Structural mechanics

E x a m p l e. Calculate the frame shown in Figure 14.6. The ratio of the rigidity of the bars in tension (compression) and bending is taken equal to

EA h2 1, ( h = 1 m).

EJ

Figure 14.6

First of all, we transform the given load to the nodal one. The end reactions in single-span statically indeterminate beams loaded with a distributed load, and the outline of the bending moment diagrams in them (Figure 14.7), we will find using Table. 9.1. Then the design scheme of the frame with a nodal load can be represented as it is shown in Figure 14.8.

Figure 14.7

The primary system for calculating the frame, taking into account the longitudinal deformations of the bars, is shown in Figure 14.9. The posi-

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tive directions of the primary unknowns comply with the sign rule specified in section 14.2.

Using the equilibrium matrix and the matrix of internal stiffness of the frame, we calculate the matrix of external stiffness:

R AK AT .

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The load vector F in the equation of the form R z F corresponds to the load shown in Figure 14.8:

Having solved the system of equations of the displacement method, we obtain:

The force vector calculated by the expression S K AT z can be written as:

Figures 14.10, a, b show the diagrams of the frame efforts corresponding to this vector.

Figure 14.10

By superimposing the diagrams of bending moments in the beams (Figure 14.7) on diagram M (Figure 14.10, b) we obtain the final diagram M in the frame (Figure 14.11).

Naturally for a different initial ratio of rigidities EA / EJ , the ordi-

nates of the diagram M will differ from those found.

To assess the effect of longitudinal deformations on the distribution of displacements and forces in the frame, we will perform its calculation taking into account only bending deformations (Figure 14.12, a). Ne-

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glecting longitudinal deformations, we choose the primary system of the displacement method with two unknowns (Figure 14.12, b).

Figure 14.11

Figure 14.12

Having completed the necessary steps of the calculations, we find the values of the primary unknowns:

The diagram of bending moments is shown in Figure 14.13.

Figure 14.13

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The vector-matrix recording of these operations is reduced to two subsystems of equations:

A0 z D0 S0 0 , Ax z Dx X x .

Thus, a system of equations of mixed form can be written in the form of three equations:

constructed from the action of unit forces oriented in the directions of the primary unknowns;

Lx X is efforts in the bars of the primary system from X .

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From the second equation we find z:

Substituting S0 into the last expression and then z into the third equation, we obtain the equations of the force method in the following form:

Having determined X , one can find the forces in the bars belonging

to the primary system by the formula (14.12), and then, by the formula (14.13), the vector of nodal displacements z.

E x a m p l e. We’ll show the calculation of the truss (Figure 14.14) by the force method. The cross-sectional areas of all the rods are as-

sumed to be the same and equal to A 0.25 m2 . The elastic modulus of the material E 210 GPa .

Figure 14.14

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The equilibrium matrix of this truss, taking into account the accepted numbering of nodes and rods, has the following form:

7 6 1.

One of the main conditions for choosing the primary system of the method of forces is, as you know, the condition for its geometric invariability. The determinant of the equilibrium matrix of the primary system should not be zero.

We take the force in the 4th rod as the primary unknown. Then the equilibrium matrix of the primary system (Figure 14.15) will have the following form:

Figure 14.15

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The efforts NF0 in the rods of the primary system from the load

F 0; 10.0; 0; 10.0; 0; 0 T

will have the following values:

The matrix of internal compliance of the rods of the primary system is diagonal and is represented in such a record:

0.7619; 0.7619; 0.7619; 0.9524; 0.5714; 0.9524 10 4.

The compliance of the rod, the force in which is taken as the primary

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The matrix of compliance of the primary system in the directions of the primary unknowns in the case of one unknown is represented by one element:

D LTx D0 Lx Dx 30.857 10 5 .

The displacement in the direction of the primary unknown, caused by a given load, that is, a free term in the canonical equation of the method of forces, is:

LTx D0 NF0 514.286 10 5 .

From equation (14.14), which can be written in the form:

D X1 LTx D0 NF0 0 ,

we find

X1 16.667 kN.

Using expression (14.12), we determine the final forces in all the rods of the primary system of the given truss:

N0 Lx X1 NF0

26.67; 13.33; 26.67; 16.67; 0;16.67 T кN,

and according to the expression (14.13) – the displacements of its nodes:

z A0 1 T D0 N0

0.2032; 0.5355; 0.3048; 1.477; 0.2032 ; 0.5355 T 10 2 m.

14.12. Statically Determinate Systems

In a statically determinate system, the number of independent equilibrium equations is equal to the number of unknown efforts; therefore, the equilibrium matrix A is square. In this case, the system of equations (14.7) splits into two independent groups of equations.

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