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SUMMARY OF STRUCTURAL CALCULATION OF 1-BARREL BOX CULVERT
1 Design Dimensions and Bar Arrangements Class I Road (BM100)
Type of box culvert B0,4 x H0,4
Clear width m 0,40
Clear height m 0,40
Height of fillet m 0,15
Thickness Side wall cm 20,0
Top slab cm 20,0
Bottom slab cm 20,0
Cover of reinforcement bar (between concrete surface and center of reinforcement bar)
Side wall Outside cm 6,0
Inside cm 6,0
Top slab Upper cm 6,0Lower cm 6,0
Bottom slab Lower cm 6,0
Upper cm 6,0
Bar arrangement (dia - spacing per uni t length of 1.0 m)
Side wall Lower outside Tensile bar mm 12@150
Distribution bar mm 12@250
Middle inside Tensile bar mm 12@250
Distribution bar mm 12@250
Upper outside Tensile bar mm 12@150
Distribution bar mm 12@250
Top slab Upper edge Tensile bar mm 12@150
Distribution bar mm 12@250
Lower middle Tensile bar mm 12@150Distribution bar mm 12@250
Bottom slab Lower edge Tensile bar mm 12@150
Distribution bar mm 12@250
Upper middle Tensile bar mm 12@150
Distribution bar mm 12@250
Fillet Upper edge Fillet bar mm 12@250
Lower edge Fillet bar mm 12@250
2 Design Parameters
Unit Weight Reinforced Concrete c= 2,4 tf/m3
Backfill soil (wet) s= 1,8 tf/m3
(submerged) s'= 1,0 tf/m3
Live Load Class of road Class I (BM100)
Truck load at rear wheel P= 10,0 tf
Impact coefficien (for Class I to IV road Ci= 0,3 (D4.0m)
Pedestrian load (for Class V roads) 0 tf/m2
Concrete Design Strength ck= 175 kgf/cm2
(K175) Allowable Compressive Stress ca= 70 kgf/cm2
Allowable Shearing Stress a= 5,5 kgf/cm2
Reinforcement Bar Allowable Tensile Stress sa= 2.100 kgf/cm2
(U24, deformed bar) Yielding Point of Reinforcement Bar sy= 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: B0,40m x H0,40m Class I Road
Soil Cover Depth: 0,4 m
1 Dimensions and ParametersBasic Parameters
Ka: Coefficient of static earth pressure 0,5
w: Unit weight of water (t/m3) 1,00 t/m3
d: Unit weight of soil (dry) (t/m3) 1,80 t/m3
s: Unit weight of soil (saturated) (t/m3) 2,00 t /m3
c: Unit weight of reinforced concrete (t/m3) 2,40 t/m3ck: Concrete Design Strength 175 kgf/m2
ca Allowable Stress of Concrete 70 kgf/m2
sa: Allowable Stress of Reinforcement Bar 2100 kgf/m2
a: Allowable Stress of Shearing (Concrete) 5,5 kgf/m2
sy: Yielding Point of Reinforcement Bar 3000 kgf/m2
n: Young's Modulus Ratio 24
Fa: Safety factor against uplift 1,2
Basic Dimensions
H: Internal Height of Box Culvert 0,40 m
B: Internal Width of Box Culvert 0,40 m
Hf: Fillet Height 0,15 m
t1: Thickness of Side Wall 0,20 m (> 0.25m)
t2: Thickness of Top Slab 0,20 m (> 0.25m)
t3: Thickness of Invert (Bottom Slab) 0,20 m (> 0.25m)
BT: Gross Width of Box Culvert 0,80 m
HT: Gross Height of Box Culvert 0,80 m
D: Covering Depth 0,41 mGwd: Underground Water Depth for Case 1, 2 0,41 m (= D)
hiw: Internal Water Depth for Case 1, 2 0,00 m
for Case 3, 4 0,40 m
Cover of R-bar Basic Conditions
Top Slab d2 0,06 m Classification of Live load by truck Class 1
Side Wall d1 0,06 m PTM: Truck load of Middle Tire 10,00 t
Bottom Slab d3 0,06 m Ii: Impact coefficient (D4.0m:0, D
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Load distribution of truck tire
(1) Middle tire's acting point: center of the top slab
a) distributed load of middle tire
Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 8,3030 tf/m2, B = 0,600 m
am': length of distributed load 2D+1.75+bm = 3,070 m
bm': width of distributed load 2D+am = 1,020 m
b) distributed load of rear tire
Pvtr: distributed load of rear tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
ar': length of distributed load 2D+1.75+br = 3,070 mbr': width of distributed load 2D+ar = 1,020 m
c) distributed load of front tire
Pvtf: distributed load of front tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
af': length of distributed load 2D+1.75+bf = 3,070 m
bf': width of distributed load 2D+af = 1,020 m
(2) Middle tire's acting point: on the side wall
a) distributed load of middle tire
Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 8,3030 tf/m2, B = 0,600 m
am': length of distributed load 2D+1.75+bm = 3,070 m
bm': width of distributed load 2D+am = 1,020 m
b) distributed load of rear tire
Pvtr: distributed load of rear tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
ar': length of distributed load 2D+1.75+br = 3,070 m
br': width of distributed load 2D+ar = 1,020 m
c) distributed load of front tire
Pvtf: distributed load of front tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
af': length of distributed load 2D+1.75+bf = 3,070 m
bf': width of distributed load 2D+af = 1,020 m
(3) Rear tire's acting point: on the side wall
a) distributed load of rear tire
Pvtr: distributed load of rear tire 2PTR(1+Ii)/(ar'br') = 8,3030 tf/m2, B = 0,600 m
ar': length of distributed load 2D+1.75+br = 3,070 m
br': width of distributed load 2D+ar = 1,020 m
b) distributed load of middle tire
Pvtm: distributed load of middle tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
am': length of distributed load 2D+1.75+bm = 3,070 m
'' ,
c) distributed load of front tire
Pvtf: distributed load of front tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
af': length of distributed load 2D+1.75+bf = 3,070 m
bf': width of distributed load 2D+af = 1,020 m
(4) Combination of load distribution of track tire
Case.L1: Pvt1 = 8,3030 tf/m2, B = 0,600 m Combination for Case.L2 (2) (2) (3) (3)Pvt2 = 0,0000 tf/m2, B = 0,000 m a) + b) a) + c) a) + b) a) + c)
Case.L2: Pvt1 = 8,3030 tf/m2, B = 0,600 m Distributed load total 8,3030 8,3030 8,3030 8,3030
Pvt2 = 0,0000 tf/m2, B = 0,000 m Select the combination case of 8,3030 tf/m2,
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= 2,8913 > 1,2 ok
where, Vd: Total dead weight (t/m) Vd= 1,850 tf/m
U: Total uplift (t.m)
U=BT*HT*w U= 0,640 tf/m
Ws: Weight of covering soil Ws = BT*{(D-Gwd)*(sw)+Gwd*d} = 0,590 tf/m
Wc: Self weight of box culvert Wc = (HT*BT-H*B+2*Hf^2)*c = 1,260 tf/m
Fa: 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)*c/B0 Wtop= 0,7300
Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*gd*Gwd P3= 0,3690
P4=Ka*gs*(D1-Gwd) P4= 0,1000
P5=gw*(D1-Gwd) P5= 0,1000
Ph1= 4,7205
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*Gwd P3= 0,3690
P4=Ka*s*(D1+H0-Gwd) P4= 0,7000
P5=w*(D1+H0-Gwd) P5= 0,7000Ph2= 5,9205
4) self weight of side wall
Acting Load (tf/m)
Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
Pvd Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,0000 hiw: internal water depth 0,00 m
Up=-U/B0 U= -1,0667
Q= 10,0743
summary of resistance moment
Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,300 m
side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 10,0743 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 10,0743 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 3,1923 - 0,2887 0,9217
side wall (right) - -3,1923 - 0,2887 -0,9217
internal water 0,0000 - 0,3000 - 0,0000
uplift -0,6400 - 0,3000 - -0,1920
total 6,0446 1,8134
6) load against invert
Acting Load (tf/m2)Pvd 0,7380
Pvt1 8,3030
Pvt2 0,0000
Wtop 0,7300
Ws 0,6400
Pq= 10,4110
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Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L2
1) vertical load against top slab
Acting Load (tf/m2)
Wtop= (t2*BT+Hf^2)*c/B0 Wtop= 0,7300
Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*gd*Gwd P3= 0,3690
P4=Ka*gs*(D1-Gwd) P4= 0,1000
P5=gw*(D1-Gwd) P5= 0,1000
Ph1= 4,7205
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*Gwd P3= 0,3690
P4=Ka*s*(D1+H0-Gwd) P4= 0,7000
P5=w*(D1+H0-Gwd) P5= 0,7000
Ph2= 5,9205
4) self weight of side wall
Acting Load (tf/m)
Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
v v = ,
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,0000 hiw: internal water depth 0,00 m
Up=-U/B0 U= -1,0667
Q= 10,0743
summary of resistance moment
Item V H x y M(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,3000 m
side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,0000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 10,0743 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 10,0743 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 3,1923 - 0,2887 0,9217
side wall (right) - -3,1923 - 0,2887 -0,9217
internal water 0,0000 - 0,3000 - 0,0000
uplift -0,6400 - 0,3000 - -0,1920
total 6,0446 1,8134
6) load against invert
Acting Load (tf/m2)
Pvd 0,7380
Pvt1 8,3030Pvt2 0,0000
Wtop 0,7300
Ws 0,6400
total Pq= 10,4110
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Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L1
1) vertical load against top slab
Acting Load (tf/m2)
Wtop= (t2*BT+Hf^2)*c/B0 Wtop= 0,7300
Pvd=D*d Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*d*D1 P3= 0,4590
WP=-w*0 P4= 0,0000
Ph1= 4,6105
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*(D1+H0) P3= 0,9990
WP=-w*H P4= -0,4000
Ph2= 4,7505
4) self weight of side wall
Acting Load (tf/m)Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
Pvd Pvd= 0,7380
Pvt1 Pvt1= 8,3030
v v = ,
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,1917 hiw: internal water depth 0,400 m
Up=0 U= 0,0000
Q= 11,3327
summary of resistance moment
Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,3000 m side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,0000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 11,3327 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 11,3327 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 2,8083 - 0,2985 0,8383
side wall (right) - -2,8083 - 0,2985 -0,8383
internal water 0,1150 - 0,3000 - 0,0345
uplift 0,0000 - 0,3000 - 0,0000
total 6,7996 2,0399
6) load against invert
Acting Load (tf/m2)
Pvd 0,7380
Pvt1 8,3030
Pvt2 0,0000
Wtop 0,7300Ws 0,6400
total Pq= 10,4110
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Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2
1) vertical load against top slab
Acting Load (tf/m2)
Wtop= (t2*BT+Hf^2)*c/B0 Wtop= 0,7300
Pvd=D*d Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*d*D1 P3= 0,4590
WP=-w*0 P4= 0,0000
Ph1= 4,6105
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*(D1+H0) P3= 0,9990
WP=-w*H P4= -0,4000
Ph2= 4,7505
4) self weight of side wall
Acting Load (tf/m)Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
Pvd Pvd= 0,7380
Pvt1 Pvt1= 8,3030
v v = ,
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,1917 hiw: internal water depth 0,400 m
Up=0 U= 0,0000
Q= 11,3327
summary of resistance moment
Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,3000 m side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,0000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 11,3327 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 11,3327 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 2,8083 - 0,2985 0,8383
side wall (right) - -2,8083 - 0,2985 -0,8383
internal water 0,1150 - 0,3000 - 0,0345
uplift 0,0000 - 0,3000 - 0,0000
total 6,7996 2,0399
6) load against invert
Acting Load (tf/m2)
Pvd 0,7380
Pvt1 8,3030
Pvt2 0,0000
Wtop 0,7300Ws 0,6400
total Pq= 10,4110
Summary of Load Calculation
Item Pv1 Ph1 Ph2 Pq Wsw q1
Case (tf/m2) (tf/m2) (tf/m2) (tf/m2) (tf/m) (tf/m2)
Case.1 9,7710 4,7205 5,9205 10,4110 0,1920 10,0743
Case.2 9,7710 4,7205 5,9205 10,4110 0,1920 10,0743
Case.3 9,7710 4,6105 4,7505 10,4110 0,1920 11,3327
Case.4 9,7710 4,6105 4,7505 10,4110 0,1920 11,3327
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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 Term
Ph1 Horizontal Pressure at top of side wall 4,720 tf/m2
Ph2 Horizontal Pressure at bottom of side wall 5,920 tf/m2
Pv1 Vertical Pressure(1) on top slab 9,771 tf/m
2
Pv2 Vertical Pressure(2) on top slab 0,000 tf/m2
Pq Reaction to bottom slab 10,411 tf/m2
a Distance from joint B to far end of Pv2 0,600 m
b Distance from joint B to near end of Pv2 0,000 m
H0 Height of plane frame 0,600 m
B0 Width of plane frame 0,600 m
t1 Thickness of side wall 0,200 m
t2 Thickness of top slab 0,200 m
t3 Thickness of invert (bottom slab) 0,200 m
CAB= CDC= (2Ph1+3Ph2)H02/60 = 0,16321 tf m
CBA= CCD= (3Ph1+2Ph2)H02/60 = 0,15601 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 = 0,29313 tf m
CDA= CAD= PqB02/12 = 0,31233 tf m
2) Calculation of Bending Moment at joint
k1 = 1,0
k2 = H0t23/(B0t1
3) = 1,0000
k3 = H0t33/(B0t1
3) = 1,0000
2(k1+k3) k1 0 k3 -3k1 A CAB- CAD
k1 2(k1+k2) k2 0 -3k1 B CBC- CBA
0 2 2(k1+k2 k1 -3k1 = C - C
B
A
(t2)
(t1)
B0
(t3)
(t1)
C
H0
D
k3 0 k1 2(k1+k3) -3k1 D CDA- CDC
k1 k1 k1 k1 -4k1 R 0
As load has bilateral symmetry, the equation shown below is formed.
A= -D B= -C R =0
2k1+k3 k1 A
k1 2k1+k2 B
3,0000 1,0 A1,0 3,0000 B
By solving above equation, the result is led as shown below.
A = -0,07306 C = -0,07006
B = 0,07006 D = 0,07306
=CAB- CAD
CBC- CBA
=-0,14911493
0,13711493
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MAB= k1(2A+B) - CAB = -0,2393 tf m
MBA= k1(2B+A)+CBA = 0,2231 tf m
MBC= k2(2B+C) - CBC = -0,2231 tf m
MCB = k2(2C+B)+CCB = 0,2231 tf m
MCD= k1(2C+D) - CCD = -0,2231 tf m
MDC=k1 (2D+ C)+CDC = 0,2393 tf m
MDA= k3(2D+A) - CDA = -0,2393 tf m
MAD= k3(2A+D)+CAD = 0,2393 tf m
2) Calculation of Design Force
2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 5,920 tf/m2
w2 Load at end B 4,720 tf/m2
MAB Bending moment at end A -0,2393 tf m
MBA Bending moment at end B 0,2231 tf m
L Length of member (=H0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SAB= (2w1+w2)L/6 - (MAB+MBA)/L= 1,683 tf
SBA= SAB- L(w1+w2)/2 = -1,509 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SAB- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,104 tf
w1
w2
A
B
Lx
MAB
MBA
= ,
Sx2 = -0,109 tf
c) Bending Moment
MA = MAB = -0,239 tf m
MB = -MBA
= -0,223 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SAB- w1x - (w2 - w1)x2/(2L)
= 1,6831 -5,9205 x + 1,0000 x2 , x = 5,621
0,299
Bending moment at x 0,2994 m is;
Mmax = SABx - w1x2/2 - (w2-w1)x
3/(6L) + MAB = 0,008 tf m
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2-2) Top Slab
a) Shearing Force at joint
w1 Uniform load 9,771 tf/m2
w2 Uniform load 0,000 tf/m2
a Distance from end B to near end of 0,000 m
b Length of uniform load w2 0,600 m
MBC Bending moment at end B -0,2231 tf m
MCB Bending moment at end C 0,2231 tf m
L Length of member (=Bo) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SBC= (w1L+w2b)/2-(MBC+MCB)/L = 2,931 tf
SCB= SBC-w1L - w2b = -2,931 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SBC- w1x - w2(x-a) in case of 0,000 m
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b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SCD- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,109 tf
(ii) In case of x2 = 0,320 m
Sx2 = -0,104 tf
c) Bending Moment
MC = MCD = -0,223 tf m
MD = -MDC = -0,239 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SCD- w1x - (w2 - w1)x2/(2L)
= 1,5091 -4,7205 x -1,0000 x2 , x = -5,021
0,301
Bending moment at x 0,3006 m is;
Mmax = SCDx - w1x2/2 - (w2-w1)x
3/(6L) + MCD = 0,00825 tf m
2-4) Bottom Slab
a) Shearing Force at joint
w1 Reaction at end D 10,411 tf/m2
w2 Reaction at end A 10,411 tf/m2
MDA Bending moment at end B -0,23927 tf m
MAD Bending moment at end C 0,23927 tf m
L Length of member (=B0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SDA= (2w1+w2)L/6 - (MDA+MAD)/L
L
x
D
MDA
w1w2
A
MAD
= 3,123 tf
SAD= SDA- L(w1+w2)/2 = -3,123 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SDA- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,208 tf (ii) In case of x2 = 0,320 m
Sx2 = -0,208 tf
c) Bending Moment
MD = MDA = -0,239 tf m
MA = -MAD = -0,239 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SDA- w1x - (w2 - w1)x2/(2L)
= 3,1233 -10,4110 x , x = 0,300
Bending moment at x 0,3000 m is;
Mmax = SDAx - w1x2/2 - (w2-w1)x
3/(6L) + MDA = 0,229 tf m
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Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L2
1) Calculation of Load Term
Ph1 Horizontal Pressure at top of side wall 4,720 tf/m2
Ph2 Horizontal Pressure at bottom of side wall 5,920 tf/m2
Pv1 Vertical Pressure(1) on top slab 9,771 tf/m2
Pv2 Vertical Pressure(2) on top slab 0,000 tf/m2
Pq Reaction to bottom slab 10,411 tf/m2
a Distance from joint B to far end of Pv2 0,600 mb Distance from joint B to near end of Pv2 0,000 m
H0 Height of plane frame 0,600 m
B0 Width of plane frame 0,600 m
t1 Thickness of side wall 0,200 m
t2 Thickness of top slab 0,200 m
t3 Thickness of invert (bottom slab) 0,200 m
CAB= CDC= (2Ph1+3Ph2)H02/60 = 0,16321 tf m
CBA= CCD= (3Ph1+2Ph2)H02/60 = 0,15601 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= 0,29313 tf m
CDA= CAD= PqB02/12 = 0,31233 tf m
2) Calculation of Bending Moment at joint
k1 = 1,0
k2 = H0t23/(B0t13) = 1,00000
k3 = H0t33/(B0t1
3) = 1,00000
2(k1+k3) k1 0 k3 -3k1 A CAB- CAD
k1 2(k1+k2) k2 0 -3k1 B CBC- CBA
0 k2 2(k1+k2) k1 -3k1 C = CCD- CCB
k3 0 k1 2(k1+k3) -3k1 D CDA- CDC
k1 k1 k1 k1 -4k1 R 0
B
A
(t2)
(t1)
B0
(t3)
(t1)
C
H0
D
As load has bilateral symmetry, the equation shown below is formed.
A= -D B= -C R =0
2k1+k3 k1 A
k1 2k1+k2 B
3,0000 1,0 A1,0 3,0000 B
By solving above equation, the result is led as shown below.
A = -0,07306 C = -0,07006
B = 0,07006 D = 0,07306
=CAB- CAD
CBC- CBA
=-0,14911493
0,13711493
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MAB= k1(2A+B) - CAB = -0,2393 tf m
MBA= k1(2B+A)+CBA = 0,2231 tf m
MBC= k2(2B+C) - CBC = -0,2231 tf m
MCB = k2(2C+B)+CCB = 0,2231 tf m
MCD= k1(2C+D) - CCD = -0,2231 tf m
MDC=k1 (2D+ C)+CDC = 0,2393 tf m
MDA= k3(2D+A) - CDA = -0,2393 tf m
MAD= k3(2A+D)+CAD = 0,2393 tf m
2) Calculation of Design Force
2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 5,920 tf/m2
w2 Load at end B 4,720 tf/m2
MAB Bending moment at end A -0,2393 tf m
MBA Bending moment at end B 0,2231 tf m
L Length of member (=H0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SAB= (2w1+w2)L/6 - (MAB+MBA)/L
= 1,6831 tf SBA= SAB- L(w1+w2)/2 = -1,5091 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SAB- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,104 tf
(ii) In case of x2 = 0,32 m
= -
w1
w2
A
B
Lx
MAB
MBA
,
c) Bending Moment
MA = MAB = -0,239 tf m
MB = -MBA = -0,223 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SAB- w1x - (w2 - w1)x2/(2L)
= 1,6831 -5,9205 x + 1,0000 x2 , x = 5,621
0,299
Bending moment at x 0,2994 m is;
Mmax = SABx - w1x2/2 - (w2-w1)x
3/(6L) + MAB = 0,008 tf m
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2-2) Top Slab
a) Shearing Force at joint
w1 Uniform load 9,771 tf/m2
w2 Uniform load 0,000 tf/m2
a Distance from end B to near end of 0,000 m
b Length of uniform load w2 0,600 m
MBC
Bending moment at end B -0,2231 tf m
MCB Bending moment at end C 0,2231 tf m
L Length of member (=Bo) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SBC= (w1L+w2b)/2-(MBC+MCB)/L = 2,931 tf
SCB= SBC-w1L - w2b = -2,931 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SBC- w1x - w2(x-a) in case of 0,000 m
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b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SCD- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,109 tf
(ii) In case of x2 = 0,320 m
Sx2 = -0,104 tf
c) Bending Moment
MC = MCD = -0,223 tf m
MD = -MDC = -0,239 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SCD- w1x - (w2 - w1)x2/(2L)
= 1,5091 -4,7205 x -1,0000 x2 , x = -5,0211
0,3006
Bending moment at x 0,3006 m is;
Mmax = SCDx - w1x2/2 - (w2-w1)x
3/(6L) + MCD = 0,0083 tf m
2-4) Bottom Slab
a) Shearing Force at joint
w1 Reaction at end D 10,411 tf/m2
w2 Reaction at end A 10,411 tf/m2
MDA Bending moment at end B -0,2393 tf m
MAD Bending moment at end C 0,2393 tf m
L Length of member (=B0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SDA= (2w1+w2)L/6 - (MDA+MAD)/L
= 3,123 tf
L
x
D
MDA
w1w2
A
MAD
SAD= SDA- L(w1+w2)/2 = -3,123 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SDA- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,208 tf
(ii) In case of x2 = 0,320 mSx2 = -0,208 tf
c) Bending Moment
MD = MDA = -0,239 tf m
MA = -MAD = -0,239 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SDA- w1x - (w2 - w1)x2/(2L)
= 3,1233 -10,4110 x , x = 0,3000
Bending moment at x 0,3000 m is;
Mmax = SDAx - w1x2/2 - (w2-w1)x
3/(6L) + MDA = 0,229 tf m
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Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L1
1) Calculation of Load Term
Ph1 Horizontal Pressure at top of side wall 4,610 tf/m2
Ph2 Horizontal Pressure at bottom of side wall 4,750 tf/m2
Pv1 Vertical Pressure(1) on top slab 9,771 tf/m2
Pv2 Vertical Pressure(2) on top slab 0,000 tf/m2
Pq Reaction to bottom slab 10,411 tf/m2
a Distance from joint B to far end of Pv2 0,600 mb Distance from joint B to near end of Pv2 0,000 m
H0 Height of plane frame 0,600 m
B0 Width of plane frame 0,600 m
t1 Thickness of side wall 0,200 m
t2 Thickness of top slab 0,200 m
t3 Thickness of invert (bottom slab) 0,200 m
CAB= CDC= (2Ph1+3Ph2)H02/60 = 0,14083 tf m
CBA= CCD= (3Ph1+2Ph2)H02/60 = 0,13999 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= 0,29313 tf m
CDA= CAD= PqB02/12 = 0,31233 tf m
2) Calculation of Bending Moment at joint
k1 = 1,0
k2 = H0t23/(B0t13) = 1,00000
k3 = H0t33/(B0t1
3) = 1,00000
2(k1+k3) k1 0 k3 -3k1 A CAB- CAD
k1 2(k1+k2) k2 0 -3k1 B CBC- CBA
0 k2 2(k1+k2) k1 -3k1 C = CCD- CCB
k3 0 k1 2(k1+k3) -3k1 D CDA- CDC
k1 k1 k1 k1 -4k1 R 0
B
A
(t2)
(t1)
B0
(t3)
(t1)
C
H0
D
As load has bilateral symmetry, the equation shown below is formed.
A= -D B= -C R =0
2k1+k3 k1 A
k1 2k1+k2 B
3,0000 1,0 A
1,0 3,0000 B
By solving above equation, the result is led as shown below.
A = -0,08345 C = -0,07886
B = 0,07886 D = 0,08345
=CAB- CAD
CBC- CBA
=-0,17149493
0,15313493
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MAB= k1(2A+B) - CAB = -0,22888 tf m
MBA= k1(2B+A)+CBA = 0,21427 tf m
MBC= k2(2B+C) - CBC = -0,21427 tf m
MCB = k2(2C+B)+CCB = 0,21427 tf m
MCD= k1(2C+D) - CCD = -0,21427 tf m
MDC=k1 (2D+ C)+CDC = 0,22888 tf m
MDA= k3(2D+A) - CDA = -0,22888 tf m
MAD= k3(2A+D)+CAD = 0,22888 tf m
2) Calculation of Design Force
2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 4,750 tf/m2
w2 Load at end B 4,610 tf/m2
MAB Bending moment at end A -0,2289 tf m
MBA Bending moment at end B 0,2143 tf m
L Length of member (=H0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SAB= (2w1+w2)L/6 - (MAB+MBA)/L
= 1,435 tf SBA= SAB- L(w1+w2)/2 = -1,373 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SAB- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,115 tf
(ii) In case of x2 = 0,320 m
w1
w2
A
B
Lx
MAB
MBA
x = - ,
c) Bending Moment
MA = MAB = -0,229 tf m
MB = -MBA = -0,214 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SAB- w1x - (w2 - w1)x2/(2L)
= 1,4355 -4,7505 x + 0,1167 x2 , x = 40,414
0,304
Bending moment at x 0,3045 m is;
Mmax = SABx - w1x2/2 - (w2-w1)x
3/(6L) + MAB = -0,011 tf m
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2-2) Top Slab
a) Shearing Force at joint
w1 Uniform load 9,771 tf/m2
w2 Uniform load 0,000 tf/m2
a Distance from end B to near end of 0,000 m
b Length of uniform load w2 0,600 m
MBC
Bending moment at end B -0,214 tf m
MCB Bending moment at end C 0,214 tf m
L Length of member (=Bo) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SBC= (w1L+w2b)/2-(MBC+MCB)/L = 2,931 tf
SCB= SBC-w1L - w2b = -2,931 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SBC- w1x - w2(x-a) in case of 0,000 m
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b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SCD- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,073 tf
(ii) In case of x2 = 0,320 m
Sx2 = -0,115 tf
c) Bending Moment
MC = MCD = -0,214 tf m
MD = -MDC = -0,229 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SCD- w1x - (w2 - w1)x2/(2L)
= 1,3728 -4,6105 x -0,1167 x2 , x = -39,81
0,296
Bending moment at x 0,2955 m is;
Mmax = SCDx - w1x2/2 - (w2-w1)x
3/(6L) + MCD = -0,011 tf m
2-4) Bottom Slab
a) Shearing Force at joint
w1 Reaction at end D 10,411 tf/m2
w2 Reaction at end A 10,411 tf/m2
MDA Bending moment at end B -0,229 tf mMAD Bending moment at end C 0,229 tf m
L Length of member (=B0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SDA= (2w1+w2)L/6 - (MDA+MAD)/L
= 3,123 tf
SAD= SDA- L(w1+w2)/2 = -3,123 tf
L
x
D
MDA
w1w2
A
MAD
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SDA- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,208 tf
(ii) In case of x2 = 0,320 m
Sx2 = -0,208 tf
c) Bending Moment
MD = MDA = -0,229 tf m
MA = -MAD = -0,229 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SDA- w1x - (w2 - w1)x2/(2L)
= 3,1233 -10,4110 x , x = 0,300
Bending moment at x 0,3000 m is;
Mmax = SDAx - w1x2/2 - (w2-w1)x
3/(6L) + MDA = 0,240 tf m
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Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2
1) Calculation of Load Term
Ph1 Horizontal Pressure at top of side wall 4,610 tf/m2
Ph2 Horizontal Pressure at bottom of side wall 4,750 tf/m2
Pv1 Vertical Pressure(1) on top slab 9,771 tf/m2
Pv2 Vertical Pressure(2) on top slab 0,000 tf/m2
Pq Reaction to bottom slab 10,411 tf/m2
a Distance from joint B to far end of Pv2 0,600 m
b Distance from joint B to near end of Pv2 0,000 m
H0 Height of plane frame 0,600 m
B0 Width of plane frame 0,600 m
t1 Thickness of side wall 0,200 m
t2 Thickness of top slab 0,200 m
t3 Thickness of invert (bottom slab) 0,200 m
CAB= CDC= (2Ph1+3Ph2)H02/60 = 0,14083 tf m
CBA= CCD= (3Ph1+2Ph2)H02/60 = 0,13999 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= 0,29313 tf m
CDA= CAD= PqB02/12 = 0,31233 tf m
2) Calculation of Bending Moment at joint
k1 = 1,0
k2 = H0t23/(B0t1
3) = 1,00000
k3 = H0t33/(B0t1
3) = 1,00000
2(k1+k3) k1 0 k3 -3k1 A CAB- CAD
k1 2(k1+k2) k2 0 -3k1 B CBC- CBA
0 k2 2(k1+k2) k1 -3k1 C = CCD- CCB
k3 0 k1 2(k1+k3) -3k1 D CDA- CDC
k1 k1 k1 k1 -4k1 R 0
B
A
(t2)
(t1)
B0
(t3)
(t1)
C
H0
D
As load has bilateral symmetry, the equation shown below is formed.
A= -D B= -C R =0
2k1+k3 k1 A
k1 2k1+k2 B
3,0000 1,0 A
1,0 3,0000 B
By solving above equation, the result is led as shown below.
A = -0,08345 C = -0,07886
B = 0,07886 D = 0,08345
=CAB- CAD
CBC- CBA
=-0,17149493
0,15313493
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MAB= k1(2A+B) - CAB = -0,22888 tf m
MBA= k1(2B+A)+CBA = 0,21427 tf m
MBC= k2(2B+C) - CBC = -0,21427 tf m
MCB = k2(2C+B)+CCB = 0,21427 tf m
MCD= k1(2C+D) - CCD = -0,21427 tf m
MDC=k1 (2D+ C)+CDC = 0,22888 tf m
MDA= k3(2D+A) - CDA = -0,22888 tf m
MAD= k3(2A+D)+CAD = 0,22888 tf m
2) Calculation of Design Force
2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 4,750 tf/m2
w2 Load at end B 4,610 tf/m2
MAB Bending moment at end A -0,229 tf m
MBA Bending moment at end B 0,214 tf m
L Length of member (=H0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SAB= (2w1+w2)L/6 - (MAB+MBA)/L
= 1,435 tf
SBA= SAB- L(w1+w2)/2 = -1,373 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SAB- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,115 tf
=
w1
w2
A
B
Lx
MAB
MBA
,
Sx2 = -0,073 tf
c) Bending Moment
MA = MAB = -0,229 tf m
MB = -MBA = -0,214 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SAB- w1x - (w2 - w1)x2/(2L)
= 1,4355 -4,7505 x + 0,1167 x2 , x = 40,414
0,304
Bending moment at x 0,3045 m is;
Mmax = SABx - w1x2/2 - (w2-w1)x
3/(6L) + MAB = -0,011 tf m
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2-2) Top Slab
a) Shearing Force at joint
w1 Uniform load 9,771 tf/m2
w2 Uniform load 0,000 tf/m2
a Distance from end B to near end of 0,000 m
b Length of uniform load w2 0,600 m
MBC
Bending moment at end B -0,214 tf m
MCB Bending moment at end C 0,214 tf m
L Length of member (=Bo) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SBC= (w1L+w2b)/2-(MBC+MCB)/L = 2,931 tf
SCB= SBC-w1L - w2b = -2,931 tf
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SBC- w1x - w2(x-a) in case of 0,000 m
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b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SCD- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,0727 tf
(ii) In case of x2 = 0,320 m
Sx2 = -0,1145 tf
c) Bending MomentMC = MCD = -0,214 tf m
MD = -MDC = -0,229 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SCD- w1x - (w2 - w1)x2/(2L)
= 1,3728 -4,6105 x -0,1167 x2 , x = -39,8
0,296
Bending moment at x 0,2955 m is;
Mmax = SCDx - w1x2/2 - (w2-w1)x
3/(6L) + MCD = -0,011 tf m
2-4) Bottom Slab
a) Shearing Force at joint
w1 Reaction at end D 10,411 tf/m2
w2 Reaction at end A 10,411 tf/m2
MDA Bending moment at end B -0,229 tf mMAD Bending moment at end C 0,229 tf m
L Length of member (=B0) 0,600 m
ch Protective covering height 0,060 m
t Thickness of member (height) 0,200 m
d Effective height of member 0,140 m
SDA= (2w1+w2)L/6 - (MDA+MAD)/L
= 3,123 tf
SAD= SDA- L(w1+w2)/2 = -3,123 tf
L
x
D
MDA
w1w2
A
MAD
b) Shearing Force at 2d point from joint
Shearing force at the point with a distance of 2d from joint is calculated by following equation.
Sx = SDA- w1x - (w2 - w1)x2/(2L)
(i) In case of x1 = 0,280 m
Sx1 = 0,208 tf
(ii) In case of x2 = 0,320 m
Sx2 = -0,208 tf
c) Bending Moment
MD = MDA = -0,229 tf m
MA = -MAD = -0,229 tf m
The maximum bending moment occurs at the point of that shearing force equal to zero.
Sx = 0 = SDA- w1x - (w2 - w1)x2/(2L)
= 3,1233 -10,4110 x , x = 0,300
Bending moment at x 0,3000 m is;
Mmax = SDAx - w1x2/2 - (w2-w1)x
3/(6L) + MDA = 0,240 tf m
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Summary of Internal forces
M N
(tfm) (tf) at joint at 2d
Side wall A -0,239 3,123 1,683 0,104 (left) Case.1 Middle 0,008 3,027 0,000 -
B -0,223 2,931 -1,509 -0,109
A -0,239 3,123 1,683 0,104
Case.2 Middle 0,008 3,027 0,000 -
B -0,223 2,931 -1,509 -0,109
A -0,229 3,123 1,435 0,115
Case.3 Middle -0,011 3,026 0,000 -
B -0,214 2,931 -1,373 -0,073
A -0,229 3,123 1,435 0,115
Case.4 Middle -0,011 3,026 0,000 -
B -0,214 2,931 -1,373 -0,073
Top slab B -0,223 1,509 2,931 0,195
Case.1 Middle 0,217 1,509 0,000 -
C -0,223 1,509 -2,931 -0,195
B -0,223 1,509 2,931 0,195
Case.2 Middle 0,217 1,509 0,000 -
C -0,223 1,509 -2,931 -0,195
B -0,214 1,373 2,931 0,195
Case.3 Middle 0,225 1,373 0,000 -
C -0,214 1,373 -2,931 -0,195
B -0,214 1,373 2,931 0,195
Case.4 Middle 0,225 1,373 0,000 -
C -0,214 1,373 -2,931 -0,195
Side wall C -0,223 2,931 1,509 0,109
(right) Case.1 Middle 0,008 3,027 0,000 -
D -0,239 3,123 -1,683 -0,104
C -0,223 2,931 1,509 0,109
Case.2 Middle 0,008 3,027 0,000 -
D -0,239 3,123 -1,683 -0,104
C -0,214 2,931 1,373 0,073
Case.3 Middle -0,011 3,026 0,000 -
D -0,229 3,123 -1,435 -0,115
C -0,214 2,931 1,373 0,073
Case.4 Middle -0,011 3,026 0,000 -
D -0,229 3,123 -1,435 -0,115
Invert D -0,239 1,683 3,123 0,208
Case.1 Middle 0,229 1,683 0,000 -
A -0,239 1,683 -3,123 -0,208D -0,239 1,683 3,123 0,208
Case.2 Middle 0,229 1,683 0,000 -
A -0,239 1,683 -3,123 -0,208
D -0,229 1,435 3,123 0,208
Case.3 Middle 0,240 1,435 0,000 -
A -0,229 1,435 -3,123 -0,208
D -0,229 1,435 3,123 0,208
Case.4 Middle 0,240 1,435 0,000 -
A -0,229 1,435 -3,123 -0,208
S (tf)Member Case
Check
Point
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5 Calculation of Required Reinforcement Bar5-1 Calculation of Required Reinforcement Bar
1) At Joint "A"of side wall
Case.1
M= 0,2393 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,1233 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,6831 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1038 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,66 cm
Solving the formula shown below, c 20,727 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +125,68 c^2 -975,55 c -42680,20
s = nc/(nc+sa) = 0,1915
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
Case.2
M= 0,2393 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,1233 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,6831 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1038 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,66 cm
Solving the formula shown below, c 20,727 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +125,68 c^2 -975,55 c -42680,20
s = nc/(nc+sa) = 0,1915
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
Case.3
M= 0,2289 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,1233 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,4355 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1145 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,33 cm
Solving the formula shown below, c 20,389 kgf/cm2 ( 0 kgf/cm2) o.k.
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
d
h d
0 = c^3 +125,83 c^2 -947,7 c -41462,03
s = nc/(nc+sa) = 0,1890
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
Case.4
M= 0,2289 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,1233 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,4355 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1145 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,33 cm
Solving the formula shown below, c 20,389 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +125,83 c^2 -947,7 c -41462,03
s = nc/(nc+sa) = 0,1890
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
The maximum requirement of reinforcement bar is 0,0000 cm2 in Case1 from above calculation.
Case. 1 2 3 4
Requirement 0,0000 0,0000 0,0000 0,0000 (cm2)
2) At Joint "B"of side wall
Case.1
M= 0,2231 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 2,9313 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)S0= 1,5091 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1090 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,61 cm
Solving the formula shown below, c 19,943 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,04 c^2 -911,58 c -39881,76
s = nc/(nc+sa) = 0,1856
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
Case.2
M= 0,2231 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 2,9313 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,5091 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1090 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,61 cm
Solving the formula shown below, c 19,943 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,04 c^2 -911,58 c -39881,76
s = nc/(nc+sa) = 0,1856
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
d
h d
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Case.3
M= 0,2143 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 2,9313 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,3728 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,0727 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,31 cm
Solving the formula shown below, c 19,648 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,18 c^2 -888 c -38849,92s = nc/(nc+sa) = 0,1834
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
Case.4
M= 0,2143 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 2,9313 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 1,3728 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,0727 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 7,31 cm
Solving the formula shown below, c 19,648 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,18 c^2 -888 c -38849,92
s = nc/(nc+sa) = 0,1834
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2
The maximum requirement of reinforcement bar is 0,0000 cm2 in Case1 from above calculation.
Case. 1 2 3 4Requirement 0,0000 0,0000 0,0000 0,0000 (cm2)
3) At Joint "B"of top slab
Case.1
M= 0,2231 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,5091 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 2,9313 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1954 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 14 78 cm ,
Solving the formula shown below, c 17,985 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,91 c^2 -759,21 c -33215,43
s = nc/(nc+sa) = 0,1705
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3035 cm2
Case.2
M= 0,2231tf
mca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,5091 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 2,9313 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1954 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 14,78 cm
Solving the formula shown below, c 17,985 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,91 c^2 -759,21 c -33215,43
s = nc/(nc+sa) = 0,1705
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3035 cm2
Case.3
M= 0,2143 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,3728 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 2,9313 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1954 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 15,61 cm
Solving the formula shown below,
c 17,472 kgf/cm2 ( 0 kgf/cm2) o.k. 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 +127,13 c^2 -721,02 c -31544,46
s = nc/(nc+sa) = 0,1664
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3156 cm2
Case.4
M= 0,2143 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,3728 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 2,9313 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,1954 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 15,61 cm
Solving the formula shown below, c 17,472 kgf/cm2 ( 5,1E-07 kgf/cm2) o.k.
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 +127,13 c^2 -721,02 c -31544,46
s = nc/(nc+sa) = 0,1664
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3156 cm2
The maximum requirement of reinforcement bar is 0,3156 cm2 in Case3 from above calculation.
Case. 1 2 3 4
Requirement 0,3035 0,3035 0,3156 0,3156 (cm2)
d
h d
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4) At Joint "A"of invert
Case.1
M= 0,2393 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,6831 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 3,1233 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,2082 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 14,22 cm
Solving the formula shown below, c 18,799 kgf/cm2 ( 4,7E-08 kgf/cm2) o.k.
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 +126,56 c^2 -821,25 c -35929,50
s = nc/(nc+sa) = 0,1768
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3067 cm2
Case.2
M= 0,2393 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,6831 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 3,1233 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,2082 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 14,22 cm
Solving the formula shown below, c 18,799 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,56 c^2 -821,25 c -35929,50
s = nc/(nc+sa) = 0,1768
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3067 cm2
Case.3
M= 0,2289 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,4355 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 3,1233 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,2082 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 15,94 cm
Solving the formula shown below, c 18,087 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,87 c^2 -766,87 c -33550,47
s = nc/(nc+sa) = 0,1713
* -
d
h d
- ,
Case.4
M= 0,2289 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,4355 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
S0= 3,1233 tf n = 24 d' = 6 cm (protective covering depth)
S2d= 0,2082 tf c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 15,94 cmSolving the formula shown below, c 18,087 kgf/cm2 ( 1,7E-07 kgf/cm2) o.k.
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 +126,87 c^2 -766,87 c -33550,47
s = nc/(nc+sa) = 0,1713
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3492 cm2
The maximum requirement of reinforcement bar is 0,3492 cm2 in Case3 from above calculation.
Case. 1 2 3 4
Requirement 0,3067 0,3067 0,3492 0,3492 (cm2)
5) At Middle of side wall
Case.1
M= 0,0083 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,0275 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 0,27 cm
Solving the formula shown below, c 11,6723 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +129,27 c^2 -346,47 c -15158,20
s = nc/(nc+sa) = 0,1177
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2 Tensile is on inside of member
Case.2
M= 0,0083 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,0275 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 0,27 cm
Solving the formula shown below, c 11,672 kgf/cm2 ( 0,0 kgf/cm2) o.k.
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 +129,27 c^2 -346,47 c -15158,20
s = nc/(nc+sa) = 0,1177Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2 Tensile is on inside of member
d
h d
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Case.3
M= 0,0109 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,0259 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 0,36 cm
Solving the formula shown below, c 11,798 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +129,23 c^2 -353,41 c -15461,54s = nc/(nc+sa) = 0,1188
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2 Tensile is on outside of member
Case.4
M= 0,0109 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 3,0259 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 0,36 cm
Solving the formula shown below, c 11,798 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +129,23 c^2 -353,41 c -15461,54
s = nc/(nc+sa) = 0,1188
Asreq = (c*s/2 - N/(bd))bd/sa = 0,0000 cm2 Tensile is on outside of member
The maximum requirement of reinforcement bar is 0,0000 cm2 in Case1 from above calculation.
Case. 1 2 3 4
Requirement 0,0000 0,0000 - - (cm2)Side inside inside outside outside
6) At Middle of top slab
Case.1
M= 0,2166 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,5091 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 14,35 cm
Solving the formula shown below, c 17,754 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +127,01 c^2 -741,93 c -32459,57
s = nc/(nc+sa) = 0,1687
Asreq = (c*s/2 - N/(bd))bd/sa = 0,2796 cm2 Tensile is on inside of member
Case.2
M= 0,2166 tfm ca = 70 kgf/m2 h = 20 cm (height of member)N= 1,5091 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 14,35 cm
Solving the formula shown below, c 17,754 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +127,01 c^2 -741,93 c -32459,57
s = nc/(nc+sa) = 0,1687
Asreq = (c*s/2 - N/(bd))bd/sa = 0,2796 cm2 Tensile is on inside of member
Case.3
M= 0,2254 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,3728 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 16,42 cm
Solving the formula shown below, c 17,875 kgf/cm2 ( 0 kgf/cm2) o.k. 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 +126,96 c^2 -750,91 c -32852,27
s = nc/(nc+sa) = 0,1696
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3570 cm2 Tensile is on inside of member
Case.4
M= 0,2254 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,3728 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 16,42 cm
Solving the formula shown below, c 17,875 kgf/cm2 ( 2,5E-07 kgf/cm2) o.k.
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 +126,96 c^2 -750,91 c -32852,27
s = nc/(nc+sa) = 0,1696
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3570 cm2 Tensile is on inside of member
The maximum requirement of reinforcement bar is 0,3570 cm2 in Case3 from above calculation.
Case. 1 2 3 4
Requirement 0,2796 0,2796 0,3570 0,3570 (cm2)
Side inside inside inside inside
d
h d
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7) At Middle of invert
Case.1
M= 0,2292 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,6831 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 13,62 cm
Solving the formula shown below, c 18,4488 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,71 c^2 -794,33 c -34751,76
s = nc/(nc+sa) = 0,1741
Asreq = (c*s/2 - N/(bd))bd/sa = 0,2693 cm2 Tensile is on inside of member
Case.2
M= 0,2292 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,6831 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 13,62 cm
Solving the formula shown below, c 18,449 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,71 c^2 -794,33 c -34751,76
s = nc/(nc+sa) = 0,1741
Asreq = (c*s/2 - N/(bd))bd/sa = 0,2693 cm2 Tensile is on inside of member
Case.3
M= 0,2396 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,4355 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 16,69 cm
Solving the formula shown below, c 18,466 kgf/cm2 ( 8,7E-08 kgf/cm2) o.k.
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 +126,7 c^2 -795,64 c -34809,07
s = nc/(nc+sa) = 0,1743
* -
d
h d
- ,
Case.4
M= 0,2396 tfm ca = 70 kgf/m2 h = 20 cm (height of member)
N= 1,4355 tf sa = 2100 kgf/m2 d = 14 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)
c = 4,00 cm (distance from neutral axis)
b = 100 cm
e = M/N = 16,69 cmSolving the formula shown below, c 18,466 kgf/cm2 ( 0 kgf/cm2) o.k.
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 +126,7 c^2 -795,64 c -34809,07
s = nc/(nc+sa) = 0,1743
Asreq = (c*s/2 - N/(bd))bd/sa = 0,3891 cm2 Tensile is on inside of member
The maximum requirement of reinforcement bar is 0,3891 cm2 in Case3 from above calculation.
Case. 1 2 3 4
Requirement 0,2693 0,2693 0,3891 0,3891 (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 0,000 0,000 0,316 0,349 0,000 0,357 0,389 (cm2)
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6 Bar Arrangement and Calculation of StressType: B0,40m x H0,40m
bottom middle top end middle end middle
outside inside outside outside inside outside inside
Bending moment M kgfcm 23.927 825 22.307 21.427 22.543 22.888 23.962
Shearing force (joint) S kgf 1.683 0 1.509 2.931 0 3.123 0Shearing force (2d) S2d kgf 115 - 109 195 - 208 -
Axial force N kgf 3.123 3.027 2.931 1.373 1.373 1.435 1.435
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 0,00 0,00 0,00 0,32 0,36 0,35 0,39
R-bar arrangement 12@150 12@250 12@150 12@150 12@150 12@150 12@150
Reinforcement As cm2 7,54 4,52 7,54 7,54 7,54 7,54 7,54
Perimeter of R-bar U 25,13 15,08 25,13 25,13 25,13 25,13 25,13
M/N e cm 7,661 0,273 7,610 15,608 16,421 15,944 16,692
Dist. from neutral axis c cm 4,00 4,00 4,00 4,00 4,00 4,00 4,00
a' -7,0 -29,2 -7,2 16,8 19,3 17,8 20,1
b' 126,6 27,8 126,1 212,9 221,7 216,5 224,7
' - - - - - - -
Side wall Top slab Invert
, , , , , , ,
x 10,69 28,69 10,74 7,52 7,40 7,47 7,37
0,000 0,001 0,000 0,000 0,000 0,000 0,000
(check) ok ok ok ok ok ok ok
Compressive stress c kgf/cm2 6,5 2,0 6,1 6,2 6,6 6,7 7,0
Allowable stress ca kgf/cm2 70,0 70,0 70,0 70,0 70,0 70,0 70,0ok ok ok ok ok ok ok
Tensile stress s kgf/cm2 48,6 0,0 44,4 128,6 140,3 139,5 150,9
Allowable stress sa kgf/cm2 2100,0 2100,0 2100,0 2100,0 2100,0 2100,0 2100,0
ok ok ok ok ok ok ok
Shearing stress at joint kgf/cm2 1,20 0,00 1,08 2,09 0,00 2,23 0,00
Allowable stress a kgf/cm2 11,00 11,00 11,00 11,00 11,00 11,00 11,00
ok ok ok ok ok ok ok
Shearing stress at 2d 2d kgf/cm2 0,08 - 0,08 0,14 - 0,15 -
Allowable stress 2da kgf/cm2 5,50 - 5,50 5,50 - 5,50 -
ok - ok ok - ok -
Resisting Moment Mr kgfcm 235.438 199.883 235.421 235.378 235.378 235.377 235.377
Mr for compression Mrc kgfcm 235.438 199.883 235.421 235.378 235.378 235.377 235.377 x for Mrc cm 5,884 4,897 5,862 5,685 5,685 5,692 5,692 s for Mrc kgf/cm2 2317,2 3122,5 2332,3 2457,0 2457,0 2451,9 2451,9
Mr for tensile Mrs kgfcm 379.690 218.260 375.059 338.900 338.900 340.307 340.307
x for Mrs cm 7,789 6,329 7,749 7,420 7,420 7,434 7,434 c for Mrs kgf/cm2 109,7 72,2 108,5 98,7 98,7 99,1 99,1
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 4,52
ok 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
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DATA PLOT PEMBESIAN
Section of Culvert
b1b1 b2
h3
h2
h1
Bb
Ha
.
b1 = m b2 = m
h1 = m h2 = m h3 = m d = m
Bb = m Ha = m Fillet = m
2. Data Pembesian
Bottom slab : Tulangan bagi :As1 (cm ) : O 12 @ O 12 @ O 12 @
As2 (cm ) : O 12 @ O 12 @
Side wall : Tulangan bagi :
As1 (cm ) : O 12 @ O 12 @ O 12 @
As2 (cm ) : O 12 @ O 12 @
Top slab : Tulangan bagi :
As1 (cm ) : O 12 @ O 12 @ O 12 @
As2 (cm ) : O 12 @ O 12 @
3. Tulangan Miring (fillet) :
Bottom slab : O 12 @Top slab : O 12 @
4. Nama Bangunan : Culvert Type 2 (0,4 m x 0,4 m)
Lokasi : Kantor Bupati
0,200 0,400
0,200 0,400 0,200 0,06
150 150 250
150 150
0,800 0,800 0,15
Tumpuan Lapangan
250
250 150
Tumpuan Lapangan
Tumpuan Lapangan
150 250
250250
150 150 250
150 150
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SUMMARY OF STRUCTURAL CALCULATION OF 1-BARREL BOX CULVERT
1 Design Dimensions and Bar Arrangements Class I Road (BM100)
Type of box culvert B0,4 x H0,4
Clear width m 0,40
Clear height m 0,40
Height of fillet m 0,15
Thickness Side wall cm 20,0
Top slab cm 20,0
Bottom slab cm 20,0
Cover of reinforcement bar (between concrete surface and center of reinforcement bar)
Side wall Outside cm 6,0
Inside cm 6,0
Top 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 12@150
Distribution bar mm 12@250
Middle inside Tensile bar mm 12@250
Distribution bar mm 12@250
Upper outside Tensile bar mm 12@150
Distribution bar mm 12@250
Top slab Upper edge Tensile bar mm 12@150
Distribution bar mm 12@250
Lower middle Tensile bar mm 12@150
Distribution bar mm 12@250
Bottom slab Lower edge Tensile bar mm 12@150Distribution bar mm 12@250
Upper middle Tensile bar mm 12@150
Distribution bar mm 12@250
Fillet Upper edge Fillet bar mm 12@250
Lower edge Fillet bar mm 12@250
2 Design Parameters
Unit Weight Reinforced Concrete c= 2,4 tf/m3
Backfill soil (wet) s= 1,8 tf/m3
(submerged) s'= 1,0 tf/m3
Live Load Class of road Class I (BM100)
Truck load at rear wheel P= 10,0 tf Impact coefficient (for Class I to IV road Ci= 0,3 (D4.0m)
Pedestrian load (for Class V roads) 0 tf/m2
Concrete Design Strength ck= 175 kgf/cm2
(K175) Allowable Compressive Stress ca= 70 kgf/cm2
Allowable Shearing Stress a= 5,5 kgf/cm2
Reinforcement Bar Allowable Tensile Stress sa= 2.100 kgf/cm2
(U24, deformed bar) Yielding Point of Reinforcement Bar sy= 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: B0,40m x H0,40m Class I Road
Soil Cover Depth: 0,4 m
1 Dimensions and ParametersBasic Parameters
Ka: Coefficient of static earth pressure 0,5
w: Unit weight of water (t/m3) 1,00 t/m3
d: Unit weight of soil (dry) (t/m3) 1,80 t/m3
s: Unit weight of soil (saturated) (t/m3) 2,00 t /m3
c: Unit weight of reinforced concrete (t/m3) 2,40 t/m3ck: Concrete Design Strength 175 kgf/m2
ca Allowable Stress of Concrete 70 kgf/m2
sa: Allowable Stress of Reinforcement Bar 2100 kgf/m2
a: Allowable Stress of Shearing (Concrete) 5,5 kgf/m2
sy: Yielding Point of Reinforcement Bar 3000 kgf/m2
n: Young's Modulus Ratio 24
Fa: Safety factor against uplift 1,2
Basic Dimensions
H: Internal Height of Box Culvert 0,40 m
B: Internal Width of Box Culvert 0,40 m
Hf: Fillet Height 0,15 m
t1: Thickness of Side Wall 0,20 m (> 0.25m)
t2: Thickness of Top Slab 0,20 m (> 0.25m)
t3: Thickness of Invert (Bottom Slab) 0,20 m (> 0.25m)
BT: Gross Width of Box Culvert 0,80 m
HT: Gross Height of Box Culvert 0,80 m
D: Covering Depth 0,41 mGwd: Underground Water Depth for Case 1, 2 0,41 m (= D)
hiw: Internal Water Depth for Case 1, 2 0,00 m
for Case 3, 4 0,40 m
Cover of R-bar Basic Conditions
Top Slab d2 0,06 m Classification of Live load by truck Class 1
Side Wall d1 0,06 m PTM: Truck load of Middle Tire 10,00 t
Bottom Slab d3 0,06 m Ii: Impact coefficient (D4.0m:0, D
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Load distribution of truck tire
(1) Middle tire's acting point: center of the top slab
a) distributed load of middle tire
Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 8,3030 tf/m2, B = 0,600 m
am': length of distributed load 2D+1.75+bm = 3,070 m
bm': width of distributed load 2D+am = 1,020 m
b) distributed load of rear tire
Pvtr: distributed load of rear tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
ar': length of distributed load 2D+1.75+br = 3,070 mbr': width of distributed load 2D+ar = 1,020 m
c) distributed load of front tire
Pvtf: distributed load of front tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
af': length of distributed load 2D+1.75+bf = 3,070 m
bf': width of distributed load 2D+af = 1,020 m
(2) Middle tire's acting point: on the side wall
a) distributed load of middle tire
Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 8,3030 tf/m2, B = 0,600 m
am': length of distributed load 2D+1.75+bm = 3,070 m
bm': width of distributed load 2D+am = 1,020 m
b) distributed load of rear tire
Pvtr: distributed load of rear tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
ar': length of distributed load 2D+1.75+br = 3,070 m
br': width of distributed load 2D+ar = 1,020 m
c) distributed load of front tire
Pvtf: distributed load of front tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
af': length of distributed load 2D+1.75+bf = 3,070 m
bf': width of distributed load 2D+af = 1,020 m
(3) Rear tire's acting point: on the side wall
a) distributed load of rear tire
Pvtr: distributed load of rear tire 2PTR(1+Ii)/(ar'br') = 8,3030 tf/m2, B = 0,600 m
ar': length of distributed load 2D+1.75+br = 3,070 m
br': width of distributed load 2D+ar = 1,020 m
b) distributed load of middle tire
Pvtm: distributed load of middle tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
am': length of distributed load 2D+1.75+bm = 3,070 m
'' ,
c) distributed load of front tire
Pvtf: distributed load of front tire not reach to top slab 0,0000 tf/m2, B = 0,000 m
af': length of distributed load 2D+1.75+bf = 3,070 m
bf': width of distributed load 2D+af = 1,020 m
(4) Combination of load distribution of track tire
Case.L1: Pvt1 = 8,3030 tf/m2, B = 0,600 m Combination for Case.L2 (2) (2) (3) (3)Pvt2 = 0,0000 tf/m2, B = 0,000 m a) + b) a) + c) a) + b) a) + c)
Case.L2: Pvt1 = 8,3030 tf/m2, B = 0,600 m Distributed load total 8,3030 8,3030 8,3030 8,3030
Pvt2 = 0,0000 tf/m2, B = 0,000 m Select the combination case of 8,3030 tf/m2,
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= 2,8913 > 1,2 ok
where, Vd: Total dead weight (t/m) Vd= 1,850 tf/m
U: Total uplift (t.m)
U=BT*HT*w U= 0,640 tf/m
Ws: Weight of covering soil Ws = BT*{(D-Gwd)*(sw)+Gwd*d} = 0,590 tf/m
Wc: Self weight of box culvert Wc = (HT*BT-H*B+2*Hf^2)*c = 1,260 tf/m
Fa: 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)*c/B0 Wtop= 0,7300
Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*gd*Gwd P3= 0,3690
P4=Ka*gs*(D1-Gwd) P4= 0,1000
P5=gw*(D1-Gwd) P5= 0,1000
Ph1= 4,7205
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*Gwd P3= 0,3690
P4=Ka*s*(D1+H0-Gwd) P4= 0,7000
P5=w*(D1+H0-Gwd) P5= 0,7000Ph2= 5,9205
4) self weight of side wall
Acting Load (tf/m)
Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
Pvd Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,0000 hiw: internal water depth 0,00 m
Up=-U/B0 U= -1,0667
Q= 10,0743
summary of resistance moment
Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,300 m
side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 10,0743 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 10,0743 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 3,1923 - 0,2887 0,9217
side wall (right) - -3,1923 - 0,2887 -0,9217
internal water 0,0000 - 0,3000 - 0,0000
uplift -0,6400 - 0,3000 - -0,1920
total 6,0446 1,8134
6) load against invert
Acting Load (tf/m2)Pvd 0,7380
Pvt1 8,3030
Pvt2 0,0000
Wtop 0,7300
Ws 0,6400
Pq= 10,4110
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Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L2
1) vertical load against top slab
Acting Load (tf/m2)
Wtop= (t2*BT+Hf^2)*c/B0 Wtop= 0,7300
Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*gd*Gwd P3= 0,3690
P4=Ka*gs*(D1-Gwd) P4= 0,1000
P5=gw*(D1-Gwd) P5= 0,1000
Ph1= 4,7205
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*Gwd P3= 0,3690
P4=Ka*s*(D1+H0-Gwd) P4= 0,7000
P5=w*(D1+H0-Gwd) P5= 0,7000
Ph2= 5,9205
4) self weight of side wall
Acting Load (tf/m)
Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
v v = ,
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,0000 hiw: internal water depth 0,00 m
Up=-U/B0 U= -1,0667
Q= 10,0743
summary of resistance moment
Item V H x y M(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,3000 m
side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,0000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 10,0743 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 10,0743 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 3,1923 - 0,2887 0,9217
side wall (right) - -3,1923 - 0,2887 -0,9217
internal water 0,0000 - 0,3000 - 0,0000
uplift -0,6400 - 0,3000 - -0,1920
total 6,0446 1,8134
6) load against invert
Acting Load (tf/m2)
Pvd 0,7380
Pvt1 8,3030Pvt2 0,0000
Wtop 0,7300
Ws 0,6400
total Pq= 10,4110
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Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L1
1) vertical load against top slab
Acting Load (tf/m2)
Wtop= (t2*BT+Hf^2)*c/B0 Wtop= 0,7300
Pvd=D*d Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*d*D1 P3= 0,4590
WP=-w*0 P4= 0,0000
Ph1= 4,6105
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*(D1+H0) P3= 0,9990
WP=-w*H P4= -0,4000
Ph2= 4,7505
4) self weight of side wall
Acting Load (tf/m)Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
Pvd Pvd= 0,7380
Pvt1 Pvt1= 8,3030
v v = ,
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,1917 hiw: internal water depth 0,400 m
Up=0 U= 0,0000
Q= 11,3327
summary of resistance moment
Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,3000 m side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,0000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 11,3327 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 11,3327 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 2,8083 - 0,2985 0,8383
side wall (right) - -2,8083 - 0,2985 -0,8383
internal water 0,1150 - 0,3000 - 0,0345
uplift 0,0000 - 0,3000 - 0,0000
total 6,7996 2,0399
6) load against invert
Acting Load (tf/m2)
Pvd 0,7380
Pvt1 8,3030
Pvt2 0,0000
Wtop 0,7300Ws 0,6400
total Pq= 10,4110
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Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2
1) vertical load against top slab
Acting Load (tf/m2)
Wtop= (t2*BT+Hf^2)*c/B0 Wtop= 0,7300
Pvd=D*d Pvd= 0,7380
Pvt1 Pvt1= 8,3030
Pvt2 Pvt2= 0,0000
Pv1= 9,7710
2) horizontal load at top of side wall
Acting Load (tf/m2) Horizontal pressure by track tire
P1=Ka*we1 P1= 4,1515 we1= 8,3030 tf/m2
P2=Ka*we2 P2= 0,0000 we2= 0,0000 tf/m2
P3=Ka*d*D1 P3= 0,4590
WP=-w*0 P4= 0,0000
Ph1= 4,6105
3) horizontal load at bottom of side wall
Acting Load (tf/m2)
P1=Ka*we1 P1= 4,1515
P2=Ka*we2 P2= 0,0000
P3=Ka*d*(D1+H0) P3= 0,9990
WP=-w*H P4= -0,4000
Ph2= 4,7505
4) self weight of side wall
Acting Load (tf/m)Wsw=t1*H*c Wsw= 0,1920
5) ground reaction
Acting Load (tf/m2)
Wbot=(t3*BT+Hf^2)*c/B0 Wbot= 0,7300
Wtop Wtop= 0,7300
Ws=Wsw*2/B0 Ws= 0,6400
Pvd Pvd= 0,7380
Pvt1 Pvt1= 8,3030
v v = ,
Wiw=(hiw*B-2Hf^2)*w/B0 Wiw= 0,1917 hiw: internal water depth 0,400 m
Up=0 U= 0,0000
Q= 11,3327
summary of resistance moment
Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant force
Self weight top slab 0,4380 - 0,3000 - 0,1314 X = M/V = 0,3000 m side wall (left) 0,1920 - 0,0000 - 0,0000 e = B0/2 - X = 0,0000 m
side wall (right) 0,1920 - 0,6000 - 0,1152
invert 0,4380 - 0,3000 - 0,1314 ground reaction
load on top slab Pvd 0,4428 - 0,3000 - 0,1328 q1 = V/Bo + 6Ve/Bo^2 = 11,3327 tf/m2
Pvt1 4,9818 - 0,3000 - 1,4945 q2 = V/Bo - 6Ve/Bo^2 = 11,3327 tf/m2
Pvt2 0,0000 - 0,3000 - 0,0000
soil pressure side wall (left) - 2,8083 - 0,2985 0,8383
side wall (right) - -2,8083 - 0,2985 -0,8383
internal water 0,1150 - 0,3000 - 0,0345
uplift 0,0000 - 0,3000 - 0,0000
total 6,7996 2,0399
6) load against invert
Acting Load (tf/m2)
Pvd 0,7380
Pvt1 8,3030
Pvt2 0,0000
Wtop 0,7300Ws 0,6400
total Pq= 10,4110
Summary of Load Calculation
Item Pv1 Ph1 Ph2 Pq Wsw q1
Case (tf/m2) (tf/m2) (tf/m2) (tf/m2) (tf/m) (tf/m2)
Case.1 9,7710 4,7205 5,9205 10,4110 0,1920 10,0743
Case.2 9,7710 4,7205 5,9205 10,4110 0,1920 10,0743
Case.3 9,7710 4,6105 4,7505 10,4110 0,1920 11,3327
Case.4 9,7710 4,6105 4,7505 10,4110 0,1920 11,3327
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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 Term
Ph1 Horizontal Pressure at top of side wall 4,720 tf/m2
Ph2 Horizontal Pressure at bottom of side wall 5,920 tf/m2
Pv1 Vertical Pressure(1) on top slab 9,771 tf/m
2
Pv2 Vertical Pressure(2) on top slab 0,000 tf/m2
Pq Reaction to bottom slab 10,411 tf/m2
a Distance from joint B to far end of Pv2 0,600 m
b Distance from joint B to near end of Pv2 0,000 m
H0 Height of plane frame 0,600 m
B0 Width of plane frame 0,600 m
t1 Thickness of side wall 0,200 m
t2 Thickness of top slab 0,200 m
t3 Thickness of invert (bottom slab) 0,200 m
CAB= CDC= (2Ph1+3Ph2)H02/60 = 0,16321 tf m
CBA= CCD= (3Ph1+2Ph2)H02/60 = 0,15601 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 = 0,29313 tf m
CDA= CAD= PqB02/12 = 0,31233 tf m
2) Calculation of Bending Moment at joint
k1 = 1,0
k2 = H0t23/(B0t1
3) = 1,0000
k3 = H0t33/(B0t1
3) = 1,0000
2(k1+k3) k1 0 k3 -3k1 A CAB- CAD
k1 2(k1+k2) k2 0 -3k1 B CBC- CBA
0 2 2(k1+k2 k1 -3k1 = C - C
B
A
(t2)
(t1)
B0
(t3)
(t1)
C
H0
D
k3 0 k1 2(k1+k3) -3k1 D CDA- CDC
k1 k1 k1 k1 -4k1 R 0
As load has bilateral symmetry, the equation shown below is formed.
A= -D B= -C R =0
2k1+k3 k1 A
k1 2k1+k2 B
3,0000 1,0 A1,0 3,0000 B
By solving above equation, the result is led as shown below.
A = -0,07306 C = -0,07006
B = 0,07006 D = 0,07306
