US Wall Cantilever w

7/30/2019 US Wall Cantilever w

1/13

source: This copy will expires in -1970 days

Project: hhghh General Notes:

Job No.: hhghh gkg

Client: yyy uyuu

Design by: uuu hghDate: tyyuy

Sheet: 2 Of 5

Ref: Reinforced Concrete Design, 6th ed., pg. 430, by Wang and Salmon*Input data in green cells

CANTILEVER RETAINING WALL

Design under strength method ACI Code

Design notes:

a. Overload factor for earth and concrete 1.4 and for surcharge 1.7

b. Factor of Safety against overturning: 2

c. Factor of Safety against sliding: 1.5

d. It is recommended that a batter of 1/4 in/ft. height be provided on the front face to offset

deflection or forward tilting of the structure

e. Where weep holes are utilized use minimum 2 diameter pipe at 6-0 spacing maximum,and provide continuous gravel pocket behind retaining wall, wrapped in geo-fab to prevent

clogging of pipes.

Cantilever Wall no. 5

surcharge 960 psf.

wt. of retained materials 120 pcf.

angle of internal friction 35 deg.

max. soil pressure 5000 psf.

cofficient of friction bet. masonry and soil.0.4

concrete cover at stem 2 in.

concrete cover at footing 3 in.

try thickness of heel 7-10% of overall height 2.5 ft. OK

try thickness of toe 7-10% of overall height 1.75 ft. OK

= 0 degrees

Reinforcement

front face fc' 3000 psibank of back face fy 40000 psi

earth ft. 16 type of bar

retained r-bar at heel r-bar at stem

material; ft. heel # 8 # 8

4 toe r-bar at toe temperature bar

# 8 # 5

Jerv Works Const.

http://www.reinforced-concrete.com

non-epoxy coated

XEXPIRES!! Pls. visit

www.reinforced-concrete.com

and download your new copy.

course-grained soils (with silt)
http://www.reinforced-concrete.com/http://www.reinforced-concrete.com/


2/13

Results:

FS against overturning OK

Soil pressure OK

Frictional sliding sliding fails, put shear key!

Shear at base of stem OK

Dimension Output

10 in. top of stem

17.5 ft.

3.96 ft. 5.95 ft.

1.58 ft.

5 in. 14"

Shear key needed

14"

11.5 ft.

Reinforcement Output

# 8 @ 0 in. c.c.

# 5 @ 0 in. c.c. temp. bars

# 5 @ 0 in. c.c. temp. bars

# 5 @ 0 in. c.c

#DIV/0! # 8 @ 0 in. c.c

# 8 @ 0 in. c.c #DIV/0!

#DIV/0!

1.75 ft. 2.5 ft.


3/13

solution:

surcharge 8 ft.

Caw = equivalent fluid pressure

Rankine Theory

C = cos[(cos-cos^2-cos^2)/(cos+cos^2-cos^2)]cos 1cos^2 1cos^2 0.6710101

0.5735764

0.2709901

Caw = equivalent fluid pressure 32.518806 pcf

B1 0.85

establish the limits within the reinforcement ratio can be chosen

for maximum b = .85B1f'c/fy(87000/87000+fy) 0.0371206for adequate deflection control on the cantilever wall

max = .75b/2 0.0139202m=fy/(.85f'c) 15.686275

Rn=fy(1-1/2m) 496.01723 psi

Preliminary Proportioning

Base length calculation

P = Cwh where:

C = coefficient of materials

w = unit wt. of retained material

h = distance below the earth surface

Pa = Ca*wh^2/2

where:

Ca*w = equivalent fluid pressure

height of wall 20 ft.stem wall height 17.5 ft.

a d

8 surcharge

260 psf

W

x/216

P1

10 P2 active pressure

4 6.67

b

R c 650 psf

x


4/13

11.5

base=(Cw)(h)

surcharge 260.1505 psf

The force Pa caused by active pressure on a wall of height h is base =(Cw)(h)

P1surcharge = ( base)(height)(1 foot strip) 5203.009 lbs

P2earth retained = (1/2)(base)(height)*1 ft. strip 6503.761 lbs

wt. of materials bounded by a,b,c,d

W= weight of surcharge+weight of retained materials 3360 x in lbs

M at point bW(x/2) = P1d/2 +P2d/3 95388.5

substitute W

x^2 = 56.77887

x 7.535175 ft.

base length = 1.5x 11.30276 ft.

rounding off base length 11.5 ft.

Stem thickness calculation

stem wall height 17.5 ft.

surcharge

a d

17.5

2.5 b 260.15 650.38

c

x

by bending moment general expression

My = (1/2)(surcharge base)(y^2) + (1/6)(retained mat. Base)(y^3) ,where y = 17.5

My= 39.83554 + 29.04675 68.88228

using overload factor of 1.7

Mu =1.7(My) 117.0999 ft-kips

required d = sqrt(Mu/oRnb) 16.19603 in.

say use r-bar # 8 radius = 0 in.

total stem base thickness 18.19603 in.

stem base thickness 19 in.

calculation of batter

usual batter is 1/4 in per foot height of stem 4.5 in. each side

front and back of stem 9 in.

hence, top of stem wall will be 10 in.


5/13

check for shear strength

Vy = 0.26015 y + 0.01626 y^2 where y = 17.5

using overload factor of 1.7

actual Vu = 1.7(Vy) 16.2045 kips

allowable Vu = Vc = (2sqrtf'c)bd) 19.5537 kips >>>> 16.20453

conditional OK

if conditon is met no need for shear reinforcement for slab (ACI-11.5.5.1)

check against overturning 0.83 6.70

8

w1 17.5

20

w4

w2

w3 2.5

toe

0.75 heel

1.58

3.96 7.54

11.5

ref. heel force arm moment

lb. ft. ft-lb

w1 20507.64 3.35092104 68719.47w2 984.375 6.45184208 6351.032

w3 4312.5 5.75 24796.88

w4 2187.5 7.11850874 15571.74

total 27992.01 115439.1

resultant, from heel 4.124002 ft.

resisting moment 206469 ft-lbs

overturning moment = P1*x + P2*x 95388.5 ft-lbs

FS against overturning 2.164506 >>>> 2

conditional OK

if condition is met FS against overturning OK

otherwise, increase footing base/thickness

location of resultant and footing soil pressures

R = 27992.01 lbs.

x = (moment + overturning moment)/R 7.531706 ft. from heel


6/13

Pmax = 4868.17596

R = 27992.012

R =1/2(Pmax)(base length)

Pmax = 2R/base length 4868.18 lbs. > 1.5

conditional sliding fails, put shear key!

if condition is met, against sliding OK,otherwise increase footing width, thickness or put key

Design of key

** Generally it seems desirable to place the front face of the key about5 in. in front of the back

face of the stem.

5"

neglect 1' 11.5

5.95

h1= 3

u1 = 0.700207538

F H

a Pp inner block G uR2

C D

uR1

4868.17596 psf 2695.9149 psf u2 = 0.4

5.13 6.37

Method 1 - Inert-block concept of frictional resistance

using the inert-block concept, the passive force Pp developed over the distance BC

C = 3.690172

Passive pressure Efp = equivalent fluid pressure 442.8207 pcf

Pp = equivalent fluid press(h1+a)^2(1/2) - equivalent fluid press(h)^2(1/2)

h1 = height of earth retained - 1 ft.(neglect 1 ft.) 3 ft.

by inducing inert-block, the frictional coefficient over the region CD tan a = 35 degrees

tan 35 0.700208

while coefficient 0.4 applies at DG, FH


7/13

Force equilibrium

P = Pp + F

where: P = ( P1+P2)1.5 17560.16 lbs.

where: F = u1R1 + u2R2 13589.3 + 3433.7915 17023.09 lbs.

hence, F - P = -537.063

where: Pp = efp(h1+a^2)/2 - efp(h1^2)/2

simplify Pp =efp/2[2h1a + a^2]

hence Pp + (F - P) = 0

a^2 + a 6 -2.4256 = 0

a = 0.380184 ft.

Method 2 - Passive resistance concept of frictional resistance

neglecting the frictional force in front of the key and considering the passive resistance developed below the

toe, the depth of key required may be computed as

5"

neglect 1' 11.5

5.951842078

h1= 3

F H

a Pp D G

C u = 0.4

b Pp angle of internal friction

soil pressure distribution

2695.91488 psf

C5.13 6.37

b = dist tan a 3.5931089 ft.

force equilibrium

(P1 +P2)1.5 = Pp + u2R2

where: (P1+P2)1.5 = 17560.156 lbs.

where: u2R2 = 3433.7915

where: Pp=efp(1/2)(h1+a+b)^2-efp(1/2)h1^2

efp(1/2)(h1+a+b)^2

Pp= efp * a^2(1) 221.41034 a^2 a

efp * a(2hi+2b) 2919.5649 a b

efp * 2h1b+b^2 7631.8117 c

-6494.552 final c

hence, a = 1.9392845 ft.

average between Method 1 and Method 2 a = 14 in.

DESIGN OF HEEL CANTILEVER

2 " + 1/2bar

wt. of overburden


8/13

critical section for cracking wt. of concrete

due to shear 28

neglect upward pressure

for design of heel

#DIV/0! 5.95

bar embedment

wu = weight of overburden + wt. of concrete

where: overload factor for surcharge = 1.7

overload factor for earth retained and concrete = 1.4

surcharge = 960 plf

earth overburden = 2100 plf

concrete = 375 plf

wu = 5097 plf

EM @ the center of stem steelcenter of steel + concrete cover 2 in.

Mu downwards = 95406.026 lbs-ft.

Check for shear

Vu = 30336.539 lbs.

the design shear strength, unless shear reinforcement is used, is

effective d 28 in.

Vc = (2(f'c)bd 31285.912 lbs. >>>> 30336.54conditional OK

if condition is met, shear OK; otherwise increase footing depth

required Rn =Mu/bd^2 135.21262 psi.m=fy/(.85f'c) 15.686275

required reinforcement ratio to satisfy the factored loading = 1/m(1-1-(2mRn/fy) 0.003475min. reinforcement requirement is, according to ACI-10.5.1

Asmin = 3(f'c)bwd/fy 1.3802608 in^2but not less than Asmin = 200bwd/fy 1.68 in^2

hence min = Asmin/bwd1min = 3(f'c)/fy 0.0041079min2 = 200/fy 0.005

0.005

hence, 0.0046218required As = bwd 1.5529205 in^2/ft.use # 8 @ 0 in c.c

Check development length

general equation; Ld/db = 3/40((fyY)/f'c(c+Kr/db))

= modification factor for reinforcement location1.3 for top bars, when cast with more than 12 in. of concrete beneath them

1 for other bars

cast of concrete beneath the bars 28 in.

hence = 1.3 = modification factor for epoxy-coated reinforcement

1.5 when cover


9/13

epoxy coated

3db 0 >>> 2

6db 0 >>> 0

hence, if epoxy coated 1.2final 1

need not exceed 1.7 1.3 = modification factor for bar size

0.8 for # 6 and smaller

1 for # 7 and larger bars

hence, 0.8 = modification factor for lightweight concrete

1.3 for lightweight aggregate concrete

1 for normal-weight concrete 1

c = clear cover + 1/2 radius of r-bar 2

Kr = since no shear reinforcement 0

but, (c+Kr)/db


10/13

use # 8 @ 0 in.

Check for shear

Vc =2(f'c)bd 20112.372 lbs >>> 15576.21conditional OK

if condition is met shear OK, otherwise increase toe depth


general equation; Ld/db = 3/40((fyABYS)/f'c(c+Kr/db))


1 for other bars

cast of concrete beneath the bars 3 in.

hence = 1 = modification factor for epoxy-coated reinforcement

1.5 when cover


11/13

required Rn =Mu/bd^2 450.21101where: m=fy/(.85f'c) 15.686275

= 1/m(1-1-(2mRn/fy) 0.01247611min = 3(f'c)/fy 0.0041079min2 = 200/fy 0.005

0.005

min 0.0124761required As = bwd 2.5451202 in^2/ft.use # 8 @ 0 in. c.c

vertical reinforcement at the front face to support the horizontal temperature

and shrinkage steel ACI-14.3.5

use # 5 @ 0 in.

Check shear

the shear has been previously checked and found satisfactory


general equation; Ld/db = 3/40((fyY)/f'c(c+Kr/db))


1 for other barscast of concrete beneath the bars

hence = 1 = modification factor for epoxy-coated reinforcement

1.5 when cover > 2

6db 0 >>> 0

hence, if epoxy coated 1.2final 1

need not exceed 1.7 1 = modification factor for bar size0.8 for # 6 and smaller

1 for # 7 and larger bars

hence, 1 = modification factor for lightweight concrete

1.3 for lightweight aggregate concrete

1 for normal-weight concrete 1

c = clear cover + 1/2 radius of r-bar 2

Kr = since no shear reinforcement 0

but, (c+Kr)/db


12/13

DESIGN OF WEEPHOLE

say 4" diameter every 10 to 15 ft. along the wall

cofficient of friction bet. masonry and soil.

silt 0.3

course-grained soils (with silt) 0.4

course-grained soils (without silt) 0.5

sound rock (with rough surface 0.6

0.4

r-bar at he area in^2 diameter in. no. r-bar at heel

# 3 0.110 0.375 3 type of bar

# 4 0.200 0.500 4 epoxy coated 1.2

# 5 0.310 0.625 5 non-epoxy coated 1

# 6 0.440 0.750 6 1

# 7 0.600 0.875 7

# 8 0.790 1.000 8 DATE

# 9 1.000 1.128 9 41270

# 10 1.270 1.270 10 39300# 11 1.560 1.410 11 X

0 0 0

r-bar at toe area in^2 diameter in. no. r-bar at toe

# 3 0.110 0.375 3 type of bar

# 4 0.200 0.500 4 epoxy coated 1.2

# 5 0.310 0.625 5 non-epoxy coated 1

# 6 0.440 0.750 6 1

# 7 0.600 0.875 7

# 8 0.790 1.000 8

# 9 1.000 1.128 9# 10 1.270 1.270 10

# 11 1.560 1.410 11

0 0 0

r-bar at ste area in^2 diameter in. no. r-bar at toe

# 3 0.110 0.375 3 type of bar

# 4 0.200 0.500 4 epoxy coated 1.2

# 5 0.310 0.625 5 non-epoxy coated 1

# 6 0.440 0.750 6 1

# 7 0.600 0.875 7

# 8 0.790 1.000 8

# 9 1.000 1.128 9

# 10 1.270 1.270 10

# 11 1.560 1.410 11

0 0 0

temperatur area in^2 diameter in.

# 3 0.110 0.375

# 4 0.200 0.500

# 5 0.310 0.625

# 6 0.440 0.750


13/13

# 7 0.600 0.875

# 8 0.790 1.000

# 9 1.000 1.128

# 10 1.270 1.270

# 11 1.560 1.410

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