1. Collection of loads

Before starting the calculation of a steel beam, it is necessary to collect the load acting on the metal beam. Depending on the duration of the action, the load is divided into permanent and temporary.

• own weight of a metal beam;
• own weight of the floor, etc.;

• long-term load (payload, taken depending on the purpose of the building);
• short-term load (snow load, taken depending on the geographical location of the building);
• special load (seismic, explosive, etc. This calculator does not take into account);

The loads on the beam are divided into two types: design and standard. Design loads are used to calculate the beam for strength and stability (1 limit state). The normative loads are established by the norms and are used to calculate the beam for deflection (limit state 2). Design loads are determined by multiplying the standard load by the reliability load factor. Within the framework of this calculator, the design load is applied when determining the deflection of the beam to the margin.

After collecting the surface load on the floor, measured in kg / m2, it is necessary to calculate how much of this surface load the beam takes on. To do this, you need to multiply the surface load by the step of the beams (the so-called cargo lane).

For example: We calculated that the total load turned out to be Qsurface = 500kg / m2, and the step of the beams was 2.5m. Then the distributed load on the metal beam will be: Qdistribution = 500kg/m2 * 2.5m = 1250kg/m. This load is entered into the calculator

2. Plotting

Next, the diagram of the moments, the transverse force is plotted. The diagram depends on the beam loading scheme, the type of beam support. The plot is built according to the rules of structural mechanics. For the most commonly used loading and support schemes, there are ready-made tables with derived formulas for diagrams and deflections.

3. Calculation of strength and deflection

After plotting the diagrams, the strength (1st limit state) and deflection (2nd limit state) are calculated. In order to select a beam for strength, it is necessary to find the required moment of inertia Wtr and select a suitable metal profile from the assortment table. The vertical limit deflection fult is taken according to Table 19 of SNiP 2.01.07-85* (Loads and impacts). Paragraph 2.a depending on the span. For example, the maximum deflection is fult=L/200 with a span of L=6m. means that the calculator will select the section of the rolled profile (an I-beam, a channel or two channels in a box), the maximum deflection of which will not exceed fult=6m/200=0.03m=30mm. To select a metal profile according to the deflection, the required moment of inertia Itr is found, which is obtained from the formula for finding the ultimate deflection. And also from the assortment table, a suitable metal profile is selected.

4. Selection of a metal beam from the assortment table

From the two selection results (limit state 1 and 2), a metal profile with a large section number is selected.

eccentric compression. Construction of the section kernel. Bending with twist. Calculations for strength under complex stress state.

Eccentric compression is a type of deformation in which the longitudinal force in cross section the rod is not attached to the center of gravity. At eccentric compression, in addition to the longitudinal force (N), there are two bending moments (M x and M y).

It is considered that the rod has a high bending rigidity in order to neglect the deflection of the rod under eccentric compression.

Let's transform the formula of moments for eccentric compression by substituting the values ​​of bending moments:

Let us denote the coordinates of some point of the neutral (zero) line under eccentric compression xN, yN and substitute them into the formula for normal stresses under eccentric compression. Considering that the stresses at the points of the neutral line are equal to zero, after the reduction by P/F, we obtain the equation of the neutral line under eccentric compression:

(35)

The zero line for eccentric compression and the point of application of the load are always located along different sides from the center of gravity of the section.

Rice. 43. Eccentric compression

The segments cut off by the zero line from the coordinate axes, denoted ax and ay, can be easily found from the zero line equation for eccentric compression. If we first take xN = 0, yN = ay, and then take yN = 0, xN = ax, then we find the points of intersection of the zero line under eccentric compression with the main central axes:

Rice. 44. Neutral line with eccentric tension - compression

The neutral line under eccentric compression will divide the cross section into two parts. In one part, the stresses will be compressive, in the other - tensile. The strength calculation, as in the case of oblique bending, is carried out according to the normal stresses that occur at the dangerous point of the cross section (the furthest from the zero line).

(36)

Section core - a small area around the center of gravity of the cross section, characterized by the fact that any compressive longitudinal force applied inside the core causes compressive stresses at all points of the cross section.

Examples of the section kernel for rectangular and circular bar cross sections.

Rice. 45. Section Kernel Shape for Rectangle and Circle

Bending with twist. Shafts of machines and mechanisms are often subject to such loading (simultaneous action of torques and bending moments). To calculate the beam, it is necessary first of all to establish dangerous sections. To do this, plots of bending and torque moments are built.

Using the principle of independence of the action of forces, we determine the stresses that arise in the beam separately for torsion and for bending.

During torsion, tangential stresses arise in the cross sections of the beam, reaching the greatest value at the points of the contour of the section When bending in the cross sections of the beam, normal stresses arise, reaching the highest value in the extreme fibers of the beam.

Eccentric tension this type of loading of a beam is called, in which external forces act along the longitudinal axis of the beam, but do not coincide with it (Fig. 8.4). Stresses are determined using the principle of independence of the action of forces. Eccentric tension is a combination of axial tension and oblique (in particular cases - flat) bending. The formula for normal stresses can be obtained as the algebraic sum of normal stresses arising from each type of loading:

where ; ;

y F , z F– coordinates of the force application point F.

To determine the dangerous points of the section, it is necessary to find the position of the neutral line (n.l.) as the locus of points at which the stresses are equal to zero.

.

Equation n.l. can be written as the equation of a straight line in segments:

,

where and are segments cut off by n.l. on the coordinate axes,

, are the main radii of inertia of the section.

The neutral line divides the cross section into zones with tensile and compressive stresses. The diagram of normal stresses is presented in fig. 8.4.

If the section is symmetrical about the main axes, then the strength condition is written for plastic materials, in which [ s c] = [s p] = [s], as

. (8.5)

For brittle materials with [ s c]¹[ s p], the strength condition should be recorded separately for the dangerous point of the section in the tension zone:

and for the dangerous point of the section in the compressed zone:

,

where z1, y 1 and z2, y2- the coordinates of the points of the section most distant from the neutral line in the stretched 1 and compressed 2 zones of the section (Fig. 8.4).

Zero line properties

1. The zero line divides the entire section into two zones - tension and compression.

2. The zero line is straight, since the x and y coordinates are in the first degree.

3. The zero line does not pass through the origin (Fig. 8.4).

4. If the point of application of force lies on the main central inertia section, then the zero line corresponding to it is perpendicular to this axis and passes on the other side of the origin (Fig. 8.5).

5. If the point of application of the force moves along the ray emerging from the origin, then the zero line corresponding to it moves behind it (Fig. 8.6):

n.l

Rice. 8.5 Fig. 8.6

a) when the point of force application moves along the beam emanating from the origin from zero to infinity (y F ®∞, z F ®∞), a at ®0; a z ®0. The limit state of this case: the zero line will pass through the origin (bend);

b) when the point of application of the force (t. K) moves along the beam emanating from the origin from infinity to zero (y F ® 0 and z F ® 0), a y ®∞; a z ®∞. The limiting state of this case: the zero line is removed to infinity, and the body will experience a simple stretching (compression).

6. If the point of force application (point K) moves along a straight line intersecting the coordinate axes, then in this case the zero line will rotate around a certain center located in the opposite quadrant from point K.

8.2.3. Section kernel

Some materials (concrete, masonry) can absorb very small tensile stresses, while others (such as soil) cannot resist stretching at all. Such materials are used for the manufacture of structural elements in which tensile stresses do not occur, and are not used for the manufacture of instruction elements that experience bending, torsion, central and eccentric tension.

Only centrally compressed elements can be made from these materials, in which tensile stresses do not occur, as well as eccentrically compressed elements, if tensile stresses do not form in them. This occurs when the point of application of the compressive force is located inside or on the border of some central region of the cross section, called the core of the section.

Section kernel a beam is called its some central area, which has the property that the force applied at any of its points causes stresses of the same sign at all points of the cross section of the beam, i.e. the zero line does not pass through the section of the beam.

If the point of application of the compressive force is located outside the core of the section, then compressive and tensile stresses arise in the cross section. In this case, the zero line crosses the cross section of the beam.

If the force is applied at the boundary of the core of the section, then the zero line touches the contour of the section (at a point or along a line); at the point of contact, the normal stresses are equal to zero.

When calculating eccentrically compressed rods made from a material that poorly perceives tensile stresses, it is important to know the shape and dimensions of the section core. This allows, without calculating stresses, to establish whether tensile stresses arise in the cross section of the beam (Fig. 8.7).

It follows from the definition that the kernel of a section is some area that is inside the section itself.

For brittle materials, a compressive load should be applied in the core of the section in order to exclude tensile zones in the section (Fig. 8.7).

To build the core of the section, it is necessary to sequentially combine the zero line with the contour of the cross section so that the zero line does not intersect the section, and at the same time calculate the corresponding point

application of compressive force K with

Rice. 8.7 dinami y F and z F according to the formulas:

; .

The resulting points of force application with coordinates y F , z F must be connected by straight lines. The area bounded by the resulting polyline will be the core of the section.

The sequence of constructing the section kernel

1. Determine the position of the center of gravity of the cross section and the main central axes of inertia y and z, as well as the values ​​of the squared radii of inertia i y , i z .

2. Show all possible positions of the n.l. relating to the contour of the section.

3. For each position of n.l. define segments a y and a z, cut off by it from the main central axes of inertia y and z.

4. For each position of n.l. set the coordinates of the center of pressure y F, and z F .

5. The obtained centers of pressure are connected by line segments, inside which the core of the section will be located.

Torsion with bend

The type of loading in which the bar is subjected to the action of twisting and bending moments at the same time is called bending with torsion.

When calculating, we use the principle of independence of the action of forces. Let's determine the stresses separately during bending and torsion (Fig. 8.8) .

When bending in the cross section, normal stresses arise, reaching a maximum value in the outermost fibers

.

During torsion in the cross section, shear stresses arise, reaching the highest value at the points of the section near the shaft surface

.

s

t

C

B

x

y

z

Rice. 8.9

s

s

t

t

Rice. 8.10

C

x

z

y

M

T

Rice. 8.8

Normal and shear stresses simultaneously reach their maximum value at the points FROM and AT shaft section (Fig. 8.9). Consider the stress state at the point FROM(Fig. 8.10). It can be seen that the elementary parallelepiped selected around the point FROM, is in a plane stress state.

Therefore, to test the strength, we apply one of the strength hypotheses.

Strength condition according to the third strength hypothesis (the hypothesis of the largest shear stresses)

.

Given that , , we obtain the condition of the shaft strength

. (8.6)

If the shaft bends in two planes, then the strength condition will be

.

Using the fourth (energy) strength hypothesis

,

after substitution s and t we get

. (8.7)

Questions for self-examination

1. What kind of bend is called oblique?

2. What combination of types of bend is an oblique bend?

3. What formulas are used to determine the normal stresses in the cross sections of a beam during oblique bending?

4. How is the position of the neutral axis in oblique bending?

5. How are dangerous points determined in a section with oblique bending?

6. How are the displacements of the beam axis points determined during oblique bending?

7. What kind of complex resistance is called eccentric tension (or compression)?

8. What formulas are used to determine the normal stresses in the cross sections of the rod during eccentric tension and compression? What is the form of the diagram of these stresses?

9. How is the position of the neutral axis determined in eccentric tension and compression? Write down the corresponding formulas.

10. What stresses occur in the cross section of the beam when bending with torsion?

11. How are the dangerous sections of a round beam in bending with torsion?

12. Which points of a circular cross section are dangerous when bending with torsion?

13. What stress state occurs at these points?

Eccentric tension or compression this type of rod deformation is called, in which a longitudinal force and bending moments (and, perhaps, transverse forces) arise in its cross section.

The longitudinal force and bending moments can be considered as the result of an eccentrically applied force acting on the rod (Fig. 25). That is why this kind of complex resistance is called eccentric tension or compression.

Bending moments are related to the coordinates of the force application point by the relations Therefore, from (1), formula (1) Ch. 3 and the principle of independence of the action of forces for normal stresses at an arbitrary point of any cross section with coordinates x, y, we obtain

Neutral axis in eccentric tension or compression. The equation of the neutral axis of the cross section, at the points of which the stresses are equal to zero, in this case has the form

It is easy to see that the neutral axis does not pass through the center of gravity of the section. The remaining properties are the same as for oblique bending. In addition, we indicate one more property of the neutral axis in eccentric tension or compression: the neutral axis does not intersect the quarter of the section in which the force is applied

Section kernel. The position of the neutral axis, as can be seen from equation (4), depends on the coordinates of the force application point. . all points of the section experience normal stresses of the same sign. On fig. 26 shows the kernels for rectangular and circular sections.

The strength conditions for eccentric tension or compression have the form of restrictions on the maximum normal stresses.

Example. Calculate the maximum normal stresses in the cross section of an eccentrically compressed rectangular rod at (Fig. 27). Point K of force application has coordinates (Fig. 27, b).

Solution. Compute geometric characteristics sections:

The equation of the neutral axis (4) takes the form From its location (Fig. 27, b) it can be seen that B and C are the most stressed points