11 th INTERNA TIONAL BRICKlBLOCK MASONR Y CONFERENCE

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

TEMPERA TURE STRESS CALCULATION

OF BRICK MASONRY BUILDINGS

Yangming Xiao

l.ABSTRACT

Based on tests, this paper proposes the approximate method of calculating temperature

stress and deformation of brick masonry structures according to the principie of con­

straint and deformation compatibility among brick masonry members. The results of

calculation present a satisfactory agreement with tests and show that formulas are

sim pie and useful. Design engineers can use the formu las for reference.

2. INTRODUCTION

Brick masonry buildings are space structure composed of concrete top pia te , floor and

block. Because the physical properties of matereials are different, slabs and walls of

each floor bear different temperature difference. Each membere of buildings will pro­

ducedifferent deformation when outside temperature changes and material shrinks. In

the meantime, the structural interior will produce temperature shrinkang stress be­

cause of the constraint among members. When the stress is beyond ultimate tensile

strength or shearing strength, varioÍls types of cracks will be appeared in the struc­

ture. Since there are so many factors that affect temperature stress of brick masonry

structure, such as tPmperature difference, material properties, the pattern and size of

Keywords: Brick masonry structure;Temperature stress; Wall cracks

Dep. of Building Engineering, Hefei U niversityof Technology, Hefei 230009, China.

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structure and construction, it is difficult to find precise methods to calculate tempera­

ture stress at present. So, it is necessary that we provide several simplified and usefu(

methods for engineering designo

3. BASIC ASSUMPTION

Fig. 1. shows the work of the slabs and walls of structure under temperature differ­

ence. When initial temperature is To; slabs and walls are locater at position 1; when

temperature rises to T ,we assume that slabs freeelongate to 4 and walls to 2. Because

of constraint between slabs and walls, they should finally be at postion 3 which is the

equilibrious position. Thus the slabs shorten Ec and walls extend Eb'

The calculation element which consists of top plate,floor and wall taken from the ac­

tual structure is shown in Fig. 2. Seperate the top plate,wall and floor and mark out

constraint shear between each part: the shear between top plate and wall is Qp and

shear between wall and floor is Qz. Then, we assume initial temperatureof structure is

T o, After temperature changing, temperature difference of top plate is TI' and temper­

ature difference of walls and floor is T z.

4. COMPATIBILlTY EQUATION OF DEFORMATION

To consider the effect of material shrinkage,we substitute the shrinkage of wall and

slab for equivalent temperature difference:

AoI =a" T I-e,

Âbl =abTz-eb

\-____ ~"r_-_\-----_+--.<--Fl__,OO<"

Fig. 1

Âcz=a" Tz-e,

ÂbZ=abTZ-eb

.-______ ,.~~T_' __________ _+~~ROOf OI

Fig. 2

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The compatibility relation of deformation between slab and wall as shown in Fig. 1 is:

(1) D.c2 - D.b2 = ec2 + eb2

in which ac, ab are coefficient of li~ear expansion of slab and wall respectively; C::.c, D.b

are free deformation of unit length of slab and wall respectively; ec ,eb are shrinkage of

slab and wall respectively ; ec, eb are strain capacity of slab and wall respectively; TI' T 2

are temperature difference of top pia te ,wall and floor respectively.

We can derive ecl (rooO ,ec2 (floor) ,e:bl (top of wall) and eb2 (bottom of wall)from the

calculation diagramo They can be expressed as:

e 1= (Ql + M I )l= (Qbl + Qbl~I /2 )l= 4Qbl c ACl W cl Ec b l ~ 1 bl~1 2/ 6 Ec EcAcI

2Qb2 ~2= EcAc2

ebl = (Q2+ M 2)l = [Qbl +Qb2+ (Qbl -Qb2)h/2 Jl= 4Qbl-2Qb2 Ab Wb Eb th th2/ 6 Eb EbAb

eb - (Q2_ M2)l_ 4Qb2- 2Qbl 2- Ab Wb Eb - EbAb

in which Ec, Eb are elastic modulus of slab and wall respectively; Ac, Ab are effective

area of slab and wall respectively ;~I '~2 are thickness of top plate and floor respective­

ly; b l , b2 are effective width of top plate and floor respectively; h is height of top wall;

t is thickness of top wall.

5. STRAIN AND STRESS CALCULATION

Substituting those parameters in the equation ( 1), we can obtain

setting:

4Qbl 4Qbl - 2Qb2 acTI-abT2-ec+eb=E A + E A

c c l b b

81 = acT I- abT2-ec+eb

82 =aCT 2 -abT 2 -ec +eb

EbAb 1)1=--

ECACl

EbAb 1)2= E A c c2

in which 81,82 stand for temperature coefficient of shrinkage; 1)1,1)2 stand for rigidity

ratio of wall to slab. The above formulas can be simplified as:

81E bA b 2 ( 1)1 + 1 )Qbl -Qb2 = -2-

82EbAb - Qbl+ (1)2+ 2)Qb2=-2-

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setting

w= 2 (1)1 + 1) (1)2+2)-1

The edge stress, strain and shear of wall can be obtained:

Qbl = EbA b[61 (1)2+ 2) +62J/ 2w Qb2 = EbAb[61 + 262 ( ~ 1 + 1) J/2 w

[ 1)162 J Eb l =61 3+2(1)2-~) Iw

[ 1)261 J Eb2 = 62 3 + ( 41)1 - e;-) I w

[ 1)/12 J Obl = Eb61 3+2(1)2-~) Iw

[ 1)2 61 J Ob2 = E b62 3 + 41)1 - ----e;- I w

(2)

(3)

(4)

The tests and theoretical analysis illustrate that the shear stress between topplate and

wall varies curvilinearly within the length of buildings, and the maximum shear stress

Lmax is at a position which is 0/5 -li 4)a from the end of buildings[l]. To simplify

calculation, we assume that shear stress is linearly distributed and the maximum value

appears at the ends of structure. Because the vertical stress of upper part of the wall is

very weak, the maximum shear of this region approximately equal to the main tensile

stress. Thus, it can be show:n as:

(5)

Since shear stress varies linearly along the length of structure, the strain of upper part

of the wall varies linearly too. Setting Eb(X) = XEbl , we and obtain the constraining de­a

formation of the end of longitudinal wall:

u-fa ()d _faxEbld _ aEbl_ aObl - O Ebl X X - O a x - 2 - 2Eb

The horizontal deformation of the end of structure is the sum of free and constraining

deformation which can be expressed as:

aObl Um .. =t.b1 a+ 2Eb

[ 1)162 JI Um .. =a(ab T 2-eb) +a61 3+ 2 (1)2-~) 2w (6)

6. DISCUSSION OF RESUL TS

We have insitu surveyed a fourstory composite building. The plate of each floor is

made of precast concrete slab (Ec = 2. 6 X 1 04MPa, ac = 1. O X 10- 5) and the wall is

made of concrete block CEb = 2. 8 X 1 03MPa, ab = O. 7 X 10- 5). The specific parame­

ters are as follows:h=3. 3m,b1 =3. 3m,b2=2. 9m,81=120mm,82=80mm,t=

200mm,a=61. 4m,b=13. 8m,c=13. 2m. in which h stands for height of f]oor;-bl'

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U(mm)

2.0

1.5

1.0

0.5

'----'---4'---'---'---1.1..0--

1.1..2--

1-'--4 T("C)

Fig. 3

b2 stand for effective width of top plate and floor respectively; ap az stand for thick­

ness of top plate and floor respectively ; t stands for thickness of wall; The dimensions

of the building are 61. 4 X 13.8 X 13. 2m.

According to regression analysis[Z]of test results, under shortterm temperature differ­

ence,the deformation of building (U)is 1. 52mm. If applying formulas,the deforma­

tion andshear stress are U=1. 62mm and T=O.178MPa which dose to the test re­

sults. Contrasting the results of calculation with tests (showing in Fig. 3), we can

find a better agreement between them. Under yearly temperature difference , the test­

ed TI and T 2 were 44. 8 'C and 31. 5·C. Considering factors of creep and a long time of

yearly temperature difference,we choose H(t)=O. 75 and calculate the stress i!nd de­

formation are T=O. 373 MPa and U=7. 121mm which dose to the test result of U =

6. 81mm.

To those contractile products,such as concrete blocks and silicate bricks ,etc,it is es­

sentia I to take dried contraction of materials into account. When calculating the value

of yearly temperature difference, we may choose rate of contraction eb = O. 1mm/m

and obtain the temperature stress of contraction T is O.527MPa which increasing

41.3%. Thus,it is dear that the stress of structure is greatly increasd owing to ma.te­

rial contraction.

7. SUMMARY AND CONCLUSION

(1) We can find the formulas of calculating temperature stress and deformation of

brick masonry buildings are easy and useful,and the calculating results have a better

agreement with correspondi~gtest values,so that design,engineers can use those for­

mulas for reference.

(2) It is illustrated by calculating that temperatue stress of buildings are so great. We

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also notice cracks are likely to appear in severa I frail parts of structure first such as

the corner of windows and doors, under shortterm temperature difference. For exam­

pie, there are some inclined cracks in the undercorner of windows after having com­

pleted construction in ten days as a result of survey. Simultaneously, the calculating

values show that temperature stress of structure is so great thea is enough to have the

wall cracked if we do not adopt effective crackproof measure under yearly temperature

difference. So in engineering, temperature cracks in walls in common.

(3) To those products whose resistant strength of tensile and shear stress are rela­

tively low, such as concrete blocks and si li cate brecks, the material contraction causes

wall stress further increased, so the cracking degree of this sort of products in struc­

ture is more serious than that of clay bricks, the classification of cracks is much more

complicated.

(4) From the formulas, we can obtain that temperature stress is effected by various

factors like the pattern, size of buildings, material properties and time ,but the most

important factor is the contracted coefficient of temperature e1&.e2 which is in propor­

tion to stress. Therefore, lessening temperature difference and contraction is the best

method of reducing temperature stress and preventing wall cracking.

8. REFERENCES

(1) Wang, T. M . ,"Crack Control of Buildings", Shanghai Scie and Tech. Press,

1992.

( 2 ) Xiao, Y. M . ," An Experimental research on Small-Size Bloch Houses under the

Action of Temperature" ,Jounal of Southeast University, Vol. 11. 1995 ,pp. 640-

645.

( 3) Xiao, Y. M . ," A Summary of the Research on the problem of Masonry Structural

Cracks and Control", Supplement to Engineering Mechanics, 1994, pp. 1447-

1452.

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