Design of Cantilever Retaining WallCreated by: David Taylor

Given: UMKC SCE f'c= 3 ksi Senior Designfy= 60 ksi

4000 psf

100 psf19 feet h1= 2.5 feet

h2= 1.5 feeth3= 8.5 feeth4= 12.5 feet

2.5 feet h5= 1 feetΦ= 31μ= 0.5 (friction coeff)

300 psfFreeze depth= 5

Ca= 0.32Cp= 3.12

1.1 Estimate Footing length and stem position688 lb

7398 lb

2065 lbW= 2150XX= 8.36 feet (distance from heel to face of stem)

12.55 feet

1.2 Overturning MomentOverturning Moment

53021.038 75215.89 ft lb

22194.853Righting Moment

Force Moment Arm Moment

4688 lb * 6.3 ft = 29296.88 ft-lb

713 lb * 2.8 ft = 1959.375 ft-lb

2850 lb * 3.5 ft = 9975 ft-lb

18700 lb * 8.3 ft = 154275 ft-lb

26950 lb 195506 ft-lb

Safety Factor against overturni = 2.60 >2.0 => OKAYSafety Factor against sliding = 2.98 >1.5 => OKAY

*Assumes soils on both sides are the same

qa=

үsoil

=

Surcharge=feet (make sure you can meet this)

ρa=

H1=

H2=

Lfooting

=

MH1

= Moverturn

=

MH2

=

W1=

W2=

W3=

W4=

Rv= M

Righting =

h1 h2 h3

h4

h5

2.1 Footing Soil PressuresRv= 26950 lb located @ x distance from the toe of footing

x= 4.46 ft *Verify that it is in the middle third of footing

-4004.85 psf

-307.15 psf

ftoe

=

fheel

=

3 Stem Design3.1 Design of stem for moment

86.28 ft-k*Using a load factor of 1.6 with forces

ρ= 0.009 ** Find the equilalent Mu/bd^2 from Table A.12 of Book "Design of Reinf. Concrete"

482.6 psi

2383.798 h= 16.59 in

*assume b= 12 ROUND UP to next best number

h= 18 in(d= 15.5 in)

Therefore,

399.04

ρ= 0.0073 ** Find the equilalent ρ from Table A.12 of Book "Design of Reinf. Concrete"

1.3578 Use #8 @ 6" (1.57 in^2)

3.2 Minimum Vertical ρ 0.0015 OKAY(ACI 318 sect. 14.3)

0.45

*Assume one third inside face and two-thirds are outside face.USE #4 @ 7.5" outside face and #4 @ 15" inside face.

3.4 Checking Shear Stress in Stem

12163.76 lbs

15281.4594 lbs => Therefore, OKAY φVc>Vu

Mu=

bd2=

As= in2 =>

3.3 Minimum Horiz. As= in2

Vu=

φVc=

MU

bd 2=

MU

bd 2=

4 Footing Design4.1 Design of Heel*The upward soil pressure is conservatively neglected, and a load factor of 1.2 is used for calculating the shear and

moment because soil and concrete make up the load. Assume d = footing depth - 3.5"

24735 lb

26126.366 lb => Therefore, OKAY φVc>Vu

105124 ft-lb

166

ρ= 0.00333 ** Find the equilalent ρ from Table A.12 of Book "Design of Reinf. Concrete"

Using ρ,

1.06 => Use #9 @ 11"

81 in. available.Note: Temperature and shrinkage steel is normally considered unnecessary in the heel and toe.However, #4 bars @ 18" in. O/C in the long direction are used to serve as spacers for the flexuralsteel and to from mats out of the reinforcing.

4.2 Design of Toe* For service loads, the soil pressures previously determined (in sect. 2.1) are multiplied by 1.6 by a load factor of 1.6

because they are primarily caused by lateral forces.

1281.55 + 5598.157 = 6880 lb

6801 ft-lb

285.2

ρ= 0.00512 ** Find the equilalent ρ from Table A.12 of Book "Design of Reinf. Concrete"

Using ρ,

1.63 => Use #9 @ 6"

48 in. available.

Vu=

φVc=

Mu at the face of stem =

[ρmin

=0.0033]

As= in2/ft

ℓd required calculated with ACI Equation 12-1 for #9 top bars with c=2.50 in. and K

tr = 0 is 49 in. <

Vu=

Mu at the face of stem =

[ρmin

=0.0033]

As= in2/ft

ℓd required calculated with ACI Equation 12-1 for #9 top bars with c=2.50 in. and K

tr = 0 is 38 in. <

MU

bd 2=

MU

bd 2=

4.3 Selection of Dowels and Lengths of Vertical Stem Reinforcing

ρ

5 2987.589 11.08 24.34 0.0033 0.44 #8@ 18"10 16218.34 12.66 101.22 0.0033 0.50 #8 @ 18"15 46094.23 14.24 227.42 0.00388 0.66 #8 @ 12"19 86281.57 15.50 359.13 0.00648 1.21 #8 @ 6"

Distance from top of stem (ft)

Mu (ft-lb)

Effective stem d

(in.)M

u/bd2 A

s req'd

(in2/ft)Bars

needed