Tension - compression This type of deformation is called in which only a longitudinal force N occurs in the cross section of the beam.

Straight bars working in tension - compression are called rods.

Longitudinal force is called the resultant of all internal normal forces arising in this section.

The longitudinal force in any stressed section of a beam is determined by the method of sections: it is equal to the algebraic sum of the projections of all external forces applied on one side of the section under consideration onto the longitudinal axis.

If the longitudinal force along the entire length of the beam is not constant, then plot “N” is constructed. Diagram is a graph of changes in the internal force factor along the length of the beam.

Rules for constructing diagrams of longitudinal forces:

We divide the beam into sections, the boundaries of which are the sections where external forces are applied.

Within each section, the section method is used and the longitudinal force is determined. Moreover, if an external force stretches the remaining part of the rod, i.e. directed away from the section - the longitudinal force is positive;

if an external force compresses the remaining part of the rod, i.e. directed towards the section - the longitudinal force is negative.

We set aside the obtained values ​​and construct a diagram of longitudinal forces. If a uniformly distributed load does not act on the section, then the diagram is limited to a straight line parallel to the zero line.

The correctness of constructing diagrams of longitudinal forces is determined as follows: in sections where an external force is applied, there are “jumps” on the diagram, equal in magnitude to the applied force.

Rules for constructing normal stress diagrams:

We divide the beam into sections, the boundaries of which are the points of application of external forces and sections where the area changes.



At each section we calculate normal stresses using the formula We construct a diagram of normal stresses, from which we determine the dangerous section.

In tension-compression, the dangerous section is the one in which the magnitude of the normal stresses is greatest.

When stretched, the length of the part increases and the cross-section decreases; when compressed, the opposite is true.

∆l = l – l 0 - absolute elongation.

relative elongation or longitudinal deformation.

Hooke's law in tension - compression: 

The absolute elongation value is calculated using Hooke's formula:

Algorithm for solving problems of constructing diagrams of longitudinal forces and

normal stresses, calculation of absolute elongation of the rod

Divide the zero line into sections to construct a diagram of longitudinal forces. Draw the boundaries of the sections in sections where external forces are applied.

At each section, calculate the longitudinal force using the section method.

Set aside the obtained values ​​and construct a diagram of longitudinal forces. Correctness is controlled as follows: in sections where external forces are applied to the rod, there are “jumps” on the diagram of longitudinal forces that are numerically equal to these forces.

Divide the zero line into sections to construct a normal stress diagram. The boundaries of the sections are sections in which the area changes and external forces are applied.

At each section, calculate the normal stress using the formula

Set aside the obtained values ​​and construct a diagram of normal stresses. Using the diagram, determine the dangerous section of the part. Dangerous sections are the sections at which the normal stresses are greatest.

For each section on the normal stress diagram, calculate the absolute elongation using Hooke’s formula.

Determine the total value of absolute elongation for the entire part as a whole: find the algebraic sum of the absolute elongations of all sections. Moreover, if the total value is positive, the rod has lengthened; if it is negative, the rod has shortened.

Analysis of the most common errors.

It should be remembered that on the diagram of longitudinal forces the boundaries of the sections pass at the points of application of external forces, and on the diagram of normal stresses - at the points of application of external forces and in sections where the area of ​​the rod changes.

In order to correctly substitute the values ​​into the normal stress formula, you need to go from the section of the stress diagram for which the calculation is being carried out to the normal force diagram and see what the value of the longitudinal force is in this particular section. Then go up to the drawing of the part and see what the cross-sectional area of ​​the rod is in this particular area.

When calculating absolute elongation, the longitudinal force should be substituted into Hooke’s formula from the longitudinal force diagram, and the cross-sectional area and length of a given section should be substituted from the part drawing.

The value of the longitudinal force for a given section should be substituted into the normal stress formula and Hooke’s formula.

Calculation of stress diagrams is a basic task of such a discipline as strength of materials. In particular, only with the help of a diagram is it possible to determine the maximum permissible load on the material

Also, a diagram is a schematic drawing or graph. It is practically not used in this meaning, see diagram.

Wikimedia Foundation.

2010.:

Synonyms

See what “Epyura” is in other dictionaries:

A working drawing in which the depicted structure is shown in the simplest and clearest way possible, indicating only those dimensions that are needed to carry out the work. For example, on the E. laying of a turnout there are switches, crosspieces and guardrails... ... Noun, number of synonyms: 1 diagram (2) Dictionary of synonyms ASIS. V.N. Trishin. 2013…

Synonym dictionary diagram

Diagram- Graphic representation of the law of change of function depending on change in argument Synonyms of diagrams EN diagramepure DE Zustandslinie FR diagrammeépure ... - – graph of parameter change along the considered axis of the element. [Polyakova, T.Yu. Road bridges: educational English-Russian and Russian-English terminological dictionary minimum / T.Yu. Polyakova, N.G. Karaseva, D.V. Polyakov. – M.: MADI, 2015. – …

Synonym dictionary Encyclopedia of terms, definitions and explanations of building materials - EPUR a, m., EPUR s, f épure f. 1. special A drawing of the projections of a figure obtained by combining the projection planes. BAS 1. The professor was told about him that he measured out his wife’s figure with a compass and, according to diagrams, tailored her a ball gown, which... ...

Historical Dictionary of Gallicisms of the Russian Language

Diagram, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams, diagrams (Source: “Full accentuated paradigm according to A. A. Zaliznyak”) ... Forms of words - [EPUR] graphic image law of change of function depending on change of argument (Bulgarian language; Български) diagram; schedule (Czech language; Čeština) obrazec vnitřních sil ( German ; Deutsch) Zustandslinie (Hungarian language;… …

Construction dictionary diagram of the main vectorial areas - Nrk. single diagram of normal stresses under constrained torsion. Diagram of sectorial areas enclosed between specially selected fixed and moving radius vectors. [Collection of recommended terms. Issue 82. Construction... ...

Technical Translator's Guide

Layout of sleeper axes on a rail link depending on the length of the rails and the number of sleepers per 1 km. According to E. u. w. Markings are made on the neck of the rails for laying sleepers. Technical railway dictionary. M.: State transport... ... Technical railway dictionary

Books

• Descriptive geometry and engineering graphics (for technical areas of training) (bachelor's degree) Textbook. , Georgievsky Oleg Viktorovich. Considered various ways diagram transformations, numerous examples of solving positional and metric problems are given. Information is provided on the types of products and design documents,…
• Descriptive geometry and engineering graphics (for technical areas of training) (bachelor's degree). Textbook, O. V. Georgievsky, V. I. Veselov, G. I. Nichugovsky. Various ways of transforming a diagram are considered, and numerous examples of solving positional and metric problems are given. Information is provided on the types of products and design documents,…

3. RULES FOR CONSTRUCTING DIAGRAMS OF INTERNAL FORCES M, Q, N

3.1. Bending moment diagram M

The procedure for constructing the ordinates of the diagram M

numerical value of the bending moment in the section.

2. Plot the found numerical value as an ordinate perpendicular to the axis of the rod from the side of the stretched fiber of the rod.

The numerical value of the bending moment in the section is equal to the numerical value of the algebraic sum of the moments of all forces acting on the rod system on either side of the section, taken relative to a point on the section axis.

How a stretched fiber is installed in a cross-section is demonstrated using the example of a broken-shaped console when it is loaded with three types of load (Fig. 3.1). The ordinates of the corresponding three diagrams M are plotted on the stretched side of the rods forming the console.

Signs the right kind diagrams M

With the specified rule for constructing the ordinates of the diagram M, this diagram has the following properties.

1. In a section of a straight rod free from load, the diagram is straight.

2. In the area of ​​distributed load, it is outlined by a curved line, convex in the direction of the load. When the load is uniformly distributed, the curve is a parabola of the second degree.

3. At the point of application of a concentrated force, the diagram has a kink, the tip of which is directed in the direction of the force.

4. At the point of application of the concentrated moment, the diagram has a jump in ordinates equal to the magnitude of the moment.

5. In the section located on the border of the unloaded section of the rod and the section loaded with a distributed load, the curved line of the diagram smoothly (without a break) turns into a straight line, which is tangent

To curved section.

These properties are used to control the constructed diagrams M.

Rule of signs for ordinates of diagrams M

When constructing the ordinates of the diagram M from the side of the stretched fiber of the rod manually, the ordinate sign was not required. However during numerical calculation on a PC, each ordinate of the diagram M a sign is assigned. The plot sign is used M and when constructing a diagram Q from it.

In this textbook The rule of signs adopted for the ordinates of diagrams M in the SCAD program is given.

If the “lower” fiber of the rod is stretched, then the ordinate is laid off from the axis of the rod “down” and is assigned the sign “+”

If the “upper” fiber of the rod is stretched, then the ordinate is plotted “up” from the axis of the rod and the sign “–” is assigned to it (Fig. 3.3).

The “bottom” fiber of a rod in the SCAD program is considered to be the fiber of a rod finite element (FE) of the “Flat frame rod” type, located on the side of the negative ordinates of the Z1 axis of the local coordinate system (MCS), and the “top” – on the side of the positive ordinates of the Z1 axis ( see Fig. 3.2, 3.3).

Note. When manually calculating the algebraic sum of the moments of all forces on one side of the section to determine the bending moment in the section of the rod, it is recommended to immediately put the signs of the moments in accordance with this sign rule. Then the ordinate of the bending moment will have its own sign in accordance with the accepted rule and can be plotted from the axis of the rod according to this rule.

Construction of diagram M on a rod element free from load

From the above properties of diagram M (signs of a correct diagram)

it is known that if there is no external load on the final element of the rod, then the diagram of bending moments on it will be rectilinear. To construct it, it is enough to calculate the ordinates only in the finite sections of such an element.

Note. In the SCAD program, to obtain the ordinates of bending moments on FEs loaded with a distributed load “by default,” a calculation can be assigned for several, for example, three FE sections: at the beginning (n), in the middle (s) and at the end (k) of finite elements ( the initial section “n” is associated with the beginning of the X1 axis in the MSC).

Then, in order to reduce the output results for FE without load within their limits

in the Assignments section on the toolbar, click the “Assignment of intermediate sections for force calculation” button. The Calculate Forces... dialog box will open (see SCAD help for this window). In the dialog box, enter the number 2 in the “number of sections” field. Next, you need to close the window and mark the finite elements on the diagram of the rod system, on which linear diagrams M are expected. How this is done is shown in the manuals.

In Fig. 3.2, 3.3, the end sections of the rod are indicated by the nodes “n” and “k” of the MSK. After assigning only two sections for the calculation of forces in the marked elements, the SCAD program in the corresponding force table will display the values ​​of bending moments Mn (M1) and Mk (M2) only in nodes “n” (1) and “k” (2) (with their

signs in MSC).

When digitizing the ordinates of a moment diagram, which is performed when the button is pressed

Display filter, within each finite element of the indicated two moments (M 1, M 2), the moment with the maximum value is given.

Construction of diagram M on a rod element under the action of a uniformly distributed load along its length

If a uniformly distributed load is located along the entire length of the FE, then the diagram of bending moments on it will have the form of a parabola with a convexity directed towards the action of the load.

Note. In the SCAD program, using the procedure that has just been discussed for calculating bending moments in only two FE sections, you can assign the calculation of moments in a number of sections between the nodes “n” and “k” of the element in the MSK.

To approximate the construction of a parabola, it is enough to calculate the ordinates of the diagram M in three sections of the FE: at the beginning of “n”, in the middle of “c” and at the end of “k”. In the resulting force table of the SCAD program, these sections are designated 1, 2, 3, respectively. In the SCAD program, the calculation of moments in the specified sections can be provided by default. However, if for some reason the designer only knew two ordinates of the diagram M at the ends of the element (M n and M k ), then you can easily calculate

ordinate M c in the middle section, applying the principle of independence of forces.

Example. Let us cut out (at nodes “n” (1) and “k” (3) MSK) from the rod system an element loaded with a uniformly distributed load of intensity q (Fig. 3.4,a).

Let's consider it as a beam on two supports, under the influence of internal forces at the ends of the element and a distributed load (Fig. 3.4, b). Adding the indicated three support links does not affect the forces in the element, since in the cut state it is in equilibrium, therefore, in the added links the forces (reactions) will be zero.+ M c o .

Both summed ordinates in the example considered are positive, since they are located below the beam axis. In Fig. 3.5 shows the option when the ordinate

M c (crowbar) = 0.5(M n + M k) is negative (the ordinate M c o = ql 2 / 8 with the indicated load direction q is positive). Here is also a graphical-analytical method for constructing a parabolic diagram using its three total ordinates (Mn, Ms, Mk) and three

tangent to the parabola at the corresponding ends of the ordinates (marked with a cross).

The meaning of this graphic-analytical method will be clear if we look at Fig. 3.4, d diagram M (R) of a triangular shape, shown with dashed lines. Diagram

Occurring in various cross sections rod are not identical, the law of their change along the length of the rod is presented in the form of a graph N(z), called diagram of longitudinal forces. The diagram of longitudinal forces is necessary to evaluate the rod and is constructed in order to find the dangerous section (the cross section in which the longitudinal force takes highest value ).

How to build a diagram of longitudinal forces?

To construct the diagram N is used. Let's demonstrate its application with an example (Fig. 2.1).

Let us determine the longitudinal force N arising in the cross section we have planned.

Let's cut the rod in this place and mentally discard its lower part (Fig. 2.1, a). Next we must replace the action of the discarded part with top part rod internal longitudinal force N.

To make it easier to calculate its value, let’s cover the upper part of the rod we are considering with a piece of paper. Let us recall that N arising in the cross section can be defined as the algebraic sum of all longitudinal forces acting on the rejected part of the rod, that is, on the part of the rod that we see.

In this case, we apply the following: the forces causing tension in the remaining part of the rod (covered by us with a piece of paper) are included in the mentioned algebraic sum with a “plus” sign, and the forces causing compression – with a “minus” sign.

So, to determine the longitudinal force N in the cross section we have planned, we simply need to add up all the external forces that we see. Since the force kN stretches the upper part, and the force kN compresses it, then kN.

The minus sign means that in this section the rod experiences compression.

You can find the support reaction R (Fig. 2.1, b) and create an equilibrium equation for the entire rod to check the result.

When calculating strength, it is necessary to know the law of change in internal forces in the cross sections of a beam along its length, arising from the load acting on the beam. This law can be expressed in the form of analytical dependencies and depicted using special graphs called diagrams.

A diagram of bending moments (diagram) is a graph depicting the law of change in the magnitudes of these moments along the length of the beam. Similarly, a diagram of transverse forces (diagram Q) or a diagram of longitudinal forces (diagram N) is a graph depicting the change in transverse or longitudinal forces along the length of a beam.

Each ordinate of the diagram M (or Q, or N) represents the magnitude of the bending moment (or transverse force, or longitudinal force) in the corresponding cross section of the beam.

Let us use specific examples to analyze the construction of diagrams for beams under the influence of a system of forces located in one plane (parallel to the plane of the drawing).

Let us construct diagrams Q and M for the cantilever beam, embedded at the right end, shown in Fig. 10.7, a.

Let us call each part of a beam a section within which the laws of change in shear force and bending moment remain constant. The boundaries of the sections are the cross sections of the beam in which concentrated loads are applied to it (including support reactions) or in which a distributed load begins or ends, or in which the intensity of this load begins to change according to a new law.

The beam under consideration has four sections I, II, III and IV, shown in Fig. 10.7, a.

Let us compile [based on formulas (3.7) and (2.7)] expressions for the transverse force and bending moment in the cross section of the beam at a distance x from its left end.

Plot:

Here, is the resultant of a uniformly distributed load within a segment of length section I. It is applied in the middle of this segment, and therefore its moment relative to the section under consideration is equal to. The sign of the transverse force is negative because the projection of the resultant is directed downward; the sign of the bending moment is negative because the moment acts counterclockwise.

In the final expressions, the value is substituted in meters, since the intensity q is expressed in

The resulting expressions Q and are valid within section I, i.e., at a distance ranging from 0 to

The dependence on is linear, and therefore to construct a diagram of Q in the area it is enough to determine the values ​​​​at two values

at (at the beginning of section I)

at (at the end of section I)

The dependence of M on is not linear, but quadratic. To construct a diagram of M on a section, we calculate the values ​​​​at three values

According to the obtained values ​​in Fig. 10.7, b, c, diagrams of Q and M were constructed for the section of the beam (straight and curved)

Ordinates of diagrams corresponding positive values internal forces, we put it upward from the axes of these diagrams, and negative - down (the axes of the diagrams are parallel to the axis of the beam). With this construction, the ordinates of the M diagrams are located on the side of the compressed fibers of the beam.

where the distance is expressed in meters.

At (at the beginning of section II)

at (at the end of the section)

According to the obtained values ​​in Fig. 10.7, b, c plots Q and M are plotted for section II of the beam (straight lines to and Section III:

At (at the beginning of section III)

at (at the end of section III)

According to the obtained values ​​in Fig. 10.7, b, c, diagrams of Q and M are plotted for section III of the beam (straight lines and c). Section IV:

According to the obtained values ​​in Fig. 10.7, b, c plots Q and M are plotted for section IV of the beam (straight lines).

Bending moments and shear forces in cross sections can also be determined through the right external forces, using dependencies. But for this it is necessary to find the values ​​of the support reactions in the embedment B of the beam.

Let us now select a part CD of length from the beam (Fig. 10.7, a) and apply all external forces acting on it (Fig. 10.7, d). These include force and moment, as well as forces and moments applied to the part in question in cross sections C, and these forces and moments are equal to the shear forces and bending moments in sections C and D and represent the effect of parts AC and DB on the part

The transverse force in section C of the beam, as can be seen from diagram Q (Fig. 10.7, b), is equal and negative; in accordance with the accepted rule of signs, it tends to rotate part of the CD beam counterclockwise, relative to some point E of the beam (Fig. 10.7, d) and, therefore, should be directed downward. The transverse force QD in section D is positive, equal to (Fig. 10.7, b) and, therefore, tends to rotate part CD of the beam clockwise relative to the point?; therefore it should be directed downwards (Fig. 10.7, d).

Bending moments and MD in sections C and D are equal, respectively, i.e. they are negative (Fig. 10.7, c); therefore, both cause compression of the lower fibers and tension of the upper fibers of the beam. In accordance with this, the moment is directed counterclockwise, and the moment is directed clockwise.

Let's make sure that the selected part of the CD beam is in equilibrium. To do this, we will compile three equilibrium equations for all forces acting on it (see Fig. 10.7, d):

The equality of the values ​​to zero indicates the equilibrium of the CD part of the beam.

In Fig. 10.7, (3 shows the internal forces acting in section B of the beam, coinciding with its embedded end. Their magnitudes and directions are established according to diagrams Q and M (Fig. 10.7, b, c). They represent pinching reactions of beam B.

From diagram Q (Fig. 10.7, b) it is clear that in section F of the beam, in which a concentrated force is applied to it, the value of the transverse force changes abruptly from to, i.e., by the value P.

This is a consequence of the fact that this force is not included in the expression compiled for the section located at a distance to the left of the force P; it enters into the expression compiled for the section located at a distance to the right of the force P.

So, in a section in which a concentrated external force is applied to the beam, perpendicular to the axis of the beam (including the support reaction in the form of a concentrated force), the value of the transverse force Q changes abruptly by the magnitude of the applied force. When the concentrated external force is directed upward, there is an upward jump in the Q diagram (when moving from left to right), and when the force is directed downward, there is a downward jump.

Similarly, in a section in which a concentrated external moment is applied to the beam (including the support reaction in the form of a concentrated moment), the value of the bending moment M changes abruptly by the value of the applied moment. When the concentrated external moment acts clockwise, there is an upward jump on the diagram M (when moving from left to right); and when the moment acts counterclockwise - a downward jump. So, for example, in section G of a beam, in which a concentrated moment is applied to it (Fig. 10.7, a), in diagram M (Fig. 10.7, c) there is an upward jump (when moving from left to right), equal to and in section B- downward jump, equal (i.e. equal to the reaction of support B in the form of a concentrated moment directed counterclockwise).

Let us now construct diagrams Q and M for a simple beam on two supports, shown in Fig. 11.7, a. The beam consists of two sections.

Let us determine the vertical support reactions RA and RB of the beam. A horizontal reaction may also occur in support A, but for a given vertical load it is zero. To determine reactions and RB, we draw up equilibrium equations in the form of sums of the moments of all forces relative to points A and B.