Design a RCC Retaining Wall to Retain Earth Up To
-
Upload
muhammad-farooq-zia -
Category
Documents
-
view
4.499 -
download
8
description
Transcript of Design a RCC Retaining Wall to Retain Earth Up To
![Page 1: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/1.jpg)
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
![Page 2: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/2.jpg)
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
![Page 3: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/3.jpg)
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
![Page 4: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/4.jpg)
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.
![Page 5: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/5.jpg)
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
![Page 6: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/6.jpg)
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
![Page 7: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/7.jpg)
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
![Page 8: Design a RCC Retaining Wall to Retain Earth Up To](https://reader036.fdocuments.us/reader036/viewer/2022082412/546ad496af795962298b48d5/html5/thumbnails/8.jpg)
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