Structural mechanics

To build unit diagram M1 (Figure 9.20, e) we use the previously found values of the mutual linear displacements of the ends of the bars. Diagrams M2 and M F are shown in Figures 9.20, f, g.

The coefficients and free terms of the canonical equations are determined by static or kinematic methods. We show, for example, the

The same value can be obtained by the static method from the equilibrium equation of node 2 (Figure 9.20, h). We obtain the transverse and longitudinal forces in rod 1–2 from the equilibrium equation for node 1. The multiplier EJ in the notation in the figure of the transverse and longitudinal forces is omitted.

To calculate R1F by the kinematic method, we use the diagram of

bending moments M F0 (Figure 9.20, i), built in the primary system of the method of forces:

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From the equilibrium equation Mc 0 for frame fragment (Figure 9.20, j) we find the same value R1F .

We write the canonical equations in numerical form.

The final diagram of bending moments (Figure 9.20, k) is obtained according to the formula:

M M F M1Z1 M2Z2.

E x a m p l e 2. Let us build diagrams of internal forces for the frame (Figure 9.21, a), assuming that the stiffnesses of all bars on bending are constant and the same. As in the previous examples, we do not take into account the longitudinal deformation of the bars ( EA ).

The degree of kinematic indeterminacy of the frame is three. We select the main system (Figure 9.21, b) and build the load diagrams of bending moments MF (Figure 9.21, e) and unit diagrams of bending moments M1 and M2 from angular unknowns Z1 = 1 and Z2 = 1 (Figure 9.21, f, g).

To build a unit diagram from linear displacement Z3 = 1, it is necessary to determine the mutual displacements of the ends of the bars.

First, a given system (Figure 9.21, a) is converted into a hinge-rod system (Figure 9.21, c). Then the hinge-rod system (Figure 9.21, c) receives an offset along the entered additional linear link by Z3 = 1. The nodes 3, 4, 5 are moved to a new position 3', 4', 5'. Next a displacement diagram is constructed (Figure 9.21, d).

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equal to the displacement of the node 3 in the direction perpendicular to the rod 0–3; mutual vertical displacement of the ends of the rod 3–4 is determined by the length of the segment 3'–4'.

Unit diagram M3 is presented in Figure 9.21, h.

Coefficient r33 of the canonical equation, as well as free term R3F, is conveniently calculated by the kinematic method. One of the possible

variants of diagram M F0 for determining R3F is shown in Figure 9.21, i.

After determining the coefficients and free terms, the system of canonical equations is written in the form:

The final diagrams of the internal forces M , Q and N are shown in Fiures 9.21, j, k, l.

9.8. Using System Symmetry

It is known that any load acting on a symmetric system (Figure 9.22, a) can be represented as the sum of symmetric and inverse-symmetric components.

The first of them corresponds to the symmetric form of deformation of the frame (Figure 9.22, b), and the second to the inverse-symmetric (Figure 9.22, c). Therefore, the rotation angles of the nodes of the given

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frame (Figure 9.22, a) can be found as the sum or difference of symmetric and inverse-symmetric unknowns:

Figure 9.22

Similar relations hold for linear displacements. For example, for а frame (Figure 9.23, a) it is possible to calculate its linear displacements 1 and 2, using symmetric (Figure 9.23, b) and inverse-symmetric

(Figure 9.23, c) loading, by the expressions:

and the rotation angles of the nodes as:

1 Z3 Z4 , 2 Z3 Z4.

These relations show that the unknown movements of nodes 1, 2, φ1 and φ2 (“old” unknowns) can be expressed in terms of the “new” unknowns Z1, Z2, Z3, Z4, which are grouped displacements of symmetrically located nodes. Introduction to the calculation of new unknowns leads to significant simplifications in the calculation. Unit bending moments diagrams from grouped displacements are divided only into symmetric or inverse-symmetric ones. Such diagrams have the property of mutual orthogonality, and therefore the system of canonical equations splits into two independent subsystems of equations, one of which contains only symmetric unknowns, and the second – inverse symmetric. The de-

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scribed method for calculating frames is called the method of grouping unknown displacements.

Figure 9.23

Note that if there is a bar on the frame symmetry axis (its position coincides with the axis of symmetry), then the symmetric unknowns do not cause bending moments in it. Therefore, the result of “multiplying” the symmetric diagram by inverse-symmetric one will be zero.

Coefficients and free terms in canonical equations are generalized reactions caused by the displacement of grouped (pair) unknowns. They are determined in a static or kinematic way.

E x a m p l e 1. It is necessary to build the final diagram of bending moments in a three-span symmetrical frame (Figure 9.24, a), assuming the rigidity in bending of all the bars is equal to EJ.

The degree of kinematic indeterminacy of the frame is n na nl

2 3 5 . The primary system and positive directions of grouped unknowns are shown in Figure 9.24, b. Unit diagrams of bending moments and the load diagram of bending moments are presented in Figu-

res 9.24, c–g. Diagrams M1 and M3 are symmetric, and diagrams M2 , M4 and M5 are inversely symmetric. Due to their orthogonality, the

system of five canonical equations of the displacement method is divided into two subsystems. One contains symmetric unknowns:

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r11Z1 r13Z3 R1F 0,

r31Z1 r33Z3 R3F 0,

and the second contains inversely symmetric unknowns:

r22Z2 r24Z4 r25Z5 R2F 0,

r42Z2 r44Z4 r45Z5 R4F 0,

r52Z2 r54Z4 r55Z5 R5F 0.

After determining unit and load reactions (we advise the reader to find them independently), these subsystems of equations will have the following form:

The final diagram of bending moments is based on the expression:

M MF M1Z1 M2Z2 M3Z3 M4Z4 M5Z5 .

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It is depicted in Figure 9.24, i.

Figure 9.24

Analyzing the subsystems of equations recorded in this example, we can draw the following conclusions.

1. If a symmetric exposure has an effects on a symmetrical frame, then the free terms in the system of equations with inversely symmetric

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unknowns will be equal to zero. Therefore, inversely symmetric unknowns will also be equal to zero (from the solution of the system of homogeneous equations).

2. If the external action is inversely symmetric, then the symmetric unknowns become zero.

As a consequence of the method of grouping unknowns, it should be noted that the calculation of a symmetric system for a symmetric or inversely symmetric load can be performed for one half of the design scheme. Depending on the loading in the second half of the system, the distribution of forces will be symmetrical or inversely symmetric compared to the first half.

In particular, if the axis of symmetry crosses a certain bar, then in this section it is necessary to set movable pinching, when a load is symmetrical. For example, for the frame (Figure 9.22, a), the corresponding “half frame” is shown in Figure 9.25, a. Under the action of the inverse symmetric load in such section, the deformed axis of the bar has an inflection point and, in addition, the vertical displacement of the section (in the direction of the axis of symmetry) is zero. Therefore, in the design scheme of the “half frame” in the indicated section, a hinged movable support is placed (Figure 9.25, b).

Figure 9.25

9.9. Calculating Frames Subjected to Thermal Effects

The calculation is carried out to change the temperature of the system with respect to the temperature of its initial state. Accepting the linear law of temperature change along the height of the cross section of the bar, the thermal effect can be represented as the sum of the symmetric and inverse symmetric components of this effect.

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Suppose, for example, that a bar has a symmetric cross section of height d (Figure 9.26, a) and t1 t2 , i. e., the lower fibers of the bar are

“warm”, but the upper ones are “cold”.

Let us decompose this effect into a symmetric and inversely symmetric. At symmetric exposure the bar is uniformly heated (Figure 9.26, b). The temperature of the upper and lower fibers is the same and equal to

t t1 t2 . 2

The elongation of the bar in this case is equal to t l. At inversely symmetric exposure (Figure 9.26, c), the temperature of the upper fiber

ric heating, the temperature along the axis of the bar is zero. The bar from such exposure does not elongate, but only bends. The value of the displacement of any of its points is determined according to the rules set out in section 7.8.

Figure 9.26

A similar decomposition of the thermal effect can be done for the bars with the conditions for the fastening of their ends corresponding to the fastening of the bars in the primary system of the displacement method.

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