11 th INTERNA TIONAL BRICKJBLOCK MASONRY CONFERENCE

TONGJI UNIVERSITY, SHANGHAI, CHINA, 14 - 160CTOBER 1997

THE STUDY ON LOAD-BEARlNG CAPACITY OF MASONRY SLENDER COLUMN UNDER BI-ECCENTRIC COMPRESSION

Huaqingyul Chuxian shi 2 Zhibo song 3

l.ABSTRACT I In this paper ,a new calculated formula for eccentric compression, which is based on

cross section average strain keeps Ipan deformation,static equilibrium and the relatiomship of bending curvature,is founded through the experimental results of 24 masonry columns under biaxial compression.And,on this principle,the author put forward a new calculated method about biaxial compression by utilizing the mutul\l relationship simplifY bimoments. The comparison shows that the calculated results of eccentric compression march the code methed and biaxial compression is also inarch experimental results.So they can be directly used in engineering design.The paper analyses the limit value of eccentricity and put forward suggested value of eccentricity. 2.INTRODUCTION

The research work as described in this paper was proposed by the ''Revision Group Of National Design Code For Masonry Structure" for the purpose of revising the said cide.This paper is one of the research results of the task .. The formula of load-bearing capacity under unilateral eccentric compression was deduced according to theoritic calculation of load-bearing capacity under unilateral eccentric compression. Then it was applied to formula of load-bealing capacity under bi-ecceotric compression .. This makes formula of load-bearing capacity under unilateral and bi-e.;centric compression correspond with each other.

Keywords : Masonry; Biaxial Compression; Static Equilibrium; Radual Eccentric Rate.

~ . Dept. of Agri. Engineeririg,Laiyang Agri. College ,Laiyang 265200,China

2 Dept. ofCivil Engineering,Hunan University,Cbangsha,China

3 Laiyang Shenglong Real Estate Development CO.,LTD.Shandong,China

116

3.MATERIAL AND METHOD In this experiment ,common bricks were used .. The cross section dimension of

masonrywas 240x370mm,height was 4 .08m, 3.6m, 2.88m and 1.92m respectively.

There were six. reputic members at every height, the total number is 24. Experiment was conducted on static platform with large-bearing capacity frame.

Load-bearing capacity was recorded by compression sensor.During experiment,in two directions at two height,lateral deflection at different height was determined.At about half height,the strain a10ng cross section perimeter to some extend was determined. 4.INDUCTION OF MASONRY LOAD-BEARING CAPACITY FORMULA

UNDER UNlLATERAL . ECCENTRIC COMPRESSION '.'

BasaI hypothesis: , (1). The average straio of tb'e cross section corresponds to the hypothesis of plane

deformatioD for masonry under unilateral eccentric compression (2). The relationship between strain and stress of compression masonry is shown by

the following equation [I] :

I (}" 8=--ln(I--)

460 j ... (1)

(3).Strength oftension region in fault cross section ofmasonry is neglected. (4). On fringe of compression region in failure cross section. When maximum compression strain iS&1l ,the stress arrived at average compression

strength j. ..

O H HlII,H K [2] Suggested that u1timate compression strain on compression fringe was

approximately 50% greater than that of under axial compression.Under eccentric compression,especially bi-eccentric compression,in order to apply the formulae of strain­stress more resonablely, &11 must use much bigger constant,it is:

8 = 0.015$ $>;$ 8.0 x 10-3 (2)

According to levei cross section hypothesis(see fig.l),the strain of compression region at any heigbt is:

~r I I

fc h -,l/HJy . I-

7J 1 Fig.

117

S.II S=-y

y' (3)

Thus,curvature offault cross section is:

; = Sull (4) y'

According to the hypothesis of small deflection and linear elastic in mechanics of materials,the formulae of maximum center deflection of U ofartieulation joint column in

fault cross~section is as follows[3] :

~=e(sec KHo -1),., NeH; = Ne. H: (5) 2 8E! E! 8

Referring to formula (4) . (5),we get the equation of, :

; = (s.u - 2p X 10--4) (6) I y'

The formulae of u in elastic stage is:

u = (8od1 - 2p X 10--4) . H; (7)

y' '" Where, \If is formal parameter for deflection curve along length . This curve has two

kinds of ultimateshape[4] : triangular and rectangular shapes.When triangular section

\If=12;rectangular seetion, ~8;parabola section, ~9. 6;sine section ,\If=7t I .we get 'I'=11(approxirnate triangula)in this paper.

As shown in fig.l, the statie equílibrium ean be shown by the fowllowing equation: y' y'

N = f a(s)bdy = f (1- e-MIJ.ffo.(SuIt/Y·)Y')bf..dy = 0.855bf .. y' (8) o

y' y'

N(~+ e + u) = f a(s)(h - y'+y)bdy = f (1 - e--460.ffm(s.'lIy·)y')(h - y'+y)bf",dy 2 o o

= (0.855hy· -0.375(y)2)f.. (9)

According to equation (8). (9) and (7),we ean get the formula ofN and u ,they are

N = [1.03(1- 2e) - (40 - P>P1]Afm (10) h 32164(1-

2e)

h

u= (40-P)p1h (11)

66248(1 - ~) AIthough having considerated' the non-linear effect of material during deduetion,there

_ would be some other faetors still not to be considerated on determinging the value of '-&nt and (jl and geometric oon·linear,thusformula (lO), (11) must be correctoo according

to experiment results or code value, the corrected formu1ae is shdwn as follows:

N = [0.75(1- 1.5e) _ (40 - p>pl ]Afm (12)

h 4000

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u = 0.065(1- 1.5e )h + (40 - f3)f32 h h 55000(1- 1.5e)

h

(13)

Where,load-bearing capacity can be standed by its affecting coefficient cp :

.. = 0.75(1- 1.5e) _ (40 - f3)f32 (fJ h 4000

(14)

the comparing calculated value according to formula (13), (14) with code value is

shown in table 1.1 and table 1.2 5.THE INDUCTION OF MASONRY LOAD-BEARING CAPACITY

FORMULA UNDER BI-ECCENTRIC tOMPRESSION Of varied calculated methods of steel reforced concrete under bi-eccentric

compreesion,a series methods based on mutual relation curved surface have considerate

representative. Gouwens(6)suggested a series simplely calculat methods ofdesign,which

adopting mutual relation oftwo direction moment(see fig.2).

When Mz.~MJ'Il , Mx..+(I-Bb).MJ'Il=I .0 (15) Mox Moy Mox Bb Moy

When Mz. ~ MYI' Mz. (I-Bb)+ MyN = 1.0 (16) Mox Moy MO% Bb Muj.

Where, value of coefficient Bb is affected by many factors. Gouwens suggested that

Bb be 0.8 by studying effect factors on steel reinforced concrete.Considerating

characterestics ofmasonry, Bb is 0.998 in this paper. When applying mutual relation of

Mz. (1- Bb) + MYI' = 1.0 MO% Bb Moy

Mz. +(I-Bb). MYI' = l.0 Mox Bb Moy

~

MYI' Moy

..... ~sJt::-----=~7I'

o '" ~-----,( o

o 1.00

'E 'I)

J ..

'I) '1)"

Fig.

eb

eró ebu

2

119

;.

two direction moment as above,how to corisider the influnce of eccentric moment in two directions each other can directly affect the accuracy of calculated result.In gererally, first calculating accentricity to one direction ,then revising according to another eccentric direction.But it is difficult to find a suitable formulae,because the value is also afifected by the ratio of the height to the thickness of member. Referring to document[4] ,firstly, calcu1ating radial compressive capacity of unilateral eccentricity in two axial directions.Its eccenticity ali adopted the value of radial eccentricity ,using the ratio of the height to thickness in two directions respectively.

Revising according to experiment results ,we can not only get a ~es methods of directly calculating load-bearing capacity llJl(l laterài deflection,but al!!O the formulae conforms to boundary condition under unilateral eccentric co~ression,its calculated methods areshown as follows: .

Sopposing prirnitive eccentricity is ( e", e h ),eccentric angle is

Q = tg-I (~) (17) eb

Prirnitive radial eccentricity is:

erO = ~e; + e; (18)

When eccentricity is e rO ,load-bearing capacity(N and M),and lateral deflection u

under unilateral compression in two derections are:

Nb = [0.75(1- l~ro) _ (40~::;;P! ]Af .. (19)

N = [0.75(1- l.Sero) _ (40- !3h)!3! JAflfl h h 4000

(20)

ub =0.065(1- ero)+ (40-Pb)P!

b 55000(1- ero ) b

(21)

Uh = 0.065(1- ero) + (40- Ph)P!

h 55000(1- ero ) h

(22)

Mo" = N b (Uh + eoQ.) = Nb - erb (23) .Mo>, = Nh(u/t+erO ) = N h -erh (24)

When failure,eccentricity.eccntric angle and radial eccentricity individualy are : eb. ' = (ub +erO )' cosa = erb -cosa (25) ehu =(uh+ero)-sina=erh -sina (26)

e e tgr = .2!.. = -!!!.. -tga (27)

eb• e",

e", = ~e;. +e!. (28) Supposing ultimate moment isM,. ,then

M,. = N. -e ..

M"" = M,. -cosr MJII' = M,. -sinr

120

(29) (30)

(31)

When r ~ 45° N = 500 • 500cosa sina ---+--

Nb N h

o 500 When r ~ 45 N. = -=-50::-::0:--:-' --­sma cosa

---+--N h N b

(32)

(33)

Ub. =(e'O~ub) · cosa-eb (34) u1DI =(e,o+uh )·sina-eh (35)

Design formula

When the probability-based limit statt:s design method is used [7] ,the load-~ capacity for compressive members of brick masonry under bi-eccentric loads can be calculated by the following formula:

N ~ qJAf (36)

6. RESULTS AND DlSCUSSION We choose the comparative situation on condition two in order to analyse the

discrepancy of the value cpl-q>4.The result is shown as table 2. cpl is the results of NiUchin formula; cp2 is the results of equivalent eccentricity method; cp3 is the results of additional deflection method; cp4 is the results of this paper results; cp , is the results of

experimental. The results of this experiment suggested that,when 13~12,eb ~ 0.4r and

eh ~ 0.4y ,not only vertical cracks appear,also horizontal cracks produce.The horizontal

cracks along with vertical cracks become wide and extend continuously. In partial experiment,horizontal cracks have developed much wider when failure of brickwork,it is

about 1/2 edge Iong in one derecti~n and approximately 1/3 edge long in another dTection, load-bearing capacity of brickwork reduces significantly about 35%(in comparlsion with eb ~ 0.4r and eh ~ 0.4y).When ex ~ 0.4r and ey ' ~ 0.6y ,calculated value of load-

bearing capacity reduces significantly,gernerally about 300Io.Because brickwork cross section is rectangular, load-bearing capacity when ex ~ 0.6r and ey ~ 0.4yis different

from that when ex ~ 0.4r and ey :<! 0.6y silently,but total tendency is alI to decreue

significantly. All above,this paper suggested that ultimate value of eccentric moment eb :<! 0.4r

and eh ~ 0.4yaccording to the foUowing formulae to calculate under bi-eccentric

compression can be determined according to radual eccentric rate e p :

ep = ~(ex / r)2 + (ey / y)2 = 2~(eb / b)2 + (eh / h)2 ~ 0.7 (37)

Thisvaluemeanstocontrol ex(y) :<!0.4x(y) and ey(x) 2!O.4y(x)[ep =.J0.42 +0.62

=0.721). When,O. 7<e p ~O. 8,according to the following formulae to calculate:

Aj"".K N K ~ _--'--,,=c-.-_

Ae Aey _ X+ _ _ l (38)

~ ~ When e p >0.8 • according to the following formulae to calculate:

121

Afim. N ~----'~-­Ae Ae, - "+--1 JJ: Jfy

7. CONCLUSION

(39)

(1).Although brickwork is made of non-even materials,dispersion rate of deformation is according to changing tendency of the strain obtained by ten strain determing poins in masonry cross section,average strain in cross-section of masonry s1ender column under bi-eccentric compression still accords with two hypotheses of levei cross section.

(2).According to the experimental results, N cr / N. is approximately 70 -

80%(mean value 72.1%),this indicated that compressive member under bi-eccentric compression will rapidly arrlve at ultimate failure after cracks,having obvious brittleness.

(3).Experimental results suggested that only vertieal cracks appear but no horizontal cracks produce,under relative less eccentricity.Thus,we considerated that there is only compressive failure in our experiment.Referring to some documents and failure process of experiment,in order to avoid producing tension failure,and load-bearlngs decreasing rapidly,we should limit its maximum eccentricity by adopting radial eccentric rate,it is ep~0 . 7 .

(4).According to levei cross section hypothses,static balance and curature calculation formulae,this paper supposed that a series of compressive formulae under bi-

t "'~~.

unilaterd compression. Then set up compressive member under bi-eccentric compression calcu1arated method by using relationship oftwo direction momento This method accords with the boundary condition under unilaterd cómpression.

REFERENCE l.The National Standard ofthe People's Republic ofChina,Desisn Code For Masonry

Structure (O BJ3-88),Building Industry Publishing House ofChi.ruhBeijing, 1989. 2.Brick Masonry Structure Study ,Chiefly Edited By O H H~H K,Publishing Scientific. 3.Chuxian Shi.Analysis of the Strength for Compressive Members of Brick Masonry Under Eccentric loads. Third intemational Symposium On Wall Structures, Vol.l .Warsaw: 1984,7 4.Reinforcement Structure Theory.Chiefly By Chuanzhi Wang.Building Industry Publishing House of China. 1985 5.Aodong BO.The Study and Design of Vertical Reinforced Composite Bric\cwork Column.Proceedings of Masonry Structure Research,Publishing House of Hunan University,1989 6.R.Park And T.Paulay,Reinforced Concrete Structures,Newyork,1975 7.The National Standard of the People's Republic of China,Common Unified Standard

.for Building Structures Design(OBJ68-84),Building Industry Publishing House of China,Beijing, 1984.

122

T bl 11Th a e f h e companson o t e va ue cp

~ 0.00 0.05 010. 0..15 02:> 0.25 0..30 0.35

60. 1297* 1.273 !1213 1.134 1.051 0.978 ú.9Z3 0.8Çfj

8.0. 12m 1230 1163 1.00 Hm 0.934 0.883 0.855 10.0. 1235 1.188 1.12:> 1.00 o.~ o.sm 0.858 0.&38 12.0. 1.1% 1147 HEI HXE 0..939 0.882 0.845 0.837 140. 1.154 1.107 LO:16 o.S'OO 0..918 o.rn 0.843 0.849 160. 1100 HXJ8 LCl4 0..956 o.~ 0.864 0.849 0.875

180. 1~1 H128 o.~ 0.933 0.889 0.861 0.861 0.912 2:>'0. 1.010 o.~ 0.<}t9 o.~ 0.874 0.857 0.873 0.954 220. 0.955 0..939 0.911 o.m 0.854 0.848 o.m 0.9)2

240. o.~ 0.886 0.8(ij 0.842 0.824 0.826 0.88) laE average value 0.978

coefficient of variation 0.12& *:value of codc/value ofthis paper

Table 1.2 The comparison ofthe value u

~/h 0..0.0. 0. .0.5 0. .10 0..15 0..20. 0. .25 0..30. 0..35

6.0. 0. .888* 0..879 0..894 0..935 1.002 1.0.92 1.199 1.310. 8.0. 1.010. 0. .983 0. .980. 1.0.0.2 1.0.46 1.10.8 1.178 1.243 10.0. 1.0.80. 1.0.35 1.0.16 1.0.21 1.0.46 1.0.85 1.129 1.165 12.0. 1.120. 1.0.61 1.0.29 1.0.21 1.031 1.0.55 1.0.81 1.10.0.

14.0. 1.146 1.0.77 1.0.34 1.0.16 1.0.16 1.0.29 1.0.44 1.0.52 16.0. 1.169 1.0.91 1.0.40. 1.0.14 1.0.0.7 1.0.12 1.0.20. 1.0.21 18.0. 1.194 1.10.8 1.0.52 1.0.20. 10..0.8 1.0.0.7 1.0.10. 1.006 20..0. 1.227 1.134 1.0.72 1.0.36 1.0.19 1.0.15 1.0.15 1.0.0.7 22.0. 1.271 1.172 1.10.5 1.0.65 1.0.45 1.0.38 1.0.35 . 1.0.25 24.0. 1.332 - 1.226 1154 1.110 1.0.88 1.0.79 1.0.74 1.0.62

average value 1.0.68 coefficient of variation 0..0.79

*: value of code/value of this paper

123

Table 2. The companson ofvalue <j> on condition one TetN>. ffiB FHH <j>' <j>1 <p2 <j>3 <p4 <j>'/<j>1 <j>'/<j>2 <j>'/<j>3 <j>'/<p4

17-1 0.05 0.048 o.m .0195 0.523 0.õT3 0.719 1649 1J<Xl 100 1011

17-2 0.05 0.048 0.686 0.495 0.523 0.õT3 0.719 0.386 0.312 L019 0.954

17-3 0.10 mo. 0.589 0376 o.tfJl 0.685 0.628 15ffi 0.119 o.sro 0.938

174 0..10. mo. 0.404 0.376 o.tfJl 0.685 0.628 L076 o.<m 0.s<Xl 0.6t3

17-5 0..10. 0.15 0.484 0..324 0350 o.5~ o.5Xl 1494 1385 0.810 0.931

17-ó 0.10 m5 0.400 0.323 0.349 o.m 0.520 1262 1170 0.683 0.784

15-1 0..10. 0..10. 0.588 0.405 0.439 0.678 0.6t8 L450 1339 0.868 o.<ul

15-2 0.10 mo. 0.756 0.405 0.439 0.678 0.6t8 186t L721 1.116 1.167

15-3 0..15 m5 o.~ 0.300 0.332 0.572 0.5ffi 1638 1519 0.881 0.888

154 m5 m5 0.647 0310 0.334 0.572 0.5ffi 2WJ 1937 1.129 1140

15-5 o.Xl 0.2) 0..685 o..Z35 0.248 0.406 0.488 2913 27ffJ 1689 1405

1S-ó 0.2) 0.2) 0.444 o..Z35 0.248 0.406 0.488 1.888 L789 W)5 0.911

15-7 0.2) 0.2) o.ffJS o..Z35 0.248 0.405 0.488 2573 2437 1491 1241

12-1 0.2) 0.2) 0.402 o.2ffJ 0.284 0.471 0.515 14<Xi 1415 0.854 0.781

12-2 0.2) 0.2) 0.362 o.2ffJ 0.284 0.471 0.515 1.347 1174 0..7(1) o.ím

12-3 0.2) 0.2) 0.362 o.2fl> 0.284 0.471 0.515 1.347 1275 0..7(1) 0.703

124 0.2) 0.2) 0.S<t2 o.2fl> 0.284 0.471 0.515 2017 l<ul 1.151 1053

12-5 0.30. 030. 0.282 mS9 0..158 0.244 0.352 1m 1783 1.155 0.8)2

12-ó 0.2) 0.2) 0.756 o.2fl> 0.284 0.471 0.515 2814 2<Xl2 lffJ6 1468

12-7 0.2) 0.30 o.2A4 o..XlO o.XlS 0.304 0.362 1223 1.190 mm 0.674

}2.8 020. 0.2) o.f92 0.268 0.284 0.471 0.515 2500 2441 14Xl 1344

12-9 0.2) 0.30 0.293 o..XlO 0.2)5 0.304 0.362 1468 1429 L%5 o.W

12-10. 0.2) 0.30 0.489 o..XlO 0.2)5 0.304 0.362 2450 2386 1611 1351

8-1 mo. 0.2) o.:;Q5 0.412 0.435 0.544 0.515 1715 1<Xl2 1~ 1372

8-2 mo. 0.2) o.n; 0.412 0.435 0.544 0.515 1715 L<Xl2 1~ 1372

8-3 0.05 0.2) o.!m 0.447 0.465 0.5ffi 0.505 2010. L993 1582 1m

84 0.2) 0.2) o.~ 0.326 0.343 0.467 0.545 L5S9 L482 LOOl 0.934

8-5 0.05 0.2) 0.578 0.446 0.464 0.567 0.505 1295 1246 1.019 1.145

average value L767 L674 L~ 1.0t3

coefficient of variation 0.277 o.E 0.275 o.2ffJ

124