Structural mechanics

To check the coefficients and free terms, a total diagram of the unit moments is constructed (Figure 8.14, e). Using the formula (8.20), we obtain:

Indeed:

11 12 21 22 3203EJ EJ66 EJ66 144EJ 3239EJ . By the formula (8.21) we have:

sF 2E1J 320 4 5 6EJ6 320 3 20 3 4100EJ ,

that is equal to 1F 2F 3160EJ 7260EJ 4100EJ . We record the system of equations in numerical form:

Having solved this system of equations, we find:

To construct the final moment diagrams, we use the formula (8.15). The diagrams M1X1 and M2 X2 are shown in Figures 8.14, g, h, and the

final diagram M is shown in Figure 8.14, i. Its static verification is performed (The reader is advised to conduct its own verification). We perform a kinematic check:

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The relative error of the calculations is:

0.02 100 0.01 % 140.55

which is less than the acceptable value.

The diagram Q (Figure 8.14, k) is constructed in accordance with the

diagram M. Once again, we note that a simpler way of constructing is based on dependency

Q dMdx .

We use the formula (8.17).

Considering the element 2–3 as a simple beam loaded with a uniformly distributed load, we construct a diagram of the shear forces (diagram of the shear forces for the beam). It is shown in Figure 8.14, l.

Given the distribution of moments on this element (Figure 8.14, i) using the formula (8.17), we find that in the cross-section adjacent to the node 2:

Q2 30 33.43 18.62 27.53 kN,

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and in the cross-section adjacent to the node 3:

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The Q diagram for the cantilever 3–4 is constructed as for a statical-

ly determinable fragment of a frame. However, in this case we can also use the formula (8.17), if you consider section 3–4 as a beam with two supports (Figure 8.14, m).

Then in the cross-section adjacent to the node 3:

and in the cross-section adjacent to the node 4:

constructed on the stretched fibers of the element. For horizontal elements, the positive ordinates of the bending moments must be located below the axis of the element. Therefore, the sign of the transverse force in a given cross-section “k” of the horizontal bar can be determined as follows. Drawing a tangent to the line bounding the diagram M , at a

point, corresponding to the position of the cross-section k (Figure 8.14, n), it is necessary to find the intersection point of this tangent and the axis of the element (point O).

If the axis of the element must be rotated around the point O until it coincides with the tangent in the shortest way clockwise, then the shear

(transverse) force in the cross-section k will be positive Q 0 . When

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the axis of the element moves anticlockwise, then the shear force in the cross-section will be negative (Q 0).

On the linear zones of the diagram of bending moments, the position of the tangent coincides with the line bounding the diagram. The shear force along the entire length of this section will be constant. For the element 3–5

Q 13.43 4.48 kN, 3

and for the element 1–2

Q 18.62 0.71 4.48 kN. 4

After determining the shear forces in the frame elements the longitudinal forces N are determined from the equilibrium equations of the nodes. The calculations begin with a node in which the elements with no more than two unknown forces are joined, and then, sequentially cutting out the nodes, determine the efforts in all other bars. The equilibrium equations are written as the sum of the projections of all the forces (both internal and external forces applied to the nodes, if any) on the vertical and horizontal axes. In the presence of inclined bars, if the calculations may be simplified, the forces projections can be performed to the axes perpendicular to the bars directions.

Composing equations for node 2 (Figure 8.14, o) X 0, Y 0, we find N2 3 4.48 kN, N1 2 27.53 kN.

From the equation Y 0 for node 3 (Figure 8.14, p) we get N3 5 52.47 kN.

The equation X 0 for node 3 is a test one. The diagram N is

shown in Figure 8.14, p.

For carrying out a static check of the diagrams Q and N we cut off

the frame from the support connections, load it by a given load and shear and longitudinal forces in the cross-sections separating the rods from the support connections (Figure 8.14, c). Composing the equations X 0,

Y 0 and М 0, we make sure that the frame is in equilibrium.

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8.8. The Concept of Rational Primary System

and Methods of Its Choice

A rational primary system is such a system for which in the canonical equations greatest possible number of secondary coefficients is zero. At the same time, it is very important to set zero coefficients only on the basis of a visual analysis of the outline of the force diagrams, without spending time on their numerical determination. Zeroing secondary coefficients leads to significant simplifications in the calculation.

If some coefficient ik is equal to zero, the corresponding diagrams Mi and Mk are usually called mutually orthogonal. An analogy with

the scalar product of mutually orthogonal vectors is used.

The most commonly used methods for obtaining rational primary systems include: using the symmetry of the system, grouping unknowns, transforming of the load, breaking up multi-span frames.

1. Using the symmetry of the system. The primary system for a frame which has a symmetric geometric dimensions and symmetric rigidity of the elements should be taken symmetrical. If the primary unknowns can be positioned on the axis of symmetry, then some of them will be symmetric, and the other – inversely symmetric (or skewsymmetric). Due to the action of a symmetrical load on the symmetrical frame, the distribution of forces in its elements will be symmetric, and vice versa: inverse-symmetrical loading of the symmetrical frame causes inverse-symmetrical forces in its elements. Therefore, the diagrams of bending moments in the primary system will be either symmetrical or inversely-symmetrical. Symmetrical and inverse-symmetrical diagrams are mutually orthogonal.

For example, taking for the frame (Figure 8.15, a) the primary system shown in Figure 8.15, b, we obtain symmetrical diagrams M1 , M2 , M4

(Figures 8.15, c, d, f) and inverse-symmetrical M3 (Figure 8.15, e). Therefore, the coefficients 13, 31, 23, 32 , 34 , 43 are equal

to zero.

Crossing out in the system of equations (the reader should write them down) the terms including the listed coefficients, we see that it has decomposed into a subsystem containing only symmetrical unknowns and one equation with inverse-symmetrical unknown.

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unknowns Xi corresponds to the operation of addition or subtraction of symmetrical and inverse-symmetrical group unknowns Zi .

Figure 8.16

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The diagrams of efforts caused by group unknowns are shown in Figure 8.16, d–i. Due to the mutual orthogonality of symmetrical and in- verse-symmetrical diagrams, the system of canonical equations decomposes into two independent ones: one of them will include only symmetrical unknowns Z1, Z3, Z5, and the other will include only inverse-

symmetrical Z2 , Z4 , Z6.

3. Transforming of the load. Further simplifications in the calculation of symmetric systems (Figure 8.17, a) are associated with the decomposition of the load into symmetrical and inverse-symmetrical components.

Using the property of mutual orthogonality of the diagrams, it is easy to show that, when a symmetrical load is applied to a symmetrical system, inverse-symmetrical unknowns become zero, and when a inversesymmetrical load acts, symmetrical unknowns turn out to be zero. In relation to the design scheme of the frame shown in Figure 8.17, b, this means that it should be calculated as systems with three unknowns X1,

X2 , X4 (the primary system is shown in Figure 8.15, b), and the frame

calculation for the action of inverse-symmetrical load (Figure 8.17, c) as systems with one unknown X3 .

Figure 8.17

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4. Breaking up multi-span frames. This method is used for both symmetrical and asymmetrical frames. Less computational work to define iк, will be if the diagrams of the internal forces in the primary sys-

tem extend to small fragments of the frame, i. e., they are “localized” in the vicinity of the load.

For a frame (Figure 8.18, a) with four unknowns in Figures 8.18, b, c, two variants of the primary system are presented. Analyzing the distribution of bending moments due to Xi 1 in the frame shown in Figu-

re 8.18, b, we can verify that none of the coefficients iк is equal to zero.

In the system shown in Figure 8.18, c, bending moment diagrams occur only on columns directly perceiving the action Xi 1. Therefore,

Figure 8.18

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8.9. Determining Displacements

in Statically Indeterminate Systems

To determine the displacements using the Mohr formula, described in section 7.6, it is necessary to construct in the system the bending moment diagrams caused by the given loading (Figure 8.19, a) and the auxiliary loading (Figure 8.19, b). Then the required displacement will be calculated by the formula (8.24):

However, this method of calculation is not entirely convenient, since it will be necessary to calculate the statically indeterminate system twice.

A simpler calculation method can be obtained from the following reasoning. If you load the primary system with a given load and primary unknowns, which have been determined from the canonical equations, then the diagram of bending moment in this statically determinate system (Figure 8.19, c) will completely coincide with the final moment diagram (Figure 8.19, a). Therefore, if we consider the state of the frame in Figure 8.19, c as the initial one, then to determine the displacement of the point k it is possible to take a statically determinate system (Figure 8.19, d) as an auxiliary state. In this case:

where Mk0 – is the bending moments in a statically determinate system due to Fk 1.

Another method can be used to calculate the same displacement: the diagram of bending moments caused by given load can be constructed in

the primary system, and the diagram caused by Fk 1 – in a given stat-

ically indeterminate system. We will show this.

Applying reciprocity theorem to the states of the frame shown in Figures 8.19, a, b, we get:

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