cantilever wall

Syed Saqib Mehboob 06-CE-50

Cantilever Retaining Wall Page 1

DESIGN OF A CANTILEVER WALL

GIVEN DATA:

Height of stem above G.L. =30

Bearing Capacity of soil = 5 ksf

fc = 5000psi & fy = 60,000psi

= 30O

AND Overburden pressure = 4 height

CALCULATIONS:

HEIGHT OF WALL:

Allowing 4 for the frost penetration to the bottom of the footing in front of

the wall, so total height becomes;

h = 30 + 4 = 34

THICKNESS OF BASE:

At this, it may be assumed 7 to 10% of the overall height of wall.

Assume a uniform thickness = t = 3.5 (i.e. 10% of h)

BASE LENGTH:

h =34 and h = 4

So, P= Cahwh (h+ 2h)

P =0.5(0.333)(120)(34)[34 + 2(4)] where Cah=1-Sin/1+sin

P=28560 lb..(1)

And y= h/3 [(h+3h)/ (h+2h)] = 34/3 [(34+ 3X4)/(34+2X4)] = 12.41 ..(2)



Taking moment about point a

W=(120)(x)(34+8)=5040x lb.

So; W(x/2)=(P)(y)

2520x2 = (28560)(12.41)

x= 11.86

So the base length =1.5X11.86 =17.79

Use 17-9 with toe =4-4 and x=13-5

STEM THICKNESS:

Prior computing stability factors, more accurate knowledge of the concrete dimensions is

necessary.

The thickness of the base of stem is selected with regard of the bending and shear requirement.

P for 30 height and h = 4

P= Cahwh (h+ 2h)

= 0.5(0.3333)(120)(30.5)[30.5 +2(4)]

=23,461.5 lb.



y= h/3 [(h+3h)/(h+2h)] = 30.5/3 [ (30.5+ 3X4)/(30.5+2X4)] = 11.22

and Mu =1.7(P)(y) =447504.65 lb.ft.

b = 0.85 fc/fy [87000/87000+fy] = ..

, X

, = 0.0356

max = 0.75 b =0.0267

For adequqte deflection control, choose = max 0.01335

Then Rn = fy [1-

m] And; m= fy/0.85fc

= (0.01335)(60,000) [1-

(0.01335) 14.12] = 60,000/(0.85)(5000)

= 726 =14.12

Required bd2

= Required Mn/ Rn

Or bd2 =/

d2 =

/

= . .

= 26.17

Total thickness =26.17 + 0.5 +3 =29.7 30

Try 30 of base of stem and select 12 for top of the wall.

Note: By taking these calculated value, when we proceed for the shear atd and afterwards

For F.O.S. against overturning; F.O.S comes less than 2 (so its not ok).

Also for same values, the settlement check is also not ok.

So; by inspection/ by trial & error method.

Revise the following dimensions as given below:

.Total base length = 23 where; x=19 and toe= 4

.Thickness of stems base = 32

SHEAR AT d

d used now = 32-0.5-3 = 28.5 or 2.375



At 30.5 2.375 = 28.125 from top.

P = x 0.333 x120 x 28.125 [28.125+ 2(4)]

= 20,299.99 lb.

& Vu= 1.7P =34509.98 lb.

& Vu = 2bd = 0.85 x 2x 5000 x 12 x 28.5 = 41,111.18 lb.

Since, Vu > Vu; so no shear reinforcement is required.

FOS AGAINST OVERTURNING:



Components Force (lb.) Arm(ft) Moments(lb.ft)

W1 (16.33)(30.5)(120) = 59767.8 14.825 886057.635

W2 (1.66)(30.5)(150) = 3797.25 5.55 21074.7375

W3 (1)(30.5)(150) = 4575 4.5 20587.5

W4 (23)(3.5)(150) = 12075 11.5 138862.5

W5 (1.66)(30.5)(120)= 3037.8 6.106 18548.80

W6 (18)(4)(120) = 8640 14 120960

TOTAL F = 91892.85 lb M=1206091.173 lbS

From eq. (1) & (2); we have:

P = 28560 lb

y = 12.41

Overturning moment = 28560 x 12.41 = 354429.6 lb.ft

FOS against overturning = .173

$. = 3.4 > 2 so ok.

LOCATION OF RESULTANT AND FOOTING SOIL PRESSURE:

Distance of the resultant fro edge is:

a = %&'()&*' +,+-*).,/-0)0*&*' +,+-*)),)12 2,13 righting

= 1206091.$.$.),)12 2,13 0&'()&*'

= 9.27

Middle third = 7.66; so resultant is within the middle third.



Therefore q1 = RV/l2 (4l 6a) =

91892.$$

[4(23) 6(9.27)] = 6.31 > 5 ksf so not ok..

So revise; Readjust the stem location over base

Now take toe length = 7 8

And length x = 15 4

Then recalculating the F & M for new values we get;

RV = 76752.65 lb.

& M = 112570.86 lb. S

Now; FOS comes 3.1> 2.0 0k

Resultants location = a= 10.0

Thus;

q1 = RV/l2 (4l 6a) =

.$$

[4(23) 6(10.0)] = 4642.8 < 5 ksf

q2 = RV/l2 (6a 2l) = 2031.26 psf < 5 ksf so ok.

FOS AGAINST SLIDING:

Force causing sliding = P = 28560 lb.

Frictional resistance= RV = (0.4)(76752.65) = 30701.06 lb.

Passive earth pressure against 3.5 height of footing = wh2

Cph = (120)(3.52)(3) = 2205 lb.

FOS = $.

= 1.2 < 1.5 SO KEY IS

REQUIRED.

The front of key is 5 in front of back face of stem. This will permit anchoring the stem

reinforcement in the key.

>

$. =

.

$



? = 1486.31 lb/ft

Total ordinate = 3517.57 lb/ ft

Frictional resistance b/w soil to soil = R = (tan) [.$.

](9.91) = 23344.95 lb.

Frictional resistance b/w heel concrete to soil = R = (0.4) [$.$.

](13.09) = 14526.84

lb.

Passive earth pressure = wh2

Cph = (120)(h2

)(3) = 180 h2 lb.

FOS against sliding = 1.5

1.5=

$$..@A

h= 5.25, so key height = 53 3.5 = 1 9

And the heel height =

DESIGN OF HEEL CANTILEVER:

WU =

(1.7)(120)(4)+(1.4)(30.5X120+3.5X150)

= 6675 lb/ft



Mu = wl2/2

= (6.675)(12.673)2 = 536.0 k. ft

Vu = Factored shear at a joint of stem and heel ..(3)

Note: When the support reaction introduces compression into the end region, then critical

shear is at a distance d from the face of support. However, the support is not producing

compression; therefore critical shear is at a joint of stem and heel.

Now Vu = (12.673)(6675) = 84592.27 lb.

Vu = 2bd = 0.85x2x5000 x 12 x 39.5 = 56978.66 < Vu So; depth is required to be increased.

d =BC

EFGH3 = .

,.() = 58.6 60 (i.e. 5) and d = 57.5

Now; Wu= 1.7(120)(4) + 1.4[(29)(120) + (5)(150)] = 6738 lb/ft

Mu = (6738)(12.673)2 = 541078.0 lb.ft

Required Rn = IJ

H3A = (541078.)( ) (.)()(.)A = 181.83 psi.

= M[1 N1 MPEQ ] = 0.00309

min =

EQ = 0.00333

As= minbd = (0.00333)(12)(57.5) = 2.297 in2

Use # 8 @4 c/c (As = 2.36 in2)

Development length required:

Since in cantilever heel main bars are on top;

So;



ld = (1.3).ER ST

EU = (1.3)

.,.

= 34.85 where; V = r2

)

Available length = 152.07 -3 = 149.076 so ok

DESIGN OF TOE:

>

. =

15.$

$

? = 1741.79 lb/ft

Total ordinate = 1741.79 + 2031.26 = 3773.05 lb/ ft

Self load = (0.9)(1x3.5x150) = 472.5 lb/ ft

Wu= (1.7)(.$

$.

= 7153.47 lb/ ft

Wu= 7153.47- 472.5 = 6680.97 lb/ ft

Mu = wl2/2 = = (6680.97)(7.66)

2 = 196004.96 lb. ft

Required Rn= IJ

H3A =

. .$.A

= 139.58 psi

= 0.00236

min =

W

= 0.00333 (holds)

As= min bd = (0.00333)(12)(39.5) = 1.578 in2

Use # 8 @ 51 2X c/c (As = 1.71 in2)

Development length required:



ld = .04ER ST

EU = 26.80

Available length = 7.66-( $

12 ) = 7.41 or 88.92

At distance d= 39.5= 3.29 = 7.66 -3.29

= 4.37

>

. =

.$

$

? = 2115.36 lb/ft

Total ordinate = 4146.62

Earth pressure = Y..

](4.37)(1.7) = 32648.30 lb.

Vu = 32648.30 (472.5 -4.37) =30583.47 lb.

Vc = 2bd = 0.85500012 (39.5) = 56978.66 lb.

Therefore; Vc > Vu. so no shear reinforcement is required.

REINFORCEMENT FOR STEM:

P= Cahwh (h+ 2h) Note: (30.5-1.5 = 29) As; 3.5+1.5 =5,

= 0.5(0.3333)(120)(29)[29 +2(4)] which was new heel base thickness.

=21438.54 lb.

y= h/3 [(h+3h)/(h+2h)] = 10.71

Mu = (1.7)(21438.54)(10.71) = 390331.49 lb.ft

Rn = IJ

H3AK

$$$. ..A

= 533.95 psi.

= 0.00954

As= bd = (0.00954)(12)(28.5) = 3.26 in2

Use # 10 @ 41 2X c/c (As = 3.37 in2)

At 5 from top:



P=1298.7 lb. y= 2.179 Mu=4.81 k.ft

At 10 from top:

P=3596.4 lb. y= 4.07 Mu=14.65 k.ft

At 15 from top:

P=6893.1 lb. y= 5.87 Mu=68.79 k.ft

At 20 from top:

P=11188.8 lb. y= 7.62 Mu=144.9 k.ft

At 25 from top:

P=16483.5 lb. y= 9.34 Mu=261.72 k.ft

At base: Mu=390.33 k.ft

WITH FULL REINFORCEMENT:

C = 0.85fcba = 0.85 x 5000 x 12 x a =51000a

T = Asfy = 3.37 x 60,000 = 202200

a = 3.96 in. (C=T)

At top of wall d = 8.5

Mn = 0.9Asfy (d - \

) = (0.9)(202200)(8.5 3.96/2) = 1186509.6 lb.in or 98.87 k.ft

At base of stem d = 28.5

Mn = 402.176 k.ft

WITH HALF REINFORCEMENT:

C = 0.85fcba = 0.85 x 5000 x 12 x a =51000a

T = Asfy = 1.685 x 60,000 = 101100

a = 1.98 in. (C=T)


Mn = 56.94 k.ft




Mn = 208.59 k.ft

WITH HALF REINFORCEMENT:

C = 0.85fcba = 0.85 x 5000 x 12 x a =51000a

T = Asfy = 0.8425 x 60,000 = 505500

a = 0.99 in. (C=T)


Mn = 30.34 k.ft


Mn = 106.17 k.ft

THEORATICAL CUT OFF POINTS:

Theoretical cutoff points should be cut off at 7 distance from bottom and further half bars

should be cut off at 14 theoretically from bottom.



The actual termination is found out by extending beyond the intersection of capacity moment

line with the factored moment diagram a distance of either the effective depth d or 12 bar

diameters, whichever is greater.

(a) Effective Depth:

>Q

=

$.

? = 20

$. x y

d at 7 from bottom (y=23.5) where; y = 30.5 7 = 23.7

? = 15.4

Therefore; 8.5+15.4 = 23.9= 24

Also; ? =

$. x y = 10.81 where; d at 14 from bottom

(y=16.5)

Therefore; 8.5+10.81 = 19.3= 24

(b) 12 bar diameters:

bar used #10 of diameter = 1.25

So; 12db = 15

Conclusions from (a) & (b):

Since; we have to use the greater value thus; add 24 in all termination point lengths.

Actual Cut off points:

Therefore half bars should be terminated actually at 7 + 2 = 9 from bottom.

And further half bars should be terminated actually at 14 + 2 = 16.

For tension bars to be terminated in the tension zone, one of the following conditions must be

satisfied:

(1) Vu at cut off points must not exceed two third of the shear strength Vn.



(2) Continuing bars must provide at least twice the area required for bending moment at the

cutoff point.

(3) Excess shear reinforcement is provided.

Condition No: 01. (Check whether theoretical cutoff point is enough or not).

At 23.5 from top (for half reinforcement)

Vc = 2bd = 0.855000(12) (24) x = 34.62 kips. ]^ Vc = 23.08 kips.

& Vu = 1.7 [ Cah wh(h+ 2h) ] x

= 25.14 kips .(4) (Not Satisfied)

Vc = 2bd = 0.855000(12) (19.3) x = 27.84 kips. ]^ Vc = 18.56 kips.

Now; Vu = 1.7 [ Cahwh (h+ 2h) ] x

= 13.7 kips ( Satisfied)

Now; these bars can be terminated.

Stress concentration check:

Shear at bottom = Vu = (1.7)(21.43854) = 36.45 kips.

Vc = 2bd = 0.855000(12) (28.5) x 1000 = 41.11 kips. Vc > Vu ; so; no need of shear reinforcement.

Note: thus; from eq. (4) results we cannot cutoff the bars at 7 from bottom as the condition is

not satisfied. Thats why we had found out the actual cutoff points i.e. 9 & 16 respectively.

Check for :

The used should not be less than

W at any point. This minimum limit, strictly speaking,

does not hold true for retaining walls. However; because the integrity of retaining wall depend

absolutely on the vertical walls, it appears to be suitable to use this limit in such cases:

First termination point is 9 from bottom where d = 22.59; As = 1.685 in2 (half case)



Since; As= bd Consider fig. 1 and relation

So; =0.0062 >

EQ =

0.0033 (ok) ? =

$. x y ; here y= 30.5-9=21.5

Then; ? = 14.09 & d =8.5+ 14.09 = 22.59

= 0.0039 >

EQ

(ok)

Another requirement is that maximum spacing of the primary flexural reinforcement exceed

neither 3 times the wall thickness nor 18 in.

These restrictions are satisfied as well.

Splice lengths:

For the splice of deformed bars in tension; at sections where the ratio of steel provided to the

required is less than 2 and where no more than 50% of the steel is spliced; the ACI code

requirement, a class B splice of length 1.3 ld

ld for #10 bars = 42 (from table A.10.)

Splice Length = 1.3x42 = 54.6 or 4-6 (ok)

TEMPERATURE & SHRINKAGE REINFORCEMENT:

Total amount of horizontal bars (h is average thickness)

As = 0.002bh =0.002x 12 x $

= 0.528 in2/ft

Since; front face is more exposed to temperature changes; therefore

$

rd of this amount is

placed in front face and one third in rear.

Accordingly,

$

As = 0.352 in2/ft Use # 10 @ 61 2X c/c (As = 0.36 in

2)

Also; $

As = 0.176 in2/ft Use # 3 @ 18 c/c (As = 0.36 in

2)

For vertical reinforcement on the front face; use any nominal amount; Use # 10 @ 18 c/c

Since; base is not subjected to extreme temperature changes, therefore; use 4@12c/c just for

spacers will be sufficient.

cantilever wall

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