Structural mechanics

the beam around the point А at an angle φ (Figure 3.6, b). We write

down the sum of the forces acting on the system on the considered infinitesimal displacements:

F(x) VB B 0.

From this equation we get:

B B

Different positions of the force F = 1 lead to a change in the value of the corresponding displacement (x). In this case, all possible values of

(x) along the length of the beam show a diagram of the vertical displacements of the beam points. The denominator in the formula (3.5) is a

Consequently, the outline of the influence line coincides with the diagram of the vertical displacements of the points of the beam (Figure 3.6, c).

From the ordinate ratios in Figure 3.6,b we get (x) x , which, forB l

F = 1, corresponds to the expression (3.4) obtained by the static method. Construct influence lines for bending moment in the section k.

The design scheme of the beam is shown in Figure 3.7, a. We eliminate the constraint in the cross-section k through which the moment is transmitted (we set the hinge), replacing its action with the moment M K

(Figure 3.7, b). The figure shows the interaction forces of the left and right parts of the beam. We will set the possible displacements to the obtained mechanism in the direction of the moments M K action, taking

the angle of mutual rotation of the end cross-sections equal to unity. The ordinates between the initial position of the beam and the new (broken) form a diagram of the beam displacement (Figure 3.7, b).

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At F = 1 we get M K δ(x), that corresponds to the above conclusion: influence line M K (Figure 3.7, d) coincides with the diagram of

the vertical displacements of the beam points.

Using the notation given in Figure 3.7,a, shows that it is exactly coincides with the influence line previously constructed by a static method (Figure 3.4, e).

Possible displacements, in fact, are infinitesimal. Therefore, when analyzing the relations in Figure 3.7, you can use simplifications of the form:

tg1 1; tg2 2.

From the data in Figure 3.7, c, provided that φ1 φ2 1, we obtain:

l 1 b 2 b 1; l 1 b; 1 bl ; k a 1 abl . l 2 a 1 a 2; l 2 a; 2 al ; k b 2 abl .

The ordinate of influence line in cross-section k is equal to the ordinate obtained by the static method (Figure 3.4, f).

Let us construct the influence lines for shear force in the section k (Figure 3. 8, а).

We eliminate the constraint in this cross-section, in which a shear force can arise. The connection of the left and right parts of the beam after this is carried out by means of two horizontally arranged links through which longitudinal forces and bending moments can be transmitted. On the newly formed design scheme, we show in the cross-section the positive directions of the shear forces for both parts of the beam (Figure 3.8, b). Giving the unity value for mutual displacement of the beam ends along the directions of the shear forces, we obtain a diagram of the beam's displacements (Figure 3.8, c), the outline of which completely corresponds to influence line for shear force (Figure 3.8, d).

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If n concentrated vertical forces act on the beam (Figure 3.9), then the force S, based on the principle of superposition, should be calculated by the formula:

i n

S F1 y1 F2 y2 ... Fn yn Fi yi . (3.7)

i 1

In this expression, the value of the looking downward force is taken with the plus sign, the value of the looking upward force is taken with the minus sign.

Figure 3.9

Consider the action on the beam of a load distributed according to an arbitrary law q(x), (Figure 3.10, a). On this beam, we select a section of

infinitely small length dx. The concentrated force replacing the distributed load on this section is equal to dF q(x)dx. (Figure 3.10, a). The

elementary effort dS from the action of the force dF is:

dS dFy q(x) y(x)dx.

Integrating this expression along the length of the loading section, we find:

a

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Figure 3.10

If a uniformly distributed load acts on the beam q(x) q (Figure 3.10, b), then

Here is the area of influence line S corresponding the uniformly distributed load action site. In figure 3.10, b the area is highlighted by hatching. It should be kept in mind that the ordinates of the influence lines located above the axis of the beam are positive, the ordinates of the lines of influence located below the axis of the beam are negative. The area below the axis is negative.

Let us consider the action on the beam of a concentrated moment M . (Figure 3.11). Replace the moment with a couple of forces F with

arm x : F Mx . With the help of the formula (3.7) we find:

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We find the efforts:

Mk1 4 ( 1) 4 ( 4 / 3) 101 10 2 2 ( 1/ 213)

2 ( 1/ 2 2 4 / 3) 5 1/ 2 2 9 1/ 2 0 1/ 2 2 61kNm;

Mk2 Mk3 4 ( 3) 2 ( 1/ 2 3 3) 21kNm;

Qkleft1 4 1/ 3 4 (2 / 9) 10 (1/ 3) 10 1/ 3 2 (1 / 2 1/ 3 3)2 (1/ 2 2 / 9 2) 5 (1/ 2 2 / 3 6 1 / 2 1/ 3 3)

1/ 2 0 1/ 2 (2 / 3) 6.833kN;

Qkright1 4 1/ 3 4 ( 2 / 9) 10 ( 1/ 3) 10 ( 2 / 3) 2 (1/ 2 1/ 3 3)2 ( 1/ 2 2 / 9 2) 5 ( 1/ 2 2 / 3 6 1/ 2 1/ 3 3)

1/ 2 0 1/ 2 ( 2 / 3) 16.833kN;

Qk2 4 1/ 3 4 ( 2 / 9) 10 2 / 3 10 1/ 3 2 (1/ 2 1/ 3 3)2 ( 1/ 2 2 / 9 2) 5 (1/ 2 1 9) 1/ 31 1/ 3 0 33.167kN;

Qk3 4 1 2( 1 3) 10kN.

Note. Other factors can be defined similarly if the corresponding influence lines are constructed for them.

Let us turn to Figure 3.3, which shows influence line of the vertical displacement of the beam point with a coordinate x 7.5m .

Using the displacements given in Table 3.1 for characteristic points, we find an approximating polynomial that describes the outline of influence line and the first derivative of it:

p(x) [1.59222x 0.130259x2 0.0190617x3 0.0115391x40.000768176x5] E1I .

dp(x) [1.59222 0.260519x 0.0571852x2 0.0461564x3 dx

0.00384088x4 ] E1I .

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Figure 3.13

Find the displacement, knowing the outline of influence line and the first derivative:

3.5. Influence Lines for Efforts in Case of the Nodal Transfer of the Load

Consider the construction design scheme shown in Figure 3.14, a. The main bearing element of this scheme is the beam AB. It is called the main beam. The main beam bears cross beams. They are presented on the design scheme in the form of support rods for short longitudinal beams located at the upper level. Short beams are essentially flooring performed in the simplest case of planks. The load (force F = 1 is shown on the design scheme) applied to the upper short beams is transferred to the main beam at specific points, which are called nodes.

Hence the name follows: nodal transfer of the load.

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