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

JAMBI MERANG DEVELOPMENT PROJECT

EPC SALES GAS PIPELINE SKN-GRISSIK

SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

1. DESIGN DATA1.1. Material

SoilThe soil parameter to be used in the design shall be as follows :

Average of qc = 19.29

= 1890.00 depth 1-2.2 m)

Angle of internal friction, φ = 1.48

= 61.94 KPa (base on Triaxial UU test HB-26)

= 19.74 (base on HB-26, depth 1-4 m)

Soil bearing capacity of structural fill base on Terzaghi Formula:

= 1.21

= 8.26

= 10.83

= 0.13

615.32 kPa

= 615.32 kPa

= 818.38 kPa (increased 33%)

Soil bearing capacity of structural fill base on Soil Lab Investigation DataAllow. Bearing Capacity Q = 129.97 kN/m2 (based on S26, depth 2m)Bottom Area of Box Culvert A = B x L = 24.00 m2Allow. Bearing Capacity, Qa = Q x A = 3,119.28 kNAllow. Bearing capacity for temp., Qat = 4,148.64 kN (increased 33%)

Soil Bearing Capacity to be used = minimum based from Terzaghi formula & Soil Lab InvestigationAllow. Bearing Capacity, Qa = Q x A = 615.32 kNAllow. Bearing capacity for temp., Qat = 818.38 kN

ConcreteCement shall conform to SII 0013-81 or ASTM C150-89.

= 28.0 MPa

= 23.5

= 24,870 MPa

Reinforcing Steel BarDeformed bar & plain bar shall conform to ASTM A-615-89, 5110136-84.

= 413.0 MPa

= 200000 MPa

Sand Bedding in Box Culvert

= 1400 kg/m3 = 13.7

Box Culvert will be fill with sand 0.15m height = 0.15 m

1.2. Equipment DataBox Culvert Equipment load = load of 16" Sales Gas pipeline16" Sales Gas pipeline weight = 93 kg/m

1.3. Box Culvert DimensionLength net Ln = 12.000 m Height net Hn = 1.000 mthick of wall = 0.250 m thick of soil, t = 1.656 mLength gross, L = 12.000 m thick of slab = 0.250 m (top & bottom)

Height to Bottom, H = 3.156 mWidth net Bn = 1.500 mthick of wall = 0.250 m Depth of soil cover, Hds = 1.656 mWidth, B = 2.000 m Height of Foundation, Df = 1.500 m

kg/cm2 (considered from qcavr for BH-26

kN/m2

o

Cohesion, c = su

Soil density, gs kN/m3

Nq = e(0.75π-φ/2) tanf / (2 cos2(45o+φ/2))

Nc = (Nq-1) cotg φ

Kpg = 3 tan2 {45o+1/2(φ+33o)}

Ng = 1/2 tanφ (Kpg /cos2φ - 1)

For square footing: qult = c Nc [1+0.3(B/L)]+ q Nq + 0.5 γ' D Ng [1-0.2(B/L)] =

Ultimate soil bearing capacity, Qa

Ultimate soil bearing capacity for temp., Qtemp

Concrete strength, f'c

Concrete unit weight, gc kN/m3

Modulus Elasticity of concrete, Ec = 4700 √f'c

Reinforcement yield strength, fy

Modulus elasticity of reinforcement steel, Es

Sand unit weight, gsand gsand kN/m3

hsand

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

Section View Longitudinal View

1*2.000*0.250*23.50 = 11.75 kN

1*2.000*0.250*23.50 = 11.75 kN

1*1.500*0.250*23.50* 2 = 17.63 kN

2.000*1.000*1.656*"19.74 = 65.37 kN

Weight of sand in Box Culvert, Wgsand = (Ln Bn hsand) gsand = 12.000*1.500* 0.15* 13.72 = 3.09 kNWeight of Box Culvert, WFD = Wgr + WgS + Wglw + Wgsw + Wgs + Wgsand = = 109.58 kNnote : 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 = 100 kg/m2 LR = B x L x 100 = 2*1*100 = 2400 kg = 23.52 kNSlab = 400 kg/m2 LS = B x L x 400 = 2*1*400 = 9600 kg = 94.08 kN

2.3. Equipment Loads (E)Box Culvert Equipment load = load of 16" Sales Gas pipeline16" Sales Gas pipeline weight (E) = 12 x 93 = 1116 kg = 10.94 kN

Equipment LoadFx Fz Fy Mx Mz

(Kn) (kN) (kN) (kN) (kNm)

EmptyCondition 0.000 0.000 10.944 0.000 0.000

Operating Condition 0.000 0.000 10.944 0.000 0.000

2.4. Wind Loads (W: Wx = Wz))The Wind Load is not aplicable to Box Culvert analysis because location of Box Culvert is in underground

2.5 Seismic 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 = 1.25 Table 16-K Occupancy CategoryR : Seismic reduction factor = 2.2 Table 16-PWi : Weigth of Box Culvert = 109.58 kN from Dead Load of Foundation

= 46.70 kN

Weight of top slab, WgR = L B t top slab gsteel =

Weight of slab, WgS = L B t slab gc =

Weight of Long Wall , Wglw = L H t wall gc *2 =

Weight of soil above roof, Wgs = (B L)t. gs =

Vs= (2.5 x Ca x I )/R x ∑▒Wi

Seismic Load(Fsl)= (2.5 x 0.3 x1.25)/2.2 x 109,58

2000

15

00

80

61

94

50

0

2400

1900

200

25

0

200

120002000

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

2.6 Earth Pressure

`

Specific gravity Gs = 2.591 LLs = 3.9 kPavoid ratio e = 0.688 Htota = 3.156 mSaturated soil density 34.97 H = 1.500 m

Soil density 19.74 d = 1.656 mShear angle 1.48 Undrained Shear Strength c = 0.632 kg/cm2Ground Water Level GWL = 0Back Fill Density 18

= 0.950

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

48.99 kN

Type LoadHorizontal (Fx) Vertical (Fy) Moment (Mz)

(kN) (kN) (kNm)Structure Dead Load D 0.000 109.582 0.000Structure Live Load L 0.000 117.600 0.000Equipment Empty Weight E(E) 0.000 10.944 0.000Equipment Operating Weight E(O) 0.000 10.944 0.000Wind Load W 0.000 0.000 0.000 Not AplicableSeismic Load EQ 46.697 0.000 0.000Soil Load S 48.990 0.000 0.000

kg/m3

g sat = kg/m3

g s = kN/m3

f =

g b = kN/m3

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

Seismic Load(Fsl)= (2.5 x 0.3 x1.25)/2.2 x 109,58

2000

15

00

16

56

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

(kN) (kN) (kNm)LC 01 D + L + E(E) + S 48.990 238.126 0.000LC 02 D + L + E(E) + EQ + S 95.687 238.126 0.000

` LC 03 D + L + E(O) + S 48.990 238.126 0.000LC 04 D + L + E(O) + EQ + S 95.687 238.126 0.000

(kN) (kN) (kNm)LC 11 1.4 [D+L+E(E)+S] 68.586 333.376 0.000LC 12 1.2 [D+L+E(E)+S] + 1.4 EQ 124.163 285.751 0.000LC 13 1.4 [D+L+E(O)+S] 68.586 333.376 0.000LC 14 1.2 [D+L+E(O)+S] + 1.4 EQ 124.163 285.751 0.000

4. FOUNDATION DESIGN4.1 Soil Stress

Bearing capacity of soil is taken from Geotechnical Survey Report.For exmaple LC 01,

= 0/238.13 = 0 m

= 0/238.13 = 0 m

238.13/(2.000*12.000) * (1 + 6*0.000/12.000 + 6*0.000/2.00) = 9.92 kPa

238.13/(2.000*12.000) * (1 - 6*0.000/12.000 - 6*0.000/2.00) = 9.92 kPa

= 615.32 / 9.92 = 62.017 > 3.00 OK !

The results for all Load Combinations are presented in the following Table

Load Combination SFm m kPa kPa

LC 01 - - 9.92 9.92 62.02 3.00 OKLC 02 - - 9.92 9.92 82.48 2.00 OKLC 03 - - 9.92 9.92 82.48 2.00 OKLC 04 - - 9.92 9.92 82.48 2.00 OK

LC 11 - - 13.89 13.89 58.92 2.00 OKLC 12 - - 11.91 11.91 68.74 2.00 OKLC 13 - - 13.89 13.89 58.92 2.00 OKLC 14 - - 11.91 11.91 68.74 2.00 OK

4.2 Stability Against Overturning and SlidingExample For LC 01,The stability against sliding:

= 238.13*tan(2/3*1.48) + 61.94*2.000*12.000 = 1,490.56 kN

= 1,490.56 / 48.99 = 30.43 > 2.00 OK !

The results for all Load Combinations are presented in the following Table

SFResult

Actual CriteriaLC 01 30.4 2.0 OKLC 02 15.6 1.5 OKLC 03 30.3 1.5 OKLC 04 15.5 1.5 OK

LC 11 21.8 1.5 OKLC 12 12.0 1.5 OKLC 13 21.8 1.5 OKLC 14 12.0 1.5 OK

eX = MZ / FY

eZ = MX / FY

q = (FY / BF LF) (1 ± 6 eX / LF ± 6 eZ / BF)

qmax =

qmin =

SF = Qa / qmax

eX eZ qmax qmin

FR = FY tan 2/3 f + c B L

Safety factor, SF = FR / H

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

4.3 SettlementThe 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 = 9450 kPaB/L = m' = 0.2H/(L/2) = n' = 0.5

= m' ln((1+sqrt(m'2+1)sqrt(m'2+n'2)/m'(1+sqrt(m'2+n'2+1)) = 0.245709

= ln((m'+sqrt(m'2+1))sqrt(1+n'2)/(m'+sqrt(m'2+n'2+1)) = -0.697819

= m'/(n'sqrt(m'2+n'2+1)) = 0.277427

= = -0.144

= = 0.023

= = -0.131

= 0.750

= 0.98

a = 4 (center of foundation)

B' = 1 (center of foundation)

= 13.9 KPa

Elastic settlement, Se = = -0.00069 m = -0.687 mm

For consolidation settlement:This consolidation settlement considering normally consolidation as follow:

Sc = HCc

log

Distribution of ∆p will be describe as fogure below.

0

3.16 m

-3.2 m1

2 0.1 m-3.3 m

0.2 m-3.5 m

z Hi ∆pCc log

Sc

(m) (m) mm

24.00 1.500 0 29.606 13.891 0.168 1 0.167 024.70 1.550 0.1 30.593 13.50 0.168 1 0.159 1.3326.84 1.650 0.2 32.566 12.42 0.168 1 0.140 2.36

Sc tot = 3.691

Note:

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

4.4 Anti Buoyancy Check

Min Dead Load (Va) = 109.58 kN = 11,181.80 kg Weight of Structurre

Bouyancy (uplift) force = = 1 * 2 * 1.5 * 1000 = 3000 kg

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

(from 5 qc avr)

A0

A1

A2

F1 1/p (A0 + A1)

F2 n' / 2p tan-1 A2

Is F1 + (1-2m)F2/(1-m)

Df/B

If

qn

qnaB'(1-m2)IsIf/Es

p0 + ∆p

1+e0 p0

BF'xLF' p0e0

p0+∆p

(m2) (kN/m2) (kN/m2) p0

Depth of settle layer is assumed until p0> max ∆p.

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

Lw * Bw * h2 * g w

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

5 CONCRETE REINFORCEMENT DESIGN

5.1 WALL DESIGN5.1.1 Wall thickness bearing control

2000 mm

12000 mmLength, L = 12.000 mWidth, B = 2.000 m

Area of slab, A = 24.000 Area a1 & a2 = 11.198 m2Area a3 & a4 = 4.000 m2

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 = 44,506 kgLL = LL roof slab x LB x RL = 450 * 0.47 *12*2 = 5,039 kg

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

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

ɸ = 0.6Axial long wall capacity = ɸ.Aw1.fc' = 0.6*30000*344 = 6192010 kg = 6192.010 TAxial short wall capacity = ɸ.Aw2.fc' = 0.6*5000*344 = 1032002 kg = 1032.002 T

P wall > P ult 6192.0 > 61.5 = 250 mm wall is adequate for axial load

5.1.2 Wall Flexural Design

5.1.2.1 For the Long WallL = 12.0 mH = 1.0 mL/H = 12.00 < 2.5 the wall assumed as 2 ways slab

Wall thickness, tw = 250 mmCover, c = 40 mm

(kN)LC 11 1.4 [D+L+E(E)+S] 68.59 LC 12 1.2 [D+L+E(E)+S] + 1.4 EQ 124.16 LC 13 1.4 [D+L+E(O)+S] 68.59 LC 14 1.2 [D+L+E(O)+S] + 1.4 EQ 124.16

Fhor max = 124.16 kN = 12670 kg= L H = 12 * 3.156 = 37.9 = B H = 2 * 3.156 = 6.3

qu of Long wall = F/A1 = 12669.72 / 37.87 = 334.5 `

From table above, table for the analysis of plates, slabs, and diaphragms

m2

Area of Long wall A1 m2

Area of Short wall A2 m2

kg/m2

Tw o W a y s S la b C a lc u la t io n

Structural Model :

Type S lab = S1 ly 4.2 1.4 < 2.5 ( two way s lab )

Thicknes s , tr = 180 m m lx 3.2

Material :

fc' = 282

fy = 4000

C o ve r, c = 50 m m Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2̂2 * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2̂2 * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

a2

a4a3

a1

Two Ways Slab Calculation

Structural Model :

Type Slab = S1 ly 4.2 1.4 < 2.5 ( two way slab )

Thickness, tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

Cover, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2 2̂ * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2 2̂ * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

H/L = 12.0Mx max = 0.001*qu*h^2*81 = 0.001*334.55*1*81 = 27 kg.mMy max = 0.001*qu*h^2*54 = 0.001*334.55*1*54 = 18 kg.m

R01 = 1/2*qu*h = 0.5*12669.72*1 = 167 kgR01 = 1/2*qu*h = 0.5*12669.72*1 = 167 kg

Rebar Required:(x-dir)Re-bar to be applied = 13 mmd = tw - c - (reinf dia*0.5) = 203.5 mmb = 1000 mmɸ = 0.8

= 13.68 = 0.1 kg/cm2

0.0000

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0290 fy 600 + fy

ρ max 0.0218

Check : > < OK0.0018

As = r min x b x d = 3.66 cm2 / m'

(y-dir)d = tw - c - (reinf dia*1.5) = 203.5 mmb = 1000 mmɸ = 0.8

= 13.68 0.1 kg/cm2

0.000

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0290 fy 600 + fy

ρ max 0.0218

Check : < > OK0.0018

As = r min x b x d = 3.66 cm2 / m'

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r min r req r maxr required =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r min r req r maxr required =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

Re-bar selectionD = 13 mm ; Ab = 1.3267 cm2Vertical Re-bar (for shorter span) S max = 3 x tw = 750 mmHorizontal Re-bar (for longer span) S max = 5 x tw = 1250 mm

No. of Re-bar per metern = As o / Abn ver = 2.8n hor = 2.8

spacing of Re-bar, s = 1000/(n-1)sx = 200 mm use D 13 @ 200 mm = 6.63 > 3.7 cm2 / m'sy = 200 mm use D 13 @ 200 mm = 6.63 > 3.7 cm2 / m'

Sketch: c = 75mm

inner side outer side

c = 40 mmDia. 13 @ 200mmVer and Hor

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

ɸVc = 1/3 √fc' bo.d

ɸ = 0.6bo = 1000 mmd = tw - d'd' = 0 mmd = 250 mm

ɸVc = 264,575 NɸVc = 26,997 kg

ɸVc > Vu OK!

5.1.2.2 For the Short WallB = 2.000 mH = 3.156 mH/B = 0.63 < 2.5 the wall assumed as 2 ways slab

Wall thickness, tw = 250 mmCover, c = 40 mm

(kN)LC 11 1.4 [D+L+E(E)+S] 68.59 LC 12 1.2 [D+L+E(E)+S] + 1.4 EQ 124.16 LC 13 1.4 [D+L+E(O)+S] 68.59 LC 14 1.2 [D+L+E(O)+S] + 1.4 EQ 124.16

Fhor max = 124.16 kN = 12670 kg= B H = 2 * 3.156 = 6.31

qu of Long wall = F/A1 = 12669.72 / 6.31 = 2,007.2 `

From table above, table for the analysis of plates, slabs, and diaphragms

Area of Short wall A2 m2

kg/m2

Two Ways Slab Calculation

Structural Model :

Type Slab = S1 ly 4.2 1.4 < 2.5 ( two way slab )

Thickness, tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

Cover, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2 2̂ * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2 2̂ * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

H/B = 0.63Mx max = 0.001*qu*h^2*63 = 0.001*12669.72*9.97*63 = 1260 kg.mMy max = 0.001*qu*h^2*54 = 0.001*0*9.97*54 = 1080 kg.m

R01 = 1/2*qu*h = 0.5*12669.72*3.16 = 3167 kgR01 = 1/2*qu*h = 0.5*12669.72*3.16 = 3167 kg

Rebar Required:(x-dir)Re-bar to be applied = 13 mmd = tw - c - (reinf dia*0.5) = 203.5 mmb = 1000 mmɸ = 0.8

= 13.68 = 3.8 kg/cm2

0.0010

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0290 fy 600 + fy

ρ max 0.0218

Check : > < OK0.0018

As = r min x b x d = 3.66 cm2 / m'

(y-dir)d = tw - c - (reinf dia*1.5) = 203.5 mmb = 1000 mmɸ = 0.8

= 13.68 3.3 kg/cm2

0.0008

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0290 fy 600 + fy

ρ max 0.0218

Check : < > OK0.0018

As = r min x b x d = 3.66 cm2 / m'

Re-bar selectionD = 13 mm ; Ab = 1.3267 cm2Vertical Re-bar (for shorter span) S max = 3 x tw = 750 mmHorizontal Re-bar (for longer span) S max = 5 x tw = 1250 mm

No. of Re-bar per metern = As o / Abn ver = 2.8n hor = 2.8

spacing of Re-bar, s = 1000/(n-1)sx = 200 mm use D 13 @ 200 mm = 6.63 > 3.7 cm2 / m'sy = 200 mm use D 13 @ 200 mm = 6.63 > 3.7 cm2 / m'

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r min r req r maxr required =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r min r req r maxr required =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

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SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

Sketch: c = 75mm

inner side outer side

c = 40 mmDia. 13 @ 200mmVer and Hor

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

ɸVc = 1/3 √fc' bo.d

ɸ = 0.6bo = 1000 mmd = tw - d'd' = 0 mmd = 250 mm

ɸVc = 264,575 NɸVc = 26,997 kg

ɸVc > Vu OK!

5.2 SLAB DESIGN

Lx = 2.000

Ly = 12.000

Since the slab ratio is Ly/Lx < 2.5 we can assumed as two ways actionLy / Lx = 6.0 < 2.5Two way slab calculationType slab = S1Thickness of Slab = 250 mmMaterial =fc' = 28 Mpa = 344 kg/cm2fy = 413 Mpa = 4,000 kg/cm2

concrete cover = 40 mmfrom SKSNI T-15-1991-03, we find:MIx = -Mtx = 0.001.q.Ix² * 72MIy = -Mty = 0.001.q.Ix² * 55

Two Ways Slab Calculation

Structural Model :

Type Slab = S1 ly 4.2 1.4 < 2.5 ( two way slab )

Thickness, tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

Cover, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2 2̂ * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2 2̂ * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

JAMBI MERANG DEVELOPMENT PROJECT

EPC SALES GAS PIPELINE SKN-GRISSIK

SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

0.250 * 2400 = 600 kg/m2

Weight of sand in Box Culvert, Wgsand = hsand * gsan0.15*1400 = 210 kg/m2q DL = 810 kg/m2

E(O) = ### kgFloor Area = B L = 2.000 m2Equiment Pressure= = #REF! kg/m2

c. Ultimate load q u = 1.2q (DL + E(O)+ 1.6 q LL = = #REF! kg/m2MomentMIx = -Mtx = #REF! kg.mMIy = -Mty = #REF! kg.m

Rebar Required:(x-dir)Re-bar to be applied = 13 mmd = tr - c - (reinf dia*0.5) = 204 mmb = 1000 mmɸ = 0.8

= 13.68 = #REF! kg/cm2

#REF!

r min = 1.4/fy = 0.00339 ρ min #REF!r min = 0,18% = 0.0018

ρ b 0.0376

fy 600 + fyρ max 0.0282

#REF!

As = r min x b x d = ### cm2 / m'(Y-dir)Re-bar to be applied = 16 mmd = tr - c - (reinf dia*0.5) = 202 mmb = 1000 mmɸ = 0.8

= 13.68 = #REF! kg/cm2

#REF!

r min = 1.4/fy = 0.00339 ρ min #REF!r min = 0,18% = 0.0018

ρ b 0.0447

fy 600 + fyρ max 0.0335

#REF!

As = r min x b x d = ### cm2 / m'

Weight of slab, WgR = t slab gc =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r required =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r required =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

JAMBI MERANG DEVELOPMENT PROJECT

EPC SALES GAS PIPELINE SKN-GRISSIK

SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013

Re-bar selectionD = 13 mm ; Ab = 1.3267 cm2Longitudinal Re-bar (for shorter span) S max = 3 x tr = 750 mmTransversal Re-bar (for longer span) S max = 5 x tr = 1,250 mm

No. of Re-bar per meter, n = As o / Abn x = ###n y = ###

spacing of Re-bar, s = 1000/(n-1)sx = 200 mm use D 13 @ 200 mm = 6.63 > #REF! cm2 / m'sy = 200 mm use D 13 @ 200 mm = 6.63 > #REF! cm2 / m'

Sketch:

250

Dia. 13 @ 200mmtop and bottom

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

1. DESIGN DATA1.1. Material

SoilThe soil parameter to be used in the design shall be as follows :

Average of qc = 4.57

= 448.00 depth 1-2.2 m)

Angle of internal friction, φ = 1.48

= 61.94 KPa (base on Triaxial UU test DB-3)

= 19.74 (base on DB-3, depth 1-4 m)

Soil bearing capacity of structural fill base on Terzaghi Formula:

= 1.21

= 8.26

= 10.83

= 0.13

600.97 kPa

= 600.97 kPa

= 799.29 kPa (increased 33%)

Soil bearing capacity of structural fill base on Soil Lab Investigation DataAllow. Bearing Capacity Q = 129.97 kN/m2 (based on S26, depth 2m)Bottom Area of Box Culvert A = B x L = 5.20 m2Allow. Bearing Capacity, Qa = Q x A = 675.84 kNAllow. Bearing capacity for temp., Qat = 898.87 kN (increased 33%)

Soil Bearing Capacity to be used = minimum based from Terzaghi formula & Soil Lab InvestigationAllow. Bearing Capacity, Qa = Q x A = 600.97 kNAllow. Bearing capacity for temp., Qat = 799.29 kN

ConcreteCement shall conform to SII 0013-81 or ASTM C150-89.

= 28.0 MPa

= 23.5

= 24,870 MPa

Reinforcing Steel BarDeformed bar & plain bar shall conform to ASTM A-615-89, 5110136-84.

= 413.0 MPa

= 200000 MPa

Sand Bedding in Box Culvert

= 1400 kg/m3 = 13.7

Box Culvert will be fill with sand 0.15m height = 0.15 m

1.2. Equipment DataBox Culvert Equipment load = load of Electrical & Instrument Cable Electrical & Instrument Cable weight = 100 kg/m

1.3. Box Culvert DimensionLength net Ln = 4.000 m Height net Hn = 1.080 mthick of wall = 0.150 m thick of soil, t = 0.200 mLength gross, L = 4.000 m thick of slab = 0.150 m (top )

thick of slab = 0.150 m ( bottom)Width net Bn = 1.000 m Height to Bottom, H = 1.580 mthick of wall = 0.150 m thick of slab wall = 0.150 m (side)Width, B = 1.300 m Height of Foundation, Df = 1.380 m

kg/cm2 (considered from qcavr for S3

kN/m2

o

Cohesion, c = su

Soil density, gs kN/m3

Nq = e(0.75π-φ/2) tanf / (2 cos2(45o+φ/2))

Nc = (Nq-1) cotg φ

Kpg = 3 tan2 {45o+1/2(φ+33o)}

Ng = 1/2 tanφ (Kpg /cos2φ - 1)

For square footing: qult = c Nc [1+0.3(B/L)]+ q Nq + 0.5 γ' D Ng [1-0.2(B/L)] =

Ultimate soil bearing capacity, Qa

Ultimate soil bearing capacity for temp., Qtemp

Concrete strength, f'c

Concrete unit weight, gc kN/m3

Modulus Elasticity of concrete, Ec = 4700 √f'c

Reinforcement yield strength, fy

Modulus elasticity of reinforcement steel, Es

Sand unit weight, gsand gsand kN/m3

hsand

CABLE DUCT 1 PLAN

SECTION - 1B

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

Section View Longitudinal View

CABLE DUCT 1 PLAN

SECTION - 1B

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

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

10.8 kN/m2

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

Equipment LoadFx Fz Fy Mx Mz

(Kn) (kN) (kN) (kN) (kNm)

EmptyCondition 0.000 0.000 3.923 0.000 0.000

Operating Condition 0.000 0.000 3.923 0.000 0.000

2.4. Wind Loads (W: Wx = Wz))The Wind Load is not aplicable to Box Culvert analysis because location of Box Culvert is in underground

2.5 Seismic 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 = 1.25 Table 16-K Occupancy CategoryR : Seismic reduction factor = 2.2 Table 16-PWi : Weigth of Box Culvert = 10.80 kN from Dead Load of Foundation

= 4.60 kN

2.6 Earth Pressure

`

𝑉𝑠= (2.5 𝑥 𝐶𝑎 𝑥 𝐼 )/𝑅 𝑥 ∑▒𝑊𝑖

𝑆𝑒𝑖𝑠𝑚𝑖𝑐 𝐿𝑜𝑎𝑑(𝐹𝑠𝑙)= (2.5 𝑥 0.3 𝑥1.25)/2.2 𝑥 35,81

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

Specific gravity Gs = 2.591 LL = 10.00 kPavoid ratio e = 0.688 Htota = 1.580 mSaturated soil density 34.97 H = 1.500 m

Soil density 19.74 d = 0.200 mShear angle 1.48 Undrained Shear Strength c = 0.632 kg/cm2Ground Water Level GWL = 0Back Fill Density 18

= 0.950

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

33.09 kN/m'

2.7 Base Pressure

Dead Load from top slab and wall including back fill = 14.331 kN/m2Sand bedding = 2.058 kN/m2Equipment W / = 0.33/(2*pi()*0.406*1.0m) = 0.128 kN/m2Live Load = 10.000 kN/m2Base Pressure 26.517 kN/m2

A E B

10001080

D F C

Type LoadHorizontal (Fx) Vertical (Fy) Moment (Mz)

kN/m' kN/m' kNm/m'Dead Load (D) 0 + 3.6x1.3 + 3.6x (1.38-2x0.15) = 13.25 0.00

Live Load (L) 0 10 x 1.3 = 13 0.00

Equipment ( E) 0 3.92 / 2πr = 2.502 0.00

Seismic Load (V) 4.60 0 0.00

Soil Load (S) 33.09 0 x 1.3 = 0 0.00

(kN) (kN) (kNm)Structure Dead Load D 0.000 13.25 0.000Structure Live Load L 0.000 13.00 0.000Equipment Operating Weight E(O) 0.000 2.50 0.000Seismic Load EQ 4.602 0.00 0.000Soil Load S 33.095 0.00 0.000

(kN) (kN) (kNm)LC 01 D + L + E(E) + S 33.095 30.171 0.000LC 02 D + L + E(E) + EQ + S 37.697 30.171 0.000

kg/m3

g sat = kg/m3

g s = kN/m3

f =

g b = kN/m3

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

2prL

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

(kN) (kN) (kNm)LC 11 1.4 [D+L+E(E)+S] 46.332 42.239 0.000LC 12 1.2 [D+L+E(E)+S] + 1.4 EQ 46.157 36.205 0.000

4. FOUNDATION DESIGN4.1 Soil Stress

Bearing capacity of soil is taken from Geotechnical Survey Report.For example LC 01,

= 0/30.17 = 0 m

= 0/30.17 = 0 m

30.17/(1.300*4.000) * (1 + 6*0.000/4.000 + 6*0.000/1.30) = 5.80 kPa

30.17/(1.300*4.000) * (1 - 6*0.000/4.000 - 6*0.000/1.30) = 5.80 kPa

= 600.97 / 5.80 = 103.58 > 3.00 OK !

The results for all Load Combinations are presented in the following Table

Load Combination SFm m kPa kPa

LC 01 - - 5.80 5.80 103.58 3.00 OKLC 02 - - 5.80 5.80 137.76 2.00 OK

LC 11 - - 8.12 8.12 98.40 2.00 OKLC 12 - - 6.96 6.96 114.80 2.00 OK

4.2 Stability Against Overturning and SlidingExample For LC 01,The stability against sliding:

= 30.17*tan(2/3*1.48) + 61.94*1.300*4.000 = 322.59 kN

= 322.59 / 33.09 = 9.75 > 2.00 OK !

The results for all Load Combinations are presented in the following Table

SFResult

Actual CriteriaLC 01 9.7 2.0 OKLC 02 8.5 1.5 OK

LC 11 7.0 1.5 OKLC 12 7.0 1.5 OK

4.3 SettlementThe 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 = 2240 kPaB/L = m' = 0.3H/(L/2) = n' = 0.8

= m' ln((1+sqrt(m'2+1)sqrt(m'2+n'2)/m'(1+sqrt(m'2+n'2+1)) = 0.300741

= ln((m'+sqrt(m'2+1))sqrt(1+n'2)/(m'+sqrt(m'2+n'2+1)) = -0.332821

= m'/(n'sqrt(m'2+n'2+1)) = 0.312801

= = -0.010

= = 0.038

= = 0.012

= 1.062

= 0.98

a = 4 (center of foundation)

B' = 0.65 (center of foundation)

= 8.1 KPa

Elastic settlement, Se = = 9.73E-05 m = 0.097 mm

eX = MZ / FY

eZ = MX / FY

q = (FY / BF LF) (1 ± 6 eX / LF ± 6 eZ / BF)

qmax =

qmin =

SF = Qa / qmax

eX eZ qmax qmin

FR = FY tan 2/3 f + c B L

Safety factor, SF = FR / H

(from 5 qc avr)

A0

A1

A2

F1 1/p (A0 + A1)

F2 n' / 2p tan-1 A2

Is F1 + (1-2m)F2/(1-m)

Df/B

If

qn

qnaB'(1-m2)IsIf/Es

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

For consolidation settlement:This consolidation settlement considering normally consolidation as follow:

Sc = HCc

log

Distribution of ∆p will be describe as fogure below.

0

1.58 m

-1.6 m1

2 0.1 m-1.7 m

0.2 m-1.9 m

z Hi ∆pCc log

Sc

(m) (m) mm

5.20 1.380 0 27.237 8.123 0.3 1 0.113 05.47 1.430 0.1 28.224 7.73 0.3 1 0.105 1.586.30 1.530 0.2 30.198 6.70 0.3 1 0.087 2.61

Sc tot = 4.188

Note:

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

5.0 Moment Calculation

5.1 Top 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 = 0 kNm (pinned - pinned)Mid Span Moment due to top slab load = (1.4 x (DLtop+ LL + E)/8 x L^2 = 4.2579 kNm (pinned - pinned)

5.2 Bottom 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 = 3.0937 kNm (fixed - fixed)Mid Span Moment due to bottom slab load = (1.4 x (DLbot+ LL + E)/24 x L^2 = 1.5468 kNm (fixed - fixed)

5.5 Side 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 = 4.9664 kNm (fixed - pinned)Mid Span Moment due to side wall load = (1.4 x (DLtop+ LL + E)*9/128 x L^2 = 2.7936 kNm (fixed - pinned)

Thickness Control Top Slab minimum = √(4.26x1e6/1000) = 65.253 mm << 150 mmBottom Slab minimum = √(3.09x1e6/1000) = 55.621 mm << 150 mmSide Wall minimum = √(4.97x1e6/1000) = 70.473 mm << 150 mm

6 CONCRETE REINFORCEMENT DESIGN

6.1 WALL DESIGN6.11 Wall thickness bearing control

Calculation per 1 meter of wall

1300 mm

1000 mm

p0 + ∆p

1+e0 p0

BF'xLF' p0e0

p0+∆p

(m2) (kN/m2) (kN/m2) p0

Depth of settle layer is assumed until p0> max ∆p.

Tw o W a y s S la b C a lc u la t io n

Structural Model :

Type S lab = S1 ly 4.2 1.4 < 2.5 ( two way s lab )

Thicknes s , tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

C o ve r, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2̂2 * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2̂2 * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

a2

a1

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

Length, L = 1.000 mWidth, B = 1.300 m

Area of slab, A = 1.300 Area a1 & a2 = 0.650 m2

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

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

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

ɸ = 0.6Axial long wall capacity = ɸ.Aw1.fc' = 0.6*1500*344 = 309600.5 kg = 309.601 TAxial short wall capacity = ɸ.Aw2.fc' = 0.6*1950*344 = 402480.7 kg = 402.481 T

P wall > P ult 309.6 > 0.8 = 150 mm wall is adequate for axial load

6.1.2 Wall Flexural Design

6.1.2.1 WallL = 1.0 mH = 1.1 mL/H = 0.93 < 2.5 the wall assumed as 2 ways slab

Wall thickness, tw = 150 mmCover, c = 30 mm

(kN)LC 11 1.4 [D+L+E(E)+S] 46.33 LC 12 1.2 [D+L+E(E)+S] + 1.4 EQ 46.16

Fhor max = 46.33 kN = 4728 kg= L H = 1 * 1.58 = 6.3 = B H = 1.3 * 1.58 = 2.1

qu of Long wall = F/A1 = 4727.79 / 6.32 = 748.1 `

From table above, table for the analysis of plates, slabs, and diaphragms

H/L = 0.9Mx max = 0.001*qu*h^2*51 = 0.001*748.07*1.17*51 = 41 kgmMy max = 0.001*qu*h^2*51 = 0.001*748.07*1.17*51 = 41 kgm

R01 = 1/2*qu*h = 0.5*4727.8*1.08 = 404 kgR01 = 1/2*qu*h = 0.5*4727.8*1.08 = 404 kg

Rebar Required:(x-dir)Re-bar to be applied = 10 mmd = tw - c - (reinf dia*0.5) = 115 mmb = 1000 mmɸ = 0.8 Mu max = 4.9664 kNm

= 13.68 = 0.5 kg/cm2

0.0001

m2

Area of Long wall A1 m2

Area of Short wall A2 m2

kg/m2

Two Ways Slab Calculation

Structural Model :

Type Slab = S1 ly 4.2 1.4 < 2.5 ( two way slab )

Thickness, tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

Cover, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2 2̂ * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2 2̂ * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0290 fy 600 + fy

ρ max 0.0218

Check : > < OK0.0018

As = r min x b x d = 2.07 cm2 / m'

(y-dir)d = tw - c - (reinf dia*1.5) = 115 mmb = 1000 mmɸ = 0.8

= 13.68 0.4 kg/cm2

0.000

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0290 fy 600 + fy

ρ max 0.0218

Check : < > OK0.0018

As = r min x b x d = 2.07 cm2 / m'

Re-bar selectionD = 10 mm ; Ab = 0.785 cm2Vertical Re-bar (for shorter span) S max = 3 x tw = 450 mmHorizontal Re-bar (for longer span) S max = 5 x tw = 750 mm

No. of Re-bar per metern = As o / Abn ver = 2.6n hor = 2.6

spacing of Re-bar, s = 1000/(n-1)sx = 200 mm use D 10 @ 200 mm = 3.93 > 2.1 cm2 / m'sy = 200 mm use D 10 @ 200 mm = 3.93 > 2.1 cm2 / m'

Sketch: c = 75mm

inner side outer side

c = 30 mmDia. 10 @ 200mm vertical

Dia. 10 @ 200mm Horizontal

150

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r min r req r maxr required =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r min r req r maxr required =

=

=fy

Rm

mn..211

1r

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

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

ɸVc = 1/3 √fc' bo.d

ɸ = 0.6bo = 1000 mm

d = tw - d'd' = 0 mmd = 150 mm

ɸVc = 158,745 NɸVc = 16,198 kg

ɸVc > Vu OK!

6.2 BOTTOM 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 = 1.3 < 2.5Two way slab calculationType slab = S1Thickness of Slab = 150 mmMaterial =fc' = 28 Mpa = 344 kg/cm2fy = 413 Mpa = 4,000 kg/cm2

concrete cover = 40 mmfrom SKSNI T-15-1991-03, we find:MIx = -Mtx = 0.001.q.Ix² * 82MIy = -Mty = 0.001.q.Ix² * 54

0.150 * 2400 = 360 kg/m2

Weight of sand in Box Culvert, Wgsand = hsand * gsan0.15*1400 = 210 kg/m2q DL = 570 kg/m2

E = 400 kgFloor Area = B L = 1.300 m2Equiment Pressure= = 307.69 kg/m2

c. Ultimate load q u = 1.2q (DL + E(O)+ 1.6 q LL = = 1,053 kg/m2MomentMIx = -Mtx = 86.36 kg.mMIy = -Mty = 56.87 kg.m

Rebar Required:(x-dir)Re-bar to be applied = 13 mmd = tr - c - (reinf dia*0.5) = 104 mmb = 1000 mmɸ = 0.8 Mu max = 3.093653 kNm

= 13.68 = 0.036 kg/cm2

0.0000

Weight of slab, WgR = t slab gc =

Two Ways Slab Calculation

Structural Model :

Type Slab = S1 ly 4.2 1.4 < 2.5 ( two way slab )

Thickness, tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

Cover, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2 2̂ * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2 2̂ * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015=

=fy

Rm

mn..211

1r

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0376

fy 600 + fyρ max 0.0282

0.00180

As = r min x b x d = 1.86 cm2 / m'(Y-dir)Re-bar to be applied = 16 mmd = tr - c - (reinf dia*0.5) = 102 mmb = 1000 mmɸ = 0.8

= 13.68 = 0.683 kg/cm2

0.0002

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0892

fy 600 + fyρ max 0.0669

0.00180

As = r min x b x d = 1.84 cm2 / m'

Re-bar selectionD = 10 mm ; Ab = 0.785 cm2Longitudinal Re-bar (for shorter span) S max = 3 x tr = 450 mmTransversal Re-bar (for longer span) S max = 5 x tr = 750 mm

No. of Re-bar per meter, n = As o / Abn x = 2.4n y = 2.3

spacing of Re-bar, s = 1000/(n-1)sx = 200 mm use D 10 @ 200 mm = 3.93 > 1.86 cm2 / m'sy = 200 mm use D 10 @ 200 mm = 3.93 > 1.84 cm2 / m'

Sketch:

150

Dia. 10 @ 200mmtop and bottom

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r required =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r required =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

6.3 TOP 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 = 1.0 < 2.5Two way slab calculationType slab = S1Thickness of Slab = 150 mmMaterial =fc' = 28 Mpa = 344 kg/cm2fy = 413 Mpa = 4,000 kg/cm2

concrete cover = 40 mmfrom SKSNI T-15-1991-03, we find:MIx = -Mtx = 0.001.q.Ix² * 75MIy = -Mty = 0.001.q.Ix² * 55

0.150 * 2400 = 360 kg/m2

Weight of cushion above Box Culvert, Wgfill = hfill * gfil* = 360 kg/m2q DL = 720 kg/m2

E = 0 kgFloor Area = B L = 0.000 m2Equiment Pressure= = 0 kg/m2

c. Ultimate load q u = 1.2q (DL + E(O)+ 1.6 q LL = = 2,464 kg/m2MomentMIx = -Mtx = 184.80 kg.mMIy = -Mty = 135.52 kg.m

Rebar Required:(x-dir)Re-bar to be applied = 13 mmd = tr - c - (reinf dia*0.5) = 104 mmb = 1000 mmɸ = 0.8 Mu max = 4.257904 kNm

= 13.68 = 0.050 kg/cm2

0.0000

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0376

fy 600 + fyρ max 0.0282

0.00180

Weight of slab, WgR = t slab gc =

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r required =

Two Ways Slab Calculation

Structural Model :

Type Slab = S1 ly 4.2 1.4 < 2.5 ( two way slab )

Thickness, tr = 180 mm lx 3.2

Material :

fc' = 282

fy = 4000

Cover, c = 50 mm Mlx = Mtx =

Mly = Mty =

2 cm =

= 63

= 0

- M & E = 0 +

q dl = 495

b. Live Load q ll = 500

c. Ultimate Load q u = 1.2 qdl + 1.6 qll =

Moment :

Mlx = - Mtx = 0.001 * 1307.6 * 3.2 2̂ * 72 = kg m

Mly = - Mty = 0.001 * 1307.6 * 3.2 2̂ * 55 = kg m

SKSNI T-15-1991-03 table

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

51 57 63 68 72 75 78 80 81 82 82 82 83 83 83 8351 53 54 55 55 55 54 54 54 54 53 53 53 52 52 51

Ly/Lx

Mlx = Mtx

Mly = Mty

lx

2

ly

= =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

TBBM-09-CIV-CAL-010-A4 Rev:A Calculation for Underground Facilities Date:143/08/2015

As = r min x b x d = 1.86 cm2 / m'(Y-dir)Re-bar to be applied = 16 mmd = tr - c - (reinf dia*0.5) = 102 mmb = 1000 mmɸ = 0.8

= 13.68 = 1.628 kg/cm2

0.0004

r min = 1.4/fy = 0.00339 ρ min 0.00180r min = 0,18% = 0.0018

ρ b 0.0892

fy 600 + fyρ max 0.0669

0.00180

As = r min x b x d = 1.84 cm2 / m'

Re-bar selectionD = 10 mm ; Ab = 0.785 cm2Longitudinal Re-bar (for shorter span) S max = 3 x tr = 450 mmTransversal Re-bar (for longer span) S max = 5 x tr = 750 mm

No. of Re-bar per meter, n = As o / Abn x = 2.4n y = 2.3

spacing of Re-bar, s = 1000/(n-1)sx = 200 mm use D 10 @ 200 mm = 3.93 > 1.86 cm2 / m'sy = 200 mm use D 10 @ 200 mm = 3.93 > 1.84 cm2 / m'

Sketch:

150

Dia. 10 @ 200mmDia. 10 @ 200mm

; if 1.33 x r req > 1.4/fy; if 1.33 x r req < 1.4/fy

r balance = bi * 0.85 fc' * 600

r max = 0,75*r balance

r required =

2.. db

MuRn

f=

'85.0 fc

fym

=

=

=fy

Rm

mn..211

1r

DESIGN OF BOX TYPE CULVERT

1 In side dimentions 0.85 x 1.5 m

4 Weight of soil 18000 wt. of water 9800

5 Angle of repose 30 Degree

6 Nominal cover top/bottom 50 mm Nominal cover Side 50 mm

6 Concrete fc 28 wt. of concrete 24000

7 m 13

7 Steel Fy 415 190

150

1 Solution Genral

For the purpose of design , one metre length of the box is considered.

The analysis is done for the following cases.

(II) Live and dead load on top and earth pressure acting from out side, and water pressure acting from inside,

with no live load on sides

(III) Dead load and earth pressure acting from out side and water pressure from in side.

Let the thicness of Horizontal slab 320 mm = 0.32 m

Vertical wall thicness 250 mm = 0.25 m

Effective slab span 0.85 + 0.32 = 1.17 m

Effective Height of wall 1.5 + 0.25 = 1.75 m

2 Case 1 : Dead and live load from out side of while no water pressure from inside.

Self weight og top slab = 0.32 x 1 x 1 x 24000 = 7680

Total load on top = 29680

Weight of side wall = 1.75 x 0.25 x 24000 = 10500 N/m

29680 x 1.17 )+( 2 x 10500 )=47628.72

1.17

Ka =1 - sin 30

=1 - 0.5

=0.5

=1

= 0.331 + sin 30 1 + 0.5 1.5 3

p = 22000 x 0.333 = 7333.333

Latral pressure due to soil Ka x w x h = 0.333 x 18000 h = 6000 h

Hence total pressure = 7333 + 6000 h

Latral presure intencity at top = 7333.333

Latral pressure intencity at bottom = 7333 + 6000 x 1.75 = 17833.33

w = 29680

7333.33 7333.3333

A E B

h 1.17

7333.333333 1.75

6000 h

D F C

17833 7333.3333 10500

w = 47629

Fig 1

N/m3

N/m2

N/m2 N/m3

N/m3

scbc N/m2

Out side sst N/m2

water side side sst N/m2

N/m2

N/m2

N/m2

\ Upward soil reaction at base = (N/m2

N/m2

N/m2

Fig 1 show the box culvert frame ABCD, along with the external loads, Due to symmetry, half of frame (i.e. AEFD) of box culvert is considered for moment distribution. Since all the members have uniform thickness, and uniform diamentions, the relative stiffness K for AD will be equal to 1 while the relative stiffness for AE and DF will be 1/2.

N/m2

N/m2

1= 2/3

1/2= 1/3

1+1/2 1+1/2

Fix end moments will be as under : =29680 x 1.75

-7574.58 N - m12 12

+ =47629 x 1.75

12155.25 N - m12 12

+ +WL Where W is the total tringular earth pressure.

12 15

+7333.3 x 1.75 10500 x 1.75

x1.75

= 2944 N-m12 2 15

- -WL

12 15

-7333.3 x 1.75 10500 x 1.75

x1.75

= -1872 -1608 = -347912 2 10

The Moment distribution is carried out as illustrate in table

Fixed End MomentsMember DC DA AD AB 4755 29680 4755

12155.246 -3479 2944 -7575 3288

Joint D A 25970 25970

Member DC DA AD AB 32884755

Distribution factore 0.33 0.67 0.67 0.33 7333 324

Fix end moment 12155.246 -3479 2944 -7575 A A

Balance -2892 -5784 3087 1544 4755 0.875

Carry over 1544 -28924386

balance -515 -1029 1928 964 1.75 m

Carry over 964 -515

balance -321 -643 343 172 8324 0.875

Carry over 172 -321 10500 D D

balance -57 -114 214 107 17833 114358324

-174

Carry over 107 -57

balance -36 -71 38 19 27863 27863

Carry over 19 -36 11435

balance -6 -13 24 12

Carry over 12 -6 47629

balance -4 -8 4 2 Fig 2

Carry over 2 -4

balance -1 -1 3 1

Final moment 8324 -8324 4755 -4755

For horizontal slab AB, carrying UDL @ 29680

Vertical reactionat a and B = 0.5 x 29680 x 1.75 = 25970 N/m2

Similarly, for the Bottom slab DC carrying U.D.L.loads @ 47629

Vertical reaction at D and C = 0.5 x 47628.72 x 1.17 = 27862.8 N

The body diagram for various members, including loading, B.M. And reactions are shown in fig.2

For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus

( -ha x 1.17 ) + 4755 - 8324 + 7333 x 1.17 x 1.17 x 1/2

+ 1/2 x 10500 x 1.17 x 1.17 x 1/3

-ha x 1.17 + -3569 + 5019.3 + 2395.575

From which, ha = 3288

Distribution factore for AD and DA= Distribution factore for AB and DC=

MFAB=wL2 2=

Mfdc=wL2 2=

MFDA =pL2

MFDA =2--

The moment distribution carried out as per table 1 for case 1

N/m2.

N/m2

Hence , hd =( 7333 + 17833 )x 1.17 - 3288 = 11435 N

2

Free B.M. at mid point E =29680 x 1.17

5079 N-m8

Net B.M. at E = 5079 - 4755 = 324 N-m

Similarly, free B.M. at F =47628.71795 x 1.17

8149.869 N -m8

Net B.M. at F = 8149.869 - 8324 = -174 N-m

For vertical member AD , Simply supported B.M. At mid span

Simply supporetd at mid sapn =7333.333333 x 1.17

1/16 x 10500 x 1.17 2153.178

Net B.M. =8324 + 4755

= 6540 - 2153 = 4386 N-m2

3 Case 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 9800 x 1.75 = 17150

w = 29680

7333.33333 7333.3333 7333.333333

Itensity = 7333.3 A E B 6650

And = 17833 - 17150

= 683.33 1.17

1.75

D F C

17833.3333 17833.333 683.33

w = 47629

Fig 3

Fix end moments will be as under := 29680 x 1.17

-3385.75 N - m12 12

=47629 x 1.17

5433.246 N - m12 12

+ +WL Where W is the total tringular earth pressure.

12 10

+683.33 x 1.17 6650 x 1.17

x1.17

= 534 N-m12 2 10

- -WL

12 15

-683.33 x 1.17 6650 x 1.17

x1.17

= -381 N -m12 2 15

The moment distribution is carrired out as illustred in table.

Fixed End MomentsMember DC DA AD AB 1686 29680 1686

5433.246 -381 534 -3386 3622

Joint D A 25970 25970

Distribution factore 0.33 0.67 0.67 0.33 7333 3622 9676

Fix end moment 5433.246 -381 534 -3386 A A

Balance -1684 -3368 1901 951 1686 0.875

Carry over 951 -1684900

2=

2=

2+2=

N/m2

At 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 of

N/m2

N/m2 At the Top

N/m2 at the bottom

N/m2

MFAB=wL2 2=

Mfdc=wL2 2=

MFDA =pL2

MFDA =2-

The moment distribution carried out as per table 1 for case 1

balance -317 -634 1123 561 1.75900

Carry over 561 -317

balance -187 -374 211 106 3183 0.875

Carry over 106 -187 683 D D

balance -35 -70 125 62 33933183

15050

Carry over 62 -35

balance -21 -42 23 12 41675 41675

Carry over 12 -21 3393

balance -4 -8 14 7

Carry over 7 -4 47629

balance -2 -5 3 1 Fig 4

Carry over 1 -2

balance 0 -1 2 1

Final moment 3183 -3183 1686 -1686

For horizontal slab AB, carrying UDL @ 29680

Vertical reactionat a and B = 0.5 x 29680 x 1.75 = 25970 N/m2

Similarly, for the Bottom slab DC carrying U.D.L.loads @ 47629

Vertical reaction at D and C = 0.5 x 47629 x 1.75 = 41675.13 N

The body diagram for various members, including loading, B.M. And reactions are shown in fig.3

For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus

( -ha x 1.75 ) + 1686 - 3183 + 683.3 x 1.75 x 1.75 x 1/2

+ 1/2 x 6650 x 1.75 x 1.75 x 2/3

-ha x 1.75 + -1497 + 1046.354 + 6788.542

From which, ha = 3622

Hence , hd =( 683.33 + 7333 )x 1.75 - 3622 = 3393 N

2

Free B.M. at mid point E =29680 x 1.75

11362 N-m8

Net B.M. at E = 11362 - 1686 = 9676 N-m

Similarly, free B.M. at F =47629 x 1.75

18232.87 N -m8

Net B.M. at F = 18232.86859 - 3183 = 15050 N-m

For vertical member AD , Simply supported B.M. At mid span

Simply supporetd at mid sapn =683.3333333 x 1.75

1/16 x 6650 x 1.75 1534.448

Net B.M. =3183 + 1686

= 2435 - 1534 = 900 N-m2

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 '= 29680

and the bottom slab is subjected to a load w = 29680

Itensity = 47629 4000 4000

1/3 x 12000 = 4000

Lateral pressure due to soil = 1.17

1/3 x 18000 = 6000 1.75

Hence earth pressure at depth h is =

4000 + 6000 h D F C

14500 14500 6650

N/m2.

N/m2

2=

2=

2+2=

N/m2

N/m2

N/m2

N/m2

N/m2

Earth pressure intensity at top = 4000 17150 w= 47629 17150

Fig 5

Earth pressure intensity at Bottom= 4000 + 6000 x 1.75 = 14500

In addition to these, the vertical wall lslab subjectednto water pressure of intensity ZERO at top and 17150

N/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 : =29680 x 1.17

-3385.75 N - m12 12

=47629 x 1.17

5433.246 N - m12 12

+ -WL Where W is the total tringular earth pressure.

12 15

+4000 x 1.17 6650 x 1.17

x1.17

= 153 N-m12 2 15

- +WL 456.3 - 303.4

12 10

-4000 x 1.17 6650 x 1.17

x1.17

= -1 N -m12 2 10

The moment distribution is carrired out as illustred in table.

Fixed End MomentsMember DC DA AD AB 1495 29680 1495

5433.246 -1 153 -3386 =

Joint D A 25970 25970

Distribution factore 0.33 0.67 0.67 0.33 4000 3584

Fix end moment 5433.246 -1 153 -3386 A A

Balance -1811 -3621 2155 1078 1495 0.875

Carry over 1078 -18112128

balance -359 -718 1207 604 1.75

Carry over 604 -359

balance -201 -402 239 120 2993 0.875

Carry over 120 -201 0 D D

balance -40 -80 134 67 66502993

5157

Carry over 67 -40

balance -22 -45 27 13 41675 41675

Carry over 13 -22 -705

balance -4 -9 15 7

Carry over 7 -4 47629

balance -2 -5 3 1 Fig 4

Carry over 1 -2

balance 0 -1 2 1

Final moment 2993 -2993 1495 -1495

For horizontal slab AB, carrying UDL @ 29680

Vertical reactionat a and B = 0.5 x 29680 x 1.75 = 25970 N

Similarly, for the Bottom slab DC carrying U.D.L.loads @ 47629

Vertical reaction at D and C = 0.5 x 47629 x 1.75 = 41675.13 N

The body diagram for various members, including loading, B.M. And reactions are shown in fig.6

For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus

( ha x 1.75 ) + 1495 - 2993 + 4000 x 1.75 x 1.75 x 1/2

N/m2 N/m2

N/m2

MFAB=wL2 2=

Mfdc=wL2 2=

MFDA =pL2

MFDA =2-

The moment distribution carried out as per table 1 for case 1

N/m2.

N/m2

- 1/2 x 6650 x 1.75 x 1.75 x 1/3

-ha x 1.75 + -1498 + 6125 - 3394

From which, ha = -705

Hence , hd =( 6650 x 1.75 )- 4000 x 1.75 - -705 = -476.25

2

Free B.M. at mid point E =29680 x 1.17

5079 N-m8

Net B.M. at E = 5079 - 1495 = 3584 N-m

Similarly, free B.M. at F =47629 x 1.17

8149.869 N -m8

Net B.M. at F = 8149.869 - 2993 = 5157 N-m

For vertical member AD , Simply supported B.M. At mid span

Simply supporetd at mid sapn =4000 x 1.17

1/16 x 6650 x 1.17 -115.58

Net B.M. =2993 + 1495

= 2244 + -116 = 2128 N-m2

5 Design of top slab :

Mid section

The top slab is subjected to following values of B.M. and direct force

Case B.M. at Center (E) B.M. at ends (A) Direct force (ha)

(i) 324 4755 3288

(II) 9676 1686 3622

(II) 3584 1495 -705

The section will be design for maximum B.M. = 9676 N -m

for water side force

= 150 wt. of concrete = 24000

= 7 wt of water = 9800

m = 13 for water side force

m*c=

13 x 7= 0.378 K = 0.378

13 x 7 + 150

= 1 - 0.378 / 3 = 0.874 J = 0.874

= 0.5 x 7 x 0.87 x 0.378 = 1.155 R = 1.155

Provide over all thickness = 250 mm so effective thicknesss = 200 mm

= 1.155 x 1000 x 200 46209490 > 9676000 O.K.

Ast =9676000 = 369

150 x 0.874 x 200

using 16 A = =3.14 x 16 x 16

= 2014 x100 4

Spacing of Bars =Ax1000/Ast 201 x 1000 / 369 = 545 say = 540 mm

Hence Provided 16 540 mm c/c

Acual Ast provided 1000 x 201 / 540 = 372

Bend half bars up near support at distance of L/5 = 1.17 / 5 = 0.30 m

Area of distributionn steel = 0.3 -0.1 x( 250 - 100

= 0.26%

450 - 100

= 0.26 x 250 x 10 = 643 ###

using 8 A = =3.14 x 8 x 8

= 504 x100 4

Spacing of Bars = Ax1000/Ast = 50 x 1000 / 322 = 156 say = 150 mm

Hence Provided 8 150 mm c/c on each face

2=

2=

2+2=

sst = N/mm2 N/m3

scbc = N/mm2 N/mm2

k=m*c+sst

j=1-k/3R=1/2xc x j x k

Mr = R . B .D2 2=

BMx100/sstxjxD=mm2

mm F bars 3.14xdia2

mm F Bars @

mm2

Ast mm2 area on each face=

mm F bars 3.14xdia2

mm F Bars @

Section at supports :-

Maximum B.M.= 4755 N-m. There is direct compression of 3622 N also.

But it effect is not considered because the slab is actually reinforced both at top and bottom .

Since steel is at top = 190 concrete M 20

k = 0.3238 J = 0.892 R = 1.011

=4755000

= 141190 x 0.8921 x 200

Area available from the bars bentup from the middle section = 372 / 2 = 186

141 < 186.07

6 Design of bottom slab:

The bottom slab has the following value of B.M. and direct force.

Case B.M. at Center (F) B.M. at ends (D) Direct force (ha)

(i) -174 8324 11435

(II) 15050 3183 3393

(II) 5157 2993 -476

The section will be design for maximum B.M. = 15050 N -m

for water side force

= 150 wt. of concrete = 24000

= 7 wt of water = 9800

m = 13 for water side force

m*c=

13 x 7= 0.378 K = 0.378

13 x 7 + 150

= 1 - 0.378 / 3 = 0.874 J = 0.874

= 0.5 x 7 x 0.87 x 0.378 = 1.155 R = 1.155

=15049869

= 115 mm D = 165 mm1000 x 1.155

Provide thickness of bottom slab D= 170 mm so that d = 120 mm

Ast =15049869 = 956

150 x 0.874 x 120

using 20 mm bars A = =3.14 x 20 x 20

= 3144 x100 4

Spacing of Bars =Ax1000/Ast 314 x 1000 / 956 = 328 say = 320 mm

Hence Provided 20 320 mm c/c

Acual Ast provided 1000 x 314 / 320 = 981

Bend half bars up near support at distance of L/5 = 1.17 / 5 = 0.30 m

Area of distributionn steel = 0.3 -0.1 x( 170 - 100

= 0.28 %450 - 100

= 0.28 x 170 x 10 = 476 238

using 8 mm bars A = =3.14 x 8 x 8

= 504 x100 4

Spacing of Bars = Ax1000/Ast = 50 x 1000 / 238 = 211 say = 210 mm

Hence Provided 8 210 mm c/c on each face

Section at supports :-

Maximum B.M.= 8324 N-m. There is direct compression of 11435 N also.

But it effect is not considered because the slab is actually reinforced both at top and bottom .

Since steel is at top = 190 concrete M 20

k = 0.3238 J = 0.892 R = 1.011

=8324000

= 410

sst N/mm2

\ Ast mm2

mm2

Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c

sst = N/mm2 N/m3

scbc = N/mm2 N/mm2

k=m*c+sst

j=1-k/3R=1/2xc x j x k

\ d

BMx100/sstxjxD=mm2

3.14xdia2

mm F Bars @

mm2

Ast mm2 area on each face=

3.14xdia2

mm F Bars @

sst N/mm2

\ Ast mm2

=190 x 0.8921 x 120

= 410

Area available from the bars bentup from the middle section = 981 / 2 = 491

410 < 490.63

using 8 mm bars A = =3.14 x 8 x 8

= 504 x100 4

Spacing of Bars = Ax1000/Ast = 50 x 1000 / -81 = -623 say = -620 mm

Hence Provided 8 -620 mm c/c throught out the slab, at its bottom.

7 Design of side wall:

The side wall has the following value of B.M. and direct force.

Case B.M. at Center (F) B.M. at ends (D) Direct force (ha)

(i) 4386 8324 27863

(II) 900 3183 41675

(II) 2128 2993 41675

The section will be design for maximum B.M. = 8324 N -m, and direct force = 41675

Eccentricity =8324 x 1000

= 200 mm41675

proposed thickness of side wall '= 320 mm \ e / D 200 / 320 = 0.63 < 1.5

thickness of side wall is OK

Let us reinforce the section with 20 300 mm c/c provided on both faces, as shown

in fig xxx . With cover of 50 mm and D = 320 mm

Asc = Ast =1000

x3.14 x 20 x 20

= 1047300 4

The depth of N.A. is computed from following expression:

n

3 3 n= e +

D- dt

b n + (m -1) Asc n - dc- m Ast

D - dt - n 2

n n

or

1000 n320 - 50 -

n+ 12 x

1047x n - 50 x

2 3 n

1000 n+12 x

1047x n - 50 - 13 x

1047x 320 - 50

2 n n

500 n 270 -n

+ n - 50 x-1256000

3 n= 200 + 110

500 n+12560

x n - 50 -13607

x 270 - nn n

135000 n - 167 + -1256000 --62800000

n= 310

500 n + 12560 -628000

-3673800

+ 13607n n

multiply by n

135000 n2 - 167 n3 + -1256000 n - -62800000= 310

500 n2 + 12560 n- 628000 - 3673800 + 13607 n

135000 n2 - 167 n3 + -1256000 n - -62800000= 0

\ Ast mm2

mm2

Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c

mm2

3.14xdia2

mm F Bars @

mm F bars @

mm2

b n D - dt - + (m - 1)Asc 1 (n - dc)(D - dt- dc)

n2

155000 n2 + 8111667 n - 1333558000= 0

-20000 n2 - 6855667 n - -1270758000 = 167

-120 n2 + 41134 n - 7624548 =

Solwing this trial and error we get, n = 133.42 mm

= ( 500 x 133.42 +12 x 1047

( ### - 50 ) -13 x 1047

133.42 133.42

x ( 320 - 50 - ### )

or 66712 + 94.136 x 83.42 - 102 x 136.58 = 60638

=41675

= 0.69 < 7 Stress is less than permissiable 60638

Also stress in steel t =m c'

(D-dc-n) =13 x 0.69

x ( 320 - 50 - 133.42 )n 133.42

= 9.1456 N/mm2 < 190 N/mm2 O.K.

Stress in steel is less than permissiable Hence section is O.K.

n3

n3

\ c'

\ c' N/mm2

DESIGN OF BOX TYPE CULVERT

(II) Live and dead load on top and earth pressure acting from out side, and water pressure acting from inside,

-3479

4755

25970

324

-174

27863

N-m

7333.333333

Net

latr

al p

ress

ure

diag

ram

1686

25970

9676

15050

41675

N-m

Net

latr

al p

ress

ure

diag

ram

6650

17150

1495

25970

3584

5157

41675

N-m

for water side force

0.378

0.874

1.155

mm2

mm2

mm2

for water side force

0.378

0.874

1.155

Hence these bars will serve the purpose. However, provide 8 mm dia.

mm2

mm2

mm2

N

OK

mm c/c provided on both faces, as shown

-100

- n

Hence these bars will serve the purpose. However, provide 8 mm dia.

mm2

Stress is less than permissiable

2000

15

00

16

56

P3 P1 P2

P4

2000

15

00

16

56

P3 P1 P2

P4