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)

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

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

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

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

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..

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 =

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

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

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

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

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

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