1 / 114 276341623.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

0.0Pedestrian 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

c= tf/m3s= tf/m3s'= tf/m3

(D4.0m)tf/m2

ck= kgf/cm2

ca= kgf/cm2

a= kgf/cm2

sa= kgf/cm2

sy= kgf/cm2

2 / 114 276341623.xls/Summary

3/ 114 (1)276341623.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 TireBottom 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

w:d:s:c:ck:casa:a:sy:

Impact coefficient (D4.0m:0, D

4/ 114 (1)276341623.xls, Load

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*w

BT*{(D-Gwd)*(sw)+Gwd*d}(HT*BT-H*B+2*Hf^2)*c

5/ 114 (1)276341623.xls, Load

6/ 114 (1)276341623.xls, Load

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)*c/B0

P3=Ka*d*GwdP4=Ka*s*(D1+H0-Gwd)P5=w*(D1+H0-Gwd)

Wsw=t1*H*c

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

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

X = M/V =

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

7/ 114 (1)276341623.xls, Load

Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L2

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

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

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

P3= 2.7000P4= 2.4500P5= 2.4500

Ph2= 7.8664

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= 0.5328Pvt2 Pvt2= 0.0000

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

Q= 5.9206

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 1.2254 - 1.1500 - 1.4092Pvt2 0.0000 - 1.1500 - 0.0000

soil pressure side wall (left) - 12.8027 - 0.9916 12.6953side wall (right) - -12.8027 - 0.9916 -12.6953

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)*c/B0

P3=Ka*d*GwdP4=Ka*s*(D1+H0-Gwd)P5=w*(D1+H0-Gwd)

Wsw=t1*H*c

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

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

X = M/V =

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

8/ 114 (1)276341623.xls, Load

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

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

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

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

P3= 4.9050P4= -2.0000

Ph2= 3.0715

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

Q= 10.3795

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) - 6.9839 - 1.1456 8.0007side wall (right) - -6.9839 - 1.1456 -8.0007

internal water 3.9550 - 1.1500 - 4.5483uplift 0.0000 - 1.1500 - 0.0000total 23.8729 27.4538

6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.2664Pvt2 0.0666Wtop 0.8374Ws 1.2522total Pq= 7.8226

Wtop= (t2*BT+Hf^2)*c/B0Pvd=D*d

P3=Ka*d*D1WP=-w*0

P3=Ka*d*(D1+H0)WP=-w*H

Wsw=t1*H*c

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

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

X = M/V =

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

9/ 114 (1)276341623.xls, Load

Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2

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

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

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

P3= 4.9050P4= -2.0000

Ph2= 3.1714

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= 0.5328Pvt2 Pvt2= 0.0000

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

Q= 10.5793

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 1.2254 - 1.1500 - 1.4092Pvt2 0.0000 - 1.1500 - 0.0000

soil pressure side wall (left) - 7.2137 - 1.1457 8.2649side wall (right) - -7.2137 - 1.1457 -8.2649

internal water 3.9550 - 1.1500 - 4.5483uplift 0.0000 - 1.1500 - 0.0000total 24.3324 27.9823

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

Summary of Load Calculation

Item Pv1 Ph1 Ph2 Pq Wsw q1

Wtop= (t2*BT+Hf^2)*c/B0Pvd=D*d

P3=Ka*d*D1WP=-w*0

P3=Ka*d*(D1+H0)WP=-w*H

Wsw=t1*H*c

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

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

X = M/V =

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

10/ 114 (1)276341623.xls, Load

Case (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

11/ 114 (1)276341623.xls, Load

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)

12/ 114 (1)276341623.xls, Load

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

13/ 114 (1)276341623.xls, Load

14/ 114 (1)276341623.xls, Load

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 = M/V =

q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =

15/ 114 (1)276341623.xls, Load

hiw: internal water depth 0.00 m

acting point of resultant force1.1500 m

e = B0/2 - X = 0.0000 m

ground reaction5.9206 tf/m25.9206 tf/m2

X = M/V =

q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =

16/ 114 (1)276341623.xls, Load

hiw: internal water depth 2.000 m

acting point of resultant force1.1500 m

e = B0/2 - X = 0.0000 m

ground reaction10.3795 tf/m210.3795 tf/m2

X = M/V =

q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =

17/ 114 (1)276341623.xls, Load

hiw: internal water depth 2.000 m

acting point of resultant force1.1500 m

e = B0/2 - X = 0.0000 m

ground reaction10.5793 tf/m210.5793 tf/m2

X = M/V =

q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =

18/ 114 (1)276341623.xls, Load

19/114 (2)276341623.xlsMSN

4 Analysis of Plane Frame

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

1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.166Ph2 Horizontal Pressure at bottom of side wall 7.766Pv1 Vertical Pressure(1) on top slab 6.570Pv2 Vertical Pressure(2) on top slab 0.000Pq Reaction to bottom slab 7.823a 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=

-0.835846941.0 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)B02/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B02

CDA = CAD = PqB02/12

k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t13)

2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA

2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC

A = -D B = -C

A CAB - CADB CBC - CBA

AB

B

A

(t1)H0

20/114 (2)276341623.xlsMSN

= -0.39962 = -0.36301= 0.36301 = 0.39962

A CB D

21/114 (2)276341623.xlsMSN

= -3.0488

= 2.5334

= -2.5334

= 2.5334

= -2.5334

= 3.0488

= -3.0488

= 3.0488

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

a) Shearing Force at joint

w1 Load at end A 7.766w2 Load at end B 3.166

Bending moment at end A -3.0488

Bending moment at end B 2.5334L 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 jointShearing 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(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm

tf/m2

tf/m2

MAB tfmMBA tfm

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

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

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

MA = MAB tfmMB = -MBA tfm

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

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

w1w1w1

w2w1w1

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23/114 (2)276341623.xlsMSN

2-2) Top Slaba) Shearing Force at joint

w1 Uniform load 6.570w2 Uniform load 0.000a 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 -2.5334Bending 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

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

(i) In case of x1 = 0.480 mSx1 = 3.430 tf

(ii) In case of x2 = 1.820 mSx2 = -3.895 tf

c) Bending Moment= -2.533

= -3.049

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

= 5.1807 -3.1665 x -1.0000 x2 , x =

Bending moment at x = 1.1894 m is;

= 0.82787

2-4) Bottom Slaba) Shearing Force at joint

w1 Reaction at end D 7.823w2 Reaction at end A 7.823

Bending moment at end B -3.04882

Bending moment at end C 3.04882L 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

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

(i) In case of x1 = 0.480 mSx1 = 5.241 tf

(ii) In case of x2 = 1.820 mSx2 = -5.241 tf

c) Bending Moment= -3.049= -3.049

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

= 8.9959 -7.8226 x , x =

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

MC = MCD tfmMD = -MDC tfm

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

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

tf/m2

tf/m2

MDA tfmMAD tfm

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

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

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

MD = MDA tfmMA = -MAD tfm

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

L

x

w2

A

MAD

26/114 (2)276341623.xlsMSN

Bending moment at x = 1.1500 m is;

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

27/114 (2)276341623.xlsMSN

Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L2

1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.266Ph2 Horizontal Pressure at bottom of side wall 7.866Pv1 Vertical Pressure(1) on top slab 6.770Pv2 Vertical Pressure(2) on top slab 0.000Pq Reaction to bottom slab 8.022a 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= 1.00000

= 1.00000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1k1 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=

-0.879885111.0 3.0000 0.73345178

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

= -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/B0

2

CDA = CAD = PqB02/12

k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t1

3)

2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA

2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC

A = -D B = -C

A CAB - CADB CBC - CBA

AB

A CB D

B

A

(t1)H0

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= -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

a) Shearing Force at joint

w1 Load at end A 7.866w2 Load at end B 3.266

Bending moment at end A -3.1149

Bending moment at end B 2.5995L 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

= 7.5071 tf= -5.2956 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.

(i) In case of x1 = 0.480 mSx1 = 3.962 tf

(ii) In case of x2 = 1.82 mSx2 = -3.497 tf

c) Bending Moment= -3.115

= -2.599

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

= 7.5071 -7.8664 x + 1.0000 x2 , x =

Bending moment at x = 1.1113 m is;

= 0.828

MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm

tf/m2

tf/m2

MAB tfmMBA tfm

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

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

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

MA = MAB tfmMB = -MBA tfm

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

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

w1w1w1

w2w1w1

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2-2) Top Slaba) Shearing Force at joint

w1 Uniform load 6.770w2 Uniform load 0.000a 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 -2.5995

Bending moment at end C 2.5995L 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

= 7.786 tf

= -7.786 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

32/114 (2)276341623.xlsMSN

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

33/114 (2)276341623.xlsMSN

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

(i) In case of x1 = 0.480 mSx1 = 3.497 tf

(ii) In case of x2 = 1.820 mSx2 = -3.962 tf

c) Bending Moment= -2.599

= -3.115

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

= 5.2956 -3.2664 x -1.0000 x2 , x =

Bending moment at x = 1.1887 m is;

= 0.8278

2-4) Bottom Slaba) Shearing Force at joint

w1 Reaction at end D 8.022w2 Reaction at end A 8.022

Bending moment at end B -3.1149

Bending moment at end C 3.1149L 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

= 9.226 tf= -9.226 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.

(i) In case of x1 = 0.480 mSx1 = 5.375 tf

(ii) In case of x2 = 1.820 mSx2 = -5.375 tf

c) Bending Moment= -3.115

= -3.115The maximum bending moment occurs at the point of that shearing force equal to zero.

= 9.2257 -8.0224 x , x =

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

MC = MCD tfmMD = -MDC tfm

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

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

tf/m2

tf/m2

MDA tfmMAD tfm

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

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

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

MD = MDA tfmMA = -MAD tfm

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

L

x

w2

A

MAD

34/114 (2)276341623.xlsMSN

Bending moment at x = 1.1500 m is;

= 2.190Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm

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

1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.001Ph2 Horizontal Pressure at bottom of side wall 3.071Pv1 Vertical Pressure(1) on top slab 6.570Pv2 Vertical Pressure(2) on top slab 0.000Pq Reaction to bottom slab 7.823a 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= 1.00000

= 1.00000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1k1 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=

-2.106769441.0 3.0000 1.56094111

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

= -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/B0

2

CDA = CAD = PqB02/12

k2 = H0t23/(B0t1

3)

k3 = H0t33/(B0t1

3)

2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA

2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC

A = -D B = -C

A CAB - CADB CBC - CBA

AB

A CB D

B

A

(t1)H0

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= -2.46329

= 2.04774

= -2.04774

= 2.04774

= -2.04774

= 2.46329

= -2.46329

= 2.46329

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

a) Shearing Force at joint

w1 Load at end A 3.071w2 Load at end B 3.001

Bending moment at end A -2.4633

Bending moment at end B 2.0477L 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

= 3.686 tf= -3.298 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.

(i) In case of x1 = 0.480 mSx1 = 2.215 tf

(ii) In case of x2 = 1.820 mSx2 = -1.854 tf

c) Bending Moment= -2.463

= -2.048

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

= 3.6861 -3.0715 x + 0.0152 x2 , x =

Bending moment at x = 1.2073 m is;

= -0.243

MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm

tf/m2

tf/m2

MAB tfmMBA tfm

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

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

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

MA = MAB tfmMB = -MBA tfm

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

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

w1w1w1

w2w1w1

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2-2) Top Slaba) Shearing Force at joint

w1 Uniform load 6.570w2 Uniform load 0.000a 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 -2.048

Bending moment at end C 2.048L 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

40/114 (2)276341623.xlsMSN

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

41/114 (2)276341623.xlsMSN

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

(i) In case of x1 = 0.480 mSx1 = 1.854 tf

(ii) In case of x2 = 1.820 mSx2 = -2.215 tf

c) Bending Moment= -2.048

= -2.463

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

= 3.2979 -3.0015 x -0.0152 x2 , x =

Bending moment at x = 1.0927 m is;

= -0.243

2-4) Bottom Slaba) Shearing Force at joint

w1 Reaction at end D 7.823w2 Reaction at end A 7.823

Bending moment at end B -2.463

Bending moment at end C 2.463L 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

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

(i) In case of x1 = 0.480 mSx1 = 5.241 tf

(ii) In case of x2 = 1.820 mSx2 = -5.241 tf

c) Bending Moment= -2.463

= -2.463The maximum bending moment occurs at the point of that shearing force equal to zero.

= 8.9959 -7.8226 x , x =

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

MC = MCD tfmMD = -MDC tfm

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

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

tf/m2

tf/m2

MDA tfmMAD tfm

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

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

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

MD = MDA tfmMA = -MAD tfm

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

L

x

w2

A

MAD

42/114 (2)276341623.xlsMSN

Bending moment at x = 1.1500 m is;

= 2.709Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm

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Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2

1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.101

Ph2 Horizontal Pressure at bottom of side wall 3.171

Pv1 Vertical Pressure(1) on top slab 6.770

Pv2 Vertical Pressure(2) on top slab 0.000

Pq Reaction to bottom slab 8.022a 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= 1.00000= 1.00000

k1 0 k3 -3k1

k1 k2 0 -3k1

0 k2 k1 -3k1 =

k3 0 k1 -3k1k1 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=

-2.150807611.0 3.0000 1.60497928

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

= -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/B0

2

CDA = CAD = PqB02/12

k2 = H0t23/(B0t13) k3 = H0t33/(B0t13)

2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA

2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC

A = -D B = -C

A CAB - CADB CBC - CBA

AB

A CB D

B

A

(t1)H0

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= -2.52934

= 2.11380

= -2.11380

= 2.11380

= -2.11380

= 2.52934

= -2.52934

= 2.52934

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

a) Shearing Force at joint

w1 Load at end A 3.171w2 Load at end B 3.101

Bending moment at end A -2.529

Bending moment at end B 2.114L 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

= 3.801 tf= -3.413 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.

(i) In case of x1 = 0.480 mSx1 = 2.282 tf

(ii) In case of x2 = 1.82 mSx2 = -1.921 tf

c) Bending Moment= -2.529

= -2.114

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

= 3.8009 -3.1714 x + 0.0152 x2 , x =

Bending moment at x = 1.2055 m is;

= -0.243

MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm

tf/m2

tf/m2

MAB tfmMBA tfm

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

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

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

MA = MAB tfmMB = -MBA tfm

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

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

w1w1w1

w2w1w1

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2-2) Top Slaba) Shearing Force at joint

w1 Uniform load 6.770w2 Uniform load 0.000a 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 -2.114

Bending moment at end C 2.114L 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

= 7.786 tf

= -7.786 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

48/114 (2)276341623.xlsMSN

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

49/114 (2)276341623.xlsMSN

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

(i) In case of x1 = 0.480 mSx1 = 1.9206 tf

(ii) In case of x2 = 1.820 mSx2 = -2.2822 tf

c) Bending Moment= -2.114

= -2.529

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

= 3.4128 -3.1014 x -0.0152 x2 , x =

Bending moment at x = 1.0945 m is;

= -0.243

2-4) Bottom Slaba) Shearing Force at joint

w1 Reaction at end D 8.022w2 Reaction at end A 8.022

Bending moment at end B -2.529

Bending moment at end C 2.529L 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

= 9.226 tf= -9.226 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.

(i) In case of x1 = 0.480 mSx1 = 5.375 tf

(ii) In case of x2 = 1.820 mSx2 = -5.375 tf

c) Bending Moment= -2.529

= -2.529The maximum bending moment occurs at the point of that shearing force equal to zero.

= 9.2257 -8.0224 x , x =

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

MC = MCD tfmMD = -MDC tfm

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

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

tf/m2

tf/m2

MDA tfmMAD tfm

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

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

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

MD = MDA tfmMA = -MAD tfm

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

L

x

w2

A

MAD

50/114 (2)276341623.xlsMSN

Bending moment at x = 1.1500 m is;

= 2.775Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm

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2.61260

2.20703

2.89644

3.44844

tfm

tfm

tfm

tfm

(t2)

(t1)

B0

(t3)

C

D

52/114 (2)276341623.xlsMSN

53/114 (2)276341623.xlsMSN

6.6561.111

A

Bw1w1

Lxw1w1

MAB

MBAw1w1

54/114 (2)276341623.xlsMSN

55/114 (2)276341623.xlsMSN

w2

MCBx

L

C

w1

w2

MCD

MDC

C

D

56/114 (2)276341623.xlsMSN

57/114 (2)276341623.xlsMSN

-4.3561.189

1.150

L

x

D

MDA

w1

58/114 (2)276341623.xlsMSN

59/114 (2)276341623.xlsMSN

2.65664

2.25107

2.98452

3.53652

tfm

tfm

tfm

tfm

(t2)

(t1)

B0

(t3)

C

D

60/114 (2)276341623.xlsMSN

61/114 (2)276341623.xlsMSN

6.7551.111

A

Bw1w1

Lxw1w1

MAB

MBAw1w1

62/114 (2)276341623.xlsMSN

63/114 (2)276341623.xlsMSN

w2

MCBx

L

C

w1

w2

MCD

MDC

C

D

64/114 (2)276341623.xlsMSN

65/114 (2)276341623.xlsMSN

-4.45511.1887

1.1500

L

x

D

MDA

w1

66/114 (2)276341623.xlsMSN

67/114 (2)276341623.xlsMSN

1.34167

1.33550

2.89644

3.44844

tfm

tfm

tfm

tfm

(t2)

(t1)

B0

(t3)

C

D

68/114 (2)276341623.xlsMSN

69/114 (2)276341623.xlsMSN

200.6341.207

A

Bw1w1

Lxw1w1

MAB

MBAw1w1

70/114 (2)276341623.xlsMSN

71/114 (2)276341623.xlsMSN

w2

MCBx

L

C

w1

w2

MCD

MDC

C

D

72/114 (2)276341623.xlsMSN

73/114 (2)276341623.xlsMSN

-198.331.093

1.150

L

x

D

MDA

w1

74/114 (2)276341623.xlsMSN

75/114 (2)276341623.xlsMSN

1.38571

1.37954

2.98452

3.53652

tfm

tfm

tfm

tfm

(t2)

(t1)

B0

(t3)

C

D

76/114 (2)276341623.xlsMSN

77/114 (2)276341623.xlsMSN

207.2001.205

A

Bw1w1

Lxw1w1

MAB

MBAw1w1

78/114 (2)276341623.xlsMSN

79/114 (2)276341623.xlsMSN

w2

MCBx

L

C

w1

w2

MCD

MDC

C

D

80/114 (2)276341623.xlsMSN

81/114 (2)276341623.xlsMSN

-204.91.095

1.150

L

x

D

MDA

w1

82/114 (2)276341623.xlsMSN

83/114 (3)276341623.xlsSum MSN

Summary of Internal forces

Member Case M N S (tf)(tf) at joint at 2dSide 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

Check Point (tfm)

84/114 (3)276341623.xlsSum MSN

85/114 (3)276341623.xlsSum MSN

86/114 (3)276341623.xlsSum MSN

87/ 114 (4)276341623.xls, R-bar req

5 Calculation of Required Reinforcement Bar5-1 Calculation of Required Reinforcement Bar1)

Case.1M= 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.2M= 3.1149 60 kgf/m2 h = 30 cm (height of member)N= 9.2257 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.5071 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.9616 tf c = 9.00 cm (distance from neutral axis)

b = 100 cme = M/N = 33.76 cm

39.304 kgf/cm2 ( 4.11E-006 kgf/cm2) o.k.

+66.95 -2397.25 -69919.910.40266.9718 cm2

Case.3M= 2.4633 60 kgf/m2 h = 30 cm (height of member)N= 8.9959 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.6861 tf n = 24 d' = 6 cm (protective covering depth)S2d= 2.2152 tf c = 9.00 cm (distance from neutral axis)

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.4M= 2.5293 60 kgf/m2 h = 30 cm (height of member)N= 9.2257 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.8009 tf n = 24 d' = 6 cm (protective covering depth)S2d= 2.2822 tf c = 9.00 cm (distance from neutral axis)

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

M= 2.5334 60 kgf/m2 h = 30 cm (height of member)N= 7.5559 1400 kgf/m2 d = 24 cm (effective height of member)S0= 5.1807 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.4304 tf c = 9.00 cm (distance from neutral axis)

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.2M= 2.5995 60 kgf/m2 h = 30 cm (height of member)N= 7.7857 1400 kgf/m2 d = 24 cm (effective height of member)S0= 5.2956 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.4973 tf c = 9.00 cm (distance from neutral axis)

At Joint "A"of side wall

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

At Joint "B"of side wall

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

d1h d

d1h d

88/ 114 (4)276341623.xls, R-bar req

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

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

89/ 114 (4)276341623.xls, R-bar req

Case.3M= 2.0477 60 kgf/m2 h = 30 cm (height of member)N= 7.5559 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.2979 tf n = 24 d' = 6 cm (protective covering depth)S2d= 1.8537 tf c = 9.00 cm (distance from neutral axis)

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.4M= 2.1138 60 kgf/m2 h = 30 cm (height of member)N= 7.7857 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.4128 tf n = 24 d' = 6 cm (protective covering depth)S2d= 1.9206 tf c = 9.00 cm (distance from neutral axis)

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

The maximum requirement of reinforcement bar is 5.6627 cm2 in Case. 2 from above calculation.

Case. 1 2 3 4Requirement 5.5151 5.6627 3.7812 3.9250 (cm2)

3)Case.1

M= 2.5334 60 kgf/m2 h = 30 cm (height of member)N= 5.1807 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.5559 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.4022 tf c = 9.00 cm (distance from neutral axis)

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.2M= 2.5995 60 kgf/m2 h = 30 cm (height of member)N= 5.2956 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.7857 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.5360 tf c = 9.00 cm (distance from neutral axis)

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.3M= 2.0477 60 kgf/m2 h = 30 cm (height of member)N= 3.2979 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.5559 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.4022 tf c = 9.00 cm (distance from neutral axis)

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.4M= 2.1138 60 kgf/m2 h = 30 cm (height of member)N= 3.4128 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.7857 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.5360 tf c = 9.00 cm (distance from neutral axis)

b = 100 cme = M/N = 61.94 cm

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

+74.89 -1471.06 -42906.03

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

At Joint "B"of top slab

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 c

d1h d

90/ 114 (4)276341623.xls, R-bar req

0.32955.6565 cm2

The maximum requirement of reinforcement bar is 6.6367 cm2 in Case. 2 from above calculation.

Case. 1 2 3 4Requirement 6.4454 6.6367 5.4700 5.6565 (cm2)

s = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

91/ 114 (4)276341623.xls, R-bar req

4)Case.1

M= 3.0488 60 kgf/m2 h = 30 cm (height of member)N= 7.3922 1400 kgf/m2 d = 24 cm (effective height of member)S0= 8.9959 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.2411 tf c = 9.00 cm (distance from neutral axis)

b = 100 cme = M/N = 41.24 cm

37.763 kgf/cm2 ( 5.85E-005 kgf/cm2) o.k.

+68.16 -2256.85 -65824.690.39307.4396 cm2

Case.2M= 3.1149 60 kgf/m2 h = 30 cm (height of member)N= 7.5071 1400 kgf/m2 d = 24 cm (effective height of member)S0= 9.2257 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.3750 tf c = 9.00 cm (distance from neutral axis)

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.3M= 2.4633 60 kgf/m2 h = 30 cm (height of member)N= 3.6861 1400 kgf/m2 d = 24 cm (effective height of member)S0= 8.9959 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.2411 tf c = 9.00 cm (distance from neutral axis)

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.4M= 2.5293 60 kgf/m2 h = 30 cm (height of member)N= 3.8009 1400 kgf/m2 d = 24 cm (effective height of member)S0= 9.2257 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.3750 tf c = 9.00 cm (distance from neutral axis)

b = 100 cme = M/N = 66.55 cm

31.945 kgf/cm2 ( 2.75E-006 kgf/cm2) o.k.

+72.54 -1744.79 -50889.770.35396.9741 cm2

The maximum requirement of reinforcement bar is 7.6354 cm2 in Case. 2 from above calculation.

Case. 1 2 3 4Requirement 7.4396 7.6354 6.7842 6.9741 (cm2)

5) At Middle of side wallCase.1

M= 0.8279 60 kgf/m2 h = 30 cm (height of member)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.2M= 0.8278 60 kgf/m2 h = 30 cm (height of member)N= 8.5299 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 = 9.70 cm

At Joint "A"of invert

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

d1h d

d1h d

92/ 114 (4)276341623.xls, R-bar req

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

+79.19 -969.48 -28276.460.27530.0000 cm2 Tensile is on inside of member

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

93/ 114 (4)276341623.xls, R-bar req

Case.3M= 0.2426 60 kgf/m2 h = 30 cm (height of member)N= 8.2401 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 = 2.94 cm16.634 kgf/cm2 ( 0 kgf/cm2) o.k.

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

Case.4M= 0.2428 60 kgf/m2 h = 30 cm (height of member)N= 8.4710 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 = 2.87 cm16.839 kgf/cm2 ( 0 kgf/cm2) o.k.

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

The maximum requirement of reinforcement bar is 0.0000 cm2 in Case. 1 from above calculation.

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

6) At Middle of top slabCase.1

M= 1.8112 60 kgf/m2 h = 30 cm (height of member)N= 5.1807 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 = 34.96 cm27.583 kgf/cm2 ( 0 kgf/cm2) o.k.

+75.64 -1383.9 -40363.640.32103.8899 cm2 Tensile is on inside of member

Case.2M= 1.8773 60 kgf/m2 h = 30 cm (height of member)N= 5.2956 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 = 35.45 cm28.160 kgf/cm2 ( 0 kgf/cm2) o.k.

+75.24 -1430.32 -41717.600.32564.0759 cm2 Tensile is on inside of member

Case.3M= 2.2969 60 kgf/m2 h = 30 cm (height of member)N= 3.2979 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 = 69.65 cm29.939 kgf/cm2 ( 0 kgf/cm2) o.k.

+73.99 -1576.05 -45968.170.33926.3478 cm2 Tensile is on inside of member

Case.4M= 2.3630 60 kgf/m2 h = 30 cm (height of member)N= 3.4128 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 = 69.24 cm30.496 kgf/cm2 ( 1.84E-005 kgf/cm2) o.k.

+73.59 -1622.47 -47322.13

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 c

d1h d

94/ 114 (4)276341623.xls, R-bar req

0.34336.5363 cm2 Tensile is on inside of member

The maximum requirement of reinforcement bar is 6.5363 cm2 in Case. 4 from above calculation.

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

s = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

95/ 114 (4)276341623.xls, R-bar req

7) At Middle of invertCase.1

M= 2.1238 60 kgf/m2 h = 30 cm (height of member)N= 7.3922 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 = 28.73 cm31.3560 kgf/cm2 ( 0 kgf/cm2) o.k.

+72.97 -1694.79 -49431.390.34964.1161 cm2 Tensile is on inside of member

Case.2M= 2.1899 60 kgf/m2 h = 30 cm (height of member)N= 7.5071 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 = 29.17 cm31.902 kgf/cm2 ( 0 kgf/cm2) o.k.

+72.58 -1741.21 -50785.350.35354.3051 cm2 Tensile is on inside of member

Case.3M= 2.7094 60 kgf/m2 h = 30 cm (height of member)N= 3.6861 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 = 73.50 cm33.146 kgf/cm2 ( 4.80E-007 kgf/cm2) o.k.

+71.66 -1847.9 -53897.220.36237.6614 cm2 Tensile is on inside of member

Case.4M= 2.7754 60 kgf/m2 h = 30 cm (height of member)N= 3.8009 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 = 73.02 cm33.682 kgf/cm2 ( 0 kgf/cm2) o.k.

+71.26 -1894.33 -55251.190.36607.8531 cm2 Tensile is on inside of member

The maximum requirement of reinforcement bar is 7.8531 cm2 in Case. 4 from above calculation.

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)

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

tfm ca =tf sa =

Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =

d1h d

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

Side wall Top slabbottom middle top end middleoutside inside outside outside inside

Bending moment M kgfcm 311,488 82,787 259,949 259,949 236,298 Shearing force (joint) S kgf 7,507 0 5,296 7,786 0 Shearing force (2d) S2d kgf 3,962 - 3,497 4,536 -Axial force N kgf 9,226 8,301 7,786 5,296 3,413 Height of member h cm 30 30 30 30 30Covering depth d' cm 6 6 6 6 6Effective height d cm 24 24 24 24 24Effective width b cm 100 100 100 100 100Effective area bd cm2 2400 2400 2400 2400 2400Young's modulus ratio n - 24 24 24 24 24

Required R-bar Asreq cm2 6.97 0.00 5.66 6.64 6.54

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

Reinforcement As cm2 8.04 4.52 8.04 8.04 8.04Perimeter of R-bar U 20.11 15.08 20.11 20.11 20.11M/N e cm 33.763 9.974 33.388 49.088 69.239Dist. from neutral axis c cm 9.00 9.00 9.00 9.00 9.00

a' 56.3 -15.1 55.2 102.3 162.7b' 495.1 123.5 490.8 672.5 905.8c' -11882.3 -2963.9 -11778.0 -16140.4 -21739.8x 10.16 17.62 10.19 9.38 8.92

0.000 0.000 0.000 0.000 0.000(check) ok ok ok ok ok

Compressive stress kgf/cm2 37.7 9.9 31.5 31.4 28.5Allowable stress kgf/cm2 60.0 60.0 60.0 60.0 60.0

ok ok ok ok okTensile stress kgf/cm2 1232.9 85.7 1023.7 1174.4 1155.0Allowable stress kgf/cm2 1400.0 1400.0 1400.0 1400.0 1400.0

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

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

ok - ok ok -

Resisting Moment Mr kgfcm 496,869 362,613 497,615 459,243 408,023 Mr for compression Mrc kgfcm 496,869 410,229 497,615 499,428 501,254 x for Mrc cm 9.240 7.521 9.013 8.634 8.359

kgf/cm2 2300.3 3155.3 2394.6 2562.9 2694.5 Mr for tensile Mrs kgfcm 577,461 362,613 532,234 459,243 408,023 x for Mrs cm 12.982 10.971 12.517 11.670 10.986

kgf/cm2 68.7 49.1 63.6 55.2 49.2

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

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

cca

ssa

a

2d2da

s for Mrc

c for Mrs

106/114 (5)276341623.xlsR-bar stress

Minimum requirement of reinforcement bar As min = 4.5 cm2

107/114 (5)276341623.xlsR-bar stress

Invertend middle

outside inside311,488 277,544

9,226 0 5,375 -7,507 3,801

30 306 6

24 24100 100

2400 240024 24

7.64 7.85

16@250 12@125

8.04 9.0520.11 30.16

41.492 73.0209.00 9.0079.5 174.1

584.6 1068.9-14030.0 -25653.1

9.69 9.27 0.000 0.000

ok ok

37.7 32.260.0 60.0

ok ok1336.1 1227.41400.0 1400.0

ok ok3.84 0.00

11.00 11.00ok ok

2.24 -5.50 -

ok -

497,785 469,708 497,785 521,262

8.969 8.7872413.1 2492.9

523,749 469,708 12.426 11.642

62.6 55.0

12@250 12@2504.52 4.52

ok ok

108/114 (5)276341623.xlsR-bar stress

DATA PLOT PEMBESIAN

Section of Culvert

1. Design Datab1 = 0.300 m b2 = 2.000 mh1 = 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 PembesianBottom 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:504Top 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 Datab1 = 0.300 m b2 = 2.000 mh1 = 0.300 m h2 = 2.000 m h3 = 0.300 mBb = 2.300 m Ha = 2.300 m

2. Result CalculationSide wall (left) Ma Mmax Mb Sa Sb NaCase 1 -3.049 0.828 -2.533 7.392 -5.181 8.996Case 2 -2.463 -0.243 -2.048 3.686 -3.298 8.996

Side wall (right) Mc Mmax Md Sd Sc NdCase 1 -2.533 0.828 -3.049 -7.392 5.181 8.996Case 2 -2.048 -0.243 -2.463 -3.686 3.298 8.996

Top slab Mb Mmax Mc Sb Sc NbCase 1 -2.533 1.811 -2.533 7.556 -7.556 5.181Case 2 -2.048 2.297 -2.048 7.556 -7.556 3.298

Bottom slab Ma Mmax Md Sa Sd NaCase 1 -3.049 2.124 -3.049 -8.996 8.996 7.392Case 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 0.863 1.150 1.438 1.725

Bb = 2.300 0.288 0.575 0.863 1.150 0.863 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 0.863 1.150 1.438 1.725

Bb = 2.300 0.288 0.575 0.863 1.150 0.863 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 MBA8.996 7.556 -3.049 2.5338.996 7.556 -2.463 2.048

Nd Nc MCD MDC8.996 7.556 -2.533 3.0498.996 7.556 -2.048 2.463

Nb Nc MBC MCB5.181 5.181 -2.533 2.5333.298 3.298 -2.048 2.048

Na Nd MAD MDA7.392 7.392 3.049 -3.0493.686 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 2.013

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 2.013

0.575 0.288

x6 x7-1.57 -2.43

SummaryLoadMSNSum MSNR-bar reqR-bar stressPlotRebarPlotMoment