appendix d preliminary basement calculations

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SUBTERRANEAN CONSTRUCTION METHOD STATEMENT

FOR

YORK HOUSE, TURKS ROW, LONDON SW3 4TH

APPENDIX D: PRELIMINARY CALCULATIONS

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Calculated 275 mm

coverlink

lstlayerspacer2ndlayerbreadth% of min rebarmin rebar

mm10000.13

422.5

moment max 236.0lkNm

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RklaTedds

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Ca.lu.LI.cih8w $r 200mM '7+tl aC WCU.CRETAINING WALL ANALYSIS -

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

incorporating Corrigendum No.1

2009 and the UK National Annex

Tedds calculation version 2.6.11

Retaining wall detailsStem type

Stem height

Prop heightStem thickness

Angle to rear face of stem

Stem density

ToelengthBase thickness

Base density

Heightofretained soil

Angle of soil surface

Depth ofcover

Height of water

Water density

Propped cantileverh;t.. = 3500 mm

hprop = 3500 mmt:t.. = 200 mma = 90 degYstem : 25 kN/m3

It.. = 1 500 mm

tb;;. = 350 mm'ybase : 25 kN/m3h,.t = 3500 mm

l3 = 0 degd...., = 0 mm

h.,t., = 2500 mm

yw = 9.8 kN/m3

Retained soilpropertiesSoiltype

Moist density

Saturated density

Characteristic effective shear resistance angle

Characteristic wallfriction angle

Medium dense well graded sand

ymr : 21 kN/m3

ysr : 23 kN/m3

0'r.k : 30 deg8r.k : 0 deg

Base soilpropertiesSoiltype

Soil densityCharacteristic cohesion

Characteristic effective shear resistance angle

Characteristic wallfriction angle

Characteristic base friction angle

Medium dense well graded sand

yb = 18 kN/m3c'b.k = 0 kN/m2

@'b.k : 30 deg

8b.k : 15 deg

8bb.k = 30 deg

Loading details

Permanent surcharge loadVariable surcharge loadVertical line load at 1600 mm

SurchargeG = 10 kN/m2

Surcharges = 10 kN/m2

PGI = ll kN/m

PQI = 4.5 kN/m

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1509 >:200

160

1700

General arrangement

Calculate retaining wallgeometry

BaselengthSaturated soilheight

Moist soil heightLength ofsurchargeload

- Distance to vertical component

Effective height of wall- Distance to horizontal component

Area of wall stem

- Distance to vertical componentArea of wall base

- Distance to vertical component

lbase =

h:,t =

hmoist

Is.r =

Xsur v

h.tf=Xsur h

A;t..Xstem

Abase

Xbase

: it.. + tst.. = 1700 mm

h.,t., + d...., = 2500 mm

= h..t - h.,t., = 1000 mmIheel = 0 mm

= lb,se - Iheei / 2 = 1700 mm

hb,s. + dc.,.r + hr.t = 3850 mm

= h.ff / 2 = 1925 mm= hst.m X tst.m = 0.7 m2

= it.. + t;t.. / 2 = 1600 mm= lease X tease = 0.595 m2= lb;s. / 2 = 850 mm

Partial factors on actions - Table A.3

Permanent unfavourable action

Permanent favourable action

Variable unfavourable action

Variable favourable action

Combination I

'yG : 1.35

'yGt: 1.00

'yQ = 1.50

yQt= 0.00

Partial factors for soil parameters - Table A.4 - Combination I

Angle of shearing resistance .b = 1.00

Effective cohesion 'c = 1.00

Weight density yv = 1.00

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Water properties

Design water density 'yw ' = 'w / '. = 9.8 kN/m3

Retained soil properties

Design moist density

Design saturated density

Design effective shear resistance angle

Design wall friction angle

'ymr : 'mr / y. = 21 kN/m3

'ysr : 'sr / 'v = 23 kN/m3

$',.d = atan(tan(+'r.k) / '.b.) = 30 deg

8,d = atan(tan(8rk) / Y.b.) = 0 deg

Base soil properties

Design soil density



Design base friction angle

Design effective cohesion

'yb ' = 'b / 'v = 1 8 kN/m3

4'bd = atan(tan(+'b.K) / 'r.P.) = 30 deg

8b d : atan(tan(8b k) / 'r.b.) = 15 deg

8bb d : atan(tan(8bb.K) / 'r.P.) = 30 degc'b.d = c'b k / 'c = 0 kN/m2

Using Coulomb theory

Active pressure coefficient KA= sin(a + 0'.d):/(sin(a): x sin(CE - 8,d) xll+ Vjsin(+',d + 8.d) x sin(+',d

- l3) / (sin(a - 8.d) x sin(a + l3))lj:) = 0.333

Kp= sin(90 - +'bd)2/(sin(90 + 8bd) xtl- Vjsin(+'bd + 8bd) x sin(+'b.d) /

(sin(90 + 8bd))ll2) = 4.977

Passive pressure coefficient

Bearing pressure check

Vertical forces on wall

Wall stem

Wall base

Line loads

Total

Fstem = 'G X Astem X 'stem = 23.6 kN/m

Fbase = 'G X Ab,se X 'base = 20.1 kN/m

FP.«= 'yG X PGI+ 'yQ X PQI= 21.6 kN/mFt.t,I . = Fst.. + Fb,s. + F.,t.r . + Fp . = 65.3 kN/m

Horizontal forces on wall

SurchargeloadSaturated retained soil

Water

Moist retained soil

Fsuc..h = KA x ('yG x SurchargeG + 'a x Surcharges) x h.ft = 36.6 kN/m

Fsat.h = 'G X KA X ('sr ' - 'w ') X (hsat + hbase)2 / 2 = 24.1 kN/m

Fwatel.h = 'G X 'w ' X (hwater + d.o«er + hbase)2 / 2 = 53.8 kN/m

Fmoist.h = 'G X KA X 'mr ' X ((h.ft - hsat - haase)2 / 2 + (h.ff - heat - hbase) X (h:

+ haase)) = 31 .7 kN/m

Fpass.h : -'Gt X Kp X COS(8bd) X 'b ' X (dco«er + haase)2 / 2 = -5.3 kN/m

FtotaLh : Fsat.h + Fmoist.h + Fpass.h + Fwatel..h + Fsur:h = 140.8 kN/m

tt

Base soil

Total

Moments on wall

Wall stem

Wall base

SurchargeloadLineloads

Saturated retained soil

Water

Moist retained soil

Total

M;t.. = F;t.. x x;t.. = 37.8 kNm/m

Mb,se = Fb,s. X Xbase = 17.1 kNm/m

Ms.r = -Fs.r h X Xs.r h = -70.4 kNm/m

MP=('yG X PGI+ 'yQ X PQI) X PI= 34.6 kNm/m

Ms,t = -Fs,t h x xs,t h = -22.9 kNm/m

Mw,t.r = -Fw,ter h X Xw,ter h = -51.1 kNm/m

Mmoist = -Fmoist h X Xmoist h = -53.4 kNm/m

Mtot,t = Mst.m + Mb,s. + Ms,t + Mmoist + Mw,t.r + M: +Mp .1 08.4 kNm/m

Check bearing pressure

Propping force to stem Fprop.stem : (FtotaLv X lb: / 2 - Mtotal) / (hprop + tease) = 42.6 kN/m

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Propping force to base Fprop.base : FtotaLh - Fprop.stem : 98.3 kN/m

Moment from propping force Mprop = Fprop.stem X (hprop + tease) = 163.9 kNm/m

Distance to reaction x = (Mtotal + Mprop) / Ftotal.v : 850 mm

Eccentricity of reaction e = x - lease / 2 = 0 mm

Loaded length of base li.,d = lease = 1700 mm

Bearing pressure at toe qt.. = Ftotal.v / lb,s. = 38.4 kN/m2Bearing pressure at heel qhe.i = Ftotal.v / lease = 38.4 kN/m2

Effective overburden pressure q = max((tease + d.o«er) X 'b ' - (tbase + d.o«er + hwater) X 'w ', O kN/m2) = 0kN/m2

Design effective overburden pressure q ' = q / 'v = 0 kN/m2

Bearing resistance factors Nq = Exp(7t x tan($'b.d)) x (tan(45 deg + 0'bd / 2))2 = 18.401

Nc = (Nq - 1) X COt($'b.d) = 30.14

N. = 2 x(Nq - 1) x tan(+'bd) = 20.093

H = FsuLh + Fsat.h + FwateLh + Fmoist.h + Fpass.h ' Fprop.stem ' Fprop.base = 0kN/m

V = Ft.t,1 . = 65.3 kN/mm=2iq =]1- H /(V + lload X C'bd X COt(+'bd))]" = I

iv =ll- H/(V + Itoad X C'bd X COt(+'b.d))I(m'D = I

i. = iq -(l- iq)/(N. x tan(0'b.d)) = I

nt = c'b.d x Nc x sc x ic + q ' x Nq x Sq X iq + 0.5 X ('b ' - 'w ') X lload X N.r X S.x

nt = 139.9 kN/m2

FoSbp = nt / max(qtoe, qheel) = 3.641PASS - Allowable bearing pressure exceeds maximum applied bearing pressure

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

Permanent unfavourable action 'G = 1.00

Permanent favourable action 'af = 1.00

Variable unfavourable action 'o = 1.30

variable favourable action 'Qf = 0.00

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

Angle of shearing resistance .}. = 1.25

Effective cohesion 'c = 1.25

Weight density 'v = 1 .00

Water properties

Design water density

Retained soil properties

Design moist density 'mr ' = 'mr / 'v = 21 kN/m3

Design saturated density Ysr ' = 'sr / yv = 23 kN/m3

Design effective shear resistance angle 0'rd = atan(tan(+'r.k) / '.>.) = 24.8 deg

Design wall friction angle 6rd = atan(tan(8r.k) / Y.b) = 0 deg

lSq

lSv

lSc

I.r

rt)iJri(]HiiorisriHI)H iH(:ions

Load inclination factors

Net ultimate bearing capacity

Factorofsafety

'yw ' = 'w / '* = 9.8 kN/m3

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Base soil properties

Design soil density



Design base friction angle

Design effective cohesion

'yb ' :

@'b.d

8bd

8bb.d

c'b.d

'yb / 'v = 18 kN/m3

: atan(tan(4,'. k) / y.b)

: atan(tan(8. k) / '.b) :

= atan(tan(8b. k) / '+-)= c'b.k / 'c ' = 0 kN/m2

24.8 deg

12.1 deg

= 24.8 deg

Using Coulomb theory

Active pressure coefficient KA = sin(a + b'.d):/(sin(ct): x sin(a - 6rd) xll+ Vlsin(+',d + 8,d) x sin(+'

- l3) / (sin(a - 8rd) x sin(a + l3))ll:) = 0.409Kp= sin(90 - +'bd)2/(sin(90 + 8b.d) xll- Vjsin(b'b.d + 8bd) x sin(+'b.d) /

(sin(90 + 8b d))ll2) = 3.473

d

Passive pressure coefficient

Bearing pressure check

Vertical forces on wall

Wall stem

Wall base

Lineloads

Total

Fstem = 'G X Astem X 'st.« = 17.5 kN/m

Fbase = 'G X Abase X 'base = 14.9 kN/m

Fp.« = 'G x PGI + 'Q x Pal = 16.9 kN/mFt.t,I . = Fst.m + Fb,s. + F.,t.r . + Fp , = 49.2 kN/m

Horizontal forces on wall

SurchargeloadSaturated retained soil

Water

Moist retained soil

Fs.[..h = KA.x ('yG x SurchargeG + 'Q x Surcharges) x h.tf = 36.2 kN/m

Fsat..h = 'G X KA X ('sr ' - Yw ') X (heat + hbase)2 / 2 = 21 .9 kN/m

Fwatel..h = 'G X 'w ' X (h«ater + d.o«er + hbase)2 / 2 = 39.8 kN/m

Fmoist.h : 'G X KA X 'mr ' X ((h.fr - hsat - hbase)2 / 2 + (h.ff - hsat - haase) X (h:

+ haase)) = 28.8 kN/m

Fpass.h : ''yGf X Kp X COS(8b.d) X 'b ' X (dco,er + haase)2 / 2 = -3.7 kN/mFtotal.h : Fsat.h + FmoisLh + Fpass.h + Fwatel..h + Fsul.h = 1 23 kN/m

it

Base soil

Total

Moments on wall

Wall stem

Wall base

SurchargeloadLineloads

Saturated retained soil

Water

Moist retained soil

Total

M;t.. = F;t.. x x;t.. = 28 kNm/m

Mbas. = Fb,s. X Xbase = 12.6 kNm/m

Ms.r = -Fs.r h X Xs.r h = -69.7 kNm/m

MP=('yG X PGI+ 'yQ X PQI) X PI= 27 kNm/m

Ms;t = -Fs,t h x xs,t h = -20.8 kNm/m

M.;t.r = -F.;t.r h X X.,t.r h = -37.8 kNm/m

Mmoist = -Fmoist h X Xmoist h : -48.6 kNm/m

Mtotal = Mstem + Mb.se + Msat + Mmoist + Mwater + M: + Mp .1 09.4 kNm/m

Check bearing pressure

Propping force to stem

Proppingforce to base

Moment from propping forceDistance to reaction

Eccentricity ofreaction

Loadedlength ofbaseBearing pressure attoe

Bearing pressure atheel

Fprop.stem : (FtotaLv X lease / 2 - Mtotal) / (hp,op + tease)

Fprop.base : FtotaLh ' Fprop.stem : 83.7 kN/m

Mprop = Fprop.stem X (hprop + tease) = 151 .2 kNm/m

X =(Mtotat+ Mprop)/ FtotaLv= 850 mm

e = x - lb,se/2 = 0 mm

lload = lease = 1700 mm

qt.. = Ftotal.v / lb,s. = 29 kN/m2

qh..i = FtotaLv / leas. = 29 kN/m2

39.3 kN/m

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Effective overburden pressure q = max((tease + d.o«er) X 'b ' - (tbase + d.o«er + hwater) X Yw ', 0 kN/m2) = 0kN/m2

q' = q / 'v = 0 kN/m2

Nq = Exp(7t x tan(+'bd)) x(tan(45 deg + +'bd/ 2))2 = l0.431

Nc = (Nq - 1) X COt(+'b.d) = 20.418

N. = 2 x(Nq - 1) x tan(+'b.d) = 8.712sq = I

s. = ISc = I

H = Fsur:h + Fsat..h + FwateLh + Fmoist.h + Fpass.h ' Fprop.stem ' Fprop.base = 0kN/m

V = Ft.t,1 . = 49.2 kN/mm=2

iq =]1- H/(V + li.,d X C'b.d X cot(0'b.d))]" = I

iv = [1- H/(V + lload X C'bd X COt(+'bd))]("'D = ]

ic = iq -(l- iq) /(Nc x tan($'b.d)) = I

nt = c'b.d x Nc x sc x ic + q ' x Nq x Sq X iq + 0.5 X ('b ' - 'w ') X lioad X Ny X S.x

nt = 60.6 kN/m2

FoSbp = nt / max(qtoe, qheel) = 2.094Allowable bearing pressure exceeds maximum applied bearing pressure

I.r

Design effective overburden pressure

Bearing resistancefactors

Foundation shape factors

Load inclination factors

Net ultimate bearing capacity

FactorofsafetyPASS

RETAINING WALL DESIGN

In accordance with EN1992-1-1 :2004 incorporating Corrigendum dated January 2008 and the UK National Annexincorporating NationaIAmendment No.1

Teddy calculation version 2.6.11

Concrete details - Table 3.1 - Strength and deformation characteristics for concreteConcrete strength class C32/40

Characteristic compressive cylinder strength fck = 32 N/mm2

Characteristic compressive cube strength fck,c.be : 40 N/mm2

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

Mean value of axial tensile strength fct« = 0.3 N/mm2 x (fck / I N/mm2)2/3 = 3.0 N/mm2

5% fractile of axial tensile strength f.tk,o.os : 0.7 x f.t« = 2.1 N/mm2

Secant modulus of elasticity of concrete E.« = 22 kN/mm2 x (f.m / 10 N/mm2)03 = 33346 N/mm2Partial factor for concrete - Table 2. 1 N 'c = 1.50

Compressive strength coefficient - cl.3.1 .6(1) CEcc = 0.85

Design compressive concrete strength - exp.3.15 fcd = CEcc x fck / 'c = 18.1 N/mm2

Maximum aggregate size h,gg = 20 mm

Reinforcement details

Characteristic yield strength of reinforcementModulus of elasticity of reinforcement

Partial factor for reinforcing steel - Table 2. 1 N

Design yield strength of reinforcement

fyk = 500 N/mm2Es = 200000 N/mm2

ys = 1.15

fyd = fvk / 's = 435 N/mm2

Cover to reinforcementFront face of stem

Rear face of stemcsf = 40 mmCsr = 50 mm

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Topface ofbaseBottom face of base

cbt = 50 mrn

cbb = 75 mm

ILoading details - Combination No.1 - kN/m Shear force - Combination No.1 - kN/m Bending moment - Combination No.1 - kNm/m

Loading details-Combination No.2-kN/mi :Shearforce-Combination No.2-kN/m Bending moment - Combination No.2 - kNm/m

30.3

45.€

Check stem design at 1956 mmDepth of section h = 200 mm

Rectangular section in flexure - Section 6.1

Design bending moment combination I

Depth to tension reinforcement

M

d

K

K'

24.9 kNm/m

h - csf - bsx - dstM / 2 = 144 mm

M / (d2 x f.k) = 0.0370.207

K' > K - No compression reinforcement is requiredmin(0.5 + 0.5 x(1- 3.53 x K)'', 0.95) x d = 137 mm

2.5 x (d -- z) = 18 mm

Lever arm

Depth of neutral axis

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Area of tension reinforcement required

Tension reinforcement provided

Area of tension reinforcement provided

Minimum area of reinforcement - exp.9.1 N

Maximum area of reinforcement - cl.9.2.1.1(3)

AstM req : M / (fyd x z) = 41 8 mm2/m

12 dia.bars @ 200 c/c

AsfM.prov : n X QstM2 / (4 x SsfM) = 565 mm2/m

AsfM min : max(0.26 X fctm / fyk, 0.0013) x d = 226 mm2/m

AsfM.m,x : 0.04 x h = 8000 mm2/m

max(AsfM.req, AsfM.min) / AsfM.pro" : 0.74

reinforcement provided is greater than area of reinforcement requiredPASS-.Areaof

Deflection control - Section 7.4

Reference reinforcement ratio

Required tension reinforcement ratio

Required compression reinforcement ratio

Structural system factor - Table 7.4N

Reinforcement factor - exp.7. 17

Limiting span to depth ratio - exp.7.16.a

po: V(f.k/ IN/mm2)/ 1000 = 0.006P = AsfM.req / d = 0.003

P' = AsfM.2.req / d2 = 0.000Kb= I

Ks = min(500 N/mm2 / (fyk X AsfM.req / AsfM.prev), 1.5) = 1.352

Ks x Kb xl11+ 1.5 x V(f.k/ IN/mm2) x Po/ P + 3.2 x V(f.k/ IN/mm2) x

(PO / P - 1)3n] = 59.8hprop / d = 24.3

PASS - Span to depth ratio is less than deflection control limitActual span to depth ratio

Crack control - Section 7.3

Limiting crack widthVariable load factor - EN1990 - Table AI.I

Serviceability bending momentTensile stress in reinforcement

Load duration

Load duration factor

Effective area of concrete in tension

Mean value of concrete tensile strengthReinforcement ratio

Modular ratio

Bond property coefhcientStrain distribution coefficient

w.,x = 0.3 mm

V2 = 0.6M;i; = 17 kNm/m

a; = Msts / (AsfM pro" X Z) = 219.9 N/mm2

Long termkt= 0.4

A.eff = min(2.5 x(h - d),(h -- x)/ 3, h/ 2) = 60667 mm2/mfct..tt = fctm = 3.0 N/mm2

Pp.eff = AsfM.prev / Ac.eff = 0.009

cl. = E;/ E.. = 5.998

kl = 0.8

k2 = 0.5

k3 = 3.4

k4 = 0.425

Sr.max = k3 X Csf + kl X k2 X k4 X @sfM / Pp.eff = 355 mm

Wk: Sr.ma x max(as -- kt X(f.teff/ Ppeff) X(l+ Qe X pp.eff), 0.6 X as)/ Es

wk = 0.234 mm

Wk / Wm,* = 0.78

PASS - Maximum crack width is less than limiting crack width

Maximum crack spacing - exp.7.1 1

Maximum crack width - exp.7.8

Check stem design at base of stemDepth ofsection

Rectangular section in flexure - Section 6.1Design bending moment combination I


h = 200 mm

M

d

K

K'

55.3 kNm/m

h - Csr - Osr / 2 = 144 mm

M / (d2 x f.k) = 0.0830.207

K' > K - No compression reinforcement is required

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Lever arm







z = min(0.5 + 0.5 x(1- 3.53 x K)o ', 0.95) x d = 132 mm

x = 2.5 x (d -- z) = 29 mm

Asr req = M / (fyd x z) = 959 mm2/m12 dia.bars @ 200 c/c

Asr prev : 7t X Qsr2 / (4 X Ssr) = 565 mm2/m

Asrmin = max(0.26 X fctm / fyk, 0.0013) x d = 226 mm2/mAsr.max : 0.04 X h = 8000 mm2/m

max(A:r.req, Asr.min) / Asr p,o« : 1 .696of reinforcement provided is less than area of reinforcement requiredFAIL - Area

Deflection control - Section 7.4

Reference reinforcement ratio

Required tension reinforcement ratio

Required compression reinforcement ratio

Structural system factor - Table 7.4N

Reinforcement factor - exp.7. 1 7

Limiting span to depth ratio - exp.7.16.b

po : V(f.k/ IN/mm2)/ 1000 = 0.006P : Asr.req / d = 0.007

P' = Asr.2 req / d2 = 0.000Kb= I

Ks = mid(500 N/mm2 / (fvk X Asrreq / Asrp,ov), 1 .5) = 0.59Ks x Kb xl11+ 1.5 x V(f.k/ IN/mm2) x Po/(P - P ') + V(fck/ IN/mm2) x

'V(P '/ PO)/ 12] = 10.7hprop/ d = 24.3

FAIL - Span to depth ratio exceeds deflection control limit

Actual span to depth ratio



Serviceability bending momentTensile stress in reinforcementLoad duration




Modular ratio

Bond property coefficientStrain distribution coefficient

w.,x = 0.3 mm

V2 = 0.6M;i; = 38.3 kNm/m

Gs = Msls/(Asrpro« X Z) = 511.5 N/mm2

Long termkt= 0.4

A.eff = min(2.5 x(h - d),(h -- x)/ 3, h/ 2) = 57082 mm2/mfct .ff = fct. = 3.0 N/mm2

Pp.eff = Asr.prov / Ac.eff = 0.010ct. = E;/ E.. = 5.998

kl = 0.8

k2 = 0.5

k3 = 3.4k4= 0.425

Sr.max = k3 X Csr + kl X k2 X k4 X Osr / Ppeff = 376 mm

Wk: Srm«. x max(as -- kt X(fdeff/ Ppeff) X(l+ CEe X Ppeff), 0.6 X as)/ Es

wk = 0.71 8 mmWk / Wm,* = 2.394

FAIL - Maximum crack width exceeds limiting crack width



Rectangular section in shear - Section 6.2

Design shear force V = 94.2 kN/m

CRd,. = 0. 18 / YC = 0.120

k = min(l+ V(200 mm/ d), 2) = 2.000Pi : min(A;r.p,o" / d, 0.02) = 0.004

Vmin = 0.035 NI/2/mm x k3/2 x fckos = 0.560 N/mm2

Longitudinal reinforcement ratio

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York House


Calcs date Checked by14/11/2017

Checked date

Design shear resistance - exp.6.2a & 6.2b vRdc = max(Croc x k x (100 N2/mm4 x Pi x fck)1/3, Vmin) X dvRd.c = 80.6 kN/mV / VRd.. = 1 .1 68

FAIL - Design shear resistance is less than design shear force

Check stem design at propDepth of section h = 200 mm

Rectangular section in shear - Section 6.2Design shear force V = 27.4 kN/m

CRd,. = 0. 18 / 'C = 0.120

k = min(l+ V(200 mm/ d), 2) = 2.000

Longitudinal reinforcement ratio pi = min(Asrl.pr '" / d, 0.02) = 0.004

Vmin = 0.035 NI/2/mm x k3/2 x fck0.5 = 0.560 N/mm2

vnd c= max(CRd c x k x(100 N2/mm4 x pix fck)1/3, Vmin) X d

vRd.c = 80.6 kN/mV / VKd.. = 0.339

PASS - Design shear resistance exceeds design shear forceHorizontal reinforcement parallel to face of stem - Section 9.6

Minimum area of reinforcement - cl.9.6.3(1) Asxreq = max(0.25 X Asrpro«, 0.001 X tste«) = 200 mm2/mMaximum spacing of reinforcement -- cl.9.6.3(2) Ss*..m,. = 400 mmTransverse reinforcement provided 10 dia.bars @ 200 c/c

Area of transverse reinforcement provided Asx pro« : 7t X 0s*2 / (4 x ssx) = 393 mm2/m

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

Design shear resistance - exp.6.2a & 6.2b

Check base design at toeDepth of section h = 350 mm

Rectangular section in flexure - Section 6.1Design bending moment combination I


M = 29.9 kNm/m

d = h - cbb - obb / 2 = 269 mm

K = M / (d2 x f.k) = 0.013K': 0.207

K' > K - No compression reinforcement is required

z = min(0.5 + 0.5 x(1- 3.53 x K)'5, 0.95) x d = 256 mm

x = 2.5 x (d -- z) = 34 mm

Abb.req : M / (fyd x z) = 269 mm2/m

12 dia.bars @ 200 c/c

Abb.prev : it X Qbb2 / (4 x sbb) = 565 mm2/m

Abb.min : max(0.26 X fctm / fyk, 0.0013) x d = 423 mm2/mAbb.max : 0.04 X h = 14000 mm2/m

max(Abb.req, Abb.min) / Abb.prev : 0.748


Lever arm









Serviceability bending momentTensile stress in reinforcement

Load duration

w.,x = 0.3 mm

V2 = 0.6M;i: = 21 .8 kNm/m

as = Msis / (Abb.prev X Z) = 151.1 N/mm2Long term

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Modular ratio

Bond property coefficientStrain distribution coefficient

kt= 0.4

A.etr = min(2.5 x(h - d),(h - x)/ 3, h/ 2) = 105458 mm2/mfct..tf : fctm = 3.0 N/mm2

Pp.eff = Abb.prov / Ac.eff = 0.005

ct. = E;/ E.. = 5.998

kl = 0.8

k2 = 0.5k3 = 3.4

k4 = 0.425

Sr.max = k3 X Cbb + kl X k2 X k.4 X $bb / Pp.eff = 635 mm

Wk: Sr.m«. x max(as -- kt X(f.teff/ Ppeff) X(l+ CEe X Ppeff), 0.6 X as)/ Es

wk = 0.288 mm

Wk / Wm,* = 0.96

PASS - Maximum crack width is less than limiting crack width



Rectangular section in shear - Section 6.2Design shear force V = 39.9 kN/m

CRd,c = 0. 18 / 'C = 0.120

k = min(l+ V(200 mm/ d), 2) = 1.862

Longitudinal reinforcement ratio Pi = min(Abb.pro" / d, 0.02) = 0.002

Vmin = 0.035 NI/2/mm x k3/2 x fck0.5 = 0.503 N/mm2

vRd.c = max(CRd.c x k x (100 N2/mm4 x Pi x fck)1/3, Vmin) X d

vRd.c = 135.3 kN/mV / VRd.. = 0.295

PASS - Design shear resistance exceeds design shear force

Secondary transverse reinforcement to base - Section 9.3

Minimum area of reinforcement -- cl.9.3.1 .1 (2) Abx.req : 0.2 X Abb.prev : 1 13 mm2/m

Maximum spacing of reinforcement -- cl.9.3.1 .1 (3) Sb*...max : 450 mmTransverse reinforcement provided 10 dia.bars @ 200 c/c

Area of transverse reinforcement provided Ab*.prev : 7t X obx2 / (4 x sbx) = 393 mm2/m


Design shear resistance - exp.6.2a & 6.2b

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Start page no./Revision14200mm RC Basement wall

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York House

Predicted Movement 17151

Assessment of predicted ground movement Introduction Predicted settlement of adjacent buildings is calculated on the following pages. This settlement is due to ground movements from excavations and propping at York House basement The ground conditions are 'fine to coarse SAND and fine to coarse FLINT and GRAVEL" (refer to Site Investigation in Appendix B). The graph from CIRIA C512 for settlements in sand has been used in calculations. Long term deflection is not predicted to be minimal, due to the nature of underlying soil (sand/gravel).

23

appendix d preliminary basement calculations

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