chapter 8 imposed deformations - applications

7/23/2019 Chapter 8 Imposed Deformations - Applications

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8 Imposed deformation: Practical applications

8.1 Mechanics of composite structures1

8.1.1 Introduction

In the preceding chapters, structures are often split into elements or layers, having different

properties. Fig. 8.1 shows a typical example. The example concerns a wall monolithically

connected to a rigid base slab. The base slab is below ground level and has a more or less

constant temperature. The wall rises up to above ground level and is, therefore, subjected to

temperature variations and drying. A temperature drop causes shortening of the wall. The

shortening is restrained at the construction joint with the base slab. The structure is schematised

as a system consisting of two layers: the wall and the slab. For such systems, the mechanics of

composite structuresapplies when it comes to the quantification of the cross-sectional forces,stresses and the risk of cracking.

Fig. 8.1 Cooling of a wall (hardened concrete), monolithically connected to a foundation

slab

Very often, the individual layers have different properties. This is the case, for example, in the

early stage of a wall cast on a rigid foundation (Chapter 7). In the latter case, the E-modulus of

the cooling wall is, for example, 20000 MPa, whereas the E-modulus of the slab is 35000 MPa.

In this chapter, problems of imposed deformations are solved using the mechanics of composite

structures. Many of these problems are the same as the ones from a set of standard cases,

namely:

young to old thin to thick cold to warm dry to wet inside outside

In the following sections, the expressions given are needed for the calculation of stresses in

composite structures subjected to imposed deformations.

1 ) The term Composite structure refers to (concrete) structures that consist of rigidly connected

elements having different properties. The term composite refers to the macro-level. The term compositeis used for materials made from different constituents too, in which case it refers to the micro- or mesolevel.


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8.1.2 Cross sectional properties

3

2

1 Zi

Zs

E A1 1

E A2 2

E A3 3

Axial stiffness of one single layeri

Ki=EiAi (8.1)

Axial stiffness of a system consisting ofn layers

Position of the elastic centre of gravity of a composite structure

Distance of the centre of gravity of an individual layer to that of the composite structure

ai=zs-zi (8.4)

Flexural stiffness of an individual layer i

Si=EiIi (8.5)

Flexural stiffness of a system consisting ofn layers

AE=)EA( iin

1=is (8.2)

)(EA

AE.z=z

s

iii

n

1=is

(8.3)

ii2

n

1isii

n

1iis AE.)z(zIE(EI)

==

+= (8.6)


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8.1.3 Response under external loading

External axial force in the elastic centre of gravity of a composite cross section:

N0

E , Ai i

Flexural moment on composite cross section:

M0

1

2

aih

Force in individual layer:

0( )

i ii

s

E AN N

EA= (8.7)

Curvature of composite structure:

)(EI

M=h

+||=s

o21s (8.8)

Strain and axial force in individual layeri :

Strain:)(EI

M.a=.a=s

oisii (8.9)

Normal force:)(EI

)AE(.a.M=A.E.=Ns

iiioiiii (8.10)

Curvature and flexural moment in individual layeri :

Curvaturei o

ii i s

M M = =

( ) (EI)E I (8.11)

Bending moment)(EI

)(EI.M=M

s

ioi (8.12)


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8.1.4 Response under internal forces

Assume a system built up of different layers. One of these layers experiences a change of

temperature (Fig. 8.2a). For the determination of the stresses in the cross-section, the standard

procedure explained in section 4.2.4 is used.

First, the layer of which the temperature changes, is assumed to be able to deform freely.The shortening of the layer, element k, is k. We then apply an axial force N

* that

eliminates the temperature-induced deformation (Fig. 8.2b). The force is:

N*= k. (EkAk) (8.13a)

After application of this force, the layers are again connected to each other.

The forceN*is now applied to the composite cross-section, but with the reverse sign (Fig.8.2c).

By moving the force N* to the elastic centre of gravity and by inserting a compensatingmomentM

*we obtain (Fig. 8d):

M*=N

*. e (8.13b)

Forces and moments in all layers can now be determined. In formula (8.13b) e is the

distance of the axis along which the force N* works to the centre of gravity of the

composite cross-section.

For the forces in the different layers, the following expressions hold:

N*

N*

N*

M*N*

1

k

b

d

a

c

k redured by T

Fig. 8.2 Calculation procedure in case of a temperature drop of one single layer


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Axial force in arbitrary layer i :

Due to force N*(formula (8.7)):

N.)(EA

AE=N*

s

iii (8.14)

Due to momentM*(formula (8.10)):

a.)(EA.)(EI

M=N ii

s

*

i (8.15)

Bending moment in layer i

From formula (8.12):

)(EI

IE.M=Ms

ii*i (8.16)

Axial force in layer k (directly loaded, Fig. 8.2a,b):

a.)(EI

)(EA.M+N.

)(EA

)(EA-N=N k

s

k**

s

k*k (8.17)

Bending moment in layer k (directly loaded, Fig. 8.2a,b)

From formula (8.16):

IE.)(EI

M=M kk

s

*

k (8.18)

8.2 Calculating with composite cross sections - examples

8.2.1 Long wall, monolith ically connected to a foundation slab

a. Stress distribution shortly after cooling of the wall

To illustrate the foregoing, a case of a long wall monolithically connected to a rigid base slab is

considered. The height of the wall is 8,0 m and wall thickness is 1,0 m. The width of the base

slab is 6,0 m and the thickness is 2,0 m. (Fig. 8.3). This situation is a typical stage in the cooling

process of a hardening concrete wall cast on a rigid slab. In chapter 7, we have considered this

case in more detail. In the following, a more general, less sophisticated calculation procedure is

shown. In this procedure, many non-linearities that play in role in the stress development in

hardening concrete are not accounted for. As a consequence, predicted stresses are less accurate.

The procedure, however, has the charm of its simplicity.


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6

1

6

23.52

8

(dimensions in m)

( T)

N*

N*

M*

Fig. 8.3 Wall, cast on a rigid base slab.

First, the situation shortly after casting of the wall is considered. In that stage, the temperature ofthe wall increases because of the liberation of the heat of hydration. The strength and the

stiffness of the concrete increase. After some time, the tensile strength and the E-modulus have

increased substantially and the wall starts to cool down. The wall is in axial direction restrained

by the base slab and, as a result, tensile stresses develop in the wall. In order to calculate stresses

in the concrete, some simplifications are required.

The E-modulus of the hardened slab is assumed to be 30,000 N/mm2. Furthermore, it is

assumed that one week after casting of the wall, theEcof its concrete is 20000 N/mm2and its

mean tensile strength is 2,0 N/mm2. The temperature drop due to cooling of the wall is

estimated at 20C.

For the structure, the following quantities hold:

Axial stiffness

Wall : (EcA)1= 20000 8 = 0,16 106MN

Base slab : (EcA)2= 30000 12 = 0,36 106MN

Flexural stiffness

Wall : (EcI)1= (1/12) 1 8

3 20000 = 0,85 10

6MNm

2

Base slab : (EcI)2= (1/12) 6 2

3 30000 = 0,12 10

6MNm

2

Position of the centre of gravity, measured from the bottom side of the slab (formule (8.3)):

z= (6 8 20000 + 1 12 30000)/((0,16 + 0,36) 106) = 2,53 m

(EI)s= (0,85 + 0,12 + 3,472 0,16 + 1,53

2 0,36) 10

6= 3,74 MNm

2

Assume that the wall is not connected to the base slab. The wall then shortens due to a

temperature-induced shrinkage strain (T) = Tc= 20 10-5 = 0,2 mm/m. To restore its

original length, a force is required (compare Fig. 8.2b):

N*= 0,2 10-3 20000 8,0 1,0 = 32 MN

Now, the wall is again connected to the foundation slab. The force N* is applied to thecomposite structure at the same height, but with a reverse sign. In order to simplify the


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calculation, this force is replaced by a combination of a force N*in the centre of gravity of the

composite structure and a flexural moment (Fig. 8.2c):

M*= 32 (6,0 2,53) = 111 MNm

It is assumed that bending of the wall is not prevented.

With the expressions (8.14) to (8.18) the following stresses in the wall and base slab are found:

Wall - top : wb= - 1,65 N/mm2; bottom : wo= + 3,10 N/mm

2

Base slab - top : fb= - 1,38 N/mm2; bottom : fo= + 0,40 N/mm

2

The stress diagram is in Fig. 8.4.

-1.65 N/mm

+3.1

-1.38

+0.400.4

5

1.5

5

5.2

2

2.7

8

Fig. 8.4 Distribution of stresses over the height of the wall

The diagram is easily checked by multiplication of the areas of the stress diagram with the

width of the corresponding element in the structure. The sum of the internal forces should then

be zero.

b. Determination of the crack width Short term

Because the mean tensile strength of the wall in this stage of hardening was assumed to be

2,0 N/mm2, it is clear that cracking occurs at the bottom part of the wall. For reasons of

simplicity, it is assumed that cracking occurs over the full height of the tensile zone, i.e. over a

height of 5,22 m; see Fig. 8.4. Now, the questions are how the crack pattern looks like and atwhat spacing cracks appear. Fig. 8.5 shows the crack pattern in an indicative way. If a crack

forms over the assumed height, the concrete immediately adjacent to the crack relaxes. The next

crack, having the same length as the previous crack, occurs only if the tensile stresses are

present again at some distance from the first crack. Assuming that the stresses spread into the

wall under an angle of about 45, each next crack has the same length and occurs at a distanceof about 5,22 m from the previous crack. Between those cracks, new cracks can occur which are

shorter in length. In Fig. 8.5, it is indicated how the crack pattern can be constructed.

N/mm2


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45

5

.22m

8

2

Fig. 8.5 Crack pattern from the stress diagram from Fig. 8.4.

Although the calculated tensile stresses in the wall are the highest directly above the

construction joint between wall and slab, the highest crack width occurs a little higher in the

wall. This is because the stiff base slab has a crack width reducing and a crack distributing

effect (as if it were a stiff reinforcing bar). Let us assume that the largest crack width occurs half

way the length of the crack. If we assume that the shrinkage strain at that position in the wallmanifests itself as a crack, the mean crack width is:

c wom

c

3,1 = = = 5220 = 0,40 mmw2 2 E 2 20000

l l

where lis the crack spacing. If no measures to prevent cracking (like cooling or isolating) are

taken, reinforcing steel has to be used to control the crack width.

For the case considered, it is assumed that the curvature (bending) of the structure is not

restrained. For the stress diagram of Fig. 8.4, the curvature is:

*-6 -1

6

s

111M = = = 30 10 m

(EI 3,74 10)

For a length of the structure of 40 m, this results in a deflection at the middle of the wall relative

to the ends of the wall of 5 mm. In a relatively soft soil, this does not result in high bending

moments in the structure.

In the case of very stiff soil, or a rigid foundation slab or a slab founded on piles, the situation in

terms of risk of cracking becomes more severe. This is since in those cases, the wall cannot

bend and, as a result, the tensile stresses have the same magnitude over the full height of thewall. Cracks, once they occur, then develop over the full height of the wall. If bending is fully

restrained, these major cracks develop at a mutual distance of 1,0 to 1,5 times the height of the

wall. In Fig. 8.6, a crack pattern is shown where the crack spacing of the major cracks is 1,5 h.

For a height of the wall of 8,0 m and a temperature-induced deformation of 0,2 mm/m, the crack

width wm= 1,5 8 0,2 = 2,4 mm if no reinforcement for crack width control is used.


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1:1.5

1.5 h

h

Fig. 8.6 Crack pattern in a wall in case of fully restrained bending

c. Effect of creep on long-term behaviour

The same wall is now assumed to experience both temperature-induced shrinkage and drying

shrinkage. The shrinkage strains are generated very slowly and the corresponding stresses are,

therefore, subjected to relaxation. For the calculation of the stresses, we start from a temperature

differential between wall and slab of 20C and a difference in shrinkage of 0,2010-3. In a wall

free to deform, this results in a difference in strain between wall and slab of:

= Tc+ shr= 20 10-5

+ 0,20 10-3= 0,40 10

-3m/m

The effect of relaxation on the magnitude of the stresses is approximated by using a reduced

modulus of elasticity. The reduced modulus of elasticity is calculated as a function of the creep

coefficient. For the wall a creep coefficient = 3,1 is adopted; for the slab it is = 1,9. At a

short-term modulus of elasticity of the concrete of 30000 N/mm2, the long-term moduli can be

calculated:

wall : Ec,=Eco/(1 + ) = 30,000 / (1 + 0,8 3,1) = 8620 N/mm2

foundation slab : Ec,=Eco/(1 + ) = 30,000 / (1 + 0,8 1,9) = 11900 N/mm2

For the long-term tensile strength of the concrete, it holds (see section 4.3.6.2):

fct,= 0,7 (1 + 0,05fccm)

where 0,7 is a long-term factor. For an estimated value of the compressive strength at later ages

offccm= 40 N/mm2, we obtainfct, = 2,1 N/mm

2.

In the same way as in the previous sections, the stiffness of the wall and slab, including creep

effects, are calculated:

Axial stiffness

wall : (EA)1= 8620 8,0 = 0,069 106MN base slab : (EA)2= 11900 12,0 = 0,144 106MN composite structure : (EA)s= (0,069 + 0,144) 106= 0,213 106MN

Bending stiffness

wall : (EI)1= (1/12) 1,0 8,03 8620 = 0,368 106MNm2 base slab : (EI)2= (1/12) 1,0 2,03 11900 = 0,048 106MNm2


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10

2.6

1m

N =27.6 MN*

N = -27.6 MN*

1

2

M =N x e* *

Fig. 8.7 ForcesN*andM

*on the composite structure

The position of the centre of gravity from the bottom side of the base slab is (Fig. 8.7):

z= (6,0 8,0 8620 + 1,0 12,0 12000) / ( (0,069 + 0,144) 106) = 2,61 m

For the bending moment on the composite structure is follows:

(EI)s= (0,368 + 0,048 + 3,392 0,069 + 1,612 0,144) 106= 1,58 MNm2

The same calculation procedure as presented under a) is now followed. Starting point for the

calculation are the forcesN*andM

*(see also Fig. 8.7):

N*= EcA1= 0,40 10

-3 8620 8,0 = 27,6 MN

M*=N

*(2,0 + 8,0/2 -z) = 27,4 (6,0 2,61) = 27,4 3,39 = 93,6 MNm

For the stresses in critical points of the cross-section, it is found:

1. Due to the axial tensile force in the wall:

wb= wo=N*/A1= 27,6 10

6/ 8,0 10

3 10

3= +3,45 N/mm

2

2. Due to N*on the composite cross section (formula (8.14)):

wall6

*11 6

s

(EA) 0,069 10= . = - 27,6 = - 8,9 MNN N

(EA 0,213 10)

wb= wo=N1/A1= - 8,9 10-6/ ( 8,0 10

-3 10

-3) = -1,1 N/mm

2

base slab N2=N*-N1= - 27,6 - (- 8,9) = - 18,7 MN

fb= fo=N2/A2= - 18,7 106/ ( 12,0 106) = -1,56 N/mm2

3. Due to M*on the composite structure (formula (8.15)):

wall *11 1s

(EA) 0,069= . . = 93,6 3,39 = 13,86 MNN aM

(EI 1,58)

wb= wo= - 13,86 106/ (8 103 103) = - 1,73 N/mm2


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base slab *22 2s

(EA) 0,144= = 93,6 1,61 = 13,73 MNN aM

(EI 1,58)

fb= fo= + 13,73 106/ (2 6 10

6) = + 1,14 N/mm

2

4. Stresses due to momentMper layer (formula (8.16)):

wall 1 1*1s

93,6E I= . = 0,368 = 21,8 MNmM M(EI 1,58)

2wb 21

6

21,8= - = - 2,04 N/mm

1, 0 8,0

wo= - wb= + 2,04 N/mm2

base slab 2 2*fs

0,048E I= . = 93,6 = 2,84 MNmM M(EI 1,58)

2fb 21

6

2,84= - = - 0,71 N/mm

6,0 2,0

fo= - fb= + 0,71 N/mm2

The resulting stress distribution is in Fig. 8.8. A comparison of Fig. 8.8 with Fig. 8.4 shows that

the first case (linear elastic, no creep) is dominant. If the structure is to remain uncracked aftercasting, for example by artificial cooling of the concrete, thef2


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5.

19

1.

56

0.

44

2.

81

- 1.42 N/mm

-1.11

+2.64

+0.31

Fig. 8.8 Distribution of stresses over the height of the structure from a temperature drop

and drying shrinkage of the wall. Stresses include the effect of relaxation.

8.2.2 Hollow core slab with compressive top layer

We assume a hollow core slab with a compressive top layer as shown in Fig. 8.9. The slab has

an E-modulus of 40000 N/mm2, whereas the top layer has an E-modulus of 3000 N/mm

2. The

shrinkage differential between the hollow core slab and the compressive layer is estimated at

shr= 20 10-5. For both elements, a creep factor of = 2,5 is adopted. For this composite

system, the stresses in the top layer are to be determined.

For the determination of the stresses, we start by disconnecting the slab from the top layer.

Because of shrinkage of the top layer, a difference in length of the two elements occurs. This

has to be eliminated by introducing a tensile force N1 (Fig. 8.10a). Then, the top layer is

connected again to the slab (Fig. 8.10b). The force N1, indicated with a dashed line in Fig.

8.10b, is now replaced by a force No= N1and a moment Mo=N1a, acting in the centre ofgravity of the composite structure.

200

50

1200 mm

compression layer

A = 165800 mmc 2

I = 1427 10 mmc 4. 6

Fig. 8.9 Hollow core slab with compressive top layer

N/mm2


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N1N1A1

a

N =|N |0 1

M0

0 1 1= N /A

a. Disconnecting compressive layer b. Forces on composite structure

Fig. 8.10 Procedure for determining stresses in a hollow core slab with top layer

In this example, we have a system in which the stresses develop slowly. Stresses are, therefore,

subjected to relaxation. For calculating the stresses, a reduced modulus of elasticity is used:

+1

E="E

coc

For the compressive top layer we get:

2c1

30000" = = 10000 N/mmE

1 + 0,8 2,5

and for the hollow core slab:

2c2

40,000" = = 13333 N/mmE

1 + 0,8 2,5

For the tensile force N1it holds:

'' -5 31 r c1 1N = E A = 20 10 10000 (50 1200) = 120 10 N

Directly after stretching the top layer we get (Fig. 8.10a):

o=N1/A1= 120 103/ (50 1200) = 2,0 N/mm2

For the momentMoit holds:

Mo=N1a= 120 0,096 = 11,5 kNm

While ignoring the intermediate steps, the result obtained is in Fig. 8.11. The maximum tensile

stress in the top layer is 1,0 N/mm2. At the bottom side of the hollow core slab, a tensile stress

of 0,6 N/mm2occurs.


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ForceN1, required to bring the shortened

top layer back to its original length

ForceNo= N1applied in opposite direction in thecentre of gravity of the composite structure

MomentMo=Noaon the composite structure

Resulting stress diagram

M = 11.5 kNm0

a

N =|N |0 1

N = 120 kN1

+0.70

+0.60

+0.10

+1.20

-0.66

-0.49

-0.84

= -0.46

= +0.20

= -0.64

-1.30

Fig. 8.11 Steps for calculating stresses in a composite structure

8.2.3 Thermal stresses in a chimney

In chimneys, hot smoke discharged can heat up the inside of the chimney shaft. Consider a

segment of a cylindrical wall, heated from the inside. Due to inside heating, the segment has a

temperature differential between inside and outside of the wall. The segment would like to bend

as shown in Fig. 8.12a. Since bending is restrained due to the cylindrical shape of the wall, a

momentMooccurs which fully eliminates the curvature caused by the temperature differential.

This moment causes tensile stresses at the outside part of the wall, possibly leading to vertical

and horizontal cracks. Fig. 8.12b shows an example of vertical cracks. This example dates backto the seventies of the last century. The installed reinforcing steel was plain steel and appeared

to be not sufficient to control crack widths properly: the reinforcing steel was beyond the yield

stress.

+2 0

+1,05


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M0

M0

M0

a b

a. Bending of segments at inside heating b. Cracking of factory chimney [2]

(only bending of the segments is shown)

Fig. 8.12 Deformational behaviour of horizontal segments of a chimney at heating from the

inside (a) and corresponding crack pattern (b).

In case the wall is composed of several layers, the calculation scheme from Fig. 8.12 must be

extended, see Fig. 8.13. Suppose that the chimney wall consists of: 1 = steel outer liner, 2 =

insulation, 3 = fire proof brick. During warming up of the wall, the temperature distribution is

non-linear. When ignoring the eigen temperatures, the temperature profile can be split up in a

part that leads to extension of the wall (TN) and a part that results in bending (TM) (Fig. 8.13). If

TNand TMare restrained, (fully or partly), stresses occur.

h1

h2

h3

TMTNT1

T0

2

3

1

~= +

Fig. 8.13 Temperature distribution and splitting up in components

For each of the profiles (TNand TM), corresponding stresses can be calculated. For that purpose,

the layers are first disconnected, so that they can deform freely (Fig. 8.14a). Subsequently,

forcesNiand momentsMieliminate the deformations. It holds:

Ni= iTNiEiAi

Mi= iEiIi

where:

mii

i

T=

h

i


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16

M1

M2

M3

M1

M2

M3

N1

N2

N3

N1

N2

N3

+

+

+

a. Free deformation of individual layers b. Eliminating deformations of individual layers

N +N +N1 2 3

M +M +M +1 2 3 N ai i.

c. Forces in composite structure

Fig. 8.14 Calculation procedure for internal forces due to a temperature differential over the

wall thickness.

The resulting axial force is applied as a tensile force in the elastic centre of gravity of the

composite structure (Fig. 8.14c). Furthermore, a resulting bending moment

Mo= Mi+ Ni ai

is applied, where ai is the distance from the centre of gravity of an individual layer i to the

centre of gravity of the composite structure. Then, the cross-sectional forces and thecorresponding stresses are determined, which are added to the stresses (per layer) found earlier.

Fig. 8.15 shows a chimney in which the tensile stresses should be smaller than 2 N/mm2 (or

2 MN/m2). Fig. 8.15a shows how the wall is constructed and gives a global indication of the

temperature and stress distribution. In Fig. 8.15b, the heating curves for the wall are given. Fig.

8.15c shows the calculated development of the tensile stresses as a function of time.


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17

50

100

150

180

T ( C)

T

0 2 4 6 8 10 12t (h)

0

1 2 3

maximum heating rate

b. Temperature distribution in a chimney as a function of time

0 2 4 6 8 10 12t (h)

11.6

allowable stress = 2.0

3.0

2

2.2

3

0

10

2

3

ct 2(MN/m )

12

5ct

T

foam glass

fire resistand stones

T

a. Side view and cross-section c. Maximum stresses ctin chimney shaft .

Fig. 8.15 Development of tensile stresses in a chimney wall as a function of time. The rate of

heating of the wall is the parameter varied [2].

Heating curve 2 results in maximum tensile stresses of 2,2 N/mm2 and does not meet the

requirements. Heating curve 3 results in maximum tensile stresses of 1,6 N/mm2and does meet

the requirements. In order to avoid cracking, the heating rate should, therefore, be limited. This,

however, does introduce a risk. Recently, chimneys are provided with a system to remove SO2

from the gasses. As a consequence of this, the temperature control is less efficient and the

temperature can increase rapidly. In Fig. 8.15b, this is indicated with temperature curve 1. The

temperature can increase up to 180C in 5 minutes. The maximum tensile stress in the chimneywall then is 3,0 N/mm2. At this stress, cracking is very likely to occur. For that reason, it is

important to provide a concrete chimney with hoop reinforcement. In a chimney constructed

from brick, an outer steel liner or steel rings can be used.

8.2.4 Cracking in a prestressed concrete bridge due totemperature effects

In the wall of a prestressed concrete box girder bridge (spans 32,3 m and 23,0 m, see Fig. 8.16)

a 17 m long crack was observed in September 1964. Over a long section of the crack, the crack

width was 5-6 mm (Fig. 8.17) [3]. The bridge was designed such that the shear stresses in the

walls would be below the shear stress capacity of the concrete (in EN 1992-1-1- notation: below

the vRd,ccapacity). As a result, only minimum shear reinforcement was required: stirrups 12-

250 (sw= 0,12%) were applied. According to the calculations, no cracks were expected underthe external loading. The question was, therefore, which load had caused cracking.


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18

cracked area

23.00 m 32.20 m

a. Side view of the box girder bridge

510/m 510/m2.510/m

0.1

4

0.

28

0.1

1

0.6

0

50 mm asphalt50 mm concrete10 mm sealing

prestress cable

damaged body

verticale shear reinforcementin crack area, stirrups4 12/m,FeB 240

2.000.750.752.00 0.750.75 1.00

1.50 6.009.00 m

1.50

1.00

b. Cross section of the box girder

Fig. 8.16 Side view and cross-section of the Jagstbrcke in Untergriesheim [3].

Fig. 8.17 Crack pattern in box girder bridge (side view).

50 mm asphalt50 mm concrete10 mm sealing

220 mm reinforced concrete

37C

70C

Fig. 8.18 Temperature distribution in the bridge deck due to solar radiation

At the time the crack formed (in September), it was an exceptionally bright day. In the shadow,

the temperature raised up to 28C, while at night it was below zero at ground level. Due to theintensive solar radiation, the temperature of the deck (dark asphalt) could reach 70C. Thecorresponding temperature profile in the deck is in Fig. 8.18. Due to the warm deck, also the

temperature inside the box increased. In the late afternoon, the wall of the box girder was


19/19

subjected to solar radiation for another 2 hours. The temperature of the wall, therefore, reached

a value of 37C. At night, the temperature at ground level dropped to about 0C. For thecalculation of thermal stresses, a temperature differential between the centre of the wall and the

environment of 30C was assumed. Furthermore, it was assumed that the temperature in the boxdropped to 15C. Fig. 8.19a shows the subsequent calculated temperature distributions in the

wall. Fig. 8.19b shows the corresponding strains after 10 hours. The stresses follow from:

ct=EccT=Ec(T)

In case of anEcof 35,000 N/mm2and a maximum strain (T) = ~ 7,5 10

-5(see Fig. 8.19b),

the maximum tensile stress is ct(T) = 2,6 N/mm2. The thermal load described before and its

resulting stress, were not considered in the design. In combination with stresses from other

loads, the calculated stresses were found to exceed the tensile strength of the concrete. Since the

amount of reinforcing steel (stirrups) was very low, cracking of the concrete immediately

caused yielding of the steel. For repair, vertical holes were drilled in the webs of the box girder

and a vertical prestressing was installed.

0.75 ms-0.175 m

s-0.175 m

R

R

T -0outside

T -15inside

t-0h

T10h

1

510

14

T C

30

15

0.75 m

~0.16 m~7.5

10.

-5

T

T

15

0

3010-5

+

+

-

a. Temperature distribution in the wall of a box girder b. Temperature-strains at t = 10 hours.

Fig. 8.19 Temperature distribution in the web of a box girder due to cooling from the outside

at different stages (a) and temperature-induced strains (eigen strains) at a

temperature distribution after 10 hours (in case of free deformation of the web)

8.3 Literature

1 Noakowski, P., van Dornick, K., 1992. "Bewertung der Bausubstanz anhand der zugrundegelegte Bemessungsvorschriften", VGB Kraftwerkstechnik, 72. Jahrgang, Heft 3, blz. 251-

259.

2 Noakowski, P., 1987. "Einige statische Aufgaben im Feuerfestbau", Die Bautechnik, No. 2,

blz. 51-57.

3 Leonhardt, F., 1965. "Temperaturunterschiede gefhrden Spannbetonbrcke", Beton- und

Stahlbetonbau, Heft 7, blz. 157-163.

chapter 8 imposed deformations - applications

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