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Struct Multidisc Optim 22, 394406 Springer-Verlag 2001

Optimal plastic design of a bar under combined torsion, bendingand shearW. Egner and M. Zyczkowski

Abstract Optimal plastic design of a bar or beam undertorsion with superposed bending and shear is obtainedvia the boundary perturbation method. Effects of bend-ing and shear are expressed by a small parameter . Theresults of optimization of solid bars and of hollow bars areshown.

Key words plastic design, combined loadings, solidbars, hollow bars

1Introduction

In the present paper we consider optimal plastic design of a bar under torsion with superposed simultaneous small

bending and shear. This problem will be solved by meansof the boundary perturbation method (BPM) with a barof constant circular or annular section assumed as thebasic solution (optimal in the case of pure torsion) anda small perturbing parameter describing the effect of bending and shear.

The boundary perturbation method is now widelyused in various branches of solid and uid mechanics.BPM was applied to determination of shapes of bodiesshowing full plastication at the stage of collapse rstby Kordas and Zyczkowski (1970). Full plastication, if possible at all, was then regarded as the rst step to-wards subsequent optimal plastic design. Ample surveyof the relevant literature was given in the paper by Egneret al. (1994) and will not be repeated here. We mention just some more recent related papers by Zyczkowski andEgner (1995) and by Egner (1996, 2000a,b), devoted tooptimal plastic design of rotationally symmetric elements

Received May 15, 2000

W. Egner and M. Zyczkowski

Institute of Mechanics and Machine Design, Cracow Universityof Technology (Politechnika Krakowska), ul. Warszawska 24,PL 31-155 Krakow, Polande-mail: m-1@institute.pk.edu.pl

and to plane heads of tension members. General approachto perturbation analysis of optimization problems withsome reference to optimization in plasticity is presentedin a recent monograph by Bonnans and Shapiro (2000).

The problem of combined plastic torsion with bend-ing (but without shear) was formulated by Handelman

(1944)and Hill (1948,1950).They used the Prandtl stressfunction. The relevant equation in terms of displacementswas derived by Piechnik (1961), Piechnik and Zyczkowski(1961). Piechnik applied the perturbation method (butwithout boundary perturbation and optimal design) toobtain effective formulae for a circular cross-section. Ad-vantages of numerical integration of Piechniks equationwere pointed out by Miller and Malvern (1967). A moredetailed survey is given by Zyczkowski (1981).

The present paper uses three displacements and meanstress as the dependent variables. In comparison to Piech-niks paper two essential generalizations are introduced:the effect of shear resulting in the loss of longitudinalhomogeneity, and the boundary perturbation leading tooptimal design of the bar.

Plastic optimization of beams under bending withshear was considered by Heyman (1959) and Rozvany(1973, 1976) for some relatively simple cost functions.Rozvany and Hill (1976) proved that even for grillages inwhich the beams are permitted to resist torsion, the opti-mal solution often leads to torsion-free beams. However,in some structures or structural elements torsion cannotbe eliminated and must be taken into account ( Zyczkow-ski 1997, 1998). The present paper is devoted to bars orbeams with prevailing torsion.

2General perturbations for a plastic circular bar undertorsion

2.1Assumptions, basic solution

As the basic solution (zero-th approximation) we con-sider a prismatic bar of the length under pure torsion.Optimal shape of the cross-section is then circular (withinthe class of solid, simply connected sections) or annular

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(within the class of doubly connected sections). Subse-quent optimization of annular shapes leads to thin-walledsections, but then a constraint imposed on the loss of sta-bility of the wall must additionally be introduced. Thisproblem will not be considered in the present paper: wesimply assume that the ratio of the internal radius b to theexternal radius a is prescribed and the wall stability is notlost. The limit twisting moment M equals

M t = 23

0 a3 b3 , (1)where 0 is the yield-point stress in pure shear. For a solidsection we may substitute b = 0. In the following we as-sume the material to be perfectly plastic, isotropic, in-compressible, obeying the HuberMisesHencky (HMH)yield condition. The strains are assumed to be small. For-mally, the HenckyIlyushin deformation theory is used,but it may also be considered as the LevyMises theoryof plastic ow if strains are replaced by strain rates. Oneend of the bar is assumed to be clamped and the otherfree. The cylindrical coordinate system r, , z is intro-duced with the origin in the centroid of free cross-section(Fig. 1). The moments are shown as vectors with doublearrows.

In the basic state the bar is under twisting momentM z = M t which causes full yielding of the bar (limitload carrying capacity). The following equations show thenon-zero stresses, strains and displacements, denoted byan additional subscript 0:

z 0 = r , z 0 = 0 , 0 = r2 0

= Cr ,

C = 2 0 , u

0 = r ( z) , (2)where denotes unit angle of twist (arbitrary in the limitstate), and is the plastic modulus.

Fig. 1 Scheme of the bar

2.2Governing equations in the general case

In the general case, under additional perturbing loadingswhich cause mainly bending and shear we obtain multi-axial state of stress and strain. Next we assume change of the shape of the bar in order to satisfy the yield condition.So, we can write the plasticity condition in the followingform:

(r z )2 + ( r )

2 + ( z )2 +

6 2r + 2rz + 2z = 2 20 , where 0 = 3 0 . (3)The equations of internal equilibrium and the strain-displacement relations in cylindrical coordinates have theform

rr

+ 1r

r

+ rz

z +

r r

= 0 ,

rr

+ 1r

+ z

z + 2

rr

= 0 ,

rzr

+ 1r

z

+ zz

+ rz

r = 0 , (4)

r = ur

r , =

1r

u

+ urr

, z = uz

z ,

r = u

r

u

r

+ 1

r

u r

, rz = u r

z

+ uz

r

,

z = 1r

u z

+ uz

. (5)

The physical equations can be written in the followingform:

k = (k k m ) , m = 13

jj , (6)

where k denotes Kroneckers symbol and Einsteinssummation convention is introduced. Five of the equa-tions (6) are independent. Together with the incompress-ibility condition

r + + z = 0 , (7)

we obtain a set of 16 equations with 16 unknowns.According to the boundary perturbation method the

unknowns (including perturbed shape a = a(, z) andb = b(, z) are expanded into series of certain small pa-rameters i , i = 1 , 2, . . . , n ,

X =

j =0

k =0

. . .

m =0X jk...m j1

k2 . . .

mn , (8)

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where

X = r , , z , r , rz , z , r , , z , r , rz , z ,

u r , u , u z , a ,bT

, (9)

and individual parameters i correspond to various per-turbing loadings, for example to concentrated momentM x , concentrated force P 1 = P , uniformly distributedloading q , etc. In what follows, we conne our consid-erations just to one perturbing loading, namely concen-trated force. The only small parameter, proportional tothis force, will be simply denoted by . Optimal designwith perturbation due to concentrated bending moment(much simpler because of absence of shear and unifor-mity along the axis z) was considered by Bochenek et al.(1983). Moreover, the exact solution to this plastic op-timization problem, was derived by Zyczkowski (1997),who proved that the solution obtained earlier by Ober-weis and Zyczkowski (1997) for a certain family of elliptic

sections satises the optimality condition. On the otherhand, perturbation due to uniformly distributed loadingcan go along the lines of the present paper, being slightlymore complicated. Some equations given below will begeneral, valid for any perturbing loading.

Making use of the expansions (8) we realise that tenlinear equations (4), (5), and (7) retain their form, just in-dividual quantities are labelled with relevant additionalsubscripts. On the other hand, six remaining non-linearequations (3) and (6) are subject to changes resultingin their linearization. From (3) with substituted (2) weobtain

z i = f i k ( i 1) , k ( i 2) , . . . , i = 1 , 2, . . . , (10)

where

f 1 = 0 ,

f 2 = 112 0

(r 1 1 )2 + ( 1 z 1 )

2 +

(z 1 r 1 )2 + 6 2r 1 + 6

2rz 1 ,

f 3 = 16 0

[(r 1 1 ) (r 2 2 ) +

(1 z 1 ) (2 z 2 )+ ( z 1 r 1 ) (z 2 r 2 ) +

6 r 1 r 2 + 6 rz 1 rz 2 ] , (11)

and so on, and from (6)

k i =i

j =0

j k ( i j ) k m ( i j ) . (12)

2.3Basic equations of the rst correction

On the rst level of correction we obtain from (10) and(11) z 1 = 0. The equations of internal equilibrium aretherefore simplied. Making use of (2), (5), and (12) weexpress the stresses in terms of displacements and meanstress m 1 as follows:

r 1 = m 1 + r 1Cr

= m 1 + 1Cr

u r 1r

,

1 = m 1 + 1Cr

= m 1 + 1Cr 2

u 1

+ 1Cr

u r 1r

,

z 1 = m 1 + z 1Cr

= m 1 + 1Cr

u z 1z

,

rz 1 = 12Cr

rz 1 = 12Cr

u r 1z

+ uz 1

r ,

r 1 = 12Cr

r 1 = 12Cr

u 1r

u1r

+ 1r

u r 1

. (13)

Substituting the above relations into internal equilibriumequations (4) and incompressibility condition (7) we ob-tain the following set of 4 partial differential equationswith 4 unknowns:

u r 1r

+ 1r

u 1

+ ur 1

r +

uz 1z

= 0 ,

m 1r +

1

Cr

2u r 1r 2 +

12Cr 2

u 1r

u1r

+ 1r

u r 1

+

12Cr

z

u r 1z

+ uz 1

r 1Cr 3

u 1

+ u r 1 = 0 ,

m 1

+ 12C

r

u 1r

u1r

+ 1r

u r 1

+

1

2Cr

u 1

r u 1

r +

1

r

u r 1

+

1Cr 2

u 1

+ u r 1 = 0 ,

m 1z

+ 12Cr

r

u r 1z

+ uz 1

r+

1Cr

2uz 1z 2

= 0 . (14)

The solution of the above set of equations depends onthe additional loading of the bar (which is already pre-loaded by twisting moment). If we assume that loadingacts in the yz plane, we can look for the solution in thefollowing form:

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u r 1 =

n =1

f rn (r, z )sin( n) ,

u1 =

n =1f n (r, z )cos( n) ,

uz 1 =

n =1

f zn (r, z )sin( n) ,

m 1 =

n =1

f mn (r, z )sin( n) . (15)

In the case of the horizontal or skew loadings the sineas well as cosine functions had to be introduced to theabove equations, but for the concentrated force P 1 shownin Fig. 1, (15) are sufficient. Substituting (15) into (14) weobtain

rf rnr

+ f rn nf n + rf znz

= 0 ,

2Cr 3f mn

r + 2 r 2

2f rnr 2 nr

f nr

+ 3 nf n

n2 + 2 f rn + r 2 2f rnz 2

+ r 2 2f znrz

= 0 ,

2nCr 2 f mn + r 2 2f nr 2 2n

2f n + 2 nf rn + nrf rnr

= 0 ,

2Cr 3f mn

z + r 2

2f rnrz

+ r 2 2f znr 2

+ 2 r 2 2f znz 2

= 0 . (16)

These equations may easily be reduced to a set of 2 n equa-tions by eliminating f n , and f mn . Namely, from the rstequation (16) we can calculate f n

f n = rn

f rnr

+ f rn

n +

rn

f znz

, (17)

then from the third equation

f mn = 3

2Cn 2 2f rnr 2

r2Cn 2

3f rnr 3

1Cn 2

2f znzr

r2Cn 2

3f znzr 2

+ 12Cr

f rnr

+ 1Cr

f znz

. (18)

As the result we obtain the following set of 2 n equationswith unknowns f rn and f zn :

r 4 4f rnr 4

+ 4f znr 3z

+ r 3 4 3f rnr 3

+ 3 3f znr 2z

r 2n2 2 2f rnr 2

+ 2f rn

z 2 + 2

2f znzr

+

n2(n2

1)f rn = 0 ,

r 4 4f rnr 3z

+ 4f znr 2z 2

+ r 3 3 3f rnr 2z

+ 2 3f znrz 2

r 2n2 2 2f rnrz

+ 4 2f znz 2

+ 2f zn

r 2= 0 . (19)

2.4Boundary conditions

Boundary conditions for a contour in cylindrical coordi-nates r = a(, z) were derived by Egner et al. (1994). Fora lateral surface free of loadings they may be written inthe form

a r a

r aaz

zr = 0 ,

a r a

aaz

z = 0 ,

a zr a

z aaz

z = 0 . (20)

Similar conditions hold for the internal surface r = b(, z)of a hollow bar. General expansion of (20) into power se-ries of allowing also for the expansion of a, was alsogiven in the above-mentioned paper.

3Effective solution for a solid bar under twistingmoment and concentrated vertical force at the free

end3.1The solution of governing equations for the rstcorrection

The differential equations (19) are close to ordinary dif-ferential equations of Eulers type. Hence, we can look forthe solution in the class of homogenous polynomials of mth-degree, where m can be any real or complex number.Here we restrict m to integers. Then the solution takes theform

f rn =

m =0

m

i =0B (n )i,m i r i zm i ,

f zn =

m =0

m

i=0

C (n )i,m i ri zm i . (21)

For a concentrated force it is sufficient to consider n =1, 3; m = 0 , 1, 2, 3, 4. For example, for n = 1 and m = 0 , 1,we look for the solution in the form

f r 1 = B(1)00 + B

(1)10 r + B

(1)01 z ,

f z 1 = C (1)00 + C

(1)10 r + C

(1)01 z , (22)

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and obtain the following general solutions:

u r 1 = B(1)00 + B

(1)10 r + B

(1)01 z sin ,

u 1 = B(1)00 + 2 B

(1)10 r + B

(1)01 z + C

(1)01 r cos ,

uz 1 = C (1)00 + C

(1)10 r + C

(1)01 z sin ,

r 1 = B(1)10 sin , 1 = B

(1)10 + C

(1)01 sin ,

z 1 = C (1)01 sin ,

r 1 = B(1)10 cos , rz 1 = B

(1)01 + C

(1)10 sin ,

z 1 =C (1)00

r + C (1)10 + C

(1)01

zr + B (1)01 cos ,

r 1 = 1

Cr

3

2B (1)10 + C

(1)10 sin ,

1 = 12C

B (1)10r

sin ,

z 1 = 1Cr

12

B (1)10 + 2 C (1)01 sin ,

rz 1 = 12Cr

B (1)01 + C (1)10 sin ,

r 1 = 12Cr

B (1)10 cos . (23)

Similar solutions can be derived for higher coefficients nand exponents m. They will be given in Sect. 3.2.

3.2Boundary conditions for the rst correction

For the rst correction the boundary conditions obtainedfrom the expansions of (20) into power series take theform:

r 1 |r = a 0 = 0 ,

r 1 |r = a 0 0a 1z = 0 ,

a0 rz 1 |r = a 0 0a 1

= 0 . (24)

In order to satisfy the rst equation of (24) we write theformulae for r 1 using (23) and similar solutions for n = 1,m = 2 , 3, 4 and n = 3, m = 0 , 1, 2, 3, 4,

r 1 = 1C 24B

(1)40 + 5 C

(1)31 r

2 +12

B (1)30 + 8 C (1)03 r +

3 3C (1)40 72

B (1)13 rz + 32

B (1)10r

+ B (1)01

r +

32

B (1)11 + 2 C (1)02

zr + 3 B (1)30 C

(1)03

z2

r +

C (1)40 + 14

B (1)13 z3

rsin +

1C

403

C (3)13 r2 +

2910

B (3)30 r + 348 C (3)04 rz +

409

B (3)20 +

409

C (3)12 z + 203

C (3)13 z2 +

C (3)01r

+ 2 C (3)02zr

+

110

B (3)30z2

r + 4 C (3)04

z3

rsin3 . (25)

The requirement to satisfy the rst of the above condi-tions for any and z gives eight relations:

C (3)04 = 0 , B(3)30 =

2003

C (3)13 a0 , C (1)40 =

14

B (1)13 ,

C (1)03 = B(1)30 , C

(3)02 = 209 C

(3)12 ,

C (1)02 = 518

B (1)13 a20

34

B (1)11 ,

C (3)01 = 409

B (3)20 a0 +180 C (3)13 a

30 ,

B (1)01 = 32

B (1)10 172

B (1)30 a20 + 24 B

(1)40 a

30 + 5 C

(1)31 a

30 . (26)

In the remaining two conditions we have the same func-tion a1 . In order to satisfy both of them the followingcondition (Schwarz condition) must hold:

r 1 r = a 0

= a0 rz 1

z r = a 0. (27)

The formulae for r 1 and rz 1 are written below:

r 1 = 1C

8B (1)40 + 32

C (1)31 r2 +

12

B (1)30 4C (1)03 r +

9

2

B (1)13

3C (1)40 rz + B

(1)20

1

2

B (1)02 +

B (1)21 32

B (1)03 z +874

C (1)31 +120 B(1)40 z

2 + 12

B (1)10r

+

12

B (1)11zr + B (1)30 2C

(1)03

z2

r +

12

B (1)13z3

rcos +

1C

20C (3)13 r2 +

2710

B (3)30 r +324 C (3)04 rz +

73

B (3)20 +

73

C (3)12 z + 72

C (3)13 z2 cos3 , (28)

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rz 1 = 1C

3C (1)40 12

B (1)13 r2+

212

C (1)31 + 48 B(1)40 rz +

2B (1)21 + 32

B (1)03 r 4C (1)02 +

12

B (1)11 +

2 B (1)30 + 4 C (1)03 z + 6 B

(1)13 C

(1)40 z

2 +

12

B (1)01 + B(1)10

1r +

12

B (1)02 B(1)20

zr

+

12

32

B (1)03 B(1)21

z2

r 294

C (1)31 + 40 B(1)40

z3

rsin +

1C

20C (3)04 r2

209

C (3)12 r 203

C (3)13 rz 4C (3)02

25B

(3)30 z 24C

(3)04 z

2

+ 12

C (3)10r +

B (3)20zr +

12

C (3)12z2

r +

12

C (3)13z3

rsin3 . (29)

Fullment of (27) with (26) taken into account leads tothe following relations:

C (3)13 = B(3)30 = C

(3)12 = C

(3)02 = B

(1)30 = C

(1)03 = B

(1)13 =

C (1)40 = B(3)20 = C

(3)01 = B

(1)11 = C

(1)02 = 0 ,

B (1)10 =

112B (1)40 a

30

24C (1)31 a

30 . (30)

Then the rst correction of the contour equals

a1 (, z) = a11 (z)cos + a13 (z)cos3 , (31)

where a11 (z) and a13 (z) are polynomials of the thirddegree.

3.3The condition of invariant location of the centroid

Now we demand from the rst correction not to causechange of location of centroid, otherwise the additionaleccentric twisting would appear. The radius vector de-scribing the contour of the bar in rst correction is givenby (31) with added a0 . Then the static moment with re-spect to the vertical axis y which is required to vanish wecan express in the form

S y = x dA = [a()]2 cos dr d =13

2

0 [a()]3 cos d . (32)

Substituting (31) into (32) we obtain the following inte-gral:2

0 3a20 a1 cos2 + a31 cos4 + 6 a0a1 a3 cos2 cos3 +3a31a3 cos

3 cos3 + 3 a0a23 cos cos2 3 +

3a1a23 cos2 cos2 3 + a33 cos cos3 3 d = 0 . (33)

From the above equation we obtain a11 (z) = 0, anda13 (z) may be an arbitrary constant. So we obtain theadditional conditions:

B (1)40 = 29160

C (1)31 ,

B (1)21 = 32

B (1)03 ,

B (1)20 = 12

B (1)02 + 965

C (1)31 a20 ,

B (1)01 = B(1)10 9B

(1)03 a

20 , (34)

and the function describing the contour in the rst ap-proximation has the form

a = a0 C (3)106C 0

cos3 . (35)

3.4The integral conditions of internal equilibrium

In this section we require the sectional forces resultingfrom the stress distribution to be equal to those resultingfrom external loadings. In the case under considerationwe have three kinds of forces (the condition on normalforce is satised identically): bending moment, twistingmoment and shear force. These conditions can be writtenin the following form:2

0a

0

k =0

k zk r 2 sin dr d = tz , (36)

2

0a

0

k =0 k ( rzk sin + zk cos) r dr d = t , (37)

2

0a

0

k =0

k zk r 2drd = M z = M t , (38)

where t = P and a = a(z, ). M t denotes load-carryingcapacity for twisting of circular cross-section. The upperlimit of integration a is subject to expanding into series:a(z, ) =

i =0 ai i . In the rst condition of (36) appears

the stress z . Taking into account the relations (26), (30),and (34) we obtain the following formula for z 1 :

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z 1 = 1C

4340

C (1)31 r2 3B

(1)02

3845

C (1)31 a20 9B

(1)03 z

95

C (1)31 z2 +

21120

C (1)31 a30

1r

sin . (39)

Substituting the above formula into the condition (36) weobtain the following relations:

C (1)31 = B(1)02 = 0 , B

(1)03 = Ct3a 30 . (40)

Taking them into account we obtain from (29)

rz 1 = 3t2a 30

a20r r sin +

12

C (3)10r

sin3 . (41)

Substituting the above result into equations (37) and (38)we nd them satised identically. Since z 1 = 0, there isno additional twisting for the rst correction.

3.5The DruckerShield condition

At the end we demand to satisfy the necessary optimalitycondition of the contour shape. This condition is knownin literature as the DruckerShield condition. Accordingto it, if the boundary of a structure in the plastic state isnot loaded, then the necessary condition of optimality of the shape is that the unit power dissipated in the LevyMises theory or energy dissipated in the HenckyIlyushintheory on that boundary is constant. It can be written inthe following form:

D = ij ij |S = const . (42)For the rst correction in view of z 1 = 0, it has the fol-lowing shape:

D = 0 z 0 |r = a 0 +

0 z 1 |r = a 0 + 0a1 z 0

r r = a 0+ . . . = const . (43)

The above condition results in the following relations:

C (1)00 = 8 B(1)03 a

30 , C

(3)00 = 89C

(3)10 a0 . (44)

Finally, we must require to satisfy the boundary con-dition connected with clamping of the bar: u (r = a0; = 0 , ; z = )=0 , uz (r = a0 ; = / 2; z = ) = 0.Theseconditions give certain relations involving the constantsB (1)10 , and B

(1)00 not yet used. The conditions are not shown

here, because these constants will not appear in the fur-ther analysis. On the other hand, the constant C (3)10 in(35) is not determined by the DruckerShield conditionin the linear approximation (43). However, calculatingthe cross-sectional area by integration of (35) squared

and leaving the term with 2 we nd that the minimumis obtained for C (3)10 = 0 and in the following we assumea1 = 0.

3.6List of formulae for the rst correction for solid bar

Finally for the rst correction we have the following for-mulae:

r 1 = 1 = 0 , z 1 = 3ta 30

z sin ,

r 1 = z 1 = 0 , rz 1 = 32

ta 30

a20r r sin ,

a1 = 0 , 1 = Ct6a 30 0

a20 r 2 cos . (45)

3.7The second correction for a solid bar

For the second correction the yield condition (3) gives

2z 1 + 3 2rz 1 + 6 0 z 2 = 0 . (46)

Making use of (45) we present z 2 in the form

z 2 = 34

t2

2a60 034

a20r r

2

+ z2 (cos21) . (47)

Similarly as for the rst correction the number of equa-tions can be reduced by elimination of stresses and strainsby displacements and mean stress. Making use of (4)(7)and (45) we eliminate the remaining stress components

r 2 = m 2 + 14

H 0

3a20r

r3

z sin2 + 1Cr

u r 2r

,

2 = m 2 + 14

H 0

3a20r

r3

z sin2 + 1Cr 2

u 2

+

1Cr 2

u r 2 ,

z 2 = m 2 12

H 0

3a20r

r3

z sin 2 + 1Cr

u z 2z

, (48)

rz 2 = 12Cr

u r 2z

+ uz 2

r 38

H 0

r 2

3 23

a20 + a403r 2

sin2 ,

r 2 = 12Cr

u 2r

u2r

+ 1r

u r 2

. (49)

In the above equations H is proportional to the concen-trated force squared, H = t2 / 2a60 .

Substituting these stresses to the equilibrium equa-tions, together with incompressibility condition we ob-

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tain the following set of four linear nonhomogenous par-tial differential equations

u r 2r

+ 1r

u 2

+ ur 2

r +

uz 2z

= 0 ,

m 2r

+ 1Cr

2u r 2r 2

+ 12Cr 2

2u2r

32Cr 3

u 2

+

12Cr 3

2u r 22

+ 12Cr

2u r 2z 2

+ 12Cr

2uz 2rz

ur 2Cr 3

=

14

H 0

13

+ 3a20r 2

z sin2 ,

1r

m 2

+ 12Cr

2u2r 2

+ 1Cr 3

2u22

+ 1Cr 3

u r 2

+

12Cr 2

2u r 2r

= 32

H 0

z 189

+ a20r 2

cos2 ,

m 2z

+ 12Cr

r

u r 2z

+ uz 2r

+ 1Cr

2uz 2z 2

=

H 0

43

r a20r

+ 32

z2

r +

a40r 3

sin2 . (50)

The particular solution can be searched within the func-tions:

u r 2 = f r 2 (r, z )sin2 ,

u2 = f 2 (r, z )cos2 + 12

CH 0

r 3z ,

uz 2 = f z 2 (r, z )sin2 , m 2 = f m 2 (r, z )sin2 . (51)

Hence we obtain the following equations:

f m 2r

+ 1Cr

2f r 2r 2

1Cr 2

f 2r

+ 3Cr 3

f 2 f r 2 +

12Cr

f r 2z 2

+ 2f z 2rz

= 14

H 0

z13

+ 3a20r 2

,

2

rf m 2 +

1

2Cr

2f 2

r2 +

2

Cr3 f r 2

2f 2 +

1Cr 2

f r 2r

= 32

H 0

z89

+ a20r 2

,

12Cr

2f r 2zr

+ 12Cr

2f z 2r 2

+f m 2

z +

1Cr

2f z 2z 2

=

H 0

43

r a20r

+ 32

z2

r +

a40r 3

,

f r 2r

2r

f 2 + f r 2

r +

f z 2z

= 0 . (52)

Similarly as for the rst correction we eliminate f 2 andf m 2 and obtain the following two equations:

r 4 4f r 2r 4

+ 4f z 2r 3z

+ r 3 4 3f r 2r 3

+ 3 3f z 2r 2z

4r 2 8 2f r 2r 2

+ 4 2f r 2z 2

+ 8 2f z 2zr

+ 12 f r 2 =

6CH 0

r 3z ,

r 4 4f r 2r 3z

+ 4f z 2

r 2z 2+ r 3 3

3f r 2r 2z

+ 2 3f z 2rz 2

r 2 8 2f r 2rz

+ 16 2f z 2z 2

+ 4 2f z 2r 2

=

8CH 0

14

a20r2 2r 4

32

r 2z2 a40 . (53)

We assume the following form of particular solution forthe function f r 2 :

f r 2 = 12

CH 0

r 3z . (54)

Now, the function f z 2 must satisfy simultaneously twoequations:

z

r 4 3f z 2r 3

+ 3 r 3 2f z 2r 2 8r

2 f z 2r

= 0 ,

2

z2 r

4 2f z 2

r2 + 2 r

3 f z 2

r 16r 2f z 2

4r 2

2f z 2

r2 =

8CH 0

14

a20r2 2r 4

32

r 2z2 a40 . (55)If the particular solution will be written in the followingshape:

f z 2 = D40 r4 + D20 r 2 + D 02 z2 + D04 z4 + D ln

ra0

, (56)

then the rst equation (55) is satised automatically, andthe second one results in

D 40 = 1

3

CH

0, D04 =

1

16

CH

0,

D = 2CH 0

a40 , D02 = 14

D 20 116

CH 0

a20 . (57)

Finally we have the particular solution in the followingform:

f r 2 = 12

H 1r 3z ,

f z 2 = D20 r2

14

D20 + 14

H 1a20 z2 +

13

H 1r 4+

116

H 1z4 2H 1a40 ln ra0

. (58)

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where the new notation has been introduced: H 1 =CH/ 0 = Ct2 / 02a60 .

Now, we are looking for the general solution of the ho-mogeneous part of (53). We restrict general solution tothe following functions:

u r 2 = f r 2 (r, z )sin(2 ) , u2 = f 2 (r, z )cos(2) ,

uz 2 = f z 2 (r, z )sin(2 ) , m 2 = f m 2 (r, z )sin(2 ) .(59)

Using the way shown in (21) we assume

f r 2 =4

m =0

4

i =0

E i (m i ) r i zm i ,

f z 2 =4

m =1

4

i =1

F i (m i ) r i zm i , (60)

and obtain

f r 2 = 29

E 22 r 4 + E 30 r 3 + E 20 r 2 + E 21 r 2z + E 22 r 2z2+

E 31 r 3z + E 00 ,

f z 2 = 14

E 31 r 4 18

E 21 r 3 14

E 22 r 3z + F 20 r 265

E 30 r 2z 35

E 31 r 2z2 + F 10 r 12

E 20 rz 14

E 21 rz 2

16E 22 rz3 + F 01 z 14F 20 z2 + 110E 30 z3 + 140E 31 z4 . (61)

Finally, the second corrections of displacements andmean stress have the form:

u r 2 =29

E 22 r 4 + E 30 r 3 + E 31 r 3z + 12

H 1r 3z +

E 20 r 2 + E 21 r 2z + E 22 r 2z2 + E 00 sin2 ,

u2

=31

72E

22r 4 +

7

5E

30r 3 +

7

5E

31+ H

1r 3z +

54

r 2 (E 20 + E 21 z + E 22z2) 14

E 20 + D 20 + 14

H 1a20 rz + 320

E 30 rz 2+

14

15

E 31 + 12

H 1 rz 3 + 12

E 00 cos2 + 12

H 1r 3z ,

uz 2 =14

E 31 r 4 + 13

H 1r 4 18

E 21 r 3 14

E 22 r 3z +

E 20 r 2 + D 20 r 2 65

E 30 r 2z 35

E 31 r 2z2 12

E 20 rz 14

E 21 rz 2 16

E 22 rz 3 + F 01 z 14

D20 + 14

H 1a20 z2

14

E 20 z2 + 110

E 30 z3 + 116

H 1z4 + 140

E 31 z4 2H 1a40 ln ra0

sin2 ,

m 2 = 1C

7972

E 22 r 2 95

E 30 r 95

E 31 + 1712

H 1 rz 18

E 20 18

E 21 z 18

E 22 z2 12

E 20 + D 20 +

74

H 1a20zr +

310

E 30z2

r +

1

2

1

5E 31 +

1

2H 1

z3

rsin2 . (62)

3.8Boundary conditions for the second correction

Boundary conditions for the second correction take thefollowing form resulting from expansion of (20):

r 2 |r = a 0 = 0 ,

r 2 |r = a 0 0a 2z = 0 ,

a0a1 rz 1

r r = a 0+ a0 rz 2 |r = a 0 + a1 rz 1 |r = a 0

0a 2

= 0 . (63)

The rst equation of the set (63) is of the same type as forthe rst correction. In the third equation the terms con-nected with a1 are equal to zero and from the Schwarzcondition for a2 we obtain similar equation as for the rstcorrection:

r 2 r = a 0

= a0 rz 2

z r = a 0. (64)

Since the way of satisfying the boundary conditions forthe second corrections is the same as for the rst one, wewill not give here the details and write the nal form of the second correction for the external contour:

a2 = H 1a0

C 01748

a20 F 104H 1a0

58

z2 cos2 +

H 1a0C 0

z2

4 + C a . (65)

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The integral equations of the internal equilibrium forshear forces and bending moments are satised automat-ically. The equilibrium condition for the twisting momentwhich can be written in the following form:

M z = A

z r dA =2

0a

0 0 + 2 z 2 r 2 dr d =M t + 0 + 0 2 + . . . (66)(where a = a0 + 2 2) allows us to calculate constant C a :

C a = 310

H 1a30C 1 0

. (67)

Now, the formula for a2 can be written in the followingform:

a2 = H 1a0

C 01748

a20 F 104H 1a0

58

z2 cos2 +

14 z2 +

310a20 . (68)

We should satisfy the Drucker-Shield condition. For thesecond correction it can be written in the form (non-zeroterms only):

(z 1 z 1 + 0 z 2 + z 2 z 0 + rz 1 rz 1 )r = a 0 +

2C 20 a2 = const . (69)

It turns out that this condition can be satised only ap-proximately (terms of type z0 cos2 only). Using it we

can calculate the constant F 10 .

F 10 = 1312

H 1a30 . (70)

After taking into account the above constant, dening thedimensionless small parameter in the form:

= t

a 20 0, (71)

and taking into account the results of the rst correctionwe can write the function describing lateral surface in thefollowing dimensionless form:

aa0

= 1 + 2 58

1z2

a20cos2 + 1

4z2a20

+ 310

. (72)

The conditions of support for the second correctionare of the same form as for the rst correction: u 2(r =a0 ; = 0 , ; z = ) = 0, uz 2(r = a0 ; = / 2; z = ) = 0.The condition for displacement uz 2 is satised automat-ically, thus the condition for ur 2 makes it possible to cal-culate the constant E 00 .

At the end we compare the obtained shape with theexact optimal shape of a bar under torsion and bendingin plastic range obtained by Zyczkowski (1998), who con-sidered the similar problem but without shear effect and

used this shape as an approximation if the bending mo-ment is variable along the axis z. As a result the ellipsewith the axes given by the following relation:

axa0

= 1 2P 2z2

9M 2t+ . . . ,

aya0 = 1 +

4P 2z2

9M 2t + . . . , (73)

was obtained. Axes ax and ay are placed in the planez=const (Fig. 2), and a0 is a radius of a bar under puretorsion at the stage of collapse:

a0 = 3 3M t2 0 . (74)When using boundary perturbation method with sheartaken into account, the axes of a cross-section are given bythe following formulae resulting from (72):

axa0

= 1 + 3790

P 2a20M 2t

318

P 2z2

M 2t+ . . . ,

aya0

= 1 1390

P 2a20M 2t

+ 718

P 2z2

M 2t+ . . . . (75)

Comparing the relations (73) and (75) we can see that inequation (73) all correcting terms depend on z. It is theresult of neglecting the effects of shear, taken into accountin (75). Bending effect, which is described by the termsproportional to z2 is for both solutions similar.

3.9Numerical example

The example wascalculated for /a 0 = 4 and =0.15. Thefollowing interpretation for the above data can be pre-sented. The value of small parameter can be understoodas a ratio of bending moment in clamped cross-sectionM b to the torque:

M b

M t=

t2

3

0a3

0

= 32 a0

. (76)

For the above data this ratio is equal M b /M t = 0 .9.Figure 2 shows the corresponding optimal shape.

Figure 3 shows the shape of the longitudinal contourof the bar, obtained as sections in the planes xz and yz.

It is seen from the third of equations (11) that thethird correction governed by 3 must contain the termswith cos3 which cannot be avoided as in the rst cor-rection. The symmetry of the cross-section with respectto the axis y is then lost. However, the calculations of this correction are very cumbersome and will not be givenhere.

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Fig. 2 Optimal shape of the solid bar

Fig. 3 Variation of semiaxes along the bar

4Effective solution for a hollow bar

4.1Boundary conditions for the rst correction for a barwith annular cross-section

For bending and torsion the layers which are close to theaxis of symmetry of a bar are not appropriately exploited.So we may expect further prots for annular cross-sectionof bars (doubly connected cross-sections). Such an ap-proach was applied by Bochenek et al. (1983).

In the case of an annular cross-section of a bar theform of the boundary conditions given by (20) and theirexpansion into power series do not change. However,we must remember that they must be satised at bothboundaries a(z, ) and b(z, ). It turns out that the exactfullment of all equations is impossible. Therefore we de-mand for these equations to be satised identically for theterms z0 and z1 . For the terms with z2 and z3 which ap-

pear for example in the equation for r 1 , the fullment ispossible if we take into account higher powers of variablesr and z. On the other hand, taking into account the higherpowers makes impossible to satisfy the boundary condi-tions for the terms connected with these higher powers.It is clear that we can nd the solution with any preci-sion we want, but always the terms with highest powerswould not full some equations. Hence, the obtained solu-tion can be understood as an approximate solution. Thisapproximation is better if we consider sections closer tothe free end of a bar. For z 0 the solution leads to theexact one.

It should be mentioned that in the case of the annu-lar cross-section under consideration, only the solutionconnected with trigonometric functions of a single angle will be taken into account. The calculations are morecomplicated here, since we have two independent surfacesto design. This justies neglecting the terms with tripleangle. The experiences from the previous problem showthat these terms make only the volume bigger. Moreover

the analysis for a single angle gives chances to nd a non-trivial correction for a = a(r, z ) and b = b(r, z ) as earlyas for the rst correction. We have two independent sur-faces to design, so the rigid movement of these surfacesdescribed by the solution with trigonometric functions of a single angle is possible. The condition of invariant lo-cation of centroid will not cause a zero solution. Thenfurther considerations differ from the previous one onlyin the range of neglected terms with triple angle, fulll-ing of boundary conditions in the terms with z0 and z1and the necessity of satisfying the boundary conditions atboth lateral surfaces.

After satisfying boundary conditions at both surfaceswe obtain the following relations for the rst correction of the shape:

a1 = 1c 0 3B13 a

30 2B21 +

32

B03 a20 52

B11 a0 12

(B01 + B10 ) cos ,

b1 = 1c 0 3B13 b

30 2B21 +

32

B03 b20 52

B11 b0

12

(B01 + B10 ) cos . (77)

The next step is demanding the invariant location of centroid. This condition can be written in the followingform:

S y = 13

2

0 a3 b3 cos d =13

2

0 a30 + 3 a20a1 +. . .

b30

3b20b1

. . . cos d = 0 , (78)

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where

a1 = C a cos ,

b1 = C b cos . (79)

The fullment of the abovecondition results in the follow-ing relation:

a20C a b20C b = 0 . (80)Next, we demand of satisfying the condition of the bend-ing moment equilibrium. In the rst approximation it hasthe form shown below:2

0a 0

b0 z 1 r2 sin dr d = tz . (81)

The remaining conditions, describing the equilibrium of the twisting moment and shear force, are satised identi-cally. The Drucker-Shield conditions are the same as for

the previous example, but they have to be satised atboth surfaces

z 1 + a1 z 0

r r = a 0= const ,

z 1 + b1 z 0

r r = b0= const . (82)

After the above considerations we obtain the solutionwith two arbitrary constants. However, detailed analysishas shown that both of the constants in linear way in-creased the dimensions of the bar. Therefore in further

considerations we assume them equal to zero. Finally wecan write the rst corrections of the shape in the followingform:

a1 = 32

b20b30 a30

t 0

cos ,

b1 = 32

a20b30 a30

t 0

cos . (83)

We dene the dimensionless small parameter in the sameway as for the rst example (71) introduce the followingnotation: a0 /b 0 = k, and obtain the following formulae:

aa0= 1 + 32 k(1 k3)

cos ,

ba0

= 1k

+ 32

k3

(1 k3) cos . (84)

The following relation gives the dimensionless area of a cross-section:

AA0

= 1 + 298

1k41 1k 2 1k k2

2 . (85)

In the rst approximation it is constant along the axis of the bar. The second correction will not be given here.

4.2Numerical example

Figure 4 shows the shape of the contour for the small pa-rameter = 0 .15 and ratio of radii k=2.

It is seen, that as a result of design the wall at the rightside is thicker, and at the left side is thinner. It is causedby the accumulation of shear stresses at the right-handside (from torsion and shear).

Figure 5 shows the relation between dimensionlessarea (85) and the ratio of radii k.

Fig. 4 Optimal cross-section of a hollow bar

Fig. 5 Dimensionless cross-sectional area in terms of theratio of radii k

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Three curves are shown which correspond to variousvalues of small parameter , that means to various con-tributions of bending and shearing in relation to the basicload (twisting). The above picture shows that the greateris the perturbation (force P ) the greater are the prots involume.

5Conclusions

1. Effective solutions for optimization of a bar under tor-sion with superposed bending and shear are obtainedvia the boundary perturbation method.

2. The rst correction of cross-sectional shape vanishesfor a solid bar, but is different from zero for a hollowbar.

3. The more thin-walled is the doubly connected cross-section, the greater is the prot in volume. How-ever, we should remember that for thin-walled sec-tions the condition of stability should be introduced.Such a condition was not considered in the presentpaper.

References

Bochenek, B.; Kordas, Z.; Zyczkowski, M. 1983: Optimal plas-tic design of a cross section under torsion and small bending.J. Struct. Mech. 11 , 383400

Bonnans, J.F.; Shapiro, A. 2000: Perturbation analysis of op-

timization problems . Berlin, Heidelberg, New York: Springer

Egner, W. 1996: Polynomial boundary perturbation for opti-mal plastic design of heads of plane tension members. Eng.Trans. 44 , 261281

Egner, W. 2000a: Optimal plastic shape design of heads of plane tension members with skew bearing surfaces. Eng. Opt.32 , 463483

Egner, W. 2000b: Optimal plastic shape design of rotationallysymmetric elements with conical bearing surfaces. J. Theor.Appl. Mech. 38 , 747765

Egner, W.; Kordas, Z.; Zyczkowski, M. 1994: Optimal plasticshape design via the boundary perturbation method. Struct.

Optim. 8 , 145155

Handelman, G.H. 1944: A variational principle for a state of combined plastic stress. Quart. Appl. Math. 1 , 351353

Heyman, J. 1959: On the absolute minimum weight design of framed structures. Quart. J. Mech. Appl. Math. 12 , 314324

Hill, R. 1948: A variational principle of maximum plastic workin classical plasticity. Quart. J. Mech. Appl. Math. 1 , 1828

Hill, R. 1950: The mathematical theory of plasticity . Oxford:Clarendon Press

Kordas, Z.; Zyczkowski, M. 1970: Investigation of the shapeof non-circular thick-walled cylinders subject to full yielding

at the stage of plastic collapse. Bull. Acad. Pol. Sci., Ser. Sci.Techn. 18 , 839847

Miller, P.M.; Malvern, L.E. 1967: Numerical solution of Piechniks equation for the combined bending and torsion of rigid-plastic bars. Proc. 10-th Midwestern Conf. Sol. Mech .,pp. 571580, Colorado State University

Oberweis, P.; Zyczkowski, M. 1997: On Palas method forplastic bending with torsion. J. Appl. Mech. 64 , 435437

Piechnik, S. 1961: The inuence of bending on the limitstate of a circular bar subject to torsion. Archiwum Mech.Stosowanej 13 , 77106

Piechnik, S.; Zyczkowski, M. 1961: On the plastic interactioncurve for bending and torsion of a circular bar. Archiwum Mech. Stosowanej 13 , 669692

Rozvany,G.I.N. 1973: Optimum design for bending and shear.J. Eng. Mech. Div. ASCE 99 , 11071109

Rozvany, G.I.N. 1976: Optimal design of exural systems . Ox-ford, New York: Pergamon Press

Rozvany,G.I.N.; Hill, R. 1976: General theory of optimal forcetransmission by exure. Advances Appl. Mech. 16 , 183308

Zyczkowski, M. 1981: Combined loadings in the theory of plas-ticity . Warszawa-Alphen aan den Rijn: PWNNijhoff

Zyczkowski, M. 1997: Optimal plastic design for combinedtorsion with bending. In: Gutkowski, W.; Mr oz, Z. (eds.):Proc. 2nd World Congr. on Structural and Multidisciplinary Optimization (held in Zakopane), pp. 809814, Warsaw: Pol-ish Academy of Sciences

Zyczkowski, M. 1998: Optimal plastic design of statically in-determinate beams under bending with torsion. In: Ober-weis, P. (ed.): Mechanik, Festschrift Peter Gummert , pp. 251262. Berlin: Institut f ur Mechanik, TU Berlin

Zyczkowski, M.; Egner, W. 1995: Boundary perturbationmethod applied to optimal plastic shape design of rotationallysymmetric elements. In: Olhoff, N.; Rozvany, G.I.N. (eds.):Proc. 1st World Congr. on Structural and Multidsicplinary Optimization WCSMO-1 (held in Goslar), pp. 511516. Ox-ford: Pergamon Press