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1 / 114 276341623.xls/Summary
SUMMARY OF STRUCTURAL CALCULATION OF 1-BARREL BOX CULVERT
1 Design Dimensions and Bar Arrangements Class III Road (BM50)
B2.0 x H2.0
Clear width m 2.00Clear height m 2.00Height of fillet m 0.15
Thickness Side wall cm 30.0Top slab cm 30.0Bottom slab cm 30.0
Cover of reinforcement bar (between concrete surface and center of reinforcement bar)Side wall Outside cm 6.0
Inside cm 6.0Top slab Upper cm 6.0
Lower cm 6.0Bottom slab Lower cm 6.0
Upper cm 6.0
Bar arrangement (dia - spacing per unit length of 1.0 m)
Side wall Lower outside Tensile bar mm 16@250Distribution bar mm 12@250
Middle inside Tensile bar mm 12@250Distribution bar mm 12@250
Upper outside Tensile bar mm 16@250Distribution bar mm 12@250
Top slab Upper edge Tensile bar mm 16@250Distribution bar mm 12@250
Lower middle Tensile bar mm 16@250Distribution bar mm 12@250
Bottom slab Lower edge Tensile bar mm 16@250Distribution bar mm 12@250
Upper middle Tensile bar mm 12@125Distribution bar mm 12@250
Fillet Upper edge Fillet bar mm Err:504Lower edge Fillet bar mm Err:504
2 Design Parameters
Unit Weight Reinforced Concrete 2.4
Backfill soil (wet) 1.8
(submerged) 1.0
Live Load Class of road Class III (BM50)Truck load at rear wheel P= 5.0 tfImpact coefficient (for Class I to IV road) Ci= 0.3
0.0Pedestrian load (for Class V roads) 0
Concrete Design Strength 175(K175) Allowable Compressive Stress 60
Allowable Shearing Stress 5.5
Reinforcement Bar Allowable Tensile Stress 1,400 (U24, deformed bar) Yielding Point of Reinforcement Bar 3,000
Young's Modulus Ratio n= 24
Coefficient of static earth pressure Ka= 0.5
Type of box culvert
c= tf/m3s= tf/m3s'= tf/m3
(D4.0m)tf/m2
ck= kgf/cm2
ca= kgf/cm2
a= kgf/cm2
sa= kgf/cm2
sy= kgf/cm2
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STRUCTURAL CALCULATION OF BOX CULVERT Type: B2.00m x H2.00mSoil Cover Depth:
1 Dimensions and ParametersBasic ParametersKa: Coefficient of static earth pressure
Unit weight of water (t/m3)Unit weight of soil (dry) (t/m3)Unit weight of soil (saturated) (t/m3)Unit weight of reinforced concrete (t/m3)Concrete Design StrengthAllowable Stress of ConcreteAllowable Stress of Reinforcement BarAllowable Stress of Shearing (Concrete)Yielding Point of Reinforcement Bar
n: Young's Modulus RatioFa: Safety factor against uplift
Basic DimensionsH: Internal Height of Box CulvertB: Internal Width of Box CulvertHf: Fillet Heightt1: Thickness of Side Wallt2: Thickness of Top Slabt3: Thickness of Invert (Bottom Slab)BT: Gross Width of Box CulvertHT: Gross Height of Box CulvertD: Covering DepthGwd: Underground Water Depth for Case 1, 2hiw: Internal Water Depth for Case 1, 2
for Case 3, 4
Cover of R-bar Basic ConditionsTop Slab d2 0.06 m Classification of Live load by truck ClassSide Wall d1 0.06 m PTM: Truck load of Middle TireBottom Slab d3 0.06 m Ii:
am: Ground contact length of Middle Tirebm: Ground contact width of Middle TirePTR: Truck load of Rear Tirear: Ground contact length of Rear Tirebr: Ground contact width of Rear TirePTF: Truck load of Front wheelaf: Ground contact length of Front Tirebf: Ground contact width of Front Tireqp: Pedestrian load
Dimension of frame
H0: Height of frame t2/2 + H + t3/2B0: Width of frame B + t1
D1: Covering depth at middle of top slab D + t2/2
w:d:s:c:ck:casa:a:sy:
Impact coefficient (D4.0m:0, D
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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)/(am'bm') = 0.2664am': length of distributed load 2D+1.75+bm = 8.000bm': width of distributed load 2D+am = 6.100
b) distributed load of rear tirePvtr: distributed load of rear tire not reach to top slab 0.0000ar': length of distributed load 2D+1.75+br = 8.000br': width of distributed load 2D+ar = 6.100
c) distributed load of front tirePvtf: distributed load of front tire 2PTF(1+Ii)/(af'bf') 0.0666af': length of distributed load 2D+1.75+bf = 8.000bf': width of distributed load 2D+af = 6.100
(2) Middle tire's acting point: on the side walla) distributed load of middle tire
Pvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') = 0.2664am': length of distributed load 2D+1.75+bm = 8.000bm': width of distributed load 2D+am = 6.100
b) distributed load of rear tirePvtr: distributed load of rear tire 2PTR(1+Ii)/(ar'br') 0.2664ar': length of distributed load 2D+1.75+br = 8.000br': width of distributed load 2D+ar = 6.100
c) distributed load of front tirePvtf: distributed load of front tire 2PTF(1+Ii)/(af'bf') 0.0666af': length of distributed load 2D+1.75+bf = 8.000bf': width of distributed load 2D+af = 6.100
(3) Rear tire's acting point: on the side walla) distributed load of rear tire
Pvtr: distributed load of rear tire 2PTR(1+Ii)/(ar'br') = 0.2664ar': length of distributed load 2D+1.75+br = 8.000br': width of distributed load 2D+ar = 6.100
b) distributed load of middle tirePvtm: distributed load of middle tire 2PTM(1+Ii)/(am'bm') 0.2664am': length of distributed load 2D+1.75+bm = 8.000bm': width of distributed load 2D+am = 6.100
c) distributed load of front tirePvtf: distributed load of front tire not reach to top slab 0.0000af': length of distributed load 2D+1.75+bf = 8.000bf': width of distributed load 2D+af = 6.100
(4) Combination of load distribution of track tire
Case.L1: Pvt1 = 0.2664 tf/m2, B = 2.300 m Combination for Case.L2Pvt2 = 0.0666 tf/m2, B = 2.300 m
Case.L2: Pvt1 = 0.5328 tf/m2, B = 2.300 m Distributed load totalPvt2 = 0.0000 tf/m2, B = 0.000 m Select the combination case of
for Case.L2, which is the largest load to the top slab.
In case of covering depth (D) is over 3.0m, uniform load of 1.0 tf/m2 is applied on the top slab of culvert instead of live load calculated above.
Distribution load by pedestrian load
Pvt1 = 0.000 tf/m2
2 Stability Analysis Against Uplift
Analysis is made considering empty inside of box culvert.Fs=Vd/U > Fa Fs= 3.0728 > 1.2 ok
where, Vd: Total dead weight (t/m) Vd= 20.772 tf/mU: Total uplift (t.m)
U= 6.760 tf/m
Ws: Weight of covering soil Ws = = 14.040Wc: Self weight of box culvert Wc = = 6.732Fa: Safety factor against uplift Fa= 1.2
U=BT*HT*w
BT*{(D-Gwd)*(sw)+Gwd*d}(HT*BT-H*B+2*Hf^2)*c
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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 slabActing Load (tf/m2)
Wtop= 0.8374Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 5.4000Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666
Pv1= 6.5704
2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.0333 we2= 0.0666 tf/m2P3=Ka*gd*Gwd P3= 2.7000P4=Ka*gs*(D1-Gwd) P4= 0.1500P5=gw*(D1-Gwd) P5= 0.1500
Ph1= 3.1665
3) horizontal load at bottom of side wallActing Load (tf/m2)P1=Ka*we1 P1= 0.1332P2=Ka*we2 P2= 0.0333
P3= 2.7000P4= 2.4500P5= 2.4500
Ph2= 7.7665
4) self weight of side wallActing Load (tf/m)
Wsw= 1.4400
5) ground reactionActing Load (tf/m2)
Wbot= 0.8374Wtop Wtop= 0.8374Ws=Wsw*2/B0 Ws= 1.2522Pvd Pvd= 5.4000Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666
Wiw= 0.0000 hiw: internal water depthUp=-U/B0 U= -2.9391
Q= 5.7208
summary of resistance moment Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant forceSelf weight top slab 1.9260 - 1.1500 - 2.2149 side wall (left) 1.4400 - 0.0000 - 0.0000 e = B0/2 - X =
side wall (right) 1.4400 - 2.3000 - 3.3120invert 1.9260 - 1.1500 - 2.2149 ground reaction
load on top slab Pvd 12.4200 - 1.1500 - 14.2830Pvt1 0.6127 - 1.1500 - 0.7046Pvt2 0.1532 - 1.1500 - 0.1762
soil pressure side wall (left) - 12.5729 - 0.9887 12.4310side wall (right) - -12.5729 - 0.9887 -12.4310
internal water 0.0000 - 1.1500 - 0.0000uplift -6.7600 - 1.1500 - -7.7740total 13.1579 15.1316
6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.2664Pvt2 0.0666Wtop 0.8374Ws 1.2522
Pq= 7.8226
Wtop= (t2*BT+Hf^2)*c/B0
P3=Ka*d*GwdP4=Ka*s*(D1+H0-Gwd)P5=w*(D1+H0-Gwd)
Wsw=t1*H*c
Wbot=(t3*BT+Hf^2)*c/B0
Wiw=(hiw*B-2Hf^2)*w/B0
X = M/V =
q1 = V/Bo + 6q2 = V/Bo - 6
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7/ 114 (1)276341623.xls, Load
Case 2: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L2
1) vertical load against top slabActing Load (tf/m2)
Wtop= 0.8374Pvd=Gwd*gd+(D-Gwd)*gs Pvd= 5.4000Pvt1 Pvt1= 0.5328Pvt2 Pvt2= 0.0000
Pv1= 6.7702
2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.1332 we2= 0.2664 tf/m2P3=Ka*gd*Gwd P3= 2.7000P4=Ka*gs*(D1-Gwd) P4= 0.1500P5=gw*(D1-Gwd) P5= 0.1500
Ph1= 3.2664
3) horizontal load at bottom of side wallActing Load (tf/m2)P1=Ka*we1 P1= 0.1332P2=Ka*we2 P2= 0.1332
P3= 2.7000P4= 2.4500P5= 2.4500
Ph2= 7.8664
4) self weight of side wallActing Load (tf/m)
Wsw= 1.4400
5) ground reactionActing Load (tf/m2)
Wbot= 0.8374Wtop Wtop= 0.8374Ws=Wsw*2/B0 Ws= 1.2522Pvd Pvd= 5.4000Pvt1 Pvt1= 0.5328Pvt2 Pvt2= 0.0000
Wiw= 0.0000 hiw: internal water depthUp=-U/B0 U= -2.9391
Q= 5.9206
summary of resistance moment Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant forceSelf weight top slab 1.9260 - 1.1500 - 2.2149 side wall (left) 1.4400 - 0.0000 - 0.0000 e = B0/2 - X =
side wall (right) 1.4400 - 2.3000 - 3.3120invert 1.9260 - 1.1500 - 2.2149 ground reaction
load on top slab Pvd 12.4200 - 1.1500 - 14.2830Pvt1 1.2254 - 1.1500 - 1.4092Pvt2 0.0000 - 1.1500 - 0.0000
soil pressure side wall (left) - 12.8027 - 0.9916 12.6953side wall (right) - -12.8027 - 0.9916 -12.6953
internal water 0.0000 - 1.1500 - 0.0000uplift -6.7600 - 1.1500 - -7.7740total 13.6174 15.6600
6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.5328Pvt2 0.0000Wtop 0.8374Ws 1.2522total Pq= 8.0224
Wtop= (t2*BT+Hf^2)*c/B0
P3=Ka*d*GwdP4=Ka*s*(D1+H0-Gwd)P5=w*(D1+H0-Gwd)
Wsw=t1*H*c
Wbot=(t3*BT+Hf^2)*c/B0
Wiw=(hiw*B-2Hf^2)*w/B0
X = M/V =
q1 = V/Bo + 6q2 = V/Bo - 6
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8/ 114 (1)276341623.xls, Load
Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L1
1) vertical load against top slabActing Load (tf/m2)
Wtop= 0.8374Pvd= 5.4000
Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666
Pv1= 6.5704
2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.0333 we2= 0.0666 tf/m2
P3= 2.8350P4= 0.0000
Ph1= 3.0015
3) horizontal load at bottom of side wallActing Load (tf/m2)P1=Ka*we1 P1= 0.1332P2=Ka*we2 P2= 0.0333
P3= 4.9050P4= -2.0000
Ph2= 3.0715
4) self weight of side wallActing Load (tf/m)
Wsw= 1.4400
5) ground reactionActing Load (tf/m2)
Wbot= 0.8374Wtop Wtop= 0.8374Ws=Wsw*2/B0 Ws= 1.2522Pvd Pvd= 5.4000Pvt1 Pvt1= 0.2664Pvt2 Pvt2= 0.0666
Wiw= 1.7196 hiw: internal water depthUp=0 U= 0.0000
Q= 10.3795
summary of resistance moment Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant forceSelf weight top slab 1.9260 - 1.1500 - 2.2149 side wall (left) 1.4400 - 0.0000 - 0.0000 e = B0/2 - X =
side wall (right) 1.4400 - 2.3000 - 3.3120invert 1.9260 - 1.1500 - 2.2149 ground reaction
load on top slab Pvd 12.4200 - 1.1500 - 14.2830Pvt1 0.6127 - 1.1500 - 0.7046Pvt2 0.1532 - 1.1500 - 0.1762
soil pressure side wall (left) - 6.9839 - 1.1456 8.0007side wall (right) - -6.9839 - 1.1456 -8.0007
internal water 3.9550 - 1.1500 - 4.5483uplift 0.0000 - 1.1500 - 0.0000total 23.8729 27.4538
6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.2664Pvt2 0.0666Wtop 0.8374Ws 1.2522total Pq= 7.8226
Wtop= (t2*BT+Hf^2)*c/B0Pvd=D*d
P3=Ka*d*D1WP=-w*0
P3=Ka*d*(D1+H0)WP=-w*H
Wsw=t1*H*c
Wbot=(t3*BT+Hf^2)*c/B0
Wiw=(hiw*B-2Hf^2)*w/B0
X = M/V =
q1 = V/Bo + 6q2 = V/Bo - 6
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9/ 114 (1)276341623.xls, Load
Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2
1) vertical load against top slabActing Load (tf/m2)
Wtop= 0.8374Pvd= 5.4000
Pvt1 Pvt1= 0.5328Pvt2 Pvt2= 0.0000
Pv1= 6.7702
2) horizontal load at top of side wallActing Load (tf/m2) Horizontal pressure by track tireP1=Ka*we1 P1= 0.1332 we1= 0.2664 tf/m2P2=Ka*we2 P2= 0.1332 we2= 0.2664 tf/m2
P3= 2.8350P4= 0.0000
Ph1= 3.1014
3) horizontal load at bottom of side wallActing Load (tf/m2)P1=Ka*we1 P1= 0.1332P2=Ka*we2 P2= 0.1332
P3= 4.9050P4= -2.0000
Ph2= 3.1714
4) self weight of side wallActing Load (tf/m)
Wsw= 1.4400
5) ground reactionActing Load (tf/m2)
Wbot= 0.8374Wtop Wtop= 0.8374Ws=Wsw*2/B0 Ws= 1.2522Pvd Pvd= 5.4000Pvt1 Pvt1= 0.5328Pvt2 Pvt2= 0.0000
Wiw= 1.7196 hiw: internal water depthUp=0 U= 0.0000
Q= 10.5793
summary of resistance moment Item V H x y M
(tf/m) (tf/m) (m) (m) (tf.m/m) acting point of resultant forceSelf weight top slab 1.9260 - 1.1500 - 2.2149 side wall (left) 1.4400 - 0.0000 - 0.0000 e = B0/2 - X =
side wall (right) 1.4400 - 2.3000 - 3.3120invert 1.9260 - 1.1500 - 2.2149 ground reaction
load on top slab Pvd 12.4200 - 1.1500 - 14.2830Pvt1 1.2254 - 1.1500 - 1.4092Pvt2 0.0000 - 1.1500 - 0.0000
soil pressure side wall (left) - 7.2137 - 1.1457 8.2649side wall (right) - -7.2137 - 1.1457 -8.2649
internal water 3.9550 - 1.1500 - 4.5483uplift 0.0000 - 1.1500 - 0.0000total 24.3324 27.9823
6) load against invertActing Load (tf/m2)Pvd 5.4000Pvt1 0.5328Pvt2 0.0000Wtop 0.8374Ws 1.2522total Pq= 8.0224
Summary of Load Calculation
Item Pv1 Ph1 Ph2 Pq Wsw q1
Wtop= (t2*BT+Hf^2)*c/B0Pvd=D*d
P3=Ka*d*D1WP=-w*0
P3=Ka*d*(D1+H0)WP=-w*H
Wsw=t1*H*c
Wbot=(t3*BT+Hf^2)*c/B0
Wiw=(hiw*B-2Hf^2)*w/B0
X = M/V =
q1 = V/Bo + 6q2 = V/Bo - 6
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10/ 114 (1)276341623.xls, Load
Case (tf/m2) (tf/m2) (tf/m2) (tf/m2) (tf/m) (tf/m2)Case.1 6.5704 3.1665 7.7665 7.8226 1.4400 5.7208Case.2 6.7702 3.2664 7.8664 8.0224 1.4400 5.9206Case.3 6.5704 3.0015 3.0715 7.8226 1.4400 10.3795Case.4 6.7702 3.1014 3.1714 8.0224 1.4400 10.5793
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11/ 114 (1)276341623.xls, Load
Class III Road3.0 m
0.51.00 t/m31.80 t/m32.00 t/m32.40 t/m3175 kgf/m2
60 kgf/m21400 kgf/m2
5.5 kgf/m23000 kgf/m2
241.2
2.00 m2.00 m0.15 m0.30 m0.30 m0.30 m2.60 m2.60 m3.00 m3.00 m (= D)0.00 m2.00 m
35.00 t
0.30.10 m0.25 m5.00 t0.10 m0.25 m1.25 t0.10 m0.25 m0.00 t/m2
t2/2 + H + t3/2 2.300 m2.300 m
3.150 m
(> 0.25m)(> 0.25m)(> 0.25m)
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12/ 114 (1)276341623.xls, Load
tf/m2, B = 2.300 mmm
tf/m2, B = 0.000 mmm
tf/m2, B = 2.300 mmm
tf/m2, B = 2.300 mmm
tf/m2, B = 2.300 mmm
tf/m2, B = 2.300 mmm
tf/m2, B = 2.300 mmm
tf/m2, B = 2.300 mmm
tf/m2, B = 0.000 mmm
(2) (2) (3) (3)a) + b) a) + c) a) + b) a) + c)
0.5328 0.3330 0.5328 0.2664 0.5328 tf/m2,
for Case.L2, which is the largest load to the top slab.
tf/mtf/m
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hiw: internal water depth 0.00 m
acting point of resultant force1.150 m
e = B0/2 - X = 0.000 m
ground reaction5.7208 tf/m25.7208 tf/m2
X = M/V =
q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =
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hiw: internal water depth 0.00 m
acting point of resultant force1.1500 m
e = B0/2 - X = 0.0000 m
ground reaction5.9206 tf/m25.9206 tf/m2
X = M/V =
q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =
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hiw: internal water depth 2.000 m
acting point of resultant force1.1500 m
e = B0/2 - X = 0.0000 m
ground reaction10.3795 tf/m210.3795 tf/m2
X = M/V =
q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =
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hiw: internal water depth 2.000 m
acting point of resultant force1.1500 m
e = B0/2 - X = 0.0000 m
ground reaction10.5793 tf/m210.5793 tf/m2
X = M/V =
q1 = V/Bo + 6Ve/Bo^2 =q2 = V/Bo - 6Ve/Bo^2 =
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19/114 (2)276341623.xlsMSN
4 Analysis of Plane Frame
Case 1: Box Culvert Inside is Empty, Underground Water up to Top slab, Track load Case. L1
1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.166Ph2 Horizontal Pressure at bottom of side wall 7.766Pv1 Vertical Pressure(1) on top slab 6.570Pv2 Vertical Pressure(2) on top slab 0.000Pq Reaction to bottom slab 7.823a Distance from joint B to far end of Pv2 2.300 mb Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.300 mB0 Width of plane frame 2.300 mt1 Thickness of side wall 0.300 mt2 Thickness of top slab 0.300 mt3 Thickness of invert (bottom slab) 0.300 m
=
=
=
=
2) Calculation of Bending Moment at joint
k1 = 1.0= 1.0000
= 1.0000
k1 0 k3 -3k1
k1 k2 0 -3k1
0 k2 k1 -3k1 =
k3 0 k1 -3k1
k1 k1 k1 k1 -4k1 R 0
As load has bilateral symmetry, the equation shown below is formed.
R =0
2k1+k3 k1=
k1 2k1+k2
3.0000 1.0=
-0.835846941.0 3.0000 0.68941361
By solving above equation, the result is led as shown below.
tf/m2
tf/m2
tf/m2
tf/m2
tf/m2
CAB = CDC = (2Ph1+3Ph2)H02/60
CBA = CCD = (3Ph1+2Ph2)H02/60
CBC = CCB = Pv1B02/12 + {(a2-b2)B02/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B02
CDA = CAD = PqB02/12
k2 = H0t23/(B0t1
3)
k3 = H0t33/(B0t13)
2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA
2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC
A = -D B = -C
A CAB - CADB CBC - CBA
AB
B
A
(t1)H0
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20/114 (2)276341623.xlsMSN
= -0.39962 = -0.36301= 0.36301 = 0.39962
A CB D
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21/114 (2)276341623.xlsMSN
= -3.0488
= 2.5334
= -2.5334
= 2.5334
= -2.5334
= 3.0488
= -3.0488
= 3.0488
2) Calculation of Design Force2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 7.766w2 Load at end B 3.166
Bending moment at end A -3.0488
Bending moment at end B 2.5334L Length of member (=H0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 7.392 tf= -5.181 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 3.895 tf
(ii) In case of x2 = 1.820 mSx2 = -3.430 tf
c) Bending Moment= -3.049
= -2.533
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 7.3922 -7.7665 x + 1.0000 x2 , x =
Bending moment at x = 1.1106 m is;
= 0.828
MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm
tf/m2
tf/m2
MAB tfmMBA tfm
SAB = (2w1+w2)L/6 - (MAB+MBA)/L
SBA = SAB - L(w1+w2)/2
Sx = SAB - w1x - (w2 - w1)x2/(2L)
MA = MAB tfmMB = -MBA tfm
Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)
Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB tfm
w1w1w1
w2w1w1
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2-2) Top Slaba) Shearing Force at joint
w1 Uniform load 6.570w2 Uniform load 0.000a Distance from end B to near end of w2 0.000 mb Length of uniform load w2 2.300 m
Bending moment at end B -2.5334Bending moment at end C 2.5334
L Length of member (=Bo) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 7.556 tf
= -7.556 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
in case of 0.000 m
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b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 3.430 tf
(ii) In case of x2 = 1.820 mSx2 = -3.895 tf
c) Bending Moment= -2.533
= -3.049
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 5.1807 -3.1665 x -1.0000 x2 , x =
Bending moment at x = 1.1894 m is;
= 0.82787
2-4) Bottom Slaba) Shearing Force at joint
w1 Reaction at end D 7.823w2 Reaction at end A 7.823
Bending moment at end B -3.04882
Bending moment at end C 3.04882L Length of member (=B0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 8.996 tf= -8.996 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 5.241 tf
(ii) In case of x2 = 1.820 mSx2 = -5.241 tf
c) Bending Moment= -3.049= -3.049
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 8.9959 -7.8226 x , x =
Sx = SCD - w1x - (w2 - w1)x2/(2L)
MC = MCD tfmMD = -MDC tfm
Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)
Mmax = SCDx - w1x2/2 - (w2-w1)x3/(6L) + MCD tfm
tf/m2
tf/m2
MDA tfmMAD tfm
SDA = (2w1+w2)L/6 - (MDA+MAD)/L
SAD = SDA - L(w1+w2)/2
Sx = SDA- w1x - (w2 - w1)x2/(2L)
MD = MDA tfmMA = -MAD tfm
Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)
L
x
w2
A
MAD
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Bending moment at x = 1.1500 m is;
= 2.124Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm
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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 TermPh1 Horizontal Pressure at top of side wall 3.266Ph2 Horizontal Pressure at bottom of side wall 7.866Pv1 Vertical Pressure(1) on top slab 6.770Pv2 Vertical Pressure(2) on top slab 0.000Pq Reaction to bottom slab 8.022a Distance from joint B to far end of Pv2 2.300 mb Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.300 mB0 Width of plane frame 2.300 mt1 Thickness of side wall 0.300 mt2 Thickness of top slab 0.300 mt3 Thickness of invert (bottom slab) 0.300 m
=
=
=
=
2) Calculation of Bending Moment at joint
k1 = 1.0= 1.00000
= 1.00000
k1 0 k3 -3k1
k1 k2 0 -3k1
0 k2 k1 -3k1 =
k3 0 k1 -3k1k1 k1 k1 k1 -4k1 R 0
As load has bilateral symmetry, the equation shown below is formed.
R =0
2k1+k3 k1=
k1 2k1+k2
3.0000 1.0=
-0.879885111.0 3.0000 0.73345178
By solving above equation, the result is led as shown below.
= -0.42164 = -0.38503
= 0.38503 = 0.42164
tf/m2
tf/m2
tf/m2
tf/m2
tf/m2
CAB = CDC = (2Ph1+3Ph2)H02/60
CBA = CCD = (3Ph1+2Ph2)H02/60
CBC = CCB = Pv1B02/12 + {(a2-b2)B0
2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B0
2
CDA = CAD = PqB02/12
k2 = H0t23/(B0t1
3)
k3 = H0t33/(B0t1
3)
2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA
2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC
A = -D B = -C
A CAB - CADB CBC - CBA
AB
A CB D
B
A
(t1)H0
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= -3.1149
= 2.5995
= -2.5995
= 2.5995
= -2.5995
= 3.1149
= -3.1149
= 3.1149
2) Calculation of Design Force2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 7.866w2 Load at end B 3.266
Bending moment at end A -3.1149
Bending moment at end B 2.5995L Length of member (=H0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 7.5071 tf= -5.2956 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 3.962 tf
(ii) In case of x2 = 1.82 mSx2 = -3.497 tf
c) Bending Moment= -3.115
= -2.599
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 7.5071 -7.8664 x + 1.0000 x2 , x =
Bending moment at x = 1.1113 m is;
= 0.828
MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm
tf/m2
tf/m2
MAB tfmMBA tfm
SAB = (2w1+w2)L/6 - (MAB+MBA)/L
SBA = SAB - L(w1+w2)/2
Sx = SAB - w1x - (w2 - w1)x2/(2L)
MA = MAB tfmMB = -MBA tfm
Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)
Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB tfm
w1w1w1
w2w1w1
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2-2) Top Slaba) Shearing Force at joint
w1 Uniform load 6.770w2 Uniform load 0.000a Distance from end B to near end of w2 0.000 mb Length of uniform load w2 2.300 m
Bending moment at end B -2.5995
Bending moment at end C 2.5995L Length of member (=Bo) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 7.786 tf
= -7.786 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
in case of 0.000 m
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= -7.507 tfSDC = SCD - L(w1+w2)/2
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b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 3.497 tf
(ii) In case of x2 = 1.820 mSx2 = -3.962 tf
c) Bending Moment= -2.599
= -3.115
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 5.2956 -3.2664 x -1.0000 x2 , x =
Bending moment at x = 1.1887 m is;
= 0.8278
2-4) Bottom Slaba) Shearing Force at joint
w1 Reaction at end D 8.022w2 Reaction at end A 8.022
Bending moment at end B -3.1149
Bending moment at end C 3.1149L Length of member (=B0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 9.226 tf= -9.226 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 5.375 tf
(ii) In case of x2 = 1.820 mSx2 = -5.375 tf
c) Bending Moment= -3.115
= -3.115The maximum bending moment occurs at the point of that shearing force equal to zero.
= 9.2257 -8.0224 x , x =
Sx = SCD - w1x - (w2 - w1)x2/(2L)
MC = MCD tfmMD = -MDC tfm
Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)
Mmax = SCDx - w1x2/2 - (w2-w1)x3/(6L) + MCD tfm
tf/m2
tf/m2
MDA tfmMAD tfm
SDA = (2w1+w2)L/6 - (MDA+MAD)/L
SAD = SDA - L(w1+w2)/2
Sx = SDA- w1x - (w2 - w1)x2/(2L)
MD = MDA tfmMA = -MAD tfm
Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)
L
x
w2
A
MAD
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Bending moment at x = 1.1500 m is;
= 2.190Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm
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Case 3: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L1
1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.001Ph2 Horizontal Pressure at bottom of side wall 3.071Pv1 Vertical Pressure(1) on top slab 6.570Pv2 Vertical Pressure(2) on top slab 0.000Pq Reaction to bottom slab 7.823a Distance from joint B to far end of Pv2 2.300 mb Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.300 mB0 Width of plane frame 2.300 mt1 Thickness of side wall 0.300 mt2 Thickness of top slab 0.300 mt3 Thickness of invert (bottom slab) 0.300 m
=
=
=
=
2) Calculation of Bending Moment at joint
k1 = 1.0= 1.00000
= 1.00000
k1 0 k3 -3k1
k1 k2 0 -3k1
0 k2 k1 -3k1 =
k3 0 k1 -3k1k1 k1 k1 k1 -4k1 R 0
As load has bilateral symmetry, the equation shown below is formed.
R =0
2k1+k3 k1=
k1 2k1+k2
3.0000 1.0=
-2.106769441.0 3.0000 1.56094111
By solving above equation, the result is led as shown below.
= -0.98516 = -0.84870= 0.84870 = 0.98516
tf/m2
tf/m2
tf/m2
tf/m2
tf/m2
CAB = CDC = (2Ph1+3Ph2)H02/60
CBA = CCD = (3Ph1+2Ph2)H02/60
CBC = CCB = Pv1B02/12 + {(a2-b2)B0
2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B0
2
CDA = CAD = PqB02/12
k2 = H0t23/(B0t1
3)
k3 = H0t33/(B0t1
3)
2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA
2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC
A = -D B = -C
A CAB - CADB CBC - CBA
AB
A CB D
B
A
(t1)H0
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= -2.46329
= 2.04774
= -2.04774
= 2.04774
= -2.04774
= 2.46329
= -2.46329
= 2.46329
2) Calculation of Design Force2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 3.071w2 Load at end B 3.001
Bending moment at end A -2.4633
Bending moment at end B 2.0477L Length of member (=H0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 3.686 tf= -3.298 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 2.215 tf
(ii) In case of x2 = 1.820 mSx2 = -1.854 tf
c) Bending Moment= -2.463
= -2.048
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 3.6861 -3.0715 x + 0.0152 x2 , x =
Bending moment at x = 1.2073 m is;
= -0.243
MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm
tf/m2
tf/m2
MAB tfmMBA tfm
SAB = (2w1+w2)L/6 - (MAB+MBA)/L
SBA = SAB - L(w1+w2)/2
Sx = SAB - w1x - (w2 - w1)x2/(2L)
MA = MAB tfmMB = -MBA tfm
Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)
Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB tfm
w1w1w1
w2w1w1
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2-2) Top Slaba) Shearing Force at joint
w1 Uniform load 6.570w2 Uniform load 0.000a Distance from end B to near end of w2 0.000 mb Length of uniform load w2 2.300 m
Bending moment at end B -2.048
Bending moment at end C 2.048L Length of member (=Bo) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 7.556 tf
= -7.556 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
in case of 0.000 m
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= -3.686 tfSDC = SCD - L(w1+w2)/2
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b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 1.854 tf
(ii) In case of x2 = 1.820 mSx2 = -2.215 tf
c) Bending Moment= -2.048
= -2.463
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 3.2979 -3.0015 x -0.0152 x2 , x =
Bending moment at x = 1.0927 m is;
= -0.243
2-4) Bottom Slaba) Shearing Force at joint
w1 Reaction at end D 7.823w2 Reaction at end A 7.823
Bending moment at end B -2.463
Bending moment at end C 2.463L Length of member (=B0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 8.996 tf= -8.996 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 5.241 tf
(ii) In case of x2 = 1.820 mSx2 = -5.241 tf
c) Bending Moment= -2.463
= -2.463The maximum bending moment occurs at the point of that shearing force equal to zero.
= 8.9959 -7.8226 x , x =
Sx = SCD - w1x - (w2 - w1)x2/(2L)
MC = MCD tfmMD = -MDC tfm
Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)
Mmax = SCDx - w1x2/2 - (w2-w1)x3/(6L) + MCD tfm
tf/m2
tf/m2
MDA tfmMAD tfm
SDA = (2w1+w2)L/6 - (MDA+MAD)/L
SAD = SDA - L(w1+w2)/2
Sx = SDA- w1x - (w2 - w1)x2/(2L)
MD = MDA tfmMA = -MAD tfm
Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)
L
x
w2
A
MAD
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Bending moment at x = 1.1500 m is;
= 2.709Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm
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Case 4: Box Culvert Inside is Full, Underground Water up to invert, Track load Case. L2
1) Calculation of Load TermPh1 Horizontal Pressure at top of side wall 3.101
Ph2 Horizontal Pressure at bottom of side wall 3.171
Pv1 Vertical Pressure(1) on top slab 6.770
Pv2 Vertical Pressure(2) on top slab 0.000
Pq Reaction to bottom slab 8.022a Distance from joint B to far end of Pv2 2.300 mb Distance from joint B to near end of Pv2 0.000 mH0 Height of plane frame 2.300 mB0 Width of plane frame 2.300 mt1 Thickness of side wall 0.300 mt2 Thickness of top slab 0.300 mt3 Thickness of invert (bottom slab) 0.300 m
=
=
=
=
2) Calculation of Bending Moment at joint
k1 = 1.0= 1.00000= 1.00000
k1 0 k3 -3k1
k1 k2 0 -3k1
0 k2 k1 -3k1 =
k3 0 k1 -3k1k1 k1 k1 k1 -4k1 R 0
As load has bilateral symmetry, the equation shown below is formed.
R =0
2k1+k3 k1=
k1 2k1+k2
3.0000 1.0=
-2.150807611.0 3.0000 1.60497928
By solving above equation, the result is led as shown below.
= -1.00718 = -0.87072= 0.87072 = 1.00718
tf/m2
tf/m2
tf/m2
tf/m2
tf/m2
CAB = CDC = (2Ph1+3Ph2)H02/60
CBA = CCD = (3Ph1+2Ph2)H02/60
CBC = CCB = Pv1B02/12 + {(a2-b2)B0
2/2 - 2B0(a3-b3)/3 + (a4-b4)/4}Pv2/B0
2
CDA = CAD = PqB02/12
k2 = H0t23/(B0t13) k3 = H0t33/(B0t13)
2(k1+k3) A CAB - CAD2(k1+k2) B CBC - CBA
2(k1+k2) C CCD - CCB2(k1+k3) D CDA - CDC
A = -D B = -C
A CAB - CADB CBC - CBA
AB
A CB D
B
A
(t1)H0
-
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= -2.52934
= 2.11380
= -2.11380
= 2.11380
= -2.11380
= 2.52934
= -2.52934
= 2.52934
2) Calculation of Design Force2-1) Side Wall in left
a) Shearing Force at joint
w1 Load at end A 3.171w2 Load at end B 3.101
Bending moment at end A -2.529
Bending moment at end B 2.114L Length of member (=H0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 3.801 tf= -3.413 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 2.282 tf
(ii) In case of x2 = 1.82 mSx2 = -1.921 tf
c) Bending Moment= -2.529
= -2.114
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 3.8009 -3.1714 x + 0.0152 x2 , x =
Bending moment at x = 1.2055 m is;
= -0.243
MAB = k1(2A +B) - CAB tfmMBA = k1(2B+A)+CBA tfmMBC = k2(2B+C) - CBC tfmMCB = k2(2C+B)+CCB tfmMCD = k1(2C+D) - CCD tfmMDC =k1 (2D+ C)+CDC tfmMDA = k3(2D+A) - CDA tfmMAD = k3(2A+D)+CAD tfm
tf/m2
tf/m2
MAB tfmMBA tfm
SAB = (2w1+w2)L/6 - (MAB+MBA)/L
SBA = SAB - L(w1+w2)/2
Sx = SAB - w1x - (w2 - w1)x2/(2L)
MA = MAB tfmMB = -MBA tfm
Sx = 0 = SAB - w1x - (w2 - w1)x2/(2L)
Mmax = SABx - w1x2/2 - (w2-w1)x3/(6L) + MAB tfm
w1w1w1
w2w1w1
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2-2) Top Slaba) Shearing Force at joint
w1 Uniform load 6.770w2 Uniform load 0.000a Distance from end B to near end of w2 0.000 mb Length of uniform load w2 2.300 m
Bending moment at end B -2.114
Bending moment at end C 2.114L Length of member (=Bo) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 7.786 tf
= -7.786 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
in case of 0.000 m
-
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= -3.801 tfSDC = SCD - L(w1+w2)/2
-
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b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 1.9206 tf
(ii) In case of x2 = 1.820 mSx2 = -2.2822 tf
c) Bending Moment= -2.114
= -2.529
The maximum bending moment occurs at the point of that shearing force equal to zero.
= 3.4128 -3.1014 x -0.0152 x2 , x =
Bending moment at x = 1.0945 m is;
= -0.243
2-4) Bottom Slaba) Shearing Force at joint
w1 Reaction at end D 8.022w2 Reaction at end A 8.022
Bending moment at end B -2.529
Bending moment at end C 2.529L Length of member (=B0) 2.300 mch Protective covering height 0.060 mt Thickness of member (height) 0.300 md Effective height of member 0.240 m
= 9.226 tf= -9.226 tf
b) Shearing Force at 2d point from jointShearing force at the point with a distance of 2d from joint is calculated by following equation.
(i) In case of x1 = 0.480 mSx1 = 5.375 tf
(ii) In case of x2 = 1.820 mSx2 = -5.375 tf
c) Bending Moment= -2.529
= -2.529The maximum bending moment occurs at the point of that shearing force equal to zero.
= 9.2257 -8.0224 x , x =
Sx = SCD - w1x - (w2 - w1)x2/(2L)
MC = MCD tfmMD = -MDC tfm
Sx = 0 = SCD - w1x - (w2 - w1)x2/(2L)
Mmax = SCDx - w1x2/2 - (w2-w1)x3/(6L) + MCD tfm
tf/m2
tf/m2
MDA tfmMAD tfm
SDA = (2w1+w2)L/6 - (MDA+MAD)/L
SAD = SDA - L(w1+w2)/2
Sx = SDA- w1x - (w2 - w1)x2/(2L)
MD = MDA tfmMA = -MAD tfm
Sx = 0 = SDA - w1x - (w2 - w1)x2/(2L)
L
x
w2
A
MAD
-
50/114 (2)276341623.xlsMSN
Bending moment at x = 1.1500 m is;
= 2.775Mmax = SDAx - w1x2/2 - (w2-w1)x3/(6L) + MDA tfm
-
51/114 (2)276341623.xlsMSN
2.61260
2.20703
2.89644
3.44844
tfm
tfm
tfm
tfm
(t2)
(t1)
B0
(t3)
C
D
-
52/114 (2)276341623.xlsMSN
-
53/114 (2)276341623.xlsMSN
6.6561.111
A
Bw1w1
Lxw1w1
MAB
MBAw1w1
-
54/114 (2)276341623.xlsMSN
-
55/114 (2)276341623.xlsMSN
w2
MCBx
L
C
w1
w2
MCD
MDC
C
D
-
56/114 (2)276341623.xlsMSN
-
57/114 (2)276341623.xlsMSN
-4.3561.189
1.150
L
x
D
MDA
w1
-
58/114 (2)276341623.xlsMSN
-
59/114 (2)276341623.xlsMSN
2.65664
2.25107
2.98452
3.53652
tfm
tfm
tfm
tfm
(t2)
(t1)
B0
(t3)
C
D
-
60/114 (2)276341623.xlsMSN
-
61/114 (2)276341623.xlsMSN
6.7551.111
A
Bw1w1
Lxw1w1
MAB
MBAw1w1
-
62/114 (2)276341623.xlsMSN
-
63/114 (2)276341623.xlsMSN
w2
MCBx
L
C
w1
w2
MCD
MDC
C
D
-
64/114 (2)276341623.xlsMSN
-
65/114 (2)276341623.xlsMSN
-4.45511.1887
1.1500
L
x
D
MDA
w1
-
66/114 (2)276341623.xlsMSN
-
67/114 (2)276341623.xlsMSN
1.34167
1.33550
2.89644
3.44844
tfm
tfm
tfm
tfm
(t2)
(t1)
B0
(t3)
C
D
-
68/114 (2)276341623.xlsMSN
-
69/114 (2)276341623.xlsMSN
200.6341.207
A
Bw1w1
Lxw1w1
MAB
MBAw1w1
-
70/114 (2)276341623.xlsMSN
-
71/114 (2)276341623.xlsMSN
w2
MCBx
L
C
w1
w2
MCD
MDC
C
D
-
72/114 (2)276341623.xlsMSN
-
73/114 (2)276341623.xlsMSN
-198.331.093
1.150
L
x
D
MDA
w1
-
74/114 (2)276341623.xlsMSN
-
75/114 (2)276341623.xlsMSN
1.38571
1.37954
2.98452
3.53652
tfm
tfm
tfm
tfm
(t2)
(t1)
B0
(t3)
C
D
-
76/114 (2)276341623.xlsMSN
-
77/114 (2)276341623.xlsMSN
207.2001.205
A
Bw1w1
Lxw1w1
MAB
MBAw1w1
-
78/114 (2)276341623.xlsMSN
-
79/114 (2)276341623.xlsMSN
w2
MCBx
L
C
w1
w2
MCD
MDC
C
D
-
80/114 (2)276341623.xlsMSN
-
81/114 (2)276341623.xlsMSN
-204.91.095
1.150
L
x
D
MDA
w1
-
82/114 (2)276341623.xlsMSN
-
83/114 (3)276341623.xlsSum MSN
Summary of Internal forces
Member Case M N S (tf)(tf) at joint at 2dSide wall A -3.049 8.996 7.392 3.895 (left) Case.1 Middle 0.828 8.301 0.000 -
B -2.533 7.556 -5.181 -3.430A -3.115 9.226 7.507 3.962
Case.2 Middle 0.828 8.530 0.000 -B -2.599 7.786 -5.296 -3.497A -2.463 8.996 3.686 2.215
Case.3 Middle -0.243 8.240 0.000 -B -2.048 7.556 -3.298 -1.854A -2.529 9.226 3.801 2.282
Case.4 Middle -0.243 8.471 0.000 -B -2.114 7.786 -3.413 -1.921
Top slab B -2.533 5.181 7.556 4.402Case.1 Middle 1.811 5.181 0.000 -
C -2.533 5.181 -7.556 -4.402B -2.599 5.296 7.786 4.536
Case.2 Middle 1.877 5.296 0.000 -C -2.599 5.296 -7.786 -4.536B -2.048 3.298 7.556 4.402
Case.3 Middle 2.297 3.298 0.000 -C -2.048 3.298 -7.556 -4.402B -2.114 3.413 7.786 4.536
Case.4 Middle 2.363 3.413 0.000 -C -2.114 3.413 -7.786 -4.536
Side wall C -2.533 7.556 5.181 3.430 (right) Case.1 Middle 0.828 8.301 0.000 -
D -3.049 8.996 -7.392 -3.895C -2.599 7.786 5.296 3.497
Case.2 Middle 0.828 8.530 0.000 -D -3.115 9.226 -7.507 -3.962C -2.048 7.556 3.298 1.854
Case.3 Middle -0.243 8.240 0.000 -D -2.463 8.996 -3.686 -2.215C -2.114 7.786 3.413 1.921
Case.4 Middle -0.243 8.471 0.000 -D -2.529 9.226 -3.801 -2.282
Invert D -3.049 7.392 8.996 5.241Case.1 Middle 2.124 7.392 0.000 -
A -3.049 7.392 -8.996 -5.241D -3.115 7.507 9.226 5.375
Case.2 Middle 2.190 7.507 0.000 -A -3.115 7.507 -9.226 -5.375D -2.463 3.686 8.996 5.241
Case.3 Middle 2.709 3.686 0.000 -A -2.463 3.686 -8.996 -5.241D -2.529 3.801 9.226 5.375
Case.4 Middle 2.775 3.801 0.000 -A -2.529 3.801 -9.226 -5.375
Check Point (tfm)
-
84/114 (3)276341623.xlsSum MSN
-
85/114 (3)276341623.xlsSum MSN
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86/114 (3)276341623.xlsSum MSN
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87/ 114 (4)276341623.xls, R-bar req
5 Calculation of Required Reinforcement Bar5-1 Calculation of Required Reinforcement Bar1)
Case.1M= 3.0488 60 kgf/m2 h = 30 cm (height of member)N= 8.9959 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.3922 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.8947 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 33.89 cm
38.729 kgf/cm2 ( 0 kgf/cm2) o.k.
+67.4 -2344.55 -68382.700.39906.8199 cm2
Case.2M= 3.1149 60 kgf/m2 h = 30 cm (height of member)N= 9.2257 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.5071 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.9616 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 33.76 cm
39.304 kgf/cm2 ( 4.11E-006 kgf/cm2) o.k.
+66.95 -2397.25 -69919.910.40266.9718 cm2
Case.3M= 2.4633 60 kgf/m2 h = 30 cm (height of member)N= 8.9959 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.6861 tf n = 24 d' = 6 cm (protective covering depth)S2d= 2.2152 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 27.38 cm
34.762 kgf/cm2 ( 0 kgf/cm2) o.k.
+70.45 -1988.75 -58005.350.37344.7003 cm2
Case.4M= 2.5293 60 kgf/m2 h = 30 cm (height of member)N= 9.2257 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.8009 tf n = 24 d' = 6 cm (protective covering depth)S2d= 2.2822 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 27.42 cm
35.359 kgf/cm2 ( 0 kgf/cm2) o.k.
+70 -2041.46 -59542.560.37744.8482 cm2
The maximum requirement of reinforcement bar is 6.9718 cm2 in Case. 2 from above calculation.
Case. 1 2 3 4Requirement 6.8199 6.9718 4.7003 4.8482 (cm2)
2)Case.1
M= 2.5334 60 kgf/m2 h = 30 cm (height of member)N= 7.5559 1400 kgf/m2 d = 24 cm (effective height of member)S0= 5.1807 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.4304 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 33.53 cm
34.350 kgf/cm2 ( 0 kgf/cm2) o.k.
+70.76 -1952.63 -56951.650.37065.5151 cm2
Case.2M= 2.5995 60 kgf/m2 h = 30 cm (height of member)N= 7.7857 1400 kgf/m2 d = 24 cm (effective height of member)S0= 5.2956 tf n = 24 d' = 6 cm (protective covering depth)S2d= 3.4973 tf c = 9.00 cm (distance from neutral axis)
At Joint "A"of side wall
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
At Joint "B"of side wall
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
d1h d
d1h d
-
88/ 114 (4)276341623.xls, R-bar req
b = 100 cme = M/N = 33.39 cm
34.950 kgf/cm2 ( 0 kgf/cm2) o.k.
+70.31 -2005.33 -58488.850.37475.6627 cm2
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
-
89/ 114 (4)276341623.xls, R-bar req
Case.3M= 2.0477 60 kgf/m2 h = 30 cm (height of member)N= 7.5559 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.2979 tf n = 24 d' = 6 cm (protective covering depth)S2d= 1.8537 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 27.10 cm
30.914 kgf/cm2 ( 0 kgf/cm2) o.k.
+73.29 -1657.5 -48343.890.34643.7812 cm2
Case.4M= 2.1138 60 kgf/m2 h = 30 cm (height of member)N= 7.7857 1400 kgf/m2 d = 24 cm (effective height of member)S0= 3.4128 tf n = 24 d' = 6 cm (protective covering depth)S2d= 1.9206 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 27.15 cm
31.538 kgf/cm2 ( 0 kgf/cm2) o.k.
+72.84 -1710.21 -49881.100.35093.9250 cm2
The maximum requirement of reinforcement bar is 5.6627 cm2 in Case. 2 from above calculation.
Case. 1 2 3 4Requirement 5.5151 5.6627 3.7812 3.9250 (cm2)
3)Case.1
M= 2.5334 60 kgf/m2 h = 30 cm (height of member)N= 5.1807 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.5559 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.4022 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 48.90 cm
32.854 kgf/cm2 ( 0 kgf/cm2) o.k.
+71.88 -1822.73 -53163.050.36036.4454 cm2
Case.2M= 2.5995 60 kgf/m2 h = 30 cm (height of member)N= 5.2956 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.7857 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.5360 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 49.09 cm
33.391 kgf/cm2 ( 0 kgf/cm2) o.k.
+71.48 -1869.15 -54517.010.36406.6367 cm2
Case.3M= 2.0477 60 kgf/m2 h = 30 cm (height of member)N= 3.2979 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.5559 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.4022 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 62.09 cm
28.090 kgf/cm2 ( 0 kgf/cm2) o.k.
+75.29 -1424.64 -41552.070.32505.4700 cm2
Case.4M= 2.1138 60 kgf/m2 h = 30 cm (height of member)N= 3.4128 1400 kgf/m2 d = 24 cm (effective height of member)S0= 7.7857 tf n = 24 d' = 6 cm (protective covering depth)S2d= 4.5360 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 61.94 cm
28.662 kgf/cm2 ( 0.0002919 kgf/cm2) o.k.
+74.89 -1471.06 -42906.03
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
At Joint "B"of top slab
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 c
d1h d
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90/ 114 (4)276341623.xls, R-bar req
0.32955.6565 cm2
The maximum requirement of reinforcement bar is 6.6367 cm2 in Case. 2 from above calculation.
Case. 1 2 3 4Requirement 6.4454 6.6367 5.4700 5.6565 (cm2)
s = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
-
91/ 114 (4)276341623.xls, R-bar req
4)Case.1
M= 3.0488 60 kgf/m2 h = 30 cm (height of member)N= 7.3922 1400 kgf/m2 d = 24 cm (effective height of member)S0= 8.9959 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.2411 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 41.24 cm
37.763 kgf/cm2 ( 5.85E-005 kgf/cm2) o.k.
+68.16 -2256.85 -65824.690.39307.4396 cm2
Case.2M= 3.1149 60 kgf/m2 h = 30 cm (height of member)N= 7.5071 1400 kgf/m2 d = 24 cm (effective height of member)S0= 9.2257 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.3750 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 41.49 cm
38.275 kgf/cm2 ( 0 kgf/cm2) o.k.
+67.76 -2303.27 -67178.650.39627.6354 cm2
Case.3M= 2.4633 60 kgf/m2 h = 30 cm (height of member)N= 3.6861 1400 kgf/m2 d = 24 cm (effective height of member)S0= 8.9959 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.2411 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 66.83 cm
31.398 kgf/cm2 ( 0 kgf/cm2) o.k.
+72.94 -1698.37 -49535.810.34996.7842 cm2
Case.4M= 2.5293 60 kgf/m2 h = 30 cm (height of member)N= 3.8009 1400 kgf/m2 d = 24 cm (effective height of member)S0= 9.2257 tf n = 24 d' = 6 cm (protective covering depth)S2d= 5.3750 tf c = 9.00 cm (distance from neutral axis)
b = 100 cme = M/N = 66.55 cm
31.945 kgf/cm2 ( 2.75E-006 kgf/cm2) o.k.
+72.54 -1744.79 -50889.770.35396.9741 cm2
The maximum requirement of reinforcement bar is 7.6354 cm2 in Case. 2 from above calculation.
Case. 1 2 3 4Requirement 7.4396 7.6354 6.7842 6.9741 (cm2)
5) At Middle of side wallCase.1
M= 0.8279 60 kgf/m2 h = 30 cm (height of member)N= 8.3006 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 9.97 cm21.9840 kgf/cm2 ( 0 kgf/cm2) o.k.
+79.3 -956.98 -27912.010.27370.0000 cm2 Tensile is on inside of member
Case.2M= 0.8278 60 kgf/m2 h = 30 cm (height of member)N= 8.5299 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 9.70 cm
At Joint "A"of invert
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
d1h d
d1h d
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92/ 114 (4)276341623.xls, R-bar req
22.158 kgf/cm2 ( 0.0 kgf/cm2) o.k.
+79.19 -969.48 -28276.460.27530.0000 cm2 Tensile is on inside of member
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
-
93/ 114 (4)276341623.xls, R-bar req
Case.3M= 0.2426 60 kgf/m2 h = 30 cm (height of member)N= 8.2401 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 2.94 cm16.634 kgf/cm2 ( 0 kgf/cm2) o.k.
+82.37 -598.07 -17443.750.22190.0000 cm2 Tensile is on outside of member
Case.4M= 0.2428 60 kgf/m2 h = 30 cm (height of member)N= 8.4710 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 2.87 cm16.839 kgf/cm2 ( 0 kgf/cm2) o.k.
+82.26 -610.8 -17814.870.22400.0000 cm2 Tensile is on outside of member
The maximum requirement of reinforcement bar is 0.0000 cm2 in Case. 1 from above calculation.
Case. 1 2 3 4Requirement 0.0000 0.0000 - - (cm2)Side inside inside outside outside
6) At Middle of top slabCase.1
M= 1.8112 60 kgf/m2 h = 30 cm (height of member)N= 5.1807 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 34.96 cm27.583 kgf/cm2 ( 0 kgf/cm2) o.k.
+75.64 -1383.9 -40363.640.32103.8899 cm2 Tensile is on inside of member
Case.2M= 1.8773 60 kgf/m2 h = 30 cm (height of member)N= 5.2956 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 35.45 cm28.160 kgf/cm2 ( 0 kgf/cm2) o.k.
+75.24 -1430.32 -41717.600.32564.0759 cm2 Tensile is on inside of member
Case.3M= 2.2969 60 kgf/m2 h = 30 cm (height of member)N= 3.2979 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 69.65 cm29.939 kgf/cm2 ( 0 kgf/cm2) o.k.
+73.99 -1576.05 -45968.170.33926.3478 cm2 Tensile is on inside of member
Case.4M= 2.3630 60 kgf/m2 h = 30 cm (height of member)N= 3.4128 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 69.24 cm30.496 kgf/cm2 ( 1.84E-005 kgf/cm2) o.k.
+73.59 -1622.47 -47322.13
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 c
d1h d
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0.34336.5363 cm2 Tensile is on inside of member
The maximum requirement of reinforcement bar is 6.5363 cm2 in Case. 4 from above calculation.
Case. 1 2 3 4Requirement 3.8899 4.0759 6.3478 6.5363 (cm2)Side inside inside inside inside
s = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
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7) At Middle of invertCase.1
M= 2.1238 60 kgf/m2 h = 30 cm (height of member)N= 7.3922 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 28.73 cm31.3560 kgf/cm2 ( 0 kgf/cm2) o.k.
+72.97 -1694.79 -49431.390.34964.1161 cm2 Tensile is on inside of member
Case.2M= 2.1899 60 kgf/m2 h = 30 cm (height of member)N= 7.5071 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 29.17 cm31.902 kgf/cm2 ( 0 kgf/cm2) o.k.
+72.58 -1741.21 -50785.350.35354.3051 cm2 Tensile is on inside of member
Case.3M= 2.7094 60 kgf/m2 h = 30 cm (height of member)N= 3.6861 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 73.50 cm33.146 kgf/cm2 ( 4.80E-007 kgf/cm2) o.k.
+71.66 -1847.9 -53897.220.36237.6614 cm2 Tensile is on inside of member
Case.4M= 2.7754 60 kgf/m2 h = 30 cm (height of member)N= 3.8009 1400 kgf/m2 d = 24 cm (effective height of member)
n = 24 d' = 6 cm (protective covering depth)c = 9.00 cm (distance from neutral axis)b = 100 cm
e = M/N = 73.02 cm33.682 kgf/cm2 ( 0 kgf/cm2) o.k.
+71.26 -1894.33 -55251.190.36607.8531 cm2 Tensile is on inside of member
The maximum requirement of reinforcement bar is 7.8531 cm2 in Case. 4 from above calculation.
Case. 1 2 3 4Requirement 4.1161 4.3051 7.6614 7.8531 (cm2)Side inside inside inside inside
8) Summary of required reinforcement
Required reinforcement for design is the maximum required reinforcement calculated above in 1) - 4).
Item Side wall Side wall Top slab Invert Side wall Top slab InvertPoint bottom top end end middle middle middleSide outside outside outside outside inside inside insideCalculation 1) 2) 3) 4) 5) 6) 7)Requirement 6.972 5.663 6.637 7.635 0.000 6.536 7.853 (cm2)
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
tfm ca =tf sa =
Solving the formula shown below, c = c^3+{3sa/(2n) - 3N(e+c)/(bd^2)}c^2 - 6N(e+c)sac/(nbd^2) - 3N(e+c)sa^2/(n^2bd^2) = 0 0 = c^3 c^2 cs = nc/(nc+sa) =Asreq = (c*s/2 - N/(bd))bd/sa =
d1h d
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6 Bar Arrangement and Calculation of Stress Type: B2.00m x H2.00m
Side wall Top slabbottom middle top end middleoutside inside outside outside inside
Bending moment M kgfcm 311,488 82,787 259,949 259,949 236,298 Shearing force (joint) S kgf 7,507 0 5,296 7,786 0 Shearing force (2d) S2d kgf 3,962 - 3,497 4,536 -Axial force N kgf 9,226 8,301 7,786 5,296 3,413 Height of member h cm 30 30 30 30 30Covering depth d' cm 6 6 6 6 6Effective height d cm 24 24 24 24 24Effective width b cm 100 100 100 100 100Effective area bd cm2 2400 2400 2400 2400 2400Young's modulus ratio n - 24 24 24 24 24
Required R-bar Asreq cm2 6.97 0.00 5.66 6.64 6.54
R-bar arrangement 16@250 12@250 16@250 16@250 16@250
Reinforcement As cm2 8.04 4.52 8.04 8.04 8.04Perimeter of R-bar U 20.11 15.08 20.11 20.11 20.11M/N e cm 33.763 9.974 33.388 49.088 69.239Dist. from neutral axis c cm 9.00 9.00 9.00 9.00 9.00
a' 56.3 -15.1 55.2 102.3 162.7b' 495.1 123.5 490.8 672.5 905.8c' -11882.3 -2963.9 -11778.0 -16140.4 -21739.8x 10.16 17.62 10.19 9.38 8.92
0.000 0.000 0.000 0.000 0.000(check) ok ok ok ok ok
Compressive stress kgf/cm2 37.7 9.9 31.5 31.4 28.5Allowable stress kgf/cm2 60.0 60.0 60.0 60.0 60.0
ok ok ok ok okTensile stress kgf/cm2 1232.9 85.7 1023.7 1174.4 1155.0Allowable stress kgf/cm2 1400.0 1400.0 1400.0 1400.0 1400.0
ok ok ok ok okShearing stress at joint kgf/cm2 3.13 0.00 2.21 3.24 0.00Allowable stress kgf/cm2 11.00 11.00 11.00 11.00 11.00
ok ok ok ok okShearing stress at 2d kgf/cm2 1.65 - 1.46 1.89 -Allowable stress kgf/cm2 5.50 - 5.50 5.50 -
ok - ok ok -
Resisting Moment Mr kgfcm 496,869 362,613 497,615 459,243 408,023 Mr for compression Mrc kgfcm 496,869 410,229 497,615 499,428 501,254 x for Mrc cm 9.240 7.521 9.013 8.634 8.359
kgf/cm2 2300.3 3155.3 2394.6 2562.9 2694.5 Mr for tensile Mrs kgfcm 577,461 362,613 532,234 459,243 408,023 x for Mrs cm 12.982 10.971 12.517 11.670 10.986
kgf/cm2 68.7 49.1 63.6 55.2 49.2
Distribution bar (>As/6 and >Asmin) 12@250 12@250 12@250 12@250 12@250Reinforcement As cm2 4.52 4.52 4.52 4.52 4.52
ok ok ok ok okReinforcement bar for fillet Err:504 Err:504Reinforcement As cm2 Err:504 Err:504
cca
ssa
a
2d2da
s for Mrc
c for Mrs
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Minimum requirement of reinforcement bar As min = 4.5 cm2
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107/114 (5)276341623.xlsR-bar stress
Invertend middle
outside inside311,488 277,544
9,226 0 5,375 -7,507 3,801
30 306 6
24 24100 100
2400 240024 24
7.64 7.85
16@250 12@125
8.04 9.0520.11 30.16
41.492 73.0209.00 9.0079.5 174.1
584.6 1068.9-14030.0 -25653.1
9.69 9.27 0.000 0.000
ok ok
37.7 32.260.0 60.0
ok ok1336.1 1227.41400.0 1400.0
ok ok3.84 0.00
11.00 11.00ok ok
2.24 -5.50 -
ok -
497,785 469,708 497,785 521,262
8.969 8.7872413.1 2492.9
523,749 469,708 12.426 11.642
62.6 55.0
12@250 [email protected] 4.52
ok ok
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DATA PLOT PEMBESIAN
Section of Culvert
1. Design Datab1 = 0.300 m b2 = 2.000 mh1 = 0.300 m h2 = 2.000 m h3 = 0.300 m d =Bb = 2.600 m Ha = 2.600 m Fillet = 0.15 m
2. Data PembesianBottom slab : Tumpuan Lapangan Tulangan bagi :
: O 16 @ 250 O 12 @ 125 O 12 @: O 12 @ 125 O 16 @ 250
Side wall : Tumpuan Lapangan Tulangan bagi :: O 16 @ 250 O 12 @ 250 O 12 @: O 12 @ 250 O 16 @ 250
Top slab : Tumpuan Lapangan Tulangan bagi :
: O 16 @ 250 O 16 @ 250 O 12 @: O 16 @ 250 O 16 @ 250
3. Tulangan Miring (fillet) :Bottom slab : O ### @ Err:504Top slab : O ### @ Err:504
4. Nama Bangunan : Culvert Type 1 (2 m x 2 m)Lokasi : . Irrigation Project
As1 (cm2)As2 (cm2)
As1 (cm2)As2 (cm2)
As1 (cm2)As2 (cm2)
b1b1 b2
h3
h2
h1
Bb
Ha
-
Section of Culvert
0.06 m
Tulangan bagi :250
Tulangan bagi :250
Tulangan bagi :250
-
DATA PLOT MOMENT AND SHEAR FORCE DIAGRAM
Section of Culvert
1. Design Datab1 = 0.300 m b2 = 2.000 mh1 = 0.300 m h2 = 2.000 m h3 = 0.300 mBb = 2.300 m Ha = 2.300 m
2. Result CalculationSide wall (left) Ma Mmax Mb Sa Sb NaCase 1 -3.049 0.828 -2.533 7.392 -5.181 8.996Case 2 -2.463 -0.243 -2.048 3.686 -3.298 8.996
Side wall (right) Mc Mmax Md Sd Sc NdCase 1 -2.533 0.828 -3.049 -7.392 5.181 8.996Case 2 -2.048 -0.243 -2.463 -3.686 3.298 8.996
Top slab Mb Mmax Mc Sb Sc NbCase 1 -2.533 1.811 -2.533 7.556 -7.556 5.181Case 2 -2.048 2.297 -2.048 7.556 -7.556 3.298
Bottom slab Ma Mmax Md Sa Sd NaCase 1 -3.049 2.124 -3.049 -8.996 8.996 7.392Case 2 -2.463 2.709 -2.463 -8.996 8.996 3.686
3. Nama Bangunan : Culvert Type 1 (2 m x 2 m)
b1b1 b2
h3
h2
h1
Bb
Ha
-
Lokasi : . Irrigation Project
Case 1Mmax x x1 x2 x3 x4 x5 x6Mab = 1.1106 -1.24 -0.02 0.65 0.82 0.54 -0.14Mcd = 1.1894 -1.24 -0.02 0.65 0.82 0.54 -0.14Mmax y1 y2 y3 y4 y5 y6Mbc = 1.15 -0.63 0.73 1.54 1.81 1.54 0.73Mad = 1.15 -0.79 0.83 1.80 2.12 1.80 0.83
Ha = 2.300 0.288 0.575 0.863 1.150 1.438 1.725
Bb = 2.300 0.288 0.575 0.863 1.150 0.863 0.575
x x1 x2 x3 x4 x5 x6Sab 2.300 5.24 3.26 1.44 -0.22 -1.71 -3.03
Case 2Mmax x x1 x2 x3 x4 x5 x6Mab = 1.1106 -1.53 -0.85 -0.42 -0.25 -0.32 -0.65Mcd = 1.1894 -1.53 -0.85 -0.42 -0.25 -0.32 -0.65Mmax y1 y2 y3 y4 y5 y6Mbc = 1.15 -0.15 1.21 2.03 2.30 2.03 1.21Mad = 1.15 -0.20 1.42 2.39 2.71 2.39 1.42
Ha = 2.300 0.288 0.575 0.863 1.150 1.438 1.725
Bb = 2.300 0.288 0.575 0.863 1.150 0.863 0.575
x x1 x2 x3 x4 x5 x6Sab 2.300 2.80 1.92 1.05 0.17 -0.70 -1.57
-
Section of Culvert
Na Nb MAB MBA8.996 7.556 -3.049 2.5338.996 7.556 -2.463 2.048
Nd Nc MCD MDC8.996 7.556 -2.533 3.0498.996 7.556 -2.048 2.463
Nb Nc MBC MCB5.181 5.181 -2.533 2.5333.298 3.298 -2.048 2.048
Na Nd MAD MDA7.392 7.392 3.049 -3.0493.686 3.686 2.463 -2.463
h3
h2
h1
-
x6 x7 q1 q2-0.14 -1.18 7.766 3.166-0.14 -1.18 3.166 7.766
y6 y7 q10.73 -0.63 6.5700.83 -0.79 7.823
1.725 2.013
0.575 0.288
x6 x7-3.03 -4.19
x6 x7 q1 q2-0.65 -1.22 3.071 3.001-0.65 -1.22 3.001 3.071
y6 y7 q11.21 -0.15 6.5701.42 -0.20 7.823
1.725 2.013
0.575 0.288
x6 x7-1.57 -2.43
SummaryLoadMSNSum MSNR-bar reqR-bar stressPlotRebarPlotMoment