[Foundations of Engineering Mechanics] Statics of Rods || Straight Rods

4. Straight Rods

This chapter is devoted to applied problems of statics of straight rods. A great quantity of structural elements of mechanisms and devices could be treated as straight rods. Straight rods find wide application in mechanical engineering. Some illustrative examples are given in Introduction.

Most design problems of mechanics of rods cannot be solved exactly (e.g. problems of statics and dynamics of rods of varying cross section, nonlinear problems, etc.). These problems are usually solved by means of numerical or approximate methods. Moreover, usually the explicit representation of the exact solution (if we managed to develop it) is too complicated and useless for mathematical and physical interpretation. In general, the solution must be simplified through the use of numerical methods. The most-used methods are methods based on variational principles of mechanics.

Exact and approximate methods for analysis of equilibrium equations are discussed in this chapter. The approximate methods are based on the theory of distributions; these methods allow us to find solutions for problems of statics of straight rods subjected to concentrated forces and interacting with additional lateral supports.

4.1 Rods of Straight Natural Configuration

4.1.1 Traditional Routines of Derivation of Equilibrium Equations

Suppose that a rod axis in the natural configuration is a straight line; suppose also that the axial line remains a plane curve during deformation. Let the loads applied to the rod be dead. Under these conditions, the equilibrium equations for the rod can be easily derived within the framework of university courses of the strength of materials and structural mechanics. If a rod (see Fig. 0.21) in its natural configuration is twisted and subjected to loads of complicated nature (e.g. forces and moments which follow the normal to the axial line or are permanently directed at a fixed point or depend on the displacement of the axial points), then the equilibrium equations could hardly be derived by use of the standard methods. It is more convenient to obtain these equations by direct specialization of the general equilibrium

V. A. Svetlitsky, Statics of Rods© Springer-Verlag Berlin Heidelberg 2000

130 4. Straight Rods

equations (1.31 )~(1.35) (or (1.57)~(1.61)) for the case ofrods of straight initial configuration.

(a) (b)

Fig. 4.1.

Figure 4.1 a illustrates a rod of rectangular cross section subjected to a distributed load q = -Q2e20 applied to a part of the rod 7)1 ~ 7) ~ 1. In addition, concentrated forces p(1) = P1(~) elO and p(2) = - pP) e20 and a concentrated moment T(1) = T3e30 are applied to the rod. The base vectors e20 and e30 of the attached coordinate system are directed along the principal axes of the cross section. In the natural configuration, we have e201li2 and e30 Ilh. Hence, during deformation, the rod axis lies in the plane Xl OX2.

Suppose that displacements of the axial points are small. Let us consider an element of the rod and the forces and the moments acting on this element (see Fig. 4.1b). The element is assumed to be in equilibrium. Since in the Cartesian basis the components of the applied loads remain unchanged, it is more convenient to represent the equilibrium equations in this basis. Besides, we assume that the plane cross sections of the rod orthogonal to its axial line remain plane and orthogonal to the axial line after deformation. Hence, shear strains may be neglected.

From summation of forces in the Xl and X2 directions, we obtain

( Q1 + d~l d7)) cos (193 + d193) - ( Q2 + d~2 d7)) sin (193 + d193)

- Q1 cos 193 + Q2 sin 193 + p~~) b (7) - 1) d7) = 0; (4.1)

( Q1 + dd~l d7)) sin (193 + d193) + ( Q2 + d~2 d7)) cos (193 + d19 3)

- Q1 sin 193 - Q2 cos 193 - QX2 d7) - p~;) b (7) - 7)2) d7) = O. (4.2)

The concentrated forces enter into (4.1) and (4.2) though these forces are not applied to the element under consideration. A concentrated force qp can

4.1 Rods of Straight Natural Configuration 131

Xl

X,

Fig. 4.2.

be regarded as a distributed force with a narrow zone of application (equal to 2 ~T]) as illustrated in Fig. 4.2. Such a representation of concentrated forces is discussed in Appendix 4 in greater detail. For small1'J3 , from (4.1 )-( 4.2) it follows that

dQ1 _ ~ (Q 2 1'J 3 ) + p(1) 6(T] -1) = 0; dT] dT] Xl

dQ2 d (2) dT] + dT] (Q 1 1'J3 ) - QX 2 - PX2 6 (T] - T]2) = 0,

where

1'J3 = dU2 . dT]

(4.3)

( 4.4)

These equations are nonlinear because they contain products of the sought functions Q 11'J3 and Q2 1'J 3 . Recall that (4.3) and (4.4) are the equilibrium equations in the Cartesian basis while Q1 and Q2 are the components of the vector Q in the attached basis. To obtain linear equations, we are to adopt additional assumptions.

From (4.3) it follows that

Q1 = Q21'J3 - p;~) H (T] - 1) + Q1O' (4.5)

If the forces p(2) and q result in Q2 such that p(1) and Q2 are of the same order of magnitude, then the term Q2 1'J 3 can be neglected because of the smallness of 1'J 3 ; as a result, we get

(4.6)

Since the axial force Q1 equals zero to the right of the cross section of application of the force p;~), from (4.6) we obtain Q10 = pRJ, i.e. Q1 = Q10 = const (provided the axial force qXI = 0).

Knowing Q1, we can represent (4.4) as a linear equation


dQ2 (1) d'l93 (2) dr] + PX1 d1] - QX 2 - PX2 <5 (1] - 1]2) = O. (4.7)

The moments must satisfy the relation

It gives

dM3 (1) "ili7 + Q2 + T3 <5 (1] - 1]3) = o. (4.8)

Xl

X,

Fig. 4.3.

Equation (4.8) includes the concentrated moment applied at the cross sec

tion 1]3. Arguing as in the case of the force p~;) , we can treat the moment Ti 1)

as the limit (at the cross section 1]3) of a series of distributed moments /-l3 as indicated in Fig. 4.3.

Differentiating (4.8) and eliminating the term dQdd1] by use of (4.4), we get

d2 M3 d ( dU2) (2) (1) _ ~- d1] Q1 d1] +QX 2+ PX2 <5(1]-1]2)+T3 <5(1]-1]3) - o. (4.9)

Recall that

~ _ d'l93 _ d2U2 &3 - d1] - d1]2 .

Considering small displacements of the axial points, we obtain the following equilibrium equation:


Xz

lZ

i. f

Fig. 4.4.

Another form of the equilibrium equations can be obtained if we refer the vector Q to the Cartesian basis, i.e. Q = QX! i1 + QX2h (see Fig. 4.4). In this case, we have two linear equations

(4.11)

(4.12)

From (4.8) a nonlinear equation follows:

Equation (4.11) is uncoupled from the others. This equation can be solved for QX!' As a result, we get

Q _ p(l) x!o - x!

or

QX! = p~~).

We see that (4.13) becomes a linear equation. Further, eliminating QX2l M 3 , and '193 from (4.13), we obtain an equation, which is the same as (4.10). Solving for U2 from (4.10) and determining its first, second, and third derivatives, we find '19 3, M 3, Q1, and Q2.

In Chap. 1, the following equilibrium equation in the form (1.31) was obtained:

( 4.14)


The equilibrium equations (4.11) and (4.12) for a rod element (see Fig. 4.4) are almost the same as (4.14) but written in terms of projections in the Cartesian frame.

There exists another form of representation of the equilibrium equations. In the attached basis, the basis (el , e2) in Fig. 4.4, the equilibrium equation reads

dQ 2 - + ce x Q + q + L p(i) <5 (1] -1]i) = O. d1] i=l

( 4.15)

In the basis {ei}, the force vectors are

q = -qx2 sin 't9 3el - q X2 cos 't9 3e2 ; p(2) = -p(2)sin't93 el-p(2)cos't93e2'

X2 X2 '

pel) = p~;) cos 't93 el - p~;) sin 't93 e2 . ( 4.16)

For small values of the angle 19 3 , we have

q = -Qx2 't9 3el - QX 2 e2 ;

p(2) = -p(2)'t93el - p(2)e2 . X2 X2'

p(l) = p(l)el - p(l)'t93e2' Xl Xl

(4.17)

Since

d't93 • ce = d1] 13,

we obtain two equations in the attached basis,

If the external forces are follower, i.e. q = -Q2e2, p(1) = Pl(l)el, and

p(2) = -pF)e2, then the equilibrium equations (4.18) and (4.19) are as follows:

dQl _ Q2 d't93 + pel) <5 (''7 -1]1) = 0; d1] d1] 1

dQ2 + Ql d't9 3 _ Q2 - p(2) <5 (1] - 1]2) = O. d1] d1] 2

To derive (4.18) and (4.19) on the basis of (4.11) and (4.12), one should use the formulas of transformation of coordinates. In vector notation, (4.11) and (4.12) are as follows:


For small values of the angle '19 3 , we have

on rearrangement, in view of (4.11) and (4.12), we get

b = b1 el + b2e2 = 0

or

For numerical analysis, it is more convenient to represent the equilibrium equations in the form of a system of first-order differential equations,

( 4.20)

(4.21)

(4.22)

( 4.23)

From here on, we will use either the fourth-order equilibrium equation (4.10) or the system of first-order equations (4.20)-(4.23) depending on the features of a particular problem.

In some problems of statics of rods, a rod axis may become a spatial curve, e.g. if a rod of twisted natural configuration (see Fig. 0.20) is subjected to a load which is perpendicular to the rod axis. The analysis of such problems may be performed by use of equilibrium equations of more complicated nature.

4.1.2 General Equilibrium Equations in the Case of Straight Rods

As mentioned above, it is often more convenient to derive equilibrium equations on the basis of the general nonlinear equilibrium equations.

These general equilibrium equations referred to various coordinate systems are presented in Chap. 1. These equations allow us to account for the peculiarities of the structure of a rod and the behavior of external loads.

If the principal axes of a cross section of a straight rod are directed along the Cartesian axes, we have

(i.e. ~10 = ~20 = ~30 = 0); L(O) = E.


If a straight rod in its natural configuration is twisted (e.g. a drilling bit), then we have

d19l0 eElO = d;J;

[ 1 0 0 1 L(O) = 0 cos 1910 sin 1910 , o - sin 1910 cos 1910

where 1910(1]) is a known function. The case d19lO/d1] =j:. 0 is considered in Sect. 4.3.

Nonlinear Equilibrium Equations for a Rod whose Axial Line in a Deformed Configuration is a Spatial Curve In the case of large deflections of the rod axis, the equilibrium equations can be obtained on the basis of (1.57)-(1.61).

In the attached axes, we have

dQ + ac x Q + P = 0 . d1] ,

dM d1] + ac x M + e1 x Q + T = 0 ;

M = Aac, ac = ( eEl + d:~o) e1 + eE2e2 + eE3e3 ;

L1 dt9 -A-1M=0; d1]

(4.24)

(4.25)

( 4.26)

( 4.27)

( 4.28)

This system could be also obtained from the general equilibrium equations (1.57)-(1.61) with ac~l) = O.

For naturally twisted rods, we get

M1 = All ( eEl + d:~o) ;

where the angle 1910(1]) describes the location of the principal axes with respect to the attached coordinate axes.

In the attached basis, the following equations, a special case of (1.64)(1.68), can be obtained:

dQ1 d1] + Q3eE2 - Q2 eE3 + P1 = 0 ;

dQ2 d1] + Q1 eE3 - Q3eE1 + P2 = 0;

dQ3 d1] + Q2eE1 - Q1 eE2 + P3 = 0; (4.30)


dM1 dry + M3<£2 - M 2<£3 + T1 = 0 ;

dM2 dry + M 1<£3 - M3<£1 - Q3 + T2 = 0 ;

dM3 dry + M 2 <£1 - M 1<£2 + Q2 + T3 = 0;

Mi = Aii<£i , i = 1 , 2, 3 ; dd1 dd3 .

<£1 = dry cos d 2 cos d 3 - d1] sm d 2 ;

dd2 dd1 . <£2 = - - -smd3 ;

d1] dry dd3 dd1 .

<£1 = d1] cos d 2 + d1] sm d 2 COS d 3 ;

dU1 d1] + U3C£2 - U2C£3 + ll1 - 1 = 0;

dU2 dry +U1<£3 -U3<£1 +l21 =0;

dU3 dry + U2<e1 - U1<e2 + l31 = 0,

where

n

Pj = qj + L p?) 0 (1] - 1]i); i=1

p

Tj = f.1.j + L TY)O(ry-1],,), j=1,2,3. ,,=1

(4.31)

( 4.32)

( 4.33)

( 4.34)

In the Cartesian frame, the sought equations, a special case of (1.84)-(1.88), are as follows:

dQx dry+Px=O;

dMx T· _ . "d"ry + (L ld x Qx + Tx - 0,

Mx = LTAae;

d'!9 _ L -] A -] LM = o· d1]] x,

~~'" - (LT - E) i1 = 0 ;

(4.35)

( 4.36)

( 4.37)

( 4.38)

( 4.39)

here vectors denoted by the subscript 'x' are referred to the Cartesian basis, i.e. ax = ax,i] + ax ,i2 + ax3 i3.


Nonlinear Equilibrium Equations for a Rod whose Axial Line in a Deformed Configuration is a Plane Curve In the attached basis, these equations are a special case of (4.30)-(4.34):

dQ1 dry - Q2<£3 + PI = 0 ;

dQ2 dry + Q1<£3 + P2 = 0;

dM3 "ili7 + Q2 + T3 = 0 ;

M3 = A33<£3 ;

dt93 _ M3 _ o. dry A33 - ,

dUl d7) - U2<£3 + III - 1 = 0,

dU2 dry + U1 <£3 + 121 = 0,

In the Cartesian basis, we have

dQxl + P = o. dry x, ,

dQx2 d;J + PX2 = 0;

dMx3 . dry + cos t9 3 Qx2 - sm t93Qxl + TX3 = 0;

dt93 _ MX3 = o. dry A33 '

dUxl "ili7 - cos 193 + 1 = 0;

dU x2 • "ili7 - sm 193 = 0 .

For a rod of constant cross section, we must put A33 = 1.

( 4.40)

(4.41 )

( 4.42)

( 4.43)

( 4.44)

( 4.45)

( 4.46)

(4.4 7)

( 4.48)

4.2 Equilibrium Equations for Small Displacements and Angles of Rotation

4.2.1 Vector Equations

If the displacements Ui and the angles of rotation t9 i are small, then the curvature increments ~<£j can also be considered small. Since for straight rods <£0 = 0, we have <£ = ~<£. In view of (4.24)-(4.28), we arrive at

4.2 Small Displacements and Angles of Rotation 139

~~ + 6.ce X Q + Po + 6.P = 0; (4.49)

dM d1] +6.cexM+el xQ+To +6.T=O;

M = A6.ce;

d'19 -A-1M=0. d1] ,

du d1]+A1 'l9=0,

where n

6.P = 6.q + L 6.p(i) 6 (1] - 1]i) ; i=l

p

6.T = 6.J.L + L 6.T(v) 6 (1] -1]v). v=l

( 4.50)

(4.51)

( 4.52)

( 4.53)

(4.54)

In general, the increments 6.q, 6.p(i), 6.J.L, and ~ T(v) are functions of u, U/, and '19. Eliminating u ' from (3.20)-(3.23), we get

6.q = C 1 '19 + C 2 u;

6.J.L = C 3 '19 + C 4 u ;

6.p(i) = Bil)'19 + B~i)u; 6.T(v) = B~v)'19 + Biv)u. (4.55)

Substituting (4.55) into (4.49) and (4.50), we obtain the system of the five vector equations (4.49)-(4.53) in the five unknown vectors Q, M, 6.ce, '19, and u.

In the general case, the components of the vectors Q and M (unlike the vectors u, 6.ce, and '19) cannot be considered small. Therefore, even for small u, '19, and 6.ce, (4.49)-(4.53) are nonlinear differential equations. The equations of the zeroth approximation were obtained in Sect. 1.4 under the assumption ce ~ ce61) or 6.ce = O. Therefore, all the terms containing 6.ce are assumed to be equal to zero: 6.ce x Q = 6.ce x M ~ O. The latter assumption is reasonable for curvilinear rods because CBjO » 6.CBj. For straight rods, we have CBjO = 0, hence we need additional argumentation to omit the terms 6.ce x Q and 6.ce x M in the vector equations without argumentation.

4.2.2 Equilibrium Equations in the Attached Coordinate System

First, let us estimate the nonlinear terms in the vector equations (4.49) and (4.50). To do this, consider (4.49)-(4.53) in terms of projections on the attached axes. We have


dQ] - + 6.<:£2Q3 - 6.<:£3Q2 + PIO + 6.P] = 0; dry

dQ2 - + 6.<:£3Q] - 6.<:£] Q3 + P20 + 6.P2 = 0 ; dry

dQ3 dry + 6.<:£] Q2 - 6.<:£2Q] + P30 + 6.P3 = 0 ;

dM] d;J + 6.<:£2 M 3 - 6.<:£3 M 2 + TIO + 6.T] = 0 ;

dM2 d;J + 6.<:£3 M ] - 6.<:£]M3 - Q3 + T20 + 6.T2 = 0;

dM3 d;J + 6.<:£]M2 - 6.<:£2 M ] + Q2 + T30 + 6.T.3 = 0;

d13] _ Ml = O. dry All '

d13 2 _ M2 _ O. dry A22 - ,

d13 3 _ M3 _ O. dry A33 - ,

dUI = O· dry ,

dU2 dry -133 = 0;

dU3 dry + 132 = 0;

Mi = Aii 6.<:£i ,

where

n

PjO = qjO + L p}~) 6 (ry - 1)i) ; i=1

n

6.Pj = 6.qj + L 6.p?) 6 (1) - ryi) ; i=1

p

Tjo = J1jO + LT}~) 6 (ry - ryv); v=1

p

6.Tj = 6.J1j + L 6.TY) 6 (ry - ryv). v=]

( 4.56)

( 4.57)

( 4.58)

( 4.59)

(4.60)

(4.61)

( 4.62)

Equations (4.56)-(4.60) need be treated together with (1.67) and (1.68); we must put <:£jO = 0 in (1.67) and replace <:£j by 6.cej in (1.68). From (4.56) it follows that


or

where ~QI is a small quantity. If the force projections on the tangent to the rod axis equal zero (PIO = 0), then the axial force QI can be considered small and the terms ~re3QI and ~re2QI in (4.56) may be neglected. If PIO 1= 0, then the nonlinear terms can be rearranged as follows:

i = 2,3.

Since the quantities Q2 and Q3 cannot be considered small, the terms ~re2Q3 and ~re3Q2 in (4.56) are of the first order of smallness. Therefore, these terms cannot be neglected unless they are small as compared to the other terms in (4.56) (e.g. PlO). The term ~PI depends linearly on Uj and 1')j, which are small quantities. Hence, if PlO = 0, then all the terms in (4.56) are of the same order of magnitude. Thus, the terms ~re2Q3 and ~re3Q2 cannot be neglected.

Suppose that the applied forces are follower (~PI = 0) and, moreover, PlO = 0; then the first equation of the system (4.56) is uncoupled from the other equations. Solving for ~re2' ~re3, Q2, and Q3, we arrive at

( 4.63)

From this it follows that if PlO = ~PI = 0, then QI is of the first order of smallness. Therefore, the products in (4.56) containing ~QI can be neglected.

Suppose that the applied forces are not follower (~PI 1= 0). Assume that ~PI depends linearly on Uj and 1') j. In this case, the first equation of the system (4.56) is coupled with the other equations. In this case, the first equation of (4.56) is nonlinear because the nonlinear terms ~re2Q3 and ~re3Q2 are of the same order of magnitude as the term ~PI and, therefore, they cannot be neglected. We could drop out the terms ~re2Q3 and ~re3Q2 provided the following inequality holds:

(4.64)

To check the validity of (4.64), one should solve the system (4.56)-(4.60) in which only linear terms are retained. If the inequality (4.64) is satisfied, then the nonlinear terms in (4.56) may be neglected; otherwise, the system of equations (4.56)-(4.60) is nonlinear even for small deflections.

Consider those nonlinear terms in the second and third equations of the system (4.56) which contain ~rel' From (4.57), in view of (4.62), we get


or

(4.65)

where

MlO = - 117 TlO d( ; MP) = - 117 6.T1 d( ;

(2) f17 6.M1 = - 10 (A33 - A 22 ) 6.&2 6.&3 d( .

The quantities 6.M}I) and 6.M}2) are of the first and the second order

of smallness, respectively. Therefore, we can neglect the term 6.M}2) when deriving the linear equations (4.57). For straight rods we can put

Ml = A11 6.&1 = MlO + 6.MP) .

Consequently, the terms containing 6.&1 in the second and third equations of the system (4.56 can be rewritten as follows:

Suppose that the projection TlO of the moment To on the tangent to the rod axis equals zero and the moment To is a follower one (MlO = 0 and 6.T1 = 0). Using (4.65), we get

( 4.66)

From this relation it is seen that 6.&1 is of the second order of smallness. Therefore, the terms 6.&1 Q2 and 6.&1 Q3 in (4.56) can be neglected. Particularly, for a rod of circular cross section (A22 = A33 ), we have 6.&1 == O.

If the applied moment is not a follower one (6.Tl i- 0), then, even under the condition T10 = 0, the nonlinear terms 6.&1 Q j cannot be neglected. Therefore, the second and third equations of the system (4.56) remain nonlinear.

Using (4.60), we can rewrite the first two equations of the system (4.57) as follows:

( 4.67)


If TlO = 6,TI = 0, then the term 6,<EI defined by (4.66) is of the second order of smallness. Therefore, the terms containing 6,<EI 6,<E2 and 6,<EI 6,<E3

in (4.67) are of the third order of smallness and thus can be neglected. In the case of a follower moment with a nonzero projection on the tangent

to the rod axis (TlO f::. 0), we have 6,<EI = MlO/All ; consequently, the terms containing the products 6,<EI 6,<E2 and 6,<EI 6,<E3 need be retained.

If the applied moments are not follower but still T lo = 0, then we have 6,TI f::. 0 and MI f::. O. Nevertheless, the moment MI can be considered small, i.e. MI = 6,MI = All 6,<EI. Thus, we get (see (4.57))

All - A33 6,<E3 M I - 6,<EI M 3 = A 6,<E3 6,MI ~ 0;

II

All - A33 6,<E3 M I - 6,<EI M 3 = A 6,<E3 6,MI ~ O.

II

From these relations we obtain

d6,MI A33 - AI2 -d- + A A M2 M 3 + 6,TI = 0;

'fJ 22 33

dM2 dry - Q3 + T20 + 6,T2 = 0 ;

dM3 dry + Q2 + T30 + 6,T3 = 0 .

Finally, let T lo f::. 0 and 6,TI f::. O. Then we have

1 6,<EI = -A (MlO + 6,MI ) ,

II

and

( 4.68)

From the particular cases of expressions for 6,<EI considered above, it is seen that the terms containing the products 6,<EI M j in (4.57) can be neglected; the only exception is the case TlO f::. O. Therefore, if TlO = 0, then (4.57) are linear equations.

For TlO f::. 0 (MIO = T lO ), we have 6,<EI = MlO/All . Hence, 6,<EI is known and the terms 6,<EI M j are linear in the unknown projections MI and M 2.

Let us consider two cases:

(1) QIO = MIO = 0;

(2) QlO f::. 0 and MIO f::. O.

In the first case, the equilibrium equations (4.56)-(4.60) are as follows:


dQ1 -- + 6.re2Q3 - 6.re3Q2 + 6. P1 = 0; dry

dQ2 dry + 6.re3Q1 - 6.re1 Q3 + P20 + 6.P2 = 0 ;

dQ3 dry + 6.re1 Q2 - 6.re2Q1 + P30 + 6.P3 = 0;

dM1 d";] + 6.re2 M 3 - 6.re3 M 2 + 6.T1 = 0;

dM2 d";] + 6.re3 M 1 - 6.re1 M 3 - Q3 + T20 + 6.To = 0;

dM3 d";] + 6.re1M 2 - 6. re2M 1 + Q2 + T30 + 6.T3 = O.

(4.71)

(4.72)

The presence of nonlinear terms in these equations substantially complicates the analysis. IfTlO = MlO = 0, the terms 6.rej M i can be neglected. (We will omit these terms in the derivation of the equations of the zeroth approximation.) To derive the equations of the first and higher-order approximation, we should retain the nonlinear terms in (4.72).

Equations (4.71) and (4.72) with (4.51)-(4.53) can be solved by use of approximate methods; to do this, the equations of the zeroth and the first approximation for curvilinear rods should be used (see Sect. 1.4).

Recall that the equilibrium equations of the zeroth approximation do not contain increments of the vector ee, therefore, we can put 6.rei = 0 in (4.71) and (4.72). As a result, we obtain the equations of the zeroth approximation

dQ(O) _1_ + 6.p(O) = 0 .

dry 1 ,

dQ(O) d~ + P20 + 6.piO) = 0 ;

dQ(O) _3_ + p + "p(O) - 0 . dry 30 u 3 - ,

dM (O) __ 1_ 6.T(O) = 0 .

dry + 1 ,

dM (O) __ 2 __ Q(O) + T + 6.T(O) = 0 .

dry 3 20 2 ,

M (O) ~ + Q(O) + T + 6.T(O) - 0 dry 2 30 3 - ,

where n

6.PjO) = 6.q;O) + L 6.pr)(O) 0 (ry - ryi) ; ;=1

(4.73)

(4.74)


p

6.T}O) = 6./-L;0) + 2: 6.TY)(O) J (ry - ryV)' 1.'=1

For straight rods the increments 6.q(O) 6.11(0) 6.p(i)(O) and 6.T(v)(O) , J' I""J' J' J

depend, in general, on U2, U3, 1'h, and 1'J3 and do not depend on 1'J 1 and U1

(U1 = 0). Consequently, these increments can be written as follows:

" (0) _ (2) (0) + (1) (0) + (2).0(0) + (1).0(0). uqj - Cj2 U 2 Cj3 U 3 Cj2 u2 Cj3 u3 ,

"p(i)(O) _ (b(2)). (0) + (b(2)). (0) + (b(l)) . • 0(0) + (b(l)) . • 0(0) . u j - j2' U 2 j3 'U3 j2' u2 j3 ,u3 ,

" (0) _ (4) (0) + (4) (0) + (3).0(0) + (3).0(0). U/-Lj - Cj2 U 2 Cj3 U 3 Cj2 u2 Cj3 u3 ,

"T(i)(O) _ (b(4)) (0) + (b(4)) (0) + (b(3)) .0(0) + (b(3)) .0(0) U j - j2 v U 2 j3 v U 3 j2 v u2 j3 v u3 . (4.75)

Equations (4.73) and (4.74) are to be solved together with the equations of the zeroth approximation (4.51)-(4.53). The latter equations are of the form

M(O) = A 6.a=(O) ;

d-o(O) _v _ _ A-1 M(O) = O.

dry ,

d (0) ~ + Al {J(O) = O.

dry

(4.76)

(4.77)

(4.78)

The system of equations (4.73), (4.74) and (4.76)-(4.78) can be represented as two uncoupled systems,

and

dQ (O) _1_ + 6.p(O) = 0 .

dry 1 ,

dM(O) __ 1_ + 6.T(O) = 0 .

dry 1 ,

M (O) - A ,,=(0). 1 - 11 u'-"l ,

d1'J(O) M(O) _1 ___ 1_ -0'

dry All - ,

d (0)

~=O dry

(0) dQ2 + p. + 6.p(O) = 0 .

dry 20 2 ,

(0)

d~~ + P30 + 6.pi O) = 0 ;

(4.79)


dM(O) 2 Q(O) + rp + AT(O) - 0 . dry - 3 .1 20 L.\ 2 - ,

dM(O) ~ + Q~O) + T30 + ~T~O) = 0,

d'!9(O) M(O) _2 ___ 2_-0'

dry A22 - ,

d'!9(O) M(O) _3 ___ 3_ -0'

dry A33 - ,

d (0) ~_'!9(O) =0'

dry 3 ,

d (0) ~ '!9(O)-O'

dry + 2 - ,

M (O) - A 11.=(0). 2 - 22 L.\u-2 ,

M (O) _ A A (0) 3 - 33 L.\re3 .

Taking into account (4.75), we can rewrite (4.80) in vector form

where

Al = [ ~ -1 ] . A = [ A~2 0 o ' A33

Q (O) - [ Q~O) 1 - (0) Q3

[ (0) 1 [ (0)

1 M(O) _ M2 '19(0) _ '!92

- (0) - (0) M3 '!9 3

P _ [ P20 ] . 0- P 30 ' To = [ ~:~ ] .

(4.80)

(4.81 )

The variables u(O) u(O) '!9(O) and '!9(O) in the formulas for ~p(O) and 2'3'2' 3 1

~TiO) (see (4.75)) can be found from (4.81). Then, in view of (4.79), we can . Q(O) (0) (0) ( (0) _ ) determme 1 , Ml , ~rel' and '!91 u1 = 0 .

In view of (4.75), (4.81) can be written as follows:


y(O)1 + A (O)y(O) = reO) ; (4.82)

here

[ Q(O) 1 A(O) _ [ 1,

0 C1 C2

1 (0) _ M(O) 0 C3 C4

Y - {J(O) - 0 _A~1 0 0 u(O) 0 0 A1 0

[ - prO) - B, d(O) - B,u(O) 1 reO) _ _T(O) - B 3{J(0) - B 4u(0)

- 0 .

0

The elements of the matrices B1 , B2 , B3 , and B4 contain the factors c5 (TJ - TJi) and c5 (TJ - TJv). For this reason, these terms are rearranged to the right-hand side of (4.82).

If the zeroth-approximation solution is not satisfactory or the increments ~p(O) and ~T(O) depend on the angle 191 , then (4.73), (4.74) and (4.76)~ (4.78) cannot be decomposed into two systems of equations. In this general case, by use of vector notation, we obtain

dQ(O) ~ + AlOPO + ~p(O) = 0 ;

dM(O) --;:G) + A1 Q(O) + AlO T(O) + ~ T(O) = 0 ;

M(O) = A ~ce(O) ;

d{J(O) -- - A~lM(O) = o·

dTJ ' du(O) -- + A1 ~{J(O) = 0

dTJ (4.83)

or, in view of the formulas for ~p(O) and ~T(O), this system can be written as

y(O)1 + A (O)y(O) = reO) ,

where

[ Q'O) 1 frO) ~ [

-A10 P O - B1 {J(O) - B2 u(O)

1 (0) _ M(O) -AlOTO - B3{J(O) - B4u(0)

Y - {J(O) 0 u(O) 0

A _ [ 1, 0 C1 C2

1 [ ! 0 n 0 C3 C4 AlO = 1 ( 4.84) - 0 _A~l 0 0

0 0 Al 0 0


Unlike (4.82), the vectors Q(O), M(O), '19(0), and u(O) in (4.84) have three components.

Let us derive the equations ofthe first approximation on the basis of (4.71) and (4.72). For this purpose, we put Qi = Q;O) +Q;l), 6.&i = 6.&;0) +6.&;1), 6.p. = 6.p(O) + 6.p(l) M = M(O) + M(l) and 6.T = 6.T(O) + 6.T(l).

1, 1, t'1, 1, t' t t t

Omitting terms of the second order of smallness, we arrive at

dQ(l) _1_ + A (O)Q{l) _ A (O)Q(l) + A (l)Q(O)

dry u&2 3 u&3 2 u&2 3

_ A (l)Q(O) + Ap(l) _ A (O)Q(O) _ A (O)Q(O). u&3 2 u 1 - u&3 2 u&2 3'

Q (l) ~ + A (O)Q(l) _ A (O)Q(l) + A (l)Q(O)

dry u&3 1 u&l 3 u&3 1

_ A (l)Q(O) + Ap(l) __ A (O)Q(O) + A (O)Q(O). u&l 3 u 2 - u&3 1 u&l 3'

dQ(l) _3_ + A (O)Q(l) _ A (O)Q(l) + A (l)Q(O)

dry u&l 2 u&2 1 u&l 2

_ A (1)Q(O) + Ap(l) __ A, (O)Q(O) _ A (O)Q(O). u&2 1 u 3 - u&2 1 u&l 2, (4.85)

dM(l) __ 1_ + A (O)M(l) _ A (O)M(l) + A (l)M(O)

d U&2 3 u&~ 2 U&2 3 ry ,

_ A (l)M(O) + AT _ A (O)M(O) _ A (D)lv.1(D). u&3 2 u 1 - u&3 2 u&2 3'

(1 ) dM2 + A (D)M(1) _ A (D)M(l) + A (l)M(D)

dry u&3 1 u&l 3 u&3 1

_ 6.&(1) M(D) _ Q(1) + 6.T(l) - -6.&(0) M(O) + 6.&(0) M(D) . 133 2- 31 13'

d 1{l) ~ + A (0) '1{l) _ A (D)M(l) + A (l)M(D) d u&l lV 2 u&2 1 u321 2

Tj

_ A (l)M(l) + Q(2) + AT __ A (OlM(D) _ A . (DlM(D) u&2 1 2 u 3 - u&2 1 u&l 2' ( 4.86)

The vector equations (4.85) and (4.86) together with (4.51)-(4.53) represent the complete system of equilibrium equations of the first approximation

dQ(1) __ + A(DlQ(l) + A(D) A- 1M{l) 6.p(l)

dry '" Q

= -A -lM(D) x Q(D);

dM{l) + A (D)M(l) + A (0) A-I M(l) + A Q(l) + 6. T(l) dry '" M 1

= A -lM(D) x M(D) ;

M(l) = A 6.aP) ;

d '19 (1) __ - A-I M(l) = 0 .

dry ,

4.2 Small Displacements and Angles of Rotation

d (I) ~ A ~'I9(I) = 0 d1] + I ,

where

A(O) = '"

A (O) -Q -

149

( 4.87)

Consider now the equilibrium equations (4.56)-(4.61) with QlO -I 0 and MlO -I O. From (4.56) and (4.59) it follows that

where

QlO = - 11) PlO d( + QIOO ;

~QI = -11) (~<£2Q3 - ~<£3Q2) d( + ~QlO;

MlO = -11) TlO d( + Mloo ;

~MI = -11) (~<£2M3 - ~<£3M2) d( + ~MlO .

Since ~QI and ~MI could be considered small, we replace QI and Ml in (4.57), (4.58) and (4.60), (4.61) by QlO and MlO defined by the relations

dQlO dry +PlO = 0;

dMlO dry +Tlo = 0, ( 4.88)

where n

PIO = qlO + L pJi) 6 (1] - 1]i) ; i=1

P

TlO = /-LIO + LTI(~) 6 (1] -1]v). ( 4.89) v=1


Integrating (4.88), we get

QlO = -17] QlO d( - t Pl(~) H ('TJ - 'TJi) + Q100 ; ( 4.90)

17] P (II) M lO =- J.LlO d(- L: TlO H('TJ-'TJII)+M100 .

° 11=1

(4.91)

Fig. 4.5.

The arbitrary constants QlOo and M100 are defined by the end fixity conditions. For example, in the case shown in Fig. 4.5, we have

QlO(O) = Q100 = 11 QlO d( + P;~& + P;~& ; MlO(O) = M100 = TX10 .

For MlO =J 0, we have iElO = M lO /All . By use of (4.57), (4.58), (4.60), and (4.61), we obtain the equations of the zeroth approximation

dQ(O) --irJ + PlO + ~piO) = 0 ;

dQ~O) + A (O)Q _ MlO Q(O) + p. + Ap,(O) _ O' d uiE3 10 A 3 20 u 2 - ,

'TJ II

dQ~O) + MlO Q(O) _ A (O)Q + p. + Ap,(O) _ O' d A 2 uiE2 10 30 u 3 - ,

'TJ II (4.92)

dMiO) (0) _ . dry + TlO + ~Tl - 0 ,

dMJO) + A (O)M _ MlO M(O) _ Q(O) + T + AT.(O) - O' d uiE3 10 A 3 3 20 u 2 - ,

'TJ 11

dMd ~O) + MA 10 MJO) _ ~iE~O) MlO + Q~O) + T30 + ~T~O) = O. (4.93) 'TJ 11

If ~Pi(O) and ~T?) do not depend on '!9iO) (uiO) == 0), then, using (4.92), (4.93) and (4.76)-(4.78), we obtain two systems of equations similar to (4.79) and (4.80)


Q(O) ~ + p + ~p(O) - o.

d1] 10 1 - ,

dM (O) __ 1_ + T + ~T(O) - 0 .

d1] 10 1 - ,

M (O) _ A A (0) 1 - 11 L...l.(£1 ,

d{)(O) M(O) _1 ___ 1_ - o.

d1] All - ,

durO) _1_ -0.

d1] - ,

A =(0) _ -= . L...l.Uol - <L10 ,

dQ (O) _2_ + A (O)Q _ MlO Q(O) + p + Ap(O) = o.

d L...l.(£3 10 A 3 20 L...l. 2 , 1] 11

dQ~O) _ A (O)Q + MlO Q(O) + p + Ap(O) - o. d L...l.(£2 10 A 2 30 L...l. 3 - ,

1] 11

dMJO) A (O)M MlO M(O) Q(O) T AT(O) -- + L...l.(£ 10 - -- 3 - 3 + 20 + L...l. 2 = 0 ; d1] 3 All

d (0) M ~ _ A (O)M + ~M(O) +Q(O) +T + AT(O) - o·

d L...l.(£2 10 A 2 2 30 L...l. 3 - , 1] 11

d{)(O) M(O) _2 __ (£1O{)~0) __ 2_ = 0 ;

d1] A22

,0(0) M(O) ~ {)(O) __ 3_ - o.

d + (£10 2 A - , 1] 33

du~O) (0) ,0(0) _ o. ~ - (£lOU3 - 'U3 - ,

d (0) ~ + = u(O) + ,0(0) - 0

d1] ""-'10 2 'U2 - ,

where

M (O) - A A =(0) . 2 - 22 L...l.""-'2 ,

(0) ~(£2(0) = d{)2 .

d1] ,

M (O) - A A =(0) . 3 - 33 L...l. Uo3 ,

(0) A (0) _ d{)3

L...l.(£3 -~.

(4.94)

(4.95)

The expressions for ~pjO) and ~TjO) are identical to those in (4.73) and (4.74). By use of (4.75), these expressions can be rewritten as follows (for brevity, the relations only for one concentrated force and one concentrated moment are written out):

~p(O) = C1 '19(0) + C2u(0) + Bl '19(0) + B2 u(0) ;

~T(O) = C3'19(0) + C4 u(0) + B3'19(0) + B4 u(0) ; (4.96)


here the vector '19(0) has two components 1'J~0) and 1'J~0) 1 the vector u(O) has

two components u~O) and u~O); Ci and Bi are 2 x 2 matrices. Taking into account (4.96), we can represent (4.95) in vector form (the

same vector notation was used in (4.81))

dQ(O) _ Al QlO ~a=(0) + Al MlO Q(O)

dry All + p(O) + C I 'I9(O) + C2 u(0) = - Bl '19(0) - B2 u(0) ;

dM(O) (0) (0) MlO (0) -- + Al Q - AJ MlO ~a= + Al - M dry All

+ T(O) + C3 '19(0) + C4 u(0) = -B3 '19(0) - B4 u(0) ;

d·o(O) _u __ A-IM(O) = o.

dry ,

durO) -- + A '19(0) = 0 . dry 1 ,

M(O) = A ~a=(0) . ( 4.97)

The elements of the matrices B1 , Bz, B31 and B4 contain 6-function; the terms containing these matrices are rearranged to the right-hand side of the equations. Equations (4.97) are a system of inhomogeneous linear differential equations. Methods for analysis of equations containing 6-function are presented in Chap. 2.

4.2.3 Equilibrium Equations in the Cartesian Coordinate System

Using the expressions for .6.Px and .6.T,") we can rewrite (1.84) and (1.85) as follows:

here

dQx -- + p x + .6.P x = 0 ; drl

d~x _ (1'J 3Qx3 + 1'J 2 Qx2) il + (1'J 2Qxl - Q X 3) iz

+ (Qx2 -1'J3Qxl) i3 + Txo + .6.Tx = 0;

.6.a= = A-I LMx ;

d'19 -1 dry - A LMx = 0;

dux .0. .a. dry + U3 12 - 'U2 13 = 0 ;

( 4.98)

( 4.99)

(4.100)

(4.101)

(4.102)


Although (4.100) and (4.101) are referred to the attached basis, they contain the components of the vector M in the Cartesian basis. As noted in Sect. 1.3, the use of the two bases {i}j and {ej} may be very advantageous for numerical analysis and algebraic manipulation. For dead forces, we have

.6.qx = .6./1x = .6.p~il = .6.T~vl = O.

By use of (4.98)~(4.102), we obtain the following equations in terms of projections on the Cartesian axes:

dQXl A d";]" + Px,D + uPX1 = 0;

dQ X 2 A d";]" + Px2D + UP.-r2 = 0 ;

dQ X 3 A d";]" + Px3D + uPX3 = 0 ; (4.103)

d:;, _ rhQ X3 - rhQx2 + TXID + .6.TX1 = 0; p d:x2 + 192QXl - QX3 + /Lx2D + .6./Lx2 + 2)T~~6 + .6.Ttl) <5 (7] -7]v) = 0;

7] v=l

p d:x3 +Qx2 -rhQxl +/Lx3D+.6./LX3 + 2)T~~6+.6.T~~l)<5(7]_7]v) = 0; 7] v=l

(4.104)

(4.105)

( 4.106)


dU x3 _0 - 0 d1] +112 - . ( 4.107)

If the increments 6.qXj and 6.p~;) are linear in U Xk and '!9 k , then (4.103) are linear. Equations (4.104)-(4.107) are nonlinear because they contain the products '!9 j Qx; and '!9 j M x ;, where QXj and MXj cannot be considered small. Therefore, we need to estimate the nonlinear terms. From (4.103) it follows that

QXj = QXjO + 6.QXj ,

where 6.QXj is small, hence, the relations

ihQX3 = {J3Q x30 ;

{J1Qxl = {J1Q x lO;

{J2QX2 = {J2Q x20 ;

{J3Qxl = {J3Qx 10

are to be substituted into (4.104). Suppose that the products '!9 i Mxj are small as compared to M Xk ; then

we can represent (4.105) and (4.106) as linear equations. For /-tX10 ::j:. 0 and

T.~~6 ::j:. 0, the terms '!92Q x 2' '!9 2Q x3' 6./-tXll and 6.T~~) in the first equation of (4.104) are small as compared to /-tx10 and T X10 . Thus these increments can be omitted and the equation becomes uncoupled from the other equations of the system (4.103)-(4.107). If the increments 6.qxl and 6.P;:) are dropped out, the first equation of the system (4.103) is uncoupled from the others. Hence, for any set of boundary conditions, we get QX! = QX10. Finally, using (4.103)-(4.107), we obtain two uncoupled systems of equations:

dQx,o p dry + X10 = 0;

dMx,O ~+TXIO=O,

d'!9 1 _ M X,O _ O· d1] An - ,

dQX3 A ~(p(i) A p(i») '( ) _ . d + QX30 + UQX3 + ~ X3 0 + U X3 U 1] - 1]i - 0 , 1] i=l

(4.108)

dMx2 '!9 Q Q 6. ~(T(V) A (V»)' ) --d- + 2 xlO - X3 + /-tX2 0 + /-tX2 + ~ X20 + uTX2 U(1] -1]v - 0; 1] v=l

dMx3 Q _ '!9 Q 6. ~(T(V) AT(v») '( ) - . d + X2 3 X10 + /-tX30 + /-tX3 + ~ X30 + U X3 U 1] -1]v - 0, 1] v=l

4.3 Naturally Twisted Straight Rods 155

d{)2 _ MX2 = 0 . d1} A22 '

d{)3 _ MX3 = O' d1} A33 '

dU X2 {) dry+ 3=0;

d~~3 _ {)2 = 0 . (4.109)

Equations (4.109) enable us to determine the components of the unknown vectors Q, M, ~<E, {}, and u in the Cartesian coordinate system.

4.3 Naturally Twisted Straight Rods

4.3.1 Nonlinear Vector Equations of Equilibrium

For naturally twisted rods, we have

(1) _ _ d{)10 . <Eo - <ElOe l - dry e1 ,

<Eb1) x Q = <E10(e1 x Q);

<Eb1) X M = C€lO(el x M).

In view of the general equilibrium equations (1.57)-(1.61), we get

dQ "dr]+<ExQ+P=O;

dM d1} + <E X M + e1 x Q + T = 0 ;

M = A (<E - C€lOed ;

d{) -1 L1 d1} + L2 C€lQel - A M = 0 ;

du d1} + <E X U + (lll - 1) e1 + 121 e2 + 131 e3 = O.

4.3.2 Linear Vector Equations of Equilibrium

(4.110)

(4.111)

(4.112)

(4.113)

(4.114)

When a rod slightly deflects with respect to its natural straight configuration, we have

P = Po +~P; T=To+~T.

Therefore, using (4.110)-(4.114), we arrive at


dQ - + iElO(el X Q) + 6iE X Q + Po + 6P = 0; dry dM - + iElO(el X M) + 6iE X M + el X Q + To + 6T = 0; dry

M = A6iE;

d'19 -1 dry + iElOAI'19 - A M = 0;

du - + iElOAI U + A1 '19 = 0, dry

d19l0 iElO = --.

dry

( 4.115)

( 4.116)

(4.117)

(4.118)

(4.119)

The increments 6P and 6T in (4.115) and (4.116) defined by (4.54) depend on u and '19 (see (4.55)). Consider the nonlinear terms 6iE X Q and 6ce X M in (4.115) and (4.116). If the projections of the vectors Po and To on the tangent to the rod axis do not equal zero, we get

6ce x Q = QlO(6ce x ed + 6ce x (Q2e2 + Q3e3);

6ce x M = M lO (6ce x ed + 6ce x (M2e2 + M3e3) ,

( 4.120)

(4.121)

where QlO = PlO and MlO = TlO . To obtain linear equilibrium equations, we set 6ce x (Q2e2 + Q3e3) = 0 and 6ce x (M2e2 + M 3 e:l) = O. Hence, the following approximate relations are valid:

A more detailed discussion of the role of the nonlinear terms in the equilibrium equations for small deflections is given in Sect. 4.2.

Finally, in view of (4.115) and (4.116), we obtain the following linear equations:

dQ '"dr7 + iElO(el x Q) + QlO(6ce x ed + Po + 6P = 0; (4.122)

dM dry + iEIO(el x M) + M lO (6ce x ed + el x Q + To + 6T = O. (4.123)

4.3.3 Equilibrium Equations in the Attached Coordinate System

In terms of projections on the attached axes, the system of equations (4.122), (4.123), (4.118), and (4.119) for naturally twisted rods can be rewritten as follows:

dQl dry + PIO + 6H = 0 ;

dQ2 dry - iElOQ3 + 6iE3QlO + P20 + 6P2 = 0 ;

dQ3 dry + iElOQ2 - 6iE2QlO + P30 + 6P3 = 0 ; (4.124)

4.3 Naturally Twisted Straight Rods 157

dM1 dry + TlO + t::..T1 = 0 ;

dM2 dry - <BlOM3 + t::..<B3MlO - Q3 + T 20 + t::..T2 = 0 ;

dM3 dry + <BlOM2 - t::..<B2 MlO + Q2 + T30 + t::..T3 = 0; (4.125)

d191 _ M1 _ O. d1] Au - ,

d192 M2 - - <BlO193 - - = 0 ; d1] A22

d193 M3 - + <B20192 - - = 0 ; d1] A33

(4.126)

dU1 _ O· d1] - ,

dU2 d1] - <BlOU 3 - 193 = 0 ;

dU3 d1] + <BlOU 2 + 192 = 0 . (4.127)

Being of the most general form, the system of equations (4.124)-(4.127) is useful in many applications.

Let us derive the equilibrium equations for a twisted drilling bit (see Fig. 0.21) after a loss of stability. Assume that the force P lO and the moment TlO are follower and that the static mode of loss of stability may take place. In view of (4.124)-(4.127), we get

dQ1 d1] - P lO <> (1] - 1) = 0; (4.128)

dQ2 d1] - <BlOQ3 - t::..<B3 P lO = 0; (4.129)

dQ3 d1] + <BlOQ2 + t::..<B2 P10 = 0; (4.130)

dM1 dry+TlO <>(1]-1) =0; (4.131)

dM2 (1) dry - <B1O M3 - Q3 = 0; (4.132)

dM3 (1) _ . dry + <B1O M2 + Q2 - 0, (4.133)

(1) _ d19lO TlO <B1O - dry - Au . (4.134)

(Equations (4.126) and (4.127) remain unchanged.) From (4.128) and (4.131) it follows that Q1 = -PlO and M1 = MlO . These

equalities are used in (4.129), (4.130), and (4.132)-(4.134). From the first


equations of the systems (4.126) and (4.127), we get 79~ = MIO/All = const and U1 = O. Let us eliminate .6.<£2 and .6.<£3 from the other eight equations and represent these system of equations in vector form as follows:

dY -+AY=O' dry , (4.135)

here

Y = (Q2, Q3, M 2 , M 3 , 79 2 ,793 , U2, U3)T;

0 0 FlO

0 0 -<£10 A33

FlO <£10 0

An 0 0 0

0 -1 0 TIO

0 0 - - <£10

TlO A33

A= 1 0 <£10 -- 0 0 0 A22

1 0 0

An 0 0 -<£10

1 0 0 0

A33 <£10 0

0 0

0 0

0 0

0 0

0 0

0 0

0 0 0 0 0 -1 0 -<£10

0 0 0 0 0 0 <£10 0

4.4 Straight Rods on Elastic Foundation

4.4.1 Forces Acting on a Rod

The problems of rods contacting with elastic foundations, elastic layers, etc. (sec Fig. 0.19) are of great practical importance. Interaction between a rod and a medium and between a rod and a force field is illustrated in Figs. 4.6 and 4.7, respectively.

A transmission pipeline is shown in Fig. 4.6. The pipe rests on elastic soil. As the pipe bends, a force of interaction between the pipe and the soil appears. This force depends on the displacement of the pipe in the vertical direction U X2 •

The key feature of problems of interaction of rods with an elastic medium is as follows: when a rod deflects from its initial configuration (no matter straight or curvilinear), distributed forces q appear. These forces are functions of the displacement vector u, i.e. q = q(u). In the case of a linearly deformed elastic medium, we have

For a nonlinearly deformed medium, we get

4.4 Straight Rods on Elastic Foundation 159

Xz

Fig. 4.6.

As shown in Chapters 1-3, in most applied problems we deal with forces and moments which depend on displacements of the axial points and angles of rotation of the attached coordinate axes, i.e. q = q(u, fJ) and /L=/L(u, fJ). If displacements and angles of rotation are small, then the expressions for q and /L take the following form (see (4.55)):

Xz

I

II

n'" .,,«,0

Fig. 4.7.

( 4.136)

Hence, problems of statics of rods on an elastic foundation are special cases of the problems of statics considered above with C2 = C3 = C4 = 0 and

o 0 1 -k 0 . o 0

A rod attracted by two magnets, I and II, is shown in Fig. 4.7. The magnitudes of the distributed attracting forces are denoted by q~~) and q~~), respectively.

Suppose that the rod in its natural configuration lies right in the middle between the magnets, hence, the attracting forces are equal q~~b = q~~b. If


we apply an additional load to the rod, say, a concentrated moment TX3 as indicated in Fig. 4.7, then the rod deflects and the attracting forces are no longer equal (q~~) > q~~)). Moreover, if U X2 « a, the total attracting force acting on the rod is linear in U X2 ' that is,

(4.137)

From this example we see that when the rod deflects from the initial equilibrium configuration, the additional distributed force is directed away from this initial equilibrium configuration. In the preceding example (Fig. 4.6), on the contrary, the distributed force caused by the deformation of the pipe is directed towards the initial equilibrium configuration of the pipe.

4.4.2 Equilibrium Equations

Consider a rod of varying cross section lying on an elastic foundation. Suppose that displacements of the axial points are small. Setting Txlo = 0, J.LxIO = 0, qXlo = 0, PXIO =I- 0, and !::.qX2 = -kuX2 in (4.98)-(4.102), we obtain the following equilibrium equations:

dQ x 2 () ~ + qX20 H (1- TJq - kU X2 + P.T20 0' (TJ - TJp) = 0;

dMx3 (1) _ . ~ + QX2 - 'l'hPxlo + J.Lx30H (TJ - TJ/l) + TX30 0' (TJ - TJT) - 0,

d193 _ MX30 _ 0 . dTJ A33 - ,

dU x2 _ 193 = 0 . dTJ '

(4.138)

( 4.139)

( 4.140)

(4.141)

here k = kOZ4 / A~~) is a nondimensional coefficient (ko is the dimensional elastic stiffness of the foundation).

Equations (4.138)-(4.141) can be rewritten as one equation as follows:

(A33Ux/') /I - PXIOU~2 + kUX2 = QX20H (TJ - TJq) + PX20 0' (TJ - TJp)

- T;~6 0" (TJ - TJT) - J.Lx30 0' (TJ - TJ/L)' (4.142)

For rods of constant cross section, the nondimensional bending stiffness A33 = 1. Hence, (4.142) becomes a linear differential equation with const.ant coefficient.s

(4.143)

where

b = -QX20H - PX20 O'p - 1'.'1:30 O'~ - J.L X30 O"L'

Equations (4.142) and (4.143) are useful in the derivation of approximate solutions. Approximating met.hods for analysis of equilibrium equations of straight rods are discussed in Sect. 4.5.


4.4.3 Krylov's Functions

Let us consider a simple problem of statics of rods. A straight rod of constant cross section rests on an elastic foundation. Suppose that the axial components of external forces are equal to zero; then, setting PX10 = 0 in (4.143), we obtain

(4.144)

We seek the solution of the form U X2 = Uo exp(A7]). The corresponding characteristic equation A4 + 4a1 = 0 has the roots Al,2 = ±(i - 1) al and A3,4 = ±( i + 1) al. The solution of the homogeneous equation (4.144) (b = 0) takes the form

U X2 = Cl e-"l7) cos al7] + C2 e"l7) sin al7]

+ C3 e"l'7 cos al7] + C4 e"l7) sin al7]

(O)() (O)() (O)() (0)) =c1kll 7] +c2k12 7] +c3k13 7] +c4k14 (7] .

For straight rods of constant cross section (QI0 = 0), we have

'193 = U~2 •

Hence, (4.144) is equivalent to a system offour ordinary differential equations

Q~2 + 4aiuX2 = b1 ;

M~3 + QX2 = b2 ;

'I9~ - MX3 = 0;

U~2 - '19 3 = O. (4.145)

The fundamental matrix for the homogeneous system of equations (4.145) (b 1 = b2 = 0) is as follows:

[ (0)'" (0) '" (0)'"

(0)'" I kll k12 k13 k14 (0)" (0) " (0) " (0) "

K(O)(7]) = k\~), k12 k13 k14 (0)' (0)' (0)' .

kll k12 k13 k14 k(O) k{O) k(O) k(O) II 12 13 14

The matrix K(O) (7]) for rl = 0 is not a unit matrix. To avoid further difficulties, we should select the partial solutions such that the equality K(O) (0) = E holds; to do this, let us compose a linear combination of the partial solutions

11

kij (7]) = LkiJ)(7])bvi (4.146) v=1


and choose coefficients bVi in such a way that the matrix K(O) with the components k ij (0) becomes the unit matrix.

The coefficients bVi can be found in the following way. For any matrix B with constant elements, the matrix K = K(O) (7))B satisfies the homogeneous system of equations corresponding to (4.145). Evidently, the matrix B can be so chosen as to satisfy the equality K(O) = E. If the matrix B is equal to the inverse of the matrix K(O) (0), we get

K (7)) = KO(7)) [KO(O) ]-1.

On rearrangement, we have

Kl -al K 4 -2aiK 3 K2

Kl -al K 4 2al

K (7)) = [kij ] = K3 K2 Kl

2a2 2al 1 K4 K3 K2 4a3

1 2 2 a 1 2al

where Ki are the following Krylov's functions:

Kl = cosh al7) cos al7) ;

K2 = cosh al7) sin al7) + sinh al7) cos al T) ;

K3 = sinhal7)sinal7);

K4 = COShal7)sinal7).

Krylov's functions are tabulated. The derivatives of the functions Ki are

- 2arK 2

-2aiK 3

-al K 4

Kl

K 1 ' = -al K 4;

K 2' = 2alKl ;

K3' = al K 2;

K/ = 2alK3;

K 1" = -2aiK 3; K 1 '" = -2aiK2;

K 2" = -2aiK4 ; K/" = -4aiK3;

K 3 " = 2aiK1 ; K 3 '" = -2aiK4 ;

K 4 " = 2ai K2 ; K 4 '" = 4ai Kl .

(4.147)

(4.148)

(4.149)

( 4.150)

4.4.4 Equilibrium Equations for Rods of Constant Cross Section

To use computer-oriented methods, it is convenient to represent (4.115) in vector form

y' +AY =b, (4.151)

where


l QX2

1 l -u~~

1 A~[~

0 0

-~al y= MX3 U" 0 0 X2

()3 U~2 -1 0 o ' U X2 U X2 0 -1 0

b=

[ ~ 1 [ -q",H("-",) -P",8("-"p) 1

-T", J (" - 'IT) f ~""H ('1- q,) (4.152)

Let us consider the general approach to the analysis of the inhomogeneous equation (4.151). The corresponding homogeneous equation has a solution

y(O) = Yo e).,r). ( 4.153)

The characteristic equation det [EA + A] = 0 (or A4 + 4ai = 0) has the following roots:

A3 ,4 = ± (i + 1) a 1 .

The four independent vectors are thus obtained

y(2) = y~2) e).,2r) ;

y(4) = y~4) e).,4r) .

(4.154)

(4.155)

The components of the vectors y~i) can be found from the system of

algebraic equations [EAi + A] y~i) = O. The expanded form of these equations is as follows:

\ .y;(i) _ y;(i) - 0 . A2 01 02 - ,

\ .y;(i) + y;(i) - O' A2 03 04 - ,

(i) (i) _ . Ai Y02 - Y03 - 0 ,

4 4y;(i) \ y;(i) - 0 - a 1 01 + Ai 04 - . (4.156)

A d· (4 156) h y;(i) y;(i) d y;(i) I d y;(i) ccor mg to. , t e vectors 02' 03' an 04 are re ate to 01

as follows:

(i) _ (i) . Y02 - AiY01 , y;(i) - (A )2y;(i) .

03 - 2 01' y; (i) __ ( \ .)3y;(i)

04 - A, 01' (4.157)

The factor Yo(f) may be prescribed arbitrarily. (This factor occurs in each

component of the vectors y~i).) For definiteness, we put Yo(;) = 1 + i. Taking into account the third equation of (4.157), we see that the fourth equation of (4.156) becomes an identity. Since the roots Ai are complex, we see that the

solutions of (4.156) are also complex, that is, Yo~) = (yo~)h + i (yo~)h. From (4.157) it follows that


[ 1

1 [ 1

1 [ 1

1 [ 1

1 y(1) = -20:1

+i 0 y(2) = 20:1

+i 0

20:i -20:i 20:i -20:i 0 -40:y 0 -40:y

[ 1

1 +i [

1

1 [ 1

1 [ 1

1 y(3) = 0 20:1 y(4) = 0

+i -20:1

-20:i 20:i -20:i 20:i 40:1 0 40:y 0

(4.158)

Substituting (4.157) into (4.155), we obtain the complex eigenvectors

y(i) = (y(i)h + i (y(i)h, ( 4.159)

whose real and imaginary parts satisfy the homogeneous equation (4.151). If the eigenvectors y(i) are known, we can get two fundamental matrices

whose columns are the real parts and the imaginary parts of the vectors y(j),

respectively. These matrices differ only in the arrangement of their columns. Finally, for the homogeneous equations (4.151) (b = 0), the fundamental matrix K (T/) takes the form

(4.160)

where

Kll = e-D:11J(cOS0:1T/ - sin0:1T/);

K 13 = eD:11J(coS0:1T/ - sin 0:11]) ;

K12 = eD:l1J(coSO:l1] + sin 0:11])

K14 = e-D:l1J(cOSO:l1] + sin0:1T});

K23 = 20:1 eD:11J sinO:l1]; K21 = -20:1 e-D:ll)sinO:l1];

K31 = 20:i e-D:l1J (cos 0:11] + sin 0:1 T/) ;

K32 = 20:i eD:l1J (cos 0:11] - sin 0:11]) ;

K33 = -20:i eD:11J(COSO:l1]+sinO:l1]);

K34 = 20:i e-D:l1J (sin 0:11] - COS0:1T/);

K41 = 40:{e-D:11J sinO:l1];

K43 = 40:{ eD:l1J cos 0:11] ;

K42 = -40:{ eD:l1J sin 0:11] ;

K44 = -40:{ e-D:I1J cos 0:11] .

The elements of the matrix K(1]) are linear combinations of the partial

solutions ki~) (T/) obtained above. If the matrix K( T/) (4.160) is not the unit matrix at T/ = 0, then, using the transformation (4.147), we can define a fundamental matrix Kl (T/) such that the equality Kl (0) = E is satisfied. Hence, in what follows, we assume that the fundamental matrix K is the unit matrix at 1] = O. For example, the matrix defined by (4.148) is such a matrix.

The solution of the inhomogeneous equation (4.151) can be written as


yeO) =Yo+YH=K(7])C+ 11)K(7]- Obd(. (4.161)

The components of the vector Y H are as follows:

YHj = -11) kj1H (( - 7]q) QX20 d( - 11) kj1 6 (( - 7]p) PX20 d(

-11) kj2H (( -7]/-,) fLx30 d( -11) kj2 6 (( -7]T) TX30 d( (4.162)

or

j=1,2,3, ... , (4.163)

where

(Jjdq = -11) kj1H (( - 7]q) QX20 d(;

(Jjdp = -11) kj1 6 (( - 7]p) PX20 de

(Jj2 )/-, = -11) kj2H (( - 7],J fL x30 d(;

(Jj2)r = -11) kj2 6 (( - 7]T) TX30 d(. ( 4.164)

Formulas (4.164) containing concentrated forces and moments can be rewritten as follows:

(Jjdp = -kj1(al7]p)PX20H(T/-7]p);

(Jj2)T = -kj2 (alT/) TX30H (7] -7]T);

introducing Krylov's functions, we arrive at

(Jll)P = -Px2o K 1(al7]p)H(7]-7]p);

(J2dp = - PX20 K 2(al7]p) H (7] - 7]p) ; 2a1

(J31)P = - P2X2~ K 3(al7]p) H (7] -7]p); a 1

(J4dp = - P4X230 K 4(al7]p) H (7] -7]p); a 1

(J12)r = aTx30K4 (al7]T) H (7] - 7]T) ;

(J22 )r = -Tx30 K 1(al7]T) H (7] -7]T);

(J32)r = - TX30 K 2(alT/T) H (7] -7]T); 2a1

(J42 )r = - TX30 K 3 (al7]T) H (7] -7]T). 2a1

(4.165)

(4.166)

(4.167)


Recall that (4.150) relate Krylov's functions to their derivatives. For qX 20 = const and {lX3 0 = const, (4.162) take the form

(Jll)q = - Q2x2g [K2(O'.l7)) - K 2(O'.l7)q)] H (7) - 7)q); 0'.1

(hd q = - Q2X2~ [K3(O'.l7)) - K3(O'.l7)q)] H (7) - 7)q); 0'.1

(J3dq = - Q4X2~ [K4(O'.l7)) - K4(O'.l7)q)] H (7) - 7)q); 0'.1

(J4dq = - Q4X2~ [K1(O'.l7)) - K 1(O'.l7)q)] H (7) -7)q); 0'.1

(J12)p. = _{lx3 0 [K1(O'.l7)) -K1(O'.l7)p.)]H(7)-7)p.); 0'.1

(h2)p. = _{l2X3~ [K2(O'.l7)) -K2(O'.l7)p.)]H(7)-7)p.); 0'.1

(J32)p. = _{l2X3~ [K3(O'.17)) - K 3(O'.17)p.)] H (7) -7)p.); 0'.1

(J42 )p. = _{l4X3~ [K4(O'.l7)) -K4(O'.l7)p.)]H(7)-7)p.). 0'.1

( 4.168)

Finally, assuming that there are several concentrated forces p(i) and moments T(I/) applied to the rod, we can write the general solution of (4.151) in expanded form as follows:

4 n p

Y1 = QX2 = L kijCj + (Jll)q + L(Jll)~) + (J12 )p. + L(Jd~) ; j=l ;=1 1/=1

4 n p

Y2 = MX3 = L k2j Cj + (J21 )q + L(hd~) + (h2)p. + L(J22)~) ; j=1 i=1 1/=1

4 n p

Y3 = V3 = L k3jCj + (hdq + L(hd~) + (h2)p. + L(J32)~) ; j=l i=l 1/=1

4 n p

Y4 = U X2 = L k4jCj+(J4d q+ L(J4d~)+(J42)P.+ L(J42)~); (4.169) j=l i=1 1/=1

here k ij are the elements of the fundamental matrix (4.148) of the vector equation (4.151); these elements are expressed in terms of Krylov's functions.

Arbitrary constants Cj in (4.169) are to be determined from the boundary conditions at 7) = 0 and 7) = 1. As an example, consider a case shown in Fig. 0.19. Let the terms in (4.169) depending on external loads be denoted by YHj , j = 1,2,3,4. From (4.163) we can see that YHj = 0 at 7) = 0, hence, Y2 (0) = 0, Y4 (0) = 0, and C2 = C4 = O. Since Y2 (1) = 0 and Y4 (1) = 0, we see that the constants C1 and C3 satisfy the relations


k21 (ar) Cl + k23(ar) C3 = -YH2(l);

k41 (ar) C1 + k43(ar) C3 = -YH4(1). (4.170)

4.4.5 Equilibrium Equations for Rods with Lateral Supports

A reaction force Rl applied to a support (see Fig. 4.8) can be treated as an additional external force applied to the rod. The rod is fixed as shown in Fig. 4.8. Consider the boundary conditions. At T} = 0, we have C3 = C4 = 0; at T} = 1, (4.169) yield the following relations:

k2r(a l)Cl+ k22(a1)c2 = _qX2~ [K3(ar)-K3(0.5ad]-2Rl K2(0.5ar); 2a1 a1

k41(ar) C1 + k42(ar) C2 = qX2~ [Kl(ar) - Kl(0.5ar)]- R13 K4(0.5ar); 4a1 4a1

(4.171)

here

x,

1

Fig. 4.8.

The two equations (4.171) contain three unknowns C1, C2, and R 1. The third equation can be obtained through the condition U X2 = 0 at T} = 0.5:

This approach to the solution of the equilibrium equations for a rod on an elastic foundation with a lateral hinge support is efficient and can be applied in the case of several supports of this kind. Moreover, this approach can be generalized to the case of elastic supports. In the latter case, the reaction at the support is

R(i) = -CUX2 J (T} - T}i) .

Then, additional terms of the form


(i) _ r (Jjd R - 10 kj1 c'U x2 5((-7)i)d(

appear in the general solution (4.169). These terms are similar to those in the second relation (4.164). On rearrangement, we get

(J11 ) W = C'U X2 (7)i) K1 (a1 7)i) H (7) - 7)i) ;

(J2dW = C'U x2 (7)i) K 2 (al7)i) H (7) - 7););

(J3d~) = CU x2 (7)i) K 3(al7);) H (7) -7)i);

(hdW = CU x2 (7)i) K 4(a1 7/i) H (7) -7)i),

where 4

'U x2 (7)i) = L k4j (7)i) Cj. j=l

If we replace the hinge support (see Fig. 4.8) by an elastic support, then R1 = -C'UX2 5(7) - 0.5). The constants C1 and C2 can be determined from the relations

k21(ad C1 + k22(ad C2 = - qX2~ [K3(ad - K 3 (0.5ad] 2a1

C - - K2 (0.5ad [k41 (0.5ad C1 + k42 (0.5ad C2] ;

2a1

k41 (ad C1 + k42 (ad C2 = q4x2~ [K1 (ad - K1 (0.5ad] a 1

C - -3 K4 (0.5ad [k4dO.5ad C1 + k42 (0.5ad C2].

4a1

4.4.6 Equilibrium Equations for Rods of Varying Cross Section

Let us consider the equilibrium equations (4.138)-(4.141) for a rod of varying cross section. These equations can be solved only by use of numerical methods. As noted, it is convenient to represent the equations in vector form

Y'+A(7))Y=b, (4.172)

where

r Q"

1 A ~ r ~ 0 0

~k 1 Y = NIx 3 0 -PX1O

1 0 '133 A33 (7)) UX2 0 1

r ~ 1 r

-qx20H (7) -7)q) - PX20 5 (7) - 711')

1 b=

-TX30 5 (7) - 7)1') - f..Lx30H (7) - Til,) 0 0

4.5 Application of Approximate Methods 169

Recall the basic results obtained in Sect. 1.4. The general solution of (4.172) is

Y = K (1]) (C) + l'l K (1]) K- 1(() b (() d( = y(O) + Y H.

This formula is valid for any vector b. The vector b can be written as follows:

Hence,

Y H = 1"1 G(1], ()bqd(+ 1''1 G(1], ()bl'd(+G(1], 1]p)bpH(1]-1]p) flo fI~

+ G (1] , 1]T) bT H (1] - 1]T) ,

where

G (1], () = K (1]) K- 1(();

G (1], 1]p) = K (1]) K-1(1]p);

G (1], 1]T) = K (1]) K-1(1]T).

Computer-oriented methods for determination of the matrix K(1]) are discussed in Chap. 2.

4.5 Application of Approximate Methods

4.5.1 Principle of Virtual Displacements

Methods based on the principle of virtual displacements are useful in analysis of linear and, what is more important, of nonlinear problems. In theoretical mechanics, this principle is formulated as follows:

A mechanical system with ideal constraints is in equilibrium if and only if the work done by external forces through any virtual displacement of the system is equal to zero.

Constraints are said to be ideal if the work done by the reaction forces through any virtual displacement equals zero.

The principle of virtual displacements can be written as follows:

n

8A= ~)Fi·8ri) =0; (4.173) i=l


here bA is the work done by the forces Fi through the displacement brio Let the forces F i be conservative, hence, they can be represented in the following form:

p(i) = _ aU Xj a (i) ,

Xj

j=1,2,3, i=1,2, ... ,n;

the function U(x;i)) is called the potential energy of the system. From (4.173) it follows that

~ (au (i) aU (;) aU (i)) bA = - ~ --ri) bX l + -(-i) bX2 + --ri) bX3 = -bU = 0 i=1 aX l oX2 oX3

or

bU = 0, (4.174)

where bU is the variation of the potential energy. The relation (4.174) is the necessary condition for the potential energy to have an extremum in an equilibrium configuration. Consequently, we see that the necessary and sufficient condition for a conservative system with ideal constraints to be in equilibrium are the same as the necessary condition (not a sufficient one) for the potential energy to attain an extremum.

Along with traditional methods based on analysis of differential equilibrium equations, the principle of virtual displacements is probably the most powerful tool of structural analysis, since it allows us to obtain approximate solutions. When dealing with elastic elements (rods, plates, and shells), we should take into consideration not only the work done by the external forces but also the work done by the internal forces as the structure deflects with respect to its original location. Now, let us discuss the concept of virtual displacement within the context of the theory of rods. Any small displacement of rod points away from the initial configuration that can be performed without breakdown of the constraints is termed virtual or allowable. As an example, consider a rod illustrated in Fig. 4.9. The shape of the axial line is described by a function Y(TJ). Suppose that a function b Y(TJ) is continuous in TJ and b y(O) = b y(l) = 0; then, this function defines a virtual-displacement field of the axial points.

y p

Fig. 4.9.


Consider the formulation of the principle of virtual displacements for a straight rod subjected to an arbitrary set of loads (see Fig. 4.9). Under the action of the loads P, T, and q, the rod deforms, therefore the work done by the loads is converted into the strain energy. Neglecting the energy lost in internal friction, we get U = A, where U is the strain energy and A is the work done by the external loads. Thus, for deformable systems, the principle of virtual displacements can be formulated as follows:

If a deformable system subjected to external forces is in equilibrium, then the work done by the external forces through any virtual displacement is equal to the work done by the internal forces through the same displacement, that is,

bA = bU. (4.175)

The work bA is the work done by the generalized external forces through virtual displacements of the points of application of these forces.

Upon a virtual displacement of the system, the external forces remain unchanged. Hence, the work done by an external force is equal to the magnitude of the force times the virtual displacement, i.e.,

n

bAQ = 2:(Qk bYk), k=l

where Qk is a generalized force and bYk is a generalized virtual displacement. Here the concentrated forces and moments p(i) and T(v) are termed generalized forces, the linear displacements bu(i) and angles of rotation 6{)(v) of the attached axes are called generalized displacements by k. For distri bu ted forces and moments, the work done through virtual displacements is as follows:

(4.176)

here T)qj and T)Mj are the axial coordinates of the endpoints of the interval of application of the external loads.

u

Fig. 4.10.


As an example, consider a rod on an elastic foundation (see Fig. 4.10). Let us determine the work done by the external forces. As in the previous chapter, we denote nondimensional deflections by Uj. The work done by the external forces through a virtual displacement can be written as follows:

bA = -Pbu (7)3) + Tbu' (7)d - t qbud( - t qfbud(, iT]2 io The equilibrium equations can be obtained from (4.175). Let us apply a

tensile axial force QlO. The potential energy of the rod and the variation of the energy are

1 11 112 U = - A33U dC 2 0

( 4.177)

bU = 11 A33U" bU" d( . (4.178)

The work done by the external forces through virtual displacements is as follows:

bA = -Pb·u (7)3) + Tbu' (7)d -11

qbuH ( -7)2) d(

-11

k4ubud( - b (Ql0' ~ 11 U,2 de) Introducing the b-function, we have

bA=-P 11bUb(-7)3)d(-T 11bu'b(-7)dd(

-11 QH(-7)2)bUd(-1

1 k4ubud(-Ql0 11 u'bu'd(.

Integrating (4.178) by parts, we get

(4.179)

11 A33U" bU" dh = A33U" bu'l~ - (A33U")' bul: + 11 (A33UI)" bud(.

In view of the boundary conditions, the first and the second terms in the right-hand side of this relation vanish.

Integration of the last term in the right-hand side of (4.179) yields

QlO 11 U' bu' d( = QlO(U' bu)l~ _11 U" bud(.

Equating bU and bA, we get


11 [ (A33UI)" + k4u + P 8 (( - 1]3) + T 8' (( - 1]1)

+ qH (( - 1]2) - QlOU"] 8u d( = 0. (4.180)

Taking into account the fact that 15u is an arbitrary nonzero function, from (4.180), we get

(A33UI)" - QlOU" + k4u + P 15 (1] - 1]3) + T 15' (1] - 1]1) + qH (1] - 1]2) = 0.

(4.181)

Equation (4.181) is the equilibrium equation for the case illustrated in Fig. 4.10.

In more general and compact notation, the relation (4.180) reads

11L (U)15Ud(=0, (4.182)

where L(u) is the left-hand side of the equilibrium equation (e.g. (4.181)). If u is an exact solution of the equilibrium equation, then L( u) = 0. If u is an approximate solution, then it does not satisfy the equilibrium equation; as a result, we can regard (4.182) as an additional condition (besides the boundary conditions) that u must satisfy. Suppose that U a is an approximate solution. Substituting U a into the equilibrium equation L( u) = 0, we have L( ua ) = qa =I- 0, where qa has the dimensions of distributed load. From (4.182) it follows that

11 qa 15u d( = ° , i.e. the work done by the unbalanced distributed load qa through virtual displacements must equal zero.

The relation (4.182) is the basic formula for determination of approximate solutions. Let us represent the displacement u as a series

n

U a = L aiVi(1]) , (4.183) i=1

where ai are constants and Vi(1]) are known functions, which satisfy the geometrical and the physical boundary conditions. We will seek the virtual displacements of the axial points in the form

n

bu = L bbi Vi(1]) , (4.184) i=1

where bbi are independent arbitrary quantities. Substitution of (4.183) and (4.184) into (4.182) yields


t t5bi 11 L (aiVi) Vj d( = O. i=1 0

(4.185)

Recall that t5b i are independent functions; in view of (4.185), we have

11 L(aivi)vjd(=O. (4.186)

Integration of (4.186) yields the following linear system of algebraic equations in the unknowns ai:

n

LCXijai = Cj, i=1

"~;A 1

(a) 1I;t

lb)

UZ1 -------........ (C)

j=1,2,,,.,n.

hf· '"' ~

~ ... ?

---- ::;;;01 .. ? Fig. 4.11.

During the determination of an approximate solution, we used the relation (4.183), which contains the functions Vi(T)). As these functions we can take a set of mutually orthogonal polynomials which satisfy the boundary conditions is very effective in determination of approximate solutions. Let us obtain such polynomials for a rod shown in Fig. 4.11a. It is reasonable to represent the non dimensional displacement v in the form of a finite power series

where the number n of the coefficients ai exceeds by one the number of the boundary conditions. In view of the boundary conditions, setting a4 = 1, we get

4 5 3 3 2 VI = T) - - T) + - T)

2 2 (4.187)

This function is positive for T) E (0, 1) (see Fig. 4.11b). The degree of the polynomial for V2 (T)) exceeds by one the degree of the polynomial for VI (T)); thus,


Let a5 = I, then

5 4 (9 5 ) 3 (7 3 ) 2 V2 = TJ + a4 TJ - "2 + "2 a4 TJ + "2 + "2 a4 TJ . (4.188)

From the orthogonality condition for VI and V2 on the interval (0, 1) we have

fa 1

VI V2 de = cl a4 + b1 = 0 ;

as a result, we get a4 = -2.2535. The further calculations yield

( 4.189)

As TJ runs from 0 to I, the function V2(TJ) equals zero just once as illustrated in Fig. 4.11b. The expression for V3(TJ) consists of seven summands and contains two free parameters a4 and a5 (if we put a6 = 1). These parameters are to be found from the following orthogonality conditions:

fal VI v3 de = clla4 + C12a5 + b1 = 0;

fal V2V3 de = C21a4 + C22a5 + b2 = O.

For a particular problem, the procedure of evaluation of the functions Vi can be easily realized on computer. Let us approximate the nondimensional displacement by a finite sum,

n

Ua(TJ) = L aivi(TJ) , (4.190) i=1

where ai are arbitrary constants. Now, we can estimate the accuracy of the solution and the minimum number of terms in (4.190) to be taken to satisfy the required accuracy. The absolute error cannot be estimated because the exact solution is unknown. Hence, the only thing we can do is to compare the values of the left-hand side of (4.190) for different n.

Let Ual = al VI and Ua2 = al VI + a2v2 be approximate solutions. If for any TJ such that TJ E [0,1] we have maxl~uall :s: ~Ual, where ~Ual is the allowable error, then V2 can be regarded as the sought solution. If max I~Uall > ~ua!' then we should seek the solution in the form Ua3 = al VI + a2V2 + a3V3 and check the validity of the inequality


u

Fig. 4.12.

Finally, a function Uak such that

max I~Ua(k-l) I = max IUak - Ua(k-l) I -:; ~Ual (4.191)

can be taken as the sought deflection of the rod. As an illustration, consider a simply supported rod on a linearly elastic

foundation as shown in Fig. 4.12. Let us take the functions v;(T}) in the form

v;(T}) = sin i7TT}.

If A33 = 1, the equilibrium equation is as follows:

We seek an approximate solution as a finite sum

n

U a = L a; sin i7TT}. ;=1

(4.192)

Let virtual displacements of the points of the rod axis be of the form

or

n

6u = L 6b; sin i7TT} . ;=1

According to the principle of virtual displacements, we get

11 L (ua ) sin j7TT} dT} = 0, j=1,2, ... ,n,

a ; [(j7T)4 + k4]_ Tj7T COSj7T1/J - P sinj7Tl)2 = o.

Consequently,

a = 2 P sin j7TT}2 + Tj7T cos j7TT}l J (j7T)4+k4

The two-term approximate solution of (4.192) is as follows:


U a = 2 (P sin 7r1]2 + T cos 7r1]1 ) .

4 4 sm 7r1] 7r + k

2 (P sin 27r1]2 + T cos 27r1]1) . 2 + 4 4 sm 7r1]. 167r + k

The method presented above implies reduction of a system of the firstorder differential equilibrium equations to one fourth-order equation. This reduction requires a great deal of calculation. It is especially so for rods of varying cross section or when force increments t1q, t1p" t1Pi , and t1T are not linear in the displacement U and the angle 133 . For example, a rod of varying cross section is illustrated in Fig. 4.10. The rod is subjected to an axial force PI = PlO i I (QlO = PlO i- 0). In the case of follower forces, the equilibrium equations of the zeroth approximation are

dQ~O) QlO (0) 4 (0) -d- + -A M 3 + k u 2 - q H (1] - 1]2) - P J (1] - r/3) = 0 ;

1] 33

dM(O) __ 3_ + Q~O) + T J (1] -1]d = 0;

d1] -d13(O) M(O) __ 3 ___ 3_ = O.

d1] A33 ' d (0)

~~ -13~0)=0. (4.193)

Let MlO = O. For follower forces, the first two equations of the system (4.193) are special cases of (4.92) and (4.93), respectively. Eliminating successively 13~0), MiD), and Q~O), we obtain (see (4.181))

( (D)')" (D)" 4 (0) A33U2 - QlOu2 + k u 2

+ P J (1] -1]3) + T J' (1] - 1]d + qH (1] -1]2) = O.

Recall that (4.181) was used as an illustrative example during the derivation of (4.182).

Now let us discuss the derivation of approximate solutions of the equilibrium equations (4.193) without reduction to one differential equation. In vector notation, we have

L (Y) = y' + A Y + b = 0 , ( 4.194)

where

Y=


0 QlO 0 0 A33

A= 1 0 0 0

1 0

A33 0 0

0 0 -1 0

Consider a one-term approximate solution

Y a (1]) = alv(1) (1]) , (4.195)

where al is an arbitrary constant and v(1)(1]) is a vector whose components satisfy the boundary conditions. For example, the function VI (1]) defined by (4.187) and its derivatives satisfy the boundary conditions (see Fig. 4.11). Consequently, under the same boundary conditions, an approximate solution of (4.193) may be taken in the form

v"l ~ ~r: 1 [-r 1 To improve the accuracy of the solution, we can take the vector Y as

a sum

rn

Y a = Laiv(i) ,

i=l

where

Let virtual displacements be of the form

8v = 8b l v(1) ,

where 8b l is an arbitrary factor. Consider the scalar product

J = (L . Eo 8v(1)) = 8b l (L· EOV(l)) ,

where

Eo ~ [

0 0 0 1

1 0 0 1 0 0 1 0 0 1 0 0 0

(4.196)

(4.197)


All the summands in the scalar product (4.197) have the dimension of work.

or

Let the work be equal to zero, that is,

11 [(y(I)'Eoy(l) + Ay(l)Eoy(I)) al - bEoy(l) J de·

From (4.198) we get

b1 al = -,

b2

where

Using a two-term approximation

Y a = al y(l) + a2 y (2) ,

(4.198)

we can obtain a system of two linear inhomogeneous algebraic equations in the unknowns al and a2.

Let us consider a more complicated problem assuming that a rod axis is not, in general, a plane curve. This is usually so when a torsional moment MlO is applied to a rod. The corresponding equilibrium equations were obtained in Sect. 4.2 (see (4.95)). If the applied forces are follower, then these equations can be rewritten as two vector equations

Ll = Y~ + Al Y 1 + Bl Y 2 + b 1 = 0 ; (4.199)

L2 = Y~ + A2 Y 2 + B2 Y 1 + b 2 = 0 , ( 4.200)

where

r Q\') 1 0

QlO 0 0

A33 M(O) 1 0 0 0

Y 1 = 3 A1 = '19(0) 1 3 0

A33 0 0

(0) U2 0 0 -1 0

MlO 0 0 0

l P20

j All

b 1 = T30 B1 = 0

MlO(A22 - All) 0 0

0 AllA22

0 0 0 0 0 0 0 0 0


[ QIO) 1 0 Q10

0 0 Azz M(O) -1 0 0 0

Y2 = .,')~O) A2 = 1 0 0 0

(0) A22 u3 0 0 1 0

M 10 0 0 0

r

P',0

1

All

h2 = T20 B2 = 0

M1O(A33 - All) 0 0

0 A33 A ll

0 0 0 0 0 0 0 0 0

We restrict our consideration to a one-term approximation. Let

Y 1 = al v (1); Y 2 = b1u(1) , (4.201)

where

r

V"I

1 [ u'" 1

- 1

u(l) = ~~( . v(l)= V" 1 Vi

1

VI 'Ul

The functions VI and Ul as well as their derivatives satisfy the end fixity conditions in the vertical plane X l OX2 and in the horizontal plane Xl OX3,

respectively. Let generalized virtual displacements be of the form

Then, we substitute the vectors (4.201) into (4.199) and multiply both sides of (4.199) by the vector E(l) be1 V(l). The following equality must hold:

or

or

bellI (Ll . Eov(1)) d( = 0

al 11 (v(l)1 . EOV(l) + Al v(1) . Eov(I)) d(

+ bIll (Bl u(1) . EOV(l)) d( + 11 (hI' EOV(l)) d( = O.

Multiplying both sides of (4.200) by E(1) bql u{!), we get

bql 11 (Lz . EOU(I)) d( = 0

(4.202)


a1 11 (B2 v(1) . Eou(1)) d( + b1 11 (u(1)1 . Eou(l)

+ A2 u(1) . Eou(1)) d( + 11 (b2 . Eou(1)) d( = O. (4.203)

As a result, two equations (4.202) and (4.203) in the two unknowns a1

and b1 are obtained.

4.5.2 Principle of Minimum of Potential Energy

One of the widely used approximating methods for statics of elastic structures is based on the following principle:

An elastic structure subjected to static external forces takes an equilibrium configuration (among the variety of admissible ones) such that the potential energy attains a local minimum:

J = U = min. (4.204)

As the structure deforms, the external forces do perform the work. This work is converted into the strain energy of the structure; as a result, we have an additional condition

U=A, (4.205)

where A is the work done by the external forces. The relation (4.205) is valid only for elastic systems because in this case the strain energy is not dissipated by irreversible processes. For a rod lying on an elastic foundation (see Fig. 4.12), the potential energy of the system and the work ofthe external forces may be written as follows:

U = ~ r1 (A33U/2 + k4u 2 ) d(;

2 10 A = ~Pu (7]2) + ~TUI (7]d.

( 4.206)

(4.207)

Thus we arrive at a conditional-extremum problem of calculus of variations. Instead of (4.204), let us introduce the functional

J 1 = U + A (U - A) = extr, (4.208)

where A is a Lagrangian multiplier. If an elastic system obeys Hooke's law, the potential energy can be represented as a quadratic form in displacements Ui

of the points of application of the external forces, that is,

(4.209)


where Ui are generalized displacements (linear and angular) of the cross sections of application of the external forces and moments, Cij are stiffness parameters, and n is the total number of external forces and moments.

In the case of static loading, the work done by external forces is

( 4.210)

where the components of the forces and the moments are denoted by R i .

The relation (4.210) is valid provided the forces Ri and the displacements have the same directions. In the general case, the work done by external forces can be written as follows:

It is important to note that the potential energy of a system is a homogeneous function of degree two while the work done by external forces is a homogeneous function of degree one.

Recall some basic properties of homogeneous functions. Here we restrict our consideration to polynomials. A function is said to be homogeneous if it can be represented as a sum of terms of the same degree. For example, the function

1 (x, y) = x 2 + 2xy - 3y2

is a homogeneous polynomial of degree two. The substitution of tx and ty for x and y, respectively, is equivalent to the multiplication of the polynomial by t 2 . Similarly, for homogeneous functions of degree m, the substitution of tx for x and ty for y is equivalent to the multiplication by t m , that is,

1 (tx , ty) = en 1 (x , y) .

Suppose that a homogeneous function 1 (x, y, z) of degree m has continuous partial derivatives with respect to x, y, and z. Then, for fixed values of the arguments, we have

1 (txo , tyo, tzo) = en 1(xo , Yo, zo) .

Differentiation with respect to t yields

81 81 81 m-l ) 8x Xo + 8y Yo + 8z Zo = t m1 (xo , Yo, Zo . (4.211)

Setting t = 1 in (4.211) and using the Euler theorem, we arrive at

81 81 81 8x Xo + 8y Yo + 8z Zo = m1 (xo , Yo, zo) . ( 4.212)


The necessary conditions for the functional J1 (see (4.208)) to attain a local extremum are as follows:

aJ1 = (1 + .\) aU _.\ aA = 0, aUi aUi aUi

i=1,2, ... ,n. ( 4.213)

Multiplying (4.213) by Ui and summing the modified equations, we get

= o. ( 4.214)

According to the Euler theorem,

n aU 2: au. Ui = 2U; i=l '

therefore, from (4.214) we get

(1 + .\) 2U - '\A = o. ( 4.215)

By use of the equality A = U, we obtain the Lagrangian multiplier .\ = -2. For elastic systems governed by Hooke's law and subjected to external forces satisfying (4.209) (e.g. conservative forces), we have .\ = - 2. Consequently, we have an extremum problem (not a conditional-extremum problem) for the functional,

J1 = 2A - U = extr . ( 4.216)

We can prove that the extremum of the functional (4.216) is a local maximum, hence the functional

J1 = U - 2A (4.217)

attain a minimum. The equilibrium equations can be obtained on the basis of the principle of minimum of the potential energy (see functional (4.217)) as well as on the basis of the principle of virtual displacements.

The equality J1 = U - 2A = min is efficient for derivation of approximate solutions of problems of statics. Differential equations that appear in the problems of calculus of variations (e.g. equilibrium equations) can hardly be integrated. Hence, it is important to develop methods that do not reduce the analysis of functionals to the analysis of systems of differential equations. Such methods for variational problems are termed direct methods.


4.5.3 Ritz Method

For straight rods, the functional J 1 depends on a rod deflection 11, and its derivatives 11,' and 11,":

J 1 = J 1 (11, , 11,' , 11,") = min,

Let us represent 11, as a series

11, = 2:>jVj(1]),

where aj are constants to be determined; it is sufficient that the known functions Vj (1]) satisfy only the geometrical boundary conditions. (Recall that when we used the methods based on the principle of virtual displacements, it was needed that the approximating functions Vj(1]) satisfy both geometrical and physical boundary conditions.) As such functions, we can take orthogonal polynomials similar to (4.187) and (4.188). Substituting 11, into the functional J1 (4.217), we get

J 1 = J1 (aj) = min;

as a result, we arrive at the following equations in the unknown coefficients aj:

oJl = 0 oaj ,

j=1,2, ... ,n. ( 4.218)

Solving these equations for aj, we readily get an approximate expression for u.

As an example, let us find deflections of the rod shown in Fig. 4.12. Using (4.206) and (4.207), we get

1 11 2" 4 2) () I ( J 1 = - (A33U + k u d( - Pu 1]2 - Tu 1]d. 2 0

( 4.219)

Restrict our consideration to two-term approximation

u = a1 sin 7r1] + a2 sin 27r1] . (4.220)

Substituting (4.220) into (4.219) and using (4.218), we get two independent equations in the unknowns a1 and a2:

~ (7r4 + k4) a1 = P sin 7r1]2 + T7r cos 7r1]1 ;

~ (167r4 + k4) a2 = P sin 27rT/2 + T27r cos 27r1]1

or

2 (P sin 7r1]2 + T7r cos 7rT/1) al = 7r 4 + k4

2 (P sin 27rTJ2 + T27r cos 27r1]d a2 = 167r4 + k4


4.5.4 Approximating Methods Based on Lagrangian Multipliers

Let us consider the method based on the principle of minimum of the potential energy when approximating functions satisfy just a part of geometrical boundary conditions. For a rod shown in Fig. 4.13, we seek an approximate solution as a finite sum

u

k

U = L ai sin i 7r1] .

i=l

(4.221)

Fig. 4.13.

These functions satisfy only two boundary conditions, u(O) = 0 and u(l) = 0, but they do not satisfy the third condition, u'(O) = O. Of course, we could choose the approximating functions such that all the three boundary conditions are satisfied (e.g. the functions defined by (4.187) and (4.189)). Anyway, we use the trigonometric functions (4.221) because they are convenient for algebraic manipulation. To account for the third boundary condition, we write

k

U' (0) = L ai7r cos 7ri1]i1)=o = 0, i=l

i=1,2, ... ,k. ( 4.222)

To take advantage of the method of Lagrangian multipliers, let us consider a functional

h = J l + AU' (0) ,

where A is a Lagrangian multiplier. The unknown coefficients ai can be found from the equations

aJ2 = aJl + A au'(O) = 0 aai aai aai .

The coefficients ai are linear in the multiplier A, that is, ai = aioA + ail. Substituting these expressions into (4.222), we determine A and ai.

Suppose that a rod is not simply supported but clamped at the end 1] = 1 (Fig. 4.13). In this case, the approximating functions satisfy only one of the


three boundary conditions. The function (4.221) must satisfy two additional conditions: u'(O) = 0 and u(l) = O. The functional J 2 takes the form

J 2 = J 1 + A1U'(O) + A2 u '(1).

Thus, t.he use of Lagrangian multipliers substantially extends the class of approximating funct.ions that can be used in the methods based on the principle of minimum of potential energy.

4.6 Stability of Compressed-Twisted Rods

In the preceding sections of this chapter when considering a straight rod in compression-tension, we assumed that the rod remained straight, i.e. an axial force of any direction does not change the straight-line shape of the rod. It is a well-known fact that this assumption is not always valid.

P )P . / I / 1/ i .

Fig. 4.14.

When a compressive force reaches a certain value, the straight-line configuration of a rod becomes unstable and the rod jumps into another equilibrium configuration (see Fig. 4.14). In applications, most structural rod-like elements are loaded by forces of more complicated nature than those in the Euler problem.

A rod immersed into a steady flow of liquid or gas (e.g. a blade, an aerofoil or a wing) is subjected to distributed forces q and a torsional moment /.11 as shown in Fig. 4.15. These loads depend on the angle of rotation of the cross section 19 1 . Therefore, for a critical value of the flow velocity, a loss of bending-twisting stability of t.he rod (a divergence) may occur.

A rod subjected to a magnetic field is shown in Fig. 4.16. When the rod is in the straight-line configuration, the dist.ributed attractive forces of the magnets are of the same magnitude and perpendicular to the rod axis. If the rod deflects from the straight-line configuration, then the attractive forces become different and the resultant force acting on the rod is directed away

4.6 Stability of Compressed-Twisted Rods 187

Fig. 4.15.

a

a

s Fig. 4.16.

from the initial location of the rod. The magnitude of the resultant force depends on the component U2 of the displacement vector u. Under a critical value of the strength of the magnetic field, the rod loses its stability. These examples illustrate feedback distributed forces and moments; q~j) depends only on the component U2 of the displacement vector while J-ll depends on the angle of rotation 79 1 of the cross sections.

The two examples considered above, Figs. 4.15 and 4.16, clearly show us that the loss of stability of a straight rod may occur not necessarily under the action of compressive forces.

--....'" '"

(a)

Fig. 4.17.


Straight rods subjected to a dead bending moment T and a dead force P are shown in Fig. 4.17a and Fig. 4.17b, respectively. Although no internal axial forces arise in the straight rods, a loss of stability is still possible. With an increase of the absolute values IPI and ITI, the rod axis bends into a curve lying in the plane X 10X2. Under the critical values p. and T., the plane configuration of the rod is no longer stable and, hence, the out-of-plane loss of stability of the rod takes place.

f---- 11K -----I

Fig. 4.18.

In structural design, more complicated cases of loading exist, e.g. a rod subjected to a lateral forces q and p(2) and an axial force p(l) is illustrated in Fig. 4.18. The force p(l) may be either compressive or tensile. It should be noted that a non plane mode of loss of stability of a straight rod may be caused by both compressive and tensile forces.

X 1 X 1

jP e20 e20 ~

1 9

I i1 i1

2 X3 X3

(a) (b) (c) Fig. 4.19.


In the examples considered above, the equilibrium equations are referred to the critical state of the rods, thus the equilibrium equations governing the behavior of the rods after the loss of stability can be obtained by direct specialization of the general equilibrium equations (3.5)-(3.9) and (3.10)(3.14) (see Chap. 3).

Fig. 4.20.

To illustrate numerical determination of critical loads, let us consider a straight compressed-twisted rod whose bending stiffnesses are different (A22 :j:. A33)' Some other examples of numerical determination of critical loads are presented in Sect. 5.4. Straight rods subjected to a compressive force P, a distributed axial force g, and a concentrated moment T are shown in Figs. 4.19a-c and 4.20. In the general case, each rod has different bending stiffnesses and arbitrary end fixity conditions (Fig. 4.19c). Bits for deephole drilling (Fig. 4.21) and pipe strings used in boring (Fig. 4.22) can be mathematically treated as rods subjected to compressive forces and torsional moments.

Fig. 4.21.

In spite of a great amount of publications devoted to problems of stability of straight rods, many problems in the field of static stability of rods have not been yet discussed, these being, problems of stability of compressed-twisted rods of different bending stiffnesses. In most papers on problems of stability of


straight compressed-twisted rods, the authors assume that the natural (before the loss of stability) twist of a rod can be neglected. This assumption implies that the torsional stiffness of the rod tends to infinity, while in practice the torsional and bending stiffnesses of a rod are of the same order of magnitude. Hence, torsional stiffness must be taken into consideration.

Fig. 4.22.

The loss of static stability is one of the most undesirable phenomena in applications. Consider a rod shown in Fig. 4.19b. Before the loss of stability, the rod remains straight. Hence, the state vector Z., which characterizes the critical configuration of the rod, has the following components:

Q. = -Q1oel, i.e. Q10 = p.; Q2. = Q3. = 0;

M. = M1oel, i.e. M10 = T.; M 2• = M 3• = 0;

ce. = &1oel ; i.e. &2. = &3. = 0;

here p. and T. are the critical values of compressive force and torsional moment, respectively.

Suppose that the rod is also subjected to a distributed axial force, say, the gravity force g as indicated in Fig. 4.19a. In this case, the internal axial force depends on the axial coordinate 7). Other more complicated cases may appear, e.g. a set of forces and moments is applied to a rod (Fig. 4.20). Some structures may include rigid and elastic local constraints (Fig. 4.21). These constraints may seriously affect the load critical values.

Let us analyze the stability of a compressed-twisted rod subjected to a concentrated force P and an end torsional moment T. Several types of end fixity conditions will be considered (see Fig. 4.19a-c). In addition, the rod may be loaded by the distributed axial force g.


The behavior of the rod after the loss of stability can be studied on the basis of (3.29)-(3.32). On rearrangement, we arrive at the following equilibrium equations referred to the attached coordinate system:

( 4.223)

Hence, Q10 and Mh in (4.223) can be replaced by p. and T •. When, besides the forces mentioned above, the rod is subjected to a dis

tributed axial load qXl and a distributed torsional moment fJI, then, in general, Q10 and M10 are not equal to P. and T., respectively. That is why the coefficients in (4.223) are represented in terms of Q10 and M1o . This representation of the equations is useful for analysis of more complicated problems.

The terms /:::"g2 and /:::"g3 in (4.223) are functions of the distributed axial load g = -gxl i l . These terms allows us to account for the rod weight in the stability analysis.

Recall that the equilibrium equations (4.223) were obtained under assumption that the strain-stress state of the rod before and after a loss of stability obeys the generalized Hooke's law. For M10 = 0, the system of equations (4.223) decomposes into two uncoupled systems. Each system consists of four equations and governs rod equilibrium in the planes X l OX2 and Xl OX3, respectively.

Unlike the commonly used equilibrium equations, the initial twisting is taken into consideration in (4.223).


As noted above, the equality iEh = 0 is used in most publications. Since

Mh iEl* = -A '

11 ( 4.224)

we can see that if Mh oJ. 0, then the condition iEh = 0 is equivalent to the assumption All = 00. The latter equality is in contradiction with common sense because all the three stiffnesses of a real rod (the two bending stiffnesses A22 and A33 and the torsional stiffness All) are of the same order of magnitude.

It should be emphasized that the system of equations (4.223) is valid for rods of varying and constant cross section. For rods of varying cross section, the bending stiffnesses A22 and A33 and the torsional stiffness All

are functions of the axial coordinate 1]. For rods of constant cross section, the non dimensional stiffness A33 is equal to unity (see (1.29)).

To account for the force of gravity g, we include the terms 6.g2 and 6.g3 into (4.223). Let us obtain expressions for these terms. Since the gravity force is a dead force (see Fig. 4.19a), we get

(4.225)

Assume that the angles of rotation are small. In the attached basis, we have

hence,

( 4.226)

In the case of critical load, we arrive at

where

Finally, we get

(4.227)

Because of the gravity force, the internal compressive force Qh varies along the rod as follows:

(0) Q h = p •. ( 4.228)

The first two equations of the system (4.223) take the form


dQ2 Ql* -d - -A M3 - C£hQ3 + 9Xl.1')3 = 0;

7) 33

dQ3 Ql* -d + -A M2 + C£hQ2 - 9Xl. 1')2 = o. (4.229)

7) 22

The other equations of the system (4.223) remain unchanged. Later, when solving (4.223) numerically, we put

l' ~ 0,

hence,

(4.230)

Using the formulas (4.227) for 6.92 and 6.93, we can represent (4.223) in vector form

dY AY = 0 d7) + ,

where

o -1

A= 1 0

o 0

o 0

o 0 o 0

Mh C£h --

A22 1

A22 o o o

o

o 1

A33 o o

The solution of (4.231) is of the form

Y=K(7))C, K (0) = E,

o

o

o 1

o

o

o

o -1 o

(4.231)

o 0

o 0

o 0

o 0

o 0

o 0

( 4.232)

where K(7)) is the fundamental matrix of the homogeneous system of equations (4.231) and C = (C1 , C2 , .•. , Cs)T is a vector with arbitrary constant components.

Let us consider a cantilever beam shown in Fig. 4.19a. Suppose that the beam is clamped at 7) = 0, i.e. Y5(0) = Y6 (0) = Y7 (0) = Ys(O) = O. Hence, C5 = C6 = C7 = Cs = O. The boundary conditions at 7) = 1 depend on the


type of loading. For a dead force p. = - P.i l and a dead torsional moment T. = T.i l , we have

Q2(1) = (P* . e2(1)) = -Po (h, e2(1));

Q3(1) = (P* . e3(1)) = -Po (il . e3(1));

M2(1) = (T •. e2(1)) = T. (il . e2(1));

M3(1) = (T* . e3(1)) = T. (il . e3(1)). ( 4.233)

If the angles of rotation are small, the matrix of transformation from the basis {ij } to the basis {ej} is as follows (see Appendix I, (A.46)):

-rh 1 '19 1 .

1

Consequently, the boundary conditions (4.223) take the form

Q2(1) - P.t93(1) = 0;

M 2 (1) + T.t93(1) = 0;

Q3(1) - P.t92 (1) = 0;

M3(1) - T.t9 2 (1) = 0

or, in terms of the components of the state vector Y,

Yd 1) - p. Y6 (1) = 0 ;

Y3(1) + P.Y5(1) = 0;

Y2 (1) + T. Y6 ( 1) = 0 ;

Y4 (1) - T. Y5 (1) = O.

For a follower load, the boundary conditions at 7) = 1 are

(4.234)

(4.235)

In the case of a dead force P and a follower moment T. = T.el (1), the boundary conditions at 7) = 1 are as follows:

Yl (1) - P.Y6 (1) = 0;

Y2 (1) + P.Y5(1) = 0;

Y3 (1) = Y4 (1) = O. ( 4.236)

In view of the boundary conditions at T} = 0, the solution of (4.232) in coordinate form is

4

Y j = L kjiCi , j=1,2, ... ,8. (4.237) i=l

Therefore, we can rewrite the boundary conditions (4.232) for dead loads as follows:

4

L(k1j - P.k6j ) Cj = 0; j=l

4

L(k3j + T.k6j ) Cj = 0; j=l


4

L(k2j + P.ksj ) Cj = 0; j=l

4

L(k4j - T.ksj ) Cj = O. j=l

(4.238)

For the linear algebraic system (4.238) to have nontrivial solutions, the determinant of the coefficients C i must vanish, that is,

D (P. , T.) = O. ( 4.239)

This relation allows us to determine p.j as functions of T •. The roots of (4.239) are P.j(T.). These roots are termed eigenvalues of the boundaryvalue problem. The least root is called the critical force. Equation (4.239) has an infinite set of roots; in general, these roots can be determined only by use of numerical methods.

Further, we will give examples of determination of critical loads for various types of external loads. Emphasize is on the analysis of the influence of the initial twist ceh 'I- O. Equations (4.223) are written in terms of nondimensional quantities. Hence, Po} and T. are also nondimensional. To evaluate the corresponding dimensional force and moment, we should multiply P. j

and T. by A33(0)/12 and A33 (0)/I, respectively (here, A33(0) is the dimensional bending stiffness and I is the rod length).

As a numerical example, consider a steel cantilever beam of constant cross section the following parameters: I = 0.7 m, h = 0.7 . 10-2 m, b = 0.5 . 10-2 m, E = 2· lOll N . m- 2 , and G = 8 . 1010 N . m- 2 . The values of nondimensional stiffnesses are All = 0.577, A22 = 0.5, and A33 = 1.

The plots of the first two eigenvalues P. 1 versus the magnitude of the torsional moment T = ITI are shown in Fig. 4.23a. The force P and the moment T are dead. The plots of P. 3 and p. 4 versus T are illustrated in Fig. 4.23b.

Similar plots can be obtained for other pairs of eigenvalues. The common property of the plots is that as the magnitude of T is increased, the curves get closer and finally merge at T = Tnj . In general, for different pairs, the corresponding values Tnj are also different. It should be noted that any two successive eigenvalues P.(2j-l) and P'2j ' j = 1,2, ... , are related in such a manner.

The plots for Po3 and P. 4 are of theoretical rather than of practical interes~. Nevertheless, the eigenvalues Poj , j = 2, 3, ... , are useful because they allow us to determine the eigenvectors z(j)(T/). These vector-functions satisfy the boundary conditions and can be used in the derivation of approximate solutions of linear and nonlinear problems of statics and dynamics.

The least eigenvalue p~~) corresponds to the loss of stability in the plane Xl OX3. This plane in its turn corresponds to the least bending stiffness of the rod (see Fig. 4.19a). If the rod is subjected to a compressive force P of


P A(O) *2

2.4

2.0

1.6

1.2

0.8 0

P

22

20

18

16

14

12

10

FlO) -!2

riO) *2

II F;2(7;.)

0.2

...... ~

-

a:! ** 0

"'

0.4 Tn2 Tn1

(a)

r-.... a:!1*= 0

~ ~ a:!1* * 0 ~ ~

\ \ _V -- ..--'

Tn1

o 0.5 1.0 1.5 2.0 (b) Fig. 4.23.


slow growth, then, as soon as P reaches the value p;~), the loss of stability occurs. If after the loss of stability the force P continues its growth, the rod will bend in the plane Xl OX3' Therefore, the loss of stability of a cantilever beam in the plane Xl OX2, which is the plane of the largest bending stiffness, can hardly take place in practice.

Let us consider in detail some numerical results (see Fig. 4.23a). If &b 1:- 0, then a region of values of P and T at which straight line configuration of the rod is stable is shown by hatching (region 1). The boundary points of the region correspond to the critical state of the rod when straightline configuration is no longer stable. For P > P*l and some fixed value of T, the rod jumps (bifurcates) into another equilibrium configuration (region II).

Linear equilibrium equations allow us to determine the eigenvalues P*j of a boundary-value problem; however, the parameters of a deformed configuration of the rod after the loss of stability (the displacements U2 and U3 and the angles of rotation {)2 and {)3 of the attached axes) cannot be obtained from linear equations. That is why, to investigate the strain-stress state of a rod after the loss of stability, one should use nonlinear equilibrium equations, say, (4.40)-(4.44).

As noted, for each pair of eigenvalues (P*(2j-I), P*2j) there exists a certain limit moment Tnj , say, Tnl in Fig. 4.23a. If T < Tnl , (4.239) has a nonempty set ofroots, i.e. static loss of stability is possible. IfT > Tnl (region III), there is no eigenvalues such that P*l S P S P*2. Hence, for T and P from region III, the static loss of stability can never appear, i.e. the rod has no equilibrium configurations close to the straight-line configuration. Therefore, for P and T belonging to region III, the behavior of the rod can be described by equations of small oscillations; at that, a dynamic loss of stability is possible. In greater detail, the outlined problems are discussed in Svetlitsky (1987) devoted to the dynamics of rods.

Plot 1 in Fig. 4.23a corresponds to the case of an infinitely large torsional stiffness (&b = 0). Plot 2 corresponds to the case of a finite value of torsional stiffness (&b 1:- 0). The dashed curve corresponds to the case of a dead force P and a follower moment T. For cantilever beams, the values of critical load are insensitive to the behavior of follower moments.

The fact that the real properties of a rod (&b 1:- 0) are taken into consideration enables us to determine the critical loads with a higher accuracy. For example, for T = 0.56, the critical value P*l determined for the case &h 1:- 0 is 10% less than the critical value obtained under the condition &b = O. Of special note is the fact that the value of the critical force for &h = 0 exceeds the value of the critical force for a real rod (&h 1:- 0). Such errors are impermissible in design of optimal structures, in which operating characteristics of the elements are selected to be very close to their critical values.

The plots of P*l and P*2 versus T = ITI for A22/A33 = 0.125 are shown in Fig. 4.24. Recall that P = IPI. For P and T from the hatched region, the static loss of stability of a rod do not occur. Taking into consideration


2. 0 I__----l--~~I------"'.....,.._+_---I__--__l

1.5 �__----l----I----..:~_+_--~I__--__l

1.01----~---~--~~--1---~-

o ~~~~~~LU~~~~~~~~~ o 0.2 0.4 0.6 0.8 Tn1 T

Fig. 4.24.

the initial twist, we can essentially enlarge the region of static stability, i.e. a rod remains stable under torsional moments which considerably exceed the previously obtained critical value. Indeed, the critical moment Tnl for reI< f:. 0 differs by 30% from the moment Tn2 for reI< = O. For instance, from Fig. 4.24 it follows that for the values of T and P corresponding to the point A the rods keeps its straight configuration while computations not accounting for the initial twist show us that for these values of T and P the loss of stability has already happened.

p--~-~--~

R(O) *2

3.01----+---+-=''..---+-

R(O)

* 1

1 . 5 ~-+---!-+-+---+-~-+--1 I I

,..,(0) I __ --I--r-r-'*1 1 .0 '-------'---U I

o 0.2 Tn 0.4 Tn Fig. 4.25.


Now let us consider the relationship between the stability of compressedtwisted rods and the ratio of bending stiffnesses A22 / A33 = k. For dead loads P and T, the plots of eigenvalues of P = IPI versus T = ITI for some values of k are illustrated in Fig. 4.25. An interesting feature of the plots is that as k approaches unity, the eigenvalues p;~) and P;g) get closer and merge at k = 1 while the hatched regions in Fig. 4.24 collapse into a point C. The critical value of the moment decreases continually as k tends to unity and vanishes for k = 1. Let us examine the situation for k = 1 in greater detail. In this case, the stiffnesses in the planes X l OX2 and Xl OX2 are equal, hence, the eigenvalues are also equal. Suppose that the rod is subjected to a torsional moment T such that the magnitude of the moment is as small as is wished; then, for k = 1, the rod has no static equilibrium configurations. Indeed, any infinitesimally small torsional moment does exceed the limit value because it equals zero (see Fig. 4.25). Consequently, (4.239) has no roots and thus there is no adjacent equilibrium configurations. In this case, the loss of dynamic stability of the rod occurs.

The special case k = 1, when a rod is subjected to a compressive force and a torsional moment, is known as the Nikolai paradox (Bolotin (1963)). Actually, the paradox can be explained if we assume that the concept stability implies not only the transition from an unstable equilibrium state to a stable one but also the transition from an unstable equilibrium state to a motion with respect to this equilibrium state.

Fig. 4.26.

A similar paradox appears in the stability analysis of a straight cantilever beam of an arbitrary ratio k subjected to a compressive follower force P (see Fig. 4.26). The equation, which is a counterpart of (4.239), has no roots.


As mentioned above, the roots (eigenvalues) are characteristics of possible equilibrium configurations. Thus, we have proved that the straight-line equilibrium configuration is stable irrespective of the follower force P. Considering small oscillations of a rod subjected to a follower compressive force, we can show that that at a certain value of the force the straight-line configuration becomes unstable and the rod breaks into oscillation, i.e. the loss of dynamic stability occurs. An overview of nonstandard problems of elastic stability of rods is presented in Bolotin (1963) devoted to the theory of dynamic stability of rods.

For some non dimensional values of the dead distributed load gXl (see Fig. 4.19a), the plots of the first two eigenvalues F. 1 and F. 2 are illustrated in Fig. 4.27. The curves obtained with regard for the initial twist are shown as solid lines and those obtained with no regard for the initial twist are shown as dashed lines. The loads P and T are dead.

1 . 5 1---+-+--+-----f~""'k:.~.t+-__\t-\-----1

1 . 0 f---r+--+--+-----+-..-.~=-+-+_+-+----1

0.5bd~:::±::1:I~~~ o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 T

Fig. 4.27.

Solving for the eigenvalues F. j , we obtain the corresponding eigenvec

tors ybj ). For each eigenvalue F. j , we get the following equation:

dy(j) () -dO +A[7], F.j(T')]YoJ = 0;

T)

here Ybj ) is the corresponding eigenvector.

( 4.240)

The eigenvalues F. j are determined from (4.239), which is the characteristic equation for (4.238). Substituting the roots F.j(T.) into (4.238), we get


4 4

"'(kU) - P ·k(j») C(j) = O· ~ 11/ 'J 61/ 1/ ,

"'(k(j) + P ·k(j») C(j) = O· ~ 21/ 'J 51/ 1/ ,

1/=1 1/=1

4 4

"'(k(j) +T ·k(j»)C(j) =0' ~ 31/ 'J 61/ 1/ , L(ki~ - T.jk~~) cy) = 0, (4.241) 1/=1 1/=1

where kW are the elements of the fundamental matrix of the vector equation (4.240). In view of the first three equations of the system (4.241), we can write

(4.242)

Setting ci j ) = 1 and using (4.237), we have

4

YoY)(TJ) = Lki~)(TJ)al/' i=I,2, ... ,8. ( 4.243) 1/=1

This procedure of determination of eigenvectors can be applied to any boundary conditions and loads under which the static loss of stability is possible.

Rods Simply Supported at 1] = 1 A rod can be simply supported at TJ = 1 in two ways as indicated in Fig. 4.19c. Restrict our consideration to case 1, i.e. let us study an in-plane loss of stability of a rod which is simply supported in the plane Xl OX2; let the rod be subjected to a dead load. We have, then, the following boundary-condition equations at TJ = 1:

or

Q3(1) + P.'l9 2 (1) = 0;

M3(1) - T*'l92 (1) = 0;

Y2 (1) + P.Y5(1) = 0;

Y4 (1) - T.Y5(1) = 0;

M 2 (1) + T.'l93 (1) = 0;

u2(1) = 0

Y3 (1) + T.Y6 (1) = 0;

Y7 (1) = O. (4.244)

Plots of the first two eigenvalues P*j versus T = ITI for A22/ A33 = 0.5 are shown in Fig. 4.28 for the cases reI> = 0 and reI> :f O. We clearly see that these plots P. j = p. j (T) differ qualitatively from those for a cantilever beam (see Fig. 4.23a). If a continuously growing torsional moment T is applied to the rod, then the first eigenvalue p.1 also increases at the beginning but then starts decreasing and, finally, at a certain T;~), vanishes. The value of

T;~) that corresponds to the case reI> = 0 is different from that in the case reI> :f O. Thus, the loss of stability may be caused by the torsional moment T;~) alone (for P = 0). The region of values of P and T which do not cause the loss of stability is hatched (see Fig. 4.28). If the initial twist is not taken


p~--~---.--~,----,----,

10~--+-~-P~-+~~r---+

8~--~--~----~--~----r

2~~~X+-+--~r---+----ft--~

p'(O)'~'77~~~~~ *1 i/o

2 Fig. 4.28.

into consideration, then the corresponding critical value of the moment is 25% above the real critical value (for P = 0). Another qualitative difference between the plots in Figs. 4.28 and 4.23a is that in Fig. 4.23a the eigenvalues get closer as T increases and merge at T = Tn. The fact is that a torsional moment alone cannot cause a loss of stability of a cantilever beam.

4.7 Stability of Straight Rods with Local Constraints

A straight rod with elastic and hinge local constraints is shown in Fig. 4.29. Some basic types of local constraints are illustrated in Fig. 4.29b. For simplicity, we will consider a rod with only two hinged cross sections TJl and TJ2 (variant 1). The loss of stability results in reaction forces R{l) and R(2). The directions of the reaction forces are unknown a priori. Nevertheless, for small angles of rotation of the attached coordinate axes, the reaction forces are assumed to lie in planes orthogonal to the rod axis after the loss of stability. To account for the reaction forces R{l) and R(2) in the equilibrium equations (4.231) after the loss of stability, we will use the is-functions iSj = is (TJ - TJj). Then, we can write

dY d;] + AY = b,

where

R (l) - R(l)e + R(l)e . - 2 2 3 3,

(4.245)

R (2) - R(2)e + R(2)e - 2 2 3 3·

4.7 Stability of Straight Rods with Local Constraints 203

(a) (b)

Fig. 4.29.

Equation (4.245) has a solution

Y = K (1]) C + lory G (1] , 0 b (() de (4.246)

where G(1] , 0 is a Green matrix

K (0) = E.

For rods of constant cross section, the Green matrix takes the form

(4.247)

Let us consider in detail the term containing the vector b in the right-hand side of (4.246),

Y H = IoryK(1]-()b(Od(.

The right-hand side of (4.245) can be represented as a sum

b = b 1 61 + b2 62 ,

where

_( (1) (1) )T. b 1 - - R2 ,-R3 ,0, 0, 0, 0, 0, 0 , _ ((2) (2) )T b 2 - - R2 ,-R3 ,0, 0, 0 , 0 , 0, 0 .

Substituting the vector b into (4.248) and integrating, we have

where H j = H (1] - r/j) is a Heaviside function. As a result, the vector Y takes the form

(4.248)

(4.249)

(4.250)

(4.251)

For brevity, let us denote the elements of the matrices K (1]), K (1] - 1]d, (1) (2) . and K (1] - 1]2) by k ij , k ij , and k ij , respectIvely.


The components of the vector Yare as follows:

8

Yj (7]) = L kjvCv - (kji) R~I) + kJ;) R~I)) HI - (kJ~) R~2) + kJ;) R~2)) H2 . v=1

( 4.252)

Each component Yj contains twelve unknowns: Cv (v = 1, 2, ... , 8),

R~l), R~I), R~2), and R~2). These unknowns are to be determined from the boundary conditions at 7] = 0 (four equations) and at 7] = 1 (four equations) and from the conditions at 7] = 7]1 (two equations) and at 7] = 7]2 = (two equations) .

At 7] = 7]1 and 7] = 7]2, we get

U2(7]d = Y7(7]d = 0;

U2(7]2) = Y7(7]2) = 0;

u3(7]d = YS(7]d = 0;

U3(7]2) = YS (7]2) = 0;

In expanded notation, we have

4

Y7(7]d = L k7j (7]I) Cj = 0; j=1

4

YS (7]d = L kSj(7]d Cj = 0; j=1

4

Y7 (7]2) = L k7j (7]2) Cj - kW (7]2 - 7]d R~I) - kW R~I) = 0; j=1

4

V' ( ) - "k ( ) C k(1)( ) R(I) k(l)R(l)-.l S 7]2 - L-- Sj 7]2 j - SI 7]2 - 7]1 2 - S2 3 - O.

j=1

(4.253)

( 4.254)

For a rod subjected to a dead load (Fig. 4.29), we have the following boundary conditions at 7] = 0: 1}2 = Y5 = 0, 1}3 = Y6 = 0, U2 = Y7 = 0, and U3 = Ys = 0; hence, C5 = C6 = C7 = Cs = O.

The boundary conditions at 7] = 1 have been already obtained (see (4.235)). Using (4.251), we can represent (4.235) as follows:

4

"( ) C ((1) (1)) (1) ((1) (1)) (1) L-- klj - P.k6j j - kll - P.k61 R2 - k12 - P.k62 R3 j=1

_ (k(2) _ P k(2)) R(2) _ (k(2) _ P k(2)) R(2) - O' II • 61 2 12 • 62 3 - ,

4

L(k2j + P.k5j) Cj - (kg) + p.k~i)) R~I) - (kg) + P.kg)) R~I) j=1

_ (k(2) + P k(2)) R(2) _ (k(2) + P k(2)) R(2) - O' 21 • 51 2 22 • 52 3 - ,


4

'l)k3j + T.k6j ) Cj - (kg) + T.k~i)) R~l) - (kg) + T.kg)) R~l) j=l

- (k(2) + T k(2)) R(2) - (k(2) + T k(2)) R(2) - O· 31 • 61 2 32 • 62 3 - ,

4

'l)k4j - T.k5j ) Cj - (kii) - T.kg)) R~l) - (kW - T.kg)) R~l) j=l

_ (k(2) _ T k(2)) R(2) - (k(2) - T k(2)) R(2) - 0 41 • 51 2 42 • 52 3 - .

Combining (4.254) and (4.255), we get

BZ =0,

where

16r-----r-----r-~~~--

p

14~--~----~-----+~--~----~

12~--~----~-----+~~~~--~

0.5 1.0 1.5 2.0 T Fig. 4.30.

(4.255)

( 4.256)

By use of the equation det B = 0, we can determine the eigenvalues P. j

as functions of T, TJ1, and TJ2. Equations (4.245) were solved for the numerical values of the physical

parameters taken from the previous example. The relationships between the first two eigenvalues and the moment T

were obtained numerically for the following cases: (1) the dead force and the dead moment (Fig. 4.30); (2) the dead force and the follower moment (Fig. 4.31). The plots were obtained for TJ1 = 0.1 and TJ2 = 0.9 both with regard for the initial twist (<£10 ::j:. 0) and with no regard for the initial twist (<£10 = 0). From the plots it is clearly seen that the eigenvalues are insensitive


to behavior of the torsional moment while the initial twist may seriously affect the eigenvalues. For example, the limit value Tn2 is 20% higher than the limit value Tnl (Fig. 4.30).

P

18

16

14

12

10

8 0 0.5 1.0 1.5 2.0 T Fig. 4.31.

Let us consider the values of the parameters corresponding to the point A. Considering the plots corresponding to the case <Eh = 0, we conclude that the rod is in a stable configuration, while exactly the opposite conclusion follows from the plots corresponding to the case <Eh :j:. O.

Let us consider a rod with a local elastic support (see Fig. 4.29, variant III). Of course, this elastic support is a simplified model of a real support. Assume that the radial stiffness c of the support is the same in all directions perpendicular to the rod axis. This model may give a better insight into the role of local constraints.

Considering small displacements of the axial points, we can assume that the vector u lies in a plane perpendicular to the axial line. Hence, the reaction R, which appears after the loss of stability, is proportional to the displacement vector of the point K, that is,

( 4.257)

where c is the stiffness of the elastic support. To derive the equilibrium equations, we should add the vector b(1)

bel) = (-R2 8, -R3 8, 0,0,0,0,0, O)T

= (CUk2 8, CUk3 8,0,0,0,0,0, O)T (4.258)

to the right-hand side of (4.245). The solution of (4.245) in which the vector b is replaced by the vector b(l) is identical in form to (4.251), that is,

Y = K (17) c + K (17 - 17k) b(l) H . ( 4.259)


For a cantilever beam subjected to a dead load, the boundary conditions at TJ = 0 and TJ = 1 are the same as those from the preceding example (see (4.235)).

Taking into account (4.258), we obtain the following boundary-condition equations (see (4.255)):

j=l

4

"( ) G (k(l) P k(l)) (k(l) P k(l)) - O· ~ k 2j + P.k5j j + 21 + • 51 CUk2 + 22 + • 52 CUk3 - ,

j=l

4

"(k T k ) G (k(l) T k(l)) (k(l) T k(l)) - O· ~ 3j + • 6j j - 31 + • 61 CUk2 + 32 + • 62 CU k3 - ,

j=l

( 4.260)

In the case under consideration, the conditions at TJ = TJK differ from those in the previous example of hinge supports. It is necessary to consider these conditions in greater detail. For the case of hinge supports, after the loss of stability, the displacements of the points TJ = TJ1 and TJ = TJ2 equal zero, hence, the displacements are known. In the case of elastic support, the displacements UK2 and UK3 are unknown. Consider the expressions for U2 and U3 (they coincide with the components Y7 and Ys of the vector Y, respectively). We have

or

4

Y 7 (TJk) = Uk2 = L k 7j (TJk) Gj ;

j=l

4 4

4

Ys(TJk) = Uk3 = L kSj(TJk) Gj

j=l

L k 7j (TJk) Gj - Uk2 = 0; j=l

L kSj(TJk) Gj - Uk3 = O. (4.261) j=l

As a result, we have obtained the system of six homogeneous equations (4.260) and (4.261) in the six unknowns G1 , G2 , G3 , G4 , UK2' and UK3.

For a nontrivial solution to exist, the determinant of this system must equal zero. This allows us to find the eigenvalues P. j as functions of the moment T and of the stiffness of elastic support c.

The methods presented in this section enable us to examine the stability of rods for an arbitrary set of local constraints.


4.8 Problems

4.1. A rod of constant cross section is subjected to a follower force Po permanently directed at a fixed point 0 as illustrated in Fig. 4.32. Find a one-term approximation for the displacement of the point K using the principle of virtual displacements. The magnitude of Po remains unchanged. Assume that the displacements of the axial points are small.

o Xz

i, '1

Fig. 4.32.

4.2. By use of the principle of minimum of potential energy, determine the displacements of the point K in Problem 4.1.

4.:1. A rod of constant cross section is subjected to a distributed load of constant magnitude as shown in Fig. 4.33. The load vectors are permanently directed at a fixed point 0 with coordinates '1]0 = 0.5 and X20 = 0.5. Determine the displacements of the point K. Use the principle of virtual displacements.

'10 '1

Fig. 4.33.

4.4. Determine the displacement of the point K ill Problem 4.3.' Use the principle of minimum of potential energy.

4.8 Problems 209

1 Fig. 4.34.

4.5. A rod of constant cross section resting on a linearly elastic foundation is shown in Fig. 4.34. The rod is subjected to a uniformly distributed load. Determine the displacements of the axial points.

4.6. A rod of constant cross section and a linearly elastic foundation are in contact at 0 :s: T} :s: 0.5 as shown in Fig. 4.35. Determine the displacements of the axial points. Use the principle of virtual displacements.

Fig. 4.35.

4.7. A rod of constant cross section (Fig.4.36) subjected to a distributed load rests on an elastic foundation. The nonlinear characteristic of the foundation is as follows: qr = 4o:iu X2 + 'YU~2' Obtain a one-term approximation for the rod deflections.

Fig. 4.36.

4.8. Assume that displacements of the axial points are small (see Fig. 4.32). Determine numerically the stress-strain state of the rod on the basis of the equilibrium equations. Put !Pol = 0.5. Use the equations of the zeroth approximation.

4.9. Derive the equilibrium equations of the zeroth and the first approximation for a rod shown in Fig. 4.32. Obtain the numerical solution for IPol = 0.5 and T}K = 0.5. Estimate the error.


4.10. Derive the equilibrium equations for a rod shown in Fig. 4.32 for large displacements of the axial points. Obtain the numerical solution of the nonlinear equilibrium equations for the case IPol = 2 and T/K = 0.5. Use the method of step-by-step loading.

4.11. Obtain the numerical solution of the equilibrium equations of the zeroth approximation for a rod shown in Fig. 4.33. Put qo = 1.

[Foundations of Engineering Mechanics] Statics of Rods || Straight Rods

Documents

Transcript of [Foundations of Engineering Mechanics] Statics of Rods || Straight Rods