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### Transcript of 2. Calculation for Box Culvert Rev.A

Sheet1JAMBI MERANG DEVELOPMENT PROJECTEPC SALES GAS PIPELINE SKN-GRISSIKSP-CL-C-011Rev:ACalculation for Box CulvertDate:15/07/2013

1.DESIGN DATA1.1.MaterialSoilThe soil parameter to be used in the design shall be as follows :Average of qc=19.29kg/cm2(considered from qcavr for BH-26=1890.00kN/m2depth 1-2.2 m)Angle of internal friction, =1.48oCohesion, c = su=61.94KPa(base on Triaxial UU test HB-26)Soil density, gs =19.74kN/m3(base on HB-26, depth 1-4 m)

Soil bearing capacity of structural fill base on Terzaghi Formula:Nq = e(0.75-/2) tanf / (2 cos2(45o+/2)) =1.21ERROR:#REF!Nc = (Nq-1) cotg =8.26Kpg = 3 tan2 {45o+1/2(+33o)} =10.83Ng = 1/2 tan (Kpg /cos2 - 1) =0.13

For square footing: qult = c Nc [1+0.3(B/L)]+ q Nq + 0.5 ' D Ng [1-0.2(B/L)] =615.32kPaUltimate soil bearing capacity, Qa=615.32kPaSoil DensityAngle FrictionUltimate soil bearing capacity for temp., Qtemp=818.38kPa(increased 33%)Depth 1-1.52.0141.478Depth 2.5 - 3--Soil bearing capacity of structural fill base on Soil Lab Investigation DataAllow. Bearing Capacity Q=129.97kN/m2(based on S26, depth 2m)DepthCohesionDepthFriction ResistenceBottom Area of Box Culvert A = B x L=24.00m2118311.63Allow. Bearing Capacity, Qa = Q x A=3,119.28kN1.2209112Allow. Bearing capacity for temp., Qat=4,148.64kN(increased 33%)1.425Friction depth 3-9m100.371.621Soil Bearing Capacity to be used = minimum based from Terzaghi formula & Soil Lab Investigation1.820DepthUpliftAllow. Bearing Capacity, Qa = Q x A=615.32kN21538.14Allow. Bearing capacity for temp., Qat=818.38kN2.216978.75

Concrete19.2857142857Friction depth 3-9m70.61Cement shall conform to SII 0013-81 or ASTM C150-89. Concrete strength, f'c =28.0MPaConcrete unit weight, gc =23.5kN/m3Shallow Boring CapacityModulus Elasticity of concrete, Ec = 4700 f'c =24,870MPa=264.6kN/m2(based on S26, depth 2m)=3.22m2Reinforcing Steel Bar=852.012kNDeformed bar & plain bar shall conform to ASTM A-615-89, 5110136-84. Shallow Boring Capacity=1133.17596kNReinforcement yield strength, fy =413.0MPa=852.012kNModulus elasticity of reinforcement steel, Es =200000MPa

Sand Bedding in Box Culvert=852.012kNSand unit weight, gsand =1400kg/m3gsand =13.7kN/m3Box Culvert will be fill with sand 0.15m heighthsand=0.15m

1.2. Equipment DataBox Culvert Equipment load = load of 16" Sales Gas pipeline16" Sales Gas pipeline weight=93kg/m1.3. Box Culvert DimensionLength netLn=12.000mHeight netHn=1.000mthick of wall=0.250mthick of soil, t=1.656mLength gross, L =12.000mthick of slab=0.250m(top & bottom)Height to Bottom, H =3.156mWidth netBn=1.500mthick of wall=0.250mDepth of soil cover, Hds =1.656mWidth, B =2.000mHeight of Foundation, Df = 1.500m

s

1.4. Basic AssumptionThe footing assume to be rigid. This rigidity is necessary to keep the assumption soil pressure will be linearly distributed along footing. In designing reinforcement and checking strength of foundationaccording to member forces occured in foundation, the loads shall be considered in factored loadcondition and calculation shall be done in ultimate condition.Section ViewLongitudinal View

2.1.Dead Load (D) per m'Weight of top slab, WgR = L B t top slab gsteel =1*2.000*0.250*23.50=11.75kNWeight of slab, WgS = L B t slab gc =1*2.000*0.250*23.50=11.75kNWeight of Long Wall , Wglw = L H t wall gc *2 = 1*1.500*0.250*23.50* 2=17.63kNWeight of soil above roof, Wgs = (B L)t. gs =2.000*1.000*1.656*"19.74=65.37kNWeight of sand in Box Culvert, Wgsand = (Ln Bn hsand) gsand =12.000*1.500* 0.15* 13.72=3.09kNWeight of Box Culvert, WFD = Wgr + WgS + Wglw + Wgsw + Wgs + Wgsand ==109.58kNnote : Box Culvert will be fill with sand until 15cm height

2.2.Live Loads (L)Live Load for operating & maintenance as per SKBI 1.53 1987Roof=100kg/m2LR = B x L x 100=2*1*100=2400kg=23.52kNSlab=400kg/m2LS = B x L x 400=2*1*400=9600kg=94.08kNLive Load=117.6kNSurcharge Live Load400kg/m2note: Live load surcharge will be included into lateral soil load2.3.Equipment Loads (E)Box Culvert Equipment load = load of 16" Sales Gas pipeline16" Sales Gas pipeline weight (E)=12 x 93=1116kg=10.94kN

2.4.Wind Loads (W: Wx = Wz))The Wind Load is not aplicable to Box Culvert analysis because location of Box Culvert is in undergroundVelocity Pressure (qz)qz = 0.613 * Kz * Kzt * Kd * V2 * I

Velocity pressure exposure coefficient Kz 0.85As per structure height above ground (ASCE 7/02)Topographic FactorKzt1Wind directional FactorKd0.85(ASCE 7/02 Buildings, Solid Signs,Trussed Tower)Basic Wind SpeedV35m/s Important FactorI1.15(ASCE 7/02 Category III)Directional90degreesqz = 0.613 x 0.85 x 1 x 0.85 x 35^2 x 1.15=623.924809375N/m20.6239248094kN/m2Wind Force (p)p = qz * G * CpGuest effect factorG0.85Pressure CoefficientCp(Figure 6-21 ASCE-7-02)Pedestal: square (wind normal to face)h/d=ERROR:#REF!=ERROR:#REF!, Cp=1.3Wind Load Fwl=0.624*0.85*1.3=0.69kNMoment=Fwl x (Df + Hs1)=0.69 x ( + )=0.00kNm

2.5Seismic Load (EQ) - SX, SZSeismic Load for Box Culvert Refer to UBC 1997, as shown in Attachment 3:The static seismic load are as follow:

where ;Vs: Basic shear nominal static loadCa: Seismic Coefficient=0.3I: Important factor=1.25Table 16-K Occupancy CategoryR: Seismic reduction factor=2.2Table 16-PWi: Weigth of Box Culvert=109.58kN from Dead Load of Foundation

=46.70kN

2.6Earth Pressure

`

Specific gravityGs =2.591kg/m3LLs=3.9kPavoid ratioe =0.688Htotal=3.156mSaturated soil densityg sat =34.9656398104kg/m3H=1.500mSoil densityg s =19.74kN/m3d=1.656mShear anglef =1.48Undrained Shear Strengthc =0.632kg/cm2Ground Water LevelGWL =0Back Fill Densityg b =18kN/m3

Active earth pressure coefficient, Ka = tan2 (45o - f/2) Ka=0.950

Pressure due to weight of soil & live load per meter width, P1 = Pa*d*+qLLs*Ka =32.01kNPressure due to soil lateral load per meter width, P2 = 0.5*Pa*H =14.06kNPressure due to seismic per meter width, P3 = 0.5*Pe*H =2.92kN48.99kN

LoadingFx =FzFyMx = Mz(kN)(kN)(kNm)LC 01D + L + E(E) + S48.990238.1260.000LC 02D + L + E(E) + EQ + S95.687238.1260.000`LC 03D + L + E(O) + S48.990238.1260.000LC 04D + L + E(O) + EQ + S95.687238.1260.000

LoadingFx =FzFyMx = Mz(kN)(kN)(kNm)LC 111.4 [D+L+E(E)+S]68.586333.3760.000LC 121.2 [D+L+E(E)+S] + 1.4 EQ124.163285.7510.000LC 131.4 [D+L+E(O)+S]68.586333.3760.000LC 141.2 [D+L+E(O)+S] + 1.4 EQ124.163285.7510.000

4. FOUNDATION DESIGN4.1Soil StressBearing capacity of soil is taken from Geotechnical Survey Report.For exmaple LC 01,eX = MZ / FY=0/238.13=0m4.8STABILITY CHECKeZ = MX / FY=0/238.13=0m4.8.1BEARING CAPACITYq = (FY / BF LF) (1 6 eX / LF 6 eZ / BF)qmax =238.13/(2.000*12.000) * (1 + 6*0.000/12.000 + 6*0.000/2.00)=9.92kPaThe soil bearing stress analyzed by ignoring the GWL that caused uplift, increasing the soil bearing capacity.qmin =238.13/(2.000*12.000) * (1 - 6*0.000/12.000 - 6*0.000/2.00)=9.92kPaMax Va=333.38kgSF = Qa / qmax=615.32 / 9.92=62.0165905669>3.00 OK !Base Area of valve box, A=24m2The results for all Load Combinations are presented in the following TableLoad CombinationeXeZqmaxqminSFmmkPakPaLC 010.00.09.929.9262.023.00OK =P/A =13.8907kg/m2LC 020.00.09.929.9282.482.00OK=0.0014kg/cm2< all =615.323kNLC 030.00.09.929.9282.482.00OK(Taking the lowest from pre design survey report )LC 040.00.09.929.9282.482.00OK .< all...OK!!the allowable bearing pressure is adequateLC 110.00.013.8913.8958.922.00OKLC 120.00.011.9111.9168.742.00OK4.8.2ANTI BOUYANCY CHECKLC 130.00.013.8913.8958.922.00OKLC 140.00.011.9111.9168.742.00OKThe anti bouyancy check analyzed with the worst GWL condition (at -0,5m from GL), for conservatively control the structure's stability

4.2Stability Against Overturning and SlidingTotal Dead Load (Va)=333.38kgExample For LC 01,Bouyancy (uplift) force=Lw * Bw * h2 * g w The stability against overturning about X axis:=ERROR:#REF!kgMRX = FY L / 2=238.13*12.00/2=1,428.75kN.m=ERROR:#REF!kgSafety factor, SF = MRX / MX =1,428.75 / 0.00 =->2.00 OK !The stability against overturning about Y axis:Safety Factor for Buoyance=ERROR:#REF!MRZ = FY B / 2=238.13*2.00/2=238.13kN.m=ERROR:#REF!ERROR:#REF!1.1ERROR:#REF!Safety factor, SF = MRZ / MZ =238.13/0.00=->2.00 OK !The stability against sliding:FR = FY tan 2/3 f + c B L=238.13*tan(2/3*1.48) + 61.94*2.000*12.000=1,490.56kNSafety factor, SF = FR / H =1,490.56 / 48.99=30.43>2.00 OK !The results for all Load Combinations are presented in the following Table

Load CombinationSFResultMRXSFMRZSFFRActualCriteriakN.mActualCriteriakN.mActualCriteriakN.mLC 0130.42.0OK1428.8-2.0238.1-2.01488.3LC 0215.61.5OK1428.8-1.5238.1-1.51488.5LC 0330.31.5OK1428.8-1.5238.1-1.51486.7LC 0415.51.5OK1428.8-1.5238.1-1.51486.7

LC 1121.81.5OK2000.3-1.5333.4-1.51492.2LC 1212.01.5OK1714.5-1.5285.8-1.51491.4LC 1321.81.5OK2000.3-1.5333.4-1.51492.2LC 1412.01.5OK1714.5-1.5285.8-1.51491.4

4.3SettlementThe soil parameter of fill material to be used in this settlement calculation shall be as follow:For elastic or immediate settlement:For this calculation, elastic settlement considered to be zero. For this case, the soil should be well compacted during construction in order to neglect the settlement.Poisson's ratio, m=0.3(assuming) ]=\Modulus of Elasticity, Es=9450kPa(from 5 qc avr)B/L=m'=0.2H/(L/2)=n'=0.5A0=m' ln((1+sqrt(m'2+1)sqrt(m'2+n'2)/m'(1+sqrt(m'2+n'2+1))=0.2457085589A1=ln((m'+sqrt(m'2+1))sqrt(1+n'2)/(m'+sqrt(m'2+n'2+1))=-0.6978191953A2=m'/(n'sqrt(m'2+n'2+1))=0.2774269568F1=1/p (A0 + A1)=-0.144F2=n' / 2p tan-1 A2=0.023Is=F1 + (1-2m)F2/(1-m)=-0.131Df/B=0.750If=0.98a=4(center of foundation)B'=1(center of foundation)qn=13.9KPaElastic settlement, Se=qnaB'(1-m2)IsIf/Es=-0.0006867137m=-0.687mm

For consolidation settlement:This consolidation settlement considering normally consolidation as follow:Sc=HCclogp0 + p1+e0p0Distribution of p will be describe as fogure below.

03.16m

-3.2m120.1m-3.3m

0.2m-3.5m

BF'xLF'zHip0pCce0logp0+pSc(m2)(m)(m)(kN/m2)(kN/m2)p0mm24.001.500029.60613.8910.16810.167024.701.5500.130.59313.500.16810.1591.332718774726.841.6500.232.56612.420.16810.1402.36326942Sc tot =3.6913.526432Note:Depth of settle layer is assumed until p0> max p.3.52.55824

Total settlement, Stot=Se + Sc=3.004mm(OK < 25.4 mm)

4.4Anti Buoyancy Check

The anti bouyancy check analyzed with the worst GWL condition (at -0,5m from GL), for conservatively control the structure's stability

Min Dead Load (Va)=109.58kN=11,181.80kgWeight of StructurreBouyancy (uplift) force=Lw * Bw * h2 * g w =1 * 2 * 1.5 * 1000=3000kg

Safety Factor for Buoyance=11182 / 3000=3.73>1.32OK!

5CONCRETE REINFORCEMENT DESIGN5.1ROOF DESIGN

Lx = 2.000

Ly =12.000

Since the slab ratio is Ly/Lx < 2.5 we can assumed as two ways actionLy / Lx =6.028.97cm2 / m'

Sketch:

1,656

Dia. 13 @ 200mmtop and bottom

5.1WALL DESIGN5.1.1Wall thickness bearing control

2000 mmLC 131.4 [D+L+E(O)+S]56.8124308721229.457340LC 141.2 [D+L+E(O)+S] + 1.4 EQ123.6046221598196.677720

12000 mmLength, L =12.000mWidth, B =2.000mArea of slab, A =24.000m2Area a1 & a2 =11.198m2Area a3 & a4 =4.000m2

Distribution of load to the long wallRatio the load to the long wall, RL =a1 / A =0.467Ratio the load to the short wall, RS =a3 / A =0.167DL = Weight of roof slab x LB x RL=12 * 2 *1.656*2400*0.467=44,506kgLL = LL roof slab x LB x RL =450 * 0.47 *12*2=5,039kg

Axial load, P ult = 1.2DL + 1.6LL ==1.2*44506 + 1.6*5039=61,470kg=61.470T

Area wall, Aw1 = tw . L ==25*1200=30,000cmArea wall, Aw2 = tw . B ==25*200=5,000cm

=0.6Axial long wall capacity = .Aw1.fc' =0.6*30000*344=6192010.40963856kg=6192.010TAxial short wall capacity = .Aw2.fc' =0.6*5000*344=1032001.73493976kg=1032.002T

5.1.2Wall Flexural Design

5.1.2.1For the Long WallL=12.0mH=1.0mL/H=12.00 1.4/fy min0.00180r min =0,18% =0.0018; if 1.33 x r req < 1.4/fyr balance =bi * 0.85 fc' * 600 b0.0290 fy 600 + fy

r max =0,75*r balance max0.0218

Check :r min>r req3.7cm2 / m'

Sketch:c = 75mm

inner sideouter side

c =40 mmDia. 13 @ 200mmVer and Hor

250Check shear capacity for the wallVu = max of R01&R02 =167.27kgConcrete shear capacity, Vc:

Vc = 1/3 fc' bo.d

=0.6bo =1000mmd =tw - d'd' =0mmd =250mmVc =264,575NVc =26,997kgVc > VuOK!

5.1.2.2For the Short WallB=2.000mH=3.156mH/B=0.63 1.4/fy min0.00180r min =0,18% =0.0018; if 1.33 x r req < 1.4/fyr balance =bi * 0.85 fc' * 600 b0.0290 fy 600 + fy

r max =0,75*r balance max0.0218

Check :r min>r req3.7cm2 / m'

Sketch:c = 75mm

inner sideouter side

c =40 mmDia. 13 @ 200mmVer and Hor

250Check shear capacity for the wallVu = max of R01&R02 =3,167.43kgConcrete shear capacity, Vc:

Vc = 1/3 fc' bo.d

=0.6bo =1000mmd =tw - d'd' =0mmd =250mmVc =264,575NVc =26,997kgVc > VuOK!

5.2SLAB DESIGN

Lx = 2.000

Ly =12.000

Since the slab ratio is Ly/Lx < 2.5 we can assumed as two ways actionLy / Lx =6.0ERROR:#REF!cm2 / m'

Sketch:

250

Dia. 13 @ 200mmtop and bottom

5.FOUNDATION REINFORCEMENT DESIGN5.1Footing Reinforcement DesignDistribution of loads below the footing shall be determine as follow:Z DirectionX Direction

Hd : Concrete cover =0.05md : effective height of footing = HF - Hd =ERROR:#REF!ma1 : length of bending moment in z direction = (WF-WP)/2 =ERROR:#REF!ma2 : length of shear in z direction = a1 - d =ERROR:#REF!mb1 : length of bending moment in x direction = (LF - LP)/2 =ERROR:#REF!mb2 : length of shear in x direction = b1 - d =ERROR:#REF!m

For X Direction:eX = MZ / FYqXmax = (FY / WF LF) (1 + 6 eX / LF)=q2qXmin = (FY / WF LF) (1 - 6 eX / LF)=q"2

For Z Direction:eZ = MX / FYqZmax = (FY / WF LF) (1 + 6 eZ / WF)=q1qZmin = (FY / WF LF) (1 - 6 eZ / WF)=q"1

Load CombinationX Direction = Y DirectioneXq2q"2q'2mkPakPakPaERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 010.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 020.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 030.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 040.0ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

LC 110.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 120.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 130.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 140.0ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

Base on distribution load above, moment and shear in section of footing shall be listed below.Msux=[1/2 x q1 x a12 + (1/3) x a12 x (q1-q1')] x LFVsuz=0.5*(q1+q1')*a1*LFMsuz=[1/2 x q2 x b12 + (1/3) x b12 x (q2-q2')] x WFVsux=0.5*(q2+q2')*b1*WF

Load CombinationMsuz=MsuxVsuz=VsuxkNmkNERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 01ERROR:#REF!ERROR:#REF!LC 02ERROR:#REF!ERROR:#REF!LC 03ERROR:#REF!ERROR:#REF!LC 04ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

LC 11ERROR:#REF!ERROR:#REF!LC 12ERROR:#REF!ERROR:#REF!LC 13ERROR:#REF!ERROR:#REF!LC 14ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

Vs= (2.5 x Ca x I )/R x WiSeismic Load(Fsl)= (2.5 x 0.3 x1.25)/2.2 x 109,58

Coba

TBBM-09-CIV-CAL-010-A4Rev:ACalculation for Underground FacilitiesDate:143/08/2015

1.DESIGN DATA1.1.MaterialSoilThe soil parameter to be used in the design shall be as follows :Average of qc=4.57kg/cm2(considered from qcavr for S3=448.00kN/m2depth 1-2.2 m)Angle of internal friction, =1.48oCohesion, c = su=61.94KPa(base on Triaxial UU test DB-3)Soil density, gs =19.74kN/m3(base on DB-3, depth 1-4 m)

Soil bearing capacity of structural fill base on Terzaghi Formula:Nq = e(0.75-/2) tanf / (2 cos2(45o+/2)) =1.21ERROR:#REF!Nc = (Nq-1) cotg =8.26Kpg = 3 tan2 {45o+1/2(+33o)} =10.83Ng = 1/2 tan (Kpg /cos2 - 1) =0.13

For square footing: qult = c Nc [1+0.3(B/L)]+ q Nq + 0.5 ' D Ng [1-0.2(B/L)] =600.97kPaUltimate soil bearing capacity, Qa=600.97kPaSoil DensityAngle FrictionUltimate soil bearing capacity for temp., Qtemp=799.29kPa(increased 33%)Depth 1-1.52.0141.478Depth 2.5 - 3--Soil bearing capacity of structural fill base on Soil Lab Investigation DataAllow. Bearing Capacity Q=129.97kN/m2(based on S26, depth 2m)DepthCohesionDepthFriction ResistenceBottom Area of Box Culvert A = B x L=5.20m216311.63Allow. Bearing Capacity, Qa = Q x A=675.84kN1.269112Allow. Bearing capacity for temp., Qat=898.87kN(increased 33%)1.44Friction depth 3-9m100.371.64Soil Bearing Capacity to be used = minimum based from Terzaghi formula & Soil Lab Investigation1.84DepthUpliftAllow. Bearing Capacity, Qa = Q x A=600.97kN2438.14Allow. Bearing capacity for temp., Qat=799.29kN2.24978.75

Concrete4.5714285714Friction depth 3-9m70.61Cement shall conform to SII 0013-81 or ASTM C150-89. Concrete strength, f'c =28.0MPaConcrete unit weight, gc =23.5kN/m3Shallow Boring CapacityModulus Elasticity of concrete, Ec = 4700 f'c =24,870MPa=264.6kN/m2(based on S26, depth 2m)=3.22m2Reinforcing Steel Bar=852.012kNDeformed bar & plain bar shall conform to ASTM A-615-89, 5110136-84. Shallow Boring Capacity=1133.17596kNReinforcement yield strength, fy =413.0MPa=852.012kNModulus elasticity of reinforcement steel, Es =200000MPa

Sand Bedding in Box Culvert=852.012kNSand unit weight, gsand =1400kg/m3gsand =13.7kN/m3Box Culvert will be fill with sand 0.15m heighthsand=0.15m

1.2. Equipment DataBox Culvert Equipment load = load of Electrical & Instrument Cable Electrical & Instrument Cable weight=100kg/m1.3. Box Culvert DimensionLength netLn=4.000mHeight netHn=1.080mthick of wall=0.150mthick of soil, t=0.200mLength gross, L =4.000mthick of slab=0.150m(top )thick of slab=0.150m( bottom)Width netBn=1.000mHeight to Bottom, H ==1.580mthick of wall=0.150mthick of slab wall=0.150m(side)Width, B =1.300mHeight of Foundation, Df = 1.380m

s

1.4. Basic AssumptionThe footing assume to be rigid. This rigidity is necessary to keep the assumption soil pressure will be linearly distributed along footing. In designing reinforcement and checking strength of foundationaccording to member forces occured in foundation, the loads shall be considered in factored loadcondition and calculation shall be done in ultimate condition.Section ViewLongitudinal View

2.1.Dead Load (D) Back Fill=0 m *18 kN/m3=0kN/m2SW of Top Slab=0.15 m *24 kN/m3=3.60kN/m2SW of Bottom Slab=0.15 m *24 kN/m3=3.60kN/m2SW of Side Wall=0.15 m *24 kN/m3=3.60kN/m210.8kN/m2

2.3.Equipment Loads (E)Box Culvert Equipment load = load of Electrical & Instrument Cable Electrical & Instrument Cable weight (E)=4 x 100=400kg=3.92kN

2.4.Wind Loads (W: Wx = Wz))The Wind Load is not aplicable to Box Culvert analysis because location of Box Culvert is in undergroundVelocity Pressure (qz)qz = 0.613 * Kz * Kzt * Kd * V2 * I

Velocity pressure exposure coefficient Kz 0.85As per structure height above ground (ASCE 7/02)Topographic FactorKzt1Wind directional FactorKd0.85(ASCE 7/02 Buildings, Solid Signs,Trussed Tower)Basic Wind SpeedV35m/s Important FactorI1.15(ASCE 7/02 Category III)Directional90degreesqz = 0.613 x 0.85 x 1 x 0.85 x 35^2 x 1.15=623.924809375N/m20.6239248094kN/m2Wind Force (p)p = qz * G * CpGuest effect factorG0.85Pressure CoefficientCp(Figure 6-21 ASCE-7-02)Pedestal: square (wind normal to face)h/d=ERROR:#REF!=ERROR:#REF!, Cp=1.3Wind Load Fwl=0.624*0.85*1.3=0.69kNMoment=Fwl x (Df + Hs1)=0.69 x ( + )=0.00kNm

2.5Seismic Load (EQ) - SX, SZSeismic Load for Box Culvert Refer to UBC 1997, as shown in Attachment 3:The static seismic load are as follow:

where ;Vs: Basic shear nominal static loadCa: Seismic Coefficient=0.3I: Important factor=1.25Table 16-K Occupancy CategoryR: Seismic reduction factor=2.2Table 16-PWi: Weigth of Box Culvert=10.80kN from Dead Load of Foundation

=4.60kN

2.6Earth Pressure

`

Specific gravityGs =2.591kg/m3LL=10.00kPavoid ratioe =0.688Htotal=1.580mSaturated soil densityg sat =34.9656398104kg/m3H=1.500mSoil densityg s =19.74kN/m3d=0.200mShear anglef =1.48Undrained Shear Strengthc =0.632kg/cm2Ground Water LevelGWL =0Back Fill Densityg b =18kN/m3

Active earth pressure coefficient, Ka = tan2 (45o - f/2) Ka=0.950

Pressure due to weight of soil fill P1 = Pa1*H =3.419kN/m'Pressure due to soil lateral load , P2 = 0.5*Pa2*H =12.934kN/m'Pressure due to live load , P3 = q Ka H =13.106kN/m'Pressure due to seismic per meter width, P4 = 0.5*Pe*Htot =3.636kN/m'33.09kN/m'

2.7Base Pressure

Dead Load from top slab and wall including back fill =14.331kN/m2Sand bedding =2.058kN/m2EquipmentW /2prL =0.33/(2*pi()*0.406*1.0m) =0.128kN/m2Live Load =10.000kN/m2Base Pressure26.517kN/m2

AEB

10001080

DFC

LoadingFx =FzFyMx = Mz(kN)(kN)(kNm)LC 01D + L + E(E) + S33.09530.1710.000LC 02D + L + E(E) + EQ + S37.69730.1710.000`LC 03D + L + E(O) + S33.09530.1710.000LC 04D + L + E(O) + EQ + S37.69730.1710.000

LoadingFx =FzFyMx = Mz(kN)(kN)(kNm)LC 111.4 [D+L+E(E)+S]46.33242.2390.000LC 121.2 [D+L+E(E)+S] + 1.4 EQ46.15736.2050.000LC 131.4 [D+L+E(O)+S]46.33240.2500.000LC 141.2 [D+L+E(O)+S] + 1.4 EQ46.15734.5000.000

4. FOUNDATION DESIGN4.1Soil StressBearing capacity of soil is taken from Geotechnical Survey Report.For example LC 01,eX = MZ / FY=0/30.17=0m4.8STABILITY CHECKeZ = MX / FY=0/30.17=0m4.8.1BEARING CAPACITYq = (FY / BF LF) (1 6 eX / LF 6 eZ / BF)qmax =30.17/(1.300*4.000) * (1 + 6*0.000/4.000 + 6*0.000/1.30)=5.80kPaThe soil bearing stress analyzed by ignoring the GWL that caused uplift, increasing the soil bearing capacity.qmin =30.17/(1.300*4.000) * (1 - 6*0.000/4.000 - 6*0.000/1.30)=5.80kPaMax Va=40.25kgSF = Qa / qmax=600.97 / 5.80=103.5790620738>3.00 OK !Base Area of valve box, A=5.2m2The results for all Load Combinations are presented in the following TableLoad CombinationeXeZqmaxqminSFmmkPakPaLC 010.00.05.805.80103.583.00OK =P/A =7.7403kg/m2LC 020.00.05.805.80137.762.00OK=0.0008kg/cm2< all =600.971kNLC 030.00.05.805.80137.762.00OK(Taking the lowest from pre design survey report )LC 040.00.05.805.80137.762.00OK .< all...OK!!the allowable bearing pressure is adequateLC 110.00.08.128.1298.402.00OKLC 120.00.06.966.96114.802.00OK4.8.2ANTI BOUYANCY CHECKLC 130.00.07.747.74103.262.00OKLC 140.00.06.636.63120.472.00OKThe anti bouyancy check analyzed with the worst GWL condition (at -0,5m from GL), for conservatively control the structure's stability

4.2Stability Against Overturning and SlidingTotal Dead Load (Va)=40.25kgExample For LC 01,Bouyancy (uplift) force=Lw * Bw * h2 * g w The stability against overturning about X axis:=ERROR:#REF!kgMRX = FY L / 2=30.17*4.00/2=60.34kN.m=ERROR:#REF!kgSafety factor, SF = MRX / MX =60.34 / 0.00 =->2.00 OK !The stability against overturning about Y axis:Safety Factor for Buoyance=ERROR:#REF!MRZ = FY B / 2=30.17*1.30/2=19.61kN.m=ERROR:#REF!ERROR:#REF!1.1ERROR:#REF!Safety factor, SF = MRZ / MZ =19.61/0.00=->2.00 OK !The stability against sliding:FR = FY tan 2/3 f + c B L=30.17*tan(2/3*1.48) + 61.94*1.300*4.000=322.59kNSafety factor, SF = FR / H =322.59 / 33.09=9.75>2.00 OK !The results for all Load Combinations are presented in the following Table

Load CombinationSFResultMRXSFMRZSFFRActualCriteriakN.mActualCriteriakN.mActualCriteriakN.mLC 019.72.0OK60.3-2.019.6-2.0322.3LC 028.51.5OK60.3-1.519.6-1.5322.3LC 039.71.5OK60.3-1.519.6-1.5322.1LC 048.51.5OK60.3-1.519.6-1.5322.1

LC 117.01.5OK84.5-1.527.5-1.5322.8LC 127.01.5OK72.4-1.523.5-1.5322.7LC 137.01.5OK80.5-1.526.2-1.5322.8LC 147.01.5OK69.0-1.522.4-1.5322.7

4.3SettlementThe soil parameter of fill material to be used in this settlement calculation shall be as follow:For elastic or immediate settlement:For this calculation, elastic settlement considered to be zero. For this case, the soil should be well compacted during construction in order to neglect the settlement.Poisson's ratio, m=0.3(assuming) ]=\Modulus of Elasticity, Es=2240kPa(from 5 qc avr)B/L=m'=0.3H/(L/2)=n'=0.8A0=m' ln((1+sqrt(m'2+1)sqrt(m'2+n'2)/m'(1+sqrt(m'2+n'2+1))=0.3007406319A1=ln((m'+sqrt(m'2+1))sqrt(1+n'2)/(m'+sqrt(m'2+n'2+1))=-0.332821282A2=m'/(n'sqrt(m'2+n'2+1))=0.312800716F1=1/p (A0 + A1)=-0.010F2=n' / 2p tan-1 A2=0.038Is=F1 + (1-2m)F2/(1-m)=0.012Df/B=1.062If=0.98a=4(center of foundation)B'=0.65(center of foundation)qn=8.1KPaElastic settlement, Se=qnaB'(1-m2)IsIf/Es=0.0000972784m=0.097mm

For consolidation settlement:This consolidation settlement considering normally consolidation as follow:Sc=HCclogp0 + p1+e0p0Distribution of p will be describe as fogure below.

01.58m

-1.6m120.1m-1.7m

0.2m-1.9m

BF'xLF'zHip0pCce0logp0+pSc(m2)(m)(m)(kN/m2)(kN/m2)p0mm5.201.380027.2378.1230.310.11305.471.4300.128.2247.730.310.1051.58271877476.301.5300.230.1986.700.310.0872.61326942Sc tot =4.1883.526432Note:Depth of settle layer is assumed until p0> max p.3.52.55824

Total settlement, Stot=Se + Sc=4.286mm(OK < 25.4 mm)

4.4Anti Buoyancy Check

The anti bouyancy check analyzed with the worst GWL condition (at -0,5m from GL), for conservatively control the structure's stability

Min Dead Load (Va)=ERROR:#REF!kN=ERROR:#REF!kgWeight of StructurreBouyancy (uplift) force=Lw * Bw * h2 * g w =1 * 1.3 * 1.5 * 1000=1950kg

Safety Factor for Buoyance=ERROR:#REF!=ERROR:#REF!ERROR:#REF!1.32ERROR:#REF!

5.0Moment Calculation

5.1Top SlabThe Top slab end do not fixed to the wall so the top slab can be lifted to put the gas pipe.

Fixed End Moment due to top slab load=0=0kNm(pinned - pinned)Mid Span Moment due to top slab load=(1.4 x (DLtop+ LL + E)/8 x L^2=4.257904kNm(pinned - pinned)

5.2Bottom SlabThe Bottom slab end fixed to the wall

Fixed End Moment due to bottom slab load=(1.4 x (DLbot+ LL + E)/12 x L^2=3.0936526362kNm(fixed - fixed)Mid Span Moment due to bottom slab load=(1.4 x (DLbot+ LL + E)/24 x L^2=1.5468263181kNm(fixed - fixed)

5.5Side wallThe one Side end fixed to the wall but the other end is free

Fixed End Moment due to side wall load=(1.4 x (DLtop+ LL + E)/8 x L^2=4.9664192256kNm(fixed - pinned)Mid Span Moment due to side wall load=(1.4 x (DLtop+ LL + E)*9/128 x L^2=2.7936108144kNm(fixed - pinned)

Thickness Control Top Slab minimum=(4.26x1e6/1000)=65.2526168058mm2.76cm2 / m'

Sketch:

200

Dia. 13 @ 200mmtop and bottom

6.1WALL DESIGN6.11Wall thickness bearing control

Calculation per 1 meter of wall

1300 mmLC 131.4 [D+L+E(O)+S]56.8124308721229.457340LC 141.2 [D+L+E(O)+S] + 1.4 EQ123.6046221598196.677720

1000 mm

Length, L =1.000mWidth, B =1.300mArea of slab, A =1.300m2Area a1 & a2 =0.650m2Area a3 & a4 =0.217m2

Distribution of load to the long wallRatio the load to the long wall, RL =a1 / A =0.500Ratio the load to the short wall, RS =a3 / A =0.167DL = Weight of roof slab x LB x RL=1 * 1.3 *0.2*2400*0.5=312kgLL = LL roof slab x LB x RL =450 * 0.5 *1*1.3=293kg

Axial load, P ult = 1.2DL + 1.6LL ==1.2*312 + 1.6*293=842kg=0.842T

Area wall, Aw1 = tw . L ==15*100=1,500cmArea wall, Aw2 = tw . B ==15*130=1,950cm

=0.6Axial long wall capacity = .Aw1.fc' =0.6*1500*344=309600.520481928kg=309.601TAxial short wall capacity = .Aw2.fc' =0.6*1950*344=402480.676626506kg=402.481T

6.1.2Wall Flexural Design

6.1.2.1WallL=1.0mH=1.1mL/H=0.93 1.4/fy min0.00180r min =0,18% =0.0018; if 1.33 x r req < 1.4/fyr balance =bi * 0.85 fc' * 600 b0.0290 fy 600 + fy

r max =0,75*r balance max0.0218

Check :r min>r req2.1cm2 / m'

Sketch:c = 75mm

inner sideouter side

c =30 mmDia. 10 @ 200mm vertical

Dia. 10 @ 200mm Horizontal

150Check shear capacity for the wallVu = max of R01&R02 =403.96kgConcrete shear capacity, Vc:

Vc = 1/3 fc' bo.d

=0.6bo =1000mmd =tw - d'd' =0mmd =150mmVc =158,745NVc =16,198kgVc > VuOK!

6.2BOTTOM SLAB DESIGN

Lx = 1.000

Calculation per 1 m length of the wall

Ly =1.300

Since the slab ratio is Ly/Lx < 2.5 we can assumed as two ways actionLy / Lx =1.31.84cm2 / m'

Sketch:

150

Dia. 10 @ 200mmtop and bottom

6.3TOP SLAB DESIGN

Lx = 1.000

Calculation per 1 m length of the wall

Ly =1.000

Since the slab ratio is Ly/Lx < 2.5 we can assumed as two ways actionLy / Lx =1.01.84cm2 / m'

Sketch:

150

Dia. 10 @ 200mmDia. 10 @ 200mm

5.FOUNDATION REINFORCEMENT DESIGN5.1Footing Reinforcement DesignDistribution of loads below the footing shall be determine as follow:Z DirectionX Direction

Hd : Concrete cover =0.05md : effective height of footing = HF - Hd =ERROR:#REF!ma1 : length of bending moment in z direction = (WF-WP)/2 =ERROR:#REF!ma2 : length of shear in z direction = a1 - d =ERROR:#REF!mb1 : length of bending moment in x direction = (LF - LP)/2 =ERROR:#REF!mb2 : length of shear in x direction = b1 - d =ERROR:#REF!m

For X Direction:eX = MZ / FYqXmax = (FY / WF LF) (1 + 6 eX / LF)=q2qXmin = (FY / WF LF) (1 - 6 eX / LF)=q"2

For Z Direction:eZ = MX / FYqZmax = (FY / WF LF) (1 + 6 eZ / WF)=q1qZmin = (FY / WF LF) (1 - 6 eZ / WF)=q"1

Load CombinationX Direction = Y DirectioneXq2q"2q'2mkPakPakPaERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 010.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 020.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 030.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 040.0ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

LC 110.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 120.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 130.0ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 140.0ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

Base on distribution load above, moment and shear in section of footing shall be listed below.Msux=[1/2 x q1 x a12 + (1/3) x a12 x (q1-q1')] x LFVsuz=0.5*(q1+q1')*a1*LFMsuz=[1/2 x q2 x b12 + (1/3) x b12 x (q2-q2')] x WFVsux=0.5*(q2+q2')*b1*WF

Load CombinationMsuz=MsuxVsuz=VsuxkNmkNERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!LC 01ERROR:#REF!ERROR:#REF!LC 02ERROR:#REF!ERROR:#REF!LC 03ERROR:#REF!ERROR:#REF!LC 04ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

LC 11ERROR:#REF!ERROR:#REF!LC 12ERROR:#REF!ERROR:#REF!LC 13ERROR:#REF!ERROR:#REF!LC 14ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!ERROR:#REF!

= (2.5 )/ ()= (2.5 0.3 1.25)/2.2 35,81a2a1

Let the thicness of Horizontal slab320mm=0.32mVertical wall thicness 250mm=0.25mEffective slab span 0.85+0.32=1.17mEffective Height of wall1.5+0.25=1.75m

h1.177333.33333333331.756000hD FC17833.33333333337333.333333333310500w=47628.717948718N/m2Fig 1

Distribution factore for AD and DA=1=2/3Distribution factore for AB and DC=1/2=1/31+1/21+1/2Fix end moments will be as under :MFAB=wL2=29680x1.752=-7574.5833333333N - m1212Mfdc=+wL2=47628.717948718x1.752=12155.2457264957N - m1212MFAD =+pL2+WLWhere W is the total tringular earth pressure.1215MFAD =+7333.3333333333x1.752+10500x1.75x1.75=2944N-m12215MFDA =-pL2-WL1215MFDA =-7333.3333333333x1.752--10500x1.75x1.75=-1872-1608=-347912210The Moment distribution is carried out as illustrate in tableFixed End MomentsMemberDCDAADAB475529680475512155.2457264957-34792944-75753288The moment distribution carried out as per table 1 for case 1JointDA2597025970MemberDCDAADAB32884755Distribution factore0.330.670.670.337333324Fix end moment12155.2457264957-34792944-7575AABalance-2892-57843087154447550.875Carry over 1544-28924386balance-515-102919289641.75mCarry over 964-515balance-321-64334317283240.875Carry over 172-32110500DDbalance-57-11421410717833114358324-174Carry over 107-57balance-36-7138192786327863Carry over 19-3611435balance-6-132412Carry over 12-647629balance-4-842Fig 2Carry over 2-4balance-1-131Final moment8324-83244755-4755

For horizontal slab AB, carrying UDL @29680N/m2.Vertical reactionat a and B =0.5x29680x1.75=25970N/m2Similarly, for the Bottom slab DC carrying U.D.L.loads @47628.717948718N/m2Vertical reaction at D and C =0.5x47628.717948718x1.17=27862.8NThe body diagram for various members, including loading, B.M. And reactions are shown in fig.2For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus( -ha x1.17) +4755-8324+7333.3333333333x1.17x1.17x1/2+1/2x10500x1.17x1.17x1/3-hax1.17+-3569+5019.3+2395.575From which, ha =3288Hence , hd =(7333.3333333333+17833.3333333333)x1.17-3288=11435N2Free B.M. at mid point E =29680x1.172=5079N-m8Net B.M. at E =5079-4755=324N-mSimilarly, free B.M. at F =47628.717948718x1.172=8149.869N -m8Net B.M. at F =8149.869-8324=-174N-mFor vertical member AD , Simply supported B.M. At mid span Simply supporetd at mid sapn =7333.3333333333x1.172+1/16x10500x1.172=2153.1656258Net B.M. =8324+4755=6540-2153.165625=4386N-m23Case 2 : Dead load and live load from out side and water pressure from inside.In this case , water pressure having an intensity of zero at A and 9800x1.75=17150N/m2At D, is acting, in addition to the pressure considered in case 1. The various pressures are marked in fig 3 .The vertical walls will thus be subjected to a net latral pressure ofw =29680N/m27333.33333333337333.33333333337333.3333333333Itensity=7333.3333333333N/m2 At the TopA EB6650Net latral pressure diagramAnd =17833.3333333333-17150=683.3333333333N/m2 at the bottom1.171.75

D FC17833.333333333317833.3333333333683.3333333333w=47628.717948718N/m2Fig 3Fix end moments will be as under :MFAB=wL2=29680x1.172=-3385.746N - m1212Mfdc=wL2=47628.717948718x1.172=5433.246N - m1212MFAD =+pL2+WLWhere W is the total tringular earth pressure.1210MFAD =+683.3333333333x1.172+6650x1.17x1.17=534N-m12210MFDA =-pL2-WL1215MFDA =-683.3333333333x1.172-6650x1.17x1.17=-381N -m12215The moment distribution is carrired out as illustred in table.Fixed End MomentsMemberDCDAADAB16862968016865433.246-381534-33863622The moment distribution carried out as per table 1 for case 1JointDA2597025970MemberDCDAADAB1686Distribution factore0.330.670.670.33733336229676Fix end moment5433.246-381534-3386AABalance-1684-3368190195116860.875Carry over 951-1684900balance-317-63411235611.75Carry over 561-317balance-187-37421110631830.875Carry over 106-187683DDbalance-35-70125623393318315050Carry over 62-35balance-21-4223124167541675Carry over 12-213393balance-4-8147Carry over 7-447629balance-2-531Fig 4Carry over 1-2balance-0-121Final moment3183-31831686-1686

For horizontal slab AB, carrying UDL @29680N/m2.Vertical reactionat a and B =0.5x29680x1.75=25970N/m2Similarly, for the Bottom slab DC carrying U.D.L.loads @47629N/m2Vertical reaction at D and C =0.5x47629x1.75=41675.1282051282NThe body diagram for various members, including loading, B.M. And reactions are shown in fig.3For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus( -ha x1.75) +1686-3183+683.3333333333x1.75x1.75x1/2+1/2x6650x1.75x1.75x2/3-hax1.75+-1497+1046.3541666667+6788.5416666667From which, ha =3622Hence , hd =(683.3333333333+7333)x1.75-3622=3393N2Free B.M. at mid point E =29680x1.752=11362N-m8Net B.M. at E =11362-1686=9676N-mSimilarly, free B.M. at F =47629x1.752=18232.8685897436N -m8Net B.M. at F =18232.8685897436-3183=15050N-mFor vertical member AD , Simply supported B.M. At mid span Simply supporetd at mid sapn =683.3333333333x1.752+1/16x6650x1.752=1534.44010416678Net B.M. =3183+1686=2435-1534.4401041667=900N-m2

4Case 3 : Dead load and live load on top water pressure from inside no live load on side.in this case, it is assume that there is no latral oressure due to live load . As before . The top slab is subjected to a load of '=29680N/m2and the bottom slab is subjected to a load w =29680N/m2Itensity=47628.717948718N/m2 40004000Net latral pressure diagramLateral pressure due to dead load =A EB40001/3x12000=4000N/m2Lateral pressure due to soil =1.171/3x18000=6000N/m21.75Hence earth pressure at depth h is =4000+6000hD FC14500145006650Earth pressure intensity at top =4000N/m217150w=47628.717948718N/m217150Fig 5Earth pressure intensity at Bottom=4000+6000x1.75=14500N/m2In addition to these, the vertical wall lslab subjectednto water pressure of intensity ZERO at top and 17150N/m2 at Bottom, acting from inside . The lateral pressure on vertical walls Is shown in fig 5 and 6

Fix end moments will be as under :MFAB=wL2=29680x1.172=-3385.746N - m1212Mfdc=wL2=47628.717948718x1.172=5433.246N - m1212MFAD =+pL2-WLWhere W is the total tringular earth pressure.1215MFAD =+4000x1.172-6650x1.17x1.17=153N-m12215MFDA =-pL2+WL456.3-303.43951210MFDA =-4000x1.172-6650x1.17x1.17=-1N -m12210The moment distribution is carrired out as illustred in table.Fixed End MomentsMemberDCDAADAB14952968014955433.246-1153-3386=The moment distribution carried out as per table 1 for case 1JointDA2597025970MemberDCDAADAB1495Distribution factore0.330.670.670.3340003584Fix end moment5433.246-1153-3386AABalance-1811-36212155107814950.875Carry over 1078-18112128balance-359-71812076041.75Carry over 604-359balance-201-40223912029930.875Carry over 120-2010DDbalance-40-8013467665029935157Carry over 67-40balance-22-4527134167541675Carry over 13-22-705balance-4-9157Carry over 7-447629balance-2-531Fig 4Carry over 1-2balance-0-121Final moment2993-29931495-1495

For horizontal slab AB, carrying UDL @29680N/m2.Vertical reactionat a and B =0.5x29680x1.75=25970NSimilarly, for the Bottom slab DC carrying U.D.L.loads @47629N/m2Vertical reaction at D and C =0.5x47629x1.75=41675.1282051282NThe body diagram for various members, including loading, B.M. And reactions are shown in fig.6For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus( hax1.75) +1495-2993+4000x1.75x1.75x1/2-1/2x6650x1.75x1.75x1/3-hax1.75+-1498+6125-3394From which, ha =-705Hence , hd =(6650x1.75)-4000x1.75--705=-476.252Free B.M. at mid point E =29680x1.172=5079N-m8Net B.M. at E =5079-1495=3584N-mSimilarly, free B.M. at F =47629x1.172=8149.869N -m8Net B.M. at F =8149.869-2993=5157N-mFor vertical member AD , Simply supported B.M. At mid span Simply supporetd at mid sapn =4000x1.172+1/16x6650x1.172=-115.50093758Net B.M. =2993+1495=2244+-115.5009375=2128N-m2

5Design of top slab :Mid section The top slab is subjected to following values of B.M. and direct forceCase B.M. at Center (E)B.M. at ends (A) Direct force (ha)(i)32447553288(II)967616863622(II)35841495-705

The section will be design for maximum B.M. =9676N -mfor water side force sst ==150N/mm2wt. of concrete =24000N/m3scbc ==7N/mm2wt of water =9800N/mm2m=13for water side force k=m*c=13x7=0.378K=0.378m*c+sst13x7+150j=1-k/3=1-0.378/3=0.874J=0.874R=1/2xc x j x k=0.5x7x0.87x0.378=1.155R=1.155Required thickness Provide over all thickness =250mm so effective thicknesss=200mm100mmMr = R . B .D2=1.155x1000x2002=46209490>9676000O.K.Ast =BMx100/sstxjxD=9676000=369mm2150x0.874x200using 16mm F bars A=3.14xdia2=3.14x16x16=201mm24 x1004Spacing of Bars =Ax1000/Ast201x1000/369=545say =540mmHence Provided 16mm F Bars @540mm c/cAcual Ast provided1000x201/540=372mm2Bend half bars up near support at distance of L/5=1.17/5=0.30mArea of distributionn steel =0.3-0.1x(250-100=0.26%450-100Ast =0.26x250x10=643mm2 area on each face=322mm2using 8mm F bars A=3.14xdia2=3.14x8x8=50mm24 x1004Spacing of Bars =Ax1000/Ast =50x1000/322=156say =150mmHence Provided 8mm F Bars @150mm c/c on each faceSection at supports :-Maximum B.M.=4755N-m. There is direct compression of3622N also.But it effect is not considered because the slab is actually reinforced both at top and bottom . Since steel is at top sst=190N/mm2concrete M20k=0.3238434164J=0.8920521945R=1.0110983059\ Ast=4755000=141mm2190x0.8920521945x200

Area available from the bars bentup from the middle section =372/2=186.0740740741mm2141