bc1-01singlebox

1 / 82 document.xls/Summary

SUMMARY OF STRUCTURAL CALCULATION OF 1-BARREL BOX CULVERT

1 Design Dimensions and Bar Arrangements Class III Road (BM50)

B2.0 x H2.0

Clear width m 2.00Clear height m 2.00Height of fillet m 0.15

Thickness Side wall cm 30.0Top slab cm 30.0Bottom slab cm 30.0

Cover of reinforcement bar (between concrete surface and center of reinforcement bar)Side wall Outside cm 6.0

Inside cm 6.0Top slab Upper cm 6.0

Lower cm 6.0Bottom slab Lower cm 6.0

Upper cm 6.0

Bar arrangement (dia - spacing per unit length of 1.0 m)

Side wall Lower outside Tensile bar mm 16@250Distribution bar mm 12@250

Middle inside Tensile bar mm 12@250Distribution bar mm 12@250

Upper outside Tensile bar mm 16@250Distribution bar mm 12@250

Top slab Upper edge Tensile bar mm 16@250Distribution bar mm 12@250

Lower middle Tensile bar mm 16@250Distribution bar mm 12@250

Bottom slab Lower edge Tensile bar mm 16@250Distribution bar mm 12@250

Upper middle Tensile bar mm 12@125Distribution bar mm 12@250

Fillet Upper edge Fillet bar mm Err:504Lower edge Fillet bar mm Err:504

2 Design Parameters

Unit Weight Reinforced Concrete 2.4

Backfill soil (wet) 1.8

(submerged) 1.0

Live Load Class of road Class III (BM50)Truck load at rear wheel P= 5.0 tfImpact coefficient (for Class I to IV road) Ci= 0.3

0.0

Pedestrian load (for Class V roads) 0

Concrete Design Strength 175

(K175) Allowable Compressive Stress 60

Allowable Shearing Stress 5.5

Reinforcement Bar Allowable Tensile Stress 1,400

(U24, deformed bar) Yielding Point of Reinforcement Bar 3,000

Young's Modulus Ratio n= 24

Coefficient of static earth pressure Ka= 0.5

Type of box culvert

gc= tf/m3

gs= tf/m3

gs'= tf/m3

(D<4.0m)(D>4.0m)

tf/m2

sck= kgf/cm2

sca= kgf/cm2

ta= kgf/cm2

ssa= kgf/cm2

ssy= kgf/cm2

2 / 82 document.xls/Summary

3/ 82 (1)document.xls, Load

STRUCTURAL CALCULATION OF BOX CULVERT Type: B2.00m x H2.00mSoil Cover Depth:

1 Dimensions and ParametersBasic ParametersKa: Coefficient of static earth pressure

Unit weight of water (t/m3)Unit weight of soil (dry) (t/m3)Unit weight of soil (saturated) (t/m3)Unit weight of reinforced concrete (t/m3)Concrete Design StrengthAllowable Stress of ConcreteAllowable Stress of Reinforcement BarAllowable Stress of Shearing (Concrete)Yielding Point of Reinforcement Bar

n: Young's Modulus RatioFa: Safety factor against uplift

Basic DimensionsH: Internal Height of Box CulvertB: Internal Width of Box CulvertHf: Fillet Heightt1: Thickness of Side Wallt2: Thickness of Top Slabt3: Thickness of Invert (Bottom Slab)BT: Gross Width of Box CulvertHT: Gross Height of Box CulvertD: Covering DepthGwd: Underground Water Depth for Case 1, 2hiw: Internal Water Depth for Case 1, 2

for Case 3, 4

Cover of R-bar Basic ConditionsTop Slab d2 0.06 m Classification of Live load by truck ClassSide Wall d1 0.06 m PTM: Truck load of Middle Tire

Bottom Slab d3 0.06 m Ii:am: Ground contact length of Middle Tirebm: Ground contact width of Middle TirePTR: Truck load of Rear Tirear: Ground contact length of Rear Tirebr: Ground contact width of Rear TirePTF: Truck load of Front wheelaf: Ground contact length of Front Tirebf: Ground contact width of Front Tireqp: Pedestrian load

Dimension of frame

H0: Height of frame t2/2 + H + t3/2B0: Width of frame B + t1

D1: Covering depth at middle of top slab D + t2/2

gw:gd:gs:gc:sck:scassa:ta:ssy:

Impact coefficient (D≧4.0m:0, D<4.0m:0.3)

B0

H0

t2

H

t3

t1 B t1

D1

H

BT

Bt1 t1

t3

HT

Hf

Hf

t2

DGwd


Load distribution of truck tire

(1) Middle tire's acting point: center of the top slaba) distributed load of middle tire

Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 0.2664am': length of distributed load 2D+1.75+bm = 8.000bm': width of distributed load 2D+am = 6.100

b) distributed load of rear tirePvtr: distributed load of rear tire not reach to top slab 0.0000ar': length of distributed load 2D+1.75+br = 8.000br': width of distributed load 2D+ar = 6.100

c) distributed load of front tirePvtf: distributed load of front tire 2PTF(1+Ii)/(af'bf') 0.0666af': length of distributed load 2D+1.75+bf = 8.000bf': width of distributed load 2D+af = 6.100

(2) Middle tire's acting point: on the side walla) distributed load of middle tire

Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 0.2664am': length of distributed load 2D+1.75+bm = 8.000bm': width of distributed load 2D+am = 6.100

b) distributed load of rear tirePvtr: distributed load of rear tire 2PTR(1+Ii)/(ar'br') 0.2664ar': length of distributed load 2D+1.75+br = 8.000br': width of distributed load 2D+ar = 6.100

c) distributed load of front tirePvtf: distributed load of front tire 2PTF(1+Ii)/(af'bf') 0.0666af': length of distributed load 2D+1.75+bf = 8.000bf': width of distributed load 2D+af = 6.100

(3) Rear tire's acting point: on the side walla) distributed load of rear tire

Pvtr: distributed load of rear tire 2PTR(1+Ii)/(ar'br') = 0.2664ar': length of distributed load 2D+1.75+br = 8.000br': width of distributed load 2D+ar = 6.100

b) distributed load of middle tirePvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') 0.2664am': length of distributed load 2D+1.75+bm = 8.000bm': width of distributed load 2D+am = 6.100

c) distributed load of front tirePvtf: distributed load of front tire not reach to top slab 0.0000af': length of distributed load 2D+1.75+bf = 8.000bf': width of distributed load 2D+af = 6.100

(4) Combination of load distribution of track tire

Case.L1: Pvt1 = 0.2664 tf/m2, B = 2.300 m Combination for Case.L2Pvt2 = 0.0666 tf/m2, B = 2.300 m

Case.L2: Pvt1 = 0.5328 tf/m2, B = 2.300 m Distributed load totalPvt2 = 0.0000 tf/m2, B = 0.000 m Select the combination case of

for Case.L2, which is the largest load to the top slab.

In case of covering depth (D) is over 3.0m, uniform load of 1.0 tf/m2 is applied on the top slab of culvert instead of live load calculated above.

Distribution load by pedestrian load

Pvt1 = 0.000 tf/m2

2 Stability Analysis Against Uplift

Analysis is made considering empty inside of box culvert.Fs=Vd/U > Fa Fs= 3.0728 > 1.2 ok

where, Vd: Total dead weight (t/m) Vd= 20.772 tf/mU: Total uplift (t.m)

U= 6.760 tf/m

Ws: Weight of covering soil Ws = = 14.040Wc: Self weight of box culvert Wc = = 6.732Fa: Safety factor against uplift Fa= 1.2

U=BT*HT*gw

BT*{(D-Gwd)*(gs-gw)+Gwd*gd}(HT*BT-H*B+2*Hf^2)*gc


3 Load calculation

Case 1: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L1

1) vertical load against top slabActing Load (tf/m2)

Wtop= 0.8374Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 5.4000Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666

Pv1= 6.5704

2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.0333 we2= 0.0666 tf/m2P3=Ka*gd*Gwd P3= 2.7000P4=Ka*gs*(D1-Gwd) P4= 0.1500P5=gw*(D1-Gwd) P5= 0.1500

Ph1= 3.1665

3) horizontal load at bottom of side wallActing Load (tf/m2)P1=Ka*we1 P1= 0.1332P2=Ka*we2 P2= 0.0333

P3= 2.7000P4= 2.4500P5= 2.4500

Ph2= 7.7665

4) self weight of side wallActing Load (tf/m)

Wsw= 1.4400

5) ground reactionActing Load (tf/m2)

Wbot= 0.8374Wtop Wtop= 0.8374Ws=Wsw*2/B0 Ws= 1.2522Pvd Pvd= 5.4000Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666

Wiw= 0.0000 hiw: internal water depthUp=-U/B0 U= -2.9391

Q= 5.7208

summary of resistance moment Item V H x y M

(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant forceSelf weight top slab 1.9260 - 1.1500 - 2.2149 side wall (left) 1.4400 - 0.0000 - 0.0000 e = B0/2 - X =

side wall (right) 1.4400 - 2.3000 - 3.3120invert 1.9260 - 1.1500 - 2.2149 ground reaction

load on top slab Pvd 12.4200 - 1.1500 - 14.2830Pvt1 0.6127 - 1.1500 - 0.7046Pvt2 0.1532 - 1.1500 - 0.1762

soil pressure side wall (left) - 12.5729 - 0.9887 12.4310side wall (right) - -12.5729 - 0.9887 -12.4310

internal water 0.0000 - 1.1500 - 0.0000uplift -6.7600 - 1.1500 - -7.7740total 13.1579 15.1316

6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.2664Pvt2 0.0666Wtop 0.8374Ws 1.2522

Pq= 7.8226

Wtop= (t2*BT+Hf^2)*gc/B0

P3=Ka*gd*GwdP4=Ka*gs*(D1+H0-Gwd)P5=gw*(D1+H0-Gwd)

Wsw=t1*H*gc

Wbot=(t3*BT+Hf^2)*gc/B0

Wiw=(hiw*B-2Hf^2)*gw/B0

X = SM/SV =

q1 = SV/Bo + 6q2 = SV/Bo - 6




Wtop= 0.8374Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 5.4000Pvt1 Pvt1= 0.5328Pvt2 Pvt2= 0.0000

Pv1= 6.7702

2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.1332 we2= 0.2664 tf/m2P3=Ka*gd*Gwd P3= 2.7000P4=Ka*gs*(D1-Gwd) P4= 0.1500P5=gw*(D1-Gwd) P5= 0.1500

Ph1= 3.2664


P3= 2.7000P4= 2.4500P5= 2.4500

Ph2= 7.8664


Wsw= 1.4400



Wiw= 0.0000 hiw: internal water depthUp=-U/B0 U= -2.9391

Q= 5.9206






internal water 0.0000 - 1.1500 - 0.0000uplift -6.7600 - 1.1500 - -7.7740total 13.6174 15.6600

6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.5328Pvt2 0.0000Wtop 0.8374Ws 1.2522total Pq= 8.0224

Wtop= (t2*BT+Hf^2)*gc/B0

P3=Ka*gd*GwdP4=Ka*gs*(D1+H0-Gwd)P5=gw*(D1+H0-Gwd)

Wsw=t1*H*gc



X = SM/SV =

q1 = SV/Bo + 6q2 = SV/Bo - 6


Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L1


Wtop= 0.8374Pvd= 5.4000

Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666

Pv1= 6.5704

2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.0333 we2= 0.0666 tf/m2

P3= 2.8350P4= 0.0000

Ph1= 3.0015


P3= 4.9050P4= -2.0000

Ph2= 3.0715


Wsw= 1.4400



Wiw= 1.7196 hiw: internal water depthUp=0 U= 0.0000

Q= 10.3795






internal water 3.9550 - 1.1500 - 4.5482uplift 0.0000 - 1.1500 - 0.0000total 23.8729 27.4538


Wtop= (t2*BT+Hf^2)*gc/B0Pvd=D*gd

P3=Ka*gd*D1WP=-gw*0

P3=Ka*gd*(D1+H0)WP=-gw*H

Wsw=t1*H*gc



X = SM/SV =

q1 = SV/Bo + 6q2 = SV/Bo - 6




Wtop= 0.8374Pvd= 5.4000

Pvt1 Pvt1= 0.5328Pvt2 Pvt2= 0.0000

Pv1= 6.7702

2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.1332 we2= 0.2664 tf/m2

P3= 2.8350P4= 0.0000

Ph1= 3.1014


P3= 4.9050P4= -2.0000

Ph2= 3.1714


Wsw= 1.4400



Wiw= 1.7196 hiw: internal water depthUp=0 U= 0.0000

Q= 10.5793






internal water 3.9550 - 1.1500 - 4.5482uplift 0.0000 - 1.1500 - 0.0000total 24.3324 27.9823


Summary of Load Calculation

Wtop= (t2*BT+Hf^2)*gc/B0Pvd=D*gd

P3=Ka*gd*D1WP=-gw*0

P3=Ka*gd*(D1+H0)WP=-gw*H

Wsw=t1*H*gc



X = SM/SV =

q1 = SV/Bo + 6q2 = SV/Bo - 6


Item Pv1 Ph1 Ph2 Pq Wsw q1Case (tf/m2) (tf/m2) (tf/m2) (tf/m2) (tf/m) (tf/m2)

Case.1 6.5704 3.1665 7.7665 7.8226 1.4400 5.7208Case.2 6.7702 3.2664 7.8664 8.0224 1.4400 5.9206Case.3 6.5704 3.0015 3.0715 7.8226 1.4400 10.3795Case.4 6.7702 3.1014 3.1714 8.0224 1.4400 10.5793


Class III Road3.0 m

0.51.00 t/m31.80 t/m32.00 t/m32.40 t/m3175 kgf/m2

60 kgf/m21400 kgf/m2

5.5 kgf/m23000 kgf/m2

241.2

2.00 m2.00 m0.15 m0.30 m0.30 m0.30 m2.60 m2.60 m3.00 m3.00 m (= D)0.00 m2.00 m

35.00 t

0.30.10 m0.25 m5.00 t0.10 m0.25 m1.25 t0.10 m0.25 m0.00 t/m2

t2/2 + H + t3/2 2.300 m2.300 m

3.150 m

(> 0.25m)(> 0.25m)(> 0.25m)


tf/m2, B = 2.300 mmm

tf/m2, B = 0.000 mmm

tf/m2, B = 2.300 mmm

tf/m2, B = 2.300 mmm

tf/m2, B = 2.300 mmm

tf/m2, B = 2.300 mmm

tf/m2, B = 2.300 mmm

tf/m2, B = 2.300 mmm

tf/m2, B = 0.000 mmm

(2) (2) (3) (3)a) + b) a) + c) a) + b) a) + c)

0.5328 0.3330 0.5328 0.2664 0.5328 tf/m2,

for Case.L2, which is the largest load to the top slab.

tf/mtf/m


hiw: internal water depth 0.00 m

acting point of resultant force1.150 m

e = B0/2 - X = 0.000 m

ground reaction5.7208 tf/m25.7208 tf/m2

X = SM/SV =

q1 = SV/Bo + 6SVe/Bo^2 =q2 = SV/Bo - 6SVe/Bo^2 =




e = B0/2 - X = 0.0000 m


X = SM/SV =


16/82 (2)document.xlsMSN

4 Analysis of Plane Frame


1) Calculation of Load Term

Ph1 Horizontal Pressure at top of side wall 3.166

Ph2 Horizontal Pressure at bottom of side wall 7.766

Pv1 Vertical Pressure(1) on top slab 6.570


Pq Reaction to bottom slab 7.823

a Distance from joint B to far end of Pv2 2.300 mb Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.300 mB0 Width of plane frame 2.300 mt1 Thickness of side wall 0.300 mt2 Thickness of top slab 0.300 mt3 Thickness of invert (bottom slab) 0.300 m

=

=

=

=

2) Calculation of Bending Moment at joint

k1 = 1.0

= 1.0000

= 1.0000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1

k1 k1 k1 k1 -4k1 R 0

As load has bilateral symmetry, the equation shown below is formed.

R =0

2k1+k3 k1=

k1 2k1+k2

3.0000 1.0=

-0.83584694

1.0 3.0000 0.68941361

By solving above equation, the result is led as shown below.

tf/m2

tf/m2

tf/m2

tf/m2

tf/m2

CAB = CDC = (2Ph1+3Ph2)H02/60

CBA = CCD = (3Ph1+2Ph2)H02/60

CBC = CCB = Pv1B02/12 + {(a2-b2)B0

2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B02

CDA = CAD = PqB02/12

k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t1

3)

2(k1+k3) qA CAB - CAD

2(k1+k2) qB CBC - CBA

2(k1+k2) qC CCD - CCB

2(k1+k3) qD CDA - CDC

qA = -qD qB = -qC

qA CAB - CAD

qB CBC - CBA

qA

qB

B

A

(t1)H0


= -0.39962 = -0.36301

= 0.36301 = 0.39962

qA qC

qB qD


= -3.0488

= 2.5334

= -2.5334

= 2.5334

= -2.5334

= 3.0488

= -3.0488

= 3.0488

2) Calculation of Design Force

2-1) Side Wall in left

a) Shearing Force at joint

w1 Load at end A 7.766

w2 Load at end B 3.166

Bending moment at end A -3.0488

Bending moment at end B 2.5334

L Length of member (=H0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m

= 7.392 tf= -5.181 tf

b) Shearing Force at 2d point from joint

Shearing force at the point with a distance of 2d from joint is calculated by following equation.

(i) In case of x1 = 0.480 mSx1 = 3.895 tf

(ii) In case of x2 = 1.820 mSx2 = -3.430 tf

c) Bending Moment

= -3.049

= -2.533

The maximum bending moment occurs at the point of that shearing force equal to zero.

= 7.3922 -7.7665 x + 1.0000 x2 , x =

Bending moment at x = 1.1106 m is;

= 0.828

MAB = k1(2qA +qB) - CAB tf･mMBA = k1(2qB+qA)+CBA tf･mMBC = k2(2qB+qC) - CBC tf･mMCB = k2(2qC+qB)+CCB tf･mMCD = k1(2qC+qD) - CCD tf･mMDC =k1 (2qD+ qC)+CDC tf･mMDA = k3(2qD+qA) - CDA tf･mMAD = k3(2qA+qD)+CAD tf･m

tf/m2

tf/m2

MAB tf･mMBA tf･m

SAB = (2w1+w2)L/6 - (MAB+MBA)/L

SBA = SAB - L(w1+w2)/2

Sx = SAB - w1x - (w2 - w1)x2/(2L)

MA = MAB tf･mMB = -MBA tf･m

Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)

Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB tf・m

w1w1w1

w2w1w1


2-2) Top Slaba) Shearing Force at joint

w1 Uniform load 6.570


a Distance from end B to near end of w2 0.000 mb Length of uniform load w2 2.300 m

Bending moment at end B -2.5334

Bending moment at end C 2.5334

L Length of member (=Bo) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m

= 7.556 tf

= -7.556 tf

b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.

in case of 0.000 m <= x <= 2.300 m



c) Bending Moment

= -2.533

= -2.533

The maximum bending moment occurs at the center of member due to symmetry loading distribution.Bending moment at x = 1.150 m is;

= 1.811

2-3) Side Wall in righta) Shearing Force at joint

w1 Load at end C 3.166

w2 Load at end D 7.766

Bending moment at end C -2.5334

Bending moment at end D 3.0488


= 5.181 tf

= -7.392 tf

tf/m2

tf/m2

MBC tf･mMCB tf･m

SBC = (w1L+w2b)/2-(MBC+MCB)/L

SCB = SBC -w1L - w2b

Sx = SBC - w1x - w2(x-a)

MB = MBC tf･mMC = -MCB tf･m

Mmax = SBCx - w1x2/2 - w2(x-a)2/2 + MBC tf・m

tf/m2

tf/m2

MCD tf･mMDC tf･m

SCD = (2w1+w2)L/6 - (MCD+MDC)/L

SDC = SCD - L(w1+w2)/2

ba

w1

xMBC

B

MCD

MDC

xL

C

D





c) Bending Moment

= -2.533

= -3.049


= 5.1807 -3.1665 x -1.0000 x2 , x =


= 0.82787

2-4) Bottom Slaba) Shearing Force at joint

w1 Reaction at end D 7.823

w2 Reaction at end A 7.823



L Length of member (=B0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m

= 8.996 tf

= -8.996 tf




c) Bending Moment

= -3.049

= -3.049


= 8.9959 -7.8226 x , x =

Sx = SCD - w1x - (w2 - w1)x2/(2L)

MC = MCD tf･mMD = -MDC tf･m

Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)

Mmax = SCDx - w1x2/2 - (w2-w1)x3/(6L) + MCD tf・m

tf/m2

tf/m2

MDA tf･mMAD tf･m

SDA = (2w1+w2)L/6 - (MDA+MAD)/L

SAD = SDA - L(w1+w2)/2

Sx = SDA- w1x - (w2 - w1)x2/(2L)

MD = MDA tf･mMA = -MAD tf･m

Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)

L

x

w2

A

MAD



= 2.124Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tf・m










=

=

=

=


k1 = 1.0

= 1.00000

= 1.00000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1

k1 k1 k1 k1 -4k1 R 0


R =0

2k1+k3 k1=

k1 2k1+k2

3.0000 1.0=

-0.87988511

1.0 3.0000 0.73345178


= -0.42164 = -0.38503

= 0.38503 = 0.42164

tf/m2

tf/m2

tf/m2

tf/m2

tf/m2

CAB = CDC = (2Ph1+3Ph2)H02/60

CBA = CCD = (3Ph1+2Ph2)H02/60

CBC = CCB = Pv1B02/12 + {(a2-b2)B0

2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B02


k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t1

3)





qA = -qD qB = -qC

qA CAB - CAD

qB CBC - CBA

qA

qB

qA qC

qB qD

B

A

(t1)H0


= -3.1149

= 2.5995

= -2.5995

= 2.5995

= -2.5995

= 3.1149

= -3.1149

= 3.1149

2) Calculation of Design Force2-1) Side Wall in left







= 7.5071 tf

= -5.2956 tf





c) Bending Moment

= -3.115

= -2.599


= 7.5071 -7.8664 x + 1.0000 x2 , x =


= 0.828


tf/m2

tf/m2




Sx = SAB - w1x - (w2 - w1)x2/(2L)


Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)


w1w1w1

w2w1w1









= 7.786 tf

= -7.786 tf


in case of 0.000 m <= x <= 2.300 m



c) Bending Moment

= -2.599

= -2.599


= 1.877







= 5.296 tf

tf/m2

tf/m2







tf/m2

tf/m2



ba

w1

xMBC

B

MCD

MDC

xL

C

D


= -7.507 tfSDC = SCD - L(w1+w2)/2





c) Bending Moment

= -2.599

= -3.115


= 5.2956 -3.2664 x -1.0000 x2 , x =


= 0.8278







= 9.226 tf

= -9.226 tf




c) Bending Moment

= -3.115

= -3.115


= 9.2257 -8.0224 x , x =

Sx = SCD - w1x - (w2 - w1)x2/(2L)


Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)


tf/m2

tf/m2




Sx = SDA- w1x - (w2 - w1)x2/(2L)


Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)

L

x

w2

A

MAD










=

=

=

=


k1 = 1.0

= 1.00000

= 1.00000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1

k1 k1 k1 k1 -4k1 R 0


R =0

2k1+k3 k1=

k1 2k1+k2

3.0000 1.0=

-2.10676944

1.0 3.0000 1.56094111


= -0.98516 = -0.84870

= 0.84870 = 0.98516

tf/m2

tf/m2

tf/m2

tf/m2

tf/m2

CAB = CDC = (2Ph1+3Ph2)H02/60

CBA = CCD = (3Ph1+2Ph2)H02/60

CBC = CCB = Pv1B02/12 + {(a2-b2)B0

2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B02


k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t1

3)





qA = -qD qB = -qC

qA CAB - CAD

qB CBC - CBA

qA

qB

qA qC

qB qD

B

A

(t1)H0


= -2.46329

= 2.04774

= -2.04774

= 2.04774

= -2.04774

= 2.46329

= -2.46329

= 2.46329








= 3.686 tf

= -3.298 tf



(i) In case of x1 = 0.480 m

Sx1 = 2.215 tf

(ii) In case of x2 = 1.820 m

Sx2 = -1.854 tf

c) Bending Moment

= -2.463

= -2.048


= 3.6861 -3.0715 x + 0.0152 x2 , x =


= -0.243


tf/m2

tf/m2




Sx = SAB - w1x - (w2 - w1)x2/(2L)


Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)


w1w1w1

w2w1w1









= 7.556 tf

= -7.556 tf


in case of 0.000 m <= x <= 2.300 m

(i) In case of x1 = 0.480 m

Sx1 = 4.402 tf

(ii) In case of x2 = 1.820 m

Sx2 = -4.402 tf

c) Bending Moment

= -2.048

= -2.048


= 2.297







= 3.298 tf

tf/m2

tf/m2







tf/m2

tf/m2



ba

w1

xMBC

B

MCD

MDC

xL

C

D


= -3.686 tfSDC = SCD - L(w1+w2)/2





c) Bending Moment

= -2.048

= -2.463


= 3.2979 -3.0015 x -0.0152 x2 , x =


= -0.243







= 8.996 tf

= -8.996 tf




c) Bending Moment

= -2.463

= -2.463


= 8.9959 -7.8226 x , x =

Sx = SCD - w1x - (w2 - w1)x2/(2L)


Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)


tf/m2

tf/m2




Sx = SDA- w1x - (w2 - w1)x2/(2L)


Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)

L

x

w2

A

MAD










=

=

=

=


k1 = 1.0

= 1.00000

= 1.00000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1

k1 k1 k1 k1 -4k1 R 0


R =0

2k1+k3 k1=

k1 2k1+k2

3.0000 1.0=

-2.15080761

1.0 3.0000 1.60497928


= -1.00718 = -0.87072

= 0.87072 = 1.00718

tf/m2

tf/m2

tf/m2

tf/m2

tf/m2

CAB = CDC = (2Ph1+3Ph2)H02/60

CBA = CCD = (3Ph1+2Ph2)H02/60

CBC = CCB = Pv1B02/12 + {(a2-b2)B0

2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B02


k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t1

3)





qA = -qD qB = -qC

qA CAB - CAD

qB CBC - CBA

qA

qB

qA qC

qB qD

B

A

(t1)H0


= -2.52934

= 2.11380

= -2.11380

= 2.11380

= -2.11380

= 2.52934

= -2.52934

= 2.52934








= 3.801 tf

= -3.413 tf





c) Bending Moment

= -2.529

= -2.114


= 3.8009 -3.1714 x + 0.0152 x2 , x =


= -0.243


tf/m2

tf/m2




Sx = SAB - w1x - (w2 - w1)x2/(2L)


Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)


w1w1w1

w2w1w1









= 7.786 tf

= -7.786 tf


in case of 0.000 m <= x <= 2.300 m



c) Bending Moment

= -2.114

= -2.114


= 2.363







= 3.413 tf

tf/m2

tf/m2







tf/m2

tf/m2



ba

w1

xMBC

B

MCD

MDC

xL

C

D


= -3.801 tfSDC = SCD - L(w1+w2)/2





c) Bending Moment

= -2.114

= -2.529


= 3.4128 -3.1014 x -0.0152 x2 , x =


= -0.243







= 9.226 tf

= -9.226 tf




c) Bending Moment

= -2.529

= -2.529


= 9.2257 -8.0224 x , x =

Sx = SCD - w1x - (w2 - w1)x2/(2L)


Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)


tf/m2

tf/m2




Sx = SDA- w1x - (w2 - w1)x2/(2L)


Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)

L

x

w2

A

MAD


2.61260

2.20703

2.89644

3.44844

tf･mtf･mtf･mtf･m

(t2)

(t1)

B0

(t3)

C

D


6.6561.111

A

Bw1w1

Lxw1w1

MAB

MBA

w1w1


w2

b

MCBx

L

C

w1

w2

MCD

MDC

C

D


-4.3561.189

1.150

L

x

D

MDA

w1


2.65664

2.25107

2.98452

3.53652


(t2)

(t1)

B0

(t3)

C

D


6.7551.111

A

Bw1w1

Lxw1w1

MAB

MBA

w1w1


w2

b

MCBx

L

C

w1

w2

MCD

MDC

C

D


-4.45511.1887

1.1500

L

x

D

MDA

w1


1.34167

1.33550

2.89644

3.44844


(t2)

(t1)

B0

(t3)

C

D


200.6341.207

A

Bw1w1

Lxw1w1

MAB

MBA

w1w1


w2

b

MCBx

L

C

w1

w2

MCD

MDC

C

D


-198.331.093

1.150

L

x

D

MDA

w1


1.38571

1.37954

2.98452

3.53652


(t2)

(t1)

B0

(t3)

C

D


207.2001.205

A

Bw1w1

Lxw1w1

MAB

MBA

w1w1


w2

b

MCBx

L

C

w1

w2

MCD

MDC

C

D


-204.91.095

1.150

L

x

D

MDA

w1

58/82 (3)document.xlsSum MSN

Summary of Internal forces

Member Case Check PointM N S (tf)

(tf) at joint at 2d

Side wall A -3.049 8.996 7.392 3.895 (left) Case.1 Middle 0.828 8.301 0.000 -

B -2.533 7.556 -5.181 -3.430A -3.115 9.226 7.507 3.962

Case.2 Middle 0.828 8.530 0.000 -B -2.599 7.786 -5.296 -3.497A -2.463 8.996 3.686 2.215

Case.3 Middle -0.243 8.240 0.000 -B -2.048 7.556 -3.298 -1.854A -2.529 9.226 3.801 2.282

Case.4 Middle -0.243 8.471 0.000 -B -2.114 7.786 -3.413 -1.921

Top slab B -2.533 5.181 7.556 4.402Case.1 Middle 1.811 5.181 0.000 -

C -2.533 5.181 -7.556 -4.402B -2.599 5.296 7.786 4.536

Case.2 Middle 1.877 5.296 0.000 -C -2.599 5.296 -7.786 -4.536B -2.048 3.298 7.556 4.402

Case.3 Middle 2.297 3.298 0.000 -C -2.048 3.298 -7.556 -4.402B -2.114 3.413 7.786 4.536

Case.4 Middle 2.363 3.413 0.000 -C -2.114 3.413 -7.786 -4.536

Side wall C -2.533 7.556 5.181 3.430 (right) Case.1 Middle 0.828 8.301 0.000 -

D -3.049 8.996 -7.392 -3.895C -2.599 7.786 5.296 3.497

Case.2 Middle 0.828 8.530 0.000 -D -3.115 9.226 -7.507 -3.962C -2.048 7.556 3.298 1.854

Case.3 Middle -0.243 8.240 0.000 -D -2.463 8.996 -3.686 -2.215C -2.114 7.786 3.413 1.921

Case.4 Middle -0.243 8.471 0.000 -D -2.529 9.226 -3.801 -2.282

Invert D -3.049 7.392 8.996 5.241Case.1 Middle 2.124 7.392 0.000 -

A -3.049 7.392 -8.996 -5.241D -3.115 7.507 9.226 5.375

Case.2 Middle 2.190 7.507 0.000 -A -3.115 7.507 -9.226 -5.375D -2.463 3.686 8.996 5.241

Case.3 Middle 2.709 3.686 0.000 -A -2.463 3.686 -8.996 -5.241D -2.529 3.801 9.226 5.375

Case.4 Middle 2.775 3.801 0.000 -A -2.529 3.801 -9.226 -5.375

(tf･m)

59/ 82 (4)document.xls, R-bar req

5 Calculation of Required Reinforcement Bar5-1 Calculation of Required Reinforcement Bar

1)Case.1

M= 3.0488 60 kgf/m2 h = 30 cm (height of member)

N= 8.9959 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.3922 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.8947 tf c = 9.00 cm (distance from neutral axis)

b = 100 cme = M/N = 33.89 cm

38.729 kgf/cm2 ( 0 kgf/cm2) o.k.

+67.4 -2344.55 -68382.700.39906.8199 cm2

Case.2



b = 100 cme = M/N = 33.76 cm

39.304 kgf/cm2 ( 4.112E-06 kgf/cm2) o.k.

+66.95 -2397.25 -69919.910.40266.9718 cm2

Case.3



b = 100 cme = M/N = 27.38 cm

34.762 kgf/cm2 ( 0 kgf/cm2) o.k.

+70.45 -1988.75 -58005.350.37344.7003 cm2

Case.4



b = 100 cme = M/N = 27.42 cm

35.359 kgf/cm2 ( 0 kgf/cm2) o.k.

+70 -2041.46 -59542.560.37744.8482 cm2

The maximum requirement of reinforcement bar is 6.9718 cm2 in Case. 2 from above calculation.

Case. 1 2 3 4Requirement 6.8199 6.9718 4.7003 4.8482 (cm2)

2)Case.1



b = 100 cme = M/N = 33.53 cm

34.350 kgf/cm2 ( 0 kgf/cm2) o.k.

+70.76 -1952.63 -56951.650.37065.5151 cm2

Case.2


At Joint "A"　of side wall

tf･m sca =

tf ssa =

Solving the formula shown below, sc = sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 0 0 = sc^3 sc^2 scs = nsc/(nsc+ssa) =Asreq = (sc*s/2 - N/(bd))bd/ssa =

tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


At Joint "B"　of side wall

tf･m sca =

tf ssa =


tf･m sca =

h

h



b = 100 cme = M/N = 33.39 cm

34.950 kgf/cm2 ( 0 kgf/cm2) o.k.

+70.31 -2005.33 -58488.850.37475.6627 cm2

tf ssa =



Case.3



b = 100 cme = M/N = 27.10 cm

30.914 kgf/cm2 ( 0 kgf/cm2) o.k.

+73.29 -1657.5 -48343.890.34643.7812 cm2

Case.4



b = 100 cme = M/N = 27.15 cm

31.538 kgf/cm2 ( 0 kgf/cm2) o.k.

+72.84 -1710.21 -49881.100.35093.9250 cm2



3)Case.1



b = 100 cme = M/N = 48.90 cm

32.854 kgf/cm2 ( 0 kgf/cm2) o.k.

+71.88 -1822.73 -53163.050.36036.4454 cm2

Case.2



b = 100 cme = M/N = 49.09 cm

33.391 kgf/cm2 ( 0 kgf/cm2) o.k.

+71.48 -1869.15 -54517.010.36406.6367 cm2

Case.3



b = 100 cme = M/N = 62.09 cm

28.090 kgf/cm2 ( 0 kgf/cm2) o.k.

+75.29 -1424.64 -41552.070.32505.4700 cm2

Case.4



b = 100 cme = M/N = 61.94 cm

28.662 kgf/cm2 ( 0.0002919 kgf/cm2) o.k.

tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


At Joint "B"　of top slab

tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =

Solving the formula shown below, sc =

h


+74.89 -1471.06 -42906.030.32955.6565 cm2



sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 0 0 = sc^3 sc^2 scs = nsc/(nsc+ssa) =Asreq = (sc*s/2 - N/(bd))bd/ssa =


4)Case.1



b = 100 cme = M/N = 41.24 cm

37.763 kgf/cm2 ( 5.853E-05 kgf/cm2) o.k.

+68.16 -2256.85 -65824.690.39307.4396 cm2

Case.2



b = 100 cme = M/N = 41.49 cm

38.275 kgf/cm2 ( 0 kgf/cm2) o.k.

+67.76 -2303.27 -67178.650.39627.6354 cm2

Case.3



b = 100 cme = M/N = 66.83 cm

31.398 kgf/cm2 ( 0 kgf/cm2) o.k.

+72.94 -1698.37 -49535.810.34996.7842 cm2

Case.4



b = 100 cme = M/N = 66.55 cm

31.945 kgf/cm2 ( 2.754E-06 kgf/cm2) o.k.

+72.54 -1744.79 -50889.770.35396.9741 cm2



5) At Middle of side wallCase.1


N= 8.3006 1400 kgf/m2 d = 24 cm (effective height of member)n = 24 d' = 6 cm (protective covering depth)

c = 9.00 cm (distance from neutral axis)b = 100 cm

e = M/N = 9.97 cm21.9840 kgf/cm2 ( 0 kgf/cm2) o.k.

+79.3 -956.98 -27912.010.27370.0000 cm2 Tensile is on inside of member

Case.2



c = 9.00 cm (distance from neutral axis)

At Joint "A"　of invert

tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =

h

h


b = 100 cme = M/N = 9.70 cm

22.158 kgf/cm2 ( 0.0 kgf/cm2) o.k.




Case.3





+82.37 -598.07 -17443.750.22190.0000 cm2 Tensile is on outside of member

Case.4





+82.26 -610.8 -17814.870.22400.0000 cm2 Tensile is on outside of member


Case. 1 2 3 4Requirement 0.0000 0.0000 - - (cm2)Side inside inside outside outside

6) At Middle of top slabCase.1






Case.2






Case.3






Case.4




e = M/N = 69.24 cm30.496 kgf/cm2 ( 1.843E-05 kgf/cm2) o.k.

tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =

Solving the formula shown below, sc =

h




Case. 1 2 3 4Requirement 3.8899 4.0759 6.3478 6.5363 (cm2)Side inside inside inside inside

sc^3+{3ssa/(2n) - 3N(e+c)/(bd^2)}sc^2 - 6N(e+c)ssasc/(nbd^2) - 3N(e+c)ssa^2/(n^2bd^2) = 0 0 = sc^3 sc^2 scs = nsc/(nsc+ssa) =Asreq = (sc*s/2 - N/(bd))bd/ssa =


7) At Middle of invertCase.1






Case.2






Case.3




e = M/N = 73.50 cm33.146 kgf/cm2 ( 4.799E-07 kgf/cm2) o.k.


Case.4







Case. 1 2 3 4Requirement 4.1161 4.3051 7.6614 7.8531 (cm2)Side inside inside inside inside

8) Summary of required reinforcement

Required reinforcement for design is the maximum required reinforcement calculated above in 1) - 4).

Item Side wall Side wall Top slab Invert Side wall Top slab InvertPoint bottom top end end middle middle middleSide outside outside outside outside inside inside insideCalculation 1) 2) 3) 4) 5) 6) 7)Requirement 6.972 5.663 6.637 7.635 0.000 6.536 7.853 (cm2)

tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


tf･m sca =

tf ssa =


h


cm (effective height of member)cm (protective covering depth)cm (distance from neutral axis)





d1

d

d1

d








d1

d








d1

d

d1

d








d1

d

74/82 (5)document.xlsR-bar stress

6 Bar Arrangement and Calculation of Stress Type: B2.00m x H2.00m

Side wall Top slabbottom middle top endoutside inside outside outside

Bending moment M kgfcm 311,488 82,787 259,949 259,949 Shearing force (joint) S kgf 7,507 0 5,296 7,786 Shearing force (2d) S2d kgf 3,962 - 3,497 4,536 Axial force N kgf 9,226 8,301 7,786 5,296 Height of member h cm 30 30 30 30Covering depth d' cm 6 6 6 6Effective height d cm 24 24 24 24Effective width b cm 100 100 100 100Effective area bd cm2 2400 2400 2400 2400Young's modulus ratio n - 24 24 24 24

Required R-bar Asreq cm2 6.97 0.00 5.66 6.64

R-bar arrangement 16@250 12@250 16@250 16@250

Reinforcement As cm2 8.04 4.52 8.04 8.04Perimeter of R-bar U ｃｍ 20.11 15.08 20.11 20.11M/N e cm 33.763 9.974 33.388 49.088Dist. from neutral axis c cm 9.00 9.00 9.00 9.00

a' 56.3 -15.1 55.2 102.3b' 495.1 123.5 490.8 672.5c' -11882.3 -2963.9 -11778.0 -16140.4x 10.16 17.62 10.19 9.38

0.000 0.000 0.000 0.000(check) ok ok ok ok

Compressive stress kgf/cm2 37.7 9.9 31.5 31.4Allowable stress kgf/cm2 60.0 60.0 60.0 60.0

ok ok ok okTensile stress kgf/cm2 1232.9 85.7 1023.7 1174.4Allowable stress kgf/cm2 1400.0 1400.0 1400.0 1400.0

ok ok ok okShearing stress at joint t kgf/cm2 3.13 0.00 2.21 3.24Allowable stress kgf/cm2 11.00 11.00 11.00 11.00

ok ok ok okShearing stress at 2d kgf/cm2 1.65 - 1.46 1.89Allowable stress kgf/cm2 5.50 - 5.50 5.50

ok - ok ok

Resisting Moment Mr kgfcm 496,869 362,613 497,615 459,243 Mr for compression Mrc kgfcm 496,869 410,229 497,615 499,428 x for Mrc cm 9.240 7.521 9.013 8.634

kgf/cm2 2300.3 3155.3 2394.6 2562.9 Mr for tensile Mrs kgfcm 577,461 362,613 532,234 459,243 x for Mrs cm 12.982 10.971 12.517 11.670

kgf/cm2 68.7 49.1 63.6 55.2

Distribution bar (>As/6 and >Asmin) 12@250 12@250 12@250 12@250Reinforcement As cm2 4.52 4.52 4.52 4.52

ok ok ok okReinforcement bar for fillet Err:504 Err:504Reinforcement As cm2 Err:504 Err:504

scsca

ssssa

ta

t2dt2da

ss for Mrc

sc for Mrs


Minimum requirement of reinforcement bar As min = 4.5 cm2


Top slab Invertmiddle end middleinside outside inside

236,298 311,488 277,544 0 9,226 0

- 5,375 -3,413 7,507 3,801

30 30 306 6 6

24 24 24100 100 100

2400 2400 240024 24 24

6.54 7.64 7.85

16@250 16@250 12@125

8.04 8.04 9.0520.11 20.11 30.16

69.239 41.492 73.0209.00 9.00 9.00

162.7 79.5 174.1905.8 584.6 1068.9

-21739.8 -14030.0 -25653.18.92 9.69 9.27

0.000 0.000 0.000ok ok ok

28.5 37.7 32.260.0 60.0 60.0

ok ok ok1155.0 1336.1 1227.41400.0 1400.0 1400.0

ok ok ok0.00 3.84 0.00

11.00 11.00 11.00ok ok ok- 2.24 -- 5.50 -- ok -

408,023 497,785 469,708 501,254 497,785 521,262

8.359 8.969 8.7872694.5 2413.1 2492.9

408,023 523,749 469,708 10.986 12.426 11.642

49.2 62.6 55.0

12@250 12@250 [email protected] 4.52 4.52

ok ok ok

DATA PLOT PEMBESIAN

Section of Culvert

1. Design Data

b1 = 0.300 m b2 = 2.000 m

h1 = 0.300 m h2 = 2.000 m h3 = 0.300 m d =

Bb = 2.600 m Ha = 2.600 m Fillet = 0.15 m

2. Data Pembesian

Bottom slab : Tumpuan Lapangan Tulangan bagi :

: O 16 @ 250 O 12 @ 125 O 12 @

: O 12 @ 125 O 16 @ 250

Side wall : Tumpuan Lapangan Tulangan bagi :

: O 16 @ 250 O 12 @ 250 O 12 @

: O 12 @ 250 O 16 @ 250

Top slab : Tumpuan Lapangan Tulangan bagi :

: O 16 @ 250 O 16 @ 250 O 12 @

: O 16 @ 250 O 16 @ 250

3. Tulangan Miring (fillet) :

Bottom slab : O ### @ Err:504

Top slab : O ### @ Err:504

4. Nama Bangunan : Culvert Type 1 (2 m x 2 m)

Lokasi : …. Irrigation Project

As1 (cm2)

As2 (cm2)

As1 (cm2)

As2 (cm2)

As1 (cm2)

As2 (cm2)

b1b1 b2

h3

h2

h1

Bb

Ha

Section of Culvert

0.06 m

Tulangan bagi :

250

Tulangan bagi :

250

Tulangan bagi :

250

DATA PLOT MOMENT AND SHEAR FORCE DIAGRAM

Section of Culvert

1. Design Data

b1 = 0.300 m b2 = 2.000 m

h1 = 0.300 m h2 = 2.000 m h3 = 0.300 m

Bb = 2.300 m Ha = 2.300 m

2. Result Calculation

Side wall (left) Ma Mmax Mb Sa Sb Na

Case 1 -3.049 0.828 -2.533 7.392 -5.181 8.996

Case 2 -2.463 -0.243 -2.048 3.686 -3.298 8.996

Side wall (right) Mc Mmax Md Sd Sc Nd

Case 1 -2.533 0.828 -3.049 -7.392 5.181 8.996

Case 2 -2.048 -0.243 -2.463 -3.686 3.298 8.996

Top slab Mb Mmax Mc Sb Sc Nb

Case 1 -2.533 1.811 -2.533 7.556 -7.556 5.181

Case 2 -2.048 2.297 -2.048 7.556 -7.556 3.298

Bottom slab Ma Mmax Md Sa Sd Na

Case 1 -3.049 2.124 -3.049 -8.996 8.996 7.392

Case 2 -2.463 2.709 -2.463 -8.996 8.996 3.686

3. Nama Bangunan : Culvert Type 1 (2 m x 2 m)

b1b1 b2

h3

h2

h1

Bb

Ha

Lokasi : …. Irrigation Project

Case 1Mmax x x1 x2 x3 x4 x5 x6Mab = 1.1106 -1.24 -0.02 0.65 0.82 0.54 -0.14Mcd = 1.1894 -1.24 -0.02 0.65 0.82 0.54 -0.14Mmax y1 y2 y3 y4 y5 y6Mbc = 1.15 -0.63 0.73 1.54 1.81 1.54 0.73Mad = 1.15 -0.79 0.83 1.80 2.12 1.80 0.83

Ha = 2.300 0.288 0.575 0.862 1.150 1.438 1.725

Bb = 2.300 0.288 0.575 0.862 1.150 0.862 0.575

x x1 x2 x3 x4 x5 x6Sab 2.300 5.24 3.26 1.44 -0.22 -1.71 -3.03

Case 2Mmax x x1 x2 x3 x4 x5 x6Mab = 1.1106 -1.53 -0.85 -0.42 -0.25 -0.32 -0.65Mcd = 1.1894 -1.53 -0.85 -0.42 -0.25 -0.32 -0.65Mmax y1 y2 y3 y4 y5 y6Mbc = 1.15 -0.15 1.21 2.03 2.30 2.03 1.21Mad = 1.15 -0.20 1.42 2.39 2.71 2.39 1.42

Ha = 2.300 0.288 0.575 0.862 1.150 1.438 1.725

Bb = 2.300 0.288 0.575 0.862 1.150 0.862 0.575

x x1 x2 x3 x4 x5 x6Sab 2.300 2.80 1.92 1.05 0.17 -0.70 -1.57

Section of Culvert

Na Nb MAB MBA

8.996 7.556 -3.049 2.533

8.996 7.556 -2.463 2.048

Nd Nc MCD MDC

8.996 7.556 -2.533 3.049

8.996 7.556 -2.048 2.463

Nb Nc MBC MCB

5.181 5.181 -2.533 2.533

3.298 3.298 -2.048 2.048

Na Nd MAD MDA

7.392 7.392 3.049 -3.049

3.686 3.686 2.463 -2.463

h3

h2

h1

x6 x7 q1 q2-0.14 -1.18 7.766 3.166-0.14 -1.18 3.166 7.766

y6 y7 q10.73 -0.63 6.5700.83 -0.79 7.823

1.725 2.012

0.575 0.288

x6 x7-3.03 -4.19

x6 x7 q1 q2-0.65 -1.22 3.071 3.001-0.65 -1.22 3.001 3.071

y6 y7 q11.21 -0.15 6.5701.42 -0.20 7.823

1.725 2.012

0.575 0.288

x6 x7-1.57 -2.43

bc1-01singlebox

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