Example of Determination of the Traction Within a Steel Retrofit Tendon for a Masonry Wall

7/24/2019 Example of Determination of the Traction Within a Steel Retrofit Tendon for a Masonry Wall

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1.1 Example of determination of the traction within a steel

retrot tendon for a masonry wall

In this paragraph an example of calculation of the force within a retrofit tendon for a masonry wall

will be shown. The tendon is meant to prevent the out-of-plane overturning of a masonry wall of a

two story masonry building, shown in Figure 1. In order to perform calculations earthquaes

referring to the following !ones will be too into consideration"

Locatio

n

Class of use Type of soil

Ferrara III #

$imini III %

&quila III %Table 1 Sesmic zones, class of use, type of soil.

Figure 1 asonry building considered in the example.

1. Calculation of the tendon


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'oad analysis

$oof"

(ermanent load" qp) * + dam)

/ariable load" qac) * 100 da m)

Floor"

(ermanent load" qp1 * 11 dam)

/ariable load" qac1 * )00 dam)

& specific weight for the masonry of 2 * )000 dam3will be assumed.

In Figure )the static scheme of the forces acting on the wall is reported.

The geometrical specifications of the wall are as follows"

b1* 0.44 m

b) * 0.50 m

d1 * 0.54 m

d) * 0.30 m

h1 * 3.30 m

h) * 3.30 m

& constant stress distribution on the overturning section will be

assumed. The stress will be equal to the crushing stress of the masonry,

i.e. 6 * )0 dacm).

The length of the stress bloc is given by the equilibrium equation"

s

Nc

k.=

'et7s consider the overturning mechanism at the first floor. For the sae of safety, only permament

loads will be considered within the calculation of 81 and 8). The load interaxes for the first floor

and the second floor are respectively"

i1 * ).)3 m

i) * 1.) m

Figure ! "alculationscheme


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Calculation of stabilizing forces.

'et7s consider a portion of wall with a length of 1 m, 81 and 8) are"

81 * qp1 9 i1 9 1 * 34: da

8) * qp) 9 i) 9 1 * 151 da

Thus the length of the stress bloc is"

2

2 0.014.

k

Sc m

ss= =

The proper weights of the two walls are"

(1 * )000 9 3.3 9 0.44 9 1 * 330 da

() * )000 9 3.3 9 0.50 9 1 * )50 da

Calculation of the unstabilizing forces.

'et7s now calculate the seismic acceleration associated to the location of Ferrara. & nominal life /n

* 40 years will be assumed. If the class of use is III, %u will be equal to 1.4. Thus it follows that /r

* 40 9 1.4 *4 anni.

The probability of exceeding the intensity of the design earthquae in the reference period /r

related to the life limit state is equal to (/$*10;, with a return period of"

and the following

elastic spectra"

Ferrara #soil $%

&imini #soil

"%

'(uila #soil

"%

ag/g 0.162 0.212 0.3

F0 2.567 2.507 2.384

Tc*

0.276 0.304 0.356

Ss 1.776219 1.3811096 1.27088


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Cc 2.379334336 1.555396139 1.476422992

St 1 1 1

S 1.776219 1.3811096 1.27088

Figure ) Elastic spectra

The 8s and %c parameters are chosen following T% )00+ prescription"

The parameter 8, is calculated as 8 * 88 9 8T. In the paragraph .+.1.4.) of T%)00+ prescribes the

following equivalent static forces"

awa

a

aa WW

q

SF .. ==


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where"

?a * weight of the element.

qa * 3

2

1

. 3(1 / )

. 0.51 (1 / )

g

a

a

a S Z H

S g T T

+= + !

@ * centroid quota of the non structural element measured from the foundation.

A * height of the building measured from the foundation.

Ta * 0

In the following table the values of 8a and Bw for the two walls in each reference location are

shown"

Ferrara Rimini Aquila

8a1 0.3: 0.503 0.4)5

8a) 0.11 0.)) 0.+10

Bw1 0.13) 0.135 0.14Bw) 0.)05 0.)0 0.)0

The masses excited by the earthquae, are given by the formula"

++ kjj!! ))1

?here the values of C)Dare defined by T%)00+"


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Thus, for the first floor it is C)D* 0.3 =category &< whilst for the roof it is C)D* 0. Easses referred

to first floor and roof floor are respectively ?1* 5:3 da e ?)* 151 da.

8tatic force to be applied to each mass is given by"

=j

jjiihi WzWzFF ..

%oefficients 2ito be applied to each force are"

#$1

0,52702

9

#%1

1,05405

7

#$2

1,58108

6

#%2

2,10811

5

Thus, corresponding forces are equal to"

F$1

1913&1

14

F%1 519&47


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11

F$2

4174&0

67

F%2

297&11

77

2. Overturning assessment

"#erturning aroun$ %oint &'(

The tractions T1and T) for the Ferrara location are being calculated. 8tabili!ing moment is given by"

))))

))

)) .)))

hTcb

Pc

$S)stab +

+

=

)stab* 151 9 0.): )50 9 0.1: T) 9 3.3 *45).4 T) 9 3.3

Gverturning moment is given by"

+=

).. )))))h

FhF) %wwrib

)rib* Bw)=):,11.3,3 515,1.1,4< * Bw)9 ++

Thus"

T** Bw)9 )3+5.) H 15.5

eeping in mind that Bw)* 0.)05, it follows that"

T** 31: da

"#erturning aroun$ %oint +'(

11)1)11

11

111)

)1

))


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T1* B 9 +331H 1131.5 * 41 da

&ssuming a steel resistance of 6f* )500 dacm)the section of the tendon for unit length of the

wall is""

&1* T1 6f* 4+)500 * 0.)3+ cm)m

&)* T) 6f* 31:)500 * 0.133 cm)m

Aere7s reported the values of the tractions in the tendons for the three locations.

Ferrara Rimini LAquila

T1 KdamL 41 00 11))

T) KdamL 31: 3)+ 5

&tir1

Kcm)mL

0.)3+ 0.)40 0.5+

&tir)

Kcm)mL

0.133 0.13 0.1::

1.2 !"ample of calculation of a CFR# reinforcement $ithin a ringing retrofit

intervention for masonry %uildings

Figure & !"ample of the use of CFR# in the retrofit of masonry %uildings

'et7s assume that the wall at ris of overturning has a length of6l =

m and eeping in mind the

values of the tractions within the tendons calculated in the previous paragraph"


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T1 KdamL 41 00 11))

T) KdamL 31: 3)+ 5

The force that the %F$( reinforcement is given by

/ 2i

F T l= '

, thus the forces acting on the

reinforcements in the three location are"


T1 KdaL 11) 1+01 33

T) KdaL :4 :+3 1531

The reinforcement for the ground floor wall in $imini and l7&quila is being calculated. The F$(

properties will be assumed as follows"

'oung (odulus !

)da*+cm2,

Traction Resistance -t

)da*+cm2,

%arbon Fiber )30 x 105 541 x 10)

The reinforcement will be designed in both cases of good and low quality masonry.

In the first case, a masonry characteristic compression resistance of40bkf =

E(a will be assumed,

with mortar of class E14 i.e. with a characteristic compression resistance of

14.3k mk

= =

E(a.

Introducing the suitable safety factors the design compression resistance will thus be

7.15bd

f =

E(a and the average traction resistance will be assumed lie 110 of the compression resistance"

1.43mtm

f =

E(a.

In the second case, a masonry characteristic compression resistance of

5bk

f =

E(a will be

assumed, with mortar of class E14 i.e. with a characteristic compression resistance of

3.5k mk

= =

E(a. Introducing the suitable safety factors the design compression resistance will


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thus be

1.75bd

f =

E(a and the average traction resistance will be assumed lie 110 of the

compression resistance"

0.35mtm

f =

E(a.

The reinforcement will be designed both considering the masonry resistance and the bric

resistance. & one-directional carbon fiber strip of Type % will be used, thus its equivalent thicness

will be =cfr. point ).3.) #T )00)005


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-

/ 2fd b mk mtm m

s c k g* = ' ' ' '

-

fs

is the interface sliding, between 0.1-0.3 mm. 'et7s assume

0.2f

s =

mm.

-

0.06lc =

from experimental tests.

-b

k

is a geometric factor, assumed 1.

-rd

g

is a partial coefficient in case of applications of type J.

The height of the reinforcement

fb

may also be obtained through the following formula giving the

average failure loadP

"

f eP b lt= ' '

wheret

may be assumed equal to the traction resistance of masonry. & further formula for the

evaluation off

b

is the following"

(a) 2

F b E t = ' ' ' '*

where mk mtmc * = ' '

. In the following tables the value offbfor three different values of

acting force 1000, )000 e 3400 da using respectively the three formulas presented"

OO/ 0ALT' (A3O*R' OO/ 0ALT' 4RC5

F )da*, %1)mm, %2 )mm, %6 )mm, %1 )mm, %2 )mm, %6 )mm,

1000 445 0 : 331 3 51

)000 110: 1)0 13+ 3 ) +)

3400 1:50 )10 )51 110 1) 155


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LO7 0ALT' (A3O*R' LO7 0ALT' 4RC5

F )da*, %1 )mm, %2 )mm, %6 )mm, %1 )mm, %2 )mm, %6 )mm,

1000 11)0 1)1 13: :3 101 11

)000 ))51 )5) )+ 1+4 )03 )33

3400 3:)1 5)5 5+ 3)+1 344 50+

Example of Determination of the Traction Within a Steel Retrofit Tendon for a Masonry Wall

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