem fill T o soil natural soil natural T sliding resist W W W W W F FS F set + + = = = = = = = = 62 . 0 ) 32 tan( tan ) (tan F friction) of fficient Force)(coe (Vertical F sliding for 1.5 Safety of Factor FS resist resist φ φ 850 200 ) 1 )( 3 12 15 )( 12 16 )( 150 ( 2810 ) 1 ( ) 12 1 2 15 12 )( 67 . 16 )( 150 ( 1670 ) 1 )( )( 67 . 16 )( 100 ( + = + + = = + = = = heel ft ft heel found lb ft in ft in in ft stem heel ft heel ft fill w plf ft w ft pcf W pcf W w ft lb w pcf W lb lb lb sliding lb ft ft sur lb ft ft fill sur fill sliding F psf P pcf P P P F 7250 2230 5020 2230 ) 1 )( 18 )( 400 31 . 0 ( 5020 ) 1 ( ) 18 )( 100 31 . 0 ( 2 1 2 = + = = × = = × = + = 18ft tf = 16in 15in 12in
ft heel ft heel heel lb lb w w usew ft lb 5 . 7 , 42 . 7 , 1870 3660 62 . 0 5 . 1 7250 5 . 1 7250 = = + = = 5. Check Overturning. ) 2 75 . 11 ( ) 2 25 . 1 3 ( 3 ), 3 12 15 2 5 . 7 ( 2 . 50 ) 9 ( 23 . 2 ) 6 ( 02 . 5 ) 2 18 ( ) 3 18 ( ft found ft ft stem ft toe ft ft fill resist ft k ft k ft k over ft sur ft fill over W W w assume ft W M M P P M + + + = + + = = + = + = − OK FS M M M M over ft k ft k over resist ft k resist ft k ft klf ft k ft ft klf resist , 0 . 2 47 . 2 2 . 50 2 . 124 2 . 124 ) 875 . 5 )( 85 . 0 5 . 7 20 . 0 ( ) 625 . 3 )( 81 . 2 ( ) 8 )( 5 . 7 67 . 1 ( = > = = = + × + + × = − − − 6. Check Bearing. ft k ft k ft k ft k found ft ft ft stem ft ft fill over k k k k T found stem fill T T v M W W W M M W W W W W bL M bL W toe of end at − − − + = = + − + + − − = = + + = + + = < + = 9 . 29 ) 25 . 2 ( 81 . 2 ) 125 . 2 ( 53 . 12 2 . 50 ) 0 ( ) 875 . 5 2 25 . 1 5 . 7 ( ) 2 5 . 7 875 . 5 ( 69 . 17 35 . 2 81 . 2 45 . 12 6 L e if only valid is equation , 6 2 σ Check that e < L/6: OK L e L W m e k kft ft ft ft T , 6 , 96 . 1 6 75 . 11 6 , 68 . 1 69 . 17 9 . 29 < ∴ = = = = = − 12.53k 2.35k 18ft tf = 16in 12in 3' 15" 7.5'
OK capacity, bearing allowable 0 . 5 80 . 2 ) 75 . 11 )( 1 ( 6 1 9 . 29 ) 75 . 11 )( 1 ( 69 . 17 2 = < = + = − ksf ksf ft ft ft k ft ft k v σ 7. Heel Design.
Max. load on heel is due to the weight of heel + fill + surcharge as the wall tries to tip over. Flexure: sur fill heel W W W W = + + ft k ft klf u u klf ft ft ft L w M W plf pcf ft pcf W − = = = = + + = 0 . 81 2 ) 5 . 7 ( 88 . 2 2 88 . 2 ) 400 ( 6 . 1 ) 1 )( 67 . 16 )( 100 ( 2 . 1 ) 1 )( 12 16 )( 150 ( 2 . 1 2 2 flexure for d d in k ft in d in k M in ft k u 0 . 13 , 71 . 5 ) 12 ( 0 . 81 71 . 5 2 2 = = = − Shear: controls shear for d d psi V setV d psi d b f V w V in in lb c u in w c c k ft klf ft u u , 9 . 21 , ) 12 ( 3000 2 ) 75 . 0 ( 600 , 21 , ) 12 ( 3000 2 ) 75 . 0 ( 2 ) 75 . 0 ( 6 . 21 ) 5 . 7 ( 88 . 2 ) 5 . 7 ( ' = = = = = = = = φ φ
Shear controls the thickness of the heel.
in heel
in in
in
heel in assume bar uset
t 24.4 ( #8 ), 21.5 2 1 cover 2 9 . 21 + + = = = Reinforcement in heel: " 8 @ 8 # , 83 . 8 ) 12 ( 07 . 1 79 . 0 07 . 1 ), 9 . 21 ( 7 . 47 ) 12 ( 0 . 81 7 . 47 2 2 2 use ft in ft in bar in in A A ft in d A M in s in s ksi ft k s ksi u = = = = − 7.5ft 16in Mu wu Vu
8. Toe Design.
Earth Pressure at Tip of Toe:
change) neglible b.c. wt foundation recalc not (did , 7 . 32 ) 1 )( 18 )( 4 . 0 ( 6 . 1 ) 35 . 2 81 . 2 53 . 12 ( 2 . 1 ) ( 6 . 1 ) ( 2 . 1 6 1 2 k ft ft ksf k k k u sur found stem fill u u u v W W W W W W bL M bL W = + + + = + + + = ± = σ
[
]
ksf ft ft ksf ksf ksf v ksf ksf ksf v ksf ksf ksf v ft ft ft k ft ft k v ft k ft k ft k ft k u ft stem ft soil over u B C A M W W M M 74 . 3 ) 75 . 8 ( 75 . 11 82 . 0 74 . 4 82 . 0 82 . 0 96 . 1 78 . 2 74 . 4 96 . 1 78 . 2 ) 75 . 11 )( 1 ( 6 1 0 . 45 ) 75 . 11 )( 1 ( 7 . 32 0 . 45 ) 1 ( 81 . 2 ) 125 . 2 ( 53 . 12 2 . 1 ) 2 . 50 ( 6 . 1 ) 0 . 1 125 . 2 ( 2 . 1 6 . 1 2 = − + = = − = = + = + = = + − = × + × − = − − − σ σ σ σ d for flexure: flexure for d d in k ft in d in k M M in ft k u ft k ft ft ft ksf ft ft ft ksf u 5 . 6 , 71 . 5 ) 12 ( 8 . 19 71 . 5 8 . 19 ) 3 3 2 )( 1 )( 3 )( 00 . 1 ( 2 1 ) 2 3 )( 1 )( 3 )( 74 . 3 ( 2 2 = = = = + = − − 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"
controls flexure for d OK V psi V ft V ft u lb in in c k ft ft ksf ksf u ksf ft ft ksf ksf ksf vcritical tion , , 750 , 17 ) 18 )( 12 ( 3000 2 ) 75 (. 74 . 6 ) 1 )( 12 18 3 )( 24 . 4 74 . 4 ( 2 1 24 . 4 ) 12 18 75 . 8 ( 75 . 11 82 . 0 74 . 4 82 . 0 sec > = = = − + = = + − + = φ σ A B C 3' 1.25' 7.5'
Reinforcement in toe: " 8 @ 4 # 6 . 8 ) 12 ( 28 . 0 20 . 0 4 # , , 33 ) 12 ( 28 . 0 79 . 0 28 . 0 ), 18 ( 7 . 47 ) 12 ( 8 . 19 7 . 47 2 2 2 2 2 use ft in ft in bar in say bars smaller try ft in ft in bar in in A A ft in d A M in in s in s ksi ft k s ksi u = = = = = −
