Solutions of the torsion problem for bars with -, [−, +-, and T-cross-section by a harmonic function continuation technique

002+-7225/8#6079I-14SU2.W0 Copyright 0 1981 Petpmon Press Ltd.

fnr. I Engng Sci. Vol. 19, pp. 791-M 1981 Printed in Great Britain. All rights reserved

~ULUT~~N~UFTHE T~R~~~NFRILLED FUR BARS WITH-l_-, [-, +-,AND ~-CR~~~-~E~TI~N BY A HARMONIC FUNCTIONCONTINUATIONTECHNIQUE CHEN YI-ZHOUt Division af Solid Mechanics, North-Western Polytechnical University, Xian, Shanxi, China and CHEN YI-HENG Division of Solid Mechanics, Xian Chiao-tung University, Xian, Shanxi, China Abstract-This paper is a continuation of the senior author’s previous papers[l-41. When solving the Dirichlet problem of the Laplace equation, we divide the original section into several rectangular regions. Using the Duhem theorem[5] (see Appendix), i.e. a theorem of continuation of harmonic function, the solution on the original section can be found, and then, the torsional rigidity of bars with I-, I-, +- and T-cross-section (see Fii. I-4) deduced easily, Numerical results are shown in Tables I-4, respectively.

1. iNTRDDUCTI~N

THCJUGH THE torsion probfem of a cylindrical bar is a rather old one in the theory of elasticity, an

effective method for solving the torsion problem of bars with I*, [I, f- and T-cross-section has not been derived at present time. For example, the torsion problem of bars with [-crass-section was considered by ArutyunyanI61. We reduced the boundary problem to a set of ordinary differential equations. The derivation of his method is rather complicated and, in addition, his method is not very clear in some points. In 1365, a harmonic function continuation technique was proposed by usff]. It is regrettable that we were not able to publish our work at that time; it was not until 1980that a paper based on this technique was published in the journal Engineering Fracture Mechanics [3]. In this paper, the Dirichlet problem for the Laplace equation, is solved by dividing the original section into several rectangular regions, and assuming undetermined functions along the dividing lines. Now the Dirichiet problem for all the rectangular regions can be solved, so the undetermined functions can be determined by requiring the normal derivatives of two harmonic functions to be equal to each other along the dividing lines. The solution so obtained satisfies ail the governing conditions of the bouridary value problem. The condition of the contjnuation of harmonic function leads in each case to the problem of solving a linear algebraic system. When the Dirichlet problem is solved, the calculation of the torsional rigidity is easily deduced. It can be proved that the method proposed is equivalent to the following method: On the dividing lines some undetermined functions are assumed, and on every rectangular region the solution of Dirichfet problem is obtained. The functional

then becomes minimum and, consequently, all the undetermined functions can be obtained and the boundary problem is solved. On the other hand, from computing practice it is proved that the technique proposed is very effective. For example, in a ~ro~amrne usinn FUSTIAN onlv about tU0 cards is needed to perform the calculation mentioned in the fo~f~wing sections, e

2. THE TORSIONAL

RIGIDITY OF BARS WITH 1J3KlSS-SECTION

It is well known[7,8], that, the governing equation of the stress function 0(x, y) in the ?Author TVwhom correspondence should b+ addressed. 7%

792

CHEN

torsion problem is

Yl-ZH~U

and CHEN

Yl-HENG

$j+$=-2,

(2)

4Il. = 0 where L is the contour of the section. The torsional rigidity I) is D = UJ,

J=2

F4(x, y) dx dy

II

(3)

where u is the shear modulus of elasticity. Instead of the stress function 4(x, y), we introduce a function U(X,y) as follows

44%Y) = - 2 + uk Y)

(4)

and by inserting this equation in eqn (2), we obtain the governing equation of u(x, y) as

$+$o,

(5)

ay

clearly, u(x, y) is a harmonic function. From eqns (3) and (4), the torsional rigidity D becomes D = u.J,

J=2

F

(u(x, y) -x2) dx dy.

(6)

In all the four problem considered later, the eqns (5) and (6) are used. Now the concrete problems are considered. For the bars with I-cross-section, referring to Fig. I, we can divide the section ~ctitiously along the lines CF, AF, OG and 01, and five rectangular regions are obtained. We denote them as FI, I$, FIIr, Frv, F,, and their corresponding harmonic functions as uI, un, um, uIv, uv, respectively. Introducing four undetermined functions j(x), g(y), h(y) and i(x) on the dividing lines CF, AF, OG and 01, respectively, we take u(x, y)lv+ = iz2+ (2a + t)(x - a) + f(x),

(a G x s CI+ t), i.e. on the line CF

4x,

(0 6 y s t), i.e. on the line AF

Y)/x=,

UC&YLo

= at+

S(Y)*

(0 9 y 6 t), i.e. on the line OG

= MY),

u(x, y)lu=o= t* - t(x + t) + i(x)

(-t 6 x s 0), i.e. on the line OI.

Fig. I. 1 -cross-section of a bar.

(7)

~~[u~i~~~ of the

torsioa problem

7%

Note that, a, b, c and t are dimensions of section as shown in Fig. i and x, y are variable. In addition, the Fourier coefficientsfm,g,,, h,, in and modifiedFourier coefficientsF,, G,, H,, 1, are written as

f(x) sin @r(x - a)!t)dx,

n = 1,2, _. . n=I,2,..,

h

n=-x-

B” 2 ’ n

t

I

o MY)

sin(nny/f)

dy,

1*_2 O. f _ t(x) sin (nnfx f t)lt) dx, ‘“=n-- I I

n=1,2,,,.

n=1,2,..,

(81

n&e that, in~oducing modi~ed Fourier coe~cients F,, G,, &, and 1, is only for the purpose of simplificationin the form written below. It can be easily found, from eqns (5) and (7), that the solutions of eqn (5) on the regions FCDE(F~), ABCF(F& OAFG(&I), ZOGH(FIv)and JKOI(&) are

(9)

(13)

194

CHEN YI-ZHDT!andCHENYI-HENG

where (14)

A, = (1 - (- 1)“)/2.

From the Duhem theorem@] (see Appendix), i.e. a theorem of harmonic function continuation, along the dividing lines we have (a

s x s u + t), i.e. on the line CY’

(0 G y s t),

i.e. on the line AF

(16)

(0 s y s t), i.e. an the tine GG &v aY

I

y=o

I

=duv JY

f-t

s x c 0), i.e. on the line 01

y=o

(18)

Substituting eqns (9)-(13) into above equations, and expanding the both sides of equal symbol in Fourier series, then an infinite linear systems for F,, G,, H, and I,, are obtained as foliows

1

m = 1,2,. , .

&tsk(mn-c/t) where Ufa, b,m), F, Q and R are

P = ctk(m?r)+ ctk(mnblt),m=l,2,... Q= cth(m~)f ctk(mnalt),m = 1,2,... m = 1,2,.... R = c~k(rn~)~~~k(rn~c~~~~

(23)

Substituting eqns (f-(13) into eqn (61, the torsional rigidity of bars with I-cross-section is finally obtained in the form

(24) We can summarize the above procedure as follows.

795

Solutions of the torsion problem

For the torsion bars whose cross-section has the form as shown in Fig. 1, we can solve the systems (19)-(22)at first and substitute the obtained solution for F,, G,, ii, and 1, in eqn (24), then the torsional rigidity is finally obtained. If we only retain the first W terms of coefficients F,, G,,H, and I, in the systems, the aF~roximatetorsional ri~dity of bar is obt~ned* In the case of W = 9, using the Seidet iterative method for solving systems (I9)-(221,we find that the calculated torsionat rigidity can be expressed as

(25) where the A-values are shown in Tables l-5.

Table 1. A-Values shown in eqn (25) (a/t = 0.2) bit c/t

0.2

2

5

3.003 2.003

4.003

1

0.5

0.2 0.5

0.7244 0.8477

t

f ,025

0.9711 I.151

~~rne~~c I.330

:

2.363 1.363

2.487 1.487

f

:E

Table 2. A-Values shown in eqn (25) (a/t = 0.5)

b/f clt

0.2

0.5

0.2 0.5

0.8242 0.9472

1.O?O

: 5

1.126 1.462 2.462

1.249 1.585 2.585

1

2

5

Symmetric

I1.764 .428 2.764

2.100 3.100

4.101

Table 3. A-Values shown in eqn (25) (u/t) = 1)

Wt 0.2

0.5

0.2 0.5

0.9907 Lif3

1.236

: 5

1.292 1.628 2.628

1.751 1.414 2.751

C/t

1

2

5

~ymrnet~~

I1.929 .593 2.929

2.263 3.265

4.266

Table 4. A-Values shown in eqn (25) (a/t = 2) Mt c/t

0.2

0.5

1

E 1’

1.447 1.324 1.625

1.569 1.747

1.926 Symmetric

:

2.961 1.961

3.084 2.084

3.262 2.262

2

3.598 2.598

5

4.598

196

CHEN YI-ZHOU and CHEN YI-HENG Table 5. A-Values shown in eqn (25) (a/t = 5)

b/t

0.2

0.5

I

2.326 2.448 2.626

2.571 2.149

Symmetric 2.921

:

2.963 3.%3

4.085 3.085

4.264 3.263

c/t 0.2 0.5

3. THE TORSIONAL

1

RIGIDITY

2

5

4.600 3.600

5.600

OF BARS WITH [-CROSS-SECTION

The derivation used in this and following sections is the same as that outlined in the above section, so we can write the whole process of solving the problem in brief. For the bars with [-cross-section, referring to Fig. 2, we can divide the section fiictitiously along the lines CF, AF, OG and GH, and five rectangular regions are obtained. We denote them as Fi, FII, FIII, FIv, F,, and their corresponding harmonic functions as ui, uII, unI, uIv, uv, respectively. Introducing four undetermined functions f(x), g(y), h(y) and i(x) on the dividing lines CF, AF, OG and GH, respectively, we take u(x, y)(,=, = a* + (2a + t)(x - a) t f(x),

(a c x G a t t), i.e. on the line CF

4x3

Y)(,=,

= a* •t g(y),

(0 c y s t), i.e. on the line AF

4%

Y&l

= h(Y),

(0 c y s t), i.e. on the line OG (- t s x s 0), i.e. on the line GH

u(x, y)l,=, = t2 - t(x t t) + i(x),

(26)

note that a, b, c and t are dimensions of section as shown in Fig. 2 and x, y are variable. Also, the Fourier coefficients f,, g., h,, inand modified Fourier coefficients H,, G,, H,, Z, are written as ~~=~=~f~~“f(x)sin(n?r(x-a)/r)dx, gn SE=2

’ o g(Y) gin (nry/t)

n=l,2,...

dy,

n=l,2,...

hn = $ = f 1’ h(y) sin (n?ry/t) dy,

n=l,2,...

f

n

I

0

: -m-z L”-

n

-

t

_

O

I I

* I(X)

n=1,2 ,...

sin (n7~(xt t)/t) dx, H

I

J

t---_rc-I I

FIpi __ _--_---,

fi

t

0

i

a

G

F,

F

“-1 _--_--

f -7b B

E I

F, ’

:

* K

I

F, -------*

C

Fig. 2. [-cross-section of a bar.

0

Y

(27)

191

Solutions of the torsion problem

It should be noted that the introduction of modified Fourier coefficients F,, G,,, H, and 1, is only for the purpose of simplification in the form written below. Since the boundary value of each rectangular region is given by eqns (5) and (26), so that their corresponding solution of Laplace equation can be found easily. For the purpose of simplication, these solutions are not written here. Using the Duhem theorem (see Appendix) along four dividing lines, we obtain, as before an infinite linear system for F,, G,, H, and I,, as follows

(28) 4(a + t)tA, -2U(t, t, m)-4A,U(a, 7r2m +2(- l)“+‘m m 2F” 2 c n=ln +m ’

m=1,2,...

TQ

Hm = Q$?k/t)t

(

t, m) $ >

4(a t t)tA, 1r2m -2U(t,t,m)-4A,U(a,t,m) m=l,2,...

i+szi

(29) m (- l)m+“ln nZtrn2 , (30) (31)

where U(a,&m)=&-

th(md2t),

-$$

P = cth(m7f) t cth(mrb/t),

m=l,2,...

Q = cth(mlr) t cth(mralt),

m = 1,2,. . .

R = cth(mlr)+ cth(mlrc/t),

m=l,2,...

(32)

Substituting the solution UI, UII, urn, UIVand uv (the concrete form of these solutions are not written out) in (6), we obtain the torsional rigidity of bars with [-cross-section in the form J = t3(b t c) t t(a3 + 2t3) 3

thy

+th~)+(G,+H.)(th~+th~)}-"~,~t~~

(33) We can summarize the above procedure as follows. For torsion bars whose cross-section has the form as shown in Fig. 2, we can solve the systems (28)-(31) first and then substitute the obtained solution of F,,, G,,, H, and 1, in eqn (33), so that the torsional rigidity is finally obtained. In the case of W = 9, using the Seidel iterative method for solving systems (28)-(31), then we calculate torsional rigidity as J=B

(

;,;,;

where the B-values are shown in Tables 6-10.

>

.t4,

(34)

CHEN YI-ZHCKJand CHEN W-HENG Table 6. B-Values shown in eqo (341

(nit = 0.2) b/t c/t :::

I 2 5

0.2

0.5

0.7205 0.8487 I .030 1.367 2.367

0.9781 t*tSO 1.497 2.497

I

2

Symmetric 1,341 1.678 2.013 2.678 3.016

5

4.016

Table 7. B-Values shown in eqn (34) (a/f = 0.5) bit

0.2

0.5

0.5

0.8185 0.9455

1.073

:. 5

1.463 1.126 2.463

1.590 1.253 2.590

clt 0.2

1

2

5

Symmetric 1.434 1.771 2.771

2.108 3.108

4.108

T&e 8. B-Values shown in eqn (34) klff = 1)

bit clt

0.2

0.2 0.5

0.5

1

1

0.9845 1.111 I.291

1.238 Symmetric 1.418 1.598

:

2.628 I.628

2.755 i.754

2.935 1.935

2

5

3.271 2.271

4.272

Table 9. B-Values shown in eqn (34) (aft = 2) b/t 0.2

0.5

0.2 0.5

I.318 I.444

1.571

: 5

1.624 1.961 2.961

2.087 1.751 3.087

c/t

1

2

5

Symmetric 2.267 1.931 3.267

2.6@4 3.604

4.604

Table 10. B-Values shown in eqn (34) (a/t = 5) tit

b/t

02

0.2 0.5 1

2.319 :.z

:

3.963 2:%2

4. THE TORSIONAL

0.5

I

2

2.572 2.752

Symmetric 2.932

4.089 3.089

4.269 3.269

4.606 3.606

5

5.606

RIGIDITY OF BARS WITH +-CROSS-SECTION

derivation used in this section is the same as that outlined in the second section, so we write the solution of the problem in brief. For bars with +-cross-section, referring to Fig. 3, we can divide the section %ztitiously along the lines GA, GD, OG and AD, and five rectangular regions are obtained. We denote them as I$, 41, Fur, J$, F,, and their corresponding harmonic function as ~1, UI~,um, uIv, UV, respectively. The

~lutions of the torsion problem

799

Fig. 3. +-cross-sectionof a bar.

Introducing four undetermined functions i(x), f(x), h(y) and g(y) placed on the dividing linesOA, GI1, OG and AD respectively, we take u(x, y)ly=lJ =

txt i(x),

(0 c x c 1), i.e. on line OA

u(x, Y)ly=r= lx + f(x),

(0 s x s t), i.e on the line GD

4% Y)jx=o= WY)*

(0 G y c t), i.e. on the line OG

Hx, Y$=l = f2 + g(Y)*

(0 G y c t), i.e. on the line AD.

(35)

It should be noted that a, b, c, d and t are the dimensions of section as shown in Fig. 3 x, y are variable. Also, the Fourier coefficients i,, fn, h,, g, and modified Fourer coefficients I,, F,,, H,, G, can be written as: 12’= i, = JL i(x) sin (n~x/~) dx, n t I0

n=1,2,...

~“=~=5Ibf(x)sin(n,xit)dr,

n=1,2,...

hn---- :

n=1,2,...

- 3 i* h(y) sin (nrylt) dy, g(y) sin (my/t)

dy,

n = 1,2, . . . .

(36)

The introduction of modified coefficients Z,, F,, ii, and G, is only for the purpose of simplifying the form written below. Since the boundary value of each rectangular region is given by eqns (5) and (39, their corresponding solution of Laplace’s equation can be found easily. For the purpose of simplification, these solutions are not written here. Just as before, using the Duhem theorem (see Appendix) along the four dividing lines, an ignite linear systems for I,, F,,,, H, and G, is obtained as follows

800

CHE~YI~~H~U~fldCHE~ YI-HENC

Substituting the solution u!, uII, ullI, qv and ~~ (the concrete form of these solutions are not written out) into eqn (S), we obtain the torsional rigidity of bars with t-cross-section in the form

We can symmarize the above procedure as follows. For the tarsion bars whose section has the form shown in Fig. 3, we can solve the systems ~~7~-~~~and substitute the s~~ut~~nof G,, I&, F, and 1, in eqn (421, to obtain the &rsianaI rigidity. fn the case of W = 9, using the Seidet iterative method far solving systems ~37~~~~~ the calculated torsional rigidity can be expressed as

where the

C-values are shswn inTables

1t and 12,

~.TH~T~~~~U~~~ RIGIDITYOF ~~~~~rT~ ~-C~~~~-~~&Tl~~ Forthebarswith T-cross-section, referring to Fig. 4, we can divide the section fictitiously along the Iines GD, UE and AD, and four rectangular regions are obtained. We denote them as F,, I$, &, &, and their corresponding harmonic function as uI, ~11~uIII, uIv, respectively, Introduces three u~determjned functions f(x), h(y) and g(y) on the dividing lines GD, QG and AD, respectively, we take

Note that, a, b, c and t ate the dimensions of section as shown in Fig, 4 and x, y are variable.

801

Solutions of the torsion problem Table 11.C-values shown in eqn (43) a/t d/t ait = 0.2 bit c/t &=0.2

a/t = 0.2 d/t=2

air = 0.2

a/t = 0.2 dit=o.s

d/t=1

a/t = 0.2 d/t=5

a/t = 0.5 d/t = 0.5

a/t = 0.5 d/t = 1

o/t = 0.5 d/t=2

u/t = 0.5 d/t = 5

b/t = 0.2 c/t = 0.2 bit = 0.2

0.4691

c/t = 0.5

0.6009

0.7448

0.7842

0.9338

1.125

1.122

1.273

1A65

1.805

2.123

2.275

2.461

2.806

3.807

0.7375

0.8946

1.090

1.431

2.430

1.059

0.9231

1.086

1.284

1.625

2.625

1.257

1.458

1.261

1.426

1.624

1.966

2.966

1.598

1.800

2.142

c/t = 5 b/t = 1 c/t= t bit= 1 c/t =2 bit = 1

2.263

2.428

2.626

2.967

3.968

2.600

2.802

3.144

4.144

1.110

1.279

1.480

1.821

2.822

1.457

1.661

2.003

3.004

1.448

1,619

1.820

2.162

3.163

1.798

2.003

2.346

3.346

c/t=5

2.450

2.621

2.822

3.164

4.164

2.800

3.005

3.348

4.348

1.787

1.959

2.161

2.503

3.503

2.140

2.345

2.688

3.688

2.788

2.961

3.163

3.505

4.505

3.142

3.341

3.690

4.690

3.790

3.963

4.165

4.507

5.507

4.144

4.349

4.692

5.692

b/t = 0.2 c/t = 1 l$ ; ;2

bit = 0.2

tit =5 b/t = 0.5 c/t = 0.5 bit = 0.5 tit- 1 b/r = 0.5 c/r =2

Symmet~c

bit = 0.5

b/t = 2 c/t =2 g; b/t = 5 c/t=5

Table 12. C-Values shown in eqn (43) a/t d/t b/f c/t

a/t = 1 d/t = 2

u/t=1 d/t=1

nit= 1

n/t =2

d/t=.5

ii/t=2

a/t=2 d/t=5

u/t=5 d/t=5

b/t = 1

c/t= 1

1.866

b/t= 1 c/t =2 b/t = 1

2.209

2.552

c/t =5

3.211

3.554

4.555

2.552

2.895

3.8%

3.238

3.554

3.897

4.898

4.240

5.241

4.556

4.899

5.900

5.242

6.243

bit=2 c/t = 2 b/t=2 c/t = 5 b/t = 5

c/t = 5

I

H C

0

-

fi!z G

-----_ I

t

t

F

fm

’

6

~-LL-_D -----A

D b

E

‘7

Fig. 4. T-Cross-section of a bar.

* Y

7.243

CHEN YI-ZH~Uand CHEN Yt.H~N~

802

Also, the Fourier coefficientsf,, h, and g,, and modified Fourier coefficients F,, If,, G, are written as:

/, =H,=: ”

n

t

I

’ o h(y) sin tn?ry/f) dy,

n=1,2,....

(45)

The introduction of the modified coefhcients F,, H, and G, is only for the purpose of simpl~yi~gform used below. Since the boundary value of each rectangular region is given by eqns (5) and (445,the corresponding solution of Laplace’s equation can be found easily. For the purpose of simp~~~cation, these solutions are not written here. Just as before, using the Duhem theorem (see Appendix) along three dividing lines, we obtain an infinite linear system for F,, H. and G, as follows

m=1,2,...

(461

m=l,2,,..

(47)

where

Substituting the solutions ur, urn uIIt and uIv (the concrete forms of these solution are not written out) into eqn (6), we obtain the following formula for the torsional rigidity of bars with T-cross-section

We can summarize the above procedure as folfows. For torsion bars whose section has the form as shown in Fig. 4, we can solve the systems (46)-(48)and then substitute the solutions F,, G, and I& in eqn (50), to obtain the torsional rigidity.

803

Solutions of the torsion problem

In the case of W = 9, using the Seidel iterative method for solving systems (44)-(48),the calculated rigitity can be expressed as

where the D-values are shown in Tables 13-17.

Table 13. D-Values shown in eqn (51) (a/t = 0.2)

b/t

dt

1

0.2

0.5

2

:: 1’

0.3599 0.4126 0.6461

0.5885 0.7635

Symmetric 0.9392

:

0.9813 1.983

2.101 1.099

2.211 1.275

2.613 1.611

5

3.614

Table 14. D-Values shown in eqn (51) (a/t = 0.5) b/t

c/t

0.2

0.5

0.2 0.5

0.4843 0.6073

0.7340

: 5

0.7858 1.122 2.124

0.9144 1.251 2.253

1

2

5

Symmetric 1.433 1.096 2.434

1.770 2.771

3.7l3

Table IS. D-Values shown in eqn (51) (a/t = 1)

b/t clt

1

0.2

0.5

2

0.2 0.5 1

0.6640 0.7919 0.9727

0.9238 1.107

Symmetric 1.290

:

2.311 1.310

::g

2.629 1.628

2.967 1.965

5

3.969

Table 16. D-Values shown in eqn (51) (a/t = 2) c/t

: 5

b/t 0.2

0.5

1.001 I.130

1.263

1.311

1.446 1.784 2.786

1

2

5

Symmetric 1.968 1.630 2.970

2.306 3.308

4.309

CHEN YI-ZHOU and CHEN Yi-HENG

804

Table 17. D-Values shown in eqn (51) (o/t = 5) c/t

b/t

0.2

0.5

1

2.001 2.130 2.311

2.263 2.447

Symmetric 2.631

:

2.648 3.650

3.786 2.784

2.968 3.970

0.2 0.5

I

2

4.308 3.306

5

5.310

REFERENCES [l] CHEN YI-ZHOU, On fhr Shear Center of u T-Form Cantilever wit&Tkick Wall (in Chinese). Technical Report, North-Western Polyte~hnical University (1965). [2] CHEN YI-ZHOU, Dn the Torsion Pfo~~e~ of a Crucked Recf~~g~~Qr Bar (in Chinese). Technical Report No. 448, North-Western Polyte~hnical University (1977). f3] CHEN YI-ZHOU, Solutions of torsion crack problems of a rectangular bar by harmonic function continuation technique. Engineering Fracture Mechanics 13, 193(1980). [4] CHEN YI-ZHOU, A method of evaluation of the Third intensity factor of prismatical bar of an angle-cross-section with crack (in Chinese). To be published in Solid Mechanics. [5] WU XIU-MOU, Methods of Mathematical Physics 2, 177(1958). [6] N. KH. ARUTYUNYAN, Prikl. Maf. Mech., AkedemiyuNuuk SSSR 13, 107(1958). [7] S. TIMOSH~NKO, Theory of EIosficity (1951). [S] I. S. SOKOLNIKOFF, ~a~~ernufic~~Theory of ~~u~~~ciry (1956).

(Received 6 June 1980) APPENDIX The Duhem theorem: Let ur(.x,y) and ua(x,y) are two harmonic functions defined in two regions o,, oa respectively. If there are: (1) the two harmonic functions have the same value on the common boundary y of two regions, (2) the two harmonic functions have the same normal derivatives along the common boundary 7, i.e. @u&n) = (&,/an), and then it can be concluded, under such condition, a function by joining u&, y) and u&x, y) together is harmonic in the region o,+y+ua. The proof of theorem is omitted here.