ABSTRACT

11th INTERNA TIONAL BRICKlBLOCK MASONRY CONFERENCE

TONGJI UNIVERSITY, SHANGHAI, CHINA, 14 - 16 OCTOBER 1997

NON - LINEAR ANAL YSIS

OF LOADED BRICK

MASONRY W ALLS

Sun Weimin Yuan Fashun

The non -linear analysis of the whole caurse for beam -end loading on brick walls is pre­

sented in this paper. The basic assumption of this topic supposes on which a compposite

planar frame is combined with two different materials,one-the reinforced concrete beam

and the other-brick masonry column. Using the method of actual rigidity with member

system as the model, the loading type is analogous to actual construction work. The analy­

sis results justified the restrained moment of the wall block is more pertinent with its up­

per pressure and upper reinforcement of the beam. The inner stress redistribution of the

wall exists in the whole caurse. The wall block with variable section may appear changing

in sign of the momento The max. moment not certainly corressponding with the max.

' loading. The max. restrained moment may appear at the construction stage.

Keywords: Non -linear; Masonry Block Frame; Rigidity ;

Sun Weimin,Msc. Associate Prvfessor,Member of China Committee for Masonry Struc­

ture Standardization, Nanjing Architectural and Civil Engineering Institute. 200 North

Zhongshan Road, N anjing ,China

709

1. CAUSE OF THE QUESTION

In the calculation of the loaded walIs of multiply stories with stiffened plan,according to

the present specification of masonry structure design in china ,the /I Hinged" assumption

always being considered, because it is convenient for practical use, so it is welicomed by

the designers, as a definite expericnce it has been prolonged for many years. ActualIy ,the

R. C. beam supported on the walIs, the beam ends would be concave upperward due to

loading itseH , With these effect, the wali would be united and being tends to rotate ,FinalIy

the walI block as a reation will produce a restrained with respects to the beam. Particulat­

ly, when adopting the more stiffened pad with much rigidity and monolithicalIy grouting,

in this case, the restrained effect could not be disregarded. Therefore, how to choose the

appropriate calculating model in considering the restrained moment at walIs, it becomes a

valuable question to discuss.

It has been analysed in reference[l] by the assupption with the linear elastic frame, but in

the opinion of this paper, as regard to this composite planar frame combined with a con­

crete beam and a brick masonry column, it exists error in analysing with the linear elastic

method. First, it is not easy to get a rationable value of masonry rigidity actual used, in

case of wall block with the same mark of brick and motar, if we take an unified value of e­

lastic modulus, if ali every stage of the stories are of the same height, it seems that the

rigidityes of the columns are the same. But it does not in fact, beca use the rigidity at the

base story is larger than that of the top story, for the axial force at base is relatively larg­

er, the base element may not crack or not so deeper in crack. Second, if the frame is com­

bined with same one material,such as alI with reinforced concrete frame,the selecting val­

ue of elastic modulus even is not exact enough, the influence with regard to the internaI

stress is not larger. When the frame combined with two different materials,it's a another

case, because the internal force is in partinent to the relative elastic modulus of two mate­

riais , therefore ,to select a more exact value of elastic modulus is obviously very important.

Owing to the brick masonry and concrete are ali of non - linear materiaIs, it is difficult to

determine a point at which we may use the secant modulus instead of elastic modulus for

elastic anlysis, it tends to make error on the calculation result of internaI force. Third, by

based on linear elastic assumption, both the behevious in non -linear and post - cracked

influences to the walI block and floor system could not be considered, therefore , it's also to

make error in calculation.

In order to solve the question previously stated in elastic analysis, we adopt a method oi

analysis based on the non -linear of the whole caurse, in connection with a plane frame

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which compoite by two different materiaIs. the brick masonry and reinforced concrete. Be­

ing anaIysed and illustrated in this paper • we write a programme and make computation as

an exampIe. It will be more significant to study and anaIyse the brick masonry structure

hereafter.

2. CALCULA TION MODEL AND BASIC ASSUMPTION

2. 1 Calculation Model

The basicassumption presented in this paper is a composite planar frame which combined

with two different materials. the reinforced concrete and the brick masonry. The analysing

modeIof nonlinear with whole caurse is adopted and the model is the member system. A

more precise method of actual rigidity is used to establish rigidity matrix with member

system • but the effect of secondary matrix of the column does not considered.

2.2 Basic Assumption

Some assumptions being introduced as following:

1. Considering the section kept in planar;

2. For brick masonry. the tensile zone of the section is not to join work. As for reinforced

concrete section. the tensile zone of the concrete is not to join work also.

3. For brick masonry .the relationship stress-strain of the compressive zone of the section

as following

in which.fm -- the average value of the compressive strength of the masonry.

4. the stress of the concrete versus to strain is :

0=00[2E/ Eo - (E/Eo)Z] E:=:;;;;EO}

0=00 Eo<E:=:;;;;Ecu

where E=O. 002.Ecu=O. 003;oo=1.lfc.

fc -- axial compressive strength of the concrete;

(1)

(2)

5. The relationship of stress --strain of steel take the ideal elastic-plastic stress versus

to strain.

3. METHOD OF CALCULA TION

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3.1 The Rigidity Calculation of the Column of the Brick mosonry.

When the actual rigidity method being used,first,the relation curve of the axial force and

moment versus curvature should be find out, for this, a method of typical relationship of

moment - curvature at the condition of which a brick masonry with eccentrical compres­

sive loaded on a rectangular section whose wide - b height - h (shown in fig 1) under a

given force N should be derived out,in general , using the method of incresing deformation

by stage and not by incresing loading by stage,in such way,it may avoid some troble in

solving the non - linear equations, the detail see reference[2].

As to the section shown in fig 1. the corresponding moment is zero,when N acting thru

the kern axis of the section, in this case, let the axial compressive strain being El' then

from equation(1)

El= - 1. 0/460J!: ·ln(1. O-N/bhfrn )

if the ultimate compressive strain of masonry being taken[l]

Emax=O. 01 J!:

(3)

(4)

Then, when the compressive strain of the section at the extreme fiber which is on the near

side of N is being arrived to Em"" ,the corresponding moment also is being arrived to max.

value Mmnx' therefore, if in incresing the strain by stage at the section fiber ,then the strain

.., will be at interval between El to Emax , but the influence of moment is more evident when

.., is less. In order to get more precision under the same job of calculation, the initial incre­

ment of strain should be taken less value ,and the remaining may increse by proper.

figl the section strain of masonry column

Let after a definite stage defomation being added, the extreme compressive fiber strain at

the near side of N being "'(El< Ec<Emax) ,suppose the relative strain at the kern axis being

Ec, (figl) ,then the section curvature:

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(5)

the distance from the neuture axis of the section to the extreme fiber on near side of N be-

mg:

(6)

from geometric relations (figl) • the strain at a distance x from the neuture axis to com­

preSSlve zone:

from condition of equilibriom:

jJ:0badX

N=

Ixo badx ·

Xo - h

substituding equation(1). (7)to equation(8-1)

(8-2)after integration and simplification:

{fN (x" -fN(O)

N-fN (xo ) -fN (xo - h)

where.function fN (t) = bfm (t+1.. • e- Pt ) p

and P=460 JÇ.tclx"

(8-1)

(8-2)

(9-1)

(9-2)

(7)

Subs. equation(5) • (6)to equation(9-1) (9 - 2) .solvin out 'êo .subs. to equation(5). then

qJ is obtained. The section moment M with correrponding to qJ will be further obtained by

follwing equations:

M={fM(Xo)-fM(O. 0)-0. 5hN

fM(x,,)-fM(xo-h)-0.5hN x" > h

in which Function fM(t)=bfm[t+ .~Pt(h-Xo+t)-O. 5t2+ \e- Pt ] p p

00-1)

00-2)

To every given N. a corresponding M - qJ curve is obtained. finally. a family of M - qJ

curves is obtained. To divide the brick masonry column to a sufficicent small and a great

number segments.assume M and N as a constant at every segment.when M and N of a

definite segment is known. the corresponding qJ by the method of double inferpolation may

be found out. then the rigidity will be found by the following equation:

B=M/qJ (1)

The rigidity calculationof brick masonry column may also be adopt with simplified method

introduced in reference[3].

3. 2 The Calculation of Rigidity of Reinforced Concrete Beam.

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The rigidity calculation of R. C. beam is wholiy similar as that of the column stated previ­

ously above, just to put N = o. Owing to limited by space, the relative calculation equations

being not listed here.

3. 3 Method of Calculation

In this paper, the non -linear analysing programme writed by the method of supposition

with variable rigidities, its calculation step is acting ali of the load at one step. Then, to ed­

just the defomation step by step until the fundamental equations is satisfied. Speaking in

detail, firstly ,from the initial tangent modolus, to find out rigidity B of every smali seg­

ment at each member (relative to initial tangent displacement {~o} =0 at the begining in

assuming and supposition), thus, further to form the total rigidity matrix [KoJ, let the

loading column matrix be [RJ, then may find the displacement of first approxmation val-

ue:

{~J=[KoJ- l { R}

Afterwords, from {~1 } to find out the moment of every smali segment of each member or

the represented value of the moment and axial force, then ,to find out its rigidity B, the

first corrective value[K1Jof the total rigidity matrix could soon be found, further, the sec­

ond approximater value of displacement is obtained:

{~2) = [K1J - l{R}

and so on until{ ~l}iS very closely near to {~1-J .

4. CALCULATION EXAMPLE

[exampleJA 5 story building,with stiffening scheme,precast slab floor construction,floor

beam bXh=200mmX500mm,the beam end is monolithicaliy casting with ring-beam.

The wide of ring beam is the same as the wali and its height is 500mm. Beams in floor

system,its spacing 3. 6m,span 5. 7m,wali thickness of the 1st story 370mm,the remain

240mm, wide of wali between windows 2. 1m, height of base story 3. 7 6m (to top of the

foundation) other story height 3. 4m, the cantilever gutter of the roof consider O. 5m as

the wide, window opening 1. 5m XL 86, the equivalent loading and reinforcement in beam

shown in fig2, (Mg ,Gg -- dead load, Mq, Qq - - live load, the loads shown are ali in

standard value), as to the class 1 steel, the yield point take 240N /mm2 ,class 2 --

340N/mm2 , brick --Mu10 ,mix mortar - M2. 5.

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2~16 (402)

2:oeD 3 20 G - 50.1 ~ctt~f----l (941) I-':

o

G=50.1 ~~:H::I------J

5700

fig 2 loading sketch

4. 1 Type of adding loads

Assume the construction tive load equal to the survice tive load, at the same time, assum­

ing the structural strength during construction reachs to 100 % of the design strength.

The whole loading of the 1st floor is divided to 10 stages added step by step,then,its live

load times O. 4, the dead load times O. 2 adding with 2 stages. The adding of second floor

is all the same as the first, etc, until up to top story.

4. 2 Calculation Results

The Calculation results as presented in fig3 to fig8, the ordinate M represent moment, the

abscissa NL represent the number of loading class.

17.40 1'7.32

16. 17.28

~ 12· 17.24

z x:. ' 6. 7.37 '- 5 ~ 4.

fig3 restrained moment at beam

end of every floor (A'. = 157)

715

4Q

E 30.

~20.

i~·

fig4 restrair ed moment at beam

encl of eVfry floor (A' .=402)

16 12

E & . 4 ~ .r..--,?-::~~~~ ....... -4 :r -8

-12 -16 NL

fig5 up section moment at wall block

oI every floor (A' 8 = 157)

27.51 25 22.&)

E 20 18!.5

~ 15 18.11 la 5

-2

Iig7 up section moment at wall

block oI every floor (A' .=402)

5. CONCLUSION

1134 E 12 '---8.68 ~ ~ 6.57 :x: ~~f--.,.:j!>.=~~~~ ........ -4 :r - 8

-12 NL

Iig6 down section moment at wall block

oI every floor (A'. = 157)

25 E20 z 15 :x: " la :r 5

..::;,

Iig8 down section moment at walI

block of every floor (A'. = 402)

This is the first time using the nonlinear analysis of whole caurse for the rigid joint with

static calculafion model to the building of stiffening seheme. Though some simplification

is made just for convence, the calculation results is still in welI corresponding with the ac­

tual internaI force distribrtion of the walI under the condition of rigid jonin stiffening

seheme. In concluding the result~ of this paper, we have:

1. The restrained moment of the walI block exists close relations with the axial force. if

the axial force is sufficienly large, it tends to restrain the rotation of R. C. beam until the

plastic hinge is formed af the beam end. As to what is the magnitude ofaxial force at

which it relate to the up reinforcement of the beam. Using the programme of this paper

suggested, it may be calculated in quantity, in general, the axial force of the top floor is

not too large ,in this case, the rotation both the beam and walI block are large ,then it is

716

nearly as approch as a hinge.

2. The restrained moment veries increasing with increasing in upper reinforcep steel of the

beam. Therefore the upper reinforcement is not suitable for large amount, otherwise it is

bad for wal1 proper.

3. Due to the base wal1 thickness from 240 increased up to 370, it makes the restrained

moment of the wal1 to produce the changing in sign, during at caurse of loading. As for

the max. moment ,it is not certanly corresponding with max. load. Care should be taken

in actual engineering.

4. In reference [2] ,a comparison befween the hinge assumption and the linear elastic

frame model has been made. With the same data as of this paper and also to check the

section, after analysis we can find that the hinge assumption is tend on safeside in gener­

al. Example in this paper, when upper reinforcement A:=157mm2 (2.010), the re­

strained moment of wal1 block is less than that of linear elastic frame calculation model,

when A:=4. 2mm2 (2.016) ,the restraine moment at each story except the third floor are

al1less than that of elastic frame models. From the check of the section to the wal1 of the

third floor ,the wal1 is safe too ,it may be further proof that the hinge assumption is appli­

cable, although its state of loading being with slightly different with the actual.

5. From conclusion, it may discover that a redistribution of inner stress is appeared in the

wall block section up and down the floor beam. Its Summation of restained moment is less

or egual to the yield momento

Owing to the programme of this paper used for quantitive analysis of a whole caurse, We

can grasp the condition of wal1 stresses at any type of loading and at any loading Stage.

Therefore, as an instrument for application ,it can be used for research in masonry struc­

tore analysis.

REFERENCES

1. Sun, W. M. ," Anlytical Models and Method of Wal1 for Multistory Structure Building

With R. C. Floors" Journal of Nanjing Architectural and Civil Engineering Institute 19~1

(4) ,pp45 - 50.

2. Zhu,B. L. ,Dong,Z.X. ,"Non-Linear Analysis of Reinforced Concrete" Tongji Uni­

vercity Press,1985,pp 50-87.

3. Sun W. M. ,"Calculation of Rigidity of compression Brick Masonry" Journal of Jiangsu

Architecture,1993(2) ,pp4-7.

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