25m single lane bridge design calculations latest
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CalculationTRANSCRIPT
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25M SPAN SINGLE LANE STEEL TRUSS BRIDGE
A. INTRODUCTION
Reference
There is a 1.8 m height hand rail at both side of the bridge.
Checked DESIGN UNIT
The span of the this bridge is 25m , width of the lane is 3.7 m and 1.2 m wide shoulders both side.
OutputCalculation
SINGLE LANE TRUSS BRIDGE- 25M SPAN Doc. No.ARJ Date
Page 01Date
This design report contains the analysis and design calculations for the single lane steel truss bridge of span 25m.
Job Code
DEC
Designed EPC DIVISION CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)
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B. GENERAL DESIGN INFORMATION
B-1. Geometry of the Steel truss bridge
Span = mMax. Truss Height = mSpacing between two truss = m
B-3 Load factors
As per Table 1 BS 5400 Part 2 : 1978
B-4 Design Codes ,Manuals and References
BS5400: 1978,British Standard institution Code of practice for bridge designRoad Development Authority (RDA) of SriLanka Bridge Design Manual 1997
B-5 Software used
Modeling, Analysis and Design are carried out using SAP2000 version 9
DESIGN UNIT ARJDEC CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)
EPC DIVISION Checked Job Code
DateDate
Output
Designed
25.03.5
SINGLE LANE TRUSS BRIDGE- 25M SPAN Doc. No.
Reference Calculation
6.0
Page 02
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C. GENERAL LOAD EVALUATION
C.1 Live Loads
Carriageway width = 3.7m
cl 3.2.9.3.1 Number of notional lanes = 1Width of a notional lane = 3.7m
C.1.1: HA LoadingNominal HA loads
cl 6.2.1 Loaded length = 25m
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C.2.3. Weight of guard railConsider longitudinally 1 meter length along the bridge laneVolume of the steel guard rail = 0.1m3Weight of the steel guard rail = 0.1x7850x9.81/1000
= kNGuard rail load as UDL = 7.7/25
= kN/m
Dead load on each joint as point load = (40.99+4.625+.31)/2*3.125= kN
C.2.2. Selfweight of truss members
Selfweights of members of main 2D truss are taken in to the analysis by softweare itself.
Selfweights of other Other transverce members
1. Bottom cross bracingLoad on a joint of a 2D main truss = (108 kg/m x 9.81 kgms-1/1000x6.0 m)/2(2/PFC 380 x100 x 54- 6 m length) = kN
2. Bottom chord wind bracing Load on a joint of a 2D main truss = (8.99kg/m x 9.81 kgms-1 /1000 x 6.76 m) (EQA 75 x 75-6.76 m length) = kN
Point load on bottom truss joint = kN
3. Top cross bracingLoad on a joint of a 2D main truss = (23.9 kg/m x 9.81 kgms-1/1000x6.0m)(PFC 150 x90 x 24- 2.2 m length) = kN
4. Top chord wind bracing Load on a joint of a 2D main truss = (8.99kg/m x 9.81 kgms-1 /1000 x 6.76 m) (EQA 75 x 75-6.76 m length) = kN
Point load on top truss joint = kN
Date Doc. No.
Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page 04
3.178
0.596
1.407
0.596
Calculation
DESIGN UNIT Designed
Output
DEC
EPC DIVISION Checked
Reference
7.70
3.775
2.003
0.31
71.8
ARJ
SINGLE LANE TRUSS BRIDGE- 25M SPAN
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C.3 Wind Loads
Wind loads on Superstructure
vc v = 22.2m/s RDA manual vc v = 22.2m/s RDA manualk1 = 5.3.2.1.2 k1 = 5.3.2.1.2s1 = 5.3.2.1.3 s1 = 5.3.2.1.3s2 = T2-5.3.2.1.4 s2 = T2-5.3.2.1.4vc = m/s vc=K1*v*s1*s2 vc = vc=K1*v*s1*s2
pt q = N/m2 q=0.613*vc2 pt q = 0.741kN/m2 0.613*vc2q=0.613*vc2Cd = 2 Cd =1.45 for all live parts5.3.3.4 T7 5.3.3.4
Bottom chordA = 0.35 m2per meter of chords Assume height of a threewheeler as 2mpt = 0.35 x 2 x 0.721
= 0.50 kN/m A = 2 x1 m2
Top chordA = 0.15 m2per meter of chords pt = q x A x Cdpt = 0.15 x 2 x 0.721
= 0.21 kN/m pt(withlive)Diagnal Bracing = 2 x 1.45 x 0.721A = 0.15 m2per meter of chords = 2.1 kN/mpt = 0.15 x 2 x 0.721
= 0.21 kN/mNominal Longitudinal wind load is not considered due to small effect to truss bridge
Pv q = 0.671kN/m2 Pv q = 0.671kN/m2A = 6m2/per meter of deck slab A = 6 m2/per meter of deck slabCl = Cl =pv = 3.0kN per meter pv = 3.0 kN per meter
C-3 Load Combinations
Ultimate Limite StateTable 1
Comb 1 = 1.15 x conc load+1.05 x steel loads+1.5 x live and parapet load
Comb 2 =
Servicibility Limit State
Com_Ser = 1.0 x conc load+1.0 x steel loads+1.0 x live and parapet load
Doc. No.
BS5400
SINGLE LANE TRUSS BRIDGE- 25M SPAN Designed
Page 05
Reference Calculation
34.76
0.741
Part 2 : 1978 1.00 1.00
Nominal Transverse wind load (pt)without LL Nominal Transverse wind load (pt)-with LL
BS5400 1.00
DateChecked Job Code
1.00
Output
1.15 x conc load+1.05 x steel loads+1.25 x live and parapet load+1.4 x wind loads
1.58
Nominal Vertical wind load (Pv)
34.76m/s
0.75 0.75
Part 2 : 1978
CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB)
DESIGN UNIT ARJ Date EPC DIVISION
DEC
1.58
Nominal vertical wind load (pv)
BS5400Part 2 : 1978
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D. STRUCTURAL ANALYSIS AND RESULTS
The main steel truss is analysised
(1) as a 2D truss for load combination1 ( Without lateral-wind forces )(2) as a 3D truss for load combination2 ( With lateral-wind forces )
D.1 Servisibility Analysis for Deflection
Servisibility analysised is used for checke the globle deflection of the entire bridge.
Maximum vertical deflection at the middle of the truss at 2D SAP model = 38 mm(For com_ser 1)
Maximum deflection at the middle of the truss at 3D SAP model(For com_ser 2) Horizontal = 25 mm
Vertical = 50 mm
D.2 Analysised Results for Combination 1
Analysised reasulta are attached as Annex ( i )
D.2.1 Maximun Axial forces values ( Com 1)
Member
Top Chord -Bottom chord -Vertical members -Diagnal members
D.3 Analysised Results for Combination 2
Analysised reasults are attached as Annex ( ii )
Member
Top Chord -Bottom chord -Vertical members 48 -Diagnal members
Tension Compression
SINGLE LANE TRUSS BRIDGE- 25M SPAN Doc. No. DESIGN UNIT Designed
Checked CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page 07
DEC
Date Date
OutputReference Calculation
All main truss members have axial forces only and no considerable bending moment generated by selfweight.
ARJ
EPC DIVISION
Tension Compression
kN kN1335
1221
360 280
159
kN kN1006
387
612 391
All 3D truss members have axial forces only and no considerable bending moment generated by selfweight..
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E. STRUCTURAL DESIGN
E.1 Top Chord Design
Comb1 is critical for Top Chord
Maximim axial force of Top chord = kN(Compression)
Length of a top Chord member = mm
Top chord is design as a compression member
Selected section for top chord is 2/PFC 200 x 90 x 30
E.1.1 Section properties
b = mm Back to back space = 10 mmd = mmT = mm py = mmt = mm
Length of a member = mmBoth ends are pin-joints
E.1.2 Section Classifications
Fig 3 b/T = 90/14 = = (275/275)^0.5BS 5950 = 1
d/t = 148/(2*7) =
Table 7 Internal element of compression flangeBS 5950
b/T = < 26
i.e.Section is plastic
web with natural axis at mid depth
d/t =
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4.7.4. Compression capacity of the member = Pc = Ag*pcBS 5950
For pc value, Table 27 ( c)(Rolled channel from table 25)
(d) Calculation of = le/= 3.125/0.0816=
(e) pc value
pc = N/mm2 (from Table 27( C) ,=38.3 and py=275)
Ag = mm2
(f) Compression Capacity
Compression capacity of members = pc*Ag= kN
Maximum design compression = kN
since1324
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E.2 Bottom Chord Design
Comb1 is critical for bottom Chord
Maximum axial force of bottom chord = kN(Tension)
Length of a top Chord member = mm
Top chord is design as a tension member
Selected section for bottom chord is 2/PFC 200 x 90 x 30
E.2.1 Section properties
b = mm Back to back space = 10 mmd = mmT = mm py = mmt = mm
Length of a member = mmBoth ends are pin-joints
E.2.2 Design of bottom chord
Tension capacity
Pt = Ae x Py
Ae = Anet connected + (3a1/(3a1+a2)) x Aoutstanduse 16mm diameter boltsa1 = mm2
(Anet connected)a2 = mm2
(Anet outstand)Ae = mm2
Py = N/mm2
Pt = kN
Pt > kN
Proposed section is adequate when the tension capacity is considered
1395
275
1212
Page 10
Date Date
1212
7
cl 3.3.2
EPC DIVISION
Reference
Table 6
3125
9020014
Doc. No.SINGLE LANE TRUSS BRIDGE- 25M SPAN
cl 4.6.3.1
4648
5071
3125
275
DEC
DESIGN UNIT Designed ARJChecked
CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code
2296
cl 4.6.1
Calculation Output
d
b
Tt
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E.3 Bottom cross bracing Design
E.3.1 Special load eveluation for Bottom cross bracing
E.3.1.1 Live loadsDefined HA load(on deck) = 30 kN/m per notional lane
Knife edge load on a member = 120kN Live load on member for 1m strip = Udl x spacing between two joints + KEL Live load on member for 1m strip = 30/3.7x3.125 + 120/6 kN/m
=
E.3.1.2 Dead LoadsSlab Load = 0.28x24x3.125x6= kNasphalt layer load = .05x25x3.7x3.125= kNload of the guard rail = 0.31x3.125 kN = kN
Consider 1 meter of the deck.Dead load on 1m of member = (126.0+14.5+0.1)/6
=
E.3.1.3 Self WeightSelf weight = kN/m
Considering 1m of the deck Self weight = 108*9.81
= kN/m
E.3.1.4 Design Load
Design load = 1.15*conc load + 1.05*self weight + 1.5*live load= kN/m= kN/m
CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code
DEC
DESIGN UNIT Designed ARJ Date EPC DIVISION Date
96.2396.2
Checked
Reference Calculation
Page 11
1.06
SINGLE LANE TRUSS BRIDGE- 25M SPAN Doc. No.
BS5400Part 2 : 1978
kN/m45.3
126.000
Output
14.4530.969
23.6
108
kN/m
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Selected section for bottom chord is 2/PFC 380 x 100 x 54
E.3.2 Design of Bottom Cross Bracing
E.3. 2.1 Section properties
b = mm Back to back space = 10d = mmT = mm py =t = mm
Length of a member = mmcl3.1.2 E = kN/mm2property table Ix = cm4 rx = cm
Both ends are pinjoints
E.3.2.2 Section classifications
Fig 3 b/T = 100/17.5 = = (275/275)^0.5BS 5950 = 1
d/t = 315/(9.5*2)=
E.3.2.3 Local buckling check
Internal element of compression flange
d/t = < 79
b/T = < 79
i.e.Section is plasticSection is non slender
4.2.3. E.3.2.4 Shear capacity checkBS 5950
Pv = 0.6*py*Av Av = txD Pv = kN= 9.5x380x2= mm2
d/t = Fv
low shear
Check is OK
E.3.2.5 Moment capacity check
Since the section is plastic and subject to low shear
Mc = Py x S but PyS S
Mc = Py x S= kNm
BMmax = Design Load x Length ^2 /8= kNm Mc > BMmax
Moment capacity check is O.K.
514.3
1896
DEC
Date DESIGN UNIT Designed ARJ
SINGLE LANE TRUSS BRIDGE- 25M SPAN Doc. No.
CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page 13
Date EPC DIVISION Checked
433
cl 4.2.5.
Reference Calculation
18701580
714.8
Output
288.7
-
E.3.2.7 Deflection check
Allowable deflection = span / 360= 6.0/360= mm
Actual deflection = 5wL4/384EI
E = 205 kN/mm2
i = mm
i < span / 360
Deflection check is satisfied
Proposed section is adequate
DESIGN UNIT Designed ARJ Date EPC DIVISION Checked
Doc. No.SINGLE LANE TRUSS BRIDGE- 25M SPAN
DEC
OutputReference Calculation
cl 3.1.2
Table 5
12.39
16.67
Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page 14
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E.4 Design of diagonals
E.4.1. Compression capacity check.Maximim axial force of diagonal = kN
(Compression)Length of maximum compression member = mmMaximim axial force of diagonal = kN
(Tension)Length of maximum tension member = mm
Selected section for diaganol is 2/PFC 150 x 75 x 18
E.1.1 Section properties
b = mm Back to back space = 10 mmd = mmT = mm py = mmt = mm
Length of a member = mmBoth ends are pin-joints
E.1.2 Section Classifications
b/T = 75/10 = = (275/275)^0.5= 1
d/t = 150/(2*5.5) =
Internal element of compression flange
b/T = < 26
i.e.Section is plastic
web with natural axis at mid depth
d/t =
-
4.7.4. Compression capacity of the member = Pc = Ag*pcBS 5950
For pc value, Table 27 ( c)(Rolled channel from table 25)
(d) Calculation of = le/= 4.650/0.0615=
(e) pc value
pc = N/mm2 (from Table 27( C) ,=75.61 and py=275)
Ag = mm2
(f) Compression Capacity
Compression capacity of members = pc*Ag= kN
Maximum design compression = kN
since408 kN
since640
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E.5 Design of vertical members
Maximum axial force of diagonal = kN(Tension)
Length of maximum tension member = mm
Vertical member is design as a tension member
Selected section for top chord is PFC 125 x 65 x 15
E.1.1 Section properties
b = mm Back to back space = 10 mmd = mmT = mm py = mmt = mm
Length of a member = mmBoth ends are pin-joints
E.4.2. Tension capacity check
Pt = Ae x Py
Ae = Anet connected + (3a1/(3a1+a2)) x Aoutstand
a1 = mm2
a2 = mm2
Ae =
Py = N/mm2
Pt = kN
Pt > kN
Since167