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 of 120 lb/ft3. Angle of internal friction of soil is 30°. Earth surface that is to be retained is horizontal. Take qall=2500 lb/ft2 Usefc'=4000 Psi & fy=40,000 Psi.
SOLUTION: Given that, Height of Retaining wall above NSL = 15’ 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'
STEP 2 DESIGN OF STEM: i.
Calculation Calculati on Of Soil Pressure: Pressur e:
Ca=1-sinφ1+sinφ Ca=1-sin30°1+sin30°=0.33
Pa=12γCaH2 Pa=12×120×0.33×15'2=4455 lb ii.
Moment & Reinforcement Calculation: Centroidal distance,y=H3=15' distance,y=H3=15'3=5' 3=5'
M=Pa×y=4455×5'=22275 lbft Mmax=1.6×M=1.6×22275 Mu=Mmax=35640 lbft=427680 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000400001-1-2×4276800.85×4000×0.9×12"×12.5"2 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.00659 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 Selectio n of bars & spacing: spacing :
Vertical Reinforcement On front face of wall: Ast=ρbd Ast=0.00659×12"×12.5"=0.9885 in2 Provide #6 bars @ 5"c/c see Nilson page 726
Temperature & Shrinkage Steel: Steel : Ast=ρbh Ast=0.002×12"×15"=0.36 in2 Provide #3 bars @3.5c/c see Nilson page 726
iv.
Check Chec k for shear: shea r:
Vu=1.6×Pa Vu=1.6×12×Ca×γ×H-d2 Vu=1.6×12×0.33×120×15'-1.042'2 Vu=6172.375 lb/ft=514.36 lb/in
v.
Capacit Cap acity y of Section: Sect ion:
=2φfc'×bd =2×0.9×4000×12"×12.5" =17076.3lbin >Vu OK
STEP 3 CURTAILMENT OF BARS: Bars are curtailed from top where B.M is Mmax2
Mu=Mmax=35640 lbft=427680 lbin Mu2=Mmax2=213840 lb in Mmax2=12×1.6×γ×Ca×H12×H13 213840=12×1.6×120×0.33×H12×H13 H1=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×fyfc'=0.04×0.44×400004000=11.18"
Maximum of above three values is selected, i.e. 12” is selected. So, Curtailment =2.25’-1’=1.25’ from top. ρ for Mmax2 , ρ=0.85×4000400001-1-2×2138400.85×4000×0.9×12"×12.5"2 ρ=0.003295>min.value of temp.& shrinkage steel i.e.0.002 Ast'=Area of curtailed steel Ast' =ρbh=0.003295×12"×15"=0.593 in2 Area of steel,As=ρbd=0.998 in2 12 As 23 As 0.494 in2 0.659 in2
0.659 in2 is more close to 0.593 in2 so curtail each 2nd alternate bar 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=17144.96 lbft
1 produces overturning moment & 2, 3 and 4 produces resisting Sr. #
Weight Area×Density "lb"
x “ft”
Resiting Moment "lbft"
01
15'×5.45'×1×120=9810
5.452+1.25+3.3=7.28'
7 1 4 1 6 .8
02
10'×1.25'×1×150=1875
102=5 '
9375
03
1.75'×3.3'×1×120=693
3.32=1.65'
1 1 4 3 .4 5
04
15'×1.25'×1×50=2812.5
1.252+3.3=3.925'
11039.06
W=15190.5 lb
Resisting momentOverturning moment>2 92974.3122275=4.17>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:
M=92974.31
Resisting forceSliding force≥1.5 Sliding force=Pa=4455 lb Resisting moment=μ×∑W=0. moment=μ×∑W=0.3×15190.5 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,
Desig Des ign n o f Key K ey:: Resisting force=1.5×Sliding force Resisting force=1.5×4455 Resisting force=6682.5 lb Additional resisting force provided,Pp=6682.5-4557.15 Additional resisting force to be provided,Pp=2125.35 lb Additional Resiting force,,Pp=12×γ×H122×1Ca force,,Pp=12×γ×H122×1Ca 2125.35=12×120×H122×10.33
H1=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 forceW a=92974.31-2227515190.5 a=4.65' Eccentricity,e=B2-a e=102-4.65' 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×eB26≯qall qmax=15190.510+15190.5×0.351026 qmax=1838.05 lb/ft2 ≯qall OK qmin=WB×1-W×eB26≮zero qmin=15190.510-15190.5×0.351026 qmin=1200.05 lb/ft 2 OK
If this check is not OK then increase “B”.
STEP 5 DESIGN OF TOE & HEEL:
63810=x15.45 x1=347.71 lb/ft2
63810=x21.25 x2=79.75 lb/ft2
63810=x33.3 x3=210.54 lb/ft2
i.
Design Of Toe: Taking moment about junction of toe and stem. Mu=1.6Moment of bearing capacity-0.9Moment of concrete in toe Mu=1.61627.51×3.3×3.32×12×210×3.3×23×3.30.93.3×1.25×1×150×3.32 Mu=8705.25 lbft=104462.97 lbin
ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000400001-1-2×104462.970.85×4000×0.9×12×12.52 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5" ρ=0.00156 ρmin=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 #5 bars @ 5" c/c
ii.
Design Desi gn of heel: heel :
Taking moment about junction of heel and stem, Mu=1.2Concrete in heel+1.6soil in heel Mu=1.21.25×5.45×150×5.452+1.615×5.45×1×120×5.452
Neglecting moment produced due to stresses. It will make our structure safer because it works as a FOS. Mu=46113.13 lbft=553357.56 lbin ρ=0.85fc'fy1-1-2Mu0.85fc'φbd2 ρ=0.85×4000400001-1-2×553357.560.85×4000×0.9×12×12.52 Where,φ=0.9 & b=12" & d=h-2.5"=15"-2.5"=12.5"
ρ=0.00863 ρmin=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=0.00863×12×12.5 Ast=1.295 in2 Provide #6 bars @ 4" c/c
Temperature & Shrinkage Steel: Ast=ρbh Ast=0.002×12×15 Ast=0.36 in2 Provide #3 bars @ 3.5" c/c
DETAILS OF STEEL IN TOE & HEEL
