On Distortion of Thin-Walled Beams with Simple Multi-Cell Closed Cross-Sections Subjected

ON THE CROSS-SECTION DISTORTION OF THIN-

WALLED BEAMS WITH MULTI-CELL CROSS-

SECTIONS SUBJECTED TO BENDING

R. PAVAZZA and B. BLAGOJEVIĆ

Faculty of Electrical Engineering, Mechanical Engineering and Naval Architecture,

University of Split, Ruđera Boskovića bb, 21000 Split, Craotia.

Abstract

This paper deals with distortion of the cross-section contour of thin-walled beams

with simple multi-cell closed rectangular cross-sections. The cross-section

distortion is considered in the limit case. It is assumed that beam plates are hinged

together along their longitudinal edges. Double symmetric three and two-cell

closed cross-sections are considered. The stresses and displacements are obtained

in the closed analytical form. The additional stresses and displacements due to

distortion with respect to the stresses and displacements of the ordinary theory of

bending are obtained. The boundary conditions are given in the general form. Some

illustrative examples are given.

Keywords: Distortion of cross-sections; Bending of thin-walled beams; Three and two cell closed sections; Hinged walls; Analytic method.

1. Introduction

The reliability of the assumption, in the theory of bending of thin-walled beams, that the shape of the cross-section is maintained depends on the stiffness of the beam transverse framing, or the cross-section itself (Vlasov, 1961); in some cases, also on the load distribution in the transverse direction (Kollbrunner and Basler, 1969).

The thin-walled beams are assembled of a number of thin plates that are restrained along their longitudinal edges. The plates can be stiffened by frames, usually in the transverse direction; sometimes, also by transverse bulkheads (“diaphragms”). Transverse framing, or the cross-section itself, cannot prevent the cross-section distortion entirely. In the limit, it may be assumed that plates are “hinged” along their longitudinal edges.

These two different types of structural behavior (with the “rigid” cross-sections and hinged cross-sections) may be considered as two limiting cases of stresses and deformations of the actual structure (Kollbrunner and Basler, 1969).

The theory that deals with such idealized beams, with hinged plates, is the folded plate theory (Schwyzer, 1920, Kollbrunner and Basler, 1969). By the theory, the longitudinal deformations and therefore the longitudinal normal stresses are equal at hinged connections. Hence, the normal stresses are linearly distributed over the cross- sections of each plate. The end cross-sections are assumed rigid. Each plate is at most preceded and followed by one plate only; the method is no longer applicable if one hinge belongs to three or more plates.

The problem of the cross-section distortion can be solved in an “exact” way, in general, by using three-dimensional models (Hughes, 1983; Bull, 1988). However, in the case of ordinary beams, with small dimensions of the cross-section contour with respect to the length of the beam, simpler analytical as well as numerical methods may also be applied (Boytzov, 1972; Boswell and Zhang, 1984; Boswell and Li, 1995; Hsu et al., 1995;

Kim and Kim, 1999, 2000, 2001; Pavazza and Matoković, 2000; Kim et al., 2002;

Pavazza, 2002). In this paper, an analytical approach to the problem of the cross-section distortion

of prismatic beams with closed rectangular thin-walled cross-sections subjected to

bending with influence of shear will be considered. The double symmetric cross-

sections with three and two closed cells will be analysed. The results will be given in

the analytical closed form, suitable for parametric studies.

2. Thin-walled beams subjected to bending with influence of shear

The forces-displacements relations for a prismatic thin-walled beam subjected to bending, by ordinary beam theories that include the shear influence (Filin, 1975), are governed by the following differential equations

dw d2w d3w dM =- b , EI=- M , EI= -y =- Q , dx ydx2 y ydx3 d x z d4wd2M dQ EI=-y = -z = q , (1) ydx4 d x 2 dx z where w= w( x ) is the displacement in the z-direction due to bending, b= b(x ) is the

angular displacement with respect to the y-axis due to bending, I y is the cross-section

moment of inertia with respect to the y-axis, My= M y ( x ) is the bending moment with

respect to the y-axis, Qz= Q z ( x ) is the shearing force with respect to the z-axis,

qz= q z ( x ) is the line load per unit length in the z-direction, Oxyz is the rectangular coordinate system, where the y and z-axis coincide with the principal axes, where the z-axis is the axis of symmetry, and E is the modulus of elasticity (Fig.1);

2 dws dws dws d Q z = - bs , GAs= Q z , GA= = - q , (2) dx dx sdx2 d x z where ws= w s ( x ) is the displacement in the z-direction due to shear, bs= b s (x ) is the

angular displacement with respect to the y-axis due to shear, As is the cross-section shear area; wt= w + w s , bt= b + b s , (3) where wt is the total displacement in the z-direction and bt is the total angular displacement with respect to the y-axis.

The stresses are given by

M y T Qz * = z xs T= S s x , t xs = , xs y , (4) I y t I y where sx= s x (x , s ) is the normal stress in the x-direction, txs= t xs (x , s ) is the shear stress

in the s-direction, Txs= T xs ( x , s ) is the shear flow in the s-direction, z= z( s ) is the cross-

* * section rectangular co-ordinate, Sy= S y ( s ) is the statical moment of the “cut-off portion” of the cross-section with respect to the y-axis, t= t( s ) is the wall thickness and s is the

cross-section curvilinear co-ordinate, with the origins K0 and K , where Txs = 0 (Fig.1);

It is assumed that the shear influence on bending is small; it will be taken into account in the calculations of displacements only, by using (2) and (3). The cross-section contour is assumed “rigid” (neglecting the contour contraction).

Thus, from Hooke’s law,

sx= E e x , (5) where ex= e x (x , s ) is the strain in x-direction. The shear flow is defined by the normal

stress s x , by the equilibrium equation in the x-direction

抖 T s xt + xs = 0 . (6) 抖x s

The normal stress in the s-direction ss= s s (x , s ) , which is ignored in Eqn (5) as a small quantity, can be obtained by the shear flow, by using the equilibrium equation in the s- direction 抖T( t ) sx+s s = 0 . (7) 抖x s where Tsx= T xs .

3. Thin-walled beams subjected to bending with the cross-section distortion

3.1. Distortion of the cross-sections of thin-walled beams subjected to bending

If the frames or diaphragms are omitted, or the cross-section is not stiff by itself, the cross-sections of thin-walled beams are no longer forced to maintain their shape. In the limit, it may be assumed that the beam walls are hinged along their longitudinal edges.

Thus, the walls will be loaded by forces per unit length along their longitudinal edges, in their planes (Pavazza, 2002).

In the case of symmetrical closed thin-walled cross-sections with three cells

(Fig.2a), the inner vertical walls and the central horizontal walls may be treated due to symmetry as a unique beam with the rigid cross section (beam component 1). Two cell cross-sections may be treated as a special case of three cell cross-sections (Fig.2b).

3.2. Beam components

The beam may be decomposed into four beam components (Fig.3). In the case of double symmetric beam cross-sections, it may be assumed that the beam components 1 and

3 are subjected to bending with influence of shear and the beam components 2 and 2 to tension and shearing due to distortion of the cross-section shape.

For the beam components 1 and 3, the following differential equations may be written (i = 1,3) (Filin, 1975; Pavazza, 1991)

2 3 dM dwi d wi d wi yi =- bi , EI=- M , EIyi3 = - = -( Q zi - m yi ) , dx yidx2 yi dx d x d4w d2M d d m EIi =-yi = -( Q - m ) = q + yi , (8) yidx4 d x 2 d x zi yi zi d x where wi= w i ( x ) is the displacement in the zi -direction, bi= b i (x ) is the angular displacement with respect to the yi -axis, I yi is the cross-section moment of inertia with

respect to the yi -axis, Myi= M yi ( x ) is the bending moment with respect to the yi -axis,

Qzi= Q zi ( x ) is the shearing force with respect to the zi -axis, qzi= q zi ( x ) is the force per

unit length in zi -direction, myi= m yi ( x ) is the moment per unit length, Oi x i y i z i is the rectangular co-ordinate system, where the yi , zi -axes coincide with principal axes of the corresponding cross-sections;

2 dwsi dwsi dwsi d Q zi = - bsi , GAsi= Q zi , GA= = - q , (9) dx dx sidx2 d x zi where wsi= w si ( x ) is the displacement in the zi -direction due to shear, bsi= b si (x ) is the

angular displacement with respect to the yi -axis due to shear, Asi is the cross-section shear area;

wti= w i + w si , bti= b i + b si , (10) where wti is the total displacement in the zi -direction and bti is the total angular

displacements with respect to the yi -axis .

For the beam component 1, the moment per unit length may be expressed as

my1 = T A h , (11) where TA= T A ( x ) is the force per unit length along the longitudinal edge A and h is the height of the cross-section. The stresses may be expressed as follows

M T (Q- m ) = y1 z = xs1 T=z1 y 1 S* + T s x1 1 , t xs1 , xs1 y 1 A A- A , (12) I y1 t1 I y1

In this equation TA is assumed to be distributed along the vertical wall of the beam components 1 only, as a constant (which is expressed symbolically by a vertical line); it represents the influence of the beam components 2 on the beam component 1. For the beam component 3, similarly, one has my3 = T B h , (13) where TB= T B ( x ) is the force per unit length along the longitudinal edge B;

* M y3 Txs3 (Qz3- m y 3 ) S y 3 s x3= z 3 , t xs3 = , Txs3 = + T B . (14) I y3 t3 I y3 Here is assumed that the origin of s is at B, i.e B´ (Fig. 1).

For the beam component 2, it may be written (Pavazza, 2000, Pavazza, 2001)

2 du2 p d u dN EA= N , EA2 p =2 = - q , (15) 2dx 2 2dx2 d x x 2 where

qx2 = T A + T B , (16)

where A2 is the cross-section area of the beam component 2, u2p= u 2 p ( x ) is the displacement of the cross-section in the x2 -direction as a plane section and N2= N 2 ( x ) is the normal force;

dQz2 dmy2 Qz2= m y 2 , = -q = , (17) dxz2 d x where b m= -( T - T ) . (18) y2 A B 2 Then

u2= u 2 p + f where u2= u 2( x , z 2 ) is the displacement in the x2 -direction; f= f( x , z ) is warping due to shear 骣 2 Qz2 q x 2 2 b f= z2 + z 2 - + g , (19) GA2 A 2 E 桫 12

where g= g( x , z ) is a function that can be obtained by assuming that f = 0 when w1= w 3 , when according to (4)

* * 2z20 SyA- S yB TA- T B = k( T A + T B ) , k = = * * (20) b SyA+ S yB

* where z20 is the coordinate of the shear flow zero point for the rigid cross-section, S yA and

* S yB are the statical moments of the cut-off portion of the rigid cross-section for the points A and B, respectively.

Thus, from (19) and (20)

骣 2 k( TA+ T B ) b TA- T B 2 b g= z2 - z 2 - , (21) 2GA2 kEA 2 桫 12 i.e. 2 b2 轾2 G骣 z z G f=轾 T - T - k( T + T ) 犏 2 + 2 - , (22) 臌A B A B 2GA犏 kE桫 b b 6 kE 2 臌犏 and b2 轾2 G骣 z2 z G u= u -轾 T - T - k( T + T ) 犏 2 + 2 - . (23) 2 2P臌 A B A B 2GA 犏 kE桫 b b 6 kE 2 臌犏 The stresses then read

2 轾骣 2 N2 Eb犏2 G玳 z 2 z 2 G d sx2 =E e x = - + - T A - T B - k( T A + T B ) , A2 GA犏 kE桫 b b 6 kE d x 臌 2 2 臌犏 T 1 z = xz2 2 t xz2 , Txz2 = -( T A - T B ) + ( T A + T B ) , (24) t2 2 b u where e = 2 . x x

The stresses given by (24) satisfy the equilibrium equation given by (6) only if d [T- T - k( T + T )] = const ; (25) dx A B A B otherwise the solution can be used only approximately. The beam component 2 will be under antisymmetric tension and shearing, with respect to the beam component 2.

3.3. Compatibility conditions

The compatibility conditions, along the edges A and B, taking into account (23), may be expressed as

h b2 骣 2 G - =u +玳1 - T - T - k( T + T ) , b1 2 p桫臌 A B A B 2 4GA2 3 kE

h b2 骣 2 G - =u -玳1 + T - T - k( T + T ) . (26) b3 2 p桫臌 A B A B 2 4GA2 3 kE Hence h b2 u= - + +轾 T - T - k T + T 2p( b 1 b 3 ) 臌 A B( A B ) , 4 6EA2 k

GA2 h TA- T B - k( T A + T B ) = -( b1 - b 3 ) . (27) b2 Then, referring to the first equation of (15) and the first and second equations of (8)

骣M M鼢A h 骣 M M GA h N = -珑 y1 + y 3鼢 2 - y 1 - y 3 2 2 珑鼢 . (28) 桫珑Iy1 I y 3鼢4 桫 I y 1 I y 3 6 kE and, referring to the second equation of (15), (16), and the third equation of (8)

骣Q- m Q - m鼢A h 骣 Q - m Q - m GA h T+ T =珑 z1 y 1 + z 3 y 3鼢 2 + z 1 y 1 - z 3 y 3 2 A B 珑鼢 , (29) 桫珑 Iy1 I y 3鼢4 桫 I y 1 I y 3 6 kE i.e. EA hd3 GA h d 3 T+ T =-2�( w - w) � 2 ( w w ) . (30) A B 4 dx31 3 6 k d x 3 1 3 The second equation of (27), taking into account the first equation of (8), may be written as

GA h d T- T + k( T + T) = - 2 � ( w w ) . (31) A B A B b2 d x 1 3

The functionsTA and TB may then be obtained from (30) and (31):

EA hd3 GA h 骣 1 d 3 T= -2(1 + k )( w + w) - 2 1 +( w - w ) + A 8 dx31 3 12桫 k d x 3 1 3 GA h d +2 ( w - w ) , 2b2 d x 1 3

EA hd3 GA h 骣 1 d 3 T= -2(1 - k )( w + w) + 2 1 -( w - w ) - B 8 dx31 3 12桫 k d x 3 1 3 GA h d -2 ( w - w ) . (32) 2b2 d x 1 3

For the case w1= w 3 , one obtains

EA hd3 w EA hd3 w T= -2(1 + k ) 1 , T= -2(1 - k ) 1 . (33) A 4 dx3 B 4 dx3

Substitution of (33) into (8), taking into account (11) and (13), yields d4w d4w EI1 = q , EI3 = q , (34) ycdx4 z1 ysdx4 z3 where A h2 A h2 I= I +(1 + k ) 2 , I= I +(1 - k ) 2 . (35) yc y1 4 ys y3 4 Since q 1 q+ q = z , I+ I = I , (36) z1 z 3 2 yc ys2 y from (34), taking into account (1), one obtains w1= w 3 = w , i.e.

q q q z1= z 3 = z . (37) Iyc I ys I y

Thus, referring to (1) and (4), TA and TB , given by (32), become

Qz * Qz * TA= S yA = - T xsA , TB= S yB = T xsB , (38) I y I y where A h A h S* =2 (1 + k) , S* =2 (1 - k) . (39) yA 4 yB 4

3.4. Internal forces and displacements

By substituting (28) into the fourth equation of (8), taking onto account (11) and

(13), the following differential equations may be written

d4w EA h 2骣 2 G d 4 GA h 2 d 2 EI1- 2(1 + k) 1 -( w - w) - 2 � ( w = w) q , ycdx4 8桫 3 kE d x 41 3 2 b 2 d x 2 1 3 z 1

d4w EA h 2骣 2 G d 4 GA h 2 d 2 EI1+ 2(1 - k) 1 +( w - w) + 2 � ( w = w) q , ysdx4 8桫 3 kE d x 41 3 2 b 2 d x 2 1 3 z 3 (40) where I yc and I ys are given by (35).

By multiplying the first equation from (36) by Iys( I yc+ I ys ) and the second by

Iyc( I yc+ I ys ) , and subtracting the second equation from the first, the following differential equation may be written d4w d 2 w EI1- 3- k 1 - 3 = q , (41) ycsdx4b d x 2 zcs where q I- q I 2 z1 ys z 3 yc GA2 骣 h w= w - w qzcs = = 1- 3 1 3 , , kb , Iyc+ I ys 2 桫b

轾骣2G 骣 2 G 犏 +珑 -鼢 + - + (1k)珑 1鼢 Iys( 1 k) 1 I yc 2 Iyc I ys 犏桫珑 3kE 桫 3 kE A h I =犏1 - 2 , (42) ycs I+ I犏 I I 8 yc ys犏 yc ys 臌犏

The following differential equations can then be plotted (see Appendix)

dw d2w d3w dM 1- 3 , 1- 3 , 1- 3 ycs , = - b1- 3 EIycs=- M ycs EIycs= - = -( Q zcs - m ycs ) dx dx2 dx3 d x d4w d2M d d m 1- 3 ycs ycs , EIycs= - = -( Q zcs - m ycs) = q zcs + (43) dx4 d x 2 d x d x where

dw1- 3 mycs = k = - k b1- 3 . (44) bdx b

Taking into account the first equation of (43), it may be written

d2w d 2 w d 2 w 1- 3 = 1- 3 , (45) dx2 d x 2 d x 2 i.e. referring to the second equation of (8) and the second equation of (43)

My1 M y 3 M ycs - = . (46) Iy1 I y 3 I ycs

From the equilibrium, it follows 1 1 M+ M - N h = M , Q+ Q = Q . (47) y1 y 3 2 2 y z1 z 3 2 z

Substitution of (28) into the first equation of (47) yields

My1 M y 3 1 Iⅱyc+ I ys = M y , (48) Iy1 I y 3 2 where A h2 骣 2 G A h2 骣 2 G I= I +2 1 + , I= I +2 1 - . (49) yc y1 4桫 3kE ys y3 4桫 3kE Here 1 Iⅱ + I = I . (50) yc ys2 y From (46) and (48), taking into account (50), it may be written

My I ys M ycs My I yc M ycs My1= I y 1 + 2 I y 1 , My3= I y 3 - 2 I y 3 , (51) Iy I ycs I y Iy I ycs I y and according to third equation of (8) and the third equation of (43)

- Qz Iys Q zcs m ycs Qz1- m y 1 = I y 1 + 2 I y 1 , Iy I ycs I y

- Qz Iyc Q zcs m ycs Qz3- m y 3 = I y 3 - 2 I y 3 . (52) Iy I ycs I y According to (51), taking into account the second equations of (8), (1) and (43), it may be written

w1= w +D w 1 , w3= w +D w 3 ;

b1= b +D b 1 , b3= b +D b 3 . (53) where

I ys I yc Dw1= 2 w 1- 3 , Dw3= - 2 w 1- 3 ; I y I y

I ys I yc Db1= 2 b 1- 3 , Db3= - 2 b 1- 3 . (54) I y I y From (28) and (51), one has

ⅱ - ⅱ + MyA2 h I yc I ys M ycs A 2 h Iyc I ys M ycs GA2 h N2 = -� 创 - 创 . (55) Iy2 I ycs I y 2 Iycs I y 3 kE From (53), it follows

1 Iycⅱ - I ys w=( w1 + w 3) + w 1- 3 . (56) 2 I y

Thus, taking into account the third equations of (1), (41) and (43), TA and TB given by (32) may be written as

Q Q Iⅱ - I T=z S* - zcs yc ys S * k b1- 3 A yA� yA b , Iy I y I ycs h Q Q Iⅱ - I T=z S* - zcs yc ys S * k b1- 3 B yB� yB b , (57) Iy I y I ycs h

* * where S yA and S yB are given by (39).

By substituting (57) into (52), taking into account (12) and (14), (35), (39) and

(49), where

* * Iy1 + S yA h = I yc , Iy3 + S yB h = I ys ,

* * 2Iysⅱ I y1+( I yc - I ⅱys) S yA h 2 I yc I y 3 +( I ycⅱ - I ys) S yB h = = I ycs , Iy I y the following relation can be obtained

Qz Qz Qz1 = I yc + Q zcs , Qz3 = I ys - Q zcs . (58) I y I y Hence, taking into account the second equation of (47)

Qz1 I ys- Q z 3 I yc Qzcs = . (59) Iyc+ I ys

The displacements due to shear ws and wsi may be obtained, according to the second equations of (2) and (9), as follows

(A ) (A ) (A ) My- M y My1- M y 1 My3- M y 3 ws = , ws1 = , ws3 = , (60) GAs GAs1 GAs3

(A ) (A ) (A ) M y3 where M y , M y1 and are the bending moments of the beam and the beam components 1 and 3 at the beam left end. From (60) and (51), it follows

(A ) ( A ) My- M y I y1 M ycs - M ycs I ys I y 1 ws1 = � 2 , GAs1 I y GA s 1 I ycs I y

(A ) ( A ) My- M y I y3 M ycs - M ycs I yc I y 3 ws3 = � 2 . (61) GAs3 I y GA s 3 I ycs I y

The unknown shear areas As1 and As3 can be obtained from the condition: when

ws1= w s 3 = w s , then w1= w 3 = w , M ycs = 0 . Thus, from (61) and (60)

I y1 I y3 As1 = A s , As3 = A s . (62) I y I y By substituting (62) into (61), one may finally write

ws1= w s + D w s 1 , ws3= w s + D w s 3 , (63) where (A ) (A ) Iys M ycs- M ycs Iyc M ycs- M ycs Dws1 = 2 , Dws3 = -2 . (64) Iycs GA s Iycs GA s 3.4. Stresses

The stresses for the beam component 1 may finally be obtained by substituting the first equation of (51) and the first equation of (52) into (12):

A M y1 Q Q =z + z* z * sx1 1D s x 1 , Txs1= S y 1 + S yA +D T xs 1 , (65) I y I I y y A where

2Iys M ycs Ds x1= z 1 , Iycs I y

A ⅱ - - A 2Iys Q zcs m ycs*Qzcs I yc I ys * m ycs DTxs1= � S y 1 � S yA . (66) I I I I h ycs y y ycs A A The stresses for the beam component 3 may be obtained by substituting the second equation of (51) and the second equation of (52) into (14):

M y3 Qz* Q z * sx3=z 3 +D s x 3 , Txs3= S y 3 + S yB +D T xs 3 , (67) I y Iy I y where

2Iyc M ycs Ds x3= - z 3 , Iycs I y

ⅱ - - 2Iyc Q zcs m ycs*Qzcs I yc I ys * m ycs DTxs3= - � S y 3 - S yB . (68) Iycs I y I y I ycs h

The stresses for the beam component 2 may be obtained by substituting (55) and

(57) into (24), taking into account the first equations of (1) and (43), and (39):

M y h Qz* * Q z * * z2 sx2= - � D s x 2 , Txz2= -( S yA - S yB) +( S yA + S yB) +D T xz 2 , I y 2 2Iy I y b (69) where ⅱ- ⅱ + ⅱ + Iyc I ys M ycsh I yc I ys M ycs Gh I yc I ys M ycs z2 Ds x2 = 创 - 创 + 创h + Iycs I y2 I ycs I y 3 kE I ycs I y b

Iⅱ + I M Gh轾骣 z 2 1 +yc ys创 ycs 犏6 2 - , I I3 kE犏桫 b 2 ycs y 臌犏 ⅱ- ⅱ - QzcsIyc I ys* * Q zcs I yc I ys * * z2 m ycs DTxz2 = �( S yA - S yB) � - ( S yA S yB ) . (70) 2Iy I ycs I y I ycs b h

The force per unit length qz2 can be obtained from (17) and (18), taking into account (57):

qz2 * * b qz2= -( S yA - S yB) +D q z 2 , (71) I y 2 where qIⅱ - I b M b q=zcs �yc ys S* - S * ycs k D z2 ( yA yB ) b . (72) Iy I ycs2 I ycs h Here

qz2 s y2 A =- , (73) t2 where s y2 A is the normal stress in the y2 -direction for y2 =- b 2 and t2 is the thickness of the beam component 2.

The first part of the expressions for stresses represent the stresses for the case of rigid cross-sections. The second part ( Ds and DT ) represent the additional stresses due

to cross-section distortion, defined by the load qzcs and internal forces Qzcs and M ycs , and

the moment mycs (see Appendix).

The same holds for the displacements given by (53) and (54), and (63) and (64),

where the additional displacements due to distortion are defined by the displacements w1- 3

and b1- 3 , and the moment M ycs (see Appendix).

3.5. Boundary conditions

The boundary conditions at the ends of a beam may be formulated as follows * * * * w1= w 1 , w3= w 3 ; b1= b 1 , b3= b 3 . (74)

Then, according to the first equation of (41),

* * * * w1- 3= w 1 - w 3 ; b1- 3= b 1 - b 3 ; (75) and according to (56)

1 * *Iⅱyc- I ys * * 1 * *Iycⅱ - I ys * * w=( w1 + w 3) +( w 1 - w 3 ) ; b=( b1 + b 3) +( b 1 - b 3 ) ; 2 I y 2 I y (76) and according to (27) h b2 u= -* + * +轾 T - T - k T + T 2( b 1 b 3 ) 臌A B( A B ) , 4 6EA2 k

* * GA2 h TA- T B - k( T A + T B ) = -( b1 - b 3 ) . (77) b2

* * * * Thus, the boundary conditions are defined by the components w1, w 3 and b1, b 3 .

In terms of forces, the boundary conditions may be formulated as follows

* * * * Qz1= Q z 1 , Qz3= Q z 3 ; My1= M y 1 , My3= M y 3 . (78)

Then, according to (59) and (46)

* * 骣 * * Qz1 I ys- Q z 3 I yc My1 M y 3 Qzcs = ; Mycs= - I ycs ; (79) Iyc+ I ys 桫Iy1 I y 3 and from (58) and (48), 骣 * * Iⅱyc* I ys * Qz=2( Q z1 + Q z 3) ; My=2 M y1 + M y 3 . (80) 桫Iy1 I y 3

* * * * Here the boundary conditions are defined by the components Qz1, Q z 3 and My1, M y 3 .

The “simply supported” ends may be defined by

wt1= w 1 = w t 3 = w 3 = 0 ; My1= M y 3 = 0 . (81)

Then, according to the first equation of (75) and the second equation of (79)

w1- 3 = 0 ; M ycs = 0 ; (82) and according to the first equation of (76) and the second equation of (80) wt = w = 0 ; M y = 0 . (83) Referring to (65), (67), (69) and (4), it follows

sx1= s x 3 = s x 2 = s x = 0 . (84) For the “built-in end”, it may be written

w1= wt 1 = w 3 = w t 3 = 0 , b1= b 3 = 0 , (85) for the left end (A);

(B ) ( A ) (B ) ( A ) My1- M y 1 My3- M y 3 wt1= w 1 + = 0 , wt3= w 3 + = 0 , b1= b 3 = 0 , (86) GAs1 GAs3 for the right end (B).

Then, according to (75),

w1- 3 = 0 , b1- 3 = 0 , (87) and according to (76),

w = 0 , b = 0 , (88) for the left end (A);

(B ) ( A ) My- M y wt = w + = 0 . (89) GAs for the right end (B).

The “free end” may be defined as

Qz1 = 0 , Qz3 = 0 ; M y1 = 0, M y3 = 0 . (90) Then, according to (79),

Qzcs = 0 , M ycs = 0 , (91) and according to (80),

Qz = 0 , M y = 0 . (92) Referring to (65), (67), (69) and (4), it follows

sx1= s x 3 = s x 2 = s x = 0 ; Txs1= T xs 3 = T xz 2 = T xs = 0 . (93)

4. Illustrative examples

In order to illustrate the application of the expressions for the stresses and displacements, a thin-walled beam with the double symmetric three-cell closed cross- section with the constant thickness is considered (see Appendix).

 Geometrical properties and the material 32 1 5 2 h= 2 b , A= 16 A , I= A b2 , S* = A b , S* = A b , k = - ; 2 y 3 2 yA 6 2 yB 6 2 3 8 2 I= A b2 , I= A b2 , I3.282 A b2 , I2.052 A b2 , y13 2 y33 2 yc 2 ys 2 7 I= 3 A b2 , I= A b2 , I0.923 A b2 , yc 2 ys 3 2 ycs 2 E l k= 2 GA , = 2.6 , v = 0.456 ; b 2 G b

I y 2I 2I Iⅱ + I 11.557 , yc 7.112 , ys 4.446 , yc ys 5.779 , I ycs I ycs I ycs I ycs

Iⅱ - I 2I 2I yc ys 1.333, yc 0.615 , ys 0.385 . I ycs I y I y Here l is the length of the beam. The shear areas are calculated as pure geometrical properties, for Poisson`s ratio equal to zero (Bhat and de Oliveira, 1985; Cowper, 1966)

As = AK , K = 0.424 ; As = AK = 6.784 A2 .  Simply supported beam under uniformly distributed load

The internal forces at x = 0 (Example 1, Appendix):

qzcs l q z l q zcs q zcs Qzcs= = � Q z , 2 2 qz q z

qzcs l q zcs Qzcs- m ycs =c3 = Q z c 3 , 2 qz

qzcs mycs= Q z (1 - c3 ) . qz The internal forces and displacements at x= l 2 (Example 1, Appendix):

2 2 qzcs l q z l q zcs q zcs Mycs=c0 = � c 0 M y c 0 , 8 8 qz q z

4 4 5qzcs l 5 q z l q zcs q zcs w1- 3=j 0 =11.557� j 0 = 11.557 w j 0 384EIycs 384 EI y q z q z 2 qzcs j 0 48EI y 骣b =11.557wt , w=h w , h =1 + = 1 + 39.245 . q t 0 0 2 琪 z h0 5GAs l桫 l

Additional stresses due to distortion at x = 0 :

A Q- mQ A m T=4.446zcs ycs S* - 1.333 zcs S * + ycs = D xs1 y 1 yA A Iy I y 2 b A

A A Qz* q zcs Q z * q zcs Q q =4.446S ( v ) - 1.333 S + z zcs y1c 3 yA A 5.333A2 b ( 1- c 3 ) , Iy q z I y q z I q y z A - Qzcs m ycs*Qzcs * m ycs DTxs3= -7.112 S y 3 - 1.333 S yB - = Iy I y 2 b

Qz* q zcs Q z * q zcs Qz q zcs = -7.112Sy3c 3 - 1.333 S yB - 5.333A2 b ( 1- c 3 ) , Iy q z I y q z Iy q z

Qzcs* * Q zcs * * z2 mycs DTxz2 =0.667( S yA - S yB) - 1.333 ( S yA + S yB ) - = Iy I y b2 b

Qz* * q zcs Q z * * z2 q zcs =0.667( SyA - S yB) - 1.333 ( S yA + S yB ) � Iy q z I y b q z

Qz q zcs -5.333A2 b ( 1 - c 3) . Iy q z

Additional stresses and displacements due to distortion at x= l 2 :

Mycs M y qzcs Dsx1=4.446z 1 = 4.446 z 1 c 0 , Iy I y q z

Mycs M y qzcs Dsx3= -7.112z 3 = - 7.112 z 3 c 0 , Iy I y q z

Mycs M ycs M ycs Ds x2=1.333b - 2.223 b + 5.779 2 z 2 + Iy I y I y

M轾骣 2 M M ycs犏3 2z2 1 y qzcs y q zcs +2.223b - = 1.333 bc0 - 2.223 b c 0 + I犏2桫 b 2 I q I q y臌犏 y z y z

Myq M y q轾3骣 2 z 1 +5.779 2zzcs + 2.223 b zcs 犏 2 - , 2c 0犏桫 c 0 Iy q z I y q z 臌2 b 2

qzcs q zcs j 0 Dw1=0.385 w 1- 3 = 4.446 wj 0 = 4.446 wt , qz q z h0

qzcs q zcs j 0 Dw3= -0.615 w 1- 3 = - 7.112 wj 0 =- 7.112 wt , qz q z h0 (A ) Mycs- M ycs qzcs q zcs c0 Dws1 =4.446 = 4.446 w sc 0 = 4.446 w t , GAs q z q z h0

(A ) Mycs- M ycs qzcs q zcs c0 Dws3 = -7.112 = - 7.112 w sc 0 = - 7.112 w t , GAs q z q z h0

2 48EI y 骣b 2 39.24琪 ws 5GAs l 桫l wt = , h = = . 0 48EI 2 h0 y 骣b 1+ 2 1+ 39.24琪 5GAs l 桫l  Built-in beam under uniformly distributed load.

Components of the internal forces at x = 0 (Example 2, Appendix):

2 2 qzcs l q z l q zcs q zcs Mycs= -c2 = - � c 2 M y c 2 , 12 12 qz q z

qzcs l q z l q zcs q zcs Qzcs= = � Q z . 2 2 qz q z Components of the internal forces and displacements at x= l 2 (Example 2, Appendix):

2 2 qzcs l q z l q zcs q zcs Mycs=c1 = � c 1 M y c 1 , 24 24 qz q z

4 4 qzcs l q z l q zcs q zcs w1- 3=j 1 =11.557� j 1 = 11.557 w j 1 384EIycs 384 EI y q z q z

2 qzcs j 1 48EI y 骣b =11.557wt , w=h w , h =1 + = 1 + 196.2 . q t 1 1 2 琪 z h1 GAs l桫 l

Additional stresses due to distortion at x = 0 :

Q QA Q q T=4.446zcs S* - 1.333 zcs S * = 4.446 z S * zcs - D xs1 y 1 yAA y 1 Iy I y I y q z QA q - 1.333 zS * zcs yA A , Iy q z

Qzcs* Q zcs * Q z * q zcs DTxs3= -7.112 S y 3 - 1.333 S yB = - 7.112 S y 3 - Iy I y I y q z Qz* q zcs - 1.333 S yB , Iy q z

Mycs M ycs M ycs Ds x2=1.333b - 2.223 b + 5.779 2 z 2 + Iy I y I y

M轾骣 2 M M ycs犏3 2z2 1 y qzcs y q zcs +2.223b - = 1.333 bc2 - 2.223 b c 2 + I犏2桫 b 2 I q I q y臌犏 y z y z

Myq M y q轾3骣 2 z 1 +5.779 2zzcs + 2.223 b zcs 犏 2 - , 2c 2犏桫 c 2 Iy q z I y q z 臌2 b 2

Qzcs* * Q zcs * * z2 DTxz2 =1.333( S yA - S yB) - 1.333 ( S yA + S yB ) = 2Iy I y b

Qz* * q zcs Q z * * z2 q zcs =0.667( SyA - S yB) - 1.333 ( S yA + S yA ) . Iy q z I y b q z

Additional stresses and displacements due to distortion at x= l 2 :

qzcs q zcs j 1 Dw1=0.385 w 1- 3 = 4.446 wj 1 = 4.446 wt , qz q z h1

qzcs q zcs j 1 Dw3= -0.615 w 1- 3 = - 7.112 wj 1 = - 7.112 wt , qz q z h1

(A ) Mycs- M ycs qzcs q zcs Dws1 =4.446 = 4.446 w s(c 1 + c 2) = 4.446 w t ( c 1 + c 2) h 1 , GAs q z q z

(A ) Mycs- M ycs qzcs q zcs Dws3 = -7.112 = - 7.112 w s(c 1 + c 2) = - 7.112 w t ( c 1 + c 2) h 1 , GAs q z q z

2 48EI y 骣b 2 196.2琪 ws GAs l 桫l wt = , h = = . 1 48EI 2 h1 y 骣b 1+ 2 1+ 196.2琪 GAs l 桫l The additional stresses and displacements due to distortion are calculated for

l b = 32 , i.e. v =14.59 (j 0 = 0.0112 , j 1 = 0.0564 , c 0 = 0.0094 , c1 = 0.0282 ,

c 2 = 0.1915, c 3 = 0.0685 , y 0 = 0.0131); h0 = 1.0383 , h0 = 0.0369 , h1 = 1.192 ,

h1 = 0.161

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

The stresses and displacements are calculated for the beam subjected to uniformly

distributed load qz1 = q z 2 , qz3 = 0 ( qzcs q z = 7 32 ), both for simply supported and built-in beam ends, for x= l 2 and x = 0 . The results are given in respect to the stresses and displacements for rigid cross-sections (Fig.4-10).

In order to check accuracy of the obtained results, the normal stresses and displacements at x= l 2 for the simply supported beam are also calculated by the finite element method; the shear stresses are calculated for x = 0 . The 3D membrane model of one-quarter of the beam is investigated. A high mesh density with 3200 quadrilateral elements and 12880 nodes is used. The displacements of the nodes are restricted at x = 0 in the cross-section contour direction and at x= l 2 in the beam longitudinal direction; at y = 0 (in the plane of symmetry) the displacements are restricted in the beam transverse direction.

5. Conclusion An analytical method has been applied to estimate the additional stresses and displacements due to distortion of the cross-sections of thin-walled beams subjected to bending. Simple rectangular cross-sections with three and two closed cells with double axes of symmetry are considered. It is assumed that beams are long enough that warping due to shearing may be ignored in the stress calculation (in the case of rigid cross-sections, i.e. when the cross-sections maintain their shape). The cross-section distortion is considered in the limit, by assuming that beam walls are hinged along their longitudinal edges.

The additional stresses and displacements due to the cross-section distortion are given in the analytical closed form and compared to the stresses and displacements of the ordinary bending theory (where the cross-sections maintain their shape).

It is shown that the additional stresses and displacements due to the cross-section distortion can be significant (compared to the stresses and displacements of the ordinary bending theory), particularly, the additional shear stresses.

A typical cross-section with three cells is analysed, where the ratio of the beam length and the cross-section breadth was equal 8 (the beam length to the cross-section height ratio equal 16) is analysed. The loads were distributed along the inner vertical walls only. The ends of the beam were simply supported and built-in, respectively.

The comparison for the simply supported beam under uniformly distributed load along the inner vertical walls to the finite element solution of the problem has shown acceptable agreements of obtained results.

Although the hinges between plates do not occur in actuality, it is important to analyse such conditions, together with the ordinary beam theory (where the cross-section is assumed rigid), as limits of the actual behaviour of the structure. In fact, there are no enough stiff cross-sections in actual thin-walled beam structures to guarantee the cross- section shape, especially under non uniform load distributions in the transverse direction.

The real stiffness of the cross-section structure can be easily included in the consideration. The torsional stiffness of the plates together with bending stiffness of transverse framing (if the structure is framed) and the shear stiffness of transverse bulkheads (if there are any) should be taken into account.

The same approach may be used in the case of thin-walled curved beams. The cross-sections of the components beams may be approximate by rectangular cross- sections; or be considered as curvilinear cross-sections. In the compatibility conditions and equilibrium equations, the cross-section properties of the beam components will be changed only. It should be noted that relative vertical displacements are small, in comparison with vertical displacement of the beams. In that case, in the finite element analysis the shell elements must be used.

Acknowledgements

This research was supported by Ministry of Science and Technology of the Republic

Croatia Contract No. 0023025.

References

Boitzov, G.V., 1972. Analiz Poperechnoi Prochnosti Krupnotonnaznih Tankerov

[Analysis of the Transverse Strength o Large Tankers]. Sudostroenie 9, 17-24.

Boswell, L.F. Zhang, S.H., 1984, The effect of distortion in thin-walled beams,

International Journal of Solids and Structures 20 (9/10), 845-862.

Boswell, L.F., Li, Q., 1995. Consideration of the Relationships Between Torsion,

Distortion and Warping of Thin-Walled Beams, Thin-Walled Structures 21, 147- 161.

Bull, J.W., 1988. Finite Element Analysis of Thin-Walled Structures, Elsevier Applied

Science, London-New York.

Filin, A.P., 1975. Prikladnaya Mehanika Tverdovo Deformiruemovo Tela, II [Applied

Mechanics of Solids, II]. Nauka, Moscow.

Hsu, Y.T., Fu, C.C., Scheling, D.R. , 1995. EBEF method for distortional analysis of steel

box girder bridges, Journal of Structural Engineering 121 (3), 557-566.

Hughes, O.W., 1983. Ship Structural Design, John Wiley, New York.

Kim, J.H., Kim, Y.Y., 1999, Analysis of thin-walled closed beams with general

quadrilateral cross-sections, ASME Journal of Applied Mechanics 66 (4), 904-912.

Kim, J.H., Kim, Y.Y., 2000. One dimensional analysis of thin-walled beams having

general cross-sections. International Journal for Numerical methods in engineering

49, 653-668.

Kim, J.H., Kim, T.S., 2000. Topology optimization of beam cross-sectiom, International

Journal of Solids and structure 37, 477-493.

Kim, J.H., Kim, Y.Y., 2001. Thin-walled multi-cell beam analysis for coupled torsion,

distortion and warping deformation, ASME Journal of Applied mechanics 68, 260-

Kim, T.S., Kim, Y.Y., 2002. Multi-objective topology optimization of a beam under

torsion and distortion. AIAA Journal 40 (2), 376-381.

Kim, J.H., Kim, H.S., Kim, D.W., Kim, Y.Y., 2002. New accurate efficient modelling

technique for the vibration analysis of T-joint thin-walled box structures,

International Journal of Solids and Structure 39, 2893-2909.

Kollbrunner, C.F., Basler, K., 1969. Torsion in Structures, Springer-Verlag, Heildeberg,

New York. Pavazza, R., 1991. Savian’e i Uvian’e stapova otvorenog tankostienog presieka na

elastichnoi oodlozi [Bending and torsion of thin-walled beams on elastic

foundation], Thesis, Faculty of Mechanical Engineering and Naval Architecture,

University of Zagreb.

Pavazza, R., Matoković, A., 2000. On the shear flow due to distortion of hull cross-

sections of tankers with two longitudinal bulkheads. Proceedings, IMAM 2000,

Vol II, Ischia, Italy, pp. F32-39.

Pavazza, R., Plazibat. B., Matoković, A., 2001. On the effective breadth problem of deck

plating of ships with two longitudinal bulkheads, International shipbuilding

progress 48, 51-85.

Pavazza, R., 2002. On the load distribution of thin-walled beams subjected to bending with

respect to the cross-section distortion, International Journal of Mechanical Sciences

44, 423-442

Schwyzer, H., 1920. Statische untersuchung der aus ebenen tragflächen

zusammengesetzten räumlichen tragwerke, Diss. ETH Zűrich.

Timoshenko, S. P. , Goodier, J. N., 1983. Theory of elasticity, McGraw-Hill, New York.

Vlasov, V.Z., 1961. Thin-walled elastic beams, Oldbourne Press, London.

APPENDIX

The solution of the equation (41) may be presented, by the method of initial parameters, as follows (Pavazza, 1991)

v= Kv0 + I , (A1) where T T v = 轾Q M w , v = 轾Q(0) M (0) (0) w (0) , 臌犏zcs ycs b1- 3 1 - 3 0臌犏zcs ycs b 1- 3 1 - 3 轾 1 0 0 0 犏 l g g g g 犏 shx ch x EIycs sh x 0 犏 g l l l l 犏 2 K = 犏 l骣 g l g g , 2 琪chx- 1 sh x ch x 0 犏 EIycs g桫 l EI ycs g l l 犏 2 2 犏 l骣 l g l骣 g l g -琪 shx - x -琪 ch x - 1 - sh x 1 犏 EI g2 g l EI g 2 l g l 臌犏 ycs桫 ycs 桫

32 2 dw(1- 3) pg d w( 1 - 3) p d w( 1 - 3) p I= -EI + EI� EI ycsdx3 ycs l 2 d x ycs d x 2 dw T - (1- 3) p w (A2) dx (1- 3) p where 1 l2 x 轾 l g k 犏 b w(1- 3) p =� 2 sh ( x -x ) - ( x x ) qzcs d x, g= l . (A3) 0 犏 EIycs g臌 g l EI ycs

For the uniformly distributed load, qzcs= q cs :

q l2轾 l 2骣 g x 2 w=cs 犏 ch x - 1 - ; (1- 3) p 2犏 2 桫 EIycs g臌 g l 2

l2骣 g q l 2 骣 l g I =赙 -q x - q ch x - 1 -cs sh x - x cs cs 2桫 2 g l EIycs g桫 g l

T q l2轾 l 2骣 g x 2 cs 犏 chx - 1 - . (A4) 2犏 2 桫 EIycs g臌 g l 2

In the case of many load distribution fields, it may be written

i v= Kv + I + K v + I i= 2,3,..., m 0( r ar- 1 , r r ) , ( ) (A5) r=2 where 轾 T va, r= Q zcs , a , r M ycs , a , r b(1- 3) a , r w( 1 - 3)a , r , r-1臌犏 r - 1 r - 1 r - 1 r- 1 轾 1 0 0 0 犏 l g g g g 犏 shx chx  EIycs sh x  0 犏 g l l l l 犏 2 K = l g l g g r 犏骣    , 2 琪chx- 1 sh x ch x 0 犏 EIycs g桫 l EI ycs g l l 犏 2 2 犏 l骣 l g l骣 g l g -琪 shx -x  -琪 ch x  - 1 - sh x  1 犏 EI g2 g l EI g 2 l g l 臌犏 ycs桫 ycs 桫

d3wg 2 d w d 2 w I = -EI(1- 3)p , r + EI� (1 - 3) p , r EI (1 - 3) p , r r ycsdx3 ycs l 2 d x ycs d x 2 dw T - (1- 3)p , r w , x = x - a . (A6) dx (1- 3)p , r r- 1

For the partly distributed uniform load, along 0 ＃x a1 ( r = 2 ):

q l2 x 轾 l g cs 犏 w(1- 3)p , r ,2 = -2 sh ( x -x ) - ( x - x ) d x= 0 犏 EIycs g臌 g l

q l2镲禳 l 2轾 g( x- a ) 2 =-cs 睚镲犏ch (x - a ) - 1 - 1 ; 2镲 2 犏 1 EIycs g铪镲 g臌 l 2

l2 轾 g q l2 轾 l g I = q ( x - a ) q犏 ch ( x - a ) - 1 cs 犏sh (x- a ) - ( x - a ) 2cs 1 cs 2 1 2 犏 1 1 g臌犏 l EIy g臌 g l

T 2禳 2 2 q l镲 l轾 g ( x- a ) -cs 睚镲犏 ch( x - a ) - 1 - 1 . (A7) EI g2镲 g 2 犏 l 1 2 ycs 铪镲臌

* For the concentrated force, Qzcs , at x= ar- 1 :

T v =轾 - Q* 0 0 0 . (A8) ar- 1, r 臌犏 zcs

Example 1: Simply supported beam under the uniformly distributed load; boundary conditions by (81):

q l骣 x Q =cs 1 - 2 , zcs 2 桫 l

qcs l骣1 x Qzcs- m ycs =y sh g - g桫2 l 轾 qcs l x2y 骣 1 x mycs =犏1 - 2 - sh g - g臌犏 l g桫2 l 2 轾 qcs l骣1 x Mycs =2 犏1 -y ch g - , g臌犏桫2 l 3 轾 qcs l x2y 骣 1 x = -犏1 - 2 - sh g - , b1- 3 2 犏桫 2EIycs g臌 l g 2 l

q l4 镲禳 x骣 x2 2轾骣 1 x cs 镲珑鼢 w1- 3 =睚 -鼢 -犏1 -y ch g - , (A9) 2EI g2镲 l桫珑 l鼢 g 2 犏桫 2 l ycs 铪镲臌 1 = y g ; ch 2

ql qcs l Qzcs(0)= - Q zcs ( l ) = , Qzcs(0)- m ycs (0) = - Q zcs ( l ) + m ycs ( l ) = c 3 ( v ) 2 2

3 qcs l b1- 3(0)= - b 1 - 3 (l ) = - y 0 ( v ) , 24EI ycs

2 4 q l 5qcs l cs w( l 2)= ( v ) Mycs ( l 2)= c 0 ( v ) , 1- 3j 0 ; (A10) 8 384EI ycs

2骣 1 24骣v2 1 3骣 th v c 0 (v )= 1 - , j 0 (v )= + - 1 , y 0 (v )= 1 - v2 桫 ch v 5v4 桫 2 ch v v2 桫 v

th v g l kb c 3(v ) = , v = = . (A11) v 2 2 EI ycs

Example 2: Built-in beam under the uniformly distributed load; boundary conditions by (85): q l骣 x Q =cs 1 - 2 , zcs 2 桫 l

qcs ly 骣1 x Qzcs- m ycs =sh g - , gc 桫2 l 轾 qcs l x2y 骣 1 x mycs =犏1 - 2 - sh g - , 2臌犏 l gc 桫 2 l q l2 轾骣1 x cs 犏 y Mycs =2 1 - ch g - , g臌犏 c 桫2 l 3 轾 qcs l x2y 骣 1 x = -犏1 - 2 - sh g - , b1- 3 2 犏桫 2Eycs g臌 l gc 2 l q l4 镲禳 x骣 x2 2轾骣 1 x cs 镲珑鼢 w1- 3 =睚 -鼢 -犏1 -y ch g - , (A12) 2E g2镲 l桫珑 l鼢 g 2 犏桫 2 l ycs 铪镲 c 臌 2 g c= y sh ; g 2

q l q l 2 = - = cs cs Qzcs(0) Q zcs ( l ) , Mycs(0)= M ycs ( l ) =- c 2 ( v ) , 2 12

2 4 q l qcs l cs w( l 2)= ( v ) Mycs ( l 2)= c1 ( v ) , 1- 3j 1 ; (A13) 24 384EI ycs

6 骣 v 3 骣v 24骣v v c1(v )= 1 - , c 2 (v )= - 1 , j 1(v )= - th . (A14) v2 桫 sh v v2 桫th v v3 桫2 2

* Example 3: Simple supported beam under the concentrated force Qzcs = F at x= l 2 ; boundary conditions by (81):

F F 1 Q(0)= - Q ( l ) = , Qzcs(0)- m ycs (0) = - Q zcs ( l ) + m ycs ( l ) = , zcs zcs 2 2 ch v

F 骣 1 Fl 2 mycs(0)=- m ycs ( l ) = 1 - , b1- 3(0)=- b 1 - 3 (l ) = - c 0 ( v ) ; 2桫 ch v 16EI ycs

Fl Fl 3 = Mycs ( l 2)= c 3 ( v ) , w1- 3( l 2)y 0 ( v ) . (A.15) 4 48EI ycs

Example 4: Built-in beam under the concentrated force at x= l 2 ; boundary conditions given by (85):

F Fl Qzcs(0)= - Q zcs ( l ) = , Mycs(0)= M ycs ( l ) = - M ycs ( l 2) =- c 4 ( v ) , 2 8

3 l r F Fl Qzcs( l 2)= - Q zcs ( l 2) = , w1- 3( l 2)= j 1 ( v ) ; (A16) 2 192EI ycs 2( ch v - 1) c 4 (v ) = . (A17) vsh v Figures

Fig. 1. Rectangular closed thin-walled cross-sections with coordinate systems: a) three cell section; b) two cell-section.

Fig. 2. Rectangular closed thin-walled sections with hinged walls: a) three cell section; b) two cell-section.

Fig. 3. Beam components of the beam with double symmetrical three cell thin-walled cross-section.

* Fig. 4. Geometrical properties of the rigid cross-section: a) z-coordinate; b) S y -coordinate

(statical moment of the cut-off portion of cross-section area).

Fig. 5. Stresses for the simply supported beam: a) additional normal stresses due to distortion at x= l 2 ; b) normal stresses due to bending with distortion at x= l 2 (FEM in brackets); c) additional shear stresses due to distortion at x = 0 ; d) shear stresses due to bending with distortion at x = 0 .

Fig. 6. Displacements for the simply supported beam at x= l 2 : a) additional displacements due to distortion ; b) displacements due to bending with distortion (FEM in brackets).

Fig. 7. Stresses for the built-in beam: a) additional normal stresses due to distortion at x= l 2 ; b) normal stresses due to bending with distortion at x= l 2 ; c) additional shear stresses due to distortion at x = 0 ; d) shear stresses due to bending with distortion at x = 0

(FEM in brackets).

Fig. 8. Stresses and displacements for the built-in beam: a) additional normal stresses at x = 0 ; b) stresses due to bending with distortion at x = 0 ; c) additional displacements due to distortion at x= l 2 ; d) displacements due to bending with distortion at x= l 2 (FEM in brackets). Dr. Radoslav Pavazza

Faculty of Electrical Engineering, Machanical Engineering and Naval Architecture

University of Split

Ruđera Boškovića, bb

21000 Split

Croatia

E-mail [email protected]

Fax +38521563877

Mr. Branko Blagojević