asc-exercise 1-solution

Advanced Structural Concrete Page

1/12

Exercise 1 Solution

mle/lg

Dimensioning of a diaphragm

Task 1

Geometry

tw =

t us =

tls =

ho = h =

bo/2 =

l = ls/2 = Figure 1: Cross-section of the bridge girder

Material Properties

Concrete C40/50 40MPa; 3.5MPack ctmf f

24MPa; 1.25MPacd cdf

3 36.3 GPacm E cmE k f

Steel B500B 500MPa; 435MPads skf f

205 GPasE

Prestressing Steel Y1860 1860MPa; 1390MPap pdkf f

195 GPapE

Loads

The load 1 2 1 2 6 MNd L L R RF V V V V is acting at both ends of the diaphragm.

The deadweight is neglected.

a) Suspension of the entire load on the diaphragm

Dimensioning of the suspension reinforcement

2

,

2 Choice: 27 x 26

k

6MN13793 mm

435MPa

14337 mm

6.2MN 6. 0MN

o

d

s erf

sd

s

Rd s sd d

FA

f

A

F A f F

tw

t =

d

10

00

mm

Figure 2: Layout of the suspension reinforcement (concentrated reinforcement)

Due to the large bending radii and anchorage length of Ø26 bars as well as spatial constraints, the

possibility of using an anchorage plate should be checked (In the sketch, the bending radii are not to

scale).

SIA 262

Tab. 3

Tab. 8

3.1.2.3.3

Tab. 5/9

3.2.2.4

Tab. 7/10

3.3.2.4.2

According to task

sheet

cnom= 45 mm

Distance between rebars

ok

8

ok

( 3 2 )

2

166mm

( 9 2 )

85mm

w nom

max

d nom

max

t ca

D

t cb

D


2/12

Exercise 1 Solution

mle/lg

Dimensioning of the diaphragm with a strut-and-tie model

The diaphragm acts like a cantilever. The upper reinforcement is distributed transversely in the upper

bridge slab. For the concrete compressive zone, only the width of the membrane element is considered.

In the lower part of the diaphragm, a bi-axial stress state with the compressive strength fcd results due to

longitudinal and transversal bending.

Note:

It is also possible to take part of the lower bridge slab into account. This would, however, require

additional reinforcement to spread the compressive force as well as a check of the interaction of the

stress field resulting from the load distribution in the longitudinal direction. This is not done in this

exercise.

Assessment of the reinforcement in the upper chord:

2

,

22

4.2m( 0.5 2.35m)

2.35m

Choice: 52 x 26@100, in two layers

6MN27340

9

7

mm0.9 0. 435MPa

2652 mm2 608

4

d

s erf us

sd

s

FA h

f

A

ld t

d

Concrete compression zone:

27608 435500mm

1000 24

s sd

d cd

A fc

t f

Inner lever arm:

5002350 2.1m

2 2

cz d

Figure 3: Stress field and strut-and-tie model for the case of the suspension of the entire load


3/12

Exercise 1 Solution

mle/lg

− Reinforcement in the tension chord

The suspension of the load leads to a concentrated load introduction in the upper chord. The load can

be carried directly with the reinforcement above the web. The shear connection of the webs to the upper

flange is ensured by a reinforcement that spreads the force. This distribution reinforcement ensures the

activation of the distributed longitudinal reinforcement over the width of 2.5 m. The distributed

reinforcement needs to be superimposed with the main bending reinforcement in the longitudinal

direction.

Figure 4: Detail of the spreading of the force in the upper bridge slab

1

2 2

9000kN 3000kN kN0.5 maximum : 0.5 1429

2.1m m

kN) 1429 tan(35 ) 1000

m

) 1tan( )

kN 2760

tan(

tan(

8mm mm 110 174

20 43

m .5m m

sup

xyd xyd

xyd f

xyd

qd xyd f

sl sd f

sl

dFn n

dx

n

nf n

a f

a

− Shear reinforcement

2

,

2

6MN mm6568

cot( ) 2100 435 cot(45 ) m

Choice 18@150 4-legged stirrups

mm6785

m

Ed

sw erf

sd

sw

aV

z f

a

− Check of compression diagonal

sin( ) cos( ) 2

k

100 sin(45 ) cos(

6MN

1000

= 5.7 MPa 13.2MPa

4

)

o

5

Ed

c

w

c cd

V

zt

fk

See also Figure 5

See Figure 4

SIA 262

4.3.3.4.3

SIA 262

4.3.3.4.6

kc = 0.55


4/12

Exercise 1 Solution

mle/lg

fqd

Figure 5: Spreading of the force in the upper bridge slab


5/12

Exercise Solution

mle/lg

b) Load transfer without suspension reinforcement

Global stress field

Figure 6: Global stress field (web and diaphragm)

Dimensioning of the diaphragm with a strut-and-tie model

The upper reinforcement is distributed in the upper bridge slab and the compression zone is limited to

the width of the diaphragm like in task 1a). The load is introduced over the height of the diaphragm.

This requires a longitudinal reinforcement in the diaphragm, which resists the longitudinal forces.

It should be noted that the strut-and-tie model only works because the dead weight of the diaphragm is

neglected. The additional load must be guided around the opening, which is shown in the lower figure.

As an alternative to the load deviation, the shear force could be guided directly into the support (dashed

line). In that case, the stress fields of the fans and the parallel fields overlap.

− Reinforcement in the tension chord

,

22

12MN

Choice: 52 x 26@100, in 2 layers

26 π 52 27608mm

4

27608 435 12MN 12MN ok

d sup

s

Rd d

T

A

T T

The shear connection is ensured with distribution reinforcement like in task 1a). The same required

maximum reinforcement content results, grading of the reinforcement content is possible according to

the flow of forces in Figure 8.

− Distribution reinforcement

kN kN1822 1740 ok

26@150, in 2 layers

m m

Rd qdf f

− Distributed longitudinal reinforcement in the diaphragm

2

,

2 2

6 mm6568

2.1 435 m

18 mm Choice: 18@150, in 4 layers 4 6768

4 0.15 m

6786 435 2.1 6.2MN 6MN ok

d

s erf

sd

s

Rd d

Ta

z f

a

T T

Stress field with

suspension reinforcement

Stress field without

suspension reinforcement


6/12

Exercise 1 Solution

mle/lg

Figure 7: Stress field and strut-and-tie model for the case without suspension reinforcement

− Check of compression diagonal (parallel field)

65.7 MPa

sin( )cos( ) 1000 2.1 sin(45°)cos(45°)

5.7 MPa 13.2 MPa ok

d

c

d

c c cd

V

t z

k f

− Stirrup reinforcement same as is task 1a)

18@150, 4-legged


7/12

Exercise 1 Solution

mle/lg

Figure 8: Spreading of the force in the upper bridge slab


8/12

Exercise 1 Solution

hs/lg

Task 2

(1) Entire suspension reinforcement in the intersection area (Task 1a))

(2) Part of the web activated for suspension

(3) Part of the diaphragm activated for suspension

(4) No suspension reinforcement (task 1b))

Längsträger =

web

Querträger =

diaphragm


9/12

Exercise 1 Solution

gat/lg

Task 3

General

− Load is directly suspended in the intersection area

− It turns out that the tension chord reinforcement as well as the width of the compression zone can

only be kept within the width of the diaphragm

− Treatment of the prestressing reinforcement as anchorage and deviation forces

Prestressing forces

2

0,1

,1 0,1

1 1

1 1

,1, 1

,

,1 1

1

,

27 0.7 5273kN ( 150 mm )

0.85 4482kN

tan 2 7

f

.0 ( 10 m, 305mm)/ 2

cos 4449kN

543

Deviation forces (simp

i

i

s n

l

p pk p

x

z

P A f A

P P

fL f

L

PP

P

,1 ,1, ,1, 12 2

0,2

,2 0,2

2 2

8 8 kN543 5 4449 0.305 108.5

2 m

6

i

15 0.7 2930kN

0.85 2491kN

cation: acting vertically):

0

15.2 with

1

z x

p pk

lP P f

l

P A f

P P

f

u

,2,

,2,

,2m

240480 mm kN

654

Deviation forces: 130.kN

7

x

zP

u

P

The nodal zones are dimensioned according to the loads resulting from the load introduction in the

anchorage zone. The resulting stresses inside the fan can be determined from the geometry of the

nodes. The stress field is idealized as point-centred fans at the supports.

The point-centred fan intersects with the opening of the bridge (Figure 9). Hence, an additional

horizontal reinforcement is needed above the opening due to the deviation forces. The loads in this area

are generally small, so the additional reinforcement that would anyways be placed around the opening

is probably sufficient to withstand them (needs to be checked).

Reinforcement in the tension chord

The required passive reinforcement is reduced due to the prestressing. Passive reinforcement is only

placed directly above the web so that no distribution reinforcement is needed.

2

,

2Choice:

N

20 30 o

57345734kN

l

13182mm435

@100, in tw ayers

5

20 707 14140mm

43 14140 6151k

d s erf

s

Rd

T A

A

T

Check of the compression chord

Height of concrete compression zone:12589

524 mm24 1000

c

The resulting height of the compression zone is slightly higher than the assumed height (500 mm). The

effective height of the compression zone can be determined by an iterative process (inner lever arm

influences the height of the compression zone). Due to the small deviation, this step is not necessary

here.


10/12

Exercise 1 Solution

mle/lg

5306

5306

2904

5809

36

191200930

90

2406

284

1

4449543

2406654

5770 57342866

9721 125894415

6000

6000

2440

1127

5931 5306

503 503 191

600

025

05

Td

Cd

Figure 9: Stress field and strut-and-tie model for the case with prestressing

Shear reinforcement in the diaphragm: same as in 1a) and b) 2

k

Choice: 18@150, 4-legged 6 8mm

m

6.2 MN 5.9 MN o

7

6sw

Rd d

a

V V

Concrete compression diagonals (parallel field)

5.1MPasin( ) cos( ) 1000 2100 sin(45 ) cos(45 )

5.1MPa 0.55 24 13.2MPa

5306

d

cd

w

cd c cd

V

b z

k f


11/12

Exercise 1 Solution

mle/lg

Superposition of a fan and a parallel field

A superposition of a fan and a parallel field is necessary in the area of the upper anchorage. The

controlling point lies at the intersection of the flattest element of the fan with the parallel field,

since the stresses in the fan originate from the nodal zone. The stress inside the fan decreases

inversely proportional to the distance from the nodal zone.

z 0 =

50

0 m

m

θ0 = 14°

ξ0 = 329 mm l 0

= 2060 mm

ξ

59°

45°

fc

σ’c c < f

Figure 11: Superposition of a fan and a parallel field

On the upper fan boundary, the stress is:

0 0 0

9.6MPasin( )cos( ) 1000 500 sin(14 )cos(14

=)

1127

d

c

w z

V

b

Due to the widening of the fan, the stress decreases inversely proportional: 1

0

1

0

1'

( ) 1

1

( 329 a

24

9.624 19.4MP20

)60

1 329

c

c

c c

c

f

fl

The two compression states intersect in an angle 0 9°5 . The superposition can be displayed

with the Mohr Circle.

=

Figure 11: Superposition of the two compression states displayed in Mohr’s circles

The resulting main stress resultant is 21.4MPacd . The check of the concrete strength can

therefore not be fulfilled with a reduction factor of 0.55ck . The concrete strength can be

increased with confinement reinforcement according to SIA 262 4.2.1.8. It is assumed that the

spiral reinforcement of the anchorage is sufficient to increase the concrete strength.

The same generally applies to the areas of the boundary of the nodal zones of the fan since fcd is

assumed there.


12/12

Exercise 1 Solution

mle/lg

Reinforcement against splitting of the load introduction zone of the prestressing reinforcement (spiral

reinforcement)

Transversal tensile force:

1

0,1

1

2

588kN4 2 2

188kN

TPb a b

T

Commonly, the spiral reinforcement is taken into consideration with 250 MPas

to avoid the

calculation of crack widths.

21

,

22

2352mm250MPa

6322 6.32 12 cut surfaces of

k

the spiral10

Spiral with 18 every 100mm (= )

20

3

2

18

12 054mm o4

s erf

s

T

A

b

A

s

ns

Reinforcement against spalling

To avoid the spalling of the concrete around the anchorage zone, the following reinforcement is

suggested by Eurocode Annex J:

20

2 18 in every direction

0.03 436mmsp p

sd

AP

f

: symbol for

flooring

Assuming a squared zone: 2

0

0 {632 ; 471}mm

Assumption: 350mm (anchorage plate)

c cd

erf

c cd

P b k f

P

k f

a

b

asc-exercise 1-solution

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