Design Of Structures

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

QUESTION: Design a RCC retaining wall to retain earth up to a height of 12’. Base of the footing is to be placed 3’ below NSL. Soil has density of 120 lb/ft3. Angle of internal friction of soil is 30°. Earth surface that is to be retained is horizontal on which a surcharge load of 300 lb/ft2 is acting. Take qall=2500 lb/ft2 Usefc'=4000 Psi & fy=40,000 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° fc'=4000 Psi fy=40,000 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' H'=Height of surcharge load H'=Surcharge loadDensity of Soil=300120=2.5'

STEP 2 DESIGN OF STEM: While we are designing the stem/arm of retaining wall, we have to consider the height of stem only (i.e. from top of base to the top of stem), because here we have to calculate the active pressure of soil on the stem only.

So use H=13.75'

i.

1

Calculation Of Soil Pressure:

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load Ca=1-sinφ1+sinφ Ca=1-sin30°1+sin30°=0.33 Pa=12γCaH(H+2H') Pa=12×120×0.33×13.75'(13.75'+2×2.5')=5104.69 lb ii.

Moment & Reinforcement Calculation: Centroidal distance,y=H2+3HH'3H+2H'=13.75'2+3×13.75'×2.5'313.75'+2×2.5'= 5.19' M=Pa×y=5104.69×5.19'=26493.34 lbft Mmax=1.6×M=1.6×26493.34 Mu=Mmax=42389.35 lbft=508672.15 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000400001-1-2×508672.150.85×4000×0.9×12"×12.5"2 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.0079 ρmin=200fy=20040000=0.005 ρmax=0.75ρb=0.75×β1×0.85fc'fy×8700087000+fy 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: Ast=ρbd Ast=0.0079×12"×12.5"=1.185 in2 Provide #6 bars @ 4"c/c see Nilson page 736

Temperature & Shrinkage Steel: Ast=ρbh Ast=0.002×12"×15"=0.36 in2 Provide #3 bars @ 3.5" c/c see Nilson page 736

iv.

Check for shear:

Vu=1.6×Pa Vu=1.6×12×Ca×γ×HH+2H' Vu=1.6×12×120×0.33×13.75'(13.75'+2×2.5') Vu=8167.5 lb/ft=680.625 lb/in

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

v.

Capacity of Section:

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

STEP 3 CURTAILMENT OF BARS: Bars are curtailed from top where B.M is Mmax2 Mu=Mmax=42389.35 lbft=508672.15 lbin Mu2=Mmax2=21194.675 lbft Mmax2=12×1.6×γ×Ca×H1×H1+2H'×H12+3H1H'3H1+2H' 21194.675=12×1.6×120×0.33×H13+3H12H'3 21194.675=12×1.6×120×0.33×H13+3H12×2.53 21194.675=10.56H13+79.2H12 H1=10.55' 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'=0.04×0.44×400004000=11.13"

Maximum of above three values is selected, i.e. 12” is selected. So, Curtailment =10.55’-1’=9.55' from top. ρ for Mmax2, ρ'=ρ2=0.00792 ρ=0.00395>min.value of temp.& shrinkage steel i.e.0.002 Ast'=Area of curtailed steel Ast' =ρbh=0.00395×12"×15"=0.711 in2 Area of steel,As=ρbd=1.185 in2 12 As 23 As 0.5925 in2 0.79 in2 0.79 in2 is more close to 0.711 in2 so curtail each 2nd alternate bar

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

PLAN & SECTION Step 4 STABILITY CHECKS: There are three stability checks, Check For Overturning Moment

i.

Resisting momentOverturning moment>2

While calculating over turning moment we have to consider total length of retaining wall (i.e. from the bottom of the base to the top of the stem) because here we have to calculate the Active force per unit length of wall. So use H=15' Pa=12γCaH2 Pa=12×120×0.33×152=4455 lb Centroidal distance,y=H3=153=5' O.T.M=Pa×y O.T.M=4455×5 O.T.M=22275 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=7.2 8'

65465.4

02

10'×1.25'×1×150=1875

102=5'

9375

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

03

1.75'×3.3'×1×120=693

3.32=1.65'

1143.45

04

13.75'×1.25'×1×150=2578.12 5

1.252+3.3=3.925'

10119.15

05

2.5’×1'×300=750

5.452+1.25+3.3=7.2 8'

5460 M=91563

W=14888.625

Resisting momentOverturning moment>2 9156322275=4.11>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≥1.5 Sliding force=Pa=4455 lb Resisting force=μ×∑W+12γCpH2=0.3×14888.625+12×120×3×1.252 Resisting force=4466.59+281.25=4747.84 lb

Here μ is the coefficient of friction between and earth and its value lies between 0.30.45. Resisting forceSliding force≥1.5 4747.844455=1.06≥1.5 NOT OK So we have to provide key under the base.

Design of Key: Resisting force=1.5×Sliding force Resisting force=1.5×4455 Resisting force=6682.5 lb Additional resisting force provided,Pp=6682.54747.84 Additional resisting force to be provided,Pp=1934.66 lb Additional Resiting force,,Pp=12×γ×H12×1Ca 1934.66=12×120×H12×10.33 H1=3.26' Depth of Key=3.26'-1.25'=2.01' iii.

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Check For Bearing Capacity Of Soil: Compiled By: Muhammad Sajid Nazir (2005-CE-38)

soil

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

a=Stablizing moment-O.T.MStablizing forceW a=91563-2227514888.625 a=4.65' Eccentricity,e=B2-a e=102-4.65' e=1.09' 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×eB26≯qall

qmax=14888.62510+14888.625×0.351026 qmax=1801.52 lb/ft2 ≯qall OK qmin=WB×1-W×eB26≮zero qmin=14888.62510-14888.625×0.351026 qmin=1176.2 lb/ft2 ≮zero OK

If this check is not OK then increase “B”. STEP 5 DESIGN OF TOE & HEEL:

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load 625.3210=x15.45 x1=340.8 lb/ft2

625.3210=x21.25 x2=78.16 lb/ft2

625.3210=x33.3 x3=206.36 lb/ft2

Design Of Toe:

i.

Taking moment about junction of toe and stem. Mu=1.6Moment of bearing capacity-0.9Moment of concrete in toe Mu=1.61595.16×3.3×3.32+12×206.36×3.3×23×3.30.93.3×1.25×1×150×3.32 Mu=14176.73 lbft=170120.75 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000400001-1-2×170120.750.85×4000×0.9×12×12.52 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.00255 ρmin=200fy=20040000=0.005 ρmax=0.75ρb=0.75×β1×0.85fc'fy×8700087000+fy ρmax=0.75×0.85×0.85×400040000×8700087000+40000 ρmax=0.037 ρmax>ρ>ρmin NOT OK

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

ii.

Design of heel:

Taking moment about junction of heel and stem,

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load Mu=1.2Concrete in heel+1.6soil in heel+Surcharge moment in heel Mu=1.21.25×5.45×1×150×5.452+1.613.75×5.45×1×120×5.452+300×5.45×5. 452 Mu=49677.43 lbft=596129.175 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000400001-1-2×596129.1750.85×4000×0.9×12×12.52 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.009 ρmin=200fy=20040000=0.005 ρmax=0.75ρb=0.75×β1×0.85fc'fy×8700087000+fy ρmax=0.75×0.85×0.85×400040000×8700087000+40000 ρmax=0.037 ρmax>ρ>ρmin OK

Selection of bars & spacing: Ast=ρbd Ast=0.009×12×12.5 Ast=1.35 in2 Provide #7 bars @ 5" c/c

Temperature & Shrinkage Steel: Ast=ρbh Ast=0.002×12×15 Ast=0.36 in2 Provide #3 bars @ 3.5" c/c

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

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

DESIGN OF STRUCTURES Design Of Retaining Wall With Surcharge Load

REINFORCEMENT DETAILS

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