Structural mechanics

With a symmetric distribution of temperatures, due to the elongation (shortening) of the bars, the nodes of the primary system move, which leads to mutual displacements of the ends of the bars in the transverse direction and causes their bending deformation.

In the case of inverse symmetric temperature distribution, the nodes of the primary system are not displaced, but since the bonds at the ends of the bars impede their free movement, forces appear in each of them. Diagrams of moments for such bars are presented in table 9.1 (lines 5, 10, 15). The technique of their construction is given in section 9.2 (example 5).

The canonical equations for calculating the frames for temperature change are as follows:

as follows from the previous reasoning, it is necessary to construct diagrams of bending moments in the primary system: Mt due to symmet-

ric termal action and Mt due to inversely symmetric action. Using them, we find that:

by these influences.

The final diagrams of bending moments is built according to the formula

M Mt Mt M1Z1 M2Z2 M n Zn .

E x a m p l e 1. Let us build the final diagram of the bending moments in the frame (Figure 9.27, a) from the specified heat exposure, taking the stiffness of the bars the same and equal to 60 MN·m2, the

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height of the cross section d = 0.6·m, the coefficient of thermal linear expansion α = 1.2·10–5·(°C)–1.

The kinematic indeterminacy of the frame is two. Figure 9.27, b shows the symmetric temperature distribution for each bar, and Figure 9.27, c – inversely symmetric.

Figure 9.27

The system of canonical equations in this case can be written as:

r11Z1 r12Z2 R1t 0,

r21Z1 r22Z2 R2t 0.

Due to the symmetric effect of temperatures, the plotting of moment diagram Mt must begin by determining the elongation of each bar ac-

cording to the formula t l. Then it is necessary to depict the new

position of the nodes and the deformed position of the bars in the primary system (Figure 9.27, d). Knowing the mutual displacements of the ends of each bar in the direction perpendicular to its axis and using the data from Table. 9.1 (lines 1, 6, 11), it is possible to make a diagram of bending moments. Final diagram Mt is shown in Figure 9.28, a.

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In this example, to build diagram Mt (Figure 9.28, b), lines 5 and 10

from table 9.1 are used. On each bar of the frame, the stretched fibers are more “cold”. It is from this side of the bar that the bending moments diagram is located.

Figure 9.28

Unit diagrams of moments are shown in Figure 9.29, a, b.

The reactive forces in additional links having been determined, the system of canonical equations can be written in numerical form:

Its solution is: Z 1 38.4636 rad; Z 2 213.1391 m.

The final bending moments diagram is shown in Figure 9.30. In parentheses there are the ordinates of the moments (in kN m) at EJ = = 60 MN m2 and α = 1.2 10–5 deg–1.

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The correctness of the plot of bending moments is checked using the equilibrium conditions of any fragments of the frame, in particular, the nodes of the frame. As a rule, this check is enough to conclude that plot M is correct.

Figure 9.29

Figure 9.30

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However, in addition, another check can be used: the result of “multiplying” the total unit diagram of the moments of the method of displacement by the final diagram should be zero, that is:

M S Mdx 0 .

EJ

9.10. Calculating Frames Subjected to Settlement of Supports

A distinctive feature of the calculation of frames at the given support settlements is associated with constructing the diagram of the bending moments in the primary system of the displacement method due to such displacements. In the future we will denote this diagram by Mc . To do

this, it must be remembered that additional floating links in the primary system prevent only the rotation of rigid nodes, while they allow linear displacements of nodes. Therefore, the effect of linear (horizontal or vertical) displacement of any support can be extended on too many frame elements adjacent to the displaced post or the displaced crossbar.

When constructing diagram Mc , it is recommended to use the princi-

ple of independence of the action of forces. First, it is necessary to construct the bending moments diagrams caused by the given displacements of each support individually. Then the total summarized diagram of bending moments Mc must be constructed, with the help of which free terms

Ric are determined. As a result, a system of canonical equations is formed:

R z Rc 0 .

The following calculation algorithm remains the same as when calculating frames for force impact.

The main verification of the final bending moments diagram is reduced to checking that the equilibrium conditions of the nodes and other parts of the frame are satisfied.

As in the calculation of the thermal effect, the result of the “multiplication” of the total unit moments diagram MS of the displacement method on the final moments diagram should be zero, that is:

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MS M dx 0. EJ

Note that no matter what method was used to construct the final diagram of bending moments, to check its correctness you can apply the kinematic check used in the force method. In particular, when calculating frames for support displacement, the result of “multiplying” the total unit moment diagram of the force method by the final moment diagram should be equal to the sum of the free terms of the canonical equations of the force method, taken with the opposite sign:

M S M dx ic , EJ

where MS is the total unit diagram in the primary system of the method of forces.

E x a m p l e 1. Let us consider the features of calculating the frame, the support of which is displaced as shown in Figure 9.31, a. Let c1 =

= 0.02 m, c2 = 0.04 m and c3 = 0.1 rad. The bending stiffnesses of all bars are assumed the same.

The primary system and the primary unknowns of the displacement method are shown in Figure 9.31, b.

Figure 9.31

Unit bending moments diagrams are shown in Figures 9.32, a, b.

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The canonical equations have the form:

After determining the values of the coefficients and free terms, we solve the system of equations and find:

The final diagram of the bending moments is shown in Figure 9.34.

Figure 9.34

9.11. Constructing Influence Lines for Efforts

The methodology for constructing the influence lines of efforts in statically indeterminate systems is reduced to the implementation of the formulas by which they are calculated. When calculating frames by the displacement method, the force in cross section”k” is calculated by the formula:

n

Sk SkF Ski Zi . i 1

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The quantities SkF and Zi given in this expression to the right of the equal sign are variables depending on the position of the external force F 1, but quantity Ski is the constant force in cross section ”k” of the

primary system caused by the offset of the i-th additional link by one. Therefore, in relation to the problem of constructing influence lines, this record can be represented as:

n

Inf. Line Sk = Inf. Line SkF Ski (Inf. Line Zi ).

i 1

In the last expression the designation “Inf. Line SkF ” is the influence line for effort Sk in the primary system of the displacement method. The

name of effort SkF in this case can be replaced by Sk0.

The construction of this influence line does not cause any particular difficulties. Indeed, in the primary system, the ends of each bar in the nodes are pinched or pivotally supported and therefore the load located on it does not affect the forces in adjacent bars. As a result, the influence line of SkF will have ordinates that are not equal to zero, only on the

bar to which section “ k ” belongs. Determining Sk according to the table 9.1 for various positions of force F 1, we will construct Inf. Line

Sk0 (Inf. Line SkF ).

It is more complex to construct influence lines Zi . Building them

based on a static method requires determination of the values of the primary unknowns for various positions of the unit force F 1.

Consider the following example. The frame shown in Figure 9.35, a, is once kinematically indeterminable. The canonical equation

r11Z1 R1F 0

for F 1 can be rewritten in the form

r11Z1 r1F 0.

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Consequently,

z1 r1F .

r11

Diagram M1 in the primary system is shown in Figure 9.35, b. From the equilibrium equation

M B 0

it follows that

r11 11lEJ .

In the loading state of the primary system at positions of force F 1 to the left of node B (Figure 9.35, c) the free term is equal to:

r1F 2l v 1 v2 .

The corresponding angle of rotation Z1 will be determined by the expression:

With the location of force F 1 on the console section of the frame

Inf. Line Z1 is shown in Figure 9.35, d.

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