A model for plane elastica with simple shear deformation pattern

Acta Mechanica 104, 241-253 (1994) ACTA MECHANICA �9 Springer-Verlag 1994

Note

A model for with simple

plane elastiea shear deformation pattern

T. M. Atanackovic and D. T. Spasic, Novi Sad, Serbia

(Received May 12, 1992; revised October 20, 1992)

Summary. Several existing theories of elastic rods that take into account shear effects are classified. Then, a new model for the influence of shearing force, based on simple shear of finite amount, is proposed. The properties of the model are examined on the stability problem for a heavy elastic rod. It is shown that the rod may exhibit sub- and super-critical bifurcation patterns at the trivial equilibrium configuration. A comparison is made between the results for the critical load of the rod predicted by the new and the three different existing models.

1 Introduction

As stated by Gjelsvik [1] the first work that generalizes the classical elastica theory as to take the shearing forces into account goes back to Engesser who treated the problem in 1889. According to the way by which the internal forces in an arbitrary cross section of the rod are decomposed, all generalizations could be classified in three different groups. In the first group, that we call Engesser's approach, the resultant force is decomposed into the "convected" direction of the sheared cross-section and into the direction of the rod axis (see Fig. 1 b). Engesser's approach was used in [2], [3] and [4], for example. In the second group, that we together with Gjelsvik call Haringx's approach, the resultant force is decomposed into the convected direction of the sheared cross section and into the direction normal to the sheared cross section (see Fig. 1 c). The second approach was used in [5], [6] and [7], for example. In the third group, that we call Timoshenko's approach, the resultant force is decomposed into the direction of the rod axis and the direction orthogonal to the rod axis (see Fig. 1 d). Timoshenko's approach was formulated in a linear version in [8, p. 132] and was used in [9], [10] and [11].

In what follows, we define the shear angle 7 to be the angle between the rotated (sheared, "convected") cross section and the direction of the normal to the rod axis in the deformed state. The central point of the generalized elastica with shear effect is in the relation that connects shearing force and shear angle. In Engesser's approach the linear relation between shearing force Qe and the shear angle ? is postulated, so that (see [2] for example)

k 7 = ~ Q~, (1)

where GA is the shear rigidity and k is Timoshenko's shear correction factor. Note that it is generally assumed that k depends on the shape of the cross section (see [8, p. 132]). Recently,

242 Y.M. Atanackovic and D. T. Spasic

Ii t ....... t I h

! 5r

c d

Fig. 1 a - d . Decomposition of resultant force in Engesser's, Haringx's and Timoshenko's approach

however, Renton [12] showed that k depends on the shape of the cross section and on the material. In Haringx's approach (see [l]) it is assumed that

k 7 = ~ Q,,. (2)

We note that Reissner postulates the same constitutive equation as (2) for small shear angles (see [5, p. 802]). A constitutive equation of type (2) for large values of 7, based on Reissner's work [5], was used in [7]. When adopted for an incompressible rod, that relation, in our notation, reads

k sin 7 = ~ Q~- (3)

Also, Reissner in [5], poses the question: should 7 be connected with Q~ or with the component of the resultant force in the direction of the normal to the deformed centerline? This leads us to Timoshenko's approach in which

k 7 = ~ Qr- (4)

Equation (4) was used in [8]. A non-linear version of (4),

k tan 7 = ~ Qr, (5)

was used in [9]. All above relations could be viewed as special cases of general constitutive equations presented in [6].

Although similar, (1) and (2) are different because Qe and QH are results of different decomposition. The constitutive equations mentioned above, when used for the stability analysis, lead to different values for critical forces. For example, in the case of an incompressible rod with an end load, Engesser's and Timoshenko's approaches give the same value (see [8] and [13]), which differs from the value obtained by Haringx's approach (see [1]). Another question concerning the constitutive equations (1)-(5) is: should the relation between 7 and Q be

A model for plane elastica 243

postulated in linear or nonlinear form? For example, in [7] it was stated that the linear relation of

type (2) when compared with the nonlinear relation of type (3), in case of stocky structures, could overestimate or underestimate the deflection of the rod.

Our intention is to generalize Engesser's approach (Eq. (1)) to be suitable for large shear

angles. The bases for our analysis will be simple shear of finite amount, as presented in [14]. We shall make precisely what are shear directions in a bent rod element. This will lead us to the

natural (in the spirit of finite elasticity theory) way to postulate a relation between shearing angle

and shearing force. Also we shall use the new model to examine the stability boundary and post-critical behavior of a heavy elastic rod in a constant gravity field.

2 A new shear model

Consider an element of the rod whose axis has length dS in the unloaded state. We assume that

the rod axis coincides with the centroidal line of the rod cross sections. Let T be a unit tangent vector at the point C of the rod element (see Fig. 2).

When loaded, the rod element displaces and deforms so that in the loaded state the unit tangent vector at the element axis at the point C becomes t. Also the length dS changes to ds. Let

be the strain of the rod axis, then

ds = (1 + e) dS, (6)

and let ~o be the rotation angle of the element defined by scalar product T - t = cos q0. In the

unloaded state the cross-sections at C and D are orthogonal to the rod axis and the angle between

them is dqS. In the loaded state the cross-sections are rotated so that they are not orthogonal to

the rod axis. We assume that they remain plane. The central assumption in formulating our

model is that the rod element KB (see Fig. 2) experiences a simple shear along the direction

parallel to the rod axis. It has as a consequence that the thickness of the rod element is not

influenced by the shear. That is similar to the deformation pattern of fibre-reinforced beams [15]. Note that this assumption is also like the theory of rods with two basic curves [16]. Note that in

[16] the position of an arbitrary point of the rod is expressed in terms of two curves (upper and lower curve of the rod). In our model the position of an arbitrary point of the rod is expressed in

terms of the quantities defined on the rod axis only (in this respect our model is the classical "one curve model").

d ~ ,

0' Fig. 2. Rod element in the unloaded and loaded state

244 T.M. Atanackovic and D. T. Spasic

Let dSy be the length of the rod fiber (KB on Fig. 2) that is on the distance y from the element

axis. Obviously

where R is the radius of curvature of the rod axis in the unloaded state, i.e., R -- dS/dO. Also, since during deformation the rod element is sheared along the lines parallel to

the rod axis, in the deformed state the curves C'D' and K'B' are the distance y apart.

Referring again to Fig. 2, we get that

jilt = - - + y +cp' --YT' 1 9' ~ +

(8)

is the length of the same element in the loaded state. In (8) we used (-)' to denote a derivative with

respect to S. From (7) and (8) we obtain the strain for the rod element in the distance y as

+ y(~0' - ~') ey = (9)

Y I + - -

R

To determine the resultant force and resultant couple of internal forces we follow the method of

[2]. Thus we assume that each layer of the rod is linearly elastic and that normal stress in it is given

as az -- Eey, where E is the modulus of elasticity. Multiplying (9) by E, and by Ey and integrating

over the original cross-section A we get

f dA f Y dA, N = Ee - - + E(q~' - 7') Y -Y 1 + ~ 1 + R

A A

= - - d A - E e Y - dA, Y M -E(9' 7'1 I + R Z 1 + g

A A

(10)

respectively. When there is no shear deformation, i.e., 7 = 0 the expressions (10) agree with the

results of Kfimmel [17]. For (y/R) ~ 1 and for small 7 our results reduce to those presented in [2]. To obtain the relation between Q and 7 one needs another assumption. As pointed out in [18]

a comparison with the plane stress theory suggests that at least three more effects should be taken into account: transverse shear, non-classical axial stress and transverse normal strain. We assume that the transverse shear is predominant, so we motivate our approach on the relations that follow from simple shear of finite amount [14]. On the basis of our assumption that the

element is sheared along the direction parallel to the rod axis (see Fig. 3) we get the components of the stress vector on the sheared planes,

T12 T~ - 1 + K ~ ' Ns -- T22 - KT~, (11.1, 2)

where K = tan 7 is the amount of shear [14, p. 280] and T~j are the components of the Cauchy stress tensor.


0 ~" X 2 N

T12

x 1 Fig. 3. Sheared element of the rod

For an isotropic solid body T~ could be expressed in terms of three material functions. Using this fact it follows that

T12 = #oK + #1 K3 q- o(K3); T22 = o(K), (12)

where #o, #1 are constants and lim (o(K)/K) --+ 0 as K ~ 0. To determine Q we integrate T~ over the sheared cross section, since T~ is determined from the Cauchy stress tensor. Thus

Q = ~ ~ d d , (13) al

where d is the sheared cross section. Since the sheared cross section remains plane we have (A /d ) = cos y. Then integration in (13) could be performed over the original cross section A. Using (12) in (13) we finally get

[ (tan 7)3 ] Q = A # o s i n T + # l l + ( t a n T ) 2 cosy +o(K3) . (14)

When (11.2) is integrated over A it will give a contribution to N defined by (10.1). We neglect this contribution to N since it is at least of the order K 2. Another argument is that there may be elastic materials for which N~ is identically equal to zero (see [19, p. 178]) and in such a case no error is introduced by neglecting N~. Also, if (14) is to represent Q for large shear angles 7 then, in simple shear, normal stress effects (Tzz + 0) should be taken into account (or otherwise the rod's thickness h will not remain constant). Since normal stresses in the direction orthogonal to the rod axis are generally neglected [18], we neglect this effect.

For simplicity we consider a linearly elastic material (#1 = 0). Then from (14) we get

K Q = A/z0 sin V = A#o (1 + K2) 1/2" (15)

Note that (15) differs from (3) since a different decomposit ion of resultant force is used (our Q is equal QE)- Since in an elastic rod the distribution of shear stresses across the cross section is not uniform, we shall take the correction factor in (15). Following [12] instead of (15) we use

Q = kA#o sin 7, (16)

where

k = B + C . (17)


In (17) B and C are positive constants depending on the geometry of the cross section and v is

Poisson's ratio.

Therefore our model of an elastic rod with axial strain and finite shear is described by constitutive equations (10) and (16). In the general context of elastic rods [6] we may say that

strains in our theory are: rotation of the cross section ~0' - 0', strain of the rod axis e and amount

of shear K.

3 An example: heavy vertical rod

To test the new model, we consider a heavy vertical rod in a constant gravity field. The rod is fixed at one and free at the other end (see Fig. 4). In the unloaded state the axis of the rod is assumed to

be straight and of length L.

Let 2, 35 be axes of a fixed rectangular coordinate system with origin at the bot tom of the rod.

We identify points of the rod in the unloaded state by the arc-length S of the point measured along

rod axis. The equilibrium equations, for an element of length dS in the unloaded state, read (see Fig. 4 b)

dH = qo dS, (18.1)

d v = o, (18.2)

d M = V dx - H dy, (18.3)

where H and V are components of the resultant force along ~ and 35 axis respectively, qo is the

weight of the rod per unit length in the unloaded state (qo dS = q:, ds), M is the resultant couple

and x and y are coordinates of an arbitrary point of the rod in the deformed state. Also from

Fig. 4 b we have the following geometrical relations:

d x = ( 1 +e) cos3 , (19)

dy = (1 + e) sin 0,

where we used (6) and where 0 = cp is the angle between the tangent to the rod axis and the 2 axis

of the coordinate system x - y. From Fig. 4 a we read the following boundary conditions:

x(0) = 0, y(0) : 0, 7(0) = 0(0), (20 .1- 3)

H(L) = O, V(L) = O, M(L) = 0. (20.4-6)

ds

y

a b Fig. 4a, b. Coordinate system and load configuration


N o t e tha t (20.3) represents the cond i t i on tha t the b o t t o m surface of the rod is fixed since its

r o t a t i o n angle ~ - 7 is equal to zero. F r o m Fig. 1 b we get Engesser ' s c o m p o n e n t s of the resu l t an t

force

cos O sin 0 V

Q cos 7 H cos 7

cos (O -- 7) sin (0 - 7) N = H - - + V

cos 7 cos 7

(21)

where the subscr ip t E is omi t ted . E q u a t i o n s (21) agree wi th [4]. Also f rom (18.1), (18.2), (20.4) and

(20.5) follows

H = - q o ( L - S)

(22) V = 0 .

Us ing (21) and (22) in (10) and (16), wi th R ~ o% we get

qo(L - S) cos (O - 7) = E A cos 7 ' (23.1)

M = - E I ( O ' - 7'), (23.2)

qo(L -- S) sin O sin 7 = (23.3)

kA#o cos 7

I n t r o d u c i n g the fo l lowing n o n - d i m e n s i o n a l quant i t ies :

2 - q~ ~ _ qoL # - qoL ~ _ M L

E1 ' k A # o ' E A ' E1 '

S x y (24)

and us ing (24) in (23.3) we get

sin 27 = 2fl(1 - t) sin O. (25)

F r o m (25), (18.3), (19), (23.1), (23.2) and (24) we get

/ ~ = 2 ( 1 - - t ) [ 1 - - # ( 1 - - t ) c ~ 7

= ~ c o s 2 7 + f i s i n O

fi(1 - t) cos O - cos 27 '

cos 27 + fi sin 0 = (26)

J ~ + fi(1 = t) cos 0 - cos 27 '

cos (0 - 7 ) ] = 1 - #(1 - t) coss~ cos 0,

ttc~ si~ cos


where (.) = d(.)/dt. The boundary conditions corresponding to (26) read

~8"(1) = 8(0) = 7(0) = x(O) = ~r = 0. (27)

Substituting (20.3) in (25) we get

sin 7(0)[cos 7 ( 0 ) - f i] = 0 . (28)

In writing (27.2), (27.3) we used the solution 7(0) = 0 of (28), that is valid for all values of ft. Equation (28) could have a non-trivial solution, say y(0) = ~). We do not treat this case here.

To study bifurcation we note that the variables a: and ~ could be omitted from the analysis. Also from (25) we could express 7 in terms of,9. Then substituting this result in (26.1), (26.2) and eliminating ~ we get a single equation in O i.e.,

i f(0, 2, fi, #, t) = 0, (29)

where

1 J ( 0 , 2, #, ~, t) = ~ +

[/30 - t) cos 0 - [1 - 4/32(1 - 02 sin 2 0] 1/2] C

x t - - 2 / 3 cos 00 - /3(1 - t) sin &~2 _ 2(1 - t)

t. [ [ sin arcs < osO+s nOcos[l ,s oO)]AJ

x sin 011 - 4/32(1 - t) 2 sin 2 0] 1/2 + 2[ - /3 sin 8 +/3(1 - t) cos 8,9]

2//(1 -- t) sin 8 t x [1 - 4/32(1 --- t ) ~ n = 81 ~/i " (30)

)

We treat J as an operator acting on X | IR where R is the set of real numbers and X is defined as the space of continuous functions mapping the interval [0, 1] into reals and having continuous derivatives up to the second order and satisfying boundary conditions 0(0) = 0, 0(1) +/3 sin 0(1) = 0. The range space of ~ is the space of continuous functions defined on the same interval. F rom (30) it follows:

J ( 0 , 2, 8, ~, t) = 0, (31)

~ (8 , 4,/~, n, t) = - ~ ( - 0 , ;,/3, #, t).

Let D Y denote the Fr~chet derivative of o~ at the point 8 = 0 c X. The linearized boundary value problem corresponding to (29), i.e., D~-8 = 0, leads to the equation

2/3 2(1 - t) [1 - #(1 - t)] 8 = 0, (32) ~ + ~ + 1 - - / 3 ( 1 - - t) 1 - - / 3 ( 1 - - t)


with the boundary conditions

,9(0) = 0, 0(1) + fl`9(1) = 0. (33.1, 2)

The boundary value problem (32), (33) could be transformed into the standard Sturm-Liouville form using the integrating factor exp (~ 2 f l / [1 - f i ( 1 - ~)] d 0. Also, (32), (33), under the condition 1 - / 3 > 0, is self adjoint and its eigenvalues are simple (see [20]). To solve (32), (33) we note that the coefficients in (32), because of 1 - / 3 > 0, are analytic so that the solution could be expressed as

n = o o

`9 = ~, a , ( 1 - t)". (34) n = 0

The radius of convergence of the series (34) is greater than or equal to 1 (see [21]). Substituting (34) into (32) and equating coefficients of equal power we get

a l = f l a o ,

a 2 = f la I ,

a 3 = - - g a o , (35)

2 2# ak = /3ak-1 - - a k - 3 + - - a k - 4 for k > 4 .

k ( k - 1) k ( k - 1)

The series (34) with coefficients (35) satisfies boundary condition (33.2) automatically. The coefficient ao remains undetermined and we take it, without loss of generality, as ao = 1. To satisfy boundary condition (33.1) we must have

`9(0)= ~ a . = O . (36) n = O

Equation (36) is the characteristic equation of the linear eigenvalue problem (32), (33). For given /3 and #, we determined the smallest root 2 = 2or of (36) numerically, by Newton's method. The results are given in Table 1.

We state now the main result of this Section as:

Theorem

The non-linear boundary value problem (29) has a bifurcation point at (0, 2or)~ X | IR. Moreover for sufficiently small A2 all solutions of (29) with 2 = 2or + A,~, which are in the neighborhood of 0 E X, are of the form

0 = m o o + 0 " , (37)

where go is eigenfunction of (32), (33) corresponding to 2or and m ~ ~ is a real number satisfying the bifurcation equation

m A2el + n~aca + h.o.t. = 0, (38)


where h.o.t, denotes terms of the order O(m 4, A2m3). Also 0* = O*(mOo) is a continuous function satisfying

1

S DYO*Oo dt = O, (39) 0 and is at most of the order m 3. Finally ca and c 2 are constants depending on 2or, fi and ft.

Proof To prove the theorem we use the standard Liapunov-Schmidt reduction [22], [23]. Thus, (37) follows from the fact that 2or is simple while (39) results from self-adjointness of (32), (33). Also the function 0* is at most of the order m 2 (see [21]). However, from (31.2) we have

O*(- moo) = -O*(mOo), ( 4 0 )

(see [23, p. 300]) and since (as a result of the application of the implicit function theorem) 0* is continuously differentiable, it follows that 0* is of the order m a. The constants cl and c3 follow after lengthy calculations and are

C 1 = CD ak(1 -- t) k 6o 1 dr, 0 k = O

ca=C3Di[(k=~ O a k ( 1 - t ) k ) 2 ( k ~ o k a k ( l - t ) k t ) 2(02

(k=~O ak(1--t)k)3 (k=~O kak(1 --t)k-1) 0)3 -~ (k=~O ak(l --t)k)4-("]"cr('94 + m5)] dr,

(41)

where C and D are constants (that could be fixed by requiring that the norms of 00 and 0*, in corresponding spaces are equal to one) and

(D 1 ( 1 - t) [ 1 - if(1 - t ) ]

- # ( 1 - t ) '

fl(1 -- t) - 4fl3(1 - 0 3 0. )2 z

I - f l (1 - t)

8 f l 3 ( 1 - - t ) 2 - - f l fi2(1 - t) - 4fl3(1 - ~)2

~o3 = 1 - / 3 ( I - t) - [1 - / 3 ( 1 - 012 '

I /#~(1 - t ) 2

o ~ 4 = - ( 1 - t ) -

# ( 1 - t) +[1--#(1- t ) ][6+2f i2(1- - t )21

+

1 - / 3 ( 1 - t)

[ 1 - # ( 1 - t)] [-fl(-l? -t) 2fl2(1 - t)2]]

[1 _~ f i - 0 7 0 ] 2-

( 4 2 )

4 / 7 3 ( i - t ) (D 5 - -

1 - # ( 1 - t)

This completes the proof of the theorem.

A model for plane elastica

'!+/ 0.9

0 .65L~-- ~ C 3 < 0 @ - @-

1,

0 0.28 0.5 fl

251

Fig. 5. The regions of c3 > 0 and c3 < 0

To determine the number of solutions to (38), we note that cl > 0 for all values of fl and

# which are of physical interest (fl, # < 1). Now, if c3 =I = 0 then (38) is contact equivalent to (see [24,

p. 41)

em 3 + m A2 = 0, (43)

where e = sgn (c3). When c3 < 0 we have super-crit ical pitchfork bifurcation, while for c3 > 0 the

bifurcation is sub-critical. In Fig. 5 we show the regions of parameter space fl - # with c3 > 0

and c3 < 0.

4 Comparison with other models and conclusions

We compare the results for the critical load obtained from our model (10), (16) with results

obta ined from the following three models:

(i) Engesser's approach, l inear version (see [2])

M = --EI(O' - y') cos 7,

NE - (44)

E A '

k 7 = ~ Qr,

(ii) Haringx's approach (see [7])

M = - E I ( 0 ' - 7 9 ,

N~ I + - -

EA - - - 1, (45)

cos 7

k sin 7 - Qn,

GA(~ + ~)

252

(iii) Timoshenko's approach (see [10])

M = - E I ( O ' - 7'),

N r E A '

k ~=~Q~.

T. M. Atanackovic and D. T. Spasic

(46)

The corresponding equilibrium equations for those three models are obtained from (18), (19), (20) and (44), (45) and (46) respectively. Equations corresponding to (26) for Timoshenko's and Haringx's approaches are obtained from Fig. 1 d and Fig. 1 c. The constitutive equations (44), (45) and (46) lead to the same boundary condition ~9(0) = 7(0) = 0. For each model the critical value of the load parameter 2 = 2or was determined by the same procedure.

We examine now the values of 2c,. As expected, the values for the first and the third model are the same as those for our model (presented in Table 1). The linearized equilibrium equation for

the second model (Haringx's approach), from which 2or was determined, reads

[1 - ~ ( 1 - t)] [1 + (/~ - ~ ) (1 - t)]

[ 2 ( 1 - - 0 2fl(fl - #) ] 0 = 0 . (47)

The boundary conditions corresponding to (47) are (33). The values of 2c~ obtained from (47), (33)

are presented in Table 2. Note that the values in Tables 1 and 2 are the same when fl = 0, and fi = p. In the case when

fl > p the values obtained from the second model are larger than those obtained from the other three models. The opposite is valid in the case when ] / < #. Also, in the case when fi = 0, we qualitatively recover the conclusion given in [1].

In concluding we state that from the study of nonlinear behavior of the rod according to the different constitutive models, one can choose a model suitable for specific application, as

Table 1

2or # 0.0 0.05 0.1 0.25

0.0 7.8373 8.0274 8.2266 8.8860 0.05 7.6508 7.8373 8.033 0 8.6810 0.1 7.4622 7.6453 7.8373 7.4838 0.25 6.8832 7.0555 7.2364 7.8373

Table 2

2or /~ 0.0 0.05 0.1 0.25

fl 0.0 7.8373 8.0274 8.2266 8.8860 0.05 7.6559 7.8373 8.0274 8.6551 0.1 7.482 4 7.655 9 7.8373 8.435 6 0.25 7.0052 7.1575 7.3164 7.8373


suggested in [1]. F o r example our model, described by (10), (16) when compared with Haringx's

model, described by (45), and for 2 -- 8.5,/3 = 0.25 and/~ = 0.1 gives larger maximum values for

gJl, 7, 0, and N.

Acknowledgement

This research was supported by S. E of Serbia.

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Authors' address: T. M. Atanackovic and D. T. Spasic, Faculty of Technical Sciences, University of Novi Sad, Trg D. Obradovica 6 Novi Sad, Serbia

A model for plane elastica with simple shear deformation pattern

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