Single Box

7/30/2019 Single Box

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SUMMARY OF STRUCTURAL CALCULATION OF 1-BARREL BOX CULVER

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

Type of box culvert B2.0 x H2.5

Clear width m 2.00Clear height m 2.50

Height of fillet m 0.20

Thickness Side wall cm 20.0

Top slab cm 20.0Bottom slab cm 20.0

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 6.0

Bottom slab Lower cm 6.0Upper 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@250

Distribution bar mm 12@250

Top slab Upper edge Tensile bar mm 16@250

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

Distribution bar mm 12@250

Bottom sla 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 12@250Lower edge Fillet bar mm 12@250

2 Design Parameters

Unit Weight Reinforced Concrete gc= 2.4 tf/m3

Backfill soil (wet) gs= 1.8 tf/m

(submerged) gs'= 1.0 tf/m3

Live Load Class of road Class III (BM50)

Truck load at rear wheel P= 5.0 tf Impact coefficien (for Class I to IV roa Ci= 0.3 (D4.0m)

Pedestrian load (for Class V roads) 0 tf/m2

Concrete Design Strength sck= 175 kgf/cm2

(K175) Allowable Compressive Stress sca= 60 kgf/cm2

Allowable Shearing Stress ta= 5.5 kgf/cm2

Reinforcement Bar Allowable Tensile Stress ssa= 1,400 kgf/cm2

(U24, deformed bar) Yielding Point of Reinforcement Bar ssy= 3,000 kgf/cm2

Young's Modulus Ratio n= 24

Coefficient of static earth pressure Ka= 0.5


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STRUCTURAL CALCULATION OF BOX CULVERT Type: B2.00m x H2.50m C

Soil Cover Depth: 1.0 m

1 Dimensions and ParametersBasic Parameters

Ka: Coefficient of static earth pressure 0.5gw: Unit weight of water (t/m3) 1.00 t/m

gd: Unit weight of soil (dry) (t/m3) 1.80 t/mgs: Unit weight of soil (saturated) (t/m3) 2.00 t/mgc: Unit weight of reinforced concrete (t/m3) 2.40 t/msck: Concrete Design Strength 175 kgsca Allowable Stress of Concrete 60 kgssa: Allowable Stress of Reinforcement Bar 1400 kgta: Allowable Stress of Shearing (Concrete) 5.5 kgssy: Yielding Point of Reinforcement Bar 3000 kgn: Young's Modulus Ratio 24Fa: Safety factor against uplift 1.2

Basic Dimensions

H: Internal Height of Box Culvert 2.50 mB: Internal Width of Box Culvert 2.00 mHf: Fillet Height 0.20 mt1: Thickness of Side Wall 0.20 mt2: Thickness of Top Slab 0.20 m

t3: Thickness of Invert (Bottom Slab) 0.20 mBT: Gross Width of Box Culvert 2.40 mHT: Gross Height of Box Culvert 2.90 mD: Covering Depth 1.00 mGwd: Underground Water Depth for Case 1, 2 1.00 mhiw: Internal Water Depth for Case 1, 2 0.00 m

for Case 3, 4 2.50 m

Cover of R-bar Basic ConditionsTop Slab d2 0.06 m Classification of Live load by truck Class 3Side Wall d1 0.06 m PTM: Truck load of Middle Tire 5.00 tBottom Slab d3 0.06 m Ii: Impact coefficient (D 4.0m:0, D


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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)(2B0-bm')/(am'B0^2) = 1.5444 tf/m2,am': length of distributed load 2D+1.75+bm = 4.000 mbm': width of distributed load 2D+am = 2.100 m

b) distributed load of rear tirePvtr: distributed load of rear tire not reach to top slab 0.0000 tf/m2,ar': length of distributed load 2D+1.75+br = 4.000 mbr': width of distributed load 2D+ar = 2.100 m

c) distributed load of front tirePvtf: distributed load of front tire not reach to top slab 0.0000 tf/m2,af': length of distributed load 2D+1.75+bf = 4.000 mbf': width of distributed load 2D+af = 2.100 m

(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') = 1.5476 tf/m2,

am': length of distributed load 2D+1.75+bm = 4.000 mbm': width of distributed load 2D+am = 2.100 m

b) distributed load of rear tirePvtr: distributed load of rear tire not reach to top slab 0.0000 tf/m2,ar': length of distributed load 2D+1.75+br = 4.000 mbr': width of distributed load 2D+ar = 2.100 m

c) distributed load of front tirePvtf: distributed load of front tire not reach to top slab 0.0000 tf/m2,af': length of distributed load 2D+1.75+bf = 4.000 mbf': width of distributed load 2D+af = 2.100 m

(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') = 1.5476 tf/m2,ar': length of distributed load 2D+1.75+br = 4.000 mbr': width of distributed load 2D+ar = 2.100 m

b) distributed load of middle tirePvtm: distributed load of middle tire not reach to top slab 0.0000 tf/m2,am': length of distributed load 2D+1.75+bm = 4.000 mbm': width of distributed load 2D+am = 2.100 m

c) distributed load of front tire

Pvtf: distributed load of front tire not reach to top slab 0.0000 tf/m2,af': length of distributed load 2D+1.75+bf = 4.000 mbf': width of distributed load 2D+af = 2.100 m

(4) Combination of load distribution of track tire

Case.L1: Pvt1 = 1.5444 tf/m2, B = 2.200 m Combination for Case.L2 (2)Pvt2 = 0.0000 tf/m2, B = 0.000 m a) + b)

Case.L2: Pvt1 = 1.5476 tf/m2, B = 2.200 m Distributed load total 1.5476Pvt2 = 0.0000 tf/m2, B = 0.000 m Select the combination case of

for Case.L2, which is the largest load to

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 calc

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= 1.3241 > 1.2 ok

where, Vd: Total dead weight (t/m) Vd= 9.216 tf/mU: Total uplift (t.m)

U=BT*HT*gw U= 6.960 tf/m

Ws: Weight of covering soil Ws = BT*{(D-Gwd)*(gs-gw)+Gwd*gd} = 4.320 tf/mWc: Self weight of box culvert Wc = (HT*BT-H*B+2*Hf^2)*gc = 4.896 tf/mFa: Safety factor against uplift Fa= 1.2


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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 slab

Acting Load (tf/m2)Wtop= (t2*BT+Hf^2)*gc/B0 Wtop= 0.5673Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 1.8000Pvt1 Pvt1= 1.5444Pvt2 Pvt2= 0.0000

Pv1= 3.9117

2) horizontal load at top of side wall

Acting Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.0000 we1= 0.0000 tf/m2P2=Ka*we2 P2= 0.0000 we2= 0.0000 tf/m2P3=Ka*gd*Gwd P3= 0.9000P4=Ka*gs*(D1-Gwd) P4= 0.1000P5=gw*(D1-Gwd) P5= 0.1000

Ph1= 1.1000

3) horizontal load at bottom of side wallActing Load (tf/m2)

P1=Ka*we1 P1= 0.0000P2=Ka*we2 P2= 0.0000P3=Ka*gd*Gwd P3= 0.9000P4=Ka*gs*(D1+H0-Gwd) P4= 2.8000P5=gw*(D1+H0-Gwd) P5= 2.8000

Ph2= 6.5000

4) self weight of side wall

Acting Load (tf/m)Wsw=t1*H*gc Wsw= 1.2000

5) ground reactionActing Load (tf/m2)Wbot=(t3*BT+Hf^2)*gc/B0 Wbot= 0.5673Wtop Wtop= 0.5673Ws=Wsw*2/B0 Ws= 1.0909

Pvd Pvd= 1.8000Pvt1 Pvt1= 1.5444Pvt2 Pvt2= 0.0000Wiw=(hiw*B-2Hf^2)*gw/B0 Wiw= 0.0000 hiw: internal water dUp=-U/B0 U= -3.1636

Q= 2.4062

summary of resistance moment

Item V H x y M(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of result

Self weight top slab 1.2480 - 1.1000 - 1.3728 X = SM/SV =side wall (left) 1.2000 - 0.0000 - 0.0000 e = B0/2 - X =side wall (right) 1.2000 - 2.2000 - 2.6400invert 1.2480 - 1.1000 - 1.3728 ground reaction

load on top slab Pvd 3.9600 - 1.1000 - 4.3560 q1 = SV/Bo + 6SVePvt1 3.3977 - 1.1000 - 3.7375 q2 = SV/Bo - 6SVe

Pvt2 0.0000 - 1.1000 - 0.0000soil pressure side wall (left) - 10.2600 - 1.0303 10.5705

side wall (right) - -10.2600 - 1.0303 -10.5705internal water 0.0000 - 1.1000 - 0.0000uplift -6.9600 - 1.1000 - -7.6560total 5.2937 5.8231

6) load against invert

Acting Load (tf/m2)Pvd 1.8000Pvt1 1.5444Pvt2 0.0000Wtop 0.5673Ws 1.0909

Pq= 5.0026


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Acting Load (tf/m2)Wtop= (t2*BT+Hf^2)*gc/B0 Wtop= 0.5673Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 1.8000Pvt1 Pvt1= 1.5476

Pvt2 Pvt2= 0.0000Pv1= 3.9149


Acting Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.7738 we1= 1.5476 tf/m2P2=Ka*we2 P2= 0.0000 we2= 0.0000 tf/m2P3=Ka*gd*Gwd P3= 0.9000P4=Ka*gs*(D1-Gwd) P4= 0.1000P5=gw*(D1-Gwd) P5= 0.1000

Ph1= 1.8738

3) horizontal load at bottom of side wall

Acting Load (tf/m2)P1=Ka*we1 P1= 0.7738P2=Ka*we2 P2= 0.0000P3=Ka*gd*Gwd P3= 0.9000P4=Ka*gs*(D1+H0-Gwd) P4= 2.8000P5=gw*(D1+H0-Gwd) P5= 2.8000

Ph2= 7.2738



5) ground reactionActing Load (tf/m2)Wbot=(t3*BT+Hf^2)*gc/B0 Wbot= 0.5673Wtop Wtop= 0.5673Ws=Wsw*2/B0 Ws= 1.0909Pvd Pvd= 1.8000Pvt1 Pvt1= 1.5476

Pvt2 Pvt2= 0.0000Wiw=(hiw*B-2Hf^2)*gw/B0 Wiw= 0.0000 hiw: internal waterUp=-U/B0 U= -3.1636

Q= 2.4094


Item V H x y M(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resu

Self weight top slab 1.2480 - 1.1000 - 1.3728 X = SM/SV =

side wall (left) 1.2000 - 0.0000 - 0.0000 e = B0/2 - X =side wall (right) 1.2000 - 2.2000 - 2.6400invert 1.2480 - 1.1000 - 1.3728 ground reaction

load on top slab Pvd 3.9600 - 1.1000 - 4.3560 q1 = SV/Bo + 6SV

Pvt1 3.4048 - 1.1000 - 3.7452 q2 = SV/Bo - 6SVe


side wall (right) - -12.3493 - 1.0844 -13.3910

internal water 0.0000 - 1.1000 - 0.0000uplift -6.9600 - 1.1000 - -7.6560total 5.3008 5.8308


Acting Load (tf/m2)Pvd 1.8000Pvt1 1.5476Pvt2 0.0000Wtop 0.5673Ws 1.0909total Pq= 5.0058


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


Acting Load (tf/m2)Wtop= (t2*BT+Hf^2)*gc/B0 Wtop= 0.5673Pvd=D*gd Pvd= 1.8000Pvt1 Pvt1= 1.5444

Pvt2 Pvt2= 0.0000Pv1= 3.9117


Acting Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.0000 we1= 0.0000 tf/m2P2=Ka*we2 P2= 0.0000 we2= 0.0000 tf/m2P3=Ka*gd*D1 P3= 0.9900WP=-gw*0 P4= 0.0000

Ph1= 0.9900


Acting Load (tf/m2)P1=Ka*we1 P1= 0.0000P2=Ka*we2 P2= 0.0000P3=Ka*gd*(D1+H0) P3= 3.4200WP=-gw*H P4= -2.5000

Ph2= 0.9200



5) ground reaction

Acting Load (tf/m2)Wbot=(t3*BT+Hf^2)*gc/B0 Wbot= 0.5673Wtop Wtop= 0.5673Ws=Wsw*2/B0 Ws= 1.0909Pvd Pvd= 1.8000Pvt1 Pvt1= 1.5444Pvt2 Pvt2= 0.0000Wiw=(hiw*B-2Hf^2)*gw/B0 Wiw= 2.2364 hiw: internal water

Up=0 U= 0.0000Q= 7.8062


Item V H x y M(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resu

Self weight top slab 1.2480 - 1.1000 - 1.3728 X = SM/SV =


load on top slab Pvd 3.9600 - 1.1000 - 4.3560 q1 = SV/Bo + 6SV

Pvt1 3.3977 - 1.1000 - 3.7375 q2 = SV/Bo - 6SVe


side wall (right) - -2.5785 - 1.3665 -3.5235internal water 4.9200 - 1.1000 - 5.4120uplift 0.0000 - 1.1000 - 0.0000

total 17.1737 18.8911


Acting Load (tf/m2)Pvd 1.8000Pvt1 1.5444Pvt2 0.0000Wtop 0.5673Ws 1.0909total Pq= 5.0026


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Acting Load (tf/m2)Wtop= (t2*BT+Hf 2)*gc/B0 Wtop= 0.5673Pvd=D*gd Pvd= 1.8000

Pvt1 Pvt1= 1.5476

Pvt2 Pvt2= 0.0000Pv1= 3.9149


Acting Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.7738 we1= 1.5476 tf/m2

P2=Ka*we2 P2= 0.0000 we2= 0.0000 tf/m2P3=Ka*gd*D1 P3= 0.9900WP=-gw*0 P4= 0.0000

Ph1= 1.7638


Acting Load (tf/m2)P1=Ka*we1 P1= 0.7738P2=Ka*we2 P2= 0.0000P3=Ka*gd*(D1+H0) P3= 3.4200WP=-gw*H P4= -2.5000

Ph2= 1.6938



5) ground reaction

Acting Load (tf/m2)Wbot=(t3*BT+Hf^2)*gc/B0 Wbot= 0.5673

Wtop Wtop= 0.5673Ws=Wsw*2/B0 Ws= 1.0909Pvd Pvd= 1.8000

Pvt1 Pvt1= 1.5476Pvt2 Pvt2= 0.0000Wiw=(hiw*B-2Hf^2)*gw/B0 Wiw= 2.2364 hiw: internal water d

Up=0 U= 0.0000Q= 7.8094


Item V H x y M

(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultSelf weight top slab 1.2480 - 1.1000 - 1.3728 X = SM/SV =


load on top slab Pvd 3.9600 - 1.1000 - 4.3560 q1 = SV/Bo + 6SVe

Pvt1 3.4048 - 1.1000 - 3.7452 q2 = SV/Bo - 6SVe/


side wall (right) - -4.6678 - 1.3591 -6.3440internal water 4.9200 - 1.1000 - 5.4120

uplift 0.0000 - 1.1000 - 0.0000

total 17.1808 18.8988


Acting Load (tf/m2)

Pvd 1.8000Pvt1 1.5476

Pvt2 0.0000Wtop 0.5673

Ws 1.0909total Pq= 5.0058

Summary of Load Calculation

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

Case.1 3.9117 1.1000 6.5000 5.0026 1.2000 2.4062

Case.2 3.9149 1.8738 7.2738 5.0058 1.2000 2.4094


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(3)a) + c)1.5476


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4 Analysis of Plane Frame


1) Calculation of Load Term

Ph1 Horizontal Pressure at top of side wall 1.100 tf/m2

Ph2 Horizontal Pressure at bottom of side wall 6.500 tf/m2

Pv1 Vertical Pressure(1) on top slab 3.912 tf/m2


Pq Reaction to bottom slab 5.003 tf/m2

a Distance from joint B to far end of Pv2 2.200 mb Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.700 mB0 Width of plane frame 2.200 mt1 Thickness of side wall 0.200 mt2 Thickness of top slab 0.200 mt3 Thickness of invert (bottom slab) 0.200 m

CAB = CDC = (2Ph1+3Ph2)H0

2

/60= 2.63655 tf m

CBA = CCD = (3Ph1+2Ph2)H02/60 = 1.98045 tf m

CBC = CCB = Pv1B02/12 + {(a

2-b

2)B0

2/2 - 2B0(a

3-b

3)/3 + (a

4-b

4)/4}Pv2/B0

2 = 1.57772 tf m

CDA = CAD = PqB02/12 = 2.01772 tf m

2) Calculation of Bending Moment at joint

k1 = 1.0

k2 = H0t23/(B0t1

3) = 1.2273

k3 = H0t33/(B0t1

3) = 1.2273

2(k1+k3) k1 0 k3 -3k1 qA CAB - CAD

k1 2(k1+k2)k2

0 -3k1 qB CBC - CBA0 k2 2(k1+k2) k1 -3k1 qC = CCD - CCB

k3 0 k1 2(k1+k3) -3k1 qD CDA - CDC

k1 k1 k1 k1 -4k1 R 0

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

qA = -qD qB = -qC R =0

2k1+k3 k1 qA

k1 2k1+k2 qB

3.2273 1.0 qA

1.0 3.2273 qB

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

qA = 0.25489 qC = 0.20377

qB = -0.20377 qD = -0.25489

=CAB - CAD

CBC - CBA

=

0.61883333

-0.40273333

B

A

(t2)

(t1)

B0

(t3)

(t1)

C

H0

D

14/41 (2)168023748.xls.ms_officeMSN


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MAB = k1(2qA +qB) - CAB = -2.3305 tfm

MBA = k1(2qB+qA)+CBA = 1.8278 tfm

MBC = k2(2qB+qC) - CBC = -1.8278 tfm

MCB = k2(2qC+qB)+CCB = 1.8278 tfm

MCD = k1(2qC+qD) - CCD = -1.8278 tfm

MDC =k1 (2qD+ qC)+CDC = 2.3305 tf m

MDA = k3(2qD+qA) - CDA = -2.3305 tfm

MAD = k3(2qA+qD)+CAD = 2.3305 tfm

2) Calculation of Design Force

2-1) Side Wall in left

a) Shearing Force at joint

w1 Load at end A 6.500 tf/m2

w2 Load at end B 1.100 tf/m2

MAB Bending moment at end A -2.3305 tfm

MBA Bending moment at end B 1.8278 tfm

L Length of member (=H0) 2.700 mch Protective covering height 0.060 mt Thickness of member (height) 0.200 md Effective height of member 0.140 m

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

= 6.531 tf

SBA = SAB - L(w1+w2)/2 = -3.729 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.

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

(i) In case of x1 = 0.280 m

Sx1 = 4.790 tf (ii) In case of x2 = 2.420 m

Sx2 = -3.342 tf

c) Bending Moment

MA = MAB = -2.331 tfm

MB = -MBA = -1.828 tfm

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

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

= 6.5312 -6.5000 x + 1.0000 x2 , x = 5.258

1.242Bending moment at x 1.2422 m is;

Mmax = SABx - w1x2/2 - (w2-w1)x

3/(6L) + MAB = 1.407 tfm

w1

w2

A

B

Lx

MAB

MBA



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2-2) Top Slaba) Shearing Force at joint

w1 Uniform load 3.912 tf/m2


a Distance from end B to near end of 0.000 m

b Length of uniform load w2 2.200 mMBC Bending moment at end B -1.8278 tf m

MCB Bending moment at end C 1.8278 tf m

L Length of member (=Bo) 2.200 mch Protective covering height 0.060 mt Thickness of member (height) 0.200 md Effective height of member 0.140 m

SBC = (w1L+w2b)/2-(MBC+MCB)/L = 4.303 tf

SCB = SBC -w1L - w2b = -4.303 tf


Sx = SBC - w1x - w2(x-a) in case of 0.000 m


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

(i) In case of x1 = 0.280 mSx1 = 3.342 tf

(ii) In case of x2 = 2.420 mSx2 = -4.790 tf

c) Bending Moment

MC = MCD = -1.828 tfm

MD = -MDC = -2.331 tfm


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

= 3.7288 -1.1000 x -1.0000 x2 , x = -2.5581.458

Bending moment at x 1.4578 m is;

Mmax = SCDx - w1x2/2 - (w2-w1)x

3/(6L) + MCD = 1.40650 tfm

2-4) Bottom Slaba) Shearing Force at joint

w1 Reaction at end D 5.003 tf/m2

w2 Reaction at end A 5.003 tf/m2

MDA Bending moment at end B -2.33054 tfm

MAD Bending moment at end C 2.33054 tfm

L Length of member (=B0) 2.200 mch Protective covering height 0.060 m

t Thickness of member (height) 0.200 md Effective height of member 0.140 m

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

= 5.503 tf

SAD = SDA - L(w1+w2)/2 = -5.503 tf


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



c) Bending Moment

MD = MDA

= -2.331 tf m

MA = -MAD = -2.331 tf m


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

= 5.5029 -5.0026 x , x = 1.100


Mmax = SDAx - w1x2/2 - (w2-w1)x

3/(6L) + MDA = 0.696 tfm

L

x

D

MDA

w1w2

A

MAD



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Pv2 Vertical Pressure(2) on top slab 0.000tf/m

2



CAB = CDC = (2Ph1+3Ph2)H02/60 = 3.10664 tfm

CBA = CCD = (3Ph1+2Ph2)H02/60 = 2.45054 tfm

CBC = CCB = Pv1B02/12 + {(a

2-b

2)B0

2/2 - 2B0(a

3-b

3)/3 + (a

4-b

4)/4}Pv2/B0

2 = 1.57901 tfm

CDA = CAD = PqB02/12 = 2.01901 tfm


k1 = 1.0

k2 = H0t23/(B0t1

3) = 1.22727

k3 = H0t33/(B0t1

3) = 1.22727

2(k1+k3) k1 0 k3 -3k1 qA CAB - CAD

k1 2(k1+k2) k2 0 -3k1 qB CBC - CBA

0 k2 2(k1+k2) k1 -3k1 qC = CCD - CCB

k3 0 k1 2(k1+k3) -3k1 qD CDA - CDC

k1 k1 k1 k1 -4k1 R 0



2k1+k3 k1 qA

k1 2k1+k2 qB

3.2273 1.0 qA1.0 3.2273 qB


qA = 0.46537 qC = 0.41425

qB = -0.41425 qD = -0.46537

=CAB - CAD

CBC - CBA

=1.08763294

-0.87153294

B

A

(t2)

(t1)

B0

(t3)

(t1)

C

H0

D



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MAB = k1(2qA +qB) - CAB = -2.5901 tfm


MBC = k2(2qB+qC) - CBC = -2.0874 tfm

MCB = k2(2qC+qB)+CCB = 2.0874 tf m

MCD = k1(2qC+qD) - CCD = -2.0874 tfm

MDC =k1 (2qD+ qC)+CDC = 2.5901 tfm

MDA = k3(2qD+qA) - CDA = -2.5901 tfm


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






L Length of member (=H0) 2.700 m

ch Protective covering height 0.060 mt Thickness of member (height) 0.200 md Effective height of member 0.140 m


= 7.5758 tf

SBA = SAB - L(w1+w2)/2 = -4.7734 tf


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



c) Bending Moment

MA = MAB = -2.590 tfm

MB = -MBA = -2.087 tfm


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

= 7.5758 -7.2738 x + 1.0000 x2 , x = 6.0141.260

Bending moment at x 1.2597 m is;Mmax = SABx - w1x

2/2 - (w2-w1)x

3/(6L) + MAB = 1.848 tfm

w1

w2

A

B

Lx

MAB

MBA



20/41





b Length of uniform load w2 2.200 mMBC Bending moment at end B -2.0874 tfm

MCB Bending moment at end C 2.0874 tfm



SCB = SBC -w1L - w2b = -4.306 tf




21/41


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



Sx2 = -5.618 tf

c) Bending Moment

MC = MCD = -2.087 tfm

MD = -MDC = -2.590 tf m


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

= 4.7734 -1.8738 x -1.0000 x2 , x = -3.3141

1.4403



3/(6L) + MCD = 1.8483 tfm






L Length of member (=B0) 2.200 mch Protective covering height 0.060 mt Thickness of member (height) 0.200 md Effective height of member 0.140 m


= 5.506 tf

SAD = SDA - L(w1+w2)/2 = -5.506 tf


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



c) Bending Moment

MD = MDA = -2.590 tfm

MA = -MAD = -2.590 tfmThe maximum bending moment occurs at the point of that shearing force equal to zero.

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

= 5.5064 -5.0058 x , x = 1.1000



3/(6L) + MDA = 0.438 tfm

L

x

D

MDA

w1w2

A

MAD



22/41






Pv2 Vertical Pressure(2) on top slab 0.000tf/m

2



CAB = CDC = (2Ph1+3Ph2)H02/60 = 0.57591 tfm

CBA = CCD = (3Ph1+2Ph2)H02/60 = 0.58442 tfm

CBC = CCB = Pv1B02/12 + {(a

2-b

2)B0

2/2 - 2B0(a

3-b

3)/3 + (a

4-b

4)/4}Pv2/B0

2 = 1.57772 tfm

CDA = CAD = PqB02/12 = 2.01772 tfm


k1 = 1.0

k2 = H0t23/(B0t1

3) = 1.22727

k3 = H0t33/(B0t1

3) = 1.22727

2(k1+k3) k1 0 k3 -3k1 qA CAB - CAD

k1 2(k1+k2) k2 0 -3k1 qB CBC - CBA

0 k2 2(k1+k2) k1 -3k1 qC = CCD - CCB

k3 0 k1 2(k1+k3) -3k1 qD CDA - CDC

k1 k1 k1 k1 -4k1 R 0



2k1+k3 k1 qA

k1 2k1+k2 qB

3.2273 1.0 qA

1.0 3.2273 qB


qA = -0.59971 qC = -0.49361

qB = 0.49361 qD = 0.59971

=CAB - CAD

CBC - CBA

=-1.44180667

0.99330167

B

A

(t2)

(t1)

B0

(t3)

(t1)

C

H0

D



23/41

MAB = k1(2qA +qB) - CAB = -1.28171 tfm


MBC = k2(2qB+qC) - CBC = -0.97193 tfm

MCB = k2(2qC+qB)+CCB = 0.97193 tf m

MCD = k1(2qC+qD) - CCD = -0.97193 tfm

MDC =k1 (2qD+ qC)+CDC = 1.28171 tfm

MDA = k3(2qD+qA) - CDA = -1.28171 tfm








L Length of member (=H0) 2.700 m

ch Protective covering height 0.060 mt Thickness of member (height) 0.200 md Effective height of member 0.140 m


= 1.388 tf

SBA = SAB - L(w1+w2)/2 = -1.190 tf


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


Sx1 = 1.130 tf

(ii) In case of x2 = 2.420 m

Sx2 = -0.914 tf

c) Bending Moment

MA = MAB = -1.282 tfm

MB = -MBA = -0.972 tfm


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

= 1.3882 -0.9200 x + -0.0130 x2 , x = -72.4501.478

Bending moment at x 1.4782 m is;Mmax = SABx - w1x

2/2 - (w2-w1)x

3/(6L) + MAB = -0.249 tfm

w1

w2

A

B

Lx

MAB

MBA



24/41









SCB = SBC -w1L - w2b = -4.303 tf




25/41


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



Sx2 = -1.130 tf

c) Bending Moment

MC = MCD = -0.972 tfm

MD = -MDC = -1.282 tf m


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

= 1.1903 -0.9900 x + 0.0130 x2 , x = 75.15

1.222



3/(6L) + MCD = -0.249 tfm








= 5.503 tf SAD = SDA - L(w1+w2)/2 = -5.503 tf


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



c) Bending Moment

MD = MDA = -1.282 tfm

MA = -MAD = -1.282 tfm


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

= 5.5029 -5.0026 x , x = 1.100



3/(6L) + MDA = 1.745 tfm

L

x

D

MDA

w1w2

A

MAD



26/41








a Distance from joint B to far end of Pv2 2.200 m

b Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.700 mB0 Width of plane frame 2.200 mt1 Thickness of side wall 0.200 mt2 Thickness of top slab 0.200 mt3 Thickness of invert (bottom slab) 0.200 m

CAB = CDC = (2Ph1+3Ph2)H02/60 = 1.04600 tfm

CBA = CCD = (3Ph1+2Ph2)H02/60 = 1.05450 tfm

CBC

= CCB

= Pv1B0

2/12 + {(a

2-b

2)B

0

2/2 - 2B

0(a

3-b

3)/3 + (a

4-b

4)/4}Pv2/B

0

2 = 1.57901 tfm

CDA = CAD = PqB02/12 = 2.01901 tfm


k1 = 1.0

k2 = H0t23/(B0t1

3) = 1.22727

k3 = H0t33/(B0t1

3) = 1.22727

2(k1+k3) k1 0 k3 -3k1 qA CAB - CAD

k1 2(k1+k2) k2 0 -3k1 qB CBC - CBA

0 k2 2(k1+k2) k1 -3k1 qC = CCD - CCB

k3 0 k1 2(k1+k3) -3k1 qD CDA - CDC

k1 k1 k1 k1 -4k1 R 0



2k1+k3 k1 qA

k1 2k1+k2 qB

3.2273 1.0 qA

1.0 3.2273 qB


qA = -0.38922 qC = -0.28313

qB = 0.28313 qD = 0.38922

=CAB - CAD

CBC - CBA

=-0.97300706

0.52450206

B

A

(t2)

(t1)

B0

(t3)

(t1)

C

H0

D



27/41

MAB = k1(2qA +qB) - CAB = -1.54132 tfm


MBC = k2(2qB+qC) - CBC = -1.23153 tfm

MCB = k2(2qC+qB)+CCB = 1.23153 tfm

MCD = k1(2qC+qD) - CCD = -1.23153 tfm

MDC =k1 (2qD+ qC)+CDC = 1.54132 tf m

MDA = k3(2qD+qA) - CDA = -1.54132 tfm








L Length of member (=H0) 2.700 mch Protective covering height 0.060 mt Thickness of member (height) 0.200 md Effective height of member 0.140 m


= 2.433 tf

SBA = SAB - L(w1+w2)/2 = -2.235 tf


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



Sx2 = -1.742 tf

c) Bending Moment

MA = MAB = -1.541 tfm

MB = -MBA = -1.232 tfm


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

= 2.4329 -1.6938 x + -0.0130 x2 , x = -132.086

1.421


Mmax = SABx - w1x2/2 - (w2-w1)x

3/(6L) + MAB = 0.193 tfm

w1

w2

A

B

L

x

MAB

MBA



28/41









SCB = SBC -w1L - w2b = -4.306 tf




29/41


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


(ii) In case of x2 = 2.420 m

Sx2 = -1.9576 tf

c) Bending Moment

MC = MCD = -1.232 tfm

MD = -MDC = -1.541 tfm


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

= 2.2349 -1.7638 x + 0.0130 x2 , x = 134.8

1.279



3/(6L) + MCD = 0.193 tfm








= 5.506 tf

SAD = SDA - L(w1+w2)/2 = -5.506 tf


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



c) Bending Moment

MD = MDA = -1.541 tfm

MA = -MAD = -1.541 tfm


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

= 5.5064 -5.0058 x , x = 1.100



3/(6L) + MDA = 1.487 tfm

L

x

D

MDA

w1w2

A

MAD



30/41

Summary of Internal forces

M N

(tfm) (tf) at joint at 2dSide wall A -2.331 5.503 6.531 4.790

(left) Case.1 Middle 1.407 4.951 0.000 -B -1.828 4.303 -3.729 -3.342A -2.590 5.506 7.576 5.618

Case.2 Middle 1.848 4.947 0.000 -B -2.087 4.306 -4.773 -4.170A -1.282 5.503 1.388 1.130

Case.3 Middle -0.249 4.846 0.000 -B -0.972 4.303 -1.190 -0.914A -1.541 5.506 2.433 1.958

Case.4 Middle 0.193 4.875 0.000 -B -1.232 4.306 -2.235 -1.742

Top slab B -1.828 3.729 4.303 3.208Case.1 Middle 0.539 3.729 0.000 -

C -1.828 3.729 -4.303 -3.208

B -2.087 4.773 4.306 3.210Case.2 Middle 0.281 4.773 0.000 -

C -2.087 4.773 -4.306 -3.210B -0.972 1.190 4.303 3.208

Case.3 Middle 1.395 1.190 0.000 -C -0.972 1.190 -4.303 -3.208B -1.232 2.235 4.306 3.210

Case.4 Middle 1.137 2.235 0.000 -C -1.232 2.235 -4.306 -3.210

Side wall C -1.828 4.303 3.729 3.342(right) Case.1 Middle 1.407 4.951 0.000 -

D -2.331 5.503 -6.531 -4.790C -2.087 4.306 4.773 4.170

Case.2 Middle 1.848 4.947 0.000 -D -2.590 5.506 -7.576 -5.618C -0.972 4.303 1.190 0.914

Case.3 Middle -0.249 4.846 0.000 -D -1.282 5.503 -1.388 -1.130C -1.232 4.306 2.235 1.742

Case.4 Middle 0.193 4.875 0.000 -D -1.541 5.506 -2.433 -1.958

Invert D -2.331 6.531 5.503 4.102Case.1 Middle 0.696 6.531 0.000 -

A -2.331 6.531 -5.503 -4.102D -2.590 7.576 5.506 4.105

Case.2 Middle 0.438 7.576 0.000 -A -2.590 7.576 -5.506 -4.105D -1.282 1.388 5.503 4.102

Case.3 Middle 1.745 1.388 0.000 -

A -1.282 1.388 -5.503 -4.102D -1.541 2.433 5.506 4.105

Case.4 Middle 1.487 2.433 0.000 -A -1.541 2.433 -5.506 -4.105

S (tf)Member Case

Check

Point

30/41 (3)168023748.xls.ms_officeSum MSN


31/41

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

1) At Joint "A" of side wall

Case.1

M= 2.3305 tfm sca = 60 kgf/m2 h = 20 cm (height of membeN= 5.5029 tf ssa = 1400 kgf/m2 d = 14 cm (effective height

S0= 6.5312 tf n = 24 d' = 6 cm (protective coverS2d= 4.7896 tf c = 4.00 cm (distance from ne

b = 100 cm

e = M/N = 42.35 cmSolving the formula shown below, sc 38.729 kgf/cm2 ( -178470 kgf/cm2) check 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 +48.46 sc^2 -4554.7 sc -132846.49s = nsc/(nsc+ssa) = 0.3990Asreq = (sc*s/2 - N/(bd))bd/ssa = 3.7960 cm2

Case.2

M= 2.5901 tfm sca = 60 kgf/m2 h = 20 cm (height of membe

N= 5.5064 tf ssa = 1400 kgf/m2 d = 14 cm (effective height

S0= 7.5758 tf n = 24 d' = 6 cm (protective coverS2d= 5.6176 tf c = 4.00 cm (distance from ne

b = 100 cm

e = M/N = 47.04 cmSolving the formula shown below, sc 39.304 kgf/cm2 ( -214195 kgf/cm2) check 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 +44.48 sc^2 -5018.6 sc -146375.06s = nsc/(nsc+ssa) = 0.4026Asreq = (sc*s/2 - N/(bd))bd/ssa = 3.9778 cm2

Case.3M= 1.2817 tfm sca = 60 kgf/m2 h = 20 cm (height of membe


S0= 1.3882 tf n = 24 d' = 6 cm (protective cover

S2d= 1.1296 tf c = 4.00 cm (distance from ne

b = 100 cme = M/N = 23.29 cmSolving the formula shown below, sc 34.762 kgf/cm2 ( -51486 kgf/cm2) check 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


Case.4




b = 100 cm



The maximum requirement of reinforcement bar is 3.9778 cm2 in Case2 from above calculation.

Case. 1 2 3 4

Requirement 3.7960 3.9778 2.5595 2.7390 (cm2)

2) At Joint "B" of side wall

Case.1




b = 100 cm



h

h


32/41

Case.3

M= 0.9719 tfm sca = 60 kgf/m2 h = 20 cm (height of memb


S0= 1.1903 tf n = 24 d' = 6 cm (protective cove

S2d= 0.9141 tf c = 4.00 cm (distance from n

b = 100 cm



Case.4





b = 100 cm




Case. 1 2 3 4

Requirement 3.2920 3.4713 2.2805 2.4577 (cm2)

3) At Joint "B" of top slab

Case.1





b = 100 cm

e = M/N = 49.02 cm

Solving the formula shown below, sc 32.854 kgf/cm2 ( -121704 kgf/cm2) check 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


Case.2





b = 100 cm


0 = sc^3 +52.63 sc^2 -4068.5 sc -118663.80

s = nsc/(nsc+ssa) = 0.3640Asreq = (sc*s/2 - N/(bd))bd/ssa = 2.6683 cm2

Case.3





b = 100 cm



Case.4

M= 1.2315 tfm sca = 60 kgf/m2 h = 20 cm (height of membN= 2.2349 tf ssa = 1400 kgf/m2 d = 14 cm (effective height


S2d= 3 2102 tf c = 4 00 cm (distance from n

h


33/41

4) At Joint "A" of invert

Case.1







Case.2





b = 100 cm

e = M/N = 34.19 cmSolving the formula shown below, sc 38.275 kgf/cm2 ( -229043 kgf/cm2) check

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 +43.22 sc^2 -5166.4 sc -150686.44s = nsc/(nsc+ssa) = 0.3962Asreq = (sc*s/2 - N/(bd))bd/ssa = 2.1706 cm2

Case.3





b = 100 cm



Case.4





b = 100 cm




Case. 1 2 3 4Requirement 2.7547 2.1706 4.5017 3.9142 (cm2)

5) At Middle of side wall

Case.1



n = 24 d' = 6 cm (protective cove

c = 4.00 cm (distance from n

b = 100 cm


0 = sc^3 +62.94 sc^2 -2865.2 sc -83569.35s = nsc/(nsc+ssa) = 0.2737

Asreq = (sc*s/2 - N/(bd))bd/ssa = 0.0000 cm2 Tensile is on inside of member

Case.2


h

h


34/41

Case.3





b = 100 cm


0 = sc^3 +80.73 sc^2 -790.27 sc -23049.47s = nsc/(nsc+ssa) = 0.2219Asreq = (sc*s/2 - N/(bd))bd/ssa = 0.0000 cm2 Tensile is on outside of member

Case.4





b = 100 cm


0 = sc^3 +81.56 sc^2 -693.38 sc -20223.53

s = nsc/(nsc+ssa) = 0.2240Asreq = (sc*s/2 - N/(bd))bd/ssa = 0.0000 cm2 Tensile is on inside of member


Case. 1 2 3 4

Requirement 0.0000 0.0000 - 0.0000 (cm2)

Side inside inside outside inside

6) At Middle of top slab

Case.1





b = 100 cm

e = M/N = 14.45 cmSolving the formula shown below, sc 27.583 kgf/cm2 ( 9834 kgf/cm2) check 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 +76.97 sc^2 -1228.4 sc -35829.57s = nsc/(nsc+ssa) = 0.3210Asreq = (sc*s/2 - N/(bd))bd/ssa = 1.7643 cm2 Tensile is on inside of member

Case.2





b = 100 cm

e = M/N = 5.89 cmSolving the formula shown below, sc 28.160 kgf/cm2 ( 37662 kgf/cm2) check 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 +80.27 sc^2 -842.93 sc -24585.42

s = nsc/(nsc+ssa) = 0.3256Asreq = (sc*s/2 - N/(bd))bd/ssa = 1.1745 cm2 Tensile is on inside of member

Case.3





b = 100 cm



Case.4

M= 1.1370 tfm sca = 60 kgf/m2 h = 20 cm (height of membN= 2.2349 tf ssa = 1400 kgf/m2 d = 14 cm (effective height


c = 4 00 cm (distance from n

h


35/41

7) At Middle of invert

Case.1







Case.2





b = 100 cm

e = M/N = 5.79 cmSolving the formula shown below, sc 31.902 kgf/cm2 ( 29116 kgf/cm2) check

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 +76.15 sc^2 -1323.9 sc -38614.44s = nsc/(nsc+ssa) = 0.3535

Asreq = (sc*s/2 - N/(bd))bd/ssa = 0.2279 cm2 Tensile is on inside of member

Case.3





b = 100 cm


0 = sc^3 +59.94 sc^2 -3215 sc -93770.34s = nsc/(nsc+ssa) = 0.3623Asreq = (sc*s/2 - N/(bd))bd/ssa = 5.0134 cm2 Tensile is on inside of member

Case.4





b = 100 cm

e = M/N = 61.13 cmSolving the formula shown below, sc 33.682 kgf/cm2 ( -67860 kgf/cm2) check

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



Case. 1 2 3 4Requirement 0.8160 0.2279 5.0134 4.4269 (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 Invert

Point bottom top end end middle middle middle

Side outside outside outside outside inside inside inside

Calculation 1) 2) 3) 4) 5) 6) 7)

Requirement 3.978 3.471 3.715 4.502 0.000 4.227 5.013 (cm2)

h


36/41

6 Bar Arrangement and Calculation of StressType: B2.00m x H2.50m

bottom middle top end middle end middle

outside inside outside outside inside outside inside

Bending moment M kgfcm 259,015 140,650 208,741 97,193 139,465 128,171 174,486

Shearing force (joint) S kgf 7,576 0 4,773 4,306 0 5,506 0

Shearing force (2d) S2d kgf 5,618 - 4,170 3,210 - 4,105 -Axial force N kgf 5,506 4,951 4,306 1,190 1,190 1,388 1,388

Height of member h cm 20 20 20 20 20 20 20

Covering depth d' cm 6 6 6 6 6 6 6

Effective height d cm 14 14 14 14 14 14 14

Effective width b cm 100 100 100 100 100 100 100

Effective area bd cm2 1400 1400 1400 1400 1400 1400 1400

Young's modulus ratio n - 24 24 24 24 24 24 24

Required R-bar Asreq cm2 3.98 0.00 3.47 3.71 4.23 4.50 5.01

R-bar arrangement 16@250 12@250 16@250 16@250 16@250 16@250 12@125

Reinforcement As cm2 8.04 4.52 8.04 8.04 8.04 8.04 9.05

Perimeter of R-bar U 20.11 15.08 20.11 20.11 20.11 20.11 30.16

M/N e cm 47.039 28.410 48.472 81.656 117.172 92.327 125.689

Dist. from neutral axis c cm 4.00 4.00 4.00 4.00 4.00 4.00 4.00

a' 111.1 55.2 115.4 215.0 321.5 247.0 347.1

b' 590.9 210.9 607.5 991.7 1402.9 1115.2 1690.1

c' -8272.7 -2953.3 -8505.1 -13883.7 -19640.3 -15613.3 -23661.5

x 10.16 17.62 10.19 9.38 8.92 9.69 9.27

######## ######## ######## ######## ######## ######## ########

(check) check check check check check check check

Compressive stress sc kgf/cm2 12.7 5.5 9.9 3.2 3.5 3.5 3.9

Allowable stress sca kgf/cm2 60.0 60.0 60.0 60.0 60.0 60.0 60.0

ok ok ok ok ok ok ok

Tensile stress ss kgf/cm2 115.0 0.0 88.6 37.6 48.4 37.2 48.3

Allowable stress ssa kgf/cm2 1400.0 1400.0 1400.0 1400.0 1400.0 1400.0 1400.0

ok ok ok ok ok ok ok

Shearing stress at joint t kgf/cm2 5.41 0.00 3.41 3.08 0.00 3.93 0.00

Allowable stress ta kgf/cm2 11.00 11.00 11.00 11.00 11.00 11.00 11.00ok ok ok ok ok ok ok

Shearing stress at 2d t2d kgf/cm2 4.01 - 2.98 2.29 - 2.93 -

Allowable stress t2da kgf/cm2 5.50 - 5.50 5.50 - 5.50 -

ok - ok ok - ok -

Resisting Moment Mr kgfcm 206,044 173,378 205,852 205,817 205,817 205,798 213,308

Mr for compression Mrc kgfcm 206,044 173,378 205,852 205,817 205,817 205,798 213,308

x for Mrc cm 6.408 5.258 6.238 5.820 5.820 5.846 6.096 ss for Mrc kgf/cm2 1706.1 2394.3 1791.9 2023.7 2023.7 2008.6 1867.3

Mr for tensile Mrs kgfcm 372,070 204,735 334,555 251,307 251,307 256,073 297,721

x for Mrs cm 8.970 7.358 8.643 7.729 7.729 7.791 8.215 sc for Mrs kgf/cm2 104.0 64.6 94.1 71.9 71.9 73.2 82.8

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

Reinforcement As cm2 4.52 4.52 4.52 4.52 4.52 4.52 4.52ok ok ok ok ok ok ok

Reinforcement bar for fillet 12@250 12@250

Reinforcement As cm2 4.52 4.52

Minimum requirement of reinforcement bar As min = 4.5 cm2

Side wall Top slab Invert

36/41 (5)168023748.xls.ms_officeR-bar stress


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DATA PLOT PEMBESIAN

1. Design Data

b1 = m b2 = m

h1 = m h2 = m h3 = m d = m

Bb = m Ha = m Fillet = m

2. Data PembesianBottom slab : Tulangan bagi :

As1 (cm ) : O 16 @ O 12 @ O 12 @

As2 (cm ) : O 12 @ O 16 @

Side wall : Tulangan bagi :

As1 (cm ) : O 16 @ O 12 @ O 12 @

As2 (cm ) : O 12 @ O 16 @

Top slab : Tulangan bagi :

As1 (cm ) : O 16 @ O 16 @ O 12 @

As2 (cm ) : O 16 @ O 16 @

3. Tulangan Miring (fillet) :

Bottom slab : O 12 @

Top slab : O 12 @

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

Lokasi : . Irrigation Project

250

125 250

Section of Culvert

0.200 2.000

0.200 2.500 0.200 0.06

Tumpuan Lapangan

2.400 2.900 0.20

Tumpuan Lapangan

250 125

Tumpuan Lapangan

250 250 250

250 250

250

250 250

250

250

250 250

b1b1 b2

h3

h2

h1

Bb

Ha


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DATA PLOT MOMENT AND SHEAR FORCE DIAGRAM

1. Design Data

b1 = m b2 = m

h1 = m h2 = m h3 = m

Bb = m Ha = m

2. Result Calculation

Side wall (left)

Case 1Case 2

Side wall (right)

Case 1

Case 2

Top slab

Case 1

Case 2

Bottom slab

Case 1

Case 2

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

Lokasi : . Irrigation Project

Case 1MmaxMab =Mcd =

Sb

Section of Culvert

0.200 2.000

0.200 2.500 0.200

Na Nb

2.200 2.700

Ma Mmax Mb Sa

-2.331 1.407 -1.828 6.531 -3.729 5.503 4.303

Nc

-1.282 -0.249 -0.972 1.388 -1.190 5.503 4.303

Mc Mmax Md Sd Sc Nd

-1.828 1.407 -2.331 -6.531 3.729 5.503 4.303

-0.972 -0.249 -1.282 -1.388 1.190 5.503 4.303

3.729

Mb Mmax Mc Sb Sc Nb Nc

-1.828 0.539 -1.828 4.303 -4.303 3.729

-0.972 1.395 -0.972 4.303 -4.303 1.190 1.190

Ma Mmax Md Sa Sd Na Nd

1.388

-2.331 0.696 -2.331 -5.503 5.503 6.531 6.531

-1.282 1.745 -1.282 -5.503 5.503 1.388

x x1 x2 x3 x4 x5 x6 x71.242 -0.48 0.70 1.30 1.38 1.04 0.34 -0.641.458 -0.48 0.70 1.30 1.38 1.04 0.34 -0.64

b1b1 b2

h3

h2

h1

Bb

Ha


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MmaxMbc =Mad =

Ha =

Bb =

Sab

Case 2MmaxMab =Mcd =MmaxMbc =Mad =

Ha =

Bb =

Sab

-0.05 -0.79y1 y2 y3 y4 y5 y6 y7

1.1 -0.79 -0.05 0.39 0.54 0.391.1 -1.01 -0.06 0.51 0.70 0.51 -0.06 -1.01

0.550 0.275

2.700 0.338 0.675 1.013 1.350 1.688 2.025 2.363

2.200 0.275 0.550 0.825 1.100 0.825

-2.53 -3.24x x1 x2 x3 x4 x5 x6 x7

2.700 4.45 2.60 0.98 -0.42 -1.59

-0.63x x1 x2 x3 x4 x5 x6 x7

-0.631.242 -0.87 -0.56 -0.35 -0.26 -0.27 -0.39

y71.458 -0.87 -0.56 -0.35 -0.26 -0.27 -0.39

0.80 0.06y1 y2 y3 y4 y5 y6

1.1 0.06 0.80 1.25 1.39 1.251.1 0.04 0.99 1.56 1.74 1.56 0.99 0.04

0.550 0.275

2.700 0.338 0.675 1.013 1.350 1.688 2.025 2.363

2.200 0.275 0.550 0.825 1.100 0.825

-0.53 -0.86x x1 x2 x3 x4 x5 x6 x7

2.700 1.08 0.76 0.44 0.12 -0.20


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MAB MBA

-1.282 0.972-2.331

MCD

1.828

MDC

-1.828 2.331

MBC MCB

-0.972

-1.828

1.282

1.828

-0.972 0.972

2.331 -2.331

MAD

1.282

MDA

-1.282

q1 q26.500 1.1001.100 6.500


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q1

q1

3.9125.003

q1 q20.920 0.9900.990 0.920

3.9125.003

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