Design Of Structures

DESIGN OF STRUCTURES Design Of Retaining Wall

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 of 120 lb/ft3. Angle of repose is 30°. Take qall=3000 lb/ft2 Use fc'=4000 Psi & fy=60,000 Psi.

SOLUTION: Given that, Height of Retaining wall above NSL = 12’ Depth of footing below NSL=3’ Density of soil,γsoil=120 lb/ft3 Density of concrete,γconc.=150 lb/ft3 Angle of internal friction, φ=30° Inclination of surface, Ѳ=20° fc'=4000 Psi fy=60,000 Psi

STEP 1 ASSUMPTION OF SIZES: Base thickness=stem thickness =H12 H10=15’×12”12=15”=1.25’ Base width=23H =23× 15'=10’ Width of toe =13×10'=3.3’ Width of heel=10'-3.3'-1.25'=5.45' Height of surcharge=5.45×tan20°=1.97'

STEP 2 DESIGN OF STEM: i.

Calculation Of Soil Pressure:

Ca=cosѲ cosѲ-cos2Ѳ-cos2φcosѲ+cos2Ѳ-cos2φ Ca=cos20° cos20°-cos220°cos230°cos20°+cos220°-cos230°=0.414 Pa=12γCaH2 Pa=12×120×0.414×13.75'2=4696.3 lb PH=Pa×cosѲ=4696.3×cos20°=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 lbft Mmax=1.6×M=1.6×20210.62

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

DESIGN OF STRUCTURES Design Of Retaining Wall Mu=Mmax=32337lbft=388044 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000600001-1-2×3880440.85×4000×0.9×12"×12.5"2 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.0039 ρmin=200fy=20060000=0.0033 ρmax=0.75ρb=0.75×β1×0.85fc'fy×8700087000+fy 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: Ast=ρbd Ast=0.0039×12"×12.5"=0.585 in2 Provide #6 bars @ 9"c/c see Nilson page 736

Temperature & Shrinkage Steel: Ast=ρbh Ast=0.0018×12"×15"=0.324 in2 Provide #3 bars @ 4"c/c see Nilson page 736

iv.

Check for shear:

Vu=1.6×Pa Vu=1.6×12×Ca×γ×H-d2 Vu=1.6×12×0.414×120×13.75'-12.5"122 Vu=6418.725 lb/ft=534.9 lb/in

v.

Capacity of Section:

=2φfc'×bd =2×0.9×4000×12"×12.5" =17076.3lb/in >Vu OK

STEP 3 CURTAILMENT OF BARS: Bars are curtailed from top where B.M is Mmax2

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

DESIGN OF STRUCTURES Design Of Retaining Wall Mu=Mmax=32337 lbft Mu2=Mmax2=16168.5 lbft Mmax2=12×1.6×γ×Ca×H12cosѲ×H13 16168.5=12×1.6×120×0.414×H12cos20°×H13 H1=10.91' 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×fyfc'=0.04×0.44×600004000=16.7"=1.39'

Maximum of above three values is selected, i.e. 12” is selected. So, Curtailment =10.91'-1.39'=9.52' from top. ρ for Mmax2, ρ=0.00392 ρ=0.00195>min.value of temp.& shrinkage steel i.e.0.0018 Ast'=Area of curtailed steel Ast' =ρbh=0.00195×12×15=0.351 in2 Area of steel,As=ρbd=0.585 in2 12 As 23 As 0.2925 in2 0.39 in2 0.39 in2 is more close to 0.585 in2 so curtail each 2nd alternate bar.

SECTION & PLAN Step 4 STABILITY CHECKS:

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

DESIGN OF STRUCTURES Design Of Retaining Wall

There are three stability checks, Check For Overturning Moment

i.

Resisting momentOverturning moment>2

Here use H=13.75'+1.97'=15.72' Pa=12γCaH2 Pa=12×120×0.414×15.72'2=6138.42 lb PH=Pa×cos20°=6138.42×cos20°=5768.23 lb PV=Pa×sin20°=6138.42×sin20°=2099.5 lb Centroidal distance,y=H3=15.72'3=5.24' O.T.M=PH×y O.T.M=5768.23×5.24' O.T.M=30225.53 lbft

Sr. #

Weight Area×Density "lb"

x “ft”

Resiting Moment "lbft"

01

13.75'×5.45'×1×120=8992. 5

5.452+1.25+3.3=7.28 '

65465.4

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

13.75'×1.25'×1×150=2578. 125

1.252+3.3=3.925'

10119.15

05

12×1.97×5.45×120=644.1 9

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+Pasin20°×10 =91372.47+6138.42sin20°×10 =112367.1 lbft Resisting momentOverturning moment>2 112367.130225.53=3.72>2 OK

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

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Check For Sliding Force: Compiled By: Muhammad Sajid Nazir (2005-CE-38)

DESIGN OF STRUCTURES Design Of Retaining Wall Resisting forceSliding force≥1.5 Sliding force=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≥1.5 5064.75768.23=0.88≥1.5 NOT OK

So we have to provide key under the base.

Design of Key: Resisting force=1.5×Sliding force Resisting force=1.5×5768.23 Resisting force=8652.345 lb Additional resisting force provided,Pp=8652.3455064.7 Additional resisting force to be provided,Pp=3587.65 lb Additional Resiting force,,Pp=12×γ×H12×1Ca 3587.65=12×120×H12×10.414 H1=4.98'≈5' Depth of Key=5'-1.25'=3.75' 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=B2-a e=102-4.87' 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=RVB×1+RV×eB26≯qall qmax=16882.31510+16882.315×0.131026 qmax=1819.91 lb/ft2 ≯qall OK

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

DESIGN OF STRUCTURES Design Of Retaining Wall qmin=RVB×1+RV×eB26≮zero qmin=16882.31510-16882.315×0.131026 qmin=1556.55 lb/ft 2 OK

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

STEP 5 DESIGN OF TOE & HEEL:

263.3610=x15.45 x1=143.53lb/ft2

263.3610=x21.25 x2=32.92 lb/ft2

263.3610=x33.3 x3=86.91 lb/ft2

Design Of Toe: Taking moment about junction of toe and stem.

i.

Mu=1.6Moment of bearing capacity-0.9Moment of concrete in toe Mu=1.61700.08×3.3×3.32+12×86.91×3.3×23×3.30.93.3×1.25×1×150×3.32 Mu=14397.03 lbft=172764.32 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000600001-1-2×172764.320.85×4000×0.9×12×12.52 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.0017 ρmin=200fy=20060000=0.0033 ρmax=0.75ρb=0.75×β1×0.85fc'fy×8700087000+fy ρmax=0.75×0.85×0.85×400040000×8700087000+60000 ρmax=0.021 ρmax>ρ>ρmin NOT OK

Selection of bars & spacing:

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

DESIGN OF STRUCTURES Design Of Retaining Wall Ast=ρminbd Ast=0.0033×12×12.5 Ast=0.495 in2 Provide #6 bars @ 10" c/c

ii.

Design of heel:

Taking moment about junction of heel and stem, Mu=1.2Concrete in heel+1.6soil in heel+Surcharge moment in heel Mu=1.21.25×5.45×150×5.452+1.613.75×5.45×1×120×5.452+12×1.97×5.45× 1×120×2×5.453 Mu=44889.38 lbft=538672.66 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000600001-1-2×538672.660.85×4000×0.9×12×12.52 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.005 ρmin=200fy=20060000=0.0033 ρmax=0.75ρb=0.75×β1×0.85fc'fy×8700087000+fy ρmax=0.75×0.85×0.85×400040000×8700087000+60000 ρmax=0.021 ρmax>ρ>ρmin OK

Selection of bars & spacing: Ast=ρbd Ast=0.005×12×12.5 Ast=0.75 in2 Provide #6 bars @ 7" c/c

Temperature & Shrinkage Steel: Ast=ρbh Ast=0.0018×12×15 Ast=0.324 in2 Provide #3 bars @ 4" c/c

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

DESIGN OF STRUCTURES Design Of Retaining Wall

REINFORCEMENT DETAILS

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