QUESTION: Design a RCC retaining wall to retain earth up to a height of 15’. Base of the footing is to be placed 3’ below NSL. Soil has density of120 lb / ft3. Angle of internal friction of soil is30°. Earth surface that is to be retained is horizontal. Take qall=2500lb / ft2 Usef c

' =4000 Psi∧ f y=40,000 Psi.

SOLUTION:

Given that,

Height of Retaining wall above NSL=15 ’

Depth of footing below NSL=3 ’Density of soil , γ soil=120lb / ft3

Density of concrete , γ conc .=150 lb / ft3

Angleof internal friction , φ=30°f c' =4000 Psi

f y=40,000 Psi

STEP 1ASSUMPTION OF SIZES:

Base thickness=stemthickness= H12

H10

=15 ’×12 ”12

=15”=1.25’

Base width=23

H=23

×15'=10 ’Widthof toe=13

× 10'=3.3 ’

Widthof heel=10'−3.3 '−1.25 '=5.45 '

STEP 2DESIGN OF STEM:

i. Calculation Of Soil Pressure:

Ca=1−sin φ1+sin φ Ca=

1−sin 30°

1+sin 30° =0.33

Pa=12

γ Ca H 2Pa=12

× 120× 0.33 ×15' 2=4455lb

ii. Moment & Reinforcement Calculation:

Centroidal distance , y= H3

=15'

3=5 '

M=Pa × y=4455 ×5'=22275lbft M max=1.6 × M=1.6 ×22275

M u=M max=35640 lbft=427680 lbin

ρ=0.85f c

'

f y [1−√1−2 M u

0.85 f c' φb d2 ]

ρ=0.85× 400040000 [1−√1− 2× 427680

0.85× 4000 × 0.9× 12 {×12.52 ¿]Where ,φ=0.9∧b=12 & d=h- 2.5=15 -2.5=12.5 ρ=0.00659

ρmin=200f y

= 20040000

=0.005

ρmax=0.75 ρb=0.75 × β1× 0.85f c

'

f y×( 87000

87000+f y)

ρmax=0.75 × 0.85× 0.85× 400040000

×( 8700087000+40000 )ρmax=0.037

ρmax >ρ>ρmin OK

iii. Selection of bars & spacing:

Vertical Reinforcement On front face of wall:

A st=ρbd A st=0.00659 ×12 ×12.5=0.9885¿2Provide ¿6bars@ 5c/c see Nilson page 726

Temperature & Shrinkage Steel:

A st=ρbh A st=0.002 ×12 ×15=0.36¿2Provide¿3bars@ 3.5 c /c see Nilson page726

iv. Check for shear:

V u=1.6 × PaV u=1.6 × 12

× Ca× γ × (H −d )2V u=1.6 × 12

× 0.33 ×120 × (15 '−1.042' )2

V u=6172.375lb / ft=514.36 lb /¿

v. Capacity of Section:

¿2 φ√ f c' ×bd¿2×0.9×√4000 × 12×12.5¿17076.3 lb

¿ >V uOK

STEP 3

CURTAILMENT OF BARS:

Bars are curtailed from top where B.M isM max

2

M u=M max=35640 lbft=427680 lbin Mu

2=

M max

2=213840 lb∈¿

M max

2=1

2×1.6 × γ ×Ca × H1

2×H 1

3213840=1

2× 1.6 ×120 ×0.33 × H 1

2 ×H1

3H 1=27.26 =2.27'≈2.25

According to ACI code, the following value should be added in the curtailed value of steel,

12 12×dia of bar=12× 1=12

0.04 × Ab× f y

√ f c'

=0.04× 0.44× 40000√4000

=11.18

Maximum of above three values is selected, i.e. 12” is selected.So, Curtailment ¿2.25 ’−1’=1.25 ’ from top.

ρ forM max

2,

ρ=0.85× 400040000 [1−√1− 2 ×213840

0.85× 4000 × 0.9× 12 {×12.52 ¿]ρ=0.003295>min . value of temp .∧shrinkage steel i . e .0 .002 A st

' =Areaof curtailed steel

A st' =ρbh=0.003295× 12×15=0.593 ¿2 Areaof steel , A s=ρbd=0.998 ¿2 1

2A s

23

A s

0.494 ¿20.659 ¿2

0.659 ¿2 ismore close ¿0.593 ¿2 so curtaileach2nd alternate ¿PLAN & SECTION

According to ACI code provide main steel on the side where there is no loading & provide same temperature and shrinkage steel on load bearing side.

Step 4

STABILITY CHECKS:

There are three stability checks,

i. Check For Overturning Moment

Resisting momentOverturning moment

>2

O .T . M=Pa× y

O .T .M=3743.44 ×4.58 '

O .T .M=17144.96lbft

Sr. # Weight ( Area× Density )lb x“ ft ” Resiting Momentlbft

01 15 ' ×5.45 ' ×1×120=9810 5.452

+1.25+3.3=7.28 ' 71416.8

02 10 ' ×1.25 ' ×1 ×150=1875 102

=5 ' 9375

03 1.75 ' ×3.3 ' ×1 ×120=693 3.32

=1.65 ' 1143.45

04 15 ' ×1.25 ' ×1 ×50=2812.5 1.252

+3.3=3.925 ' 11039.06

∑W =15190.5 lb ∑ M =92974.31

Resisting momentOverturning moment

>2 92974.3122275

=4.17>2OK

1 produces overturning moment & 2, 3 and 4 produces resisting moments.

If this check is not OK then increase the thickness of toe, heel or stem as per requirement.

ii. Check For Sliding Force:

Resisting forceSliding force

≥ 1.5Sliding force=Pa=4455 lbResisting moment=μ × ∑W=0.3× 15190.5

Resisting moment=4557.15 lb

Here μ is the coefficient of friction between soil and earth and its value lies between 0.3-0.45.

Resisting forceSliding force

≥ 1.5 4557.154455

=1.02 ≥1.5 NOT OK

So we have to provide key under the base.

Key can be provided in different states as shown below,

Design of Key:

Resisting force=1.5× Sliding forceResisting force=1.5 × 4455Resisting force=6682.5 lbAdditionalresisting force provided ,P p=6682.5−4557.15

Additionalresisting force ¿be provided , Pp=2125.35 lb

Additional Resiting force , , Pp=12

× γ ×H 1

2

2× 1

Ca2125.35=1

2× 120×

H 12

2× 1

0.33H 1=3.42' ≈3.5 '

Depth of Key=3.5'−1.25'=2.25 '

iii. Check For Bearing Capacity Of Soil:

a=Stablizing moment−O .T .MStablizing force (∑W ) a=92974.31−22275

15190.5 a=4.65 '

Eccentricity , e=B2−ae=10

2−4.65 '

e=0.35 'e≯= B6

=106

=1.67 '0.35' ≯= 1.67 ' OK

If this check is not OK then increase the length of base “B”

qmax=∑WB× 1

+∑ W ×e

B2

6

≯= qallqmax=15190.5

10+15190.5 × 0.35

102

6qmax=1838.05lb / ft 2≯= qall OK

qmin=∑WB× 1

−∑W × e

B2

6

≮= zeroqmin=15190.5

10−15190.5 ×0.35

102

6qmin=1200.05lb / ft2OK

If this check is not OK then increase “B”.

STEP 5

DESIGN OF TOE & HEEL:

63810

=x1

5.45x1=347.71lb / ft2

63810

=x2

1.25x2=79.75 lb / ft2

63810

=x3

3.3x3=210.54 lb / ft2

i. Design Of Toe:

Taking moment about junction of toe and stem.

M u=1.6 ( Moment of bearing capacity )−0.9 ( Moment of concrete∈toe )

M u=1.6 [1627.51 ×3.3× 3.32

×( 12

×210 × 3.3)× 23

× 3.3]−0.9[ (3.3× 1.25× 1× 150 )× 3.32 ]

M u=8705.25 lbft=104462.97 lbin

ρ=0.85f c

'

f y [1−√1−2M u

0.85 f c' φb d2 ]ρ=0.85× 4000

40000 [1−√1− 2 ×104462.970.85× 4000 × 0.9× 12×12.52 ]

Where , φ=0.9∧b=12 & d=h- 2.5=15 -2.5=12.5 ρ=0.00156

ρmin=200f y

= 20040000

=0.005

ρmax=0.75 ρb=0.75 × β1× 0.85f c

'

f y×( 87000

87000+f y )ρmax=0.75 × 0.85× 0.85× 4000

40000×( 87000

87000+40000 )ρmax=0.037

ρmax>ρ>ρmin NOT OK

Selection of bars & spacing:

A st=ρmin bd A st=0.005 ×12 ×12.5 A st=0.75¿2Provide¿5 bars@5 c/

ii. Design of heel:

Taking moment about junction of heel and stem,

M u=1.2 (Concrete∈heel )+1.6 (soil∈heel )

M u=1.2[ (1.25 ×5.45 ×150 ) × 5.452 ]+1.6[ (15 ×5.45 ×1 ×120 ) × 5.45

2 ]Neglecting moment produced due to stresses. It will make our structure safer because it works as a FOS.

M u=46113.13 lbft=553357.56lbin

ρ=0.85f c

'

f y [1−√1−2M u

0.85 f c' φb d2 ]ρ=0.85× 4000

40000 [1−√1− 2 ×553357.560.85× 4000 × 0.9× 12×12.52 ]

Where ,φ=0.9∧b=12 & d=h- 2.5=15 -2.5=12.5 ρ=0.00863

ρmin=200f y

= 20040000

=0.005

ρmax=0.75 ρb=0.75 × β1× 0.85f c

'

f y×( 87000

87000+f y )ρmax=0.75 × 0.85× 0.85× 4000

40000×( 87000

87000+40000 )ρmax=0.037

ρmax >ρ>ρmin OK

Selection of bars & spacing:

A st=ρbd A st=0.00863 ×12×12.5 A st=1.295 ¿2Provide ¿6bars@ 4 c/

Temperature & Shrinkage Steel:

A st=ρbh A st=0.002 ×12 ×15 A st=0.36 ¿2Provide ¿3bars@ 3.5 c/

DETAILS OF STEEL IN TOE & HEEL