Design of retaining wall with inclined surcharge loading

Compiled By: Muhammad Sajid Nazir (2005-CE-38)

DESIGN OF STRUCTURES

Design Of Retaining Wall With Surcharge Load Inclined At Some Angle

QUESTION: Design a RCC retaining wall with surface inclined at 20°and retains earth up to a height of 12’ above NSL. Base of the footing is to be placed 3’ below NSL. Soil has density of120 lb / ft3. Angle of repose is30°.

Take qall=3000lb / ft2

Usef c' =4000 Psi∧f y=60,000 Psi.

SOLUTION:

Given that,

Height of Retaining wall above NSL=12’

Depthof footing below NSL=3 ’

Density of soil , γ so il=120 lb / ft3

Density of concrete , γconc .=150 lb / ft3

Angleof internal friction ,φ=30°

Inclinationof surface ,Ѳ=20° f c' =4000 Psif y=60,000 Psi

STEP 1ASSUMPTION OF SIZES:

Base thickness=stemthickness= H12

H10

=15 ’× 12”12

=15 ”=1.25 ’

Basewidth=23

H =23

×15'=10 ’Widthof toe=13

×10'=3.3 ’

Widthof heel=10'−3.3'−1.25 '=5.45'Height of surcharge=5.45 × tan 20°=1.97 '

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STEP 2DESIGN OF STEM:

i. Calculation Of Soil Pressure:

Ca=cosѲcosѲ−√cos2 Ѳ−cos2 φ

cosѲ+√cos2 Ѳ−cos2 φCa=cos 20° cos20°−√cos220°−cos230°

cos 20°+√cos220°−cos230°=0.414

Pa=12

γ Ca H 2Pa=12

× 120 ×0.414 × 13.75'2=4696.3 lb

PH=Pa ×cosѲ=4696.3 ×cos 20°=4412.8 lb

ii. Moment & Reinforcement Calculation:

Centroidal distance , y=H3

=13.75'

3=4.58 '

M=PH × y=4412.8× 4. 58'=20210.62 lbftM max=1.6 × M=1.6 × 20210.62

M u=M max=32337 lbft=388044 lbin

ρ=0.85f c

'

f y [1−√1−2M u

0.85 f c' φb d2 ]

ρ=0.85×4000

60000 [1−√1− 2× 3880440.85× 4000 ×0.9 × 12 {×12.52 ¿]

Where ,φ=0.9∧b=12 & d=h- 2.5 =15 -2.5=12.5 ρ=0.0039

ρmin=200f y

= 20060000

=0.0033

ρmax=0.75 ρb=0.75 × β1× 0.85f c

'

f y

×( 8700087000+ f y

)ρmax=0.75 × 0.85× 0.85 ×

400040000

×( 8700087000+60000 )ρmax=0.021

ρmax >ρ>ρmin OK

iii. Selection of bars & spacing:

Vertical Reinforcement On front face of wall:

A st=ρbd A st=0.0039 ×12 ×12.5=0.585¿2Provide¿6 bars @9c/c see Nilson page 736

Temperature & Shrinkage Steel:

A st=ρbh A st=0.0018 ×12 ×15=0.324¿2Provide¿3 bars@ 4 c/c see Nilson page 736

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iv. Check for shear:

V u=1.6 × PaV u=1.6 ×12

×Ca× γ × ( H−d )2V u=1.6 ×12

×0.414 × 120× ¿¿

V u=6418.725lb / ft=534.9 lb /¿

v. Capacity of Section:

¿2 φ√ f c' ×bd¿2 ×0.9 ×√4000× 12×12.5¿17076.3 lb /¿>V u OK

STEP 3

CURTAILMENT OF BARS:

Bars are curtailed from top where B.M isM max

2

M u=M max=32337 lbftMu

2=

M max

2=16168.5 lbft

M max

2=1

2×1.6 × γ ×Ca × H 1

2 cosѲ×H 1

3

16168.5=12

× 1.6 ×120 ×0.414 × H 12 cos20° ×

H 1

3H 1=10.91 '

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

12 12 ×dia of ¿12 ×1 =12

0.04 × Ab× f y

√ f c'

=0.04× 0.44 ×60000√4000

=16.7 =1.39

Maximum of abovethree values is selected ,i . e . 12” is selected .So ,Curtailment=10.91 '−1.39'=9.52 ' ¿ top .

ρ forM max

2,

ρ=0.00392

ρ=0.00195>min . value of temp .∧shrinkage steel i . e .0 .0018

A st' =Areaof curtailed steel A st

' =ρbh=0.00195× 12×15=0.351¿2

Areaof steel , A s=ρbd=0.585¿2

12

A s23

A s0.2925¿2 0.39¿2

0.39¿2 is moreclose¿0.585¿2 socurtail each2nd alternate .

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SECTION & PLAN

Step 4

STABILITY CHECKS:

There are three stability checks,

i. Check For Overturning Moment

Resisting momentOverturning moment

>2

Here use

H=13.75 '+1.97 '=15.72'Pa=12

γ Ca H 2

Pa=12

× 120 ×0.414 × 15.72'2=6138.42lb

PH=Pa ×cos20°=6138.42 ×cos 20°=5768.23 lb

PV =Pa× sin 20°=6138.42× sin 20°=2099.5 lb

Centroidal distance , y=H3

=15.72'

3=5.24'

O .T .M =PH × yO .T .M =5768.23× 5.24 '

O .T .M =30225.53lbft

Sr. #

Weight ( Area × Density )lb x“ ft ” Resiting Moment lbft

01 13.75 ' ×5.45 ' ×1×120=8992.55.45

2+1.25+3.3=7.28 ' 65465.4

4




02 10 ' ×1.25 ' ×1×150=1875102

=5 ' 9375

03 1.75 ' ×3.3 ' ×1 ×120=6933.32

=1.65 ' 1143.45

04 13.75 ' ×1.25 ' ×1×150=2578.1251.25

2+3.3=3.925 ' 10119.15

0512

×1.97 × 5.45× 120=644.19 3.3+1.25+ 2 ×5.453

=8.18 ' 5269.47

∑W =14782.815 lb ∑ M =91372.47

Resisting moment or Stablizing moment=∑ M +Pv ×10¿91372.47+Pasin 20°× 10

¿91372.47+6138.42sin 20°× 10¿112367.1 lbft

Resisting momentOverturning moment

>2112367.130225.53

=3.72>2 OK

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

≥1.5Slidingforce=PH=5768.23 lb

Resisting force=μ (∑W +PV )=0.3× (14782.815+2099.5 )Resisting force=5064.7 lb

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

Resisting forceSliding force

≥1.55064.7

5768.23=0.88 ≥1.5 NOT OK

So we have to provide key under the base.

Design of Key:

Resisting force=1.5 × Sliding forceResisting force=1.5 ×5768.23Resisting force=8652.345 lb

Additionalresisting force provided ,Pp=8652.345−5064.7

Additionalresisting force¿be provided ,P p=3587.65 lb

Additional Resiting force , , Pp=12

× γ × H12×

1Ca

3587.65=12

× 120 × H 12×

10.414

H 1=4.98 ' ≈ 5'

Depthof Key=5 '−1.25'=3.75 '

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iii. Check For Bearing Capacity Of Soil:

a=Stablizing moment−O . T . MRV

RV =∑W +PV =14782.815+2099.5=16882.315 lb

a=112367.1−30225.5316882.315 a=4.87 'Eccentricity , e=B

2−ae=10

2−4.87 '

e=0.13 '

e≯?B6

=106

=1.67 '0.13' ≯? 1.67 ' OK

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

qmax=RV

B ×1+

RV × e

B2

6

≯? qallqmax=

16882.31510

+16882.315 × 0.13

102

6qmax=1819.91lb / ft 2≯? qall OK

qmin=RV

B ×1+

RV × e

B2

6

≮? zeroqmin=16882.315

10−16882.315 ×0.13

102

6qmin=1556.55lb / ft2OK

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

STEP 5

DESIGN OF TOE & HEEL:

263.3610

=x1

5.45x1=143.53 lb / ft2

263.3610

=x2

1.25x2=32.92 lb / ft2

263.3610

=x3

3.3x3=86.91 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 [17 00.08× 3.3 ×3.32

+( 12

×86.91 ×3.3)×23

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

M u=14 397.03 lbft=17 2764.32 lbin

ρ=0.85f c

'

f y [1−√1−2M u

0.85 f c' φb d2 ]ρ=0.85×

400060000 [1−√1− 2 ×17 2764.32

0.85× 4000 ×0.9 × 12× 12.52 ]Where ,φ=0.9∧b=12 & d=h- 2.5 =15 -2.5=12.5 ρ=0.0017ρmin=

200f y

= 20060000

=0.0033

ρmax=0.75 ρb=0.75 × β1× 0.85f c

'

f y

×( 8700087000+ f y

)ρmax=0.75 × 0.85× 0.85 ×

400040000

×( 8700087000+60000 )ρmax=0.021

ρmax>ρ>ρmin NOT OK

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Selection of bars & spacing:

A st=ρmin bd A st=0.0033 ×12 ×12.5 A st=0.495¿2Provide¿6 bars @10 c/

ii. Design of heel:

Taking moment about junction of heel and stem,

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

M u=1.2[ (1.25 ×5.45 ×150 )× 5.452 ]+1.6 [(13.75 ×5.45× 1×120 )× 5.45

2+ 1

2×1.97 × 5.45× 1× 120×

2 ×5.453 ]

M u=44889.38lbft=538672.66 lbin

ρ=0.85f c

'

f y [1−√1−2M u

0.85 f c' φb d2 ]ρ=0.85×

400060000 [1−√1− 2 ×538672.66

0.85× 4000 ×0.9 × 12× 12.52 ]Where ,φ=0.9∧b=12 & d=h- 2.5 =15 -2.5=12.5 ρ=0.005ρmin=

200f y

= 20060000

=0.0033

ρmax=0.75 ρb=0.75 × β1× 0.85f c

'

f y

×( 8700087000+ f y

)ρmax=0.75 × 0.85× 0.85 ×

400040000

×( 8700087000+60000 )ρmax=0.021

ρmax >ρ>ρmin OK

Selection of bars & spacing:

A st=ρbd A st=0.005 ×12 ×12.5 A st=0.75¿2Provide¿6 bars @7 c/c

Temperature & Shrinkage Steel:

A st=ρbh A st=0.0018 ×12 ×15 A st=0.324 ¿2Provide¿3 bars@ 4 c/

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Design of retaining wall with inclined surcharge loading

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