Structural mechanics

14.15. Influence Matrices for Displacements and Efforts

From the equality

D R 1

it follows that the displacement vector z can be expressed through the load vector F according to the formula

z DF.

Therefore, the external compliance matrix D inverse to the external rigidity matrix R , is also an influence matrix of displacements Lz , i.e.

z DF Lz F ,

where

Lz D R 1 ( AKAT ) 1 [ ij ]

node in the ith direction from the unit force Fj 1 .

The force vector S in a bars system can also be expressed in terms of the vector F . For this purpose, we write it first in the form

S KAT z ,

and then, using the expression for z, we represent it in the form

S K AT R 1 F K AT A K AT 1 F LS F,

where LS is an influence matrix of efforts.

Its size is (n m). Each element ik of this matrix determines the ith

internal force (ith element in the vector S ) from the kth external unit force ( Fk 1).

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Elements i1 of the first column of the matrix LS are the internal forces in the all bars of the system due to the unit load F1 1 . Using the-

se elements, you can plot the forces diagrams in the bars of the system from loading it by force F1 1 .

The elements of the first line show the values of the effort S1 from

the sequential loading of the system nodes by unit forces. Using these numbers, one can therefore construct a line of influence of the effort S1 .

In this operation, from the first line it is necessary to select those numbers (elements) that correspond to a given direction of movement of a unit force.

Influence matrices are very important characteristics of the calculated (investigated) system. A change in a system of a parameter will necessarily entail a change in these matrices.

The physical meaning of the elements of the influence matrices also indicates that efforts diagrams or influence lines of efforts can be used to compile them. Such calculation methods are usually used for simple (with a small number of elements) systems. In other, more complex cases, it is advisable to apply the indicated mathematical formalization of this process using equilibrium matrices A and the internal stiffness matrices K.

Between influence matrices of efforts LS and displacements Lz there is a relationship. Indeed, since

LS KAT ( AKAT ) 1 and Lz ( AKAT ) 1 ,

then

LS KAT Lz .

As for matrices of external rigidity (stiffness)

R AKAT

and external compliance

D ( AKAT ) 1 R 1 ,

they are widely used in the dynamics and stability of structures. 422

E x a m p l e. We show the use of the general equations of structural mechanics for calculating a continuous beam (Figure 15.23, a). The beam is loaded with three load options and has a constant cross section:

Dimension of forces – kN, moments – kNm, lengths – m. The total number of nodes is 9.

The boxes indicate the numbers of beam elements.

The matrix is formed “bar by bar” using the 5th option of the Table 14.2. The internal stiffness matrix is quasi-diagonal:

diagK [K1K1K1K1 | K2K2K2K2 ] 18.795 103 ,

Where

The solution of the system

AKAT z F

gives a displacement matrix z . Linear displacements are measured in meters, turning angles are in radians.

The displacements diagram (Figure 14.26, b) corresponds to the first loading of the beam.

The efforts matrix S was calculated by the expression

S KAT z .

With its help, diagrams of bending moments were plotted for each load option (Figures 14.26, c, d, e).

To construct the lines of the influence of efforts, we used the matrix of the influence of internal forces:

LS KAT ( AKAT ) 1 .

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Figure 14.26

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The elements in columns Nos. 2, 4, 6, 9, 11, 13 are the efforts in the beam caused by the vertical concentrated unit force applied, respectively, at points Nos. 2, 3, 4, 6, 7, 8. Columns Nos. 1, 3, 5, 7, 8, 10, 12, 14 contain information about the efforts due to the concentrated unit moment applied, respectively, at points Nos. 1, 2, 3, 4, 5, 6, 7, 8.

The rows of the matrix LS contain the ordinates of the influence lines

for internal forces in the corresponding cross sections of the beam. So, with the help of the elements of the 5th row, the Inf .Line M3 was built

(figure 14.26, f). The effort M3 is the bending moment in the cross sec-

tion No. 3 (at point No. 3). This moment can be considered as a bending moment M B3 at the beginning of the third element, or as a bending

moment M E2 at the end of the second element of the beam. The ordinates of M E 2 are in the 4th row of LS in columns Nos. 2, 4, 6, 9, 11, 13.

To build Inf .Line M8 (Figure 14.26, g) the ordinates from the 15th row of Ls were used.

14.16. Spatial Trusses

The coordinates of nodes of the spatial truss will be considered known. For rod P1P2 ( P1 is node at the beginning, P2 is node at the end

of the rod), as for the directed segment P1P2 , we find the direction cosines cos x , cos y , cos z by the expressions:

where

l (x2 x1)2 ( y2 y1)2 (z2 z1)2 .

cos x , cos y , cos z .

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We will consider the tensile longitudinal force in the rod to be positive. In the terminal sections of the rod, it has opposite directions. These

directions correspond to the directions of segments P2P1 and P1P2 .

The projections of the directed segment P1P2 on axis OX , OY , OZ are equal respectively:

Consequently, the projections of the longitudinal force N applied at point P2 on the same axis are equal

N cos x , N cos y , N cos z ,

and the projections of the force N applied at the point P1 must be recorded with the opposite sign:

N cos x , N cos y , N cos z .

The numbering of truss nodes determines the numbering of nodes P1 and P2 , connected by a rod. Moreover, for the beginning of the rod, that is, for point P1 , a node with a lower number is taken.

In the equilibrium matrix A , with each free node of the spatial truss, three rows are connected, in which the coefficients are written for the forces N in the rods adjacent to this node. These coefficients are factors (direction cosines) with N in the equations:

In practical problems, it is more convenient to form a matrix A not in rows but in columns. Recall once again that the direction cosines of the

rod in the equations related to the nodes P1 and P2 will have opposite

signs. When forming a matrix by columns, it is recommended to use a template vector a for each bar:

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The first three elements of the vector refer to the beginning of the rod (node P1 ), the remaining three to the end of it (node P2 ).

The general equations of structural mechanics for a spatial truss have the same notation as for a plane system.

N o t e. For plane trusses the equilibrium matrix can also be formed through the direction cosines using the above vector a , excluding the

components in it: cos z ,..., cos z .

E x a m p l e. Determine the forces in the rods of the spatial truss (Figure 14.27). The rigidity of all rods is taken equal to EA. Load vector

is F = [5.0; 10.0; –4.0]T kN.

Figure 14.27

Having determined the direction cosines of the rods, we form the equilibrium matrix:

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The truss internal stiffness matrix is diagonal:

diagK [0.15617, 0.15617, 0.15617, 0.15617]EA .

The matrix of the external rigidity of the truss is obtained by the for-

The matrix inverse to matrix R is shown below:

The vectors of displacements of node 5 and the internal forces in the rods are calculated by known formulas.

Z = [328.159; 164.079; –7.292]T 1/EA.

N = [14.941; –1.067; –17.075; –1.067]T, kN.

E x a m p l e. Determine the forces in the rods of the spatial truss (Figure 14.28). The stiffness of all rods taken equal to EA.

Figure 14.28

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The load vector is taken in the form:

F [0; 0; 0; 0; 0; 0; 0; 0; 5.0; 5.0; 50.0;5.0; 5.0; 100.0; 5.0; 5.0; 50.0 ]T kN.

The truss equilibrium matrix is given in table. 14.5.

In this example, the task of determining support reactions was not posed. Therefore, there are no rows in the matrix A corresponding to the equations of the projections of forces in the rods adjacent to the support nodes in the directions of the support links.

The truss stiffness matrix K is diagonal:

diagK= {0.2; 0.2; 0.125; 0.2; 0.2; 0.125; 0.2; 0.2; 0.125; 0.15617; 0.15617; 0.15617; 0.15617; 0.25; 0.25; 0.25; 0.25; 0.25; 0.25} EA. The solution of the system of equations Rz F gives:

zT [ 382.96, 254.51,14.27,14.27,25.73,173.99, 25.73,382.96, 236.14,114.66, 539.03, 79.32,

120.39, 826.69,158.01,126.12, 621.12] E1A m.

The efforts in the truss rods are determined by the expression:

N K AT z .

N T [ 44.79, 30.50,47.87, 71.41, 58.91,53.56, 49.25,38.54,47.87, 19,27,5.71, 10.30, 27.27,3.57,0,0, 6.43, 1.43,1.43] kN.

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