Muro L e Gravidade

7/27/2019 Muro L e Gravidade

1/55

consulte o site:

http://www.clubedoconcreto.com.br/
http://www.clubedoconcreto.com.br/http://www.clubedoconcreto.com.br/


2/55

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

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. =)

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

1. Soil or gravel without fine particles,

highly permeable.

2. Sand or gravel with silt mixture, low

permeability

3. Silty sand, sand and gravel with high

clay content

4. Medium or stiff clay

5. Soft clay, silt

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

0.5 m

qs = 20 kPa

3.50 m

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 coefficient

w = 18.85 kN/mqs = 20.00 kPa = 3.00

h' = 1.061 m

Distances computation

Active soil pressure coefficient c1 = B/2 = 2.0000 m

c2 = e/2 = 0.1250 m

= 0.3333 c3 = e + a/2 = 0.5000 m

c4 = (B - 2e - a)/3 + e + a = 1.7500 m

c5 = B - e + e/2 = 3.8750 m

c6 = (B - 2e - a)2/3 + e + a = 2.7500 m

c7 = (B - e - a)/2 + e + a = 2.3750 m

0.2

5

0.2

5

15.70 - 18.85 25 - 35 0.2 - 0.4

14.10 - 17.25 20 -25 0.2 - 0.3

17.25 - 18.85 23 - 30 0.3 - 0.4

18.85 - 20.40 25 - 35 0.4 - 0.5

w, kN/m , degrees f, coefficient

17.25 - 18.85 33 -40 0.5 -0.6

7W7c

a

3W 5c

1W

2W

4W

5W

6W

1c

2c

3c4c

6c

'h

h

e

B

d

sin1

sin1

ahC

sin1

sin1

phC


3/55

Given retaining wall dimensions:

a = 0.50 m Passive soil pressure:

b = 1.10 m h = b= 1.10 m

c = 3.50 m

d = 0.60 m = 34.213 kN

e = 0.25 m

Active soil pressure: = 0.3667 m

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

= 97.144 kN B = 4.00 m

= 1.7754 m

Check retaining wall stability wc = 23.60 kN/m

Friction coeff., f = 0.50

component weights Wi ci RM=Wici

W1 = Bdwc = 56.640 2.0000 113.280

W2 = e(b - d)ws = 2.356 0.1250 0.295

W3 = a(b + c - d)wc = 47.200 0.5000 23.600

W4 = (B - 2e - a)(b + c - d)wc/2 = 141.600 1.7500 247.800

W5 = (B - 2e - a)(b + c - d)ws/2 = 113.100 3.8750 438.263

W6 = e(b + c - d)ws = 18.850 2.7500 51.838

W7 = qs(B - e - a) = 65.000 2.3750 154.375

Wi = 444.746 Wici = 1029.450

Overturning moment: OM Factor of safety against overturning:

OM = Pahyah = 172.47 kN-m

Location of resultant with respect to toe: = 5.969008 > 2.00, ok!

= 1.9269 m Factor of safety against sliding:

= 0.0731 m = 2.641285 > 1.50, ok!

B/3 = 1.33 m

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

qmax = 121.61 kPa

qmin = 100.76 kPa

qa = 143 kPa

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

0.5 m

Retaining Wall Details qs = 20 kPa

3.50 m

1.10 m 0.60 m

4.00 m

0.2

5 0.2

5

'22

1hhwhCP ahah

'23'3

2

hh

hhhyah

2

2

1whCP phph

3hyph

iv

WR

OMRMx

xB

e 2

2minmax

61

B

e

B

Rq v

ahah

ii

goverturninyPOM

cWRMFS

ah

phiv

slid ingP

PWffRFS


4/55


5/55


a = 0.50 m Passive soil pressure:

b = 1.10 m h = b= 1.10

c = 3.50 m

d = 0.60 m = 34.2128

e = 0.25 m

Active soil pressure: = 0.36667

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

= 97.1443 kN B = 4.00 m

= 1.77536 m

Check retaining wall stability: wc = 23.60 kN/m

Friction coeff., f = 0.50

component weights Wi ci RM=Wici

W1 = Bdwc = 56.640 2.0000 113.280

W2 = e(b - d)ws = 2.356 0.1250 0.295

W3 = a(b + c - d)wc = 47.200 0.5000 23.600

W4 = (B - 2e - a)(b + c - d)wc/2 = 141.600 1.7500 247.800

W5 = (B - 2e - a)(b + c - d)ws/2 = 113.100 3.8750 438.263

W6 = e(b + c - d)ws = 18.850 2.7500 51.838

W7 = qs(B - e - a) = 65.000 2.3750 154.375

Wi = 444.746 Wici = 1029.450





= 0.0731 m = 2.641285 > 1.50, ok!

B/3 = 1.33 m

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

qmax = 121.6103 kPa

qmin = 100.7628 kPa

qa = 143 kPa

0.5 m

Retaining Wall Details qs = 20.00

3.50 m 0.2

5

0.2

5

1.10 m 0.60 m

4.00 m

'22

1hhwhCP ahah

'23'3

2

hh

hhhyah

iv WR

OMRMx

xB

e 2

2minmax

61

B

e

B

Rq v

ahah

ii

goverturninyPOM

cWRMFS

ah

phiv

sl idi ngP

PWffRFS

2

21 whCP phph

3

hy

ph

2minmax

61

B

e

B

Rq v


6/55

CARMEL B. SABADO CE-162 PROF. GERONID

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====****

Right Side Loading

Given:

fc' = 20.70 Mpa Retaining wall dimensions:

fy = 414.00 Mpa a = 0.30 m

s = 18.82 kN/m3 c = 4.50 m

= 40 o

= 0.50 Tentative dimentions:

c = 23.60 kN/m3 B = 3.20 m

qa = 143.50 kPa b = 0.40 m

qs = 19.30 kPa d = 0.50 m

3.65 m

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

b(temp) 16 mm

a

Es = 200000 Mpa

shear = 0.85 qs = 19.30 kPa

flexure

= 0.9

smax = [ 3t , 450 ]min c

h' = 1.026 m

Cantilever Retaining Wall Figure:

d

b

Property Line

a

h'

c h

d

b

B

backfill height =

W1

C1

W2

C2

W3

W4

W5

C3

C4

C5

W6

C6


7/55

Soil pressure coefficient, Rankine equation for horizontal surface:

Active soil pressure coeffecient: Cah = 0.217443

Active soil pressure: h = 5.00 m

Pah = 1 Cahh(h+2h') =

2 72.137 kN

yah = h2

+ 3hh' =

3(h + 2h') 1.9091 m

Check the retaining wall stability:

components weights Wi ci RM = Wici

W1 = 37.760 1.6000 60.4160

W2 = 31.860 0.1500 4.7790

W3 = 5.310 0.3333 1.7700

W4 = 237.132 1.8000 426.8376

W5 = 4.235 0.3333 1.4115

W6 = 55.97 1.75 97.9475

Wi = 372.267 RM = 593.1616

Overturning moment: OM = 137.7138 kN-m

Factor of safety against overturning:

FSoverturning = RM = 4.307 > 2.00 safe!!!!

OM

Factor of safety against sliding:

FSsliding = (f Wi) = 2.580 > 1.50 safe!!!!

Pah

Check for bearing pressure: qa = 143.50 kPa

Location of resultant with respect to toe:

x = RM - OM = 1.2234 m

Wi

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 = Wi 1 + 6e qmax = 142.001 kPa

B B2

qmin = 90.666 kPa

Since qma < qa, wall is safe againts soil bearing.

Design of stem:

Ve

p1=qs

P1 P2

p2 = Cahwsy

y/2

yM

V

d Vmax Mmax


8/55


9/55

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

qmax x 1.7 = 241.401 kPa V = (-Ws-Wc-qs)L = -470.11

qmin x 1.7 = 154.132 kPa M = (-Ws-Wc-qs)L2

/2 = -658.15Ws = 1.4sc = 118.566 kPa

qs x 1.7 = 32.810 kPa

Wc = 1.4cd = 16.520 kPa

e = 0.377 m

L= B - b) = 2.800 m

Try d = 400 mm

b = 1000 mm

Ru = Mu = 4.570502 As,flexure = bd = 5216.20

fbd2 s = 1000Ao = 94.11

As

= .85fc' 1 - 1 - 2Ru = 0.013041 As,temp = bd =0.002bd = 800.00

fy .85fc' stemp = 1000 Atemp = 251.33

As

Use: = 0.01304

Check for shear: Vuc = fc' bd = 257.8178 kN/m > V, safe!!!!

6

qmin

qmax

qs =


10/55

Retaining Wall Details:

Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.

0.30

25 mm @

16 mm temp. @ 16

260 oc bw mm temp @

5.00 25 mm @

m

25 mm @

0.50

16 mm temp. bars

250 oc bw

0.40 meters


11/55




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

Left Side Loading

Given:

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

fy = 275.00 Mpa a = 0.30 m B = 3.00

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

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

= 35 o

= 0.45 Surcharge load: Es = 200000

qa = 120.00 kPa qs2 = 19.50 kPa shear = 0.85b(main) = 25 mm qs1 = 19.20 kPa flexure = 0.9b(temp) = 16 mm

smax = [ 3t , 450 ]min

Minimun factor safety requirements:

Overturning = 2.00

Sliding = 1.50


qs2 = 19.50 kPa

a

h' = 1.11 m

H' = 1.13 m

H

qs1 = 19.20

b

B

Property Line

a

H' qs2

H

qs1

h'

h

D

x

b

B

W3

W4

W5

W2

W1


12/55


Active soil pressure coeffecient: Passive soil pressure coefficient:

Cah = 0.271 Cph = 3.690

h = 3.20 m h = 1.61

Active soil pressure:

Pah = 51.549 kN Passive soil pressure:

yah = 1.467 m Pph = 177.889

yph = 1.079


components weights Wi xi Mi B-xi Mx

W1 = 22.656 0.1500 3.3984 2.850 64.5696

W2 = 3.776 0.3333 1.2587 2.667 10.069333

W3 = 35.400 1.5000 53.1000 1.500 53.1

W4 = 144.690 1.7000 245.9730 1.300 188.097

W5 = 18.225 0.3780 6.8890 2.622 47.785952

Total = 202.091 310.6191 RM = 363.62189


= 0.031239833 x = 0.066


FSoverturning = RM = 4.808 > 2.00 safe!!!

OMFactor of safety against sliding:

FSsliding = (f Wi)+Pph = 5.215 > 1.50 safe!!!

Pah

Check for bearing pressure:

B / 3 = 1.000 m

x = RM - OM = 1.4250 m within 1/3 o

Wiq = 2WT , when x < 1/3 B = 94.543 kPa

3x

qmax = [4B - 6x]WT/B

2

= 77.46 kPa safe!!!qmin = [6x - 2B]WT/B

2= 57.26 kPa safe!!!

Design of stem:

Vu = 1.7 [qsH + 0.5CahwsH2

= 146.768 kN/m

Mu = 1.7 Mmax = 220.696 kN-m/m


13/55

M = qsy2/2 + 1/3Cahwsy

3

y M Amain = Dmain2

= 490.874

0.50 2.693 kN-m 4

1.00 11.797 kN-m Atemp = Dtemp2

= 201.0621.50 27.196 kN-m 4

2.00 51.466 kN-m min = 0.5 [1.4/fy, fc'/4fy] = 0.00238372.50 85.284 kN-m max = .75 .85fc'b1 600 = 0.03715713.00 129.821 kN-m fy 600+fy

Vuc = fc' = 0.74296

Depth as required by shear: b = 1000 mm

d = Vu = 197.56 mm

Vucb

Design for flexure:

Try h = = 400 mm

d = h - (100+s/2) = 287.5 mmRu = Mu = 2.966719

fbd2

= .85fc' 1 - 1 - 2Ru = 0.011576fy .85fc'

Use: = 0.011576376

As = bd = 3328.208 mm2/m

spacing, S = [1000Ao/As, 3t,450]min = 147 mm oc

Atemp = 0.0018bd = 517.50 mm2/m

spacing, S = [1000Atemp/As, 5t,450]min = 388 mm oc

Design of Toe:

Vu = 1.7b [(qmax + qmin)/2 - (wcd + ws (h+h'))]= 97.4797 kN/m

Mu = 1.7 [(B-b)2/2][qmin +2/3(B-b)(qmax - qmin) - [wc(d)-ws(h+h')]]

= 253.0727332 kN-m/m

min = 0.5 [1.4/fy, fc'/4fy] = 0.002383656max = .75 .85fc'b1 600 = = 0.037157143

fy 600+fy

Vuc = fc' = 0.7866 kN/m6

Depth as required by shear: b = 1000 mmd = Vu = 123.9243 mm

Vucb

Design for flexure:

Try h = 400 mm

d = h - (100+s/2) = 287.5 mmRu = Mu = 3.401944

fbd2

= .85fc' 1 - 1 - 2Ru = 0.013432 Use: =fy .85fc'


14/55

As = bd = 3861.697 mm2/m

spacing, S = [1000Ao/As, 3t,450]min = 127.00 mm oc

3t = 1200.00

450 = 450

Compare: 1000Ao/AS AND 3t, compare 3t and 450 whichever is smallest.

SINCE 127.00 mm < 450.00 mm

Use: 127.00 mm

Atemp = 0.0018bd = 517.50 mm2/m



0.30

25

137

16

3.70 338

25

117.00

0.50

3.00

16 mmtemp @0.40 338 oc



15/55




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

(with shear key)

fc' = 20.7 Mpa Figure: x=1.5m

fy = 414 Mpa

ws = 18.82 kN/m qa = 19.2 kPa

= 35

f= 0.5

wc = 23.6 kN/m e=3.65m

qa = 143.5 kPa

qs = 19.2 kPa

backfill height = 3.65 m

Use Wu = 1.4DL + 1.7LL + 1.7H c=1.0m

Use 25mm for main rebars, 16mm for

temperature bars.

Solution:

Composite section and location of forces

e

c

d

f


a = 0.20 m Distances:

c = 1.00 m x1 = B/2 = 1.5000

e = 3.65 m x2 =xh + a/2 = 1.6000

xh = 1.50 m x3 = xh + a + (b - a)/3 = 1.7667

x4 = xh/2 = 0.7500

Tentative dimensions: x5 = xh + b + (B - xh - b)/2 = 2.4500

B = 3.00 m x6 = xh + a + 2(b - a)/3 = 1.8333

b = 0.40 m x7 = xh + a + (B - xh - a)/2 = 2.3500

d = 0.50 m x8 = xh + g/2 = 1.7000

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

g = 0.40 m h' = qs/ws = 1.0202

h = 0.20 m

Active soil pressure coefficient Passive soil pressure coefficient

= 0.27099 = 3.69

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

h

H

ax

e

b

B

sin1

sin1

ahC

sin1

sin1

phC

7W

6W

5W

4W 3W

2W

1W

8W

ahx7

x

6x

5x

2x

4x3x

1x

8x


16/55

= 94.42831 kN = 78.13017

= 1.960 m = 0.500

Check retaining wall stability:

Component weights Wi xi RM=Wixi

W1 = Bdwc = 35.4 1.5000 53.1

W2 = a(c + e)wc = 21.948 1.6000 35.1168

W3 = 0.5(b - a)(c + e)wc = 10.974 1.7667 19.3874

W4 = c(x)ws = 28.23 0.7500 21.1725

W5 = (B - x - b)(c + e)ws = 96.2643 2.4500 235.847535

W6 = 0.5(b - a)(c + e)ws = 8.7513 1.8333 16.04405

W7 = qs(B - x - a) = 24.96 2.3500 58.656

W8 = 0.5f(g +h)qs = 2.304 1.7000 3.9168

Wi = 228.8316 Wixi = 443.241085





= 0.37192 m = 2.03907049 > 1.50, ok!

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

No tension will occur on the foundation.

qmax = 95.19002 kPa

qmin = 57.36438 kPa

qa = 143.5 kPa

qmax < qa, the wall is safe against soil bearing

Design of stem:

p1 = qs

P1

M y y/2 P2

V y/3

p2 = Cahwsx

Ve

d

Vmax Mmax

STEM P V M

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

p1 = qs = 19.2 kPa Vy = P1 + P2 = qsy + 0.5Cahwsy

p2 = Cahwsx = 5.10003 y kPa P1 = qsy = 19.2

P2 = 0.5Cahwsy = 2.55

'22

1hhwhCP ahah

2

2

1whCP phph

'23'3

2

hh

hhhyah

3

hyph

iv WR

OMRMx

x

B

e 2

ahah

ii

goverturninyPOM

cWRMFS

ah

phiv

sl id ing P

PWffR

FS

2minmax

61

B

e

B

Rq v


17/55

Moment equation at level y:

My = P1y1 +P2y2 = qsy/2 + 0.5Cahwsy/3

M1 = P1y1 = qsy/2 = 9.6 y

M2 = P2y2 = 0.5Cahwsy/3 = 0.85000547 y

Given:

Level, y Vy Vu=1.7Vy My Mu=1.7My

0.00 0.000 0.000 0.000 0.000 Es = 200

0.50 10.238 17.404 2.506 4.261 fy = 414

1.00 21.750 36.975 10.450 17.765 fc' = 20.7

1.50 34.538 58.714 24.469 41.597 fshear= 0.85

2.00 48.600 82.620 45.200 76.840 fflexure = 0.90

2.50 63.938 108.694 73.281 124.578 Db = 25

3.00 80.550 136.935 109.350 185.895 Dtemp = 16

3.50 98.438 167.344 154.044 261.875 Smax = [3t, 450]min

4.00 117.600 199.920 208.000 353.6014.50 138.038 234.664 271.857 462.156

4.65 144.418 245.510 293.039 498.167

Compute:

490.874 mm 201.062 mm

= 0.0033816

= 0.01603

try d = 400 mm

= 3.4595

= 0.00939 ok!

As,flexure = bd = 3757.84 mm/m

= 130.627 mm oc

As,temp = tempbd = 0.002bd = 800 mm/m

= 251.327 mm oc

Check for shear:

= 257.818 kN/m

At d distance from bottom of stem:

y = 4.25 m

Vud = 1.7(19.2y + 3.13667y) = 235.03537 kN/m

At 3.00 m

Try d = 300 mm

= 2.29500

= 0.00596 ok!

2

4bo DA

2

4temptemp

DA

yf

4.1min

ys

s

y

c

fE

E

f

f

003.

003.'85.75. 1max

2

/

bd

MR uu

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

s

o

A

As

1000

temps

temp

tempA

As

,

1000

bdf

Vc

vcuc6

'

2

/

bd

MR uu

'85.211'85.2111

c

u

y

c

y

u

fR

ff

fR


18/55


= 274.464 mm oc


= 335.103 mm oc

Design of heel and toe:

ws2 Qs

subject to erosion ws1 a b

heel toe

a b

L1 L2

Use load factor:

1.4 for DL qmax qmin

1.7 for ll and service load bearing pressures q1 q2

qmax x 1.7 = 161.823 kPa e = 0.371923

qmin x 1.7 = 97.51944 kPa At a, x =B/2 - xh = 0

ws1 = 1.4(ws)c = 26.348 kPa q1 = 129.6712

ws2 = 1.4(ws)(c + e) = 122.5182 kPa At b, x = [B/2 - (xh + b)] = -0.4

qs x 1.7 = 32.64 kPa q2 = 103.9498

Wc = 1.4(wc)d = 16.52 kPa L1 = xh = 1.50

L2 = B - (xh + b) = 1.10

Va = .5(qmax - q1)L1 + q1L1 - (ws1=0)L1 - WcL1 = 193.840711 kN

Ma = (qmax - q1)L1/3 + (q1 - (ws1 = 0) - Wc)L1/2 = 151.408996 kN-m

Vb = (.5(q2 - qmin)L2 = 0) + [(qmin = 0) - ws2 - wc - qs]L2 = -188.84602 kN

Mb = [(q2 - qmin) = 0]L2/6 + [(qmin = 0) - ws2 - wc - qa]L2/2 = -103.86531 kN-m

ws1 is taken equal to zero due to erosion of top soil at the heel, the expected worst condition of loading.

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

= 1.0514514

= 0.0026205 not ok!-use pmin


= 362.896026 mm oc


= 251.327412 mm oc

s

o

A

As

1000

temps

temp

tempA

As

,

1000

2/bdMR uu

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

s

o

A

As

1000

temps

temp

tempA

A

s,

1000


19/55

Check for shear:

= 257.81777 kN/m > Va, safe

At Toe:try d = 400 mm

= 0.72129



= 362.896026 mm oc

As,temp= tempbd = 0.002bd = 800 mm/m

= 251.327412 mm oc

Check for shear:

= 257.81777 kN/m > Va, safe

Design of Key:

pph1 pah1

f

pph2 h pah2

Cah = 0.27099 Cph = 3.69

At pah1: At pph1:

yah = h' + c + e + d = 6.170 m yph = c = 1.00

pah1 = Cahwsyah1 = 31.468178 kPa pph1 = Cphwsyph1 = 69.44904

At pah2: At pph2:

yah = h' + c + e + d + f = 6.5702 m yph = c + f = 1.40

pah2 = Cahwsyah2 = 33.508191 kPa pph2 = Cphwsyph2 = 97.22866

Vah = 1.7[pah1f + (pah2 - pah1)f/2] Vph = 1.7[pah1f + (pah2 - pah1)f/2]

Vah = 22.091966 kN Vph = 56.67042

Mah = 1.7[pah1f/2 + (pah2 - pah1)f

/3] Mph = 1.7[pah1f

/2 + (pah2 - pah1)f

/3]Mah = 4.465 kN-m Mph = 11.964

Net Shear:

Vu = Vph - Vah = 34.5784538 kN

Net Moment:

Mu = Mph - Mah = 7.499 kN-m

try d = 300 mm

= 0.09258


bdf

Vc

vcuc6

'

2

/

bd

MR uu

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

s

o

A

As

1000

temps

temp

tempA

As

,

1000

bdf

Vc

vcuc6

'

'hhwCp ahah whCp phph

2

/

bd

MR u

u

'85.

2

11

'85.2

11

1

c

u

y

c

y

u

f

R

f

f

f

R


20/55


= 483.861369 mm oc


= 335.103216 mm oc

Check for shear:

= 193.36332 kN/m > Va, safe

Details:


1.00 m

1.50 m

0

mmtemp @ 250 oc bw

3.650

mmtemp @ 250 oc bw

0.40

m

0.40

m

0.00 m 0.50 -0.10 m

0 mmtemp @

0.00 m 250 oc bw

s

o

A

As

1000

temps

temp

tempA

As

,

1000

bdf

Vc

vcuc6

'


21/55

ES P. ANCOG

18-Aug-09

m

m

m

m

m

m

m

- 0.4

- 0.3

- 0.4

- 0.5

fficient

-0.6


22/55

m

kN

m

!

!


23/55

ES P. ANCOG

18-Aug-09

Pah

yah


24/55


25/55

/3

y2

y3

0.003382

0.016032

safe!!!!


26/55

kN

kN-m

mm2/m

mm cc

mm2/m

mm cc


27/55

240 oc

250 oc

250 oc

360 oc

meters


28/55

ES P. ANCOG

18-Aug-09

m

m

m

Mpa

kPa

D


29/55

m

kN

m

base


30/55

mm2

mm

2

kN/m

0.013431989


31/55

mm @oc

mmtemp @oc

mm @oc

m


32/55


33/55

kN

m

y

y


34/55

GPa

MPa

MPa

mm

mm


35/55

m

m

kPa

m

m

m


36/55

m

kPa

m

kPa

kN

kN-m


37/55

mm @

25

240 oc

16

250 oc

mmtemp @

16

120 oc16

360 oc

mm @


38/55





Right Side Loading

Given:

fc' = 20.70 Mpa Retaining wall dimensions:

fy = 414.00 Mpa a = 0.30 m

s = 18.82 kN/m3 c = 4.50 m

= 40 o

= 0.50 Tentative dimentions:

c = 23.60 kN/m3 B = 3.20 m

qa = 143.50 kPa b = 0.40 m

qs = 19.30 kPa d = 0.50 m3.65 m

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

b(temp) 16 mm

a

Es = 200000 Mpa

shear = 0.85 qs = 19.30 kPa

flexure = 0.9

smax = [ 3t , 450 ]min c

h' = 1.026 m


d

b

Property Line

a

h'

c h

Pah

yah

db

B


Active soil pressure coeffecient: Cah = 0.217443

Active soil pressure: h = 5.00 m

Pah = 1 Cahh(h+2h') =

2 72.137 kN

yah =h

2+ 3hh' =

3(h + 2h') 1.9091 m

backfill height =

W1

C1

W2C2

W3

W4

W5

C3

C4

C5

W6

C6


39/55


components weights Wi ci RM = Wici

W1 = 37.760 1.6000 60.4160

W2 = 31.860 0.1500 4.7790

W3 = 5.310 0.3333 1.7700

W4 = 237.132 1.8000 426.8376

W5 = 4.235 0.3333 1.4115

W6 = 55.97 1.75 97.9475

Wi = 372.267 RM = 593.1616



FSoverturning = RM = 4.307 > 2.00 safe!!!!

OM


FSsliding = (f Wi) = 2.580 > 1.50 safe!!!!

Pah

Check for bearing pressure: qa = 143.50 kPa

Location of resultant with respect to toe:

x = RM - OM = 1.2234 m

Wi

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 = Wi 1 + 6e qmax = 142.001 kPa

B B2

qmin = 90.666 kPa

Since qma < qa, wall is safe againts soil bearing.

Design of stem:

Soil pressure at y: Moment equation at level y:

p1 = qs = 19.30 kPa My = p1y1 + p2y2 = qsy2/2 + 0.5cahwsy

3/3

p2 = cahwsy = 4.09227 y kPa M1 = qsy2/2 = 9.65 y

2

M2 = 0.5cahwsy3/3 = 0.682 y

3

Shear equation at level y:

Vy= p1 + p2 = qsy + 0.5cahwsy2

At level y = 4.50 m

p1 = qsy = 19.30 y Vu = 218.1 kN

2.04614 y2

Mu = 437.9 kN-m

Db2

= 490.874 mm2

min = 1.4 / fy = 0.003382

p2 = 0.5cahwsy2

=

Ve

p1=qs

P1 P2

p2 = Cahwsy

y/2yM

V

dVmax Mmax

Stem P MV


40/55

4 max = .75 .85fc'b1 600 = 0.016032

Atemp = Dtemp2

= 201.062 mm2

fy 600+fy

4

Try d = 400 mm ; b = 1000 mm

Ru = Mu = 3.040685

fbd2

= .85fc' 1 - 1 - 2Ru = 0.00812

fy .85fc'

Use: = 0.00812

As,flexure = bd = 3248.172 mm2/m

s = 1000Ao = 151.1231 mm cc

As

As,temp = bd =0.002bd = 800 mm2/m

stemp = 1000 Atemp = 251.3274 mm cc

As

Check for shear: Vuc = fc' bd = 257.8178 kN/m

6

At d distance from the bottom of stem:

yd = 4.10 m

Vud = 192.9935 kN/m < Vuc, safe!!!!

At y = 3.00 m

Try d = 300 mm Ru = Mu = 2.209270

fbd2

= .85fc' 1 - 1 - 2Ru = 0.005722

fy .85fc'

Use: = 0.005722

As,flexure = bd = 1716.459 mm2/m

s = 1000Ao = 285.9805 mm cc

As

As,temp = bd =0.002bd = 600 mm2/m

stemp = 1000 Atemp = 335.1032 mm cc

As

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.

qmin

qmax

qs =


41/55

Use: 1.4 for DL

1.7 for LL and service load bearing pressure

qmax x 1.7 = 241.401 kPa V = (-Ws-Wc-qs)L = -470.11 kN

qmin x 1.7 = 154.132 kPa M = (-Ws-Wc-qs)L2/2 = -658.15 kN-m

Ws = 1.4sc = 118.566 kPaqs x 1.7 = 32.810 kPa

Wc = 1.4cd = 16.520 kPa

e = 0.377 m

L= B - b) = 2.800 m

Try d = 400 mm

b = 1000 mm

Ru = Mu = 4.570502 As,flexure = bd = 5216.20 mm2/m

fbd2

s = 1000Ao = 94.11 mm cc

As

= .85fc' 1 - 1 - 2Ru = 0.013041 As,temp = bd =0.002bd = 800.00 mm2/m

fy .85fc' stemp = 1000 Atemp = 251.33 mm cc

As

Use: = 0.01304

Check for shear: Vuc = fc' bd = 257.8178 kN/m > V, safe!!!!

6



0.30

25 mm @ 240 oc

16 mm temp. @ 16

260 oc bw mm temp @ 250 oc

5.00 25 250 oc

m

25 mm @ 360 oc

0.50 meters

16 mm temp. bars

250 oc bw

0.40 meters


42/55


43/55


44/55


45/55


46/55





Left Side Loading

Given:

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

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

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

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

= 35 o

= 0.45 Surcharge load: Es = 200000 Mpaqa = 120.00 kPa qs2 = 19.50 kPa shear = 0.85

b(main) = 25 mm qs1

= 19.20 kPa flexure

= 0.9

b(temp) = 16 mm

smax = [ 3t , 450 ]min

Minimun factor safety requirements:

Overturning = 2.00

Sliding = 1.50


qs2 = 19.50 kPa

a

h' = 1.11 m

H' = 1.13 m

H

qs1 = 19.20 kPa

b

B

Property Line

a

H' qs2

H

qs1

h'

h

D

x

bB

D

W3

W4

W5

W2

W1


47/55


Active soil pressure coeffecient: Passive soil pressure coefficient:

Cah = 0.271 Cph = 3.690

h = 3.20 m h = 1.61 m

Active soil pressure:

Pah = 51.549 kN Passive soil pressure:

yah = 1.467 m Pph = 177.889 kN

yph = 1.079 m


components weights Wi xi Mi B-xi Mx

W1 = 22.656 0.1500 3.3984 2.850 64.5696

W2 = 3.776 0.3333 1.2587 2.667 10.06933

W3 = 35.400 1.5000 53.1000 1.500 53.1

W4 = 144.690 1.7000 245.9730 1.300 188.097

W5 = 18.225 0.3780 6.8890 2.622 47.78595Total = 202.091 310.6191 RM = 363.6219


= 0.03124 x = 0.066


FSoverturning = RM = 4.808 > 2.00 safe!!!

OM


FSsliding = (f Wi)+Pph = 5.215 > 1.50 safe!!!Pah

Check for bearing pressure:

B / 3 = 1.000 m

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

Wiq = 2WT , when x < 1/3 B = 94.543 kPa

3x

qmax = [4B - 6x]WT/B2

= 77.46 kPa safe!!!

qmin = [6x - 2B]WT/B2

= 57.26 kPa safe!!!

Design of stem:

Vu = 1.7 [qsH + 0.5CahwsH2

= 146.768 kN/m

Mu = 1.7 Mmax = 220.696 kN-m/m

M = qsy

2

/2 + 1/3Cahwsy

3


48/55

y M Amain = Dmain2

= 490.874 mm2

0.50 2.693 kN-m 4

1.00 11.797 kN-m Atemp = Dtemp2

= 201.062 mm2

1.50 27.196 kN-m 4

2.00 51.466 kN-m min = 0.5 [1.4/fy, fc'/4fy] = 0.002383662.50 85.284 kN-m max = .75 .85fc'b1 600 = 0.037157143.00 129.821 kN-m fy 600+fy

Vuc = fc' = 0.7429 kN/m6


d = Vu = 197.56 mm

Vucb

Design for flexure:

Try h = = 400 mmd = h - (100+s/2) = 287.5 mmRu = Mu = 2.966719

fbd2

= .85fc' 1 - 1 - 2Ru = 0.011576fy .85fc'

Use: = 0.011576

As = bd = 3328.208 mm2/m

spacing, S = [1000Ao/As, 3t,450]min = 147 mm oc

Atemp = 0.0018bd = 517.50 mm2/m


Design of Toe:

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

= 97.4797 kN/m

Mu = 1.7 [(B-b)2/2][qmin +2/3(B-b)(qmax - qmin) - [wc(d)-ws(h+h')]]

= 253.0727 kN-m/m

min = 0.5 [1.4/fy, fc'/4fy] = 0.002384max = .75 .85fc'b1 600 = 0.037157

fy 600+fy

Vuc = fc' = 0.7866 kN/m6


d = Vu = 123.9243 mm

VucbDesign for flexure:

Try h = 400 mm

d = h - (100+s/2) = 287.5 mmRu = Mu = 3.401944

fbd2

= .85fc' 1 - 1 - 2Ru = 0.013432 Use: = 0.01343199fy .85fc'

As = bd = 3861.697 mm2/m


49/55

spacing, S = [1000Ao/As, 3t,450]min = 127.00 mm oc

3t = 1200.00

450 = 450

Compare: 1000Ao/AS AND 3t, compare 3t and 450 whichever is smallest.

SINCE 127.00 mm < 450.00 mm

Use: 127.00 mm

Atemp = 0.0018bd = 517.50 mm2/m



0.30

25 mm @137 oc

16 mmtemp @3.70 338 oc

25 mm @117.00 oc

0.50 m

3.00

16 mmtemp @0.40 338 oc



50/55




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

(with shear key)

fc' = 20.7 Mpa Figure: x=1.5m

fy = 414 Mpa

ws = 18.82 kN/m qa = 19.2 kPa

= 35

f= 0.5

wc = 23.6 kN/m e=3.65m

qa = 143.5 kPa

qs = 19.2 kPa

backfill height = 3.65 m

Use Wu = 1.4DL + 1.7LL + 1.7H c=1.0m

Use 25mm for main rebars, 16mm for

temperature bars.

Solution:

Composite section and location of forces

e

c

d

f


a = 0.20 m Distances:

c = 1.00 m x1 = B/2 = 1.5000 m

e = 3.65 m x2 =xh + a/2 = 1.6000 m

xh = 1.50 m x3 = xh + a + (b - a)/3 = 1.7667 m

x4 = xh/2 = 0.7500 m

Tentative dimensions: x5 = xh + b + (B - xh - b)/2 = 2.4500 m

B = 3.00 m x6 = xh + a + 2(b - a)/3 = 1.8333 m

b = 0.40 m x7 = xh + a + (B - xh - a)/2 = 2.3500 m

d = 0.50 m x8 = xh + g/2 = 1.7000 m

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

g = 0.40 m h' = qs/ws = 1.0202 m

h = 0.20 m

Active soil pressure coefficient Passive soil pressure coefficient

= 0.27099 = 3.69

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

h

H

ax

e

b

B

sin1

sin1

ahC

sin1

sin1

phC

7W

6W

5W

4W 3W

2W

1W

8W

ah

x7

x

6x

5x

2x

4x

3x

1x

8x


51/55

= 94.4283 kN = 78.1302 kN

= 1.960 m = 0.500 m

Check retaining wall stability:

Component weights Wi xi RM=Wixi

W1 = Bdwc = 35.4 1.5000 53.1

W2 = a(c + e)wc = 21.948 1.6000 35.1168

W3 = 0.5(b - a)(c + e)wc = 10.974 1.7667 19.3874

W4 = c(x)ws = 28.23 0.7500 21.1725

W5 = (B - x - b)(c + e)ws = 96.2643 2.4500 235.848

W6 = 0.5(b - a)(c + e)ws = 8.7513 1.8333 16.0441

W7 = qs(B - x - a) = 24.96 2.3500 58.656

W8 = 0.5f(g +h)qs = 2.304 1.7000 3.9168

Wi = 228.832 Wixi = 443.241





= 0.37192 m = 2.03907 > 1.50, ok!

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

No tension will occur on the foundation.

qmax = 95.19 kPa

qmin = 57.3644 kPa

qa = 143.5 kPa

qmax < qa, the wall is safe against soil bearing

Design of stem:

p1 = qs

P1

M y y/2 P2

V y/3

p2 = Cahwsx

Ved

Vmax Mmax

STEM P V M

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

p1 = qs = 19.2 kPa Vy = P1 + P2 = qsy + 0.5Cahwsy

p2 = Cahwsx = 5.10003 y kPa P1 = qsy = 19.2 y

P2 = 0.5Cahwsy = 2.55 y

Moment equation at level y:

My = P1y1 +P2y2 = qsy/2 + 0.5Cahwsy/3

M1 = P1y1 = qsy/2 = 9.6 y

'22

hhwhCP ahah 2

2whCP phph

'23'3

2

hh

hhhyah

3

hyph

iv WR

OMRMx

xB

e 2

ahah

ii

goverturninyPOM

cWRMFS

ah

phiv

slidingP

PWffRFS

2minmax

61

B

e

B

Rq v


52/55

M2 = P2y2 = 0.5Cahwsy/3 = 0.85001 y

Given:

Level, y Vy Vu=1.7Vy My u=1.7My

0.00 0.000 0.000 0.000 0.000 Es = 200 GPa

0.50 10.238 17.404 2.506 4.261 fy = 414 MPa

1.00 21.750 36.975 10.450 17.765 fc

' = 20.7 MPa

1.50 34.538 58.714 24.469 41.597 fshear= 0.85

2.00 48.600 82.620 45.200 76.840 fflexure = 0.90

2.50 63.938 108.694 73.281 124.578 Db = 25 mm

3.00 80.550 136.935 109.350 185.895 Dtemp = 16 mm

3.50 98.438 167.344 154.044 261.875 Smax = [3t, 450]min

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 mm 201.062 mm

= 0.00338

= 0.01603

try d = 400 mm

= 3.4595

= 0.00939 ok!


= 130.627 mm oc


= 251.327 mm oc

Check for shear:

= 257.818 kN/m

At d distance from bottom of stem:

y = 4.25 m

Vud = 1.7(19.2y + 3.13667y) = 235.035 kN/m

At 3.00 m

Try d = 300 mm

= 2.29500

= 0.00596 ok!


= 274.464 mm oc

2

4bo

DA

2

4temptemp

DA

yf

4.1min

ys

s

y

c

fE

E

f

f

003.

003.'85.75. 1

max

2

/

bd

MR u

u

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

s

o

A

As

1000

temps

temp

tempA

As

,

1000

bdf

Vc

vcuc6

'

2

/

bd

MR u

u

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

oAs 1000


53/55


54/55

try d = 400 mm

= 0.72129



= 362.896 mm oc

As,temp= tempbd = 0.002bd = 800 mm/m

= 251.327 mm oc

Check for shear:

= 257.818 kN/m > Va, safe

Design of Key:

pph1 pah1

f

pph2 h pah2

Cah = 0.27099 Cph = 3.69

At pah1: At pph1:

yah = h' + c + e + d = 6.170 m yph = c = 1.00 m

pah1 = Cahwsyah1 = 31.4682 kPa pph1 = Cphwsyph1 = 69.449 kPa

At pah2: At pph2:

yah = h' + c + e + d + f = 6.5702 m yph = c + f = 1.40 m

pah2 = Cahwsyah2 = 33.5082 kPa pph2 = Cphwsyph2 = 97.2287 kPaVah = 1.7[pah1f + (pah2 - pah1)f/2] Vph = 1.7[pah1f + (pah2 - pah1)f/2]

Vah = 22.092 kN Vph = 56.6704 kN

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

Mah = 4.465 kN-m Mph = 11.964 kN-m

Net Shear:

Vu = Vph - Vah = 34.5785 kN

Net Moment:

Mu = Mph - Mah = 7.499 kN-m

try d = 300 mm

= 0.09258



= 483.861 mm oc


2

/

bd

MR uu

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

s

o

A

As

1000

temps

temp

tempA

As

,

1000

bdf

Vc

vcuc6

'

'hhwCp ahah whCp phph

2/

bdMR uu

'85.

211

'85.211

1

c

u

y

c

y

u

f

R

f

f

f

R

s

o

A

As

1000


55/55

= 335.103 mm oc

Check for shear:

= 193.363 kN/m > Va, safe

Details:


0.20 m

3.65 m 25 mm @

240 oc

16 mmtemp @

16 250 ocmmtemp @ 250 oc bw

25 mm @

1.00 120 oc

16 25

mmtemp @ 250 oc bw mm @ 360 oc

0.50

m

0.40

m

1.50 m 0.40 1.10 m

16 mmtemp @

0.20 m 250 oc bw

temps

temp

tempA

As

,

1000

bdf

Vc

vcuc6

'

Muro L e Gravidade

Documents

Transcript of Muro L e Gravidade