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INSTITUTE OF TECHNOLOGY OF CAMBODIA DEPARTMENT OF CIVIL ENGINEERING ASSIGHMENT OF REINFORCED CONCRETE SOLID SLAB OF BIRDGE DESIGN GROUP-2 SEMESTER-I LECTURER :
KAING Saoserey
STUDENTS :
I5-A-GCI
1. CHHENG Chanraksmey
e20130072
2. CHHET Hychhy
e20130077
3. CHHIN Chanbormey
e20130079
4. CHHOENG Oudom
e20130084
5. CHHOM Kanhara
e20130087
6. CHHON Chhour
e20130088
2017-2018
Institut de Technologie du Cambodge
Projet du pont CONTENTS
I. Hypothesis: ......................................................................................................................Page: Page0101 II. Determine the Section of Bridge:...................................................................................Page: Page0202 III. Calcul du hourdis: .........................................................................................................Page: Page0202 IV.Choisir les facteur de résistance : .................................................................................Page: Page0202 V. Select Load Modifier: ....................................................................................................Page: Page0202 VI. Combination: .................................................................................................................Page: Page0202 VII. Determine Live Load Strip Width: ...............................................................................Page: Page02 02 VIII. Applicability ofLive Load for Decks and Deck Systems: ...........................................Page: Page04 04 IX. Live Load Force Effect:.................................................................................................Page: Page0808 X. Force Effects Other Load ...............................................................................................Page: Page0808 XI. Investigate Service Limit State.......................................................................................Page: Page09 09 XII. Investigate Strength Limite State:..................................................................................Page: Page17 17 XIII. Distribution Reinforcement of the Deck:.....................................................................Page: Page19 19 XIV. Shrinkage or Temperature Reinforcement:.................................................................Page: Page2020 XV. Fatigue Limite State:....................................................................................................Page: Page2020 XVI. Resume the answer:....................................................................................................Page: Page2222 XVII. Conclusion:...............................................................................................................Page: Page2222 XVIII. Sketch of Bridge Section in Rebar Detail:...............................................................Page: Page2323
Page 1
Institut de Technologie du Cambodge
Projet du pont
PONT DALLE EN BÉTON ARMÉ (SOLID SLAB BRIDGE DESIGN)
I. Hypothesis Structure Clear Span Length TotalWidth W Roadway Wr Wearing Surface dw Width of Barrier B
L= W= Wr = dw = B=
Density of Wearing Surface Concrete Density Concrete Strength Reinforcement Strength Elastic Modulus of Concrete Elastic Modulus of Steel
ρw = 2250 ρc = 2400 f'c = 30 fy = 400 Ec = 27691.5 Es = 200000
14000 10760 10000 75 380
mm mm mm mm mm
Materials kg/m3 kg/m3 Mpa Mpa Mpa Mpa
L
SOIL
SOIL
ELEVATION VIEW
W B
Wr
B
h
Future Wearing Surface
B
Wr/2
W
Wr
Wr/2
B
SECTION VIEW
TOP VIEW
II. Determine the section of bridge Check minimum recommended depth [Table A2.5.2.6.3-1]
Institut de Technologie du Cambodge
hmin =
Projet du pont
1.2[S mm + 3000mm] 30
huse =
Thus, We take
680
=
680
mm
mm
III. Calcul du hourdis Il y a deux méthodes: méthode d'empérique et méthode de basée sur analyse struturale. Dans ce projet, on ultilise la méthode de basée sur analyse struturale.
IV. Choisir les facteurs de résistence, ϕ Strengt Limit State Flexural and Tension Shear and Torson Axial Compression Bearing on Concrete Compression in Strut-and-Tie Models
0.9 0.9 0.75 0.7 0.7
V. Select Load Modifier, η [A1.3.2.1]
1. Ductility, Dη 2. Redundancy, ηR 3. Importance ηI 4. η = ηD.ηR. ηI
Strength Service 0.95 1.00 1.05 1.00 1.05 N/A 1.05 1.00
Fatigue 1.00 1.00 N/A 1.00
VI. Combinaison des Charges Strength I Limit State U = η ⋅[ γ ⋅DC +γ ⋅DW +γ ⋅ (LL + IM)] Sevice I Limit State U =1×[1×DC +1×DW +1×(LL + IM)] Fatigue Limite State U = 0.75 (LL+IM)
γDC γDW γLL IM
Coefficient Strength I Limite State 1.25 1.50 1.75 0.33
Service Limite State 1.00 1.00 1.00
VII. Determine Live Load Strip Width [Table A.2.5.2.6.3-1]
Fatigue 0.75 0.75
Institut de Technologie du Cambodge A. Calculer Interior strip 1. One-Lane Loaded
Projet du pont
Multiple presence factor included [C4.6.2.3]
E = equivalent width (mm) ESint = 0.250 + 0.42* SQRT(L1 ⋅W1) Where
L1 = modified span length L1 = min(L,18m) =
14000
mm
W1 = modified edge-to-edge width W1 = min(W,9m) = ESint =
Thus
9000
mm
4964
mm
2. Mutliple Lane Loaded 𝐸 𝑀 𝑖𝑛𝑡 = 2.10 + 0.12 𝐿1 ∗ 𝑊1 ≤ Where
𝑊 𝑁𝐿
L1 = min(L,18m) =
14000
mm
W1 =min(W,18m) =
10760
mm
W = actual edge-to-edge =
10760
NL = number of design lane A [3.6.1.1.1] Wr = Clear width of road way equal to =
We take:
NL =
2.78
NL =
2
mm 𝑁𝐿 = 𝐼𝑁𝑇
10000
𝑊𝑟 3600
[3.6.1.1.1]
mm
EMint = 3572.826 mm Thus :
The Considered Interior Strip is taken
Eint =
3. Multiple Presence Factor Number of Lane Load 1 2 More than or equal 3
Multiple Presence Factor 1.2 1 0.85
3573
mm
Institut de Technologie du Cambodge
Projet du pont
B. Calculer Exterior Strip [A4.6.2.1.4] According to clause [4.6.2.1.4b], the effective width of a strip without edge beam is the sum of distance between the edge of the deck and theinside face of the barrier, plus 300mm, plus one-quarter of the strip width specified in Article 4.6.2.1.3 and also 4.6.2.3 or 4.6.2.10 but not exceeding either one-haft the full strip width or 1800mm. For simplification, we consider an exterior strip width for Edge =
Thus, the exterior strip width :
1800
Eedge =
1800
mm
mm
VIII. Applicability ofLive Load for Decks and Deck Systems Slab-type bridges shall be designed for all of vehiclar live loads sprcified in AASHTO section [A3.6.1.2], including the lane load[A3.6.1.3.3]. 1. Maximum Shear Force - Axle Loads on Slab Bridge [3.6.1.2.2]
Design Truck Load 145kN
L1
145kN
Lk=4.3to 9m
4.3m
35kN
L
design Tendem Load
[3.6.1.2.3]
110kN
110kN 1.2m
L
Design Lane Load
[3.6.1.2.4] 9.3 kN/m
L
4.3m
Observed Point at
9.3 kN/m
Institut de Technologie du Cambodge
Projet du pont L
1.1. Influence Line for Shear Force 145kN
145kN
Design for Truck Lk=4.3to 9m
L1
4.3m
35kN
4.3m
Lm1 Lm2 Lm3
Vtrappuis We take:
𝑃−
= η1*P1 + η2*P2 + η3*P3
Lk = P1= P2= P3= Vtrappuis=
4.3
m
145 145 35
kN kN kN
𝜂=
(𝑃 ∗ 𝐿𝑚 ) 𝐿𝑡 𝑃
where, Lk can vary from 4.3m ~ 9m
Figure3.6.1.2.1-1
Lm,1= Lm,2= Lm,3=
0 4.3 8.6
m m m
η1= η2= η3=
1.00 0.69 0.39
Lm1= Lm2=
0 1.2
m m
η1= η2=
1.00 0.91
258.96 kN
Design for Tendem Vtenappuis = η1*P1 + η2*P2 P1= P2=
110 110
kN kN
Vteappuis = 210.5714 kN Design For Lane q= 9.3 kN/m l 65.1 V appuis =1/2*q*L = Maximum Shear Force:
kN Vmax = 258.96 kN
The static effect design truck or tendem shall be increased by percentage of IM ( Dynamic Load Allowance), but not be applied to the design pedestrain load or to the design lane load. Table 3.6.2.1-1 Dynamic Load Allowance, IM Component IM Deck-ts_All Limit States 75% All Other Components 1/ Fatigue and Fracture Limite State 15%
Institut de Technologie du Cambodge
Projet du pont
2/ All Other Limit State 33% The factor to be applied to the static load shall be taken as: (1+IM) VLL+IM =
409.5225 kN
Where
Thus: Shear Design is taken
IM=
V = 409.5225 kN
1.2. Maximum Bending Moment at Mid-Span - Axle Load 1.2.1. Design Load of Truck L/2 = 7 m Influence Line at L/2
We observe at L/4 & L/2
145kN
145kN
Lk =4.3to 9m
L1
33%
4.3m
35kN
4.3m
Lm1 Lm2=L/2 Lm3
P1= P2= P3= M( @ L/2 )=
145 145 35
kN kN kN
750.5
kN
Influence Line at L/4
η1= η2= η3=
1.35 3.5 1.35
m
L=
14
m m m
η1= η2= η3=
0 2.625 1.55
Lm1= Lm2= Lm3=
2.7 7 11.3
m m = L/2 m
L/4 =
3.5 145kN
145kN 35kN 4.3m
Lk=4.3 ~ 9m
Lm2=L/4 Lm3
P1= P2= P3=
35 145 145
kN kN kN
M( @ L/4 )= 605.375 kN.m Mtruck =
750.5
kN.m
1.2.2. Design Load of Tendem
Lm1= Lm2= Lm3=
N/A 3.5 7.8
m
Institut de Technologie du Cambodge
Projet du pont
Influence Line at L/2 110kN
110kN 1.2m
L/2 L
P1= P2= ML/2 =
110 110 704
kN kN kN.m
Lm1= Lm2=
7 8.2
m m
η1= η2=
3.5 2.9
η1= η2=
2.63 2.33
Influence Line at L/4 110kN
110kN 1.2m
L/4 L
P1= P2=
110 110
kN kN
ML/4 =
545
kN.m
Lm1= Lm2=
3.5 4.7
m m
Maximum Moment @ L/2:
ML/2 = MAX (ML/2Ten,ML/2Tr) = 750.50 kN.m
Maximum Moment @ L/4:
ML/4 = MAX (ML/2Ten,ML/2Tr) =
605.4
kN.m
1.2.3. Design of Lane Load q= General Equation:
9.3
kN/m
M(x) = 1/2*q*L*x - 1/2*q*x2 M(L/4)=
170.89
kN.m
M(L/2)=
227.85
kN.m
The Static Effect for Moment is taken (1+IM), but not be applied for lane load deign [3.6.2.1] We use : At L/2 :
IM=
33%
MLL+IM = (1+IM) Mmax + M(L/2) =
1226.02 kN.m
Institut de Technologie du Cambodge
Projet du pont
MLL+IM = (1+IM) Mmax + M(L/4) =
At L/4 :
976.04 kN.m
IX. Live Load Force Effect 1. Interior Strip Shear and Bending Moment are given in VIII is per multiple strip Eint where Eint = 𝑉𝐿𝐿+𝐼𝑀 =
𝑉𝐿𝐿+𝐼𝑀 𝐸𝑖𝑛𝑡
VLL+IM =
114.62
kN
𝑀𝐿𝐿+𝐼𝑀 =
𝑀𝐿𝐿+𝐼𝑀 𝐸𝑖𝑛𝑡
MLL+IM =
343.15
kN.m
At L/2
MLL+IM =
273.18
kN.m
At L/4
3573 mm
2. Exterior Strip Eext =
Since
1800
mm
One lane width was considered is 3600mm, but Eedge = Eext =
𝑉𝐿𝐿+𝐼𝑀 =
1 𝑉𝐿𝐿+𝐼𝑀 𝑥1.2 2 𝐸𝑖𝑛𝑡
𝑀𝐿𝐿+𝐼𝑀 =
1 𝑀𝐿𝐿+𝐼𝑀 𝑥1.2 2 𝐸𝑖𝑛𝑡
VLL+IM =
136.51
kN
MLL+IM =
408.67
kN.m
At L/2
MLL+IM = 325.34542 kN.m
At L/4
1800
X. Force Effects Other Load 1. Interior Strip, 1m of width DC: 𝑤𝐷𝐶,𝑖𝑛𝑡 = 𝜌𝑐 ∗ 𝑔 ∗ ℎ = 𝑉𝐷𝐶 = 𝑀𝐷𝐶
1 ∗ 𝑤𝐷𝐶,𝑖𝑛𝑡 𝐿 = 2 1 = ∗ 𝑤𝐷𝐶 𝐿2 = 8
16.32
114.24 kN per m 399.84 kN.m per m
DW: Bituminous Wearing Surface, 𝑤𝐷𝑊,𝑖𝑛𝑡 = 𝜌𝐷𝑊 ∗ 𝑔 ∗ ℎ𝐷𝑊 = 1 ∗ 𝑤𝐷𝑊,𝑖𝑛𝑡 ∗ 𝐿 = 2 1 = ∗ 𝑤𝐷𝑊,𝑖𝑛𝑡 𝐿2 = 8
𝑉𝐷𝑊 = 𝑀𝐷𝑊,𝑖𝑛𝑡
2. Exterior Strip, 1m Wide
kN/m per m g = 1m of width strip
75
1.69
kN/m2
11.81
kN per m
41.34
kN.m per m
mm
mm
Institut de Technologie du Cambodge
Barrier =
4.65
Projet du pont
kN/m
Assume that barrier load spread overwidth of one-half strip at edge. DC :
𝑤𝑏𝑎𝑟𝑟𝑖𝑒𝑟 = 𝐸𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟
18.903
kN/m per m
1 ∗ 𝑤𝐷𝑐𝐶𝑖𝑛𝑡 ∗ 𝐿 = 2
132.32
kN per m
1 ∗ 𝑤𝐷𝑐𝐶𝑖𝑛𝑡 ∗ 𝐿2 8
463.13
kN.m per m
𝑤𝐷𝐶,𝑒𝑥𝑡 = 𝑤𝐷𝐶,𝑖𝑛𝑡 + 𝑉𝐷𝐶,𝑒𝑥𝑡 =
𝑀𝐷𝐶,𝑒𝑥𝑡 =
DW : 𝑤𝐷𝑊,𝑒𝑥𝑡 = 𝑤𝐷𝑤,𝑖𝑛𝑡 +
(𝐸𝑒𝑥𝑡 − 𝐷𝑏𝑎𝑟𝑟𝑖𝑒𝑟 ) = 𝐸𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟
VDW =
9.319
kN per m
MDW =
32.62
kN.m per m
1.33
kN/m
XI. Investigate Service Limit State 1. Durabilities [A5.12] Cover for Unprotected Main Reinforceing Steel(mm), Ref. [Table 5.12.3-1] such as below: 1. Deck surfaces subject to tire stud or chain wear is 2. Bottom Clear Cover of Slab 3. Bar Diameter in use Effective Depth:
30
mm
DB32
mm
634
mm
60
mm
a). Moment - Interior Strip M interior = η Σγ i Q i = 1.[1.M DC + 1.M DW + 1.M LL+IM ] =
784.33 kN.m
At L/2
M Interior = η Σγ i Q i = 1.[1.M DC + 1.M DW + 1.M LL+IM ] =
714.37 kN.m
At L/4
Trail Reinforcement: 𝐴𝑠 =
Assume:
j= fs=2/3fy=
𝑀 𝑓𝑠 𝑗𝑑
0.875 240.00 Mpa
Cracking Control
Institut de Technologie du Cambodge
Projet du pont
At L/2 :
As(L/2)= 5891.04 mm2/m n= 8 DB32 @124 As(L/4)= 5365.53 mm2/m n= 7 DB32 @141 Thus, Area of reinforcement at interior strip is taken:As,interior = 5891.04 mm2 / m b). Moment - Exterior Strip As formula above: Mexterior =
904.42 kN.m per m
at L/2
Mexterior =
821.09 kN.m per m
at L/4
As = 6792.992 mm2/m
n=
9
DB32 @110
As = 6167.138 mm2/m
n=
8
DB32 @124
As,exterior = 6792.99 mm2 / m Thus, Area of reinforcement at exterior strip is taken: [5.7.3.4]
2. Control of Cracking 𝑍
𝑓𝑠 ≤ 𝑓𝑠𝑎 =
𝑑𝑐 ∗ 𝐴
1 3
≤ 0.6𝑓𝑦
a. Interior Strip - Checking Tensile Stress Against f r [5.4.2.6, 5.7.3.4] Minterior = with
b= 𝑓𝑐 =
784.33 kN.m 1.00 m 𝑀
1 2 6∗𝑏∗ℎ
=
𝑓𝑟 = 0.8. (0.63 ∗ 𝑓 ′ 𝑐) = We find :
fc
>
10.18
Mpa
2.76
Mpa
0.8.fr ==> SECTION IS CRACKED
Elastic -Cracked Section : [5.7.1] 𝑛=
𝐸𝑠 = 𝐸𝑐
7.22
Modular ratio
As = 5891.04 mm2/m within
M=
Modulus of rupture [5.4.2.6]
784.33 kN.m
Institut de Technologie du Cambodge
Projet du pont
n.As= 42547.73 mm2/m
d-x
d
x
B= 1 m
Strain
Stress
Find the location of neutral axis (x) 1 2 𝑏𝑥 = 𝑛𝐴𝑠(𝑑 − 𝑥) 2 𝑥2 +
2𝑛𝐴𝑠 2𝑛𝐴𝑠𝑑 𝑥− =0 𝑏 𝑏 𝑥 2 + 2𝐵𝑥 − 𝑐 = 0 B = 42.54773 c = 53950.52
𝑥 = −𝐵 + 𝐵2 + 𝐶 =
193.5896 mm
Moment of Inertia of Cracked Section: 1 𝐼𝑐𝑟 = 𝑏𝑥 3 + 𝑛𝐴𝑠 𝑑 − 𝑥 3
2
= 1.067E+10 mm4 / m
Steel Stress: 𝑓𝑠 𝑀 𝑑 − 𝑥 = = 𝑛 𝐼𝑐𝑟 fs=
32.37
Mpa
233.8
Mpa
Z = 30000N/mm, It's in moderate exposure condition Z = 23000N/mm, It's in servere exposure condition We take :
Z=
23000
N/mm
dc = is concrete cover measured from the extreme tension fiber to the center of nearest bar, but not be larger than 50mm.
Institut de Technologie du Cambodge
===>
dc =
46
n=
8
b=
1000
Projet du pont
mm DB32
@124
Number of reinforcement of Interior strip
mm ( Considered strip width)
2𝑑𝑐 𝑏 = 11500 mm2 𝑛 𝑍 240 𝑓𝑠𝑎 = 1 ≤ 0.6𝑓𝑦 = 𝑑𝑐 ∗ 𝐴 3 fsa > 𝐴=
USE
DB32
Mpa fs
OK
@ 124
b. Exterior Strip, crack tensil stress Mexterior =
904.42 kN.m per m
As = 6792.992 mm2 n.As = 49061.99 mm2 B =
49.06
C = 62210.6 x=
205.14 mm
Moment of Inertia Cracked Section b=
1000
mm
Icr = 1.19E+10 mm4 / m fs = 235.3875 Mpa n=
9
DB32
A = 10222.22 mm2 fsa =
240 fs < fsa
Mpa OK
@124
Number of Reinforcement at Exterior Strip
Institut de Technologie du Cambodge USE:
DB32
Projet du pont
@124
3. Deformations [5.7.3.6]. a. Dead Load Camber [A2.5.2.6]
𝑤𝐷𝐿,𝑡𝑜𝑡𝑎𝑙 = 2 ∗ 𝐸𝑒𝑥𝑡 (𝑤𝐷𝐶,𝑒𝑥𝑡 + 𝑤𝐷𝑊,𝑒𝑥𝑡 ) + 𝐸𝑖𝑛𝑡,𝑡𝑜𝑡𝑎𝑙 (𝑤𝐷𝐶,𝑖𝑛𝑡 + 𝑤𝐷𝑊,𝑖𝑛𝑡 ) + 2 ∗ 𝑤𝑏𝑎𝑟𝑟𝑖𝑒𝑟 Eint,tot =
6.4
m
WDC,ext = 18.903 kN/m per m
Eext =
1.8
m (Fixed)
WDC,int =
16.32
kN/m per m
WDW,ext =
1.33
kN/m per m
WDW,int =
1.69
kN/m per m
Wbarrier =
4.65
kN/m per m
WDL,total = 197.3925 kN/m full width
𝑀𝐷𝐿 =
1 ∗ 𝑤𝐷𝐿,𝑡𝑜𝑡𝑎𝑙 ∗ 𝐿2 = 8
4836.1
kN.m
By using Ie [A5.7.3.6.2]
5 𝐿4 ∆𝐷𝐿 = ∗ 𝑤𝐷𝐿,𝑡𝑜𝑡𝑎𝑙 ∗ 384 𝐸𝑐 𝐼𝑒 𝑀𝑐𝑟 𝐼𝑒 = 𝑀𝑎 𝑀𝑐𝑟 = 𝑓𝑟 ∗
3
∗ 𝐼𝑔 + (1 −
𝑀𝑐𝑟 𝑀𝑎
3
𝐼𝑐𝑟 ≤ 𝐼𝑔
𝐼𝑔 𝑦𝑏
𝑓𝑟 = 0.63 ∗ 𝑓 ′ 𝑐 =
3.4507 Mpa
Ig = 2.82E+11 mm4 Mcr =
2861.4 kNm
Ma =
3401.0 kNm
fs =
240
Mpa
Eint,total =
6.4
m
Icr,interior = 1.07E+10 mm4 / m
As,inter = 5891.04 mm2 / m
Eexterior =
1.8
m
Icr,exterior = 1.19E+10 mm4 / m
As,exter = 6792.99 mm2 / m
Institut de Technologie du Cambodge
Projet du pont
𝐼𝑐𝑟,𝑡𝑜𝑡𝑎𝑙 = 2 ∗ 𝐸𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 ∗ 𝐼𝑐𝑟,𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 + 𝐸𝑖𝑛𝑡,𝑡𝑜𝑡𝑎𝑙 ∗ 𝐼𝑐𝑟,𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 Ie = 2.1E+11 mm4
ΔDL,Ig =
===>
Long-term deflection factor for A's = 0, is equal Camber = K*ΔDL = 50.25 mm
16.75
Icr,total = 1.1E+11 mm4 mm
K = 3-1.2 (A's/As)=
Short-terme 3
By Using Ig [5.7.3.6.2] ∆𝐷𝐿,𝐼𝑒 = ∆𝐷𝐿,𝐼𝑔 ∗
𝐼𝑒 = 𝐼𝑔
12.65
Long-term deflection factoris take =
mm 4
Camber = 50.5869 mm Thus: the maximum camber is 50.59
mm
b. Live Load Deflection [2.5.2.6.2]
∆𝑎𝑙𝑙𝑜𝑤 𝐿𝐿+𝐼𝑀 =
𝑠𝑝𝑎𝑛 = 800
17.5
mm
Use design truck alone or design lane load plus 25% truck load [A.6.1.3.2]. When design truck alone, it should be placed so that the distance between its resultant and the nearest wheel is bisected by the span centerline. All design lanes should be loaded: [A2.5.2.6.2]. NL =
2
m=
0.85
(1+IM) =
1.33
Number of considered lane
Impact Factor
The value of Ie changes with the magnetude of the applied moment Ma. The moment associated with the live load deflection includes the dead load moment plus the truck moment. Case1: Design for only Truck Load 𝑀𝐷𝐿+𝐿𝐿+𝐼𝑀 = 𝑀𝐷𝐿 + 𝑚 ∗ 1 + 𝐼𝑀 ∗ 𝑀𝑇𝑟𝑢𝑐𝑘 MDL = Mtruck =
4836.1 kN.m 750.5
kN.m
Institut de Technologie du Cambodge
Projet du pont
MDL+LL+IM = 6532.997 kN.m Ig = 2.82E+11 mm4
fr = 3.45065 Mpa
Mcr = 2861.41 kN.m Ma = MDL+LL+IM = 6532.997 kN.m (Mcr / Ma )3 = 0.084024 Ie = 1.3E+11 mm4 Ec*Ie = 3.5E+15 N.mm2 We take load of truck to consider For
P1,2 =
145
kN
= 327.845 kN
For
P3 =
35
kN
=
𝑥𝐺 =
79.135
kN
𝑃1 ∗ 𝑥1 + 𝑃2 ∗ 𝑥2 + 𝑃3 ∗ 𝑥3 𝑃1 + 𝑃2 + 𝑃3 145kN
145kN
Lk =4.3to 9m
4.3m
35kN
L CL
4.3m
Lk = 4.3 ~ 9m
P3
P2
R
P1
a
a xG
Mmax= ? kNm
P3
P2
R
P1
Institut de Technologie du Cambodge a
Projet du pont
a xG
Mmax= ? kNm
- For Front Wheel P3 P3=118.7kN
x
b P3 = x1 = x2 = x3 =
79.135 0 4.3 8.6
∆1 = 𝑃3 ∗ 𝑏 ∗ Δ_P3 =
0.547
kN m m m
Then
𝑥 𝐿2 − 𝑏 2 − 𝑥 2 6𝐸𝐼𝐿
xG = a= b= x=
5.76 0.728 1.972 7.73
mm
For Middle Wheel P2 P2=491.8 kN
x
b P2 = 327.845 kN
Δ_P2 =
5.28
b= x=
6.27 7.73
m m
mm
For Behind Wheel P1 P2=491.8 kN
x
b P1 =
327.85 kN
b = x=
3.43 6.27
m m
m m m m
Institut de Technologie du Cambodge Δ_P1 =
3.50
Projet du pont
mm
Thus : Deflection due to Truck : ΔmaxTruck =
9.32
mm
<
ΔLL+IMallowable
Deflection is not a problem Case 2: Design for Only Land plus 25% of Truck Load 5 𝐿4 ∆𝐿𝑎𝑛𝑒 = ∗𝑞 ∗ 384 𝐿𝑎𝑛𝑒 𝐸𝑐 𝐼𝑒
𝑞𝐿𝑎𝑛𝑒 = 1 + 𝐼𝑀 ∗ 𝑤𝐿𝑎𝑛𝑒 ∗ 𝑚 ∗ 𝑁𝐿
qLane = 21.0273 kN/m Δlane =
3.03
mm
Δ25% of Truck =
2.331
mm
Δ25% of Truck+Lane =
5.358
mm
OK
Case3: Design for Tendom Cencentrated Load at mid span We consider tendom load as a cencentrated load located at mid span for having a critical deflection P=110kN
L/2
∆ 𝑇𝑒𝑛𝑑𝑜𝑚 = qtendom = Δtendom =
L/2
1 𝐿3 ∗ 𝑞𝑡𝑒𝑛𝑑𝑜𝑚 ∗ 48 𝐸𝑐 𝐼𝑒 497.42 kN 8.2
mm
OK
XII. Investigate Strength Limite State 1. Flexural Rectagular Stress Distribution [5.7.2.2] & [5.7.3.2] f'c =
30
β=
0.84
a). Interior Strip
Mpa
Institut de Technologie du Cambodge
Projet du pont
Strength I. Minterior = η*Σγi.Qi = 1.05*[1.25*MDC + 1.5*MDW + (1+IM )*MLL+IM ] MDC =
399.84 kNm per m
MDW =
41.34
MLL+IM =
343.15 kNm per m
1+IM =
133%
kNm per m
Minterior = 1069.12 kNm per m As,inter = 5891.04 mm2 / m for service limite state b= 𝑐=
1000
mm
consider 1m of strip
𝐴𝑆 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 𝛽 ∗ 𝑏
c = 110.574 mm a = β*c = 92.40854 mm ds=
634
mm
c/ds = 0.174407 < 0.42 ratio of c and ds is =
Effective depth of reinforcement Allowable maximum reinfrocement [ 5.7.3.3.1 ]
OK
ρ=As/(bd)= 0.009292 reiforcement percentage Mn =As.fy.(ds-a/2) = 1385.092 kN.m Φ=
0.9
ΦMn= 1246.583 kNm
Thus:
Mu,interior
>
OK
Arrangement of Longitudinal Reinforcement In Interior Strip is:DB32
b). Edge Strip Mexterior = η*Σγi.Qi = 1.05*[1.25*MDC + 1.5*MDW + (1+IM )*MLL+IM ] MDC =
463.13 kNm
MDW =
32.62
kNm
@124
Institut de Technologie du Cambodge MLL+IM =
Projet du pont
408.67 kNm
Mexterior = 1229.94 kNm Asedge = b= 𝑐=
6792.992 mm2 / m 1000
mm
𝐴𝑆 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 𝛽 ∗ 𝑏
c = 127.504 mm a = β*c = 106.5567 mm ds=
634
mm
c/ds = 0.20111 < 0.42 ratio of c and ds is =
Effective depth of reinforcement Allowable maximum reinfrocement [ 5.7.3.3.1 ]
OK
ρ=As/(bd)= 0.010714 reiforcement percentage Mn =As.fy.(ds-a/2) = 1577.935 kN.m Φ=
0.9
ΦMn= 1420.14 kNm
Mu,edge
>
OK
Thus: Arrangement of Longitudinal Reinforcement at Exterior Strip is: DB32
@110
XIII. Distribution Reinforcement of the Deck The amount of bottom transverse reinforcement may be takent as a percenage of the main reinforcement required for positive moment as 14.8(%)
<=
50%
a) Interior Strip ρ= As,inter =
0.148 5891.04 mm2 / m
1750 𝐿
[5.14.4.1-1]
% ≤ 50%
Institut de Technologie du Cambodge As,transv = ρ*As,inter =
Projet du pont
871.30 mm2 / m
DB16 @188.0
b). Edge Strip ρ=
0.148
As,edge = 6792.992 mm2 / m As,transv = 1004.697 mm2 / m
DB16 @188.0
XIV. Shrinkage or Temperature Reinforcement [5.10.8] The shrinkage or temperature reinforcement shall not be spaced farther than 450mm or 3 time of compenent thickness. Ash > 0.11*Ag / fy =
187
mm2 / m
DB10 @313.3
In the both directions
XVI. Fatigue Limite State U = 0.75*(LL+IM)
IM =
15%
Fatigue load shall be one design truck with 9000mm axial spacing no multiple present factor applied (m =1).
4.3 m P1
P1 =
35
9m P2
kN
P2
`P2 =
145
kN
P3 =
145
kN
fmax < fallowable fallowable = 145-0.33*fmin +55(r/h) where fmin =0, because deck is treated as a simple beam, thus, there there no moment reversal. fmin =
0
r/h=
0.3
Institut de Technologie du Cambodge
Projet du pont
ff =
161.5
Mpa
1+IM =
1.15
for fatigue condition
η=
1.00
γ=
0.75
CL x R
P1
P3
P2
C
A
B
L/2+x/2
L/2-x/2
x1
x2
L=
14000
mm
P1=
35
kN
P2 =
145
kN
x=
0.836
m
x2 =
4.3
m
x1 =
3.12
m
ΣMA =0,
RB =
x4
x4 =
6.5819
m
84.625 kN
Mmax = Mc = 556.997 kNm Then
Mfatigue = ηΣγiQi=
480.41 kNm
- Tensile Load Stress fmax 𝑓𝑚𝑎𝑥 𝑀𝑚𝑎𝑥 𝑑 − 𝑥 = 𝑛 𝐼𝑐𝑟 d= x=
634
where n=
7.22
mm
As =
193.59 mm
b =
6792.99 mm2 / m 1000
mm
Institut de Technologie du Cambodge
Projet du pont
Cracking around reinforcement DB25, then cracking inertia is determined: Φ= 𝐼𝑐𝑟 =
25
mm
𝑏𝑥 3 12𝜋Φ4 𝑥2 + + 𝑏𝑥 + 𝑛𝐴𝑠 𝑑2 12 64 4 fmax = 80.02565 Mpa
<
= ff
2.2E+10 mm4 OK
XVI. Resume the answer Calculation Result 1. Longitudinal Reinforcement at the bottom at exterior strip: DB32 @ 110 2. Longitudinal Reinforcement at the bottom at interior strip: DB32 @ 124 3. Transversal Reinforcement at the bott at exterior&interior strip: DB16 @ 188 4. Longitudinal Reinforcement at the top at exterior strip: DB10 @ 313.3 5. LongitudinalReinforcement at the top at interior strip: DB10 @ 313.3 4. Transversal Reinforcement at the top at exterior & interior strip: DB10 @ 313.3
Actual Use @ 100 @ 120 @ 180 @ 300 @ 300 @ 300
XVII. Conclusion Après avoir fait ce projet du pont, il nous permet d'avoir le concept, le dimesionnement et le calcul sur le pont dalle de béton armé (solide slab bridge) par le norme AASHTO. En plus,il nous donne aussi un model pour calculer d'autres types du pont. Dans ce projet, on n'a calculé que les sections d'armatures de la portée L/2 et L/4. Pour qu'on peut économiser l'argent sur l'utilisation d'armatures, on doit calculer plusieurs points sur cet portée. Et on a vérifier seulement le milieu de la portée L/2, le point le plus critique qu'on doit considérer comme le flèche et le moment. On n'a pas vérifier l'effort tranchant parce qu'il n'est pas le point critique pour le pont dalle grâce au largeur de la dalle importance. En conclusion, on a adopté sur la vérification et le calcul de ce pont dalle en béton armé.
380mm
680mm
Eext = 1800mm
DB32@120
DB10@300
DB32@100
DB10@300
Eint = 6400 mm
Wearing Surface(75Thk)
DB16@180
DB10@300
Eext = 1800mm
DB32@120
DB10@300
380mm
XVIII.
Sketch of Bridge Section in Rebar Detail
Page 23
KAING Saoserey
STUDENTS :
I5-A-GCI
1. CHHENG Chanraksmey
e20130072
2. CHHET Hychhy
e20130077
3. CHHIN Chanbormey
e20130079
4. CHHOENG Oudom
e20130084
5. CHHOM Kanhara
e20130087
6. CHHON Chhour
e20130088
2017-2018
Institut de Technologie du Cambodge
Projet du pont CONTENTS
I. Hypothesis: ......................................................................................................................Page: Page0101 II. Determine the Section of Bridge:...................................................................................Page: Page0202 III. Calcul du hourdis: .........................................................................................................Page: Page0202 IV.Choisir les facteur de résistance : .................................................................................Page: Page0202 V. Select Load Modifier: ....................................................................................................Page: Page0202 VI. Combination: .................................................................................................................Page: Page0202 VII. Determine Live Load Strip Width: ...............................................................................Page: Page02 02 VIII. Applicability ofLive Load for Decks and Deck Systems: ...........................................Page: Page04 04 IX. Live Load Force Effect:.................................................................................................Page: Page0808 X. Force Effects Other Load ...............................................................................................Page: Page0808 XI. Investigate Service Limit State.......................................................................................Page: Page09 09 XII. Investigate Strength Limite State:..................................................................................Page: Page17 17 XIII. Distribution Reinforcement of the Deck:.....................................................................Page: Page19 19 XIV. Shrinkage or Temperature Reinforcement:.................................................................Page: Page2020 XV. Fatigue Limite State:....................................................................................................Page: Page2020 XVI. Resume the answer:....................................................................................................Page: Page2222 XVII. Conclusion:...............................................................................................................Page: Page2222 XVIII. Sketch of Bridge Section in Rebar Detail:...............................................................Page: Page2323
Page 1
Institut de Technologie du Cambodge
Projet du pont
PONT DALLE EN BÉTON ARMÉ (SOLID SLAB BRIDGE DESIGN)
I. Hypothesis Structure Clear Span Length TotalWidth W Roadway Wr Wearing Surface dw Width of Barrier B
L= W= Wr = dw = B=
Density of Wearing Surface Concrete Density Concrete Strength Reinforcement Strength Elastic Modulus of Concrete Elastic Modulus of Steel
ρw = 2250 ρc = 2400 f'c = 30 fy = 400 Ec = 27691.5 Es = 200000
14000 10760 10000 75 380
mm mm mm mm mm
Materials kg/m3 kg/m3 Mpa Mpa Mpa Mpa
L
SOIL
SOIL
ELEVATION VIEW
W B
Wr
B
h
Future Wearing Surface
B
Wr/2
W
Wr
Wr/2
B
SECTION VIEW
TOP VIEW
II. Determine the section of bridge Check minimum recommended depth [Table A2.5.2.6.3-1]
Institut de Technologie du Cambodge
hmin =
Projet du pont
1.2[S mm + 3000mm] 30
huse =
Thus, We take
680
=
680
mm
mm
III. Calcul du hourdis Il y a deux méthodes: méthode d'empérique et méthode de basée sur analyse struturale. Dans ce projet, on ultilise la méthode de basée sur analyse struturale.
IV. Choisir les facteurs de résistence, ϕ Strengt Limit State Flexural and Tension Shear and Torson Axial Compression Bearing on Concrete Compression in Strut-and-Tie Models
0.9 0.9 0.75 0.7 0.7
V. Select Load Modifier, η [A1.3.2.1]
1. Ductility, Dη 2. Redundancy, ηR 3. Importance ηI 4. η = ηD.ηR. ηI
Strength Service 0.95 1.00 1.05 1.00 1.05 N/A 1.05 1.00
Fatigue 1.00 1.00 N/A 1.00
VI. Combinaison des Charges Strength I Limit State U = η ⋅[ γ ⋅DC +γ ⋅DW +γ ⋅ (LL + IM)] Sevice I Limit State U =1×[1×DC +1×DW +1×(LL + IM)] Fatigue Limite State U = 0.75 (LL+IM)
γDC γDW γLL IM
Coefficient Strength I Limite State 1.25 1.50 1.75 0.33
Service Limite State 1.00 1.00 1.00
VII. Determine Live Load Strip Width [Table A.2.5.2.6.3-1]
Fatigue 0.75 0.75
Institut de Technologie du Cambodge A. Calculer Interior strip 1. One-Lane Loaded
Projet du pont
Multiple presence factor included [C4.6.2.3]
E = equivalent width (mm) ESint = 0.250 + 0.42* SQRT(L1 ⋅W1) Where
L1 = modified span length L1 = min(L,18m) =
14000
mm
W1 = modified edge-to-edge width W1 = min(W,9m) = ESint =
Thus
9000
mm
4964
mm
2. Mutliple Lane Loaded 𝐸 𝑀 𝑖𝑛𝑡 = 2.10 + 0.12 𝐿1 ∗ 𝑊1 ≤ Where
𝑊 𝑁𝐿
L1 = min(L,18m) =
14000
mm
W1 =min(W,18m) =
10760
mm
W = actual edge-to-edge =
10760
NL = number of design lane A [3.6.1.1.1] Wr = Clear width of road way equal to =
We take:
NL =
2.78
NL =
2
mm 𝑁𝐿 = 𝐼𝑁𝑇
10000
𝑊𝑟 3600
[3.6.1.1.1]
mm
EMint = 3572.826 mm Thus :
The Considered Interior Strip is taken
Eint =
3. Multiple Presence Factor Number of Lane Load 1 2 More than or equal 3
Multiple Presence Factor 1.2 1 0.85
3573
mm
Institut de Technologie du Cambodge
Projet du pont
B. Calculer Exterior Strip [A4.6.2.1.4] According to clause [4.6.2.1.4b], the effective width of a strip without edge beam is the sum of distance between the edge of the deck and theinside face of the barrier, plus 300mm, plus one-quarter of the strip width specified in Article 4.6.2.1.3 and also 4.6.2.3 or 4.6.2.10 but not exceeding either one-haft the full strip width or 1800mm. For simplification, we consider an exterior strip width for Edge =
Thus, the exterior strip width :
1800
Eedge =
1800
mm
mm
VIII. Applicability ofLive Load for Decks and Deck Systems Slab-type bridges shall be designed for all of vehiclar live loads sprcified in AASHTO section [A3.6.1.2], including the lane load[A3.6.1.3.3]. 1. Maximum Shear Force - Axle Loads on Slab Bridge [3.6.1.2.2]
Design Truck Load 145kN
L1
145kN
Lk=4.3to 9m
4.3m
35kN
L
design Tendem Load
[3.6.1.2.3]
110kN
110kN 1.2m
L
Design Lane Load
[3.6.1.2.4] 9.3 kN/m
L
4.3m
Observed Point at
9.3 kN/m
Institut de Technologie du Cambodge
Projet du pont L
1.1. Influence Line for Shear Force 145kN
145kN
Design for Truck Lk=4.3to 9m
L1
4.3m
35kN
4.3m
Lm1 Lm2 Lm3
Vtrappuis We take:
𝑃−
= η1*P1 + η2*P2 + η3*P3
Lk = P1= P2= P3= Vtrappuis=
4.3
m
145 145 35
kN kN kN
𝜂=
(𝑃 ∗ 𝐿𝑚 ) 𝐿𝑡 𝑃
where, Lk can vary from 4.3m ~ 9m
Figure3.6.1.2.1-1
Lm,1= Lm,2= Lm,3=
0 4.3 8.6
m m m
η1= η2= η3=
1.00 0.69 0.39
Lm1= Lm2=
0 1.2
m m
η1= η2=
1.00 0.91
258.96 kN
Design for Tendem Vtenappuis = η1*P1 + η2*P2 P1= P2=
110 110
kN kN
Vteappuis = 210.5714 kN Design For Lane q= 9.3 kN/m l 65.1 V appuis =1/2*q*L = Maximum Shear Force:
kN Vmax = 258.96 kN
The static effect design truck or tendem shall be increased by percentage of IM ( Dynamic Load Allowance), but not be applied to the design pedestrain load or to the design lane load. Table 3.6.2.1-1 Dynamic Load Allowance, IM Component IM Deck-ts_All Limit States 75% All Other Components 1/ Fatigue and Fracture Limite State 15%
Institut de Technologie du Cambodge
Projet du pont
2/ All Other Limit State 33% The factor to be applied to the static load shall be taken as: (1+IM) VLL+IM =
409.5225 kN
Where
Thus: Shear Design is taken
IM=
V = 409.5225 kN
1.2. Maximum Bending Moment at Mid-Span - Axle Load 1.2.1. Design Load of Truck L/2 = 7 m Influence Line at L/2
We observe at L/4 & L/2
145kN
145kN
Lk =4.3to 9m
L1
33%
4.3m
35kN
4.3m
Lm1 Lm2=L/2 Lm3
P1= P2= P3= M( @ L/2 )=
145 145 35
kN kN kN
750.5
kN
Influence Line at L/4
η1= η2= η3=
1.35 3.5 1.35
m
L=
14
m m m
η1= η2= η3=
0 2.625 1.55
Lm1= Lm2= Lm3=
2.7 7 11.3
m m = L/2 m
L/4 =
3.5 145kN
145kN 35kN 4.3m
Lk=4.3 ~ 9m
Lm2=L/4 Lm3
P1= P2= P3=
35 145 145
kN kN kN
M( @ L/4 )= 605.375 kN.m Mtruck =
750.5
kN.m
1.2.2. Design Load of Tendem
Lm1= Lm2= Lm3=
N/A 3.5 7.8
m
Institut de Technologie du Cambodge
Projet du pont
Influence Line at L/2 110kN
110kN 1.2m
L/2 L
P1= P2= ML/2 =
110 110 704
kN kN kN.m
Lm1= Lm2=
7 8.2
m m
η1= η2=
3.5 2.9
η1= η2=
2.63 2.33
Influence Line at L/4 110kN
110kN 1.2m
L/4 L
P1= P2=
110 110
kN kN
ML/4 =
545
kN.m
Lm1= Lm2=
3.5 4.7
m m
Maximum Moment @ L/2:
ML/2 = MAX (ML/2Ten,ML/2Tr) = 750.50 kN.m
Maximum Moment @ L/4:
ML/4 = MAX (ML/2Ten,ML/2Tr) =
605.4
kN.m
1.2.3. Design of Lane Load q= General Equation:
9.3
kN/m
M(x) = 1/2*q*L*x - 1/2*q*x2 M(L/4)=
170.89
kN.m
M(L/2)=
227.85
kN.m
The Static Effect for Moment is taken (1+IM), but not be applied for lane load deign [3.6.2.1] We use : At L/2 :
IM=
33%
MLL+IM = (1+IM) Mmax + M(L/2) =
1226.02 kN.m
Institut de Technologie du Cambodge
Projet du pont
MLL+IM = (1+IM) Mmax + M(L/4) =
At L/4 :
976.04 kN.m
IX. Live Load Force Effect 1. Interior Strip Shear and Bending Moment are given in VIII is per multiple strip Eint where Eint = 𝑉𝐿𝐿+𝐼𝑀 =
𝑉𝐿𝐿+𝐼𝑀 𝐸𝑖𝑛𝑡
VLL+IM =
114.62
kN
𝑀𝐿𝐿+𝐼𝑀 =
𝑀𝐿𝐿+𝐼𝑀 𝐸𝑖𝑛𝑡
MLL+IM =
343.15
kN.m
At L/2
MLL+IM =
273.18
kN.m
At L/4
3573 mm
2. Exterior Strip Eext =
Since
1800
mm
One lane width was considered is 3600mm, but Eedge = Eext =
𝑉𝐿𝐿+𝐼𝑀 =
1 𝑉𝐿𝐿+𝐼𝑀 𝑥1.2 2 𝐸𝑖𝑛𝑡
𝑀𝐿𝐿+𝐼𝑀 =
1 𝑀𝐿𝐿+𝐼𝑀 𝑥1.2 2 𝐸𝑖𝑛𝑡
VLL+IM =
136.51
kN
MLL+IM =
408.67
kN.m
At L/2
MLL+IM = 325.34542 kN.m
At L/4
1800
X. Force Effects Other Load 1. Interior Strip, 1m of width DC: 𝑤𝐷𝐶,𝑖𝑛𝑡 = 𝜌𝑐 ∗ 𝑔 ∗ ℎ = 𝑉𝐷𝐶 = 𝑀𝐷𝐶
1 ∗ 𝑤𝐷𝐶,𝑖𝑛𝑡 𝐿 = 2 1 = ∗ 𝑤𝐷𝐶 𝐿2 = 8
16.32
114.24 kN per m 399.84 kN.m per m
DW: Bituminous Wearing Surface, 𝑤𝐷𝑊,𝑖𝑛𝑡 = 𝜌𝐷𝑊 ∗ 𝑔 ∗ ℎ𝐷𝑊 = 1 ∗ 𝑤𝐷𝑊,𝑖𝑛𝑡 ∗ 𝐿 = 2 1 = ∗ 𝑤𝐷𝑊,𝑖𝑛𝑡 𝐿2 = 8
𝑉𝐷𝑊 = 𝑀𝐷𝑊,𝑖𝑛𝑡
2. Exterior Strip, 1m Wide
kN/m per m g = 1m of width strip
75
1.69
kN/m2
11.81
kN per m
41.34
kN.m per m
mm
mm
Institut de Technologie du Cambodge
Barrier =
4.65
Projet du pont
kN/m
Assume that barrier load spread overwidth of one-half strip at edge. DC :
𝑤𝑏𝑎𝑟𝑟𝑖𝑒𝑟 = 𝐸𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟
18.903
kN/m per m
1 ∗ 𝑤𝐷𝑐𝐶𝑖𝑛𝑡 ∗ 𝐿 = 2
132.32
kN per m
1 ∗ 𝑤𝐷𝑐𝐶𝑖𝑛𝑡 ∗ 𝐿2 8
463.13
kN.m per m
𝑤𝐷𝐶,𝑒𝑥𝑡 = 𝑤𝐷𝐶,𝑖𝑛𝑡 + 𝑉𝐷𝐶,𝑒𝑥𝑡 =
𝑀𝐷𝐶,𝑒𝑥𝑡 =
DW : 𝑤𝐷𝑊,𝑒𝑥𝑡 = 𝑤𝐷𝑤,𝑖𝑛𝑡 +
(𝐸𝑒𝑥𝑡 − 𝐷𝑏𝑎𝑟𝑟𝑖𝑒𝑟 ) = 𝐸𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟
VDW =
9.319
kN per m
MDW =
32.62
kN.m per m
1.33
kN/m
XI. Investigate Service Limit State 1. Durabilities [A5.12] Cover for Unprotected Main Reinforceing Steel(mm), Ref. [Table 5.12.3-1] such as below: 1. Deck surfaces subject to tire stud or chain wear is 2. Bottom Clear Cover of Slab 3. Bar Diameter in use Effective Depth:
30
mm
DB32
mm
634
mm
60
mm
a). Moment - Interior Strip M interior = η Σγ i Q i = 1.[1.M DC + 1.M DW + 1.M LL+IM ] =
784.33 kN.m
At L/2
M Interior = η Σγ i Q i = 1.[1.M DC + 1.M DW + 1.M LL+IM ] =
714.37 kN.m
At L/4
Trail Reinforcement: 𝐴𝑠 =
Assume:
j= fs=2/3fy=
𝑀 𝑓𝑠 𝑗𝑑
0.875 240.00 Mpa
Cracking Control
Institut de Technologie du Cambodge
Projet du pont
At L/2 :
As(L/2)= 5891.04 mm2/m n= 8 DB32 @124 As(L/4)= 5365.53 mm2/m n= 7 DB32 @141 Thus, Area of reinforcement at interior strip is taken:As,interior = 5891.04 mm2 / m b). Moment - Exterior Strip As formula above: Mexterior =
904.42 kN.m per m
at L/2
Mexterior =
821.09 kN.m per m
at L/4
As = 6792.992 mm2/m
n=
9
DB32 @110
As = 6167.138 mm2/m
n=
8
DB32 @124
As,exterior = 6792.99 mm2 / m Thus, Area of reinforcement at exterior strip is taken: [5.7.3.4]
2. Control of Cracking 𝑍
𝑓𝑠 ≤ 𝑓𝑠𝑎 =
𝑑𝑐 ∗ 𝐴
1 3
≤ 0.6𝑓𝑦
a. Interior Strip - Checking Tensile Stress Against f r [5.4.2.6, 5.7.3.4] Minterior = with
b= 𝑓𝑐 =
784.33 kN.m 1.00 m 𝑀
1 2 6∗𝑏∗ℎ
=
𝑓𝑟 = 0.8. (0.63 ∗ 𝑓 ′ 𝑐) = We find :
fc
>
10.18
Mpa
2.76
Mpa
0.8.fr ==> SECTION IS CRACKED
Elastic -Cracked Section : [5.7.1] 𝑛=
𝐸𝑠 = 𝐸𝑐
7.22
Modular ratio
As = 5891.04 mm2/m within
M=
Modulus of rupture [5.4.2.6]
784.33 kN.m
Institut de Technologie du Cambodge
Projet du pont
n.As= 42547.73 mm2/m
d-x
d
x
B= 1 m
Strain
Stress
Find the location of neutral axis (x) 1 2 𝑏𝑥 = 𝑛𝐴𝑠(𝑑 − 𝑥) 2 𝑥2 +
2𝑛𝐴𝑠 2𝑛𝐴𝑠𝑑 𝑥− =0 𝑏 𝑏 𝑥 2 + 2𝐵𝑥 − 𝑐 = 0 B = 42.54773 c = 53950.52
𝑥 = −𝐵 + 𝐵2 + 𝐶 =
193.5896 mm
Moment of Inertia of Cracked Section: 1 𝐼𝑐𝑟 = 𝑏𝑥 3 + 𝑛𝐴𝑠 𝑑 − 𝑥 3
2
= 1.067E+10 mm4 / m
Steel Stress: 𝑓𝑠 𝑀 𝑑 − 𝑥 = = 𝑛 𝐼𝑐𝑟 fs=
32.37
Mpa
233.8
Mpa
Z = 30000N/mm, It's in moderate exposure condition Z = 23000N/mm, It's in servere exposure condition We take :
Z=
23000
N/mm
dc = is concrete cover measured from the extreme tension fiber to the center of nearest bar, but not be larger than 50mm.
Institut de Technologie du Cambodge
===>
dc =
46
n=
8
b=
1000
Projet du pont
mm DB32
@124
Number of reinforcement of Interior strip
mm ( Considered strip width)
2𝑑𝑐 𝑏 = 11500 mm2 𝑛 𝑍 240 𝑓𝑠𝑎 = 1 ≤ 0.6𝑓𝑦 = 𝑑𝑐 ∗ 𝐴 3 fsa > 𝐴=
USE
DB32
Mpa fs
OK
@ 124
b. Exterior Strip, crack tensil stress Mexterior =
904.42 kN.m per m
As = 6792.992 mm2 n.As = 49061.99 mm2 B =
49.06
C = 62210.6 x=
205.14 mm
Moment of Inertia Cracked Section b=
1000
mm
Icr = 1.19E+10 mm4 / m fs = 235.3875 Mpa n=
9
DB32
A = 10222.22 mm2 fsa =
240 fs < fsa
Mpa OK
@124
Number of Reinforcement at Exterior Strip
Institut de Technologie du Cambodge USE:
DB32
Projet du pont
@124
3. Deformations [5.7.3.6]. a. Dead Load Camber [A2.5.2.6]
𝑤𝐷𝐿,𝑡𝑜𝑡𝑎𝑙 = 2 ∗ 𝐸𝑒𝑥𝑡 (𝑤𝐷𝐶,𝑒𝑥𝑡 + 𝑤𝐷𝑊,𝑒𝑥𝑡 ) + 𝐸𝑖𝑛𝑡,𝑡𝑜𝑡𝑎𝑙 (𝑤𝐷𝐶,𝑖𝑛𝑡 + 𝑤𝐷𝑊,𝑖𝑛𝑡 ) + 2 ∗ 𝑤𝑏𝑎𝑟𝑟𝑖𝑒𝑟 Eint,tot =
6.4
m
WDC,ext = 18.903 kN/m per m
Eext =
1.8
m (Fixed)
WDC,int =
16.32
kN/m per m
WDW,ext =
1.33
kN/m per m
WDW,int =
1.69
kN/m per m
Wbarrier =
4.65
kN/m per m
WDL,total = 197.3925 kN/m full width
𝑀𝐷𝐿 =
1 ∗ 𝑤𝐷𝐿,𝑡𝑜𝑡𝑎𝑙 ∗ 𝐿2 = 8
4836.1
kN.m
By using Ie [A5.7.3.6.2]
5 𝐿4 ∆𝐷𝐿 = ∗ 𝑤𝐷𝐿,𝑡𝑜𝑡𝑎𝑙 ∗ 384 𝐸𝑐 𝐼𝑒 𝑀𝑐𝑟 𝐼𝑒 = 𝑀𝑎 𝑀𝑐𝑟 = 𝑓𝑟 ∗
3
∗ 𝐼𝑔 + (1 −
𝑀𝑐𝑟 𝑀𝑎
3
𝐼𝑐𝑟 ≤ 𝐼𝑔
𝐼𝑔 𝑦𝑏
𝑓𝑟 = 0.63 ∗ 𝑓 ′ 𝑐 =
3.4507 Mpa
Ig = 2.82E+11 mm4 Mcr =
2861.4 kNm
Ma =
3401.0 kNm
fs =
240
Mpa
Eint,total =
6.4
m
Icr,interior = 1.07E+10 mm4 / m
As,inter = 5891.04 mm2 / m
Eexterior =
1.8
m
Icr,exterior = 1.19E+10 mm4 / m
As,exter = 6792.99 mm2 / m
Institut de Technologie du Cambodge
Projet du pont
𝐼𝑐𝑟,𝑡𝑜𝑡𝑎𝑙 = 2 ∗ 𝐸𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 ∗ 𝐼𝑐𝑟,𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 + 𝐸𝑖𝑛𝑡,𝑡𝑜𝑡𝑎𝑙 ∗ 𝐼𝑐𝑟,𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 Ie = 2.1E+11 mm4
ΔDL,Ig =
===>
Long-term deflection factor for A's = 0, is equal Camber = K*ΔDL = 50.25 mm
16.75
Icr,total = 1.1E+11 mm4 mm
K = 3-1.2 (A's/As)=
Short-terme 3
By Using Ig [5.7.3.6.2] ∆𝐷𝐿,𝐼𝑒 = ∆𝐷𝐿,𝐼𝑔 ∗
𝐼𝑒 = 𝐼𝑔
12.65
Long-term deflection factoris take =
mm 4
Camber = 50.5869 mm Thus: the maximum camber is 50.59
mm
b. Live Load Deflection [2.5.2.6.2]
∆𝑎𝑙𝑙𝑜𝑤 𝐿𝐿+𝐼𝑀 =
𝑠𝑝𝑎𝑛 = 800
17.5
mm
Use design truck alone or design lane load plus 25% truck load [A.6.1.3.2]. When design truck alone, it should be placed so that the distance between its resultant and the nearest wheel is bisected by the span centerline. All design lanes should be loaded: [A2.5.2.6.2]. NL =
2
m=
0.85
(1+IM) =
1.33
Number of considered lane
Impact Factor
The value of Ie changes with the magnetude of the applied moment Ma. The moment associated with the live load deflection includes the dead load moment plus the truck moment. Case1: Design for only Truck Load 𝑀𝐷𝐿+𝐿𝐿+𝐼𝑀 = 𝑀𝐷𝐿 + 𝑚 ∗ 1 + 𝐼𝑀 ∗ 𝑀𝑇𝑟𝑢𝑐𝑘 MDL = Mtruck =
4836.1 kN.m 750.5
kN.m
Institut de Technologie du Cambodge
Projet du pont
MDL+LL+IM = 6532.997 kN.m Ig = 2.82E+11 mm4
fr = 3.45065 Mpa
Mcr = 2861.41 kN.m Ma = MDL+LL+IM = 6532.997 kN.m (Mcr / Ma )3 = 0.084024 Ie = 1.3E+11 mm4 Ec*Ie = 3.5E+15 N.mm2 We take load of truck to consider For
P1,2 =
145
kN
= 327.845 kN
For
P3 =
35
kN
=
𝑥𝐺 =
79.135
kN
𝑃1 ∗ 𝑥1 + 𝑃2 ∗ 𝑥2 + 𝑃3 ∗ 𝑥3 𝑃1 + 𝑃2 + 𝑃3 145kN
145kN
Lk =4.3to 9m
4.3m
35kN
L CL
4.3m
Lk = 4.3 ~ 9m
P3
P2
R
P1
a
a xG
Mmax= ? kNm
P3
P2
R
P1
Institut de Technologie du Cambodge a
Projet du pont
a xG
Mmax= ? kNm
- For Front Wheel P3 P3=118.7kN
x
b P3 = x1 = x2 = x3 =
79.135 0 4.3 8.6
∆1 = 𝑃3 ∗ 𝑏 ∗ Δ_P3 =
0.547
kN m m m
Then
𝑥 𝐿2 − 𝑏 2 − 𝑥 2 6𝐸𝐼𝐿
xG = a= b= x=
5.76 0.728 1.972 7.73
mm
For Middle Wheel P2 P2=491.8 kN
x
b P2 = 327.845 kN
Δ_P2 =
5.28
b= x=
6.27 7.73
m m
mm
For Behind Wheel P1 P2=491.8 kN
x
b P1 =
327.85 kN
b = x=
3.43 6.27
m m
m m m m
Institut de Technologie du Cambodge Δ_P1 =
3.50
Projet du pont
mm
Thus : Deflection due to Truck : ΔmaxTruck =
9.32
mm
<
ΔLL+IMallowable
Deflection is not a problem Case 2: Design for Only Land plus 25% of Truck Load 5 𝐿4 ∆𝐿𝑎𝑛𝑒 = ∗𝑞 ∗ 384 𝐿𝑎𝑛𝑒 𝐸𝑐 𝐼𝑒
𝑞𝐿𝑎𝑛𝑒 = 1 + 𝐼𝑀 ∗ 𝑤𝐿𝑎𝑛𝑒 ∗ 𝑚 ∗ 𝑁𝐿
qLane = 21.0273 kN/m Δlane =
3.03
mm
Δ25% of Truck =
2.331
mm
Δ25% of Truck+Lane =
5.358
mm
OK
Case3: Design for Tendom Cencentrated Load at mid span We consider tendom load as a cencentrated load located at mid span for having a critical deflection P=110kN
L/2
∆ 𝑇𝑒𝑛𝑑𝑜𝑚 = qtendom = Δtendom =
L/2
1 𝐿3 ∗ 𝑞𝑡𝑒𝑛𝑑𝑜𝑚 ∗ 48 𝐸𝑐 𝐼𝑒 497.42 kN 8.2
mm
OK
XII. Investigate Strength Limite State 1. Flexural Rectagular Stress Distribution [5.7.2.2] & [5.7.3.2] f'c =
30
β=
0.84
a). Interior Strip
Mpa
Institut de Technologie du Cambodge
Projet du pont
Strength I. Minterior = η*Σγi.Qi = 1.05*[1.25*MDC + 1.5*MDW + (1+IM )*MLL+IM ] MDC =
399.84 kNm per m
MDW =
41.34
MLL+IM =
343.15 kNm per m
1+IM =
133%
kNm per m
Minterior = 1069.12 kNm per m As,inter = 5891.04 mm2 / m for service limite state b= 𝑐=
1000
mm
consider 1m of strip
𝐴𝑆 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 𝛽 ∗ 𝑏
c = 110.574 mm a = β*c = 92.40854 mm ds=
634
mm
c/ds = 0.174407 < 0.42 ratio of c and ds is =
Effective depth of reinforcement Allowable maximum reinfrocement [ 5.7.3.3.1 ]
OK
ρ=As/(bd)= 0.009292 reiforcement percentage Mn =As.fy.(ds-a/2) = 1385.092 kN.m Φ=
0.9
ΦMn= 1246.583 kNm
Thus:
Mu,interior
>
OK
Arrangement of Longitudinal Reinforcement In Interior Strip is:DB32
b). Edge Strip Mexterior = η*Σγi.Qi = 1.05*[1.25*MDC + 1.5*MDW + (1+IM )*MLL+IM ] MDC =
463.13 kNm
MDW =
32.62
kNm
@124
Institut de Technologie du Cambodge MLL+IM =
Projet du pont
408.67 kNm
Mexterior = 1229.94 kNm Asedge = b= 𝑐=
6792.992 mm2 / m 1000
mm
𝐴𝑆 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 𝛽 ∗ 𝑏
c = 127.504 mm a = β*c = 106.5567 mm ds=
634
mm
c/ds = 0.20111 < 0.42 ratio of c and ds is =
Effective depth of reinforcement Allowable maximum reinfrocement [ 5.7.3.3.1 ]
OK
ρ=As/(bd)= 0.010714 reiforcement percentage Mn =As.fy.(ds-a/2) = 1577.935 kN.m Φ=
0.9
ΦMn= 1420.14 kNm
Mu,edge
>
OK
Thus: Arrangement of Longitudinal Reinforcement at Exterior Strip is: DB32
@110
XIII. Distribution Reinforcement of the Deck The amount of bottom transverse reinforcement may be takent as a percenage of the main reinforcement required for positive moment as 14.8(%)
<=
50%
a) Interior Strip ρ= As,inter =
0.148 5891.04 mm2 / m
1750 𝐿
[5.14.4.1-1]
% ≤ 50%
Institut de Technologie du Cambodge As,transv = ρ*As,inter =
Projet du pont
871.30 mm2 / m
DB16 @188.0
b). Edge Strip ρ=
0.148
As,edge = 6792.992 mm2 / m As,transv = 1004.697 mm2 / m
DB16 @188.0
XIV. Shrinkage or Temperature Reinforcement [5.10.8] The shrinkage or temperature reinforcement shall not be spaced farther than 450mm or 3 time of compenent thickness. Ash > 0.11*Ag / fy =
187
mm2 / m
DB10 @313.3
In the both directions
XVI. Fatigue Limite State U = 0.75*(LL+IM)
IM =
15%
Fatigue load shall be one design truck with 9000mm axial spacing no multiple present factor applied (m =1).
4.3 m P1
P1 =
35
9m P2
kN
P2
`P2 =
145
kN
P3 =
145
kN
fmax < fallowable fallowable = 145-0.33*fmin +55(r/h) where fmin =0, because deck is treated as a simple beam, thus, there there no moment reversal. fmin =
0
r/h=
0.3
Institut de Technologie du Cambodge
Projet du pont
ff =
161.5
Mpa
1+IM =
1.15
for fatigue condition
η=
1.00
γ=
0.75
CL x R
P1
P3
P2
C
A
B
L/2+x/2
L/2-x/2
x1
x2
L=
14000
mm
P1=
35
kN
P2 =
145
kN
x=
0.836
m
x2 =
4.3
m
x1 =
3.12
m
ΣMA =0,
RB =
x4
x4 =
6.5819
m
84.625 kN
Mmax = Mc = 556.997 kNm Then
Mfatigue = ηΣγiQi=
480.41 kNm
- Tensile Load Stress fmax 𝑓𝑚𝑎𝑥 𝑀𝑚𝑎𝑥 𝑑 − 𝑥 = 𝑛 𝐼𝑐𝑟 d= x=
634
where n=
7.22
mm
As =
193.59 mm
b =
6792.99 mm2 / m 1000
mm
Institut de Technologie du Cambodge
Projet du pont
Cracking around reinforcement DB25, then cracking inertia is determined: Φ= 𝐼𝑐𝑟 =
25
mm
𝑏𝑥 3 12𝜋Φ4 𝑥2 + + 𝑏𝑥 + 𝑛𝐴𝑠 𝑑2 12 64 4 fmax = 80.02565 Mpa
<
= ff
2.2E+10 mm4 OK
XVI. Resume the answer Calculation Result 1. Longitudinal Reinforcement at the bottom at exterior strip: DB32 @ 110 2. Longitudinal Reinforcement at the bottom at interior strip: DB32 @ 124 3. Transversal Reinforcement at the bott at exterior&interior strip: DB16 @ 188 4. Longitudinal Reinforcement at the top at exterior strip: DB10 @ 313.3 5. LongitudinalReinforcement at the top at interior strip: DB10 @ 313.3 4. Transversal Reinforcement at the top at exterior & interior strip: DB10 @ 313.3
Actual Use @ 100 @ 120 @ 180 @ 300 @ 300 @ 300
XVII. Conclusion Après avoir fait ce projet du pont, il nous permet d'avoir le concept, le dimesionnement et le calcul sur le pont dalle de béton armé (solide slab bridge) par le norme AASHTO. En plus,il nous donne aussi un model pour calculer d'autres types du pont. Dans ce projet, on n'a calculé que les sections d'armatures de la portée L/2 et L/4. Pour qu'on peut économiser l'argent sur l'utilisation d'armatures, on doit calculer plusieurs points sur cet portée. Et on a vérifier seulement le milieu de la portée L/2, le point le plus critique qu'on doit considérer comme le flèche et le moment. On n'a pas vérifier l'effort tranchant parce qu'il n'est pas le point critique pour le pont dalle grâce au largeur de la dalle importance. En conclusion, on a adopté sur la vérification et le calcul de ce pont dalle en béton armé.
380mm
680mm
Eext = 1800mm
DB32@120
DB10@300
DB32@100
DB10@300
Eint = 6400 mm
Wearing Surface(75Thk)
DB16@180
DB10@300
Eext = 1800mm
DB32@120
DB10@300
380mm
XVIII.
Sketch of Bridge Section in Rebar Detail
Page 23
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