Sachpazis: Sloped rear face retaining wall example

GEODOMISI Ltd. - Dr. Costas Sachpazis Civil & Geotechnical Engineering Consulting Company for

Structural Engineering, Soil Mechanics, Rock Mechanics,

Foundation Engineering & Retaining Structures. Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 -

Mobile: (+30) 6936425722 & (+44) 7585939944, [email protected]

Project: Sloped rear face retaining wall Analysis & Design, In accordance with EN1997-1:2004 incorporating Corrigendum dated February 2009 and the recommended values

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Civil & Geotechnical Engineering Sheet no./rev. 1

Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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RETAINING WALL ANALYSIS

In accordance with EN1997-1:2004 incorporating Corrigendum dated February

2009 and the recommended values

Retaining wall details

Stem type; Cantilever with inclined rear face

Stem height; hstem = 4700 mm

Stem thickness; tstem = 300 mm

Slope length to rear of stem; lslr = 400 mm

Angle to rear face of stem; α = atan(hstem / lslr) = 85.1 deg

Stem density; γstem = 25 kN/m3

Toe length; ltoe = 3000 mm

Heel length; lheel = 2500 mm

Base thickness; tbase = 550 mm

Key position; pkey = 5700 mm

Key depth; dkey = 500 mm

Key thickness; tkey = 500 mm

Base density; γbase = 25 kN/m3

Height of retained soil; hret = 4000 mm

Angle of soil surface; β = 10 deg

Depth of cover; dcover = 700 mm

Depth of excavation; dexc = 700 mm

Height of water; hwater = 500 mm






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Dr. C. Sachpazis

Date

22/12/2013

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Water density; γw = 9.8 kN/m3

Retained soil properties

Soil type; Organic clay

Moist density; γmr = 15 kN/m3

Saturated density; γsr = 16 kN/m3

Characteristic effective shear resistance angle; φ'r.k = 29 deg

Characteristic wall friction angle; δr.k = 16 deg

Base soil properties

Soil type; Organic clay

Moist density; γmb = 17 kN/m3

Characteristic cohesion; c'b.k = 55 kN/m2

Characteristic adhesion; ab.k = 52 kN/m2

Characteristic effective shear resistance angle; φ'b.k = 26 deg

Characteristic wall friction angle; δb.k = 15 deg

Characteristic base friction angle; δbb.k = 26 deg

Loading details

Variable surcharge load; SurchargeQ = 10 kN/m2






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Calculate retaining wall geometry

Base length; lbase = ltoe + tstem + lslr + lheel = 6200 mm

Base height; hbase = tbase + dkey = 1050 mm

Saturated soil height; hsat = hwater + dcover = 1200 mm

Moist soil height; hmoist = hret - hwater = 3500 mm

Length of surcharge load; lsur = (lheel + lslr × hsoil / hstem) = 2900 mm

- Distance to vertical component; xsur_v = lbase - (lheel + lslr × hsoil / hstem) / 2 = 4750 mm

Effective height of wall; heff = hbase + dcover + hret + lsur × tan(β) = 6261 mm

- Distance to horizontal component; xsur_h = heff / 2 - dkey = 2631 mm

Area of wall stem; Astem = hstem × (tstem + lslr / 2) = 2.35 m2

- Distance to vertical component; xstem = (hstem × tstem × (ltoe + tstem / 2) + hstem × lslr / 2 × (ltoe + tstem + lslr / 3)) / Astem = 3263 mm

Area of wall base; Abase = lbase × tbase + dkey × tkey = 3.66 m2

- Distance to vertical component; xbase = (lbase2 × tbase / 2 + dkey × tkey × (pkey + tkey / 2)) / Abase = 3295 mm

Area of saturated soil; Asat = hsat × (lheel + lslr × hsat / (2 × hstem)) = 3.061 m2

- Distance to vertical component; xsat_v = lbase - (hsat × lheel2 / 2 + lslr × hsat

2 / (2 × hstem) × (lheel + lslr × hsat / (3 × hstem))) / Asat = 4924 mm

- Distance to horizontal component; xsat_h = (hsat + hbase) / 3 - dkey = 250 mm

Area of water; Awater = hsat × (lheel + lslr × hsat / (2 × hstem)) = 3.061 m2

- Distance to vertical component; xwater_v = lbase - (hsat × lheel2 / 2 + lslr × hsat

2 / (2 × hstem) × (lheel + lslr × hsat / (3 × hstem))) / Asat = 4924 mm

- Distance to horizontal component; xwater_h = (hsat + hbase) / 3 - dkey = 250 mm

Area of moist soil; Amoist = (hret - hwater) × (lheel + lslr × (hmoist + 2 × hsat) / (2 × hstem)) + tan(β) × (lheel + lslr × hsoil / hstem)2 / 2 =

10.37 m2

- Distance to vertical component; xmoist_v = lbase - (hmoist × (lheel + lslr × hsat / hstem)2 / 2 + lslr × hmoist

2 / (2 × hstem) × ((lheel + lslr × hsat / hstem) + lslr ×

hmoist / (3 × hstem)) + tan(β) × (lheel + lslr × hsoil / hstem)3 / 6) / Amoist = 4852 mm

- Distance to horizontal component; xmoist_h = ((heff - hsat - hbase) × (tbase + hsat + (heff - hsat - hbase) / 3) / 2 + (hsat + hbase) × ((hsat + hbase)/2 - dkey)) /

(hsat + hbase + (heff - hsat - hbase) / 2) = 1785 mm






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Area of base soil; Apass = dcover × ltoe = 2.1 m2

- Distance to vertical component; xpass_v = lbase - (dcover × ltoe× (lbase - ltoe / 2)) / Apass = 1500 mm

- Distance to horizontal component; xpass_h = (dcover + hbase) / 3- dkey = 83 mm

Partial factors on actions - Table A.3 - Combination 1

Permanent unfavourable action; γG = 1.35

Permanent favourable action; γGf = 1.00

Variable unfavourable action; γQ = 1.50

Variable favourable action; γQf = 0.00

Partial factors for soil parameters – Table A.4 - Combination 1

Angle of shearing resistance; γφ' = 1.00

Effective cohesion; γc' = 1.00

Weight density; γγ = 1.00


Design effective shear resistance angle; φ'r.d = atan(tan(φ'r.k) / γφ') = 29 deg

Design wall friction angle; δr.d = atan(tan(δr.k) / γφ') = 16 deg


Design effective shear resistance angle; φ'b.d = atan(tan(φ'b.k) / γφ') = 26 deg

Design wall friction angle; δb.d = atan(tan(δb.k) / γφ') = 15 deg

Design base friction angle; δbb.d = atan(tan(δbb.k) / γφ') = 26 deg

Design effective cohesion; c'b.d = c'b.k / γc' = 55 kN/m2

Design adhesion; ab.d = ab.k / γc' = 52 kN/m2

Using Coulomb theory

Active pressure coefficient; KA = sin(α + φ'r.d)2 / (sin(α)2 × sin(α - δr.d) × [1 + √[sin(φ'r.d + δr.d) × sin(φ'r.d - β) / (sin(α - δr.d) × sin(α + β))]]

2)

= 0.400






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Passive pressure coefficient; KP = sin(90 - φ'b.d)2 / (sin(90 + δb.d) × [1 - √[sin(φ'b.d + δb.d) × sin(φ'b.d) / (sin(90 + δb.d))]]

2) = 4.051

Sliding check

Vertical forces on wall

Wall stem; Fstem = γGf × Astem × γstem = 58.8 kN/m

Wall base; Fbase = γGf × Abase × γbase = 91.5 kN/m

Saturated retained soil; Fsat_v = γGf × Asat × (γsr - γw) = 18.9 kN/m

Water; Fwater_v = γGf × Awater × γw = 30 kN/m

Moist retained soil; Fmoist_v = γGf × Amoist × γmr = 155.6 kN/m

Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fwater_v = 354.8 kN/m

Horizontal forces on wall

Surcharge load; Fsur_h = KA × cos(δr.d + (90 - α)) × γQ × SurchargeQ × heff = 35.1 kN/m

Saturated retained soil; Fsat_h = γG × KA × cos(δr.d + (90 - α)) × (γsr - γw) × (hsat + hbase)2 / 2 = 7.9 kN/m

Water; Fwater_h = γG × γw × (hwater + dcover + hbase)2 / 2 = 33.5 kN/m

Moist retained soil; Fmoist_h = γG × KA × cos(δr.d + (90 - α)) × γmr × ((heff - hsat - hbase)2 / 2 + (heff - hsat - hbase) × (hsat + hbase)) =

129.3 kN/m

Total; Ftotal_h = Fsat_h + Fmoist_h + Fwater_h + Fsur_h = 205.9 kN/m

Check stability against sliding

Base soil resistance; Fexc_h = γGf × KP × cos(δb.d) × γmb × (hpass + hbase)2 / 2 = 36.7 kN/m

Base friction; Ffriction = ab.d × b + Ftotal_v × tan(δbb.d) = 225 kN/m

Resistance to sliding; Frest = Fexc_h + Ffriction = 261.7 kN/m

Factor of safety; FoSsl = Frest / Ftotal_h = 1.271

PASS - Resistance to sliding is greater than sliding force






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Overturning check

Overturning moments on wall

Surcharge load; Msur_OT = Fsur_h × xsur_h = 92.4 kNm/m

Saturated retained soil; Msat_OT = Fsat_h × xsat_h = 2 kNm/m

Water; Mwater_OT = Fwater_h × xwater_h = 8.4 kNm/m

Moist retained soil; Mmoist_OT = Fmoist_h × xmoist_h = 230.9 kNm/m

Base soil; Mexc_OT = -Fexc_h × xexc_h = 5.5 kNm/m

Total; Mtotal_OT = Msat_OT + Mmoist_OT + Mexc_OT + Mwater_OT + Msur_OT = 339.2 kNm/m

Restoring moments on wall

Wall stem; Mstem_R = Fstem × xstem = 191.7 kNm/m

Wall base; Mbase_R = Fbase × xbase = 301.5 kNm/m

Saturated retained soil; Msat_R = Fsat_v × xsat_v = 93.3 kNm/m

Water; Mwater_R = Fwater_v × xwater_v = 147.9 kNm/m

Moist retained soil; Mmoist_R = Fmoist_v × xmoist_v = 754.8 kNm/m

Total; Mtotal_R = Mstem_R + Mbase_R + Msat_R + Mmoist_R + Mwater_R = 1489.2 kNm/m

Check stability against overturning

Factor of safety; FoSot = Mtotal_R / Mtotal_OT = 4.39

PASS - Maximum restoring moment is greater than overturning moment

Bearing pressure check


Wall stem; Fstem = γG × Astem × γstem = 79.3 kN/m

Wall base; Fbase = γG × Abase × γbase = 123.5 kN/m

Surcharge load; Fsur_v = γQ × SurchargeQ × (lheel + lslr × hsoil / hstem) = 43.5 kN/m

Saturated retained soil; Fsat_v = γG × Asat × (γsr - γw) = 25.6 kN/m






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Water; Fwater_v = γG × Awater × γw = 40.5 kN/m

Moist retained soil; Fmoist_v = γG × Amoist × γmr = 210 kN/m

Base soil; Fpass_v = γG × Apass × γmb = 48.2 kN/m

Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fpass_v + Fwater_v + Fsur_v = 570.7 kN/m






129.3 kN/m

Base soil; Fpass_h = -γGf × KP × cos(δb.d) × γmb × (dcover + hbase)2 / 2 = -101.9 kN/m

Total; Ftotal_h = max(Fsat_h + Fmoist_h + Fpass_h + Fwater_h + Fsur_h - (ab.d × b + Ftotal_v × tan(δbb.d)), 0 kN/m) = 0 kN/m

Moments on wall

Wall stem; Mstem = Fstem × xstem = 258.8 kNm/m

Wall base; Mbase = Fbase × xbase = 407 kNm/m

Surcharge load; Msur = Fsur_v × xsur_v - Fsur_h × xsur_h = 114.2 kNm/m

Saturated retained soil; Msat = Fsat_v × xsat_v - Fsat_h × xsat_h = 124 kNm/m

Water; Mwater = Fwater_v × xwater_v - Fwater_h × xwater_h = 191.3 kNm/m

Moist retained soil; Mmoist = Fmoist_v × xmoist_v - Fmoist_h × xmoist_h = 788.1 kNm/m

Base soil; Mpass = Fpass_v × xpass_v - Fpass_h × xpass_h = 80.8 kNm/m

Total; Mtotal = Mstem + Mbase + Msat + Mmoist + Mpass + Mwater + Msur = 1964.1 kNm/m

Check bearing pressure

Distance to reaction; x = Mtotal / Ftotal_v = 3442 mm

Eccentricity of reaction; e = x - lbase / 2 = 342 mm

Loaded length of base; lload = 2 × (lbase - x) = 5516 mm






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Bearing pressure at toe; qtoe = 0 kN/m2

Bearing pressure at heel; qheel = Ftotal_v / lload = 103.4 kN/m2

Effective overburden pressure; q = (tbase + dcover) × γmb - (tbase + dcover + hwater) × γw = 4.1 kN/m2

Design effective overburden pressure; q' = q / γγ = 4.1 kN/m2

Bearing resistance factors; Nq = Exp(π × tan(φ'b.d)) × (tan(45 deg + φ'b.d / 2))2 = 11.854

Nc = (Nq - 1) × cot(φ'b.d) = 22.254

Nγ = 2 × (Nq - 1) × tan(φ'b.d) = 10.588

Foundation shape factors; sq = 1

sγ = 1

sc = 1

Load inclination factors; H = Ftotal_h = 0 kN/m

V = Ftotal_v = 570.7 kN/m

m = 2

iq = [1 - H / (V + lload × c'b.d × cot(φ'b.d))]m = 1

iγ = [1 - H / (V + lload × c'b.d × cot(φ'b.d))](m + 1)

= 1

ic = iq - (1 - iq) / (Nc × tan(φ'b.d)) = 1

Net ultimate bearing capacity; nf = c'b.d × Nc × sc × ic + q' × Nq × sq × iq + 0.5 × (γmb - γw) × lload × Nγ × sγ × iγ = 1482.4 kN/m2

Factor of safety; FoSbp = nf / max(qtoe, qheel) = 14.329

PASS - Allowable bearing pressure exceeds maximum applied bearing pressure

Partial factors on actions - Table A.3 - Combination 2

Permanent unfavourable action; γG = 1.00

Permanent favourable action; γGf = 1.00

Variable unfavourable action; γQ = 1.30

Variable favourable action; γQf = 0.00






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Calc. by

Dr. C. Sachpazis

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Partial factors for soil parameters – Table A.4 - Combination 2

Angle of shearing resistance; γφ' = 1.25

Effective cohesion; γc' = 1.25

Weight density; γγ = 1.00


Design effective shear resistance angle; φ'r.d = atan(tan(φ'r.k) / γφ') = 23.9 deg

Design wall friction angle; δr.d = atan(tan(δr.k) / γφ') = 12.9 deg


Design effective shear resistance angle; φ'b.d = atan(tan(φ'b.k) / γφ') = 21.3 deg

Design wall friction angle; δb.d = atan(tan(δb.k) / γφ') = 12.1 deg

Design base friction angle; δbb.d = atan(tan(δbb.k) / γφ') = 21.3 deg

Design effective cohesion; c'b.d = c'b.k / γc' = 44 kN/m2

Design adhesion; ab.d = ab.k / γc' = 41.6 kN/m2

Using Coulomb theory

Active pressure coefficient; KA = sin(α + φ'r.d)2 / (sin(α)2 × sin(α - δr.d) × [1 + √[sin(φ'r.d + δr.d) × sin(φ'r.d - β) / (sin(α - δr.d) × sin(α + β))]]

2)

= 0.489

Passive pressure coefficient; KP = sin(90 - φ'b.d)2 / (sin(90 + δb.d) × [1 - √[sin(φ'b.d + δb.d) × sin(φ'b.d) / (sin(90 + δb.d))]]

2) = 2.961

Sliding check


Wall stem; Fstem = γGf × Astem × γstem = 58.8 kN/m

Wall base; Fbase = γGf × Abase × γbase = 91.5 kN/m

Saturated retained soil; Fsat_v = γGf × Asat × (γsr - γw) = 18.9 kN/m

Water; Fwater_v = γGf × Awater × γw = 30 kN/m

Moist retained soil; Fmoist_v = γGf × Amoist × γmr = 155.6 kN/m






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fwater_v = 354.8 kN/m






119.3 kN/m

Total; Ftotal_h = Fsat_h + Fmoist_h + Fwater_h + Fsur_h = 189.3 kN/m

Check stability against sliding

Base soil resistance; Fexc_h = γGf × KP × cos(δb.d) × γmb × (hpass + hbase)2 / 2 = 27.1 kN/m

Base friction; Ffriction = ab.d × b + Ftotal_v × tan(δbb.d) = 180 kN/m

Resistance to sliding; Frest = Fexc_h + Ffriction = 207.2 kN/m

Factor of safety; FoSsl = Frest / Ftotal_h = 1.094

PASS - Resistance to sliding is greater than sliding force

Overturning check

Overturning moments on wall

Surcharge load; Msur_OT = Fsur_h × xsur_h = 99.8 kNm/m

Saturated retained soil; Msat_OT = Fsat_h × xsat_h = 1.8 kNm/m

Water; Mwater_OT = Fwater_h × xwater_h = 6.2 kNm/m

Moist retained soil; Mmoist_OT = Fmoist_h × xmoist_h = 213 kNm/m

Base soil; Mexc_OT = -Fexc_h × xexc_h = 4.1 kNm/m

Total; Mtotal_OT = Msat_OT + Mmoist_OT + Mexc_OT + Mwater_OT + Msur_OT = 324.8 kNm/m

Restoring moments on wall

Wall stem; Mstem_R = Fstem × xstem = 191.7 kNm/m






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Wall base; Mbase_R = Fbase × xbase = 301.5 kNm/m

Saturated retained soil; Msat_R = Fsat_v × xsat_v = 93.3 kNm/m

Water; Mwater_R = Fwater_v × xwater_v = 147.9 kNm/m

Moist retained soil; Mmoist_R = Fmoist_v × xmoist_v = 754.8 kNm/m

Total; Mtotal_R = Mstem_R + Mbase_R + Msat_R + Mmoist_R + Mwater_R = 1489.2 kNm/m

Check stability against overturning

Factor of safety; FoSot = Mtotal_R / Mtotal_OT = 4.584

PASS - Maximum restoring moment is greater than overturning moment

Bearing pressure check


Wall stem; Fstem = γG × Astem × γstem = 58.8 kN/m

Wall base; Fbase = γG × Abase × γbase = 91.5 kN/m

Surcharge load; Fsur_v = γQ × SurchargeQ × (lheel + lslr × hsoil / hstem) = 37.7 kN/m

Saturated retained soil; Fsat_v = γG × Asat × (γsr - γw) = 18.9 kN/m

Water; Fwater_v = γG × Awater × γw = 30 kN/m

Moist retained soil; Fmoist_v = γG × Amoist × γmr = 155.6 kN/m

Base soil; Fpass_v = γG × Apass × γmb = 35.7 kN/m

Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fpass_v + Fwater_v + Fsur_v = 428.2 kN/m






119.3 kN/m






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Base soil; Fpass_h = -γGf × KP × cos(δb.d) × γmb × (dcover + hbase)2 / 2 = -75.4 kN/m

Total; Ftotal_h = max(Fsat_h + Fmoist_h + Fpass_h + Fwater_h + Fsur_h - (ab.d × b + Ftotal_v × tan(δbb.d)), 0 kN/m) = 0 kN/m

Moments on wall

Wall stem; Mstem = Fstem × xstem = 191.7 kNm/m

Wall base; Mbase = Fbase × xbase = 301.5 kNm/m

Surcharge load; Msur = Fsur_v × xsur_v - Fsur_h × xsur_h = 79.3 kNm/m

Saturated retained soil; Msat = Fsat_v × xsat_v - Fsat_h × xsat_h = 91.5 kNm/m

Water; Mwater = Fwater_v × xwater_v - Fwater_h × xwater_h = 141.7 kNm/m

Moist retained soil; Mmoist = Fmoist_v × xmoist_v - Fmoist_h × xmoist_h = 541.8 kNm/m

Base soil; Mpass = Fpass_v × xpass_v - Fpass_h × xpass_h = 59.8 kNm/m

Total; Mtotal = Mstem + Mbase + Msat + Mmoist + Mpass + Mwater + Msur = 1407.3 kNm/m

Check bearing pressure

Distance to reaction; x = Mtotal / Ftotal_v = 3287 mm

Eccentricity of reaction; e = x - lbase / 2 = 187 mm

Loaded length of base; lload = 2 × (lbase - x) = 5826 mm

Bearing pressure at toe; qtoe = 0 kN/m2

Bearing pressure at heel; qheel = Ftotal_v / lload = 73.5 kN/m2

Effective overburden pressure; q = (tbase + dcover) × γmb - (tbase + dcover + hwater) × γw = 4.1 kN/m2

Design effective overburden pressure; q' = q / γγ = 4.1 kN/m2

Bearing resistance factors; Nq = Exp(π × tan(φ'b.d)) × (tan(45 deg + φ'b.d / 2))2 = 7.298

Nc = (Nq - 1) × cot(φ'b.d) = 16.141

Nγ = 2 × (Nq - 1) × tan(φ'b.d) = 4.915

Foundation shape factors; sq = 1

sγ = 1

sc = 1






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Calc. by

Dr. C. Sachpazis

Date

22/12/2013

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Load inclination factors; H = Ftotal_h = 0 kN/m

V = Ftotal_v = 428.2 kN/m

m = 2

iq = [1 - H / (V + lload × c'b.d × cot(φ'b.d))]m = 1

iγ = [1 - H / (V + lload × c'b.d × cot(φ'b.d))](m + 1)

= 1

ic = iq - (1 - iq) / (Nc × tan(φ'b.d)) = 1

Net ultimate bearing capacity; nf = c'b.d × Nc × sc × ic + q' × Nq × sq × iq + 0.5 × (γmb - γw) × lload × Nγ × sγ × iγ = 843 kN/m2

Factor of safety; FoSbp = nf / max(qtoe, qheel) = 11.471

PASS - Allowable bearing pressure exceeds maximum applied bearing pressure

RETAINING WALL DESIGN

In accordance with EN1992-1-1:2004 incorporating Corrigendum dated

January 2008 and the recommended values

Concrete details - Table 3.1 - Strength and deformation characteristics for concrete

Concrete strength class; C32/40

Characteristic compressive cylinder strength; fck = 32 N/mm2

Characteristic compressive cube strength; fck,cube = 40 N/mm2

Mean value of compressive cylinder strength; fcm = fck + 8 N/mm2 = 40 N/mm

2

Mean value of axial tensile strength; fctm = 0.3 N/mm2 × (fck / 1 N/mm

2)2/3 = 3.0 N/mm

2

5% fractile of axial tensile strength; fctk,0.05 = 0.7 × fctm = 2.1 N/mm2

Secant modulus of elasticity of concrete; Ecm = 22 kN/mm2 × (fcm / 10 N/mm

2)0.3 = 33346 N/mm

2

Partial factor for concrete - Table 2.1N; γC = 1.50






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Dr. C. Sachpazis

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Compressive strength coefficient - cl.3.1.6(1); αcc = 1.00

Design compressive concrete strength - exp.3.15; fcd = αcc × fck / γC = 21.3 N/mm2

Maximum aggregate size; hagg = 20 mm

Reinforcement details

Characteristic yield strength of reinforcement; fyk = 500 N/mm2

Modulus of elasticity of reinforcement; Es = 200000 N/mm2

Partial factor for reinforcing steel - Table 2.1N; γS = 1.15

Design yield strength of reinforcement; fyd = fyk / γS = 435 N/mm2

Cover to reinforcement

Front face of stem; csf = 40 mm

Rear face of stem; csr = 50 mm

Top face of base; cbt = 50 mm

Bottom face of base; cbb = 75 mm

Check stem design at base of stem

Depth of section; h = 700 mm

Rectangular section in flexure - Section 6.1

Design bending moment combination 1; M = 243.2 kNm/m

Depth to tension reinforcement; d = h - csr - φsr / 2 = 640 mm

K = M / (d2 × fck) = 0.019

K' = 0.196

K' > K - No compression reinforcement is required

Lever arm; z = min(0.5 + 0.5 × (1 – 3.53 × K)0.5, 0.95) × d = 608 mm

Depth of neutral axis; x = 2.5 × (d – z) = 80 mm

Area of tension reinforcement required; Asr.req = M / (fyd × z) = 920 mm2/m

Tension reinforcement provided; 20 dia.bars @ 200 c/c






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Area of tension reinforcement provided; Asr.prov = π × φsr2 / (4 × ssr) = 1571 mm

2/m

Minimum area of reinforcement - exp.9.1N; Asr.min = max(0.26 × fctm / fyk, 0.0013) × d = 1006 mm2/m

Maximum area of reinforcement - cl.9.2.1.1(3); Asr.max = 0.04 × h = 28000 mm2/m

max(Asr.req, Asr.min) / Asr.prov = 0.641

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

Crack control - Section 7.3

Limiting crack width; wmax = 0.3 mm

Variable load factor - EN1990 – Table A1.1; ψ2 = 0.3

Serviceability bending moment; Msls = 146.7 kNm/m

Tensile stress in reinforcement; σs = Msls / (Asr.prov × z) = 153.6 N/mm2

Load duration; Long term

Load duration factor; kt = 0.4

Effective area of concrete in tension; Ac.eff = min(2.5 × (h - d), (h – x) / 3, h / 2) = 150000 mm2/m

Mean value of concrete tensile strength; fct.eff = fctm = 3.0 N/mm2

Reinforcement ratio; ρp.eff = Asr.prov / Ac.eff = 0.010

Modular ratio; αe = Es / Ecm = 5.998

Bond property coefficient; k1 = 0.8

Strain distribution coefficient; k2 = 0.5

k3 = 3.4

k4 = 0.425

Maximum crack spacing - exp.7.11; sr.max = k3 × csr + k1 × k2 × k4 × φsr / ρp.eff = 495 mm

Maximum crack width - exp.7.8; wk = sr.max × max(σs – kt × (fct.eff / ρp.eff) × (1 + αe × ρp.eff), 0.6 × σs) / Es

wk = 0.228 mm

wk / wmax = 0.76

PASS - Maximum crack width is less than limiting crack width






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Dr. C. Sachpazis

Date

22/12/2013

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Rectangular section in shear - Section 6.2

Design shear force; V = 135.6 kN/m

CRd,c = 0.18 / γC = 0.120

k = min(1 + √(200 mm / d), 2) = 1.559

Longitudinal reinforcement ratio; ρl = min(Asr.prov / d, 0.02) = 0.002

vmin = 0.035 N1/2/mm × k3/2 × fck

0.5 = 0.385 N/mm

2

Design shear resistance - exp.6.2a & 6.2b; VRd.c = max(CRd.c × k × (100 N2/mm

4 × ρl × fck)

1/3, vmin) × d

VRd.c = 246.7 kN/m

V / VRd.c = 0.550

PASS - Design shear resistance exceeds design shear force

Horizontal reinforcement parallel to face of stem - Section 9.6

Minimum area of reinforcement – cl.9.6.3(1); Asx.req = max(0.25 × Asr.prov, 0.001 × (tstem + lslr)) = 700 mm2/m

Maximum spacing of reinforcement – cl.9.6.3(2); ssx_max = 400 mm

Transverse reinforcement provided; 16 dia.bars @ 200 c/c

Area of transverse reinforcement provided; Asx.prov = π × φsx2 / (4 × ssx) = 1005 mm

2/m


Check base design




Depth to tension reinforcement; d = h - cbb - φbb / 2 = 465 mm

K = M / (d2 × fck) = 0.020

K' = 0.196








Job Ref.

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Calc. by

Dr. C. Sachpazis

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Area of tension reinforcement required; Abb.req = M / (fyd × z) = 719 mm2/m


Area of tension reinforcement provided; Abb.prov = π × φbb2 / (4 × sbb) = 1571 mm

2/m

Minimum area of reinforcement - exp.9.1N; Abb.min = max(0.26 × fctm / fyk, 0.0013) × d = 731 mm2/m

Maximum area of reinforcement - cl.9.2.1.1(3); Abb.max = 0.04 × h = 22000 mm2/m

max(Abb.req, Abb.min) / Abb.prov = 0.465






Tensile stress in reinforcement; σs = Msls / (Abb.prov × z) = 123 N/mm2





Reinforcement ratio; ρp.eff = Abb.prov / Ac.eff = 0.010




k3 = 3.4

k4 = 0.425

Maximum crack spacing - exp.7.11; sr.max = k3 × cbb + k1 × k2 × k4 × φbb / ρp.eff = 610 mm


wk = 0.225 mm






Job Ref.

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Dr. C. Sachpazis

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wk / wmax = 0.75




CRd,c = 0.18 / γC = 0.120

k = min(1 + √(200 mm / d), 2) = 1.656

Longitudinal reinforcement ratio; ρl = min(Abb.prov / d, 0.02) = 0.003

vmin = 0.035 N1/2/mm × k3/2 × fck

0.5 = 0.422 N/mm

2


4 × ρl × fck)

1/3, vmin) × d

VRd.c = 204.3 kN/m

V / VRd.c = 0.664




Depth to tension reinforcement; d = h - cbt - φbt / 2 = 492 mm

K = M / (d2 × fck) = 0.016

K' = 0.196




Area of tension reinforcement required; Abt.req = M / (fyd × z) = 621 mm2/m


Area of tension reinforcement provided; Abt.prov = π × φbt2 / (4 × sbt) = 1005 mm

2/m

Minimum area of reinforcement - exp.9.1N; Abt.min = max(0.26 × fctm / fyk, 0.0013) × d = 774 mm2/m

Maximum area of reinforcement - cl.9.2.1.1(3); Abt.max = 0.04 × h = 22000 mm2/m

max(Abt.req, Abt.min) / Abt.prov = 0.77






Job Ref.

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Calc. by

Dr. C. Sachpazis

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22/12/2013

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Tensile stress in reinforcement; σs = Msls / (Abt.prov × z) = 146.7 N/mm2





Reinforcement ratio; ρp.eff = Abt.prov / Ac.eff = 0.007




k3 = 3.4

k4 = 0.425

Maximum crack spacing - exp.7.11; sr.max = k3 × cbt + k1 × k2 × k4 × φbt / ρp.eff = 562 mm


wk = 0.247 mm

wk / wmax = 0.825




CRd,c = 0.18 / γC = 0.120

k = min(1 + √(200 mm / d), 2) = 1.638

Longitudinal reinforcement ratio; ρl = min(Abt.prov / d, 0.02) = 0.002






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Calc. by

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vmin = 0.035 N1/2/mm × k3/2 × fck

0.5 = 0.415 N/mm

2


4 × ρl × fck)

1/3, vmin) × d

VRd.c = 204.1 kN/m

V / VRd.c = 0.443


Check key design




Depth to tension reinforcement; d = h - cbb - φk / 2 = 417 mm

K = M / (d2 × fck) = 0.001

K' = 0.196




Area of tension reinforcement required; Ak.req = M / (fyd × z) = 41 mm2/m


Area of tension reinforcement provided; Ak.prov = π × φk2 / (4 × sk) = 1005 mm

2/m

Minimum area of reinforcement - exp.9.1N; Ak.min = max(0.26 × fctm / fyk, 0.0013) × d = 656 mm2/m

Maximum area of reinforcement - cl.9.2.1.1(3); Ak.max = 0.04 × h = 20000 mm2/m

max(Ak.req, Ak.min) / Ak.prov = 0.652










Job Ref.

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Tensile stress in reinforcement; σs = Msls / (Ak.prov × z) = 17.7 N/mm2





Reinforcement ratio; ρp.eff = Ak.prov / Ac.eff = 0.007




k3 = 3.4

k4 = 0.425

Maximum crack spacing - exp.7.11; sr.max = k3 × cbb + k1 × k2 × k4 × φk / ρp.eff = 659 mm


wk = 0.035 mm

wk / wmax = 0.117



Design shear force; V = 26 kN/m

CRd,c = 0.18 / γC = 0.120

k = min(1 + √(200 mm / d), 2) = 1.693

Longitudinal reinforcement ratio; ρl = min(Ak.prov / d, 0.02) = 0.002

vmin = 0.035 N1/2/mm × k3/2 × fck

0.5 = 0.436 N/mm

2


4 × ρl × fck)

1/3, vmin) × d

VRd.c = 181.8 kN/m

V / VRd.c = 0.143






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Dr. C. Sachpazis

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Secondary transverse reinforcement to base - Section 9.3

Minimum area of reinforcement – cl.9.3.1.1(2); Abx.req = 0.2 × Abb.prov = 314 mm2/m

Maximum spacing of reinforcement – cl.9.3.1.1(3); sbx_max = 450 mm

Transverse reinforcement provided; 12 dia.bars @ 200 c/c

Area of transverse reinforcement provided; Abx.prov = π × φbx2 / (4 × sbx) = 565 mm

2/m







Job Ref.

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Dr. C. Sachpazis

Date

22/12/2013

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Sachpazis: Sloped rear face retaining wall example

Design

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