Hollow Retaining Wall.pdf

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RETAINING WALL ANALYSIS In accordance with AS4678-2002 incorporating Amendment No.2 dated August 2008 Tedds calculation version 2.6.04

Retaining wall details Stem type;

Cantilever

Stem height;

hstem = 2600 mm

Stem thickness;

tstem = 450 mm

Angle to rear face of stem;

 = 90 deg

Stem density;

stem = 23.6 kN/m3

Toe length;

ltoe = 950 mm

Heel length;

lheel = 1250 mm

Base thickness;

tbase = 350 mm

Base density;

base = 23.6 kN/m3

Height of retained soil;

hret = 1800 mm

Angle of soil surface;

 = 0 deg

Depth of cover;

dcover = 500 mm

Depth of excavation;

dexc = 200 mm

Retained soil properties Soil type;

Medium dense well graded sand

Soil conditions;

In situ

Moist density;

mr = 21 kN/m3

Saturated density;

sr = 23 kN/m3

Effective internal friction angle;

'r = 30 deg

External wall friction angle;

r = 0 deg

Base soil properties Soil type;

Medium dense well graded sand

Soil conditions;

In situ

Moist density;

mb = 18 kN/m3

Effective cohesion;

c'b = 0 kN/m2

Effective internal friction angle;

'b = 30 deg

External wall friction angle;

b = 15 deg

External base friction angle;

bb = 30 deg

Ultimate design bearing capacity;

Pbearing = 200 kN/m2

Loading details Live surcharge load;

SurchargeQ = 10 kN/m2

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23/03/2017 450

1250

300

15 kN/m2

350

500

200

2650

1800

2600

5.8 kN/m2

32.8 kN/m2

58.8 kN/m2

67.1 kN/m2

2650

Calculate retaining wall geometry Base length;

lbase = ltoe + tstem + lheel = 2650 mm

Moist soil height;

hmoist = hsoil = 2300 mm

Length of surcharge load;

lsur = lheel = 1250 mm

- Distance to vertical component;

xsur_v = lbase - lheel / 2 = 2025 mm

Effective height of wall;

heff = hbase + dcover + hret = 2650 mm

- Distance to horizontal component;

xsur_h = heff / 2 = 1325 mm

Area of wall stem;

Astem = hstem  tstem = 1.17 m2

- Distance to vertical component;

xstem = ltoe + tstem / 2 = 1175 mm

Area of wall base;

Abase = lbase  tbase = 0.927 m2

- Distance to vertical component;

xbase = lbase / 2 = 1325 mm

Area of moist soil;

Amoist = hmoist  lheel = 2.875 m2

- Distance to vertical component;

xmoist_v = lbase - (hmoist  lheel2 / 2) / Amoist = 2025 mm

- Distance to horizontal component;

xmoist_h = heff / 3 = 883 mm

Area of base soil;

Apass = dcover  ltoe = 0.475 m2

- Distance to vertical component;

xpass_v = lbase - (dcover  ltoe (lbase - ltoe / 2)) / Apass = 475 mm

- Distance to horizontal component;

xpass_h = (dcover + hbase) / 3 = 283 mm

Area of excavated base soil;

Aexc = hpass  ltoe = 0.285 m2

- Distance to vertical component;

xexc_v = lbase - (hpass  ltoe (lbase - ltoe / 2)) / Aexc = 475 mm

- Distance to horizontal component;

xexc_h = (hpass + hbase) / 3 = 217 mm

Material strength uncertainty factors for Soil - Table 5.1(A) Uncertainty factor for friction of the retained soil;

ur = 0.85

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Uncertainty factor for friction of the base soil;

ub = 0.85

Uncertainty factor for cohesion of the base soil;

uc = 0.7

Retained soil properties Design effective internal friction angle;

r = atan(ur  tan('r)) = 26.1 deg

Design external wall friction angle;

r = atan(ur  tan(r)) = 0 deg

Base soil properties Design effective internal friction angle;

b = atan(ub  tan('b)) = 26.1 deg

Design cohesion;

cb = uc  c'b = 0 kN/m2

Design external wall friction angle;

b = atan(ub  tan(b)) = 12.8 deg

Design external base friction angle;

bb = atan(ub  tan(bb)) = 26.1 deg

Using Coulomb theory Active pressure coefficient;

KA = sin( + r)2 / (sin()2  sin( - r)  [1 + [sin(r + r)  sin(r ) / (sin( - r)  sin( + ))]]2) = 0.388

Passive pressure coefficient;

KP = sin(90 - b)2  cos(b) / (sin(90 + b)  [1 - [sin(b + b)  sin(b) / (sin(90 + b))]]2) = 3.696

Load combinations for stability limit states - Appendix J3 Load combination 1;

1.25  DeadC + 1.5  LiveC < 0.8  DeadR

Sliding check Vertical forces on wall Wall stem;

Fstem = 0.8  Astem  stem = 22.1 kN/m

Wall base;

Fbase = 0.8  Abase  base = 17.5 kN/m

Moist retained soil;

Fmoist_v = 0.8  Amoist  mr = 48.3 kN/m

Base soil;

Fexc_v = 0.8  Aexc  mb = 4.1 kN/m

Total;

Ftotal_v = Fstem + Fbase + Fmoist_v + Fexc_v = 92 kN/m

Horizontal forces on wall Surcharge load;

Fsur_h = KA  1.5  SurchargeQ  heff = 15.4 kN/m

Moist retained soil;

Fmoist_h = 1.25  KA  mr  heff2 / 2 = 35.8 kN/m

Total;

Ftotal_h = Fmoist_h + Fsur_h = 51.2 kN/m

Check stability against sliding Base soil resistance;

Fexc_h = 0.8  KP  cos(b)  mb  (hpass + hbase)2 / 2 = 10.9 kN/m

Base friction;

Ffriction = Ftotal_v  tan(bb) = 45.2 kN/m

Resistance to sliding;

Frest = Fexc_h + Ffriction = 56 kN/m

Factor of safety;

FoSsl = Frest / Ftotal_h = 1.093 PASS - Resistance to sliding is greater than sliding force

Overturning check Vertical forces on wall Wall stem;

Fstem = 0.8  Astem  stem = 22.1 kN/m

Wall base;

Fbase = 0.8  Abase  base = 17.5 kN/m

Moist retained soil;

Fmoist_v = 0.8  Amoist  mr = 48.3 kN/m

Base soil;

Fexc_v = 0.8  Aexc  mb = 4.1 kN/m

Total;

Ftotal_v = Fstem + Fbase + Fmoist_v + Fexc_v = 92 kN/m

Horizontal forces on wall Surcharge load;

Fsur_h = KA  1.5  SurchargeQ  heff = 15.4 kN/m

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Moist retained soil;

Fmoist_h = 1.25  KA  mr  heff2 / 2 = 35.8 kN/m

Base soil;

Fexc_h = -0.8  KP  cos(b)  mb  (hpass + hbase)2 / 2 = -10.9 kN/m

Total;

Ftotal_h = Fmoist_h + Fexc_h + Fsur_h = 40.4 kN/m

Overturning moments on wall Surcharge load;

Msur_OT = Fsur_h  xsur_h = 20.5 kNm/m

Moist retained soil;

Mmoist_OT = Fmoist_h  xmoist_h = 31.6 kNm/m

Base soil;

Mexc_OT = -Fexc_h  xexc_h = 2.4 kNm/m

Total;

Mtotal_OT = Mmoist_OT + Mexc_OT + Msur_OT = 54.4 kNm/m

Restoring moments on wall Wall stem;

Mstem_R = Fstem  xstem = 26 kNm/m

Wall base;

Mbase_R = Fbase  xbase = 23.2 kNm/m

Moist retained soil;

Mmoist_R = Fmoist_v  xmoist_v = 97.8 kNm/m

Base soil;

Mexc_R = Fexc_v  xexc_v = 1.9 kNm/m

Total;

Mtotal_R = Mstem_R + Mbase_R + Mmoist_R + Mexc_R = 148.9 kNm/m

Check stability against overturning Factor of safety;

FoSot = Mtotal_R / Mtotal_OT = 2.736 PASS - Maximum restoring moment is greater than overturning moment

Bearing pressure check Vertical forces on wall Wall stem;

Fstem = 1.25  Astem  stem = 34.5 kN/m

Wall base;

Fbase = 1.25  Abase  base = 27.4 kN/m

Surcharge load;

Fsur_v = 1.5  SurchargeQ  lheel = 18.8 kN/m

Moist retained soil;

Fmoist_v = 1.25  Amoist  mr = 75.5 kN/m

Base soil;

Fpass_v = 1.25  Apass  mb = 10.7 kN/m

Total;

Ftotal_v = Fstem + Fbase + Fmoist_v + Fpass_v + Fsur_v = 166.8 kN/m

Horizontal forces on wall Surcharge load;

Fsur_h = KA  1.5  SurchargeQ  heff = 15.4 kN/m

Moist retained soil;

Fmoist_h = 1.25  KA  mr  heff2 / 2 = 35.8 kN/m

Base soil;

Fpass_h = -0.8  KP  cos(b)  mb  (dcover + hbase)2 / 2 = -18.6 kN/m

Total;

Ftotal_h = max(Fmoist_h + Fpass_h + Fsur_h - Ftotal_v  tan(bb), 0 kN/m) = 0 kN/m

Moments on wall Wall stem;

Mstem = Fstem  xstem = 40.6 kNm/m

Wall base;

Mbase = Fbase  xbase = 36.3 kNm/m

Surcharge load;

Msur = Fsur_v  xsur_v - Fsur_h  xsur_h = 17.5 kNm/m

Moist retained soil;

Mmoist = Fmoist_v  xmoist_v - Fmoist_h  xmoist_h = 121.2 kNm/m

Base soil;

Mpass = Fpass_v  xpass_v - Fpass_h  xpass_h = 10.3 kNm/m

Total;

Mtotal = Mstem + Mbase + Mmoist + Mpass + Msur = 225.9 kNm/m

Check bearing pressure Distance to reaction;

x = Mtotal / Ftotal_v = 1354 mm

Eccentricity of reaction;

e = x - lbase / 2 = 29 mm

Loaded length of base;

lload = lbase = 2650 mm

Bearing pressure at toe;

qtoe = Ftotal_v / lbase  (1 - 6  e / lbase) = 58.8 kN/m2

Bearing pressure at heel;

qheel = Ftotal_v / lbase  (1 + 6  e / lbase) = 67.1 kN/m2

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PASS - Allowable bearing pressure exceeds maximum applied bearing pressure RETAINING WALL DESIGN In accordance with AS3600-2009 incorporating Amendment No.1 dated November 2010 and AS3700-2011 incorporating Amendment No.1 dated September 2012 Tedds calculation version 2.6.04

Properties of concrete - Section 3.1 Characteristic compressive strength of concrete; f'c = 32 MPa Characteristic flexural tensile strength of concrete - cl.3.1.1.3 f'ct.f = 0.6 MPa  [f'c / 1 MPa] = 3.394 MPa fcv = min(f'c1/3  (1 MPa)2/3, 4 MPa) = 3.175 MPa

Concrete shear strength - cl.8.2.7.1;

Calculate compressive stress factor - exp.8.1.3(1); 2 = max(0.67, min(1.0 - 0.003  f'c / 1 MPa, 0.85)) = 0.85  = max(0.67, min(1.05 - 0.007  f'c / 1 MPa, 0.85)) = 0.826

Calculate stress block factor - exp.8.1.3(2); Reinforcement details Characteristic yield strength of reinforcement; Class of reinforcement;

fsy = 500 MPa N

Exposure classification;

A2

Shear strength of reinforcement - cl.8.8;

fvs = 17.5 MPa

Cover to reinforcement Top face of base;

cbt = 50 mm

Bottom face of base;

cbb = 75 mm

Masonry details Hollow concrete units fully bedded with M3 class mortar, grouted at 400 mm centers Characteristic unconfined compressive strength;

f'uc = 5 MPa

Characteristic lateral modulus of rupture;

f'ut = 0.8 MPa

Thickness of unit;

tb = 450 mm

Length of unit;

lb = 400 mm

Height of unit;

hb = 190 mm

Joint thickness;

tj = 10 mm

Face shell thickness;

twf = 32 mm

End shell thickness;

twe = 32 mm

Internal web thickness;

twi = 32 mm

Depth of cavity;

tc = tb - 2  twf = 386 mm

Length of cavity;

lc = (lb - twi - 2  twe) / 2 = 152 mm

Characteristic compressive strength of grout;

f'c,grout = 20 MPa 400

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Masonry design properties Compressive strength - cl.3.3.2;

f'm = 1.3  1.4  (f'uc  1 MPa) = 4.07 MPa

Tensile strength - cl.3.3.3;

f'mt = 0 MPa

Shear strength - cl.8.8;

f'vm = 0.35 MPa

Elastic moduli for short term loading - Table 3.4;

Em = 1000  f'c = 32000 MPa

Elastic moduli for long term loading - Table 3.4;

EL = 350  f'c = 11200 MPa

Load combinations for strength limit states - Appendix J2 Load combination no.1;

1.25  Dead + 1.5  Live

Load combination no.2;

0.8  Dead + 1.5  Live

Check stem design at base of stem Depth of section;

t = 450 mm

Masonry section properties Total area;

A = t - lc  tc / (lb + tj) = 306898 mm2/m

Bedded area - cl.4.5.4;

Ab = 2  twf = 64000 mm2/m

Combined cross-sectional area - cl.4.5.5;

Ac = Ab + tc  lc / (lb + tj) = 207102 mm2/m

Design cross-sectional area - cl.4.5.6;

Ad = Ac = 207102 mm2/m

Grout area - cl.4.5.7;

Ag = Ad - Ab = 143102 mm2/m

Design section modulus - cl.4.5.8;

Zd = t2 / 6 + tc2  (lc / (lb + tj) - 1) / 6 = 18123590 mm3/m

Compressive strength of grout - cl.3.5;

f'cg = min(f'c,grout, 1.3  f'uc) = 6.5 MPa

Structural design of reinforced masonry - Section 8 Factored bending moment combination 1;

Md = 36.1 kNm/m

Design compressive force;

Fd = 1.25  stem  hstem  A = 23.5 kN/m

Compressive stress on bed joints - cl.7.4.3.3; Reinforcement provided;

fd = min(Fd / Ad, 0.36 MPa) = 0.114 MPa 16 dia.bars @ 400 c/c

Area of reinforcement provided;

Asr.prov =   sr2 / (4  ssr) = 503 mm2/m

Depth of reinforcement;

d = 350 mm

Design area of reinforcement;

Asd = min(0.29  1.3  f'm  d / fsy, Asr.prov) = 503 mm2/m

Design area of tension reinforcement;

Ast = min(Asr.prov, 0.02  (lb + tj - lc) / (lb + tj)  d) = 503 mm2/m

Capacity reduction factor - Table 4.1;

 = 0.75

Basic compresive capacity - exp.8.5;

Fo =   [f'm  Ab + 1.2  (f'cg / 1.3 mm2/N)  Ag + fsy  Asr.prov] = 671.8 kN/m

Vertical slenderness coefficient;

av = 2.5

Clear height of wall;

H = hstem = 2600 mm

Thickness coefficient;

kt = 1

Slenderness ratio - cl.7.3.4.3(4);

Sr = av  H / (kt  t) = 14.444

Slenderness reduction factor;

ks = min(1.18 - 0.03  Sr, 1.0) = 0.747

Compressive strength capacity - cl.8.5;

ks  Fo = 501.6 kN/m Fd / (ks  Fo) = 0.047 PASS - Compressive strength capacity exceeds design compressive force

Bending moment capacity - exp.8.6;

Mc =   fsy  Asd  d  [1 - 0.6  fsy  Asd  (lb + tj) / (1.3  f'm  (lb + tj lc)  d)] = 57.4 kNm/m Md / Mc = 0.628 PASS - Bending moment capacity exceeds design bending moment

Design shear force;

Vd = 40.4 kN/m

Capacity reduction factor - Table 4.1;

 = 0.75

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23/03/2017 Vc =   min(f'vm  (lb + tj - lc) / (lb + tj)  d + fvs  Ast, 4  f'vm  (lb + tj - lc)

Shear capacity - exp.8.8;

/ (lb + tj)  d) = 64.4 kN/m Vd / Vc = 0.627 PASS - Shear strength capacity exceeds design shear force Check base design at toe Depth of section;

h = 350 mm

Strength of section in bending - Section 8.1 Factored bending moment combination 1;

M = 17.2 kNm/m

Depth to tension reinforcement;

d = h - cbb - bb / 2 = 267 mm

Tension reinforcement provided;

16 dia.bars @ 200 c/c

Area of tension reinforcement provided;

Abb.prov =   bb2 / (4  sbb) = 1005 mm2/m

Maximum reinforcement spacing - cl.9.1.1(b);

smax = min(2  h, 300 mm) = 300 mm

Neutral axis parameter;

ku0 = Abb.prov  fsy / (2  f'c  d  ) = 0.084

Ultimate strength in bending;

Mu0 = Abb.prov  fsy  d  (1 -   ku0 / 2) = 129.6 kNm/m

Uncracked section modulus;

Z = h2 / 6 = 20416667 mm3/m

PASS - Spacing of reinforcement is less than maximum ku0 < 0.36 - No compression reinforcement is required

Minimum strength in bending - exp.8.1.6.1(1);

Mu0_min = 1.2  Z  f'ct.f = 83.2 kNm/m

PASS - Ultimate strength in bending exceeds the minimum strength requirements Capacity reduction factor - Table 2.2.2;

 = max(0.6, min(1.19 - 13  ku0 / 12, 0.8)) = 0.8

Design strength in bending - cl.8.1.5;

Mu0 =   Mu0 = 103.7 kNm/m M / Mu0 = 0.166 PASS - Design strength in bending exceeds design bending moment

Strength of section in shear - Section 8.2 Design shear force;

V = 36.8 kN/m

Beta factors - cl.8.2.7.1;

1 = max(1.1  (1.6 - d / 1000 mm), 0.8) = 1.466 2 = 1 3 = 1

Ultimate shear strength - exp.8.2.7.1;

Vuc = 1  2  3  d  fcv  (Abb.prov / d)1/3 = 193.4 kN/m

Capacity reduction factor - Table 2.2.2;

 = 0.7

Design strength in shear;

Vuc =   Vuc = 135.4 kN/m V / Vuc = 0.272 PASS - Design shear strength exceeds design shear force

Strength of section in bending - Section 8.1 Factored bending moment combination 1;

M = 15.5 kNm/m

Depth to tension reinforcement;

d = h - cbt - bt / 2 = 292 mm

Tension reinforcement provided;

16 dia.bars @ 200 c/c

Area of tension reinforcement provided;

Abt.prov =   bt2 / (4  sbt) = 1005 mm2/m

Maximum reinforcement spacing - cl.9.1.1(b);

smax = min(2  h, 300 mm) = 300 mm PASS - Spacing of reinforcement is less than maximum

Neutral axis parameter;

ku0 = Abt.prov  fsy / (2  f'c  d  ) = 0.077 ku0 < 0.36 - No compression reinforcement is required

Ultimate strength in bending;

Mu0 = Abt.prov  fsy  d  (1 -   ku0 / 2) = 142.1 kNm/m

Uncracked section modulus;

Z = h2 / 6 = 20416667 mm3/m

Minimum strength in bending - exp.8.1.6.1(1);

Mu0_min = 1.2  Z  f'ct.f = 83.2 kNm/m

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PASS - Ultimate strength in bending exceeds the minimum strength requirements Capacity reduction factor - Table 2.2.2;

 = max(0.6, min(1.19 - 13  ku0 / 12, 0.8)) = 0.8

Design strength in bending - cl.8.1.5;

Mu0 =   Mu0 = 113.7 kNm/m M / Mu0 = 0.137 PASS - Design strength in bending exceeds design bending moment

Strength of section in shear - Section 8.2 Design shear force;

V = 25.7 kN/m

Beta factors - cl.8.2.7.1;

1 = max(1.1  (1.6 - d / 1000 mm), 0.8) = 1.439 2 = 1 3 = 1

Ultimate shear strength - exp.8.2.7.1;

Vuc = 1  2  3  d  fcv  (Abt.prov / d)1/3 = 201.4 kN/m

Capacity reduction factor - Table 2.2.2;

 = 0.7

Design strength in shear;

Vuc =   Vuc = 141 kN/m V / Vuc = 0.182 PASS - Design shear strength exceeds design shear force

Transverse reinforcement parallel to base Slab restrained in the secondary direction with moderate degree of control over cracking Minimum area of reinforcement - cl.9.4.3.4;

Abx.req = 0.0035  min(tbase / 2, 250 mm) = 613 mm2/m

Transverse reinforcement provided;

16 dia.bars @ 300 c/c

Area of transverse reinforcement provided;

Abx.prov =   bx2 / (4  sbx) = 670 mm2/m

PASS - Area of reinforcement provided is greater than area of reinforcement required 350 16 dia.bars @ 250 c/c horizontal reinforcement parallel to face of stem

16 dia.bars @ 400 c/c 16 dia.bars @ 200 c/c

16 dia.bars @ 200 c/c 16 dia.bars @ 300 c/c transverse reinforcement in base

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