Structural mechanics

2. Using the dependences (15.22), (15.23) and (15.24), we represent the displacements vector

a is the vector of unknown parameters:

a a1 a2 a3 a4 a5 a6 T .

3. For the edge cross-sections (x = 0, x = l) of the bar, using expressions (15.25) and (15.26), we obtain:

471

4. From (15.27) it follows that:

472

The expressions for u and coincide with those recorded earlier in section 15.7.

Thus, using (15.22) and the approximating polynomial (15.23), exact functions are obtained that allow one to calculate the horizontal displacement, deflection, and rotation angle of any cross-section of the bar.

6. Nodal displacements Z1 , Z2 , Z3 and Z4 , Z5 , Z6 of the bar AB correspond to the reactions R1 , R2 , R3 and R4 , R5 , R6 , allowing to find the forces N , Q and M in the edge cross-sections. To determine them, we use differential dependencies:

Differentiating in the expression (15.25) the first and third rows one time, and the second row – three times, we find the components of the vector

To determine the forces in the edge cross-sections of the bar, we form a matrix BАВ , the first three rows of which correspond to the matrix B for

x 0, and the rest – for x l .

473

Ddiag EA, EJ , EJ , EA, EJ , EJ .

7.The directions of positive efforts N , Q and M in the edge cross-

sections of the bars (vector SАВ) and positive reactions R1 , R2 , R3 , R2 , R5 , R6 (their directions coincide with the directions of the compo-

nents of the vector S) do not coincide. The relationship between them

can be established using correspondence matrix of the efforts signs by the expression:

474

From (15.33) it follows that the reactions matrix is determined by the expression:

In the case under consideration, the reactions matrix (which is also the stiffness matrix) has the form shown earlier in formula (15.21).

The second method to obtain the stiffness matrix of the bar, based on the use of the Lagrange principle, is as follows.

With the well-known equation of the bended axis (see section 15.7) of the bar, its rigidity matrix can be obtained from the total energy stationary condition.

Let us write the expression of the total energy for a bar clamped (fixed) at its ends and loaded with a distributed load:

After the necessary transformations, we obtain:

475

In this expression, the total energy is represented by a function of six independent variables.

Integrals of the form

l

q x fidx

0

give us the values of the support reactions for the beam pinched at the ends when loading it with a distributed load q x . To prove this statement, we consider the same beam (Figure 15.13) in two states: in the state a, the beam is loaded with a distributed load q x , in the state b the left end of the beam is moved by Z2 1. Remember that the positive directions of reactions correspond to the positive directions of displacements.

Figure 15.13

476

The virtual work of the forces of state a on the displacements of state b is:

The work of the forces of state b on the displacements of state a is zero:

Wba 0 .

Based on the reciprocity theorem:

When calculating M A for an auxiliary state, we will take what is

shown in table. 15.1 (see section 2). Having determined the possible work of the forces of one state on the displacements of another (in the forward and reverse directions), based on the reciprocity theorem, we obtain:

477

Note that when a different type of load (concentrated forces, moments, etc.) is applied to the bar, the support reactions can also be calculated using the reciprocity theorem.

The necessary conditions for the minimum function (15.36) of six variables are:

Applying them to the expression (15.36), we obtain a system of equations relating the displacements and reactions of the edge cross-sections:

In matrix form, the system has the form:

Another way to solve this problem is based on the general equations of structural mechanics. Given the initial conditions, we write the bar equilibrium matrix in the form:

478