2. Calculation for Box Culvert Rev.A
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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
JAMBI MERANG DEVELOPMENT PROJECT
EPC SALES GAS PIPELINE SKN-GRISSIK
SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013
Section View Longitudinal View
2. LOAD CALCULATION
2.1. Dead Load (D) per m'
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
Live Load = 117.6 kN
Surcharge Live Load 400 kg/m2
note: Live load surcharge will be included into lateral soil load
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
JAMBI MERANG DEVELOPMENT PROJECT
EPC SALES GAS PIPELINE SKN-GRISSIK
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
Summary Of Foundation Load
Type LoadHorizontal (Fx) Vertical (Fy) Moment (Mz)
kN kN kNmDead Load (D) 0.00 109.58 0.00Live Load (L) 0.00 117.60 0.00Equipment Load Empty (Ee) 0.00 10.94 0.00Equipment Load Operating( Eo) 0.00 10.94 0.00Wind Load (W) 0.00 0.00 0.00Seismic Load (V) 46.70 0.00 0.00Soil Load (S) 48.99 0.00 0.00
3.0 Loading CombinationBase on load calculation for foundation above, load summary for allowable stress and ultimatestrength design shall be as follow
3.1 Summary of Design Load for Combination Load
LoadingFx =Fz Fy Mx = 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
JAMBI MERANG DEVELOPMENT PROJECT
EPC SALES GAS PIPELINE SKN-GRISSIK
SP-CL-C-011 Rev:A Calculation for Box Culvert Date:15/07/2013
3.2 Loading Combination- Allowable Stress Design
LoadingFx =Fz Fy Mx = Mz
(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
3.3 Loading Combination-Ultimate Stress Design
LoadingFx =Fz Fy Mx = Mz
(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
Load Combination
JAMBI MERANG DEVELOPMENT PROJECT
EPC SALES GAS PIPELINE SKN-GRISSIK
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
JAMBI MERANG DEVELOPMENT PROJECT
EPC SALES GAS PIPELINE SKN-GRISSIK
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
Loading CombinationFx =Fz
(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 =
a. Dead Load
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 =
a. Dead Load
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
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
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 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
Loading CombinationFx =Fz
(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 =
a. Dead Load
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
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
JAMBI MERANG DEVELOPMENT PROJECT
EPC SALES GAS PIPELINE SKN-GRISSIK
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 =
a. Dead Load
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
Loadinga. Dead load
0.250 * 2400 = 600 kg/m2
Weight of sand in Box Culvert, Wgsand = hsand * gsan0.15*1400 = 210 kg/m2q DL = 810 kg/m2
b. Live load q LL = = 450 kg/m2c. Equipment Load
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. LOAD CALCULATION
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
2.2. Live Loads (L)Surcharge Live Load = 10.00 kN/m2note: Live load surcharge will be included into lateral soil load
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
Foundation Load
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
3.0 Loading CombinationBase on load calculation for foundation above, load summary for allowable stress and ultimatestrength design shall be as follow
3.1 Summary of Design Load for Combination Load
LoadingFx =Fz Fy Mx = Mz
(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
3.2 Loading Combination- Allowable Stress Design
LoadingFx =Fz Fy Mx = Mz
(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
3.3 Loading Combination-Ultimate Stress Design
LoadingFx =Fz Fy Mx = Mz
(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
Load Combination
(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 =
a. Dead Load
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
Loading CombinationFx =Fz
(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 =
a. Dead Load
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
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1r
2.. db
MuRn
f=
'85.0 fc
fym
=
=
=fy
Rm
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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
Loadinga. Dead load
0.150 * 2400 = 360 kg/m2
Weight of sand in Box Culvert, Wgsand = hsand * gsan0.15*1400 = 210 kg/m2q DL = 570 kg/m2
b. Live load q LL = = 0 kg/m2c. Equipment Load
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 =
a. Dead Load
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
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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
Loadinga. Dead load
0.150 * 2400 = 360 kg/m2
Weight of cushion above Box Culvert, Wgfill = hfill * gfil* = 360 kg/m2q DL = 720 kg/m2
b. Live load q LL = = 1000 kg/m2c. Equipment Load
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 =
a. Dead Load
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
2 Super imposed load 12000
3 Live load 10000
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.
(I) Live load, dead load and earth prssure acting , with no water pressure from inside.
(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
Live load and dead load = 10000 + 12000 = 22000
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
\ Latral pressure due to dead load and live load = Pv x KaN/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=
MFAD =pL2
MFAD =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
Member DC DA AD AB1686
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=
MFAD =pL2
MFAD =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
4 Case 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 '= 29680
and the bottom slab is subjected to a load w = 29680
Itensity = 47629 4000 4000
Lateral pressure due to dead load = A E B 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
Member DC DA AD AB1495
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=
MFAD =pL2
MFAD =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
Additional reinforcemet required = -80.63
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