ASC-Exercise 1-solution
Transcript of ASC-Exercise 1-solution
Advanced Structural Concrete Page
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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
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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
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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
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Exercise 1 Solution
mle/lg
fqd
Figure 5: Spreading of the force in the upper bridge slab
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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
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Exercise 1 Solution
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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
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Exercise 1 Solution
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Figure 8: Spreading of the force in the upper bridge slab
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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
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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.
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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
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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.
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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