# Retaining Wall Design_1

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Design of Retaing wall

Input

Height above GL = 3 m

Density of soil, = 18 kN/m3

Denity of Concrete = 25 kN/m3

= 30

= 18

Grade of Concrete, fck = 20 Mpa

Grade of Steel, fy = 415 Mpa

SBC of foundation strata, q0 = 100 kN/m2

Coefficient of friction, = 0.5

b5

b3

b6

H4

H3

H1

H2

b2 b4

b1

W1

W2

W3

W4

D E B C

A

TOE HEEL

SHEAR

KEY

STEM

H6

H5

H7

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Solution

1 Design constants

Xu, max = 0.48

d

2 Depth of foundation

Density of soil, = 18 kN/m3

= 18000 N/m3

ymin = (q0/) x ((1-sin)/(1+sin))2

= 0.61728 m

Keep, H3 = 1 m

Height of wall above its base

Hence, H1 = 4 m

3 Dimensions of base

"D-E" Ratio =

b

= 1-(q0/(2.2 H1))

= 0.37

ka = (1-sin)/(1+sin)

ka = 0.33333

b1 = 0.95 H Sqrt{(Ka/((1-)(1+))}

b1 = 1.90288 m

Base width from sliding condition

b1 = 0.7 H Ka/(1-)

b1 = 2.96296 m

Base width from from normal practice

b1 = 0.6 H or 0.4 H

b1 = 2.4 m

Keep, b1 = 2.4 m

Width of Toe slab, b1 = 0.9 m

Let the thickness of base, H4 = H/12

= 0.3 m

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4 Thickness of stem

Height, H5 "A-B" = 3.7 m

Consider 1m length retaining wall

Maximum Bending moment @ "B"

= Ka H53/6

= 50.653 kN-m

M @ "B" = 5.1E+07 N-mm

Mu = 7.6E+07 N-mm

Hence effective depth, b3eff = Sqrt(Mu/Ru b)

= 159.73 mm

Keep, b3eff = 240 mm

Total thickness, b3 = 300 mm

Using 16 dia bars

b5 = 200 mm

b5eff = 140 mm

Width of Heel, b4 = 1.2 m

5 Stability of wall

Designation Force, kNLever arm,

m

Moment

@ toe,

kN-m

W1 18.5 1.1 20.35

W2 4.625 0.97 4.47083

W3 18 1.2 21.6

W4 79.92 1.8 143.856

Total W = 121.045 MR = 190.30

Total resisting moment MR = 190.3 kN-m

Earth pressure, P = Ka H12/2

= 48 kN

Check for overturning

Overturning moment, M0 = 64 kN-m

F.S. againt overturning = = 190.30

64

= 3.0 should be > 2

Hence Safe

Check for Sliding

F.S. againt Sliding = W

P

= 1.3 should be > 1.5

Hence Unsafe, Provide shear key

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Pressure distribution

Net moment, M = 190.3 - 64

M = 126.30 kN-m

Distance X of the pont of application of the resultant, from toe

X = MW

X = = 126.30

121.045

X = = 1.04 m

Eccentricity, e = (b1/2) - X

e = 0.16 m Should be < b1/6

b/6 = 0.4 m

Hence No tension

Pressure p1 at toe = (W/b) x (1+(6e/b))

= 70.2 kN/m2 Should be less than SBC

= 100 kN/m2

Hence Safe

Pressure p2 at heel = (W/b) x (1-(6e/b))

= 30.6917 kN/m2

Pressure at the junction of stem with toe slab,

p = 55.4 kN/m2

Pressure at the junction of stem with heel slab,

p = 50.4458 kN/m2

b3b2 b4

b1

D E B C

A

H7

p1

p2P

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6 Design of toe slab

Downward weight of slab per unit area = 7.5 kN/m2

Hence net pressure intensity under "D" = 70.2 - 7.5 = 62.7 kN/m2

Hence net pressure intensity under "E" = 55.4 - 7.5 = 47.9 kN/m2

Total force = Shear force at E = 50 kN

X from "E" = 0.47 m

Bending moment @ "E" = 23.3955 kN-m

= 2.3E+07 N-mm

Mu @ "E" = 3.5E+07 N-mm

H4eff = Sqrt(Mu/Ru b)

= 108.555 mm

Provide, H4 = 260 mm

H4eff = 200 mm

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Steel

Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))

Ast = 513.599 mm2

This reinforcement to be provided at bottom face.

If alternate bars of stem reinforcement are to be bent & continued in the toe slab,

area available = 565.2 mm2 See step 8

Bar dia = 12 # bars

Spacing = 200 mm

Provide 12 # bars @ = 200 C/C at top of heel slab

Check for development length, Ld = 47

= 564 mm

Providing 50mm clear side cover ,

actual length available = 850 mm

Hence safe

Distribution reinforcement = 0.12 x 1000 x A /100

= 276 mm2

Bar dia= = 8 # bars

Spacing = 182 mm

Provide 8 # bars @ = 180 C/C

7 Design of heel slab

1) Total weight of soil = 79.92 kN

Lever arm @ "B" = 0.60 m

2) Total weight of heel slab = 7.80 kN

Lever arm @ "B" = 0.60 m

3) Total upward soil reaction = 48.68 kN

Lever arm @ "B" = 0.55 mTotal force = S.F. at "B" = 39.04 kN

B.M. @ "B" = 25.793 kN-m

= 2.6E+07 N-mm

Mu = 3.9E+07 N-mm

H7eff = Sqrt(Mu/Ru b)

= 113.981 mm

Provide, H6 = 260 mm

H7eff = 200 mm

Steel

Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))

= 569.737 mm2

Bar dia = 10 # bars

Spacing = 138 mm

Provide 10 # bars @ 130 C/C at top of heel slab

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Development length, Ld = 47

to the left of "B" = 470 mm

= 500 mm

Distribution reinforcement = 0.12 x 1000 x A /100

= 276 mm2

Bar dia= = 8 # barsSpacing = 182 mm

Provide 8 # bars @ = 180 C/C

Shear stress, v = 1.5 x 39.04 x 1000/(1000 x 200)

= 0.2928 N/mm2

pt Ast x 100

1000 x d

pt c = 0.28487

0.25 0.36

0.5 0.48

0.28486828 0.38674Corresponding c = 0.38674 N/mm

2

Hence Safe

8 Reinforcement in stem

Revised H5 = 3.74 m

M = Ka H53/6

= 52.3136 kN-m

= 5.2E+07 N-mm

Mu = 7.8E+07 N-mm

b3eff = Sqrt(Mu/Ru b)

= 162.327 mm

Provide, b3eff = 250 mm

b3 = 310 mm

b5 = 200 mm

b5eff = 140 mm

Ast = = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))

= 943.711939 mm2

Bar dia= = 12 # bars

Spacing = 120 mm

Provide 12 # bars @ 100 C/C

Astact = 1130.4 mm2

Continue alternate bars in the toe slab to serve as tensile reinforcement there.

Discontinue the remaining half bars after a distance of

Ld = 47

= 564 mm

= 600 mm

beyond "B" in the toe slab.

=

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Between "A" & "B" some bars can be curtiled.

Consider a section at depth h below the top of the stem.

Effective depth at that section is

d' = 140 + 29.4117647 h mm (where h in m)

Now Ast (H13/d)

or H1 = (Ast d)1/3

Hence h/H1 = [(Ast' d')/(Ast d)]1/3

where Ast' = reinforcement at depth h

Ast = reinforcement at depth H1

d' = effective depth at depth h

d = effective depth at depth H1

If Ast' = Ast/2, Ast'/Ast = 1/2

Hence h/H1 = [d'/2d]1/3

Substituting d = 250

d' = 140 + 29.4117647 h

h = H1 [d'/2d]1/3

h = H1/(2)1/3

h = 2.96844 m

Bars should be extended by a distance of 12

= 144 mm

or b3 = 250 mm

Hence h = 2.70 m

Curtail bars at this height below the top

Check for shear

Shear force = P

Ka H52/2 = 41.9628 kN

Fu = 62.9442 kN

Shear stress, v = 62.9442 x 1000/(1000 x 250)

= 0.25178 N/mm2

pt Ast x 1001000 x d

pt c = 0.37748

0.25 0.36

0.5 0.48

0.37748478 0.43119

Corresponding c = 0.43119 N/mm2

Hence Safe

Distribition & temperature reinforcement

Average thk of stem = 255 mm

Distribution reinforcement = 0.12 x 1000 x A /100

= 306 mm2

Bar dia= = 8 # bars

Spacing = 164 mm

Provide 8 # bars @ 160 C/C

at inner face of the wall, along its length.

For temperature reinforcement, provide 8 mm # bars @ 300 C/C both ways, in outer face.

=

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9 Design of shear key

Let depth of key = a

Intensity of passive pressure pp developed in front of the key depends on the soil

pp = Kp p = 166.2 kN/m2

Total passive pressure pp = pp a = 166.2 a

Dsliding force at level D1C1 = (1/3) x (/2) x (H+a)2

or pH = (1/3) x (/2) x (H+a)2

Weigth of soil between bottom of base and D1C1 = b a x

= 43.2 a

W = 121.045 + 43.2 a

F.S. against sliding = 1.5

W + pp

pH

1.5 = 1.5127695

a = 0.08 m..Adjust this value till above value = 1.5

However provide minimum

a = 0.3 m

a = H6 = 300 mm

Keep H6 = 300 mm

Keep width, b6 = 300 mm

8# @160c/c

8# @300c/c

2.7m

8# @300c/c

12# @200c/c

12# @100c/c

10# @130c/c

8# @180c/c

12# @200c/c 8# @180c/c

1.5 =

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Design of Retaing wall

Input

Height above GL = 3 m

Density of soil, = 18 kN/m3

Denity of Concrete = 25 kN/m3

= 30

= 18

Grade of Concrete, fck = 20 Mpa

Grade of Steel, fy = 415 Mpa

SBC of foundation strata, q0 = 100 kN/m2

Coefficient of friction, = 0.5

b5

b3

b6

H4

H3

H1

H2

b2 b4

b1

W1

W2

W3

W4

D E B C

A

TOE HEEL

SHEAR

KEY

STEM

H6

H5

H7

H8 H

PH

PV

H/3

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Solution

1 Design constants

Xu, max = 0.48

d

2 Depth of foundation

Density of soil, = 18 kN/m3

= 18000 N/m

3

ymin = (q0/) x ((1-sin)/(1+sin))2

= 0.61728395 m

Keep, H3 = 1 m

Height of wall above its base

Hence, H1 = 4 m

3 Dimensions of base

ka = cos (cos -sqrt(cos2

- cos2

))

(cos (cos +sqrt(cos2

- cos2

))

ka = 0.39480588

"D-E" Ratio =

b

= 1-(q0/(2.7 H1))

= 0.49

b1 = H1

b1

= 2.18384292 m

Base width from sliding condition

b1 = 0.7 H Ka/(1-)

b1 = 4.33512342 mBase width from from normal practice

b1 = 0.6 H or 0.4 H

b1 = 2.4 m

Keep, b1 = 2.4 m

Width of Toe slab, b1 = 1.2 m

Let the thickness of base, H4 = H/12

= 0.3 m

sqrt (Ka cos )

sqrt ((1-)(1+3))

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4 Thickness of stem

Height, H5 "A-B" = 3.7 m

Consider 1m length retaining wall

Earth pressure on stem is

p = Ka H52/2

= 48.6440328 kN acting at 18 to Horiz.

Hence horizontal earth pressure isPH = P cos

= 46.26 kN

M @ "B" = PH H1/3 N-mm

M @ "B" = 57.054 kN-m

= 57054000 N-mm

Mu = 85581000 N-mm

Shear force at "B" = PH

= 46.26 kN

FU = 69.39 kN

Hence effective depth, b3eff = Sqrt(Mu/Ru b)

= 169.522103 mm

Keep, b3eff = 240 mm

Total thickness, b3 = 300 mm

b3eff = 240 mm

Using 16 dia bars

b5 = 200 mm

b5eff = 140 mm

Width of Heel, b4 = 0.9 m

Shear stress, v = 1.5 x 46.26 x 1000/(1000 x 240)

= 0.289125 N/mm2

pt = 0.25 minimum

pt c

0.25 0.36

0.5 0.48

0.25 0.37

Corresponding c = 0.37 N/mm2

Hence Safe

5 Stability of wall

Height, H8 = H5 + b4 tan

Height, H8 = 3.99 m

Height, H = 4.29 m

Earth pressure, P = Ka H2/2

= 65.3944225 kN

Horizontal component

PH = P cos

= 62.1937917 kN

Vertical component

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PV = P sin

= 20.2079879 kN

Designation Force, kNLever arm,

m

Moment

@ toe,

kN-m

W1 18.5 1.4 25.9

W2 4.625 0.97 4.47083W3 18 1.2 21.6

W4 62.289 1.95 121.464

W5 = PV 20.207988 2.4 48.4992

Total W = 123.62199 MR = 221.90

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Total resisting moment MR = 221.9 kN-m

Check for overturning

Overturning moment, M0 = 88.9371221 kN-m

F.S. againt overturning = = 221.90

88.9371221

= 2.5 should be > 2Hence Safe

Check for Sliding

F.S. againt Sliding = W

P

= 1.0 should be > 1.5

Hence Unsafe, Provide shear key

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Pressure distribution

Net moment, M = 221.9 - 88.9371220988394

M = 132.96 kN-m

Distance X of the pont of application of the resultant, from toe

X = M

W

X = = 132.96

123.621988

X = = 1.08 m

Eccentricity, e = (b1/2) - X

e = 0.12 m Should be < b1/6

b/6 = 0.4 m

Hence No tension

Pressure p1 at toe = (W/b) x (1+(6e/b))

= 67.5 kN/m2 Should be less than SBC

= 100 kN/m2

Hence Safe

Pressure p2 at heel = (W/b) x (1-(6e/b))

= 35.4846746 kN/m2

Pressure at the junction of stem with toe slab,

p = 51.5 kN/m2

Pressure at the junction of stem with heel slab,

p = 47.4904216 kN/m2

b3b2

b1

D E B C

A

H7

p1

p2P

Ka H5

Ka H8

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6 Design of toe slab

Downward weight of slab per unit area = 7.5 kN/m2

Hence net pressure intensity under "D" = 67.5 - 7.5 = 60 kN/m2

Hence net pressure intensity under "E" = 51.5 - 7.5 = 44 kN/m2

Total force = Shear force at E = 62 kN

Fu = 93.6 kN

X from "E" = 0.63 mBending moment @ "E" = 39.36 kN-m

= 39360000 N-mm

Mu @ "E" = 59040000 N-mm

H4eff = Sqrt(Mu/Ru b)

= 140.80265 mm

Provide, H4 = 300 mm

H4eff = 240 mm

Shear stress, v = 93.6 x 1000/(1000 x 240)

= 0.39 N/mm2

pt = 0.3 minimum

pt c

0.25 0.36

0.5 0.48

0.3 0.394

Corresponding c = 0.394 N/mm2

Hence Safe

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Steel

Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))

Ast = 727.437498 mm2

This reinforcement to be provided at bottom face.

If alternate bars of stem reinforcement are to be bent & continued in the toe slab,

area available = 1130.4 mm2 See step 8

Bar dia = 12 # bars

Spacing = 100 mm

Provide 12 # bars @ = 100 C/C at top of heel slab

Check for development length, Ld = 47

= 564 mm

Providing 50mm clear side cover ,

actual length available = 1150 mm

Hence safe

Distribution reinforcement = 0.12 x 1000 x A /100

= 324mm

2

Bar dia= = 8 # bars

Spacing = 155 mm

Provide 8 # bars @ = 150 C/C

7 Design of heel slab

1) Total weight of soil = 62.29 kN

Lever arm @ "B" = 0.45 m

2) Total weight of heel slab = 6.75 kN

Lever arm @ "B" = 0.45 m

3) Total force due to vertical

component of earth pressure = Ka (H5+H8) b4 tan sin

2

= 2.47 kNLever arm @ "B" = 0.46 m

4) Total upward soil reaction = 37.34 kN

Lever arm @ "B" = 0.43 m

Total force = S.F. at "B" = 34.17 kN

Fu = 51.26 kN

B.M. @ "B" = 16.200581 kN-m

= 16200581 N-mm

Mu = 24300871.5 N-mm

H7eff = Sqrt(Mu/Ru b)

= 90.3334459 mm

Provide, H6 = 350 mm

H7eff = 290 mm

Steel

Ast = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))

= 236.197916 mm2

Astmin = 360 mm2

Bar dia = 8 # bars

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Spacing = 140 mm

Provide 8 # bars @ 130 C/C at top of heel slab

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Development length, Ld = 47

to the left of "B" = 376 mm

= 400 mm

Distribution reinforcement = 0.12 x 1000 x A /100

= 384 mm2

Bar dia= = 8 # barsSpacing = 131 mm

Provide 8 # bars @ = 130 C/C

Shear stress, v = 1.5 x 34.17 x 1000/(1000 x 290)

= 0.17674138 N/mm2

pt Ast x 100

1000 x d

pt c = 0.12413793

0.15 0.18

0.15 0.18

0.12413793 #DIV/0!

Corresponding c = 0.18 N/mm

2

Hence Safe

8 Reinforcement in stem

Revised H5 = 3.65 m

Shear force at "B" = PH = Ka H52/2

= 48.6440328 kN

bending moment @ "B" = 59.9943071 kN-m

Mu = 89.9914607 kN-m

= 89991460.7 N-mm

b3eff = Sqrt(Mu/Ru b)

= 0.17383543 mm

Provide, b3eff = 260 mm

b3 = 320 mmb5 = 210 mm

b5eff = 150 mm

Ast = = ((0.5 x fck x b x d)/fy)x(1-SQRT(1-((4.6xMu)/(fckxbfxd^2)))

= 1046.53945 mm2

Bar dia= = 12 # bars

Spacing = 108 mm

Provide 12 # bars @ 100 C/C

Astact = 1130.4 mm2

Continue alternate bars in the toe slab to serve as tensile reinforcement there.

Discontinue the remaining half bars after a distance of

Ld = 47

= 564 mm

= 600 mm

beyond "B" in the toe slab.

=

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Between "A" & "B" some bars can be curtiled.

Consider a section at depth h below the top of the stem.

Effective depth at that section is

d' = 150 + 30.1369863 h mm (where h in m)

Now Ast (H3/d)

or H = (Ast d)1/3

Hence h/H5 = [(Ast' d')/(Ast d)]1/3

where Ast' = reinforcement at depth hAst = reinforcement at depth H5

d' = effective depth at depth h

d = effective depth at depth H5

If Ast' = Ast/2, Ast'/Ast = 1/2

Hence h/H5 = [d'/2d]1/3

Substituting d = 260

d' = 150 + 30.1369863 h

h = H5 [d'/2d]1/3

h = H5/(2)1/3

h = 2.89701 m

Bars should be extended by a distance of 12

= 144 mmor b3 = 260 mm

Hence h = 2.60 m

Curtail bars at this height below the top

Check for shear

Shear force = P

Ka H52/2 = 48.6440328 kN

Fu = 72.97 kN

Shear stress, v = 72.97 x 1000/(1000 x 260)

= 0.28065385 N/mm2

pt Ast x 100

1000 x d

pt c = 0.4030.25 0.36

0.5 0.48

0.40 0.44

Corresponding c = 0.44320728 N/mm2

Hence Safe

Distribition & temperature reinforcement

Average thk of stem = 265 mm

Distribution reinforcement = 0.12 x 1000 x A /100

= 318 mm2

Bar dia= = 8 # bars

Spacing = 158 mmProvide 8 # bars @ 150 C/C

at inner face of the wall, along its length.

For temperature reinforcement, provide 8 mm # bars @ 300 C/C both ways, in outer face.

=

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9 Design of shear key

Let depth of key = a

Intensity of passive pressure pp developed in front of the key depends on the soil

pp = Kp p = 130.443852 kN/m2

Total passive pressure pp = pp a = 130.443852 a

Dsliding force at level D1C1 = (1/3) x (/2) x (H+a)2

or pH = (1/3) x (/2) x (H+a)2

Weigth of soil between bottom of base and D1C1 = b a x

= 43.2 a

W = 103.414 + 43.2 a

F.S. against sliding = 1.5

W + pp

pH

1.5 = 1.5009279

a = 0.1765 m..Adjust this value till above value = 1.5

However provide minimum

a = 0.3 m

a = H6 = 300 mm

Keep H6 = 300 mm

Keep width, b6 = 300 mm

8# @150c/c

8# @300c/c

2.6m

8# @300c/c12# @200c/c

12# @100c/c

8# @130c/c

8# @150c/c

12# @100c/c 8# @130c/c

1.5 =

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M-15 M-20 M-25 M-30 M-35 M-40

0.18 0.18 0.19 0.2 0.2 0.2

0.22 0.22 0.23 0.23 0.23 0.23

0.29 0.30 0.31 0.31 0.31 0.32

0.34 0.35 0.36 0.37 0.37 0.380.37 0.39 0.40 0.41 0.42 0.42

0.40 0.42 0.44 0.45 0.45 0.46

0.42 0.45 0.46 0.48 0.49 0.49

0.44 0.47 0.49 0.50 0.52 0.52

0.44 0.49 0.51 0.53 0.54 0.55

0.44 0.51 0.53 0.55 0.56 0.57

0.44 0.51 0.55 0.57 0.58 0.60

0.44 0.51 0.56 0.58 0.60 0.62

0.44 0.51 0.57 0.6 0.62 0.63

Refer IS 456-2000

fy Xu, max

d

250 0.53

415 0.48

500 0.46

2.00

2.25

2.50

2.75

3.00 and above

1.75

Permissible shear stress Table v in concrete (IS : 456-2000)

100As Permissible shear stress in concrete v N/mm2

bd

< 0.15

0.25

0.50

0.751.00

1.25

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