Sachpazis cantilever retaining wall analysis & design (en1997-1-2004)

GEODOMISI Ltd. - Dr. Costas Sachpazis
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics, Foundation
Engineering & Retaining Structures.
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6936425722 & (+44) 7585939944, costas@sachpazis.info
Project: Retaining wall Analysis & Design, In accordance with
EN1997-1:2004 incorporating Corrigendum dated February
2009 and the recommended values.
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Section
Civil & Geotechnical Engineering
Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
04/04/2014
Chk'd by
Date App'd by Date
RETAINING WALL ANALYSIS (EN1997-1:2004)
In accordance with EN1997-1:2004 incorporating Corrigendum dated
February 2009 and the recommended values
Retaining wall details
Stem type; Cantilever with inclined front face
Stem height; hstem = 4000 mm
Stem thickness; tstem = 450 mm
Slope length to front of stem; lslf = 100 mm
Angle to rear face of stem; α = 90 deg
Angle to front face of stem; αf = 88.6 deg
Stem density; γstem = 25 kN/m
3
Toe length; ltoe = 1000 mm
Heel length; lheel = 4000 mm
Base thickness; tbase = 450 mm
Base density; γbase = 25 kN/m
3
Height of retained soil; hret = 3000 mm
Angle of soil surface; β = 15 deg
Depth of cover; dcover = 500 mm
Height of water; hwater = 300 mm
Water density; γw = 9.8 kN/m
3
Retained soil properties
Soil type; Medium dense well graded sand and gravel
Moist density; γmr = 20 kN/m
3
Saturated density; γsr = 22.3 kN/m
3
Characteristic effective shear resistance angle; φ'r.k = 25 deg
Characteristic wall friction angle; δr.k = 16 deg
Base soil properties
Moist density; γmb = 20 kN/m3
Characteristic cohesion; c'b.k = 5 kN/m
2
Characteristic adhesion; ab.k = 5 kN/m
2
Characteristic effective shear resistance angle; φ'b.k = 25 deg
Characteristic wall friction angle; δb.k = 20 deg
Characteristic base friction angle; δbb.k = 25 deg
Loading details
Variable surcharge load; SurchargeQ = 10 kN/m
2

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Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
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Calculate retaining wall geometry
Base length; lbase = ltoe + lslf + tstem + lheel = 5550 mm
Saturated soil height; hsat = hwater + dcover = 800 mm
Moist soil height; hmoist = hret - hwater = 2700 mm
Length of surcharge load; lsur = lheel = 4000 mm

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- Distance to vertical component; xsur_v = lbase - lheel / 2 = 3550 mm
Effective height of wall; heff = hbase + dcover + hret + lsur × tan(β) = 5022 mm
- Distance to horizontal component; xsur_h = heff / 2 = 2511 mm
Area of wall stem; Astem = hstem × (tstem + lslf / 2) = 2 m
2
- Distance to vertical component; xstem = (hstem × tstem × (ltoe + lslf + tstem / 2) + hstem × lslf
/ 2 × (ltoe + 2 × lslf / 3)) / Astem = 1299 mm
Area of wall base; Abase = lbase × tbase = 2.498 m
2
- Distance to vertical component; xbase = lbase / 2 = 2775 mm
Area of saturated soil; Asat = hsat × lheel = 3.2 m
2
- Distance to vertical component; xsat_v = lbase - (hsat × lheel
2
/ 2) / Asat = 3550 mm
- Distance to horizontal component; xsat_h = (hsat + hbase) / 3 = 417 mm
Area of water; Awater = hsat × lheel = 3.2 m
2
- Distance to vertical component; xwater_v = lbase - (hsat × lheel
2
/ 2) / Asat = 3550 mm
- Distance to horizontal component; xwater_h = (hsat + hbase) / 3 = 417 mm
Area of moist soil; Amoist = hmoist × lheel + tan(β) × lheel
2
/ 2 = 12.944 m
2
- Distance to vertical component; xmoist_v = lbase - (hmoist × lheel
2
/ 2 + tan(β) × lheel
3
/ 6) /
Amoist = 3660 mm
- Distance to horizontal component; xmoist_h = ((heff - hsat - hbase) × (tbase + hsat + (heff - hsat
- hbase) / 3) / 2 + (hsat + tbase)
2
/2) / (hsat + tbase + (heff -
hsat - hbase) / 2) = 1757 mm
Area of base soil; Apass = dcover × (ltoe + lslf × dcover / (2 × hstem)) = 0.503
m
2
- Distance to vertical component; xpass_v = lbase - (dcover × ltoe × (lbase - ltoe / 2) + lslf ×
dcover
2
/ (2 × hstem) × (lbase - ltoe - lslf × dcover / (3 ×
hstem))) / Apass = 503 mm
- Distance to horizontal component; xpass_h = (dcover + hbase) / 3 = 317 mm
Area of excavated base soil; Aexc = hpass × (ltoe + lslf × hpass / (2 × hstem)) = 0.503
m
2
- Distance to vertical component; xexc_v = lbase - (hpass × ltoe × (lbase - ltoe / 2) + lslf ×
hpass
2
/ (2 × hstem) × (lbase - ltoe - lslf × hpass / (3 ×
hstem))) / Aexc = 503 mm
- Distance to horizontal component; xexc_h = (hpass + hbase) / 3 = 317 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

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Calc. by
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Date
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Effective cohesion; γc' = 1.00
Weight density; γγ = 1.00
Design effective shear resistance angle; φ'r.d = atan(tan(φ'r.k) / γφ') = 25 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') = 25 deg
Design wall friction angle; δb.d = atan(tan(δb.k) / γφ') = 20 deg
Design base friction angle; δbb.d = atan(tan(δbb.k) / γφ') = 25 deg
Design effective cohesion; c'b.d = c'b.k / γc' = 5 kN/m
2
Design adhesion; ab.d = ab.k / γc' = 5 kN/m
2
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
) = 0.469
Passive pressure coefficient; KP = sin(αf - φ'b.d)
2
/ (sin(αf)
2
× sin(αf + δb.d) × [1 -
√[sin(φ'b.d + δb.d) × sin(φ'b.d) / (sin(αf + δb.d) ×
sin(αf))]]
2
) = 4.403
Sliding check
Vertical forces on wall
Wall stem; Fstem = γGf × Astem × γstem = 50 kN/m
Wall base; Fbase = γGf × Abase × γbase = 62.4 kN/m
Saturated retained soil; Fsat_v = γGf × Asat × (γsr - γw) = 39.8 kN/m
Water; Fwater_v = γGf × Awater × γw = 31.4 kN/m
Moist retained soil; Fmoist_v = γGf × Amoist × γmr = 258.9 kN/m
Base soil; Fexc_v = γGf × Aexc × γmb = 10.1 kN/m
Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fexc_v +
Fwater_v = 452.6 kN/m
Horizontal forces on wall
Surcharge load; Fsur_h = KA × cos(δr.d) × γQ × SurchargeQ × heff =
33.9 kN/m
Saturated retained soil; Fsat_h = γG × KA × cos(δr.d) × (γsr - γw) × (hsat + hbase)
2
/ 2 = 5.9 kN/m
Water; Fwater_h = γG × γw × (hwater + dcover + hbase)
2
/ 2 = 10.3
kN/m

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Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
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Moist retained soil; Fmoist_h = γG × KA × cos(δr.d) × γmr × ((heff - hsat -
hbase)
2
/ 2 + (heff - hsat - hbase) × (hsat + hbase)) = 143.9
kN/m
Total; Ftotal_h = Fsat_h + Fmoist_h + Fwater_h + Fsur_h = 194.1
kN/m
Check stability against sliding
Base soil resistance; Fexc_h = γGf × KP × cos(δb.d) × γmb × (hpass + hbase)
2
/ 2
= 37.3 kN/m
Base friction; Ffriction = ab.d × b + Ftotal_v × tan(δbb.d) = 216 kN/m
Resistance to sliding; Frest = Fexc_h + Ffriction = 253.4 kN/m
Factor of safety; FoSsl = Frest / Ftotal_h = 1.306
PASS - Resistance to sliding is greater than sliding force
Overturning check
33.9 kN/m
2
/ 2 = 5.9 kN/m
Water; Fwater_h = γG × γw × (hwater + dcover + hbase)2
/ 2 = 10.3
kN/m
hbase)
2
kN/m
Base soil; Fexc_h = -γGf × KP × cos(δb.d) × γmb × (hpass + hbase)
2
/
2 = -37.3 kN/m
Total; Ftotal_h = Fsat_h + Fmoist_h + Fexc_h + Fwater_h + Fsur_h =
156.7 kN/m
Overturning moments on wall
Surcharge load; Msur_OT = Fsur_h × xsur_h = 85.2 kNm/m

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Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
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Saturated retained soil; Msat_OT = Fsat_h × xsat_h = 2.5 kNm/m
Water; Mwater_OT = Fwater_h × xwater_h = 4.3 kNm/m
Moist retained soil; Mmoist_OT = Fmoist_h × xmoist_h = 252.8 kNm/m
Total; Mtotal_OT = Msat_OT + Mmoist_OT + Mwater_OT + Msur_OT =
344.8 kNm/m
Restoring moments on wall
Wall stem; Mstem_R = Fstem × xstem = 65 kNm/m
Wall base; Mbase_R = Fbase × xbase = 173.3 kNm/m
Saturated retained soil; Msat_R = Fsat_v × xsat_v = 141.3 kNm/m
Water; Mwater_R = Fwater_v × xwater_v = 111.4 kNm/m
Moist retained soil; Mmoist_R = Fmoist_v × xmoist_v = 947.6 kNm/m
Base soil; Mexc_R = Fexc_v × xexc_v - Fexc_h × xexc_h = 16.9 kNm/m
Total; Mtotal_R = Mstem_R + Mbase_R + Msat_R + Mmoist_R +
Mexc_R + Mwater_R = 1455.4 kNm/m
Check stability against overturning
Factor of safety; FoSot = Mtotal_R / Mtotal_OT = 4.222
PASS - Maximum restoring moment is greater than overturning moment
Bearing pressure check
Wall stem; Fstem = γG × Astem × γstem = 67.5 kN/m
Wall base; Fbase = γG × Abase × γbase = 84.3 kN/m
Surcharge load; Fsur_v = γQ × SurchargeQ × lheel = 60 kN/m
Saturated retained soil; Fsat_v = γG × Asat × (γsr - γw) = 53.7 kN/m
Water; Fwater_v = γG × Awater × γw = 42.4 kN/m
Moist retained soil; Fmoist_v = γG × Amoist × γmr = 349.5 kN/m
Base soil; Fpass_v = γG × Apass × γmb = 13.6 kN/m
Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fpass_v +
Fwater_v + Fsur_v = 671 kN/m
33.9 kN/m
2
/ 2 = 5.9 kN/m
2
/ 2 = 10.3
kN/m
hbase)
2
kN/m

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Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
04/04/2014
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Base soil; Fpass_h = -γGf × KP × cos(δb.d) × γmb × (dcover + hbase)
2
/
2 = -37.3 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 = 87.7 kNm/m
Wall base; Mbase = Fbase × xbase = 233.9 kNm/m
Surcharge load; Msur = Fsur_v × xsur_v - Fsur_h × xsur_h = 127.8 kNm/m
Saturated retained soil; Msat = Fsat_v × xsat_v - Fsat_h × xsat_h = 188.3 kNm/m
Water; Mwater = Fwater_v × xwater_v - Fwater_h × xwater_h = 146.1
kNm/m
Moist retained soil; Mmoist = Fmoist_v × xmoist_v - Fmoist_h × xmoist_h = 1026.4
kNm/m
Base soil; Mpass = Fpass_v × xpass_v - Fpass_h × xpass_h = 18.7
kNm/m
Total; Mtotal = Mstem + Mbase + Msat + Mmoist + Mpass + Mwater
+ Msur = 1829 kNm/m
Check bearing pressure
Distance to reaction; x = Mtotal / Ftotal_v = 2726 mm
Eccentricity of reaction; e = x - lbase / 2 = -49 mm
Loaded length of base; lload = 2 × x = 5452 mm
Bearing pressure at toe; qtoe = Ftotal_v / lload = 123.1 kN/m
2
Bearing pressure at heel; qheel = 0 kN/m
2
Effective overburden pressure; q = (tbase + dcover) × γmb - (tbase + dcover + hwater) × γw =
6.7 kN/m
2
Design effective overburden pressure; q' = q / γγ = 6.7 kN/m2
Bearing resistance factors; Nq = Exp(π × tan(φ'b.d)) × (tan(45 deg + φ'b.d / 2))
2
=
10.662
Nc = (Nq - 1) × cot(φ'b.d) = 20.721
Nγ = 2 × (Nq - 1) × tan(φ'b.d) = 9.011
Foundation shape factors; sq = 1
sγ = 1
sc = 1
Load inclination factors; H = Ftotal_h = 0 kN/m
V = Ftotal_v = 671 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

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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 = 425.7 kN/m
2
Factor of safety; FoSbp = nf / max(qtoe, qheel) = 3.459
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
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') = 20.5 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') = 20.5 deg
Design wall friction angle; δb.d = atan(tan(δb.k) / γφ') = 16.2 deg
Design base friction angle; δbb.d = atan(tan(δbb.k) / γφ') = 20.5 deg
Design effective cohesion; c'b.d = c'b.k / γc' = 4 kN/m
2
Design adhesion; ab.d = ab.k / γc' = 4 kN/m
2
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
) = 0.590
Passive pressure coefficient; KP = sin(αf - φ'b.d)
2
/ (sin(αf)
2
× sin(αf + δb.d) × [1 -
√[sin(φ'b.d + δb.d) × sin(φ'b.d) / (sin(αf + δb.d) ×
sin(αf))]]
2
) = 3.111
Sliding check

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37.5 kN/m
2
/ 2 = 5.6 kN/m
2
/ 2 = 7.7
kN/m
hbase)
2
/ 2 + (heff - hsat - hbase) × (hsat + hbase)) = 136
kN/m
Total; Ftotal_h = Fsat_h + Fmoist_h + Fwater_h + Fsur_h = 186.8
kN/m
Check stability against sliding
Base soil resistance; Fexc_h = γGf × KP × cos(δb.d) × γmb × (hpass + hbase)
2
/ 2
= 27 kN/m
Base friction; Ffriction = ab.d × b + Ftotal_v × tan(δbb.d) = 172.8 kN/m
Resistance to sliding; Frest = Fexc_h + Ffriction = 199.8 kN/m
Factor of safety; FoSsl = Frest / Ftotal_h = 1.07
PASS - Resistance to sliding is greater than sliding force
Overturning check
37.5 kN/m
2
/ 2 = 5.6 kN/m

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2
/ 2 = 7.7
kN/m
hbase)
2
kN/m
Base soil; Fexc_h = -γGf × KP × cos(δb.d) × γmb × (hpass + hbase)
2
/
2 = -27 kN/m
Total; Ftotal_h = Fsat_h + Fmoist_h + Fexc_h + Fwater_h + Fsur_h =
159.8 kN/m
Overturning moments on wall
Surcharge load; Msur_OT = Fsur_h × xsur_h = 94.2 kNm/m
Saturated retained soil; Msat_OT = Fsat_h × xsat_h = 2.3 kNm/m
Water; Mwater_OT = Fwater_h × xwater_h = 3.2 kNm/m
Moist retained soil; Mmoist_OT = Fmoist_h × xmoist_h = 238.9 kNm/m
Total; Mtotal_OT = Msat_OT + Mmoist_OT + Mwater_OT + Msur_OT =
338.7 kNm/m
Restoring moments on wall
Wall stem; Mstem_R = Fstem × xstem = 65 kNm/m
Wall base; Mbase_R = Fbase × xbase = 173.3 kNm/m
Saturated retained soil; Msat_R = Fsat_v × xsat_v = 141.3 kNm/m
Water; Mwater_R = Fwater_v × xwater_v = 111.4 kNm/m
Moist retained soil; Mmoist_R = Fmoist_v × xmoist_v = 947.6 kNm/m
Base soil; Mexc_R = Fexc_v × xexc_v - Fexc_h × xexc_h = 13.6 kNm/m
Total; Mtotal_R = Mstem_R + Mbase_R + Msat_R + Mmoist_R +
Mexc_R + Mwater_R = 1452.2 kNm/m
Check stability against overturning
Factor of safety; FoSot = Mtotal_R / Mtotal_OT = 4.288
PASS - Maximum restoring moment is greater than overturning moment
Bearing pressure check
Wall stem; Fstem = γG × Astem × γstem = 50 kN/m
Wall base; Fbase = γG × Abase × γbase = 62.4 kN/m
Surcharge load; Fsur_v = γQ × SurchargeQ × lheel = 52 kN/m
Saturated retained soil; Fsat_v = γG × Asat × (γsr - γw) = 39.8 kN/m
Water; Fwater_v = γG × Awater × γw = 31.4 kN/m
Moist retained soil; Fmoist_v = γG × Amoist × γmr = 258.9 kN/m
Base soil; Fpass_v = γG × Apass × γmb = 10.1 kN/m

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Total; Ftotal_v = Fstem + Fbase + Fsat_v + Fmoist_v + Fpass_v +
Fwater_v + Fsur_v = 504.6 kN/m
37.5 kN/m
2
/ 2 = 5.6 kN/m
2
/ 2 = 7.7
kN/m
hbase)
2
kN/m
Base soil; Fpass_h = -γGf × KP × cos(δb.d) × γmb × (dcover + hbase)
2
/
2 = -27 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 = 65 kNm/m
Wall base; Mbase = Fbase × xbase = 173.3 kNm/m
Surcharge load; Msur = Fsur_v × xsur_v - Fsur_h × xsur_h = 90.4 kNm/m
Saturated retained soil; Msat = Fsat_v × xsat_v - Fsat_h × xsat_h = 139 kNm/m
Water; Mwater = Fwater_v × xwater_v - Fwater_h × xwater_h = 108.2
kNm/m
Moist retained soil; Mmoist = Fmoist_v × xmoist_v - Fmoist_h × xmoist_h = 708.7
kNm/m
Base soil; Mpass = Fpass_v × xpass_v - Fpass_h × xpass_h = 13.6
kNm/m
Total; Mtotal = Mstem + Mbase + Msat + Mmoist + Mpass + Mwater
+ Msur = 1298.1 kNm/m
Check bearing pressure
Distance to reaction; x = Mtotal / Ftotal_v = 2573 mm
Eccentricity of reaction; e = x - lbase / 2 = -202 mm
Loaded length of base; lload = 2 × x = 5145 mm
Bearing pressure at toe; qtoe = Ftotal_v / lload = 98.1 kN/m
2
Bearing pressure at heel; qheel = 0 kN/m
2
Effective overburden pressure; q = (tbase + dcover) × γmb - (tbase + dcover + hwater) × γw =
6.7 kN/m
2
Design effective overburden pressure; q' = q / γγ = 6.7 kN/m
2

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Calc. by
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Date
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Bearing resistance factors; Nq = Exp(π × tan(φ'b.d)) × (tan(45 deg + φ'b.d / 2))
2
=
6.698
Nc = (Nq - 1) × cot(φ'b.d) = 15.273
Nγ = 2 × (Nq - 1) × tan(φ'b.d) = 4.251
Foundation shape factors; sq = 1
sγ = 1
sc = 1
Load inclination factors; H = Ftotal_h = 0 kN/m
V = Ftotal_v = 504.6 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 = 217.7 kN/m
2
Factor of safety; FoSbp = nf / max(qtoe, qheel) = 2.22
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; C20/25
Characteristic compressive cylinder strength; fck = 20 N/mm
2
Characteristic compressive cube strength; fck,cube = 25 N/mm
2
Mean value of compressive cylinder strength; fcm = fck + 8 N/mm
2
= 28 N/mm
2
Mean value of axial tensile strength; fctm = 0.3 N/mm
2
× (fck / 1 N/mm
2
)
2/3
= 2.2 N/mm
2
5% fractile of axial tensile strength; fctk,0.05 = 0.7 × fctm = 1.5 N/mm
2
Secant modulus of elasticity of concrete; Ecm = 22 kN/mm
2
× (fcm / 10 N/mm
2
)
0.3
= 29962
N/mm
2
Partial factor for concrete - Table 2.1N; γC = 1.50
Compressive strength coefficient - cl.3.1.6(1); αcc = 1.00
Design compressive concrete strength - exp.3.15; fcd = αcc × fck / γC = 13.3 N/mm
2
Maximum aggregate size; hagg = 20 mm
Reinforcement details
Characteristic yield strength of reinforcement; fyk = 500 N/mm
2
Modulus of elasticity of reinforcement; Es = 200000 N/mm
2

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Job Ref.
Section
Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
04/04/2014
Chk'd by
Date App'd by Date
Partial factor for reinforcing steel - Table 2.1N; γS = 1.15
Design yield strength of reinforcement; fyd = fyk / γS = 435 N/mm
2
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 = 550 mm
Rectangular section in flexure - Section 6.1
Design bending moment combination 1; M = 235.8 kNm/m
Depth to tension reinforcement; d = h - csr - φsr / 2 = 488 mm
K = M / (d
2
× fck) = 0.050
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 =
463 mm
Depth of neutral axis; x = 2.5 × (d – z) = 61 mm
Area of tension reinforcement required; Asr.req = M / (fyd × z) = 1171 mm
2
/m
Tension reinforcement provided; 25 dia.bars @ 200 c/c
Area of tension reinforcement provided; Asr.prov = π × φsr
2
/ (4 × ssr) = 2454 mm
2
/m
Minimum area of reinforcement - exp.9.1N; Asr.min = max(0.26 × fctm / fyk, 0.0013) × d = 634
mm
2
/m
Maximum area of reinforcement - cl.9.2.1.1(3); Asr.max = 0.04 × h = 22000 mm
2
/m
max(Asr.req, Asr.min) / Asr.prov = 0.477
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 = 152.3 kNm/m
Tensile stress in reinforcement; σs = Msls / (Asr.prov × z) = 134 N/mm
2
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) = 156250
mm
2
/m
Mean value of concrete tensile strength; fct.eff = fctm = 2.2 N/mm
2
Reinforcement ratio; ρp.eff = Asr.prov / Ac.eff = 0.016
Modular ratio; αe = Es / Ecm = 6.675
Bond property coefficient; k1 = 0.8

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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 = 441 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 = 0.177 mm
wk / wmax = 0.59
PASS - Maximum crack width is less than limiting crack width
Rectangular section in shear - Section 6.2
Design shear force; V = 153.5 kN/m
CRd,c = 0.18 / γC = 0.120
k = min(1 + √(200 mm / d), 2) = 1.641
Longitudinal reinforcement ratio; ρl = min(Asr.prov / d, 0.02) = 0.005
vmin = 0.035 N
1/2
/mm × k
3/2
× fck
0.5
= 0.329 N/mm
2
Design shear resistance - exp.6.2a & 6.2b; VRd.c = max(CRd.c × k × (100 N
2
/mm
4
× ρl × fck)
1/3
,
vmin) × d
VRd.c = 207.2 kN/m
V / VRd.c = 0.741
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 + lslf)) =
614 mm
2
/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 × φsx
2
/ (4 × ssx) = 1005 mm
2
/m
Check base design at toe
Depth to tension reinforcement; d = h - cbb - φbb / 2 = 367 mm
K = M / (d
2
× fck) = 0.018
K' = 0.196
Lever arm; z = min(0.5 + 0.5 × (1 – 3.53 × K)
0.5
, 0.95) × d =
349 mm
Area of tension reinforcement required; Abb.req = M / (fyd × z) = 323 mm
2
/m

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Job Ref.
Section
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Dr. C. Sachpazis
Date
04/04/2014
Chk'd by
Date App'd by Date
Area of tension reinforcement provided; Abb.prov = π × φbb
2
/ (4 × sbb) = 1005 mm
2
/m
Minimum area of reinforcement - exp.9.1N; Abb.min = max(0.26 × fctm / fyk, 0.0013) × d = 477
mm
2
/m
Maximum area of reinforcement - cl.9.2.1.1(3); Abb.max = 0.04 × h = 18000 mm
2
/m
max(Abb.req, Abb.min) / Abb.prov = 0.475
Tensile stress in reinforcement; σs = Msls / (Abb.prov × z) = 100.8 N/mm
2
mm
2
/m
2
Reinforcement ratio; ρp.eff = Abb.prov / Ac.eff = 0.007
k3 = 3.4
k4 = 0.425
Maximum crack spacing - exp.7.11; sr.max = k3 × cbb + k1 × k2 × k4 × φbb / ρp.eff = 619 mm
ρp.eff), 0.6 × σs) / Es
wk = 0.187 mm
wk / wmax = 0.625
CRd,c = 0.18 / γC = 0.120
k = min(1 + √(200 mm / d), 2) = 1.738
Longitudinal reinforcement ratio; ρl = min(Abb.prov / d, 0.02) = 0.003
vmin = 0.035 N
1/2
/mm × k
3/2
× fck
0.5
= 0.359 N/mm
2
2
/mm
4
× ρl × fck)
1/3
,
vmin) × d
VRd.c = 134.9 kN/m
V / VRd.c = 0.722

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Job Ref.
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Sheet no./rev. 1
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Dr. C. Sachpazis
Date
04/04/2014
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Check base design at heel
Depth to tension reinforcement; d = h - cbt - φbt / 2 = 388 mm
K = M / (d
2
× fck) = 0.079
K' = 0.196
Lever arm; z = min(0.5 + 0.5 × (1 – 3.53 × K)
0.5
, 0.95) × d =
358 mm
Area of tension reinforcement required; Abt.req = M / (fyd × z) = 1529 mm
2
/m
Area of tension reinforcement provided; Abt.prov = π × φbt
2
/ (4 × sbt) = 2454 mm
2
/m
Minimum area of reinforcement - exp.9.1N; Abt.min = max(0.26 × fctm / fyk, 0.0013) × d = 504
mm
2
/m
Maximum area of reinforcement - cl.9.2.1.1(3); Abt.max = 0.04 × h = 18000 mm
2
/m
max(Abt.req, Abt.min) / Abt.prov = 0.623
Tensile stress in reinforcement; σs = Msls / (Abt.prov × z) = 122.5 N/mm
2
mm
2
/m
2
Reinforcement ratio; ρp.eff = Abt.prov / Ac.eff = 0.020
k3 = 3.4
k4 = 0.425
Maximum crack spacing - exp.7.11; sr.max = k3 × cbt + k1 × k2 × k4 × φbt / ρp.eff = 387 mm
ρp.eff), 0.6 × σs) / Es
wk = 0.142 mm

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Job Ref.
Section
Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
04/04/2014
Chk'd by
Date App'd by Date
wk / wmax = 0.475
CRd,c = 0.18 / γC = 0.120
k = min(1 + √(200 mm / d), 2) = 1.718
Longitudinal reinforcement ratio; ρl = min(Abt.prov / d, 0.02) = 0.006
vmin = 0.035 N
1/2
/mm × k
3/2
× fck
0.5
= 0.353 N/mm
2
2
/mm
4
× ρl × fck)
1/3
,
vmin) × d
VRd.c = 186.3 kN/m
V / VRd.c = 0.483
Secondary transverse reinforcement to base - Section 9.3
Minimum area of reinforcement – cl.9.3.1.1(2); Abx.req = 0.2 × Abt.prov = 491 mm
2
/m
Maximum spacing of reinforcement – cl.9.3.1.1(3); sbx_max = 450 mm
Transverse reinforcement provided; 16 dia.bars @ 300 c/c
Area of transverse reinforcement provided; Abx.prov = π × φbx
2
/ (4 × sbx) = 670 mm
2
/m

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Job Ref.
Section
Sheet no./rev. 1
Calc. by
Dr. C. Sachpazis
Date
04/04/2014
Chk'd by
Date App'd by Date

Sachpazis cantilever retaining wall analysis & design (en1997-1-2004)

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