es_02_courbon & engesser methods

Politecnico di TorinoDepartment of structural and geotechnical engineering Bridge design

Girder bridges 1/924

BRIDGE DESIGN

COURBON METHOD



Girder deck dimensions - Bottom view



Girder deck dimensions Interaxis between main members



Girder deck dimensions Cross section



The beams subjected to the highest bending moment are the external ones, so the other beams are designed as they were subjected to the same actions. This reduces design time and is a safe approximation.

We proceed calculating the internal actions (bending moment and shear) in the mid-span section of an external beam called beam 1.



Values of the multi component actions

Loads on carriageway Loads on footways

Vertical Horizontal Vertical

Group of

actions

Main actionLM1-2-3-4-6

Special vehicles

Crowd BrakingAccel.

Centrifugal Uniform

1 Characteristic value

2.5 kN/m2

In this exercise we will solve the structure only for the multi component action group n 1. Needless to say that the other groups have to be taken into account too.



Load analysis

Dead load g11. Longitudinal beams

2. Transverse beams

3. Slab

1. Longitudinal beam 1 30.5 1.20 15 25 225lbkNg b h l m m m kNm

2. Transverse beam 1 30.3 1.00 2.5 3 25 56tbkNg b h l m m m kNm

3. Slab 1 312 15 0.25 25 1125skNg b l h m m m kNm



Load analysis

Dead load g1

Total weight of the girder

1 1 1 14 4 4 (225 56) 1125 2249tot lb tb sg g g g kN

Dead load on the outermost beam

11,1

2249 384 4 15tot

bkNg kNg l m m

Two simplifications:

a. Dead weight uniformly distributed among beams

b. Dead weight of transverse beams taken as uniformly distributed instead of 4 concentrated forces



Load analysis

Permanent loads g21. Kerb

2. Pavement

3. Vehicle restraint system

4. Pedestrian parapet

1. Kerb 2k 31.5 0.23 15 25 129kNg b h l m m m kNm

2. Pavement

2 21.5 15 3 67.5pkNg b l m m kNm

The load value for the pavement takes into account that several layers of asphalt may be placed one over another during maintenance of the road:



Load analysis

3. V. R. S. 2 15 2 30vrskNg l m kNm

4. Pedestr. parapet 2 15 1.0 15ppkNg l m kNm

Permanent load on the outermost beam

2,1 2, 2, 2, 2,

1,1

( )

(129 68 30 15) 15 242 15 16 / 44%b k p vrs pp

b

g g g g g lkN m kN m g

One simplification:

a. The permanent load for the outermost beam is grater then for the other beams. In this example the load of kerb and barriers is fully given to the outermost beam, in reality it would distribute itself according to Courbon theory on the others beams resulting in a lesser weight for beam one.



Internal actions due to permanent loads

In the mid-span section of beam 1 we find the following internal actions:

Bending moment

21,1 2,1

1 1,1 2,1

2

838 16 15

1068 450 15188

b bb b b

g g lMg Mg Mg

kNm kNm

Shear 1 1,1 2,1 0 0 0b b bVg Vg Vg kN



Load analysis

Variable traffic load q1

We need to trace the influence lines of bending moment and shear for the mid-span cross section of the beam for moving vertical loads.

We apply a disconnection dual to the desired internal action and we calculate the function of the entity dual to the known action (vertical force).



Bending moment in mid-span

Drawing influence surface

0.54 2l l 1

0.5 One dimensional influence line for longitudinal simply supported beam

0.70.4

0.1

-0.2Transversal load repartition according to Courbon theory

,i j



Transverse distribution

1, j Is the amount of the load P=1 applied on the beam 1 that goes on the beam j (j=14)

1, j Is the amount of the load P=1 applied on the beam j (j=14) that goes on the beam 1

Or:





If we modulate the two graph seen before we obtain

0.2 4l

0.7 4l

The blue area has to be loaded to maximize the mid-span bending moment in beam 1



1.2m 6.9m6.9m

3.754l m

4 6.9 3.75 / 7.5 6.9 3.452

lml

P

, 2 3.45 6.9S PM P P

1

0.5

Longitudinal distribution (concentrated loads)



3.754l m

q

, 2 1.875 3.7 3.7 7.5 27.752S qlM Rq q q q

1.8758l m

4l

4l

Rq

1 0.5

4l

4l

Rq

Longitudinal distribution (uniformly distributed loads)



Transverse distribution

Carriageway width = 9mWidth of each notional lane = 3mNumber of notional lanes = 3



Transverse distribution (concentrated loads)

0.5 2.0 0.5

1 NotionalLane

3.0

150kN 150kN

0.5 2.0 0.5

3.0

100kN 100kN

2 NotionalLane

0.70.4

0.1

-0.2

We dont place the third notional lane because its centroid will fall inside the negative influence

line of load distribution

,i j




0.5 2.5

aF

0.70.4

0.1

-0.2

,i j1,aR

ay

1,0.5150 0.7 0.3 150 0.65 97.53a a a

R F y kN kN kN




0.52.5

bF

0.70.4

0.1

-0.2

,i j1,bR

by

1,2.5150 0.7 0.3 150 0.45 67.53b b b

R F y kN kN kN




0.5 2.5

cF

0.70.4

0.1

-0.2

,i j1,cR cy

1,0.5100 0.4 0.3 100 0.35 35.03c c c

R F y kN kN kN




0.52.5

dF

0.70.4

0.1

-0.2

,i j1,dR dy

1,2.5100 0.4 0.3 100 0.15 15.03d d d

R F y kN kN kN




1, 1, 1, 1, 1,

97.5 67.5 35 15 215concentrated a b c dR R R R R

kN



Transverse distribution (uniformly distributed loads)

1 NotionalLane

1.5 1.5

3.0

27 /kN m

1.5 1.5

3.0

2 NotionalLane

7.5 /kN m

0.70.4

0.1

-0.2


line of load distribution

,i j

29 /aq kN m 22.5 /bq kN m




1.5 1.5

3.0

27 /aq kN m

0.70.4

0.1

-0.2

,i j

1.5 1.5

3.0

7.5 /bq kN m

1R

1, . . 1, 1, 27 0.55 7.5 0.25 14.85 1.87 16.7 /u distr a b a a b bR R R q y q y kN m

ay by



Pay attention!

2 1

F

0.70.4

0.1

-0.2

If we consider a force F and we calculate its distribution using directly the influence line, or we solve the static scheme shown below and then we calculate the force in the beam 1 with the reactions, we obtain the same result.

,i j3

F 23

F

120.5 0.7 0.43 3

F FR F



Transverse distribution (crowd)

1.5

0.70.4

0.1

-0.2

We dont place crowd on this foothpath because its centroid will fall inside the negative part of the

influence surface

,i j

22.5 /q kN m

1F 2F




1.5 22.5 /q kN m3

1F 2F

qqR 1.5 3.75 /qR q m kN m

2 3.75 0.75 / 3 0.94 /F kN m 1 3.75 0.94 4.69 /F kN m

0.70.4

1,crowdR

1, 1 20.7 0.4 2.9 /crowdR F F kN m

Same result if we extrapolate the Curbon transverse line

outside beam n1




1, 215concentratedR P kN

Concentrated tandem system1.2m 6.9m6.9m

P

1

0.5

, 1,6.9 1484S concentrated concentratedM R kNm

Uniformly distributed

1, . . 16.7 /u distrR q kN m , . . 1, .27.75 463s u distr u distrM R kNm

qRq

1 0.5

Rq




Crowd

1, 2.9 /crowdR q kN m , 1,27.75 81s crowd crowdM R kNm

qRq

1 0.5

Rq

Total bending moment from vertical traffic actions

, , , . . , 1484 463 81 2028s Vtraffic s concentrated s u distr s crowdM M M M kNm



Shear in mid-span


One dimensional influence line for longitudinal simply supported beam

0.70.4

0.1

-0.2Transversal load repartition according to Courbon theory

,i j

7.5m7.5m

1



Shear in mid-span


If we modulate the two graph seen before we obtain

0.2 0.5 0.7 0.5

The blue area has to be loaded to maximize the mid-span shear in beam 1

0.7 0.5 0.2 0.5



Shear in mid-span

Variable concentrated traffic load

1.2m 7.5m6.3m

12m

6.31 0.422 7.5m

P

, (0.5 0.42) 0.92S PV P P



Shear in mid-span

Variable uniformly distributed traffic load

12m

,1 15 1.875

2 4 8S qlV q q q

14m

q

4l

4l

Rq




0.5 2.0 0.5

1 NotionalLane

3.0

150kN 150kN

0.5 2.0 0.5

3.0

100kN 100kN

2 NotionalLane

0.70.4

0.1

-0.2

,i j

0.5 2.0 0.5

3.0

50kN 50kN

3 NotionalLane



Longitudinal location of previously seen concentrated loads


1 Lane 2 Lane

3 Lane




0.5 2.0 0.5

150kN 150kN

0.5 2.0 0.5

100kN 100kN

0.70.4

0.1

-0.2

0.5 2.0 0.5

50kN 50kN

1R

1, 150 (0.65 0.45) 100 (0.35 0.15) 50 ( 0.05 0.15)165 50 5 220

concentratedRkN




1 NotionalLane

2 NotionalLane

0.70.4

0.1

-0.2

,i j

3 NotionalLane

1.5 1.5

3.0

27 /kN m

1.5 1.5

3.0

7.5 /kN m29 /aq kN m

22.5 /bq kN m

1.5 1.5

3.0

7.5 /kN m



Longitudinal location of previously seen distributed loads


1 Lane 2 Lane

3 Lane




0.70.4

0.1

-0.2

1.5 1.5

27 /kN m

1.5 1.5

7.5 /kN m

1.5 1.5

7.5 /kN m

1, . . 27 0.55 7.5 0.25 7.5 0.05 17.1 /R u distr kN m

1R



0.70.4

0.1

-0.2

Transverse distribution (crowd)1.5

22.5 /q kN m

3

1F 2F

qqR

2 3 3.75 0.75 / 3 0.94 /F F kN m 1 4 3.75 0.94 4.69 /F F kN m

1, 1 2 3 40.7 0.4 0.1 0.24.69 0.9 0.94 0.3 3.94 /

crowdR F F F FkN m

3F 4F

1.5

qqR

3

1.5 3.75 /qR q m kN m



Longitudinal location of previously seen distributed loads


1 Lane 2 Lane

3 Lane



Shear in mid-span

1, 220concentratedR P kN

Concentrated tandem system

, 1,0.92 202S concentrated concentratedV R kN

Uniformly distributed

1, . . 17.1 /u distrR q kN m , . . 1, .1.875 32s u distr u distrV R kN

1.2m

7.5m

6.3m P

Rqq



Crowd

1, 2.9 /crowdR q kN m , 1,1.875 5.4s crowd crowdV R kN

Total shear from vertical traffic actions

, , , . . , 202 32 5 239s Vtraffic s concentrated s u distr s crowdV V V V kN

Shear in mid-span

Rqq



Wind referring speed ,0 25b bmv vs

Non traffic actions: WIND

Location: Piemonte 250m o.s.l.

Kinetic referring pressure2 2

2

1 11.25 25 3912 2b b

Nq vm

Geografic zone 1Terrain roughness class D (open land without obstacles)Site exposition category II kr = 0.19

z0 = 0.05mzmin = 4 m

Maximum height of the structure z=3m



Exposure coefficient 2 min minmin0 0

ln 7 lne e rz zc z c z kz z


2 4 40.19 ln 7 ln 1.80.05 0.05e

c z Dynamic coefficient = 1

Shape coefficient = 1

Wind pressure 2391 1.8 1 1 0.74b e p dkNp q c c cm



Surface exposed to the wind

3m

1.58m

From pavement extrados to longitudinal beams intrados (13cm of pavement thickness)



Vertical position of the centroid of the deck

1. Longitudinal beams 1 225 4 0.6 540lbSg kN m kNm 2. Transverse beams 1 56 4 0.5 112tbSg kN m kNm 3. Slab 1 1125 1.325 1491 /sSg kN m kN m Total 540 112 1491 2143 /Sg kN m

Static moment of bridge masses with respect to the intrados

1. Longitudinal beams 1 225 4 900lbMg kN kN 2. Transverse beams 1 56 4 224 tbMg kN kN3. Slab 1 1125sMg kNTotal 900 224 1125 2249 Mg kN

Total mass of the bridge



Vertical position of the centroid of the deck

Vertical position of the centroid2143 0.952249

ggg

Sy m

M

gy

3.63m

0.95m

Wind MtTorque moment due to the wind




Wind resultant 20.74 4.58 3.39windkN kNq p h mm m

Torque moment 3.39 4.58 / 2 0.95 4.54 wind kN kNMt q e mm mEquivalent vertical load acting un beams 1 and 4

., / 9

4.54 / 9 0.50

Vert windq Mt mkNm

9m




Bending moment in mid-span of beam 1 due to wind action

2 2.,

,0.50 15 14.1

8 8 Vert windS wind q lM kNm

Shear in mid-span of beam 1 due to wind action

, 0S windV kN



ULS combination

Bending moment in mid-span of beam 1

, , , ,1.35 1.35 1.50

1.35 1518 1.35 2028 1.50 14 4808S tot S perm S traffic S windM M M M

kNm

Shear in mid-span of beam 1

, , , ,1.35 1.35 1.50

1.35 0 1.35 239 1.50 0 323S tot S perm S traffic S windV V V V

kN



Pay attention:

Its not possible to evaluate the internal actions in the transverse beams using Courbon, because Courbon hypothesis doesnt locate transverse beams in a specific position but smears them in the whole length of the deck.

If we want to know the internal actions in the transverse beam we have to use the Engesser model.



BRIDGE DESIGN

ENGESSER METHOD



We will analyze the same deck seen with the Courbon approach with Engesser theory.

We will calculate bending moment and shear in the mid-span of beam 1 exactly as we have done with Courbon for the same multi component actions.

For sake of simplicity we will assume for dead load and permanent actions the same values seen in Courbon example (theres very little difference as the deformation due to these loads is cylindrical).

We will then focus only on variable traffic loads.





2

2

1 2

1 010

( )1 3 325 2 2 2

a

z z for z ll

y zz lz for l z ll

1

One dimensional influence line for longitudinal beam (continuous on transverse beams)

tbR tbRz

2l

2l l

2

35

btb

EIRl

5l m





tbR

Beam 1

Beam 2

Beam 3

Beam 4

We apply the virtual reactions on the girder and we calculate with Courbon theory the global deformation of the deck.

tbR

zx

b




The equation of the surface drawn in the previous page is

3 22

33 22

3 33 22

1 6 0101( , ) 0.7 0.9 6 2

3 101 2 6 2 3

10

b

z l z for z ll

xy z x z z l l z for l z lb l

z z l z l l z for l z ll

Longitudinal direction

Transverse direction




That can become for the single beams

3 22

33 2, 1, 2

3 33 22

1 6 0101( ) 6 2

101 2 6 2 3

10

b i i

z l z for z ll

y z z z l l z for l z ll




1,

0.70.40.10.2

i

Beam 1

Beam 2

Beam 3

Beam 4




-1

-0.5

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

z

y

ya1yb1yb2yb3yb4




-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

z

y

ya1+yb1yb2yb3yb4



1. We have to distribute on the longitudinal beams the vertical loads acting on the slab using the simply supported schemes seen before

2. Once the loads are on the beams we can use the influence lines shown in the previous slide to calculate the bending moment in mid-span.

Procedure



0.70.4

0.1

-0.2

,i j1F

0.5 2.0 0.5

3.0

100kN 100kN


surface of load distribution

0.5 2.0 0.5

3.0

150kN 150kN

2F 3F 4F

Concentrated loads




1

2

3

150150 100 250100

F kNF kNF kN

2. Once the loads are on the beams we can use the influence lines shown in slide 61 to calculate the bending moment in mid-span.

iF

z

6.9z m8.1z m

iF



,1 ,1 ,1 ,1

,2 ,2

,3 ,3

(6.9) (6.9) (8.1) (8.1) 0.6 2.0 2.60(6.9) (8.1) 1.14(6.9) (8.1) 0.285

a b a b

b b

b b

y y y yy yy y

, 1 ,1 ,1 1

, 2 ,2 2

, 3 ,3 3

2 (6.9) (6.9) 2 2.60 150 780

2 (6.9) 2 1.14 250 5702 (6.9) 2 0.285 100 57

s F a b

s F b

s F b

M y y F kNm

M y F kNmM y F kNm

, 1484S concentratedM kNmWith Courbon model it was

, , 1 , 2 , 3 780 570 57 1407S concentrated S F S F S FM M M M kNm

5% difference



0.70.4

0.1

-0.2

,i j1q


surface of load distribution

2q 3q 4q

Uniformly distributed loads

1.5 1.5

27 /kN m

1.5 1.5

7.5 /kN m




1

2

3

13.5 /13.5 3.75 17.25 /3.75 /

q kN mq kN mq kN m

2. Once the loads are on the beams we can integrate the influence lines shown in slides 56 and 59 for a uniformed distributed load to calculate the bending moment in mid-span.

z

iF



a system

2

2

1 2

1 010

( )1 3 325 2 2 2

a

z z for z ll

y zz lz for l z ll

33 2 22

1 20 0

1 3( ) 2 1 210 5 2 2

ll l

al

z z z lq y z dz q dz q z dzl l

32 22

20

1 32 1 210 5 2 2

ll

l

z z z lq dz z dzl l

2 223 12 0.625

40 80 40l lq q l q



b system

3 2 33 2 3 2, 1, 2 20 0

1 1( ) 2 6 610 10

l l l

b i il

q y z dz q z l z dz q z z l l z dzl l

2 33 2 3 21, 2

0

2 6 610

l l

il

q z l z dz z z l l z dzl

3 22

33 2, 1, 2

3 33 22

1 6 0101( ) 6 2

101 2 6 2 3

10

b i i

z l z for z ll

y z z z l l z for l z ll


4 4 21, 1,2

11 11 1110 2 2 10i iq l l q ll



a + b systems

Beam 1 21 11 41 25 410.740 10 100 4

q l q q

Beam 2 211 110.4 0.4410 25

q l q q

Beam 3211 110.1 0.11

10 100 q l q q

Total, . . 1 2 3

41 410.44 0.11 13.5 0.44 17.25 0.11 3.75 1464 4

S u distrM q q q kNm

, . . 463S u distrM kNmWith Courbon model it was

37% difference



Shear in mid-span


2

2

1 32

2

1 03

( )31 1

3 2

a

z z for z ll l

y zz z z for l z ll l l

1

One dimensional influence line for longitudinal beam (continuous on transverse beams)

tbR tbRz

2l

2l l

2

6 btb

EIRl

5l m



Shear in mid-span


tbR

Beam 1

Beam 2

Beam 3

Beam 4

We apply the virtual reactions on the girder and we calculate with Courbon theory the global deformation of the deck.

tbR

zx

b




The equation of the surface drawn in the previous page is

3 2

3

3 3 2

3

3 3 3 2

3

2 06 3

2 ( )( , ) 0.7 0.9 23 6 2 3

2 ( ) ( 2 ) 2 36 2 2 3

b

z l z for z ll

x z z l l zy z x for l z lb l







That can become for the single beams



1,

0.70.40.10.2

i

Beam 1

Beam 2

Beam 3

Beam 4

3 2

3

3 3 2

1, 3

3 3 3 2

3

2 06 3

2 ( )( , ) 26 2 3

2 ( ) ( 2 ) 2 36 2 2 3

b i

z l z for z ll

z z l l zy z x for l z ll





-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

z

y

ya1

yb1

yb2

yb3

yb4




-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

z

yya1+yb1

yb2

yb3

yb4



Longitudinal position of the three tandem systems

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

z

y

ya1+yb1

yb2

yb3

yb4Lane 1 Lane 2Lane 3

Concentrated loads



Longitudinal location of previously seen concentrated loads

1 Lane2 Lane

3 Lane

Each couple of tandem systems should be treated separately

Concentrated loads



1 2 150F F kN

1F

0.5 2.0 0.5

3.0

150kN 150kN

2F

1 Lane

,1 ,1

,1 ,1

(7.5) (7.5) 0.5 0.0 0.5(8.7) (8.7) 0.27 0.13 0.40

a b

a b

y yy y

,2

,2

(7.5) 0(8.7) 0.08

b

b

yy

,1 1 0.50 0.40 0 0.08

0.98 150 147

cV F

kN

Concentrated loads : 1 lane



The z corresponding to the maximum value of has to be calculated. For sake of simplicity it is done for 0



,4

,4

(3.48) 0.070(4.68) 0.070

b

b

yy

3F

0.5 2.0 0.5

3.0

50kN 50kN

4F

3 Lane

Concentrated loads : 3 lane

,3

,3

(3.48) 0.035(4.68) 0.035

b

b

yy

3 4 50 F F kN

,3 32 0.070 0.0350.070 50 3.5

cV F

kN

The tandem loads are placed symmetrically to the ones of the 2 lane with respect to the mid-span of the bridge because of the anti-symmetry of the influence line of beams 3 and 4



Concentrated loads : total shear in mid-span

, ,1 ,2 ,3 147 35 3.5 185.5S concentrated c c cV V V V kN We add the contribution of the three lanes

, 202S concentratedV kNWith Courbon model it was

8.9% difference



Location of uniformly distributed loads

1 Lane 2 Lane

3 Lane

Each lane should be treated separately

Uniformly distributed loads



a system

3 3 32 22 2

1 2 20 0

( ) 1 1 13 3

l ll

al

z z z z zq y z dz q dz q dzl l l l l

3 33 32

2 20

1 1 13 3

ll

l

z z zq z dz z dzl l l l l

1 25 1 1 1 5 0.15612 192 64 32 32

ql ql q q

2

2

1 32

2

1 03

( )31 1

3 2

a

z z for z ll l

y zz z z for l z ll l l

N.B. For sake of simplicity the following calculations are done for 0



b system

3 33 2 3 3 22 2

1 1, 3 30 0

2 2 ( )( )6 3 6 2 3

l ll

b il

z l z z z l l zq y z dz q dz q dzl l

3 2

3

1 1, 3 3 2

3

2 06 3

( )2 ( ) 2

6 2 3

b i

z l z for z ll

y zz z l l z for l z l

l

33 3 3 22

21, 3 3

0

2 2 ( )3 2 6 2 3

ll

il

q z q z z l l zl z dz dzl l

1, 1, 1, 1,1 65 1 5 11 11 5 1.724 192 64 12 32 32i i i i

ql ql ql q q



Then is:

1,11.72 0.7 1.72 0.156 1.36 q q q q

33 2

1 13 02

( ) ( )ll

b bl

q y z dz q y z dz For beam 1:

For beam 2:

For beam 3:

For beam 4:

1,21.72 0.4 1.72 0.688q q q 1,31.72 0.1 1.72 0.172q q q

1,41.72 0.2 1.72 0.344q q q



1,1 2,1 27 / 2 13.5 /q q kN m

1,1q

3.0

1 LaneDistributed loads

1.5 1.5

27 /kN m

2,1qOn 1st beam

due to 1st laneOn 2nd beam

due to 1st lane

3.0

2 Lane

1.5 1.5

7.5 /kN m

2,2q 3,2qOn 2nd beam

due to 2nd laneOn 3rd beam

due to 2nd lane

2,2 3,2 7.5 / 2 3.75 /q q kN m



3,3 4,3 7.5 / 2 3.75 /q q kN m

3,3q

3.0

3 Lane

Distributed loads

1.5 1.5

4,3qOn 3rd beam

due to 3rd laneOn 4th beam

due to 3rd lane

7.5 /kN m



The shear due to distributed loads is then:

Beam 1 - lane 1:

Beam 2 - lane 1:

Beam 2 - lane 2:

Beam 3 - lane 2:

Beam 3 - lane 3:

Beam 4 - lane 3:

,1,1 1,11.36 1.36 13.5 18.36 dV q kN,2,1 2,10.688 0.688 13.5 9.29dV q kN ,2,2 2,20.688 0.688 3.75 2.58dV q kN ,3,2 3,20.172 0.172 3.75 0.65dV q kN ,3,3 3,30.172 0.172 3.75 0.65dV q kN ,4,3 4,30.344 0.344 3.75 1.29dV q kN

Pay attention to the signs !See next slide



Location of uniformly distributed loads

1 Lane 2 Lane

3 Lane

The influence line on beam 3 has for lane 3 opposite sign with respect to lane 2 (point a).

The influence line on beam 4 would have been negative in c but is positive in b

c

b

The distributed load on lane 3 is on the opposite side with respect to lane 1 and 2.

a



Distributed loads : total shear in mid-span

, ,1,1 ,2,1 ,2,2 ,3,2 ,3,3 ,4,3

18.35 9.29 2.58 0.65 0.65 1.29 31.5

S distributed d d d d d dV V V V V V VkN

We add the contribution of the three lanes

, 32S concentratedV kNWith Courbon model it was

0% difference

es_02_courbon & engesser methods

Documents

torinodepartment of

load analysis dead load

load analysis3

knmpermanent load

load value

tb sg g g g kn dead

p vrs ppbg g g g g lkn

lbkng b h