25690825 Retaining Walls

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CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOG

BSCE-5 2nd Excel Program ###

*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.

****====DESIGN OF GRAVITY RETAINING WALL====****

f, coefficient

1. Soil or gravel without fine particle 17.25 - 18.85 33 -40 0.5 -0.6

highly permeable.

2. Sand or gravel with silt mixture, l 18.85 - 20.40 25 - 35 0.4 - 0.5

permeability

3. Silty sand, sand and gravel with h 17.25 - 18.85 23 - 30 0.3 - 0.4

clay content

4. Medium or stiff clay 15.70 - 18.85 25 - 35 0.2 - 0.4

5. Soft clay, silt 14.10 - 17.25 20 -25 0.2 - 0.3

0.5 m

qs =20 kPa

3.50 m

0 .

2 5

0 .

2 5

1.10 m 0.60 m

Solutio (Use class 2 of the table given above)

Composite Section

3.50 m

1.10 m

Soil pressure coefficient, Rankine equation for horizontal soil surface30° Passive soil pressure coefficient

18.85 kN/m³

qs = 20.00 kPa = 3.00

h' = 1.061 m

Distances computation

Active soil pressure coefficient 2.0000 m

0.1250m

= 0.33 0.5000 m

1.7500m

3.8750m

w, kN/m³ φ, degrees

Table 1: Unit weights w, effective angles of internal friction φ, and coefficients of friction with con

φ =

w =

c1

= B/2 =

c2

= e/2 =

c3

= e + a/2 =

c4

= (B - 2e - a)/3 + e + a =

c5 = B - e + e/2 =

W 7

c7

a

W 3c5

W 1

W 2

W 4

W 5

W 6

c1

c2

c3 c

4

c6

h '

h

e

B

d

C ah=1−sinφ

1sinφ

C ph=1sinφ

1−sinφ



2.7500m

2.3750m

Given retaining wall dimensions:

a = 0.50m Passive soil pressure:

b = 1.10m h = b= 1.10 m

c = 3.50md = 0.60m = 34.21 kN

e = 0.25m

Active soil pressure: = 0.37 m

h = b + c = 4.60m Tentative wall base dimension:

= 97.14 kN B = 4.00 m

= 1.78 m

Check retaining wall sta 23.60 kN/m³

Friction coeff., f 0.50

component weights

56.640 2.0000 113.280

2.356 0.1250 0.295

47.200 0.5000 23.600

141.600 1.7500 247.800

113.100 3.8750 438.263

18.850 2.7500 51.838

65.000 2.3750 154.375444.746 1029.450

Overturning moment: OM Factor of safety against overturning:

172.47 kN-m

Location of resultant with respect to toe: = 5.97> 2.00, ok!

= 1.93m Factor of safety against sliding:

= 0.07m = 2.64 > 1.50, ok!

B/3 = 1.33 m

the middle third of the base. No tension will occur on the foundation.

121.61kPa

100.76kPa

143 kPa

qmax < qa, the wall is safe against soil bearing.

0.5 m

Retaining Wall Details qs =20 kPa

3.50 m

0 .

2 5

0 .

2 5

c6

= (B - 2e - a)2/3 + e + a

c7

= (B - e - a)/2 + e + a =

wc=

Wi

ci

RM=Wic

i

W1

= Bdwc=

W2

= e(b - d)ws=

W3

= a(b + c - d)wc=

W4

= (B - 2e - a)(b + c - d)wc/2 =

W5

= (B - 2e - a)(b + c - d)ws/2 =

W6

= e(b + c - d)ws=

W7 = qs(B - e - a) =ΣW

i= ΣW

ic

i=

OM = Pah

yah

=

qmax

=

qmin

=

qa

=

P ah=1

2C ahwh h2h '

yah=

h23hh '

3 h2h '

P ph=1

2C ph wh

2

y ph=h

3

x=RM −OM

Rv=∑W i

e= B

2

− x

max

min

qalignl ¿¿ ¿= R

v

B 1±6 e

B2 ¿

FS overturning

= RM =∑W

ic

i

OM = P ah yah

FS sliding

= fRv= f ∑W i P ph

P ah



1.10 m 0.60 m

4.00m



. =)

crete f.









****====DESIGN OF GRAVITY RETAINING WALL====****

f, coefficient

1. Soil or gravel without fine particles, 17.25 - 18.85 33 -40 0.5 -0.6

highly permeable.

2. Sand or gravel with silt mixture, low 18.85 - 20.40 25 - 35 0.4 - 0.5

permeability

3. Silty sand, sand and gravel with high 17.25 - 18.85 23 - 30 0.3 - 0.4

clay content

4. Medium or stiff clay 15.70 - 18.85 25 - 35 0.2 - 0.4

5. Soft clay, silt 14.10 - 17.25 20 -25 0.2 - 0.3

*note:the boxes in yellow should be inputed by the designer,while blue ones are c

computed by the program. 0.5 m

qs =20 kPa

3.50 m

0 . 2

5

0 . 2

5

1.10 m 0.60 m

Solution:(Use class 2 of the table given above)

Composite Section

3.50 m

1.10 m

Soil pressure coefficient, Rankine equation for horizontal soil surface

30 ° Passive soil pressure coefficient18.85kN/m³

qs = 20.00 kPa = 3.00

h' = 1.061 m

Distances computation

Active soil pressure coefficient 2.0000 m

0.1250m

= 0.33 0.5000 m

1.7500m

3.8750m

2.7500m

Table 1: Unit weights w, effective angles of internal friction φ, and coefficients of friction with concrete f.

w, kN/m³ φ, degrees

φ =w =

c1

= B/2 =

c2

= e/2 =

c3

= e + a/2 =

c4

= (B - 2e - a)/3 + e + a =

c5

= B - e + e/2 =

c6 = (B - 2e - a)2/3 + e + a =

W 7

c7

a

W 3c5

W 1

W 2

W 4

W 5

W 6

c1

c2

c3 c

4

c6

h '

h

e

B

d

C ah=1−sinφ

1sinφ

C ph=1sinφ

1−sinφ

W 7

c7

a

W 3c5

W 1

W 2

W 4

W 5

W 6

c1

c2

c3 c

4

c6

h '

h

e

B

d



2.3750m


a = 0.50 m Passive soil pressure:

b = 1.10 m h = b= 1.10 mc = 3.50 m

d = 0.60 m = 34.21 kN

e = 0.25 m

Active soil pressure: = 0.37 m

h = b + c = 4.60 m Tentative wall base dimension:

= 97.14 kN B = 4.00 m

= 1.78 m

Check retaining wall stabili 23.60 kN/m³Friction coeff., f = 0.50

component weights

56.640 2.0000 113.280

2.356 0.1250 0.295

47.200 0.5000 23.600

141.600 1.7500 247.800

113.100 3.8750 438.263

18.850 2.7500 51.838

65.000 2.3750 154.375

444.746 1029.450


172.47 kN-m

Location of resultant with respect to toe: = 5.97 > 2.00, ok!

= 1.93 m Factor of safety against sliding:

= 0.07 m = 2.64 > 1.50, ok!

B/3 = 1.33 m

the middle third of the base. No tension will occur on the foundation.

121.61 kPa

100.76 kPa

143kPa

0.5 m

Retaining Wall Details qs = 20.00

3.50 m 0 .

2 5

0 .

2 5

1.10 m 0.60 m

c7

= (B - e - a)/2 + e + a =

wc =

Wi

ci

RM=Wic

i

W1

= Bdwc=

W2

= e(b - d)ws=

W3

= a(b + c - d)wc=

W4

= (B - 2e - a)(b + c - d)wc/2 =

W5

= (B - 2e - a)(b + c - d)ws/2 =

W6

= e(b + c - d)ws=

W7

= qs(B - e - a) =

ΣWi= ΣW

ic

i=

OM = Pah

yah

=

qmax

=

qmin

=

qa

=

P ah=1

2C ah wh h2h '

yah=

h23hh '

3 h2h '

x=RM −OM

Rv=∑W

i

e= B

2

− x

max

min

qalignl ¿¿¿= R

v

B 1±6 e

B2 ¿

FS overturning =

RM =∑W ic

i

OM = P ah yah

FS sliding

= fRv= f ∑W i P ph

P ah

P ph=1

2C ph wh

2

y ph=h

3

max

min

qalignl ¿¿¿= R

v

B 1±6 e

B2 ¿



4.00m



*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)

****====Design of Cantilever Retaining Wall====****

Right Side Loading

Given:

fc' = 20.70 Mpa Retaining wall dimensions:

fy = 414.00 Mpa a = 0.30 m

18.82 kN/m3 c = 4.50 m

Φ = 40

μ = 0.50 Tentative dimentions:

23.60 kN/m3 B = 3.20 m

143.50 kPa b = 0.40 m

19.30 kPa d = 0.50 m

backfill height = 3.65 m

Øb(main) 25 mm Use Wu = 1.4DL + 1.7LL + 1.7H

Øb(temp) 16 mm

aEs = 200000 Mpa

0.85 qs = 19.30 kPa

0.9

c

h' = 1.026 m

Cantilever Retaining Wall Figure:

d

b

Property Line

a

h'

c h

d

ωs

=

o

ωc

=

qa

=

qs

=

Øshear

=

Øflexure

=

smax

= [ 3t , 450 ]min

Pah

y

ah

W1

C1

W2

C2

W3

W4

W5

C3

C4

C5

W6

C6



B

Soil pressure coefficient, Rankine equation for horizontal surface:

Active soil pressure coeffecient: 0.217443

Active soil pressure: h = 5.00 m

2 72.137 kN

3(h + 2h') 1.9091 m

Check the retaining wall stability:

components weights

37.760 1.6000 60.4160

31.860 0.1500 4.7790

5.310 0.3333 1.7700

237.132 1.8000 426.8376

4.235 0.3333 1.4115

55.97 1.75 97.9475

372.267 ∑RM = 593.1616

Overturning moment: OM = 137.7138kN-m

Factor of safety against overturning:

= 4.307 > 2.00 safe!!!!

OM

Factor of safety against sliding:

= 2.580 > 1.50 safe!!!!

Check for bearing pressure: 143.50 kPa

Location of resultant with respect to toe:

x = RM - OM = 1.2234 m

e = B - x = 0.376554 m

2

B / 3 = 1.07 m

The middle third of the base where No tension will occur on the foundation.

q = 142.001 kPa

90.666 kPa

< safe againts soil bearing.

Design of stem:

b

Cah

=

Pah

= 1 Cah

ωh(h+2h') =

y ah

= h2 + 3hh' =

W i

ci

RM = W ic

i

W 1=

W 2=

W 3=

W 4=

W 5=

W 6=

∑W i=

FSoverturning

= RM

FSsliding

= (f ∑W i)

Pah

qa =

∑W i

∑Wi 1 + 6e qmax

=

B B2 qmin

=

Since qmax

qa, wall is

p1=qs



Soil pressure at y: Moment equation at level y:

= 19.30 kPa

4.09 y kPa = 9.65

0.68

Shear equation at level y:

At level y = 4.50 m

= 19.30 y 218.08 kN2.05 437.86 kN-m

490.874 = 0

4 ρmax = = 0.02

201.062 fy 600+fy

4

Try d = 400 mm ; b = 1000 mm

Ru = Mu = 3.040685

ρ = 0.00812

Use: ρ = 0.01

= 3248.172

= 151.1231mm cc

As

800

= 251.3274mm cc

Check for shear: = 257.8178kN/m

6

At d distance from the bottom of stem:

4.10 m

192.99 kN/m < Vuc, safe!!!!

At y = 3.00 m

Try d = 300 mm

Ru = Mu = 2.209270

ρ = 0.005722

Use: ρ = 0.01

p1

= qs

My = p1 y

1+ p

2 y

2= q

s y 2/2 + 0.5c

ah w

s y 3/3

p2 = cah w s y = M 1 = qs y 2/2 y 2

M2

= 0.5cah w

s y 3/3 = y 3

V y

= p1

+ p2

= qs y + 0.5c

ah w

s y 2

p1

= qs y V

u=

p2

= 0.5cah w

s y 2 = y 2 M

u=

Ao = ∏ D b

2 = mm2 ρmin

=

1.4 / f y

.75 .85fc'b1 600

A temp

= ∏ Dtemp

2 = mm2

fbd2

.85fc' 1 - 1 - 2Ru =

fy .85f c'

A s,flexure

= ρbd mm2/m

s = 1000A o

A s,temp = ρbd =0.002bd = mm2/m

stemp

= 1000 A temp

A s

V uc

= Ø √fc' bd

y d

=

V ud

=

fbd2

.85fc' 1 - 1 - 2Ru =

fy .85f c'

Ve

1 P2

p2= C

ahw

sy

y/2V

d Vmax

Mmax

Stem P MV



= 1716.459

= 285.9805 mm cc

As

600

= 335.1032 mm cc

Design of Base:

19.30

5.00

0.50

Note: The expected worst condition of loading, the passive earth pressure of soil is generally

neglected due to erosion of top soil, and the soil bearing at the toe is neglected due to

the empending action to overturn.

Use: 1.4 for DL

1.7 for LL and service load bearing pressure

241.401 kPa = -470.11 kN

= 154.132 kPa -658.15 kN-m

118.566 kPa= 32.810 kPa

16.520 kPa

e = 0.377 m

2.800 m

Try d = 400 mm

b = 1000 mm

Ru = Mu = 4.570502 = 5216.20

= 94.11 mm cc

Asρ = 0.013041 800.00

= 251.33 mm cc

Use: ρ = 0.01

Check for shear: = 257.8178 kN/m > V, safe!!!!

6

A s,flexure

= ρbd mm2/m

s = 1000A o

A s,temp

= ρbd =0.002bd = mm2/m

stemp = 1000 A temp

A s

qmax

x 1.7 = V = (-W s-W

c-q

s)L

qmin

x 1.7 M = (-W s-W

c-q

s)L2/2 =

W s

= 1.4ωs

c =

qs x 1.7

W c

= 1.4ωcd =

L = B - b) =

A s,flexure

= ρbd mm2/m

fbd2 s = 1000A o

.85fc' 1 - 1 - 2Ru = A s,temp

= ρbd =0.002bd = mm2/m

fy .85f c' s

temp= 1000 A

temp

A s

V uc

= Ø √fc' bd

qmin

qmax

qs=



Retaining Wall Details:

Note: the spacing of the reinforcement is upon the desire of the designer based on the computed b

0.30

25mm Ø @240 oc

16 mm Ø temp. @ 16

260 oc bw mm Øtemp @ 250 oc

5.00 25 mm Ø @250 oc

m

25mm Ø @360 oc

0.50meters

16 mm Ø temp. bars

250oc bw

0.40meters




BSCE-5 2nd Excel Program ###*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)

****====Design of Cantilever Retaining Wall===****

Left Side LoadingGiven:

fc' = 27.50Mpa Retaining wall dimensions: Tentative dimentions:

fy = 275.00 Mpa a = 0.30m B = 3.00m

23.60kN/m3 H = 3.20m b = 0.40m

17.25kN/m3 h = 1.00m D = 0.50m

Φ = 35

μ = 0.45 Surcharge load: Es = 200000 Mpaqa = 120.00 kPa 19.50kPa 0.85

25 mm 19.20kPa 0.9

16 mm

Minimun factor safety requirements:

Overturning 2.00

Sliding = 1.50


19.50kPa

a

h' = 1.11m

H' = 1.13m

H

19.20kPa

D

b

B

Property Line

a

H'

ωc=

ωs=

o

qs2

= Фshear

=

Øb(main) = qs1

= Фflexure

=

Øb(temp) =s

max= [ 3t , 450 ]

min

qs2 =

qs1 =

qs2



H

h'

h

D

b

B


Active soil pressure coeffecient: Passive soil pressure coefficient:

0.271 3.690

h = 3.20m h = 1.61m

Active soil pressure:

51.549 kN Passive soil pressure:

1.467m ###kN

1.079m


components weig B-xi Mx

22.656 0.1500 3.3984 2.850 64.573.776 0.3333 1.2587 2.667 10.07

35.400 1.5000 53.1000 1.500 53.1

144.690 1.7000 ### 1.300 188.1

18.225 0.3780 6.8890 2.622 47.79

Total = 202.091 ### ∑RM = 363.62


θ = 0.03 x = 0.066


= 4.808> 2.00 safe!!!

OM


= 5.215> 1.50 safe!!!

Check for bearing pressure:

B / 3 = 1.000m

x = RM - OM = 1.4250m within 1/3 of bas

94.543kPa

3x

77.46kPa safe!!!

57.26kPa safe!!!

Design of stem:

qs1

x

Cah

= Cph

=

Pah

=

yah

= Pph

=

yph

=

Wi

xi

Mi

W1 =W

2=

W3=

W4=

W5=

FSoverturning

= RM

FSsliding

= (f ∑Wi)+P

ph

Pah

∑Wi

q = 2W T

, when x < 1/3 B =

qmax

= [4B - 6x]W T/B2 =

qmin

= [6x - 2B]W T/B2 =

W3

W4

W5

W2

W1



= 146.768kN/m

= 220.696kN-m/m

y M = ###

0.50 2.693kN-m 4

1.00 11.797 kN-m = ###

1.50 27.196 kN-m 4

2.00 51.466 kN-m = 0

2.50 85.284 kN-m ρmax = = 0.04

3.00 ### kN-m fy 600+fy

= 0.7429kN/m

6

Depth as required by shear: b = 1000mm

= 197.56mm

Design for flexure:

Try h = = 400 mm

287.5mm

Ru = Mu = 2.966719

ρ = 0.011576

Use: ρ = 0.01

= ###

= 147mm oc

= 517.50

= 388mm oc

Design of Toe:

= 97.48kN/m

= 253.07kN-m/m

= 0

= 0.04

Vu

= 1.7 [qsH + 0.5C

ahw

sH2

Mu

= 1.7 Mmax

M = qsy2/2 + 1/3C

ahw

sy3

Amain = ∏ Dmain

2mm2

Atemp

= ∏Dtemp

2mm2

ρmin

=

0.5 [1.4/f y, √f

c'/4f

y]

.75 .85fc'b1 600

Vuc

= Ф √f

d = Vu

Vuc

b

d = h - (100+Φs/2) =

fbd2

.85fc' 1 - 1 - 2Ru

fy .85f c'

As= ρbd mm2/m

spacing, S = [1000Ao/A

s, 3t,4

Atemp

= 0.0018bd mm2/m

spacing, S = [1000Atemp

/As, 5t,4

Vu

= 1.7b [(qmax + qmin)/2 - (wcd + ws (h+h'))]

Mu

= 1.7 [(B-b)2/2][qmin

+2/3(B-b)(qmax

- qmin

) - [wc(d)-w

s(h+h')]]

ρmin

=

0.5 [1.4/f y, √f

c'/4f

y]

ρmax = .75 .85fc'b1 600



fy 600+fy

= 0.7866kN/m

6

Depth as required by shear: b = 1000mm

= ###mm

Design for flexure:

Try h =400 mm

287.5mm

Ru = Mu = 3.401944

ρ = 0.013432 Use: ρ = 0.01

= 3861.697

127.00 mm oc

3t = 1200.00

450 = 450

Compare:

SINCE 127.00 mm < 450.00mm

Use: 127.00mm

= 517.50

= 489mm oc


0.30

25 mmΦ @

137oc

16 mmΦtemp

3.70 338oc

25 mmΦ @

117.00oc

0.50 m

3.00

16mmΦtemp @

0.40 338oc


Vuc

= Ф √fc'

d = Vu

Vuc

b

d = h - (100+Φs/2) =

fbd2

.85fc' 1 - 1 - 2Ru

fy .85f c'

As= ρbd mm2/m


s, 3t,450]min

1000Ao/AS

AND 3t, compare 3t and 450 whichever is smallest.

Atemp

= 0.0018bd mm2/m


/As, 5t,4





****====DESIGN OF A CANTILEVER RETAINING WALL===****

(with shear key)20.7 Mpa Figure: x=1.5m

414 Mpa

18.82 kN/m³ 19.2 kPa

φ = 35 °

0.5

23.6 kN/m³ e=3.65m

143.5 kPa

19.2 kPa

backfill height = 3.65 m

c=1.0m

Use 25mmφ for main rebars, 16mmφ for

temperature bars.

Solution:

Composite section and location of forces

e

c

d

f


f c' =

f y

=

ws= q

a=

f =

wc=

qa

=

qs=

Use Wu = 1.4DL + 1.7LL + 1.7H

h

H

a x

e

b

B

W 7

W 6

W 5

W 4

W 3

W 2

W 1

W 8

a xh

x7

x6

x5

x2

x4

x3

x1

x8



a = 0.20 m Distances:

c = 1.00 m 1.5000 m

e = 3.65 m 1.6000 m

1.50 m 1.7667 m

0.7500mTentative dimensions: 2.4500 m

B = 3.00 m 1.8333 m

b = 0.40 m 2.3500 m

d = 0.50 m 1.7000 m

f = 0.40 m H = d + c + e = 5.1500 m

g = 0.40 m 1.0202 m

h = 0.20 m

Active soil pressure coefficient Passive soil pressure coefficient

= 0.27 = 3.69

Active soil pressure: h = 5.15 m Passive soil pressure: h = 1.50 m

= 94.43 kN = 78.13 kN

= 1.960 m = 0.500 m

Check retaining wall stability:

Component weights

35.4 1.5000 53.1

21.95 1.6000 35.12

10.97 1.7667 19.39

28.23 0.7500 21.17

96.26 2.4500 235.85

8.75 1.8333 16.04

24.96 2.3500 58.662.3 1.7000 3.92

228.83 443.24


185.1kN-m



= 0.37 m = 2.04 > 1.50, ok!

B/3 = 1.00m > e, Rv will fall within the middle third of the base.

No tension will occur on the foundation.

95.19kPa

57.36kPa

143.5 kPa

qmax < qa, the wall is safe against soil bearing

x1

= B/2 =

x2

=xh

+ a/2 =

xh

= x3

= xh

+ a + (b - a)/3 =

x4 = xh/2 =x

5= x

h+ b + (B - x

h- b)/2 =

x6

= xh

+ a + 2(b - a)/3 =

x7

= xh

+ a + (B - xh

- a)/2 =

x8

= xh

+ g/2 =

h' = qs/w

s=

Wi

xi

RM=Wix

i

W1

= Bdwc=

W2

= a(c + e)wc=

W3

= 0.5(b - a)(c + e)wc

=

W4

= c(x)ws=

W5

= (B - x - b)(c + e)ws=

W6

= 0.5(b - a)(c + e)ws=

W7 = qs(B - x - a) =W

8= 0.5f(g +h)q

s=

ΣWi= ΣW

ix

i=

OM = Pah

yah

=

qmax

=

qmin

=

qa

=

C ah=1−sinφ

1sinφC ph=

1sinφ

1−sinφ

P ah=1

2C ah wh h2h ' P ph=

1

2C ph wh

2

yah=

h23hh '

3 h2h ' y ph=

h

3

x=RM −OM

Rv=∑W

i

e=

B

2− x

FS overturning

= RM =∑W

ic

i

OM = P ah yah

FS sliding =

fRv= f ∑W i P ph P ah

max

min

qalignl ¿¿ ¿= R

v

B 1±6 e

B2 ¿



Design of stem:

M y y/2

V y/3

d

STEM P V M

Soil pressure at level y: Shear equation at level y:

19.2 kPa

5.1 y kPa 19.2 y

2.55 y²

Moment equation at level y:

9.6 y²

0.85 y³

Given:

Level, y

0.00 0.000 0.000 0.000 0.000 200 GPa

0.50 10.238 17.404 2.506 4.261 414 MPa

1.00 21.750 36.975 10.450 17.765 20.7 MPa

1.50 34.538 58.714 24.469 41.597 0.85

2.00 48.600 82.620 45.200 76.840 0.90

2.50 63.938 108.694 73.281 124.578 25 mmφ

3.00 80.550 136.935 109.350 185.895 16 mmφ

3.50 98.438 167.344 154.044 261.875

4.00 117.600 199.920 208.000 353.601

4.50 138.038 234.664 271.857 462.156

4.65 144.418 245.510 293.039 498.167

Compute:

490.874 mm² 201.062 mm²

= 0

= 0.01603

try d = 400 mm

= 3.4595

p1

= qs

P1

P2

p2

= Cah

wsx

Ve

Vmax

Mmax

p1

= qs= V

y= P

1+ P

2= q

sy + 0.5C

ahw

sy²

p2

= Cah

wsx = P

1= q

sy =

P2

= 0.5Cahwsy² =

My

= P1y

1+P

2y

2= q

sy²/2 + 0.5C

ahw

sy³/3

M1

= P1y

1= q

sy²/2 =

M2

= P2y

2= 0.5C

ahw

sy³/3 =

Vy

Vu=1.7V

yM

yM

u=1.7M

y

Es=

f y

=

f c' =

f shear

=

f flexure

=

D b

=

Dtemp

=

Smax = [3t, 450]min

Ao=π

4D

b2= Atemp=

π

4D

temp2=

ρm in=1 . 4

f y

ρmax=.75[.85 f c ' β

1

f y

.003 E s

.003 E s f y ]

Ru=

M u/φ

bd 2

'



= 0.00939 ok!

3757.84 mm²/m

= 130.627 mm oc

800mm²/m

= 251.327 mm oc

Check for shear:

= 257.818 kN/m

At d distance from bottom of stem:

y = 4.25 m235.04 kN/m

At 3.00 m

Try d = 300 mm

= 2.29500

= 0.00596 ok!

1788.48 mm²/m

= 274.464 mm oc

600mm²/m

= 335.103 mm oc

Design of heel and toe:

subject to erosion a b

heel toe

a b

Use load factor:

1.4 for DL qmin

1.7 for ll and service load bearing pressures q1 q2

161.82 kPa e = 0.37 m

97.52 kPa 0 m

26.35 kPa 129.67 kPa

122.52 kPa -0.4 m

32.64 kPa 103.95

16.52 kPa 1.50 m

1.10 m

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

Vud

= 1.7(19.2y + 3.13667y²) =

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

ws2

Qs

ws1

L1

L2

qmax

qmax

x 1.7 =

qmin

x 1.7 = At a, x =B/2 - xh

=

ws1

= 1.4(ws)c = q

1=

ws2

= 1.4(ws)(c + e) = At b, x = [B/2 - (x

h+ b)] =

qsx 1.7 = q

2=

Wc

= 1.4(wc)d = L

1= x

h=

L2 = B - (xh + b) =

ρ=ω

1− 1−u

f y

=. c

f y

1− 1−u

.85 f c

'

s=1000 Ao

A s

s temp=1000 Atemp

A s , temp

V u c=φvc

f c '

6b d

Ru=M u /φ

bd 2

ρ=1

ω [1− 1−2ωR u

f y

]= .85 f c '

f y

[1− 1−2R u

.85 f c

' ]

s=1000 Ao

A s

s temp=1000 Atemp

A s , temp



193.84kN

151.41 kN-m

-188.85 kN

-103.87 kN-m

At the toe, the worst condition of loading is when the wall is at empending action to overturn, soil bearing

at the heel is assumed to be zero.

At Heel:

try d = 400 mm

= 1.05

= 0 not ok!-use pmin

1352.66 mm²/m

= 362.9 mm oc

800mm²/m

= 251.33 mm oc

Check for shear:

= 257.82 kN/m > Va, safe

At Toe:

try d = 400 mm

= 0.72129


1352.66 mm²/m

= 362.9 mm oc

800mm²/m

= 251.33 mm oc

Check for shear:

= 257.82 kN/m > Va, safe

Design of Key:

pph1 pah1

f

Va= .5(q

max- q

1)L

1+ q

1L

1- (w

s1=0)L

1- W

cL

1=

Ma= (q

max- q

1)L

1²/3 + (q

1- (w

s1= 0) - W

c)L

1²/2 =

V b

= (.5(q2- q

min)L2 = 0) + [(q

min= 0) - w

s2- w

c- q

s]L

2=

M b = [(q2 - qmin) = 0]L2²/6 + [(qmin = 0) - ws2 - wc - qa]L2²/2 =

ws1

is taken equal to zero due to erosion of top soil at the heel, the expected worst condition of loading.

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

Ru=

M u/φ

bd 2

ρ=1

ω [1− 1− 2ωR u

f y ]= .85 f c '

f y [1− 1− 2R u

.85 f c ' ]

s=1000 Ao

A s

stemp=1000 Atemp

A s , temp

V u c=φvc

f c '

6b d

Ru=

M u/φ

bd 2

ρ=1

ω [1−

1−

2ωR u

f y ]=

.85 f c

'

f y [

1−

1−

2R u

.85 f c

' ]

s=1000 Ao

A s

s temp=1000 Atemp

A s , temp

V u c=φvc

f c '

6b d



pph2 h pah2

0.27 3.69

At pph1:

6.170 m 1.00 m

31.47 kPa 69.45 kPa

At pph2:

6.5702 m 1.40 m

33.51 kPa 97.23 kPa

22.09 kN 56.67 kN

4.465 kN-m 11.964 kN-m Net Shear:

34.58 kN

Net Moment:

7.499 kN-m

try d = 300 mm

= 0.09258

= 0.00022 not ok!-use pmin

1014.49 mm²/m

= 483.86 mm oc

600mm²/m

= 335.1 mm oc

Check for shear:

= 193.36 kN/m > Va, safe

Details:


1.00 m

mm Φ @

1.50 m 25

240 oc

16

0 250 oc

mmΦtemp 250 oc bw mmΦtemp

16

3.65 120 oc

0 16

mmΦtemp 250 oc bw 360 oc

0.40 mm Φ @

m

Cah

= C ph

=

At pah1

:

yah

= h' + c + e + d = y ph

= c =

pah1

= Cahw

sy

ah1= p

ph1= C

phw

sy

ph1=

At pah2

:

yah

= h' + c + e + d + f = y ph

= c + f =

pah2

= Cahw

sy

ah2= p

ph2= C

phw

sy

ph2=

Vah

= 1.7[pah1

f + (pah2

- pah1

)f/2] V ph

= 1.7[pah1

f + (pah2

- pah1

)f/2]

Vah

= V ph

=

Mah

= 1.7[pah1

f ²/2 + (pah2

- pah1

)f ²/3] M ph

= 1.7[pah1

f ²/2 + (pah2

- pah1

)f ²/3]

Mah = M ph =

Vu= V

ph- V

ah=

Mu= M

ph- M

ah=

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

pah=C ah w hh' p p h=C p h w h

Ru=

M u/φ

bd 2

ρ=

1

ω

[1−

1−

2ωR u

f y ]=

.85 f c '

f y [

1−

1−

2R u

.85 f c

'

] s=

1000 Ao

A s

stemp=1000 Atemp

A s , temp

V u c=φvc

f c '

6b d



0.40

m

0.00 m 0.50 -0.10 m

0 mmΦtemp @

0.00 m 250 oc bw















ar spacing above.









@

ar spacing above.













ar spacing above.






BSCE-5 2nd Excel Program 18-Aug-09*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)


Right Side LoadingGiven:

fc' = 20.70 Mpa Retaining wall dimensions:fy = 414.00 Mpa a = 0.30 m

18.82 kN/m3 c = 4.50 m

Φ = 40

μ = 0.50 Tentative dimentions:

23.60 kN/m3 B = 3.20 m

143.50 kPa b = 0.40 m

19.30 kPa d = 0.50 m backfill height = 3.65 m

Øb(main) 25 mm Use Wu = 1.4DL + 1.7LL + 1.7HØb(temp) 16 mm

aEs = 200000 Mpa

0.85 qs = 19.30 kPa

0.9

c

h' = 1.026 m


d

bProperty Line

a

h'

c h

d

BSoil pressure coefficient, Rankine equation for horizontal surface:

Active soil pressure coeffecient: 0.217443

Active soil pressure: h = 5.00 m

2 72.137 kN

ωs =o

ωc

=

qa

=

qs

=

Øshear

=

Øflexure

=

smax

= [ 3t , 450 ]min

Pah

y ah

b

Cah

=

Pah = 1 Cahωh(h+2h') =

W1

C1

W2

C2

W3

W4

W5

C3

C4

C5

W6

C6



3(h + 2h') 1.9091 m


components weights

37.760 1.6000 60.4160

31.860 0.1500 4.7790

5.310 0.3333 1.7700

237.132 1.8000 426.8376

4.235 0.3333 1.4115

55.97 1.75 97.9475

372.267 ∑RM = 593.1616



= 4.307 > 2.00 safe!!!!

OM


= 2.580 > 1.50 safe!!!!

Check for bearing pressure: 143.50 kPa

Location of resultant with respect to toe:x = RM - OM = 1.2234 m

e = B - x = 0.376554 m2

B / 3 = 1.07 m

The middle third of the base where No tension will occur on the foundation.

q = 142.001 kPa

90.666 kPa

< safe againts soil bearing.

Design of stem:

Soil pressure at y: Moment equation at level y:

= 19.30 kPa

4.09 y kPa = 9.65

0.68

Shear equation at level y:

At level y = 4.50 m

y ah

= h2 + 3hh' =

W i

ci

RM = W ic

i

W 1 =

W 2=

W 3=

W 4=

W 5=

W 6=

∑W i=

FSoverturning

= RM

FSsliding

= (f ∑W i)

Pah

qa

=

∑W i

∑Wi 1 + 6e qmax

=

B B2

qmin =

Since qmax qa, wall is

p1

= qs

My = p1 y

1+ p

2 y

2= q

s y 2/2 + 0.5c

ah w

s y 3/3

p2

= cah w

s y = M

1= q

s y 2/2 y 2

M2

= 0.5cah w

s y 3/3 = y 3

V y = p1 + p2 = qs y + 0.5cah w s y 2

Ve

p1=q

s

P1 P

2

p2= C

ahw

sy

y/2yM

V

dV

maxM

max

Stem P MV



= 19.30 y 218.08 kN

2.05 437.86 kN-m

490.874 = 0

4 ρmax = = 0.02

201.062 fy 600+fy

4

Try d = 400 mm ; b = 1000 mmRu = Mu = 3.040685

ρ = 0.00812

Use: ρ = 0.01

= 3248.172

= 151.1231mm ccAs

800

= 251.3274mm cc

Check for shear: = 257.8178kN/m6

At d distance from the bottom of stem:4.10 m

192.99 kN/m < Vuc, safe!!!!

At y = 3.00 mTry d = 300 mm

Ru = Mu = 2.209270

ρ = 0.005722

Use: ρ = 0.01

= 1716.459

= 285.9805 mm ccAs

600

= 335.1032 mm cc

Design of Base:

19.30

5.00

0.50

p1

= qs y V

u=

p2

= 0.5cah w

s y 2 = y 2 M

u=

Ao = ∏ D b

2 = mm2 ρmin

=

1.4 / f y

.75 .85fc'b1 600

A temp

= ∏ Dtemp

2 = mm2

fbd2

.85fc' 1 - 1 - 2Ru =

fy .85f c'

A s,flexure = ρbd mm2/ms = 1000A

o

A s,temp

= ρbd =0.002bd = mm2/m

stemp

= 1000 A temp

A s

V uc

= Ø √fc' bd

y d

=

V ud

=

fbd2

.85fc' 1 - 1 - 2Ru =

fy .85f c'

A s,flexure

= ρbd mm2/m

s = 1000A o

A s,temp

= ρbd =0.002bd = mm2/m

stemp

= 1000 A temp

A s

qs=



Note: The expected worst condition of loading, the passive earth pressure of soil is generally neglected due to erosion of top soil, and the soil bearing at the toe is neglected due to

the empending action to overturn.

Use: 1.4 for DL1.7 for LL and service load bearing pressure

241.401 kPa = -470.11 kN

= 154.132 kPa -658.15 kN-m

118.566 kPa

= 32.810 kPa

16.520 kPae = 0.377 m

2.800 m

Try d = 400 mm b = 1000 mm

Ru = Mu = 4.570502 = 5216.20

= 94.11 mm ccAs

ρ = 0.013041 800.00

= 251.33 mm cc

Use: ρ = 0.01

Check for shear: = 257.8178 kN/m > V, safe!!!!

6


Note: the spacing of the reinforcement is upon the desire of the designer based on the comput

0.30

25mm Ø @240 oc

16 mm Ø temp. @ 16

260 oc bw mm Øtemp @ 250 oc

5.00 25 250 oc

m

25mm Ø @360 oc

0.50meters

16 mm Ø temp. bars

250oc bw

0.40meters

qmax

x 1.7 = V = (-W s-W

c-q

s)L

qmin

x 1.7 M = (-W s-W

c-q

s)L2/2 =

W s

= 1.4ωsc =

qs

x 1.7

W c

= 1.4ωcd =

L = B - b) =

A s,flexure

= ρbd mm2/m

fbd2 s = 1000A o

.85fc' 1 - 1 - 2Ru = A s,temp

= ρbd =0.002bd = mm2/m

fy .85f c' s

temp= 1000 A

temp

A s

V uc

= Ø √fc' bd

qmin

qmax









d bar spacing above.






Left Side LoadingGiven:

fc' = 27.50Mpa Retaining wall dimensionTentative dimentions:

fy = 275.00 Mpa a = 0.30m B = 3.00 m

23.60kN/m3 H = 3.20m b = 0.40 m

17.25kN/m3 h = 1.00m D = 0.50 m

Φ = 35

μ = 0.45 Surcharge load: Es = ### Mpa

qa = 120.00 kPa 19.50kPa 0.85

25mm 19.20kPa 0.916mm

Minimun factor safety requirements:

Overturning 2.00

Sliding = 1.50


19.50kPa

a

h' = 1.11m

H' = 1.13m

H

19.20kPa

D

b

B

Property Line

a

H'

H

h'

h

D

ωc=

ωs=

o

qs2

= Фshear

=

Øb(main) = qs1 = Фflexure =Øb(temp) =

smax

= [ 3t , 450 ]min

qs2 =

qs1 =

qs2

qs1

x

W3

W4

W5

W2

W1



b

B


Active soil pressure coeffecient: Passive soil pressure coefficient:

0.271 3.690h = 3.20m h = 1.61 m

Active soil pressure:

51.549 kN Passive soil pressure:

1.467m ### kN

1.079m


components wei B-xi Mx

22.656 0.1500 3.3984 2.850 64.57

3.776 0.3333 1.2587 2.667 10.07

35.400 1.5000 ### 1.500 53.1

### 1.7000 ### 1.300 188.1

18.225 0.3780 6.8890 2.622 47.79

Total = ### ### ∑RM = 363.62

Overturning moment: OM = ### kN-m

θ = 0.03 x = 0.066


= 4.808> 2.00 safe!!!OM


= 5.215> 1.50 safe!!!

Check for bearing pressure:

B / 3 = 1.000m

x = RM - OM = 1.4250 m within 1/3 of base

94.543 kPa

3x

77.46kPa safe!!!

57.26kPa safe!!!

Design of stem:

= ###kN/m

= ###kN-m/m

Cah

= Cph =

Pah

=

yah

= Pph

=

yph

=

Wi

xi

Mi

W1=

W2=

W3 =

W4=

W5=

FSoverturning = RM

FSsliding

= (f ∑Wi)

Pah

∑Wi

q = 2W T

, when x < 1/3 B

qmax

= [4B - 6x]W T/B2

qmin

= [6x - 2B]W T/B2

Vu

= 1.7 [qsH + 0.5C

ahw

sH2

Mu

= 1.7 Mmax



y M = 490.874

0.50 2.693kN-m 4

1.00 11.797 kN-m = 201.062

1.50 27.196 kN-m 4

2.00 51.466 kN-m = 0

2.50 85.284 kN-m ρmax = = 0.04

3.00 ### kN-m fy 600+fy

= 0.7429 kN/m

6

Depth as required by shear: b = 1000 mm

= 197.56 mm

Design for flexure:

Try h = = 400 mm

287.5mm

Ru = Mu = ###

ρ = ###

Use: ρ = 0.01

= ###

= 147mm oc

= 517.50

= 388mm oc

Design of Toe:

= 97.48kN/m

= 253.07 kN-m/m

= 0

= 0.04

fy 600+fy

= 0.7866 kN/m6

Depth as required by shear: b = 1000 mm

= ###mm

Design for flexure:

Try h =400 mm

287.5mm

Ru = Mu = ###

M = qsy2/2 + 1/3C

ahw

sy3

Amain = ∏ Dmain

2mm2

Atemp

= ∏Dtemp

2mm2

ρmin

=

0.5 [1.4/f y, √f

c'/4

.75 .85fc'b1 6

Vuc

= Ф √fc'

d = Vu

Vuc

b

d = h - (100+Φs/2)

fbd2

.85fc' 1 - 1 - 2Ru

fy .85f c'

As= ρbd mm2/m


s, 3

Atemp

= 0.0018bd mm2/m


/As, 5

Vu

= 1.7b [(qmax + qmin)/2 - (wcd + ws (h+h'))]

Mu

= 1.7 [(B-b)2/2][qmin

+2/3(B-b)(qmax

- qmin

) - [wc(d)-w

s(h+h')]]

ρmin

=

0.5 [1.4/f y, √f

c'/4f

ρmax =.75 .85fc'b1 6

Vuc

= Ф √fc'

d = Vu

Vuc

b

d = h - (100+Φs/2)



ρ = ### Use: ρ = 0.01

= ###

127.00 mm oc

3t = ###

450 = 450

Compare:

SINCE 127.00 mm < 450.00 mm

Use: 127.00 mm

= 517.50

= 489mm oc

Retaining Wall Details:0.30

25 mmΦ @

137 oc

16 mmΦtemp @

3.70 338 oc

25 mmΦ @

117.00oc

0.50 m

3.00

16mmΦtemp @0.40 338oc

Note: the spacing of the reinforcement is upon the desire of the designer based on the compu

fbd2

.85fc' 1 - 1 - 2Ru

fy .85f c'

As= ρbd mm2/m


s, 3t,450]m

1000Ao/AS

AND 3t, compare 3t and 450 whichever is smallest.

Atemp

= 0.0018bd mm2/m


/As, 5









ed bar spacing above.





****====DESIGN OF A CANTILEVER RETAINING WALL===****

(with shear key)

20.7 Mpa Figure: x=1.5m

414Mpa

18.82 kN/m³ 19.2 kPa

φ = 35 °

0.5

23.6 kN/m³ e=3.65m

143.5 kPa

19.2 kPa backfill height = 3.65 m

c=1.0m

Use 25mmφ for main rebars, 16mmφ for

temperature bars.

Solution:

Composite section and location of forces

e

c

d

f


a = 0.20 m Distances:

c = 1.00 m 1.5000 m

e = 3.65 m 1.6000 m

1.50 m 1.7667 m

0.7500m

Tentative dimensions: 2.4500 m

B = 3.00 m 1.8333 m b = 0.40 m 2.3500 m

d = 0.50 m 1.7000 m

f = 0.40 m H = d + c + e = 5.1500 m

g = 0.40 m 1.0202 m

h = 0.20 m

Active soil pressure coefficient Passive soil pressure coefficient

= 0.27 = 3.69

f c' =

f y

=

ws= q

a=

f =

wc=

qa=

qs =

Use Wu

= 1.4DL + 1.7LL + 1.7H

x1

= B/2 =

x2

=xh

+ a/2 =

xh

= x3

= xh

+ a + (b - a)/3 =

x4

= xh/2 =

x5

= xh

+ b + (B - xh

- b)/2 =

x6

= xh

+ a + 2(b - a)/3 =

x7

= xh

+ a + (B - xh

- a)/2 =

x8

= xh

+ g/2 =

h' = qs/w

s=

h

H

a x

e

b

B

C ah=1−sinφ

1sinφC ph=

1sinφ

1−sinφ

W 7

W 6

W 5

W 4

W 3

W 2

W 1

W 8

a xh

x7

x6

x5

x2

x4

x3

x1

x8



Active soil pressure: h = 5.15 m Passive soil pressure: h = 1.50 m

= 94.43 kN = 78.13 kN

= 1.960 m = 0.500 m

Check retaining wall stability:

Component weights

35.4 1.5000 53.1

21.95 1.6000 35.12

10.97 1.7667 19.39

28.23 0.7500 21.17

96.26 2.4500 235.85

8.75 1.8333 16.04

24.96 2.3500 58.66

2.3 1.7000 3.92

228.83 443.24


185.1kN-m



= 0.37 m = 2.04 > 1.50, ok!

B/3 = 1.00m > e, Rv will fall within the middle third of the base.

No tension will occur on the foundation.

95.19 kPa

57.36 kPa

143.5 kPaqmax < qa, the wall is safe against soil bearing

Design of stem:

M y y/2

V y/3

d

STEM P V M

Soil pressure at level y: Shear equation at level y:

19.2 kPa

5.1 y kPa 19.2 y

2.55y²

Wi

xi

RM=Wix

i

W1

= Bdwc

=

W2

= a(c + e)wc

=

W3

= 0.5(b - a)(c + e)wc=

W4

= c(x)ws=

W5

= (B - x - b)(c + e)ws=

W6

= 0.5(b - a)(c + e)ws=

W7 = qs(B - x - a) =

W8

= 0.5f(g +h)qs=

ΣWi= ΣW

ix

i=

OM = Pah

yah

=

qmax

=

qmin

=

qa =

p1

= qs

P1

P2

p2 = Cahwsx

Ve

Vmax

Mmax

p1

= qs= V

y= P

1+ P

2= q

sy + 0.5C

ahw

sy²

p2

= Cah

wsx = P

1= q

sy =

P2

= 0.5Cahwsy² =

P ah=1

2C ah wh h2h ' P ph=

1

2C ph wh

2

yah=

h23hh '

3 h2h ' y ph=

h

3

x=RM −OM

Rv=∑W

i

e= B2− x

FS overturning = RM =∑W i ci

OM = P ah yah

FS sliding = fR

v= f ∑W

i P

ph P ah

max

min

qalignl ¿¿¿= R

v

B 1±6 e

B2 ¿



Moment equation at level y:

9.6 y²

0.85 y³

Given:

Level, y

0.00 0.000 0.000 0.000 0.000 200 GPa

0.50 10.238 17.404 2.506 4.261 414 MPa

1.00 21.750 36.975 10.450 17.765 20.7 MPa

1.50 34.538 58.714 24.469 41.597 0.85

2.00 48.600 82.620 45.200 76.840 0.90

2.50 63.938 108.694 73.281 124.578 25 mmφ

3.00 80.550 136.935 109.350 185.895 16 mmφ

3.50 98.438 167.344 154.044 261.875

4.00 117.600 199.920 208.000 353.601

4.50 138.038 234.664 271.857 462.156

4.65 144.418 245.510 293.039 498.167

Compute:

490.874 mm² 201.062 mm²

= 0

= 0.01603

try d = 400 mm

= 3.4595

= 0.00939 ok!

3757.84 mm²/m

= 130.627 mm oc

800mm²/m

= 251.327 mm oc

Check for shear:

= 257.818 kN/m

At d distance from bottom of stem:

y = 4.25 m

235.04 kN/m

At 3.00 m

Try d = 300 mm

= 2.29500

My

= P1y

1+P

2y

2= q

sy²/2 + 0.5C

ahw

sy³/3

M1

= P1y

1= q

sy²/2 =

M2

= P2y

2= 0.5C

ahw

sy³/3 =

Vy

Vu=1.7V

yM

yM

u=1.7M

y

Es=

f y

=

f c' =

f shear

=

f flexure

=

D b

=

Dtemp

=

Smax

= [3t, 450]min

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

Vud

= 1.7(19.2y + 3.13667y²) =

Ao=π

4D

b2= Atemp=

π

4D

temp2=

ρm in=1. 4

f y

ρmax=. 75

[. 85 f c ' β 1

f y

. 003 E s

. 003 E s f y ]

Ru=M u /φ

bd 2

ρ=

1

ω

[1−

1−

2ωR u

f y

]=

.85 f c '

f y

[1−

1−

2R u

.85 f c '

] s=

1000 Ao

A s

stemp=1000 Atemp

A s , temp

V u c=φvc

f c '

6bd

Ru=M u /φ

bd 2

'



= 0.00596 ok!

1788.48 mm²/m

= 274.464 mm oc

600mm²/m

= 335.103 mm oc

Design of heel and toe:

subject to erosion a b

heel toe

a b

Use load factor:

1.4 for DL qmin

1.7 for ll and service load bearing pressur q1 q2

161.82 kPa e = 0.37 m

97.52 kPa 0 m

26.35 kPa 129.67 kPa

122.52 kPa -0.4 m

32.64 kPa 103.95

16.52 kPa 1.50 m

1.10m

193.84kN151.41 kN-m

-188.85 kN

-103.87 kN-m

At the toe, the worst condition of loading is when the wall is at empending action to overturn, soil bearing

at the heel is assumed to be zero.

At Heel:

try d = 400 mm

= 1.05


1352.66 mm²/m

= 362.9 mm oc

800mm²/m

As,flexure

= ρbd =

As,temp = ρtemp bd = 0.002bd =

ws2

Qs

ws1

L1

L2

qmax

qmax

x 1.7 =

qmin

x 1.7 = At a, x =B/2 - xh

=

ws1

= 1.4(ws)c = q

1=

ws2

= 1.4(ws)(c + e) = At b, x = [B/2 - (x

h+ b)] =

qsx 1.7 = q

2=

Wc= 1.4(w

c)d = L

1= x

h=

L2

= B - (xh

+ b) =

Va = .5(qmax - q1)L1 + q1L1 - (ws1=0)L1 - WcL1 =M

a= (q

max- q

1)L

1²/3 + (q

1- (w

s1= 0) - W

c)L

1²/2 =

V b

= (.5(q2- q

min)L2 = 0) + [(q

min= 0) - w

s2- w

c- q

s]L

2=

M b

= [(q2- q

min) = 0]L

2²/6 + [(q

min= 0) - w

s2- w

c- q

a]L

2²/2 =

ws1

is taken equal to zero due to erosion of top soil at the heel, the expected worst condition of loading.

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

ρ=ω

1− 1−u

f y=. c

f y1− 1−

u

.85 f c '

s=1000 A

o

A s

stemp=1000 Atemp

A s , temp

Ru=M u /φ

bd 2

ρ=1

ω [1− 1−2ωR u

f y

]=.85 f c '

f y

[1− 1−2R u

.85 f c

' ]

s=1000 Ao

A s



= 251.33 mm oc

Check for shear:

= 257.82 kN/m > Va, safe

At Toe:try d = 400 mm

= 0.72129


1352.66 mm²/m

= 362.9 mm oc

800mm²/m

= 251.33 mm oc

Check for shear:

= 257.82 kN/m > Va, safe

Design of Key:

pph1 pah1

f

pph2 h pah2

0.27 3.69

At pph1:

6.170 m 1.00 m

31.47 kPa 69.45 kPa

At pph2:

6.5702 m 1.40 m

33.51 kPa 97.23 kPa

22.09 kN 56.67 kN

4.465 kN-m 11.964 kN-m

Net Shear:

34.58 kN

Net Moment:

7.499kN-m

try d = 300 mm

= 0.09258

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

Cah

= C ph

=

At pah1

:

yah

= h' + c + e + d = y ph

= c =

pah1

= Cahw

sy

ah1= p

ph1= C

phw

sy

ph1=

At pah2

:

yah

= h' + c + e + d + f = y ph

= c + f =

pah2

= Cahw

sy

ah2= p

ph2= C

phw

sy

ph2=

Vah

= 1.7[pah1

f + (pah2

- pah1

)f/2] V ph

= 1.7[pah1

f + (pah2

- pah1

)f/2]

Vah

= V ph

=

Mah = 1.7[pah1f ²/2 + (pah2 - pah1)f ²/3] M ph = 1.7[pah1f ²/2 + (pah2 - pah1)f ²/3]

Mah

= M ph

=

Vu= V

ph- V

ah=

Mu= M

ph- M

ah=

stemp=temp

A s , temp

V u c=φvc

f c '

6bd

Ru=M u /φ

bd 2

ρ=1

ω [1− 1−2ωR u

f y ]=.85 f c '

f y [1− 1−2R u

.85 f c ' ]

s=1000 Ao

A s

stemp=1000 Atemp

A s , temp

V u c=φvc

f c '

6bd

pah=C

ahw hh' p p h=C p h w h

Ru=

M u /φ

bd 2

'



= 0.00022 not ok!-use pmin

1014.49 mm²/m

= 483.86 mm oc

600mm²/m

= 335.1 mm oc

Check for shear:

= 193.36 kN/m > Va, safe

Details:

Note: the spacing of the reinforcement is upon the desire of the designer based on the compu

0.20 m

3.65 m 25 mm Φ @

240 oc

16 mmΦtem

16 250 oc

mmΦtem 250 oc bw

25 mm Φ @

1.00 120 oc

16 25

mmΦtem 250 oc bw mm Φ @ 360 oc

0.50m

0.40

m

1.50 m 0.40 1.10 m

16 mmΦtemp @

0.20 m 250 oc bw

As,flexure

= ρbd =

As,temp

= ρtemp

bd = 0.002bd =

ρ=ω

1− 1−u

f y

=. c

f y

1− 1−u

.85 f c

'

s=1000 A

o

A s

stemp=

1000 Atemp

A s , temp

V u c=φvc

f c '

6bd













ted bar spacing above.

@

Documents

25690825 Retaining Walls