Structural mechanics

cases the initial and final positions of the beam coincide, then W1 W2. So, we have the equation:

In expanded form:

Fi ik Fk ki .

A formal record of the theorem of reciprocity work is obtained (Betty's theorem (1823–1892)): the work of the forces of the state “i” on the displacements of the state “k” is equal to the work of the forces of the state “k” on the displacements of the state “i”.

Note, that in the above formulation, the term “force” should be understood as “generalized force”, which can be a group of forces, and the term “displacement” as “generalized displacement”.

A similar dependence exists for the virtual work of internal forces on the corresponding deformations. Then the statement of the theorem of reciprocity work can be given in the following form: the virtual work of the external (internal) forces of the state i on the displacements (deformations) of the state k is equal to the work of the external (internal) forces of the state k on the displacements (deformations) of the state i.

E x a m p l e. A beam (Figure 7.12) of a constant section in state 1 is loaded with a uniformly distributed load of intensity q, and in state 2 it

is loaded with a concentrated moment M applied at the end point. Show the validity of the theorem of reciprocity work.

Figure 7.12

171

The generalized force in state 1 is the load q. Its virtual work is defined as the sum of elementary works of the forces q dx on the displacement y2 of the state 2:

l l

W12 q dx y2 q y2 dx q ,

0 0

where is the area of the diagram of the vertical displacements of the beam in the state 2.

To determine we find the equation of the bended axis of the beam. The differential equation of the bended axis is written in the form:

we find:

172

The virtual work is:

W 12 q Ml3 . 24 EJ

The virtual work of the concentrated moment M is W21 M B.

Displacements and angles of rotation of the beam in state 1 are determined from the equations:

The direction of action of the moment M coincides with the direction of displacement B , therefore:

W21 M q l3 . 24 EJ

Consequently, W12 W21.

If the generalized forces in the states “ i ” and “k” are equal to one (displacements from unit forces are indicated by the symbol δ, Figure 7.13), then it follows from theorem (7.4) that:

Figure 7.13

173

Equality (7.5) expresses one of the general properties of linearly deformable systems and is a formal record of the theorem of reciprocity displacements (Maxwell's theorem (1831–1879)): displacement in the i-th direction from the k-th unit force is equal to displacement in the k-th direction from the i-th unit force.

Remark on the dimension of displacement ik . The generalized displacement ik , caused by the generalized force Fk , is defined asik ik Fk . Therefore, the dimension of displacement ik is obtained in the form:

dimension of ik = dimension of ik . dimension of Fk

For example, when loading the beams shown in Figure 7.14, we have:

21 21 , dimension of 21 = rad/kN= kN–1; F1

12 12 , dimension of 12 = m/( kN·m) = kN –1. F2

Figure 7.14

Displacements 12 and 21 have the same dimension.

7.6. General Formula for Determining Plane Bars System Displacements

Suppose that the bars system (Figure 7.15, a) has been deformed under the influence of given actions, and it is required to determine the displacement of any of its points i in a predetermined direction that does not necessarily coincide with the true direction of displacement of this point.

174

Considered system state we denote as a “state a ”, and internal forces in

deformations in the infinitesimal element of this system in the deformed state. Here, dx is the length of the element, is the relative elongation

(shortening) of the element, 1 is the curvature of the bended axis,

– is the relative shear (angle of shear) of the edges of the element.

To determine the required displacement we consider the auxiliary (fictitious) state of the system. In this auxiliary state, we attach a unit generalized force to the same system in the direction of generalized unknown displacement (Figure 7.15, b).

Figure 7.15

The internal forces in this state (state i ) of the system are denoted by Ni , Mi , Qi . Since this state is a state of equilibrium, the principle of

virtual displacements can be applied to it. For virtual displacements, we take the displacements caused by a given action. The total work of the external and internal forces of the state i on the displacements of the state a should be equal to zero (7.3), that is:

W (virt) Aint(virt) 1 ia Ni dx Mi dx Qi dx 0

Integration is carried out along the length of each bar or section of the bar, during which the integrand is a continuous function of a certain kind.

175

Consequently,

The obtained formula allows us to express the required displacement through deformations of the system elements in the state a, and the sys-

tem itself can be both linear and physically nonlinear. The cause of the deformation of the elements is also insignificant: force impact, change in ambient temperature, creep of the material or other reasons. Therefore, formula (7.6) can be considered as a general formula for determining the displacements of bars systems.

The state of the system under the action of a given load is called loaded state (state F). From the course of resistance of materials it is known that

the deformations of elements of a linearly deformable system in this state are determined through internal forces as follows:

where EA , EJ , GA – the rigidity of the element, respectively, in tension

(compression), bending and shear.

Substituting these expressions in (7.6), we obtain a formula for determining the displacements of a plane bar system in the following form:

This formula is called the Maxwell-Mohr formula for determining the displacements of elastic systems caused by a given load.

The relative contribution of each of the three terms of formula (7.7) to the final result depends on the type of the bars system and the nature of loading. In particular, it appears that the displacements in the beams depend mainly only on the second term (bending moments); the proportion of the term, taking into account the influence of shear forces, is a negligible fraction of the final value iF . Therefore, with sufficient ac-

176

curacy for practical purposes, the displacements of systems that primarily perceive bending can be calculated by the formula:

iF Mi M F dx.

EJ

For the same reason, the calculations (especially “manually”) of the frame and arch systems neglect the influence of longitudinal and shear forces in determining displacements. At the same time, the automated calculation of these systems using computer programs is carried out, as a rule, taking into account bending moments and longitudinal forces in determining displacements.

In elements of trusses with hinged joints only longitudinal forces arise from the node loads. Therefore, the determination of the displacements of nodes in the trusses is made according to the formula:

Since with a nodal load on the truss, the longitudinal force along the length of the rod does not change, then, provided that the rigidity of each rod is constant, the formula is rewritten in the form:

where lk – the length of the k-th rod;

n – number of truss rods.

In this form (7.8), for the first time in 1864, J. Maxwell obtained a formula for determining the displacements of trusses. 10 years later, O. Mohr (1835–1918) developed a method for determining displacements for the case of arbitrary deformations of the system (see formula (7.7)).

Let us explain the features of the choice of the auxiliary state. A single generalized force must be applied to the system in the direction of the corresponding generalized displacement. Their product, as you know,

177

gives the work of force F 1 on the required unknown displacement. If, for example, for the frame in the state F (Figure 7.16, a), it is necessary to determine the angle of rotation of any cross-section of the element, for example, the cross-section D , then in the auxiliary state in this section it is necessary to apply a single concentrated moment M1 1 (Figu-

re 17.16, b), and then the virtual work of the external force in the state “1” on the displacement of the state “F” will be equal M1 1 1F .

Subsequently, the index of the unit load in the auxiliary state will determine the number of this state.

Figure 7.16

If it is necessary to determine the change in the distance between points k1 and k2, then in the auxiliary state (state 2) two unit forces directed in opposite sides should be applied along the direction of the line connecting these points (Figure 7.16, c); if it is necessary to find the

angle of mutual rotation of the cross-sections с1 and с2, then in the auxiliary state (state 3), two opposite-directional moments should be applied in these cross-sections (Figure 7.16, d) each being equal to the unit.

The directions of unit forces given in auxiliary states correspond to positive directions of displacement iF . If the result of the calculation is

iF 0, then it will mean that the required displacement is directed in the direction opposite to the direction of the force Fi 1.

178

7.7. Mohr Integrals. Ways for Calculation

The problem of calculating displacements using the Mohr's formula reduces to calculating integrals of the form

b Mi M F dx,EJ

a

which are commonly called Mohr integrals. For relatively simple problems, the integrant

f (x) Mi M F

EJ

can be such that the indefinite integral F(x) can be expressed using a

finite number of elementary functions. Then a definite integral is calculated by the formula

b

f (x) dx F (b) F(a).

a

Let us show, for example, the determination of the vertical displacement of cross-section 1 and the angle of rotation of cross-section 2 of the cantilever beam (Figure 7.17), loaded with a uniformly distributed load. The influence of only bending moments to deflection only we will take into account.

Figure 7.17

179

To determine the deflection, we use the auxiliary state 1. In the future, the designations of forces from dimensionless forces will be accompanied by the upper line. Then:

M F 0.5qx2 , M1 1 x.

To determine the angle of rotation of the cross-section in the middle of the beam, we use the auxiliary state 2. Then:

For the same example, when calculating the area of the deflection diagram using the auxiliary state 3 (the beam is loaded with a unit uniformly distributed load), we obtain:

180