Cantilever Retaining Wall Example

M ADINA FENCE Retaining wall analysis in accordance with International Building Code 2009 T edds calculation version 2.6.07

Retaining wall details Stem type;

Cantilever

Stem height;

h stem = 2430 m m

Stem thickness;

t stem = 300 m m

Angle to rear face of stem ;

 = 90 deg

Stem density;

 stem = 24 kN/m 3

Heel length;

l heel = 1000 m m

Base thickness;

t base = 400 m m

Base density;

 b ase = 24 kN/m 3

Height of retained soil;

h ret = 1150 m m

Angle of soil surface;

 = 0 deg

Depth of cover;

d cover = 1000 m m

Retained soil properties Soil type;

Medium dense well graded sand

Moist density;

 mr = 21 kN/m 3

Saturated density;

 sr = 23 kN/m 3

Effective angle of internal resistance;

r = 30 deg

Effective wall friction angle;

r = 0 deg

Base soil properties Soil type;

Medium dense well graded sand

Soil density;

 b = 18 kN/m 3

Cohesion;

c b = 0 kN/m 2

Effective angle of internal resistance;

b = 30 deg

Effective wall friction angle;

b = 15 deg

Effective base friction angle;

b b = 30 deg

Allowable bearing pressure;

P b earing = 200 kN/m 2

Loading details Vertical line load at 150 m m ;

P D1 = 15 kN/m

300

1000

280

P1

150

400

1000

2550

2430

1150

0 kN/m2

17.8 kN/m 2

133.9 kN/m 2

4.7 kN/m2

1300

General arrangement

Calculate retaining wall geom etry Base length;

l base = t stem + l heel = 1300 m m

Moist soil height;

h moist = h soil = 2150 m m

Retained surface length;

l su r = l h eel = 1000 m m

Effective height of wall;

h eff = h base + d cover + h ret = 2550 m m

Area of wall stem ;

A stem = h stem  t stem = 0.729 m 2

- Distance to vertical com ponent; Area of wall base; - Distance to vertical com ponent; Area of m oist soil;

xstem = l toe + t stem / 2 = 150 m m A b ase = l b ase  t b ase = 0.52 m 2 xb ase = l b ase / 2 = 650 mm A moist = h moist  l heel = 2.15 m 2

- Distance to vertical com ponent;

xmoist_v = l base - (h moist  l h eel2 / 2) / A moist = 800 m m

- Distance to horizontal com ponent;

xmoist_h = h eff / 3 = 850 m m

Using Coulom b theory Active pressure coefficient;

K A = sin( + r) 2 / (sin() 2  sin( -  r)  [1 + [sin(r +  r)  sin(r - ) / (sin( -  r)  sin( + ))]] 2 ) = 0.333

Passive pressure coefficient;

K P = sin(90 - b ) 2 / (sin(90 +  b )  [1 - [sin(b +  b )  sin(b ) / (sin(90 + b ))]] 2) = 4.977

From IBC 2009 cl.1807.2.3 Safety factor Load com bination 1;

1.0  Dead + 1.0  Live + 1.0  Lateral earth

Sliding check Vertical forces on wall W all stem ;

F stem = A stem   stem = 17.5 kN/m

W all base;

F b ase = A base   base = 12.5 kN/m

Line loads;

F P_v = P D1 = 15 kN/m

Moist retained soil;

F moist_v = A moist   mr = 45.1 kN/m

Total;

F total_v = F stem + F base + F moist_v + F P_v = 90.1 kN/m

Horizontal forces on wall Moist retained soil;

F moist_h = K A   mr  h eff2 / 2 = 22.8 kN/m

Total;

F total_h = F moist_h = 22.8 kN/m

Check stability against sliding Base soil resistance;

F exc_h = K P  cos(b )   b  (h p ass + h base) 2 / 2 = 84.8 kN/m

Base friction;

F friction = F total_v  tan( bb ) = 52 kN/m

Resistance to sliding;

F rest = F exc_h + F friction = 136.8 kN/m

Factor of safety;

FoS sl = F rest / F total_h = 6.012; > 1.5 PASS - Factor of safety against sliding is adequate

Overturning check Vertical forces on wall W all stem ;


W all base;


Line loads;

F P_v = P D1 = 15 kN/m



Total;




Base soil;

F exc_h = m ax(-K P  cos(b )   b  (h pass + h b ase ) 2 / 2, -(F moist_h )) = -22.8 kN/m

Total;

F total_h = F moist_h + F exc_h = 0 kN/m

Overturning m om ents on wall Moist retained soil;

M moist_OT = F moist_h  xmoist_h = 19.3 kNm /m

Total;

M total_O T = M moist_O T = 19.3 kNm /m

Restoring m om ents on wall W all stem ;

M stem_R = F stem  x stem = 2.6 kNm /m

W all base;

M base_R = F base  x base = 8.1 kNm /m

Line loads;

M P_R = abs(P D1 )  p 1 = 2.3 kNm /m


M moist_R = F moist_v  x moist_v = 36.1 kNm /m

Base soil;

M exc_R = -F exc_h  xexc_h = 10.6 kNm /m

Total;

M total_R = M stem_R + M b ase_R + M moist_R + M exc_R + M P_R = 59.7 kNm/m

Check stability against overturning Factor of safety;

FoS ot = M total_R / M total_OT = 3.087; > 1.5 PASS - Factor of safety against overturning is adequate

Bearing pressure check Vertical forces on wall W all stem ;


W all base;


Line loads;

F P_v = P D1 = 15 kN/m



Total;




Base soil;

F p ass_h = m ax(-K P  cos(b )   b  (d cover + h base) 2 / 2, -(F moist_h )) = -22.8 kN/m F total_h = m ax(F moist_h + F pass_h - F total_v  tan(b b ), 0 kN/m ) = 0 kN/m

Total; M om ents on wall W all stem ;

M stem = F stem  xstem = 2.6 kNm /m

W all base;

M base = F base  xb ase = 8.1 kNm /m

Line loads;

M P = P D1  p 1 = 2.3 kNm /m


M moist = F moist_v  xmoist_v - F moist_h  xmoist_h = 16.8 kNm /m

Base soil;

M pass = -F p ass_h  xp ass_h = 10.6 kNm /m

Total;

M total = M stem + M base + M moist + M pass + M P = 40.4 kNm /m

Check bearing pressure Distance to reaction;

x = M total / F total_v = 448 m m

Eccentricity of reaction;

e = x - l base / 2 = -202 m m

Loaded length of base;

l load = l base = 1300 m m

Bearing pressure at toe;

q toe = F total_v / l b ase  (1 - 6  e / l b ase) = 133.9 kN/m 2

Bearing pressure at heel;

q h eel = F total_v / l base  (1 + 6  e / l base ) = 4.7 kN/m 2

Factor of safety;

FoS bp = P b earing / m ax(q toe , q heel) = 1.493; PASS - Allowable bearing pressure exceeds maximum applied bearing pressure

M ADINA FENCE Retaining wall design in accordance with ACI 318-08 T edds calculation version 2.6.07

Concrete details Com pressive strength of concrete;

f'c = 27.5 N/m m 2

Concrete type;

Norm al weight

Reinforcem ent details Yield strength of reinforcem ent;

f y = 415 N/m m 2

Modulus of elasticity or reinforcement;

E s = 199948 N/m m 2

Cover to reinforcem ent Front face of stem ;

c sf = 40 mm

Rear face of stem ;

c sr = 50 mm

Top face of base;

c b t = 50 m m

Bottom face of base;

c b b = 75 m m

From IBC 2009 cl.1605.2.1 Basic load com binations Load com bination no.1;

1.4  Dead

Load com bination no.2;

1.2  Dead + 1.6  Live + 1.6  Lateral earth


1.2  Dead + 1.0  Earthquake + 1.0  Live


0.9  Dead + 1.0  Earthquake + 1.6  Lateral earth

Loading details - Combination No.1 - kN/m2

Shear force - Combination No.1 - kN/m

1

0

1

Stem

0

Bending moment - Combination No.1 - kNm/m

0.92 4.33

Heel

Z

0

2

3.42

X

8.39


1

Stem


-19.5

0.79 3.71

0.79 3.71

9.1 1.65 X

-5.8

2

13.7

2



1

Stem


1

11.72

Heel

Z

-17.6

0.59 2.78

0.59 2.78

9.4 1.65 X

8.79

2

-2.1


Z

0

1

0.92 4.33

-1.8 1.1

-5.8 Heel

2

13.7

2

Check stem design at base of stem h = 300 m m

Depth of section; Rectangular section in flexure - Chapter 10 Design bending m om ent com bination 4;

M = 18.6 kNm/m

Depth of tension reinforcem ent;

d = h - c sr - sr / 2 = 244 m m

Com pression reinforcem ent provided;

No.3 bars @ 5.906" c/c

Area of com pression reinforcem ent provided;

A sf.p rov =   sf 2 / (4  s sf) = 475 m m 2/m

Tension reinforcem ent provided;

No.4 bars @ 200 m m c/c

Area of tension reinforcem ent provided;

A sr.p rov =   sr 2 / (4  s sr) = 633 m m 2/m

Maxim um reinforcem ent spacing - cl.14.3.5;

s max = m in(18 in, 3  h) = 457 m m PASS - Reinforcement is adequately spaced

Depth of com pression block;

a = A sr.prov  f y / (0.85  f'c ) = 11 m m

Neutral axis factor - cl.10.2.7.3;

 1 = m in(m ax(0.85 - 0.05  (f'c - 28 N/m m 2) / 7 N/m m 2, 0.65), 0.85) = 0.85

Depth to neutral axis;

c = a /  1 = 13 m m

Strain in reinforcement;

t = 0.003  (d - c) / c = 0.052251

Strength reduction factor;

f = m in(m ax(0.65 + (t - 0.002)  (250 / 3), 0.65), 0.9) = 0.9

Nom inal flexural strength;

M n = A sr.prov  f y  (d - a / 2) = 62.6 kNm /m

Design flexural strength;

M n = f  M n = 56.3 kNm /m

Section is in the tension controlled zone

M / M n = 0.329 PASS - Design flexural strength exceeds factored bending moment By iteration, reinforcem ent required by analysis;

A sr.d es = 205 m m 2 /m

Minim um area of reinforcem ent - cl.10.5.3;

A sr.mod = 4  A sr.des / 3 = 274 m m 2/m

PASS - Area of reinforcement provided is greater than minimum area of reinforcement required Rectangular section in shear - Chapter 11 Design shear force;

V = 25.9 kN/m

Concrete m odification factor - cl.8.6.1;

=1

Nom inal concrete shear strength - exp.11-3;

V c = 0.17    (f' c  1 N/m m 2)  d = 217.2 kN/m


s = 0.75

Design concrete shear strength - cl.11.4.6.1;

V c = s  V c = 162.9 kN/m

V / V c = 0.159 PASS - No shear reinforcement is required Horizontal reinforcem ent parallel to face of stem Minim um area of reinforcem ent - cl.14.3.3;

A sx.req = 0.002  t stem = 600 m m 2 /m

Transverse reinforcem ent provided;

No.4 bars @ 200 m m c/c each face

Area of transverse reinforcem ent provided;

A sx.p rov = 2    sx2 / (4  s sx) = 1267 mm 2 /m

PASS - Area of reinforcement provided is greater than area of reinforcement required Check base design at heel Depth of section;

h = 400 m m

Rectangular section in flexure - Chapter 10 Design bending m om ent com bination 2;

M = 26.4 kNm/m

Depth of tension reinforcem ent;

d = h - c bt - bt / 2 = 344 m m

Com pression reinforcem ent provided;

No.4 bars @ 5.906" c/c

Area of com pression reinforcem ent provided;

A b b.p rov =   b b 2 / (4  s bb ) = 845 mm 2 /m

Tension reinforcem ent provided;

No.4 bars @ 150 m m c/c

Area of tension reinforcem ent provided;

A b t.p rov =   bt 2 / (4  s bt) = 845 m m 2/m

Maxim um reinforcem ent spacing - cl.10.5.4;

s max = m in(18 in, 3  h) = 457 m m

Depth of com pression block;

a = A b t.prov  f y / (0.85  f'c ) = 15 m m

Neutral axis factor - cl.10.2.7.3;

 1 = m in(m ax(0.85 - 0.05  (f'c - 28 N/m m 2) / 7 N/m m 2, 0.65), 0.85) =

PASS - Reinforcement is adequately spaced

0.85 Depth to neutral axis;

c = a /  1 = 18 m m

Strain in reinforcement;

t = 0.003  (d - c) / c = 0.055446


f = m in(m ax(0.65 + (t - 0.002)  (250 / 3), 0.65), 0.9) = 0.9

Nom inal flexural strength;

M n = A bt.prov  f y  (d - a / 2) = 117.8 kNm /m

Design flexural strength;

M n = f  M n = 106 kNm/m

Section is in the tension controlled zone

M / M n = 0.249 PASS - Design flexural strength exceeds factored bending moment By iteration, reinforcem ent required by analysis;

A b t.d es = 207 m m 2 /m

Minim um area of reinforcem ent - cl.7.12.2.1;

A b t.min = 0.0018  h = 720 m m 2/m

PASS - Area of reinforcement provided is greater than minimum area of reinforcement required Rectangular section in shear - Chapter 11 Design shear force;

V = 41.7 kN/m

Concrete m odification factor - cl.8.6.1;

=1

Nom inal concrete shear strength - exp.11-3;

V c = 0.17    (f' c  1 N/m m 2)  d = 306.4 kN/m


s = 0.75

Design concrete shear strength - cl.11.4.6.1;

V c = s  V c = 229.8 kN/m

V / V c = 0.182 PASS - No shear reinforcement is required Transverse reinforcem ent parallel to base Minim um area of reinforcem ent - cl.7.12.2.1;

A b x.req = 0.0018  t b ase = 720 m m 2/m

Transverse reinforcem ent provided;

No.4 bars @ 200 m m c/c each face

Area of transverse reinforcem ent provided;

A b x.p rov = 2    bx2 / (4  s bx) = 1267 m m 2/m

PASS - Area of reinforcement provided is greater than area of reinforcement required

40

50 No.4 bars @ 200 c/c horizontal reinforcement parallel to face of stem

No.3 bars @ 150 c/c

No.4 bars @ 200 c/c No.4 bars @ 150 c/c

No.4 bars @ 150 c/c No.4 bars @ 200 c/c transverse reinforcement in base

Reinforcement details

50

75

Cantilever Retaining Wall Example

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