Angle of rotation of the cross section. Scientific electronic library Calculation of beam displacement during bending

Lecture 13 (continued). Examples of solutions for calculating displacements using the Mohr-Vereshchagin method and problems for independent solution

Defining displacements in beams

Example 1.

Determine the movement of a point TO beams (see figure) using the Mohr integral.

Solution.

1) We compose an equation for the bending moment from an external force M F .

2) Apply at the point TO unit force F = 1.

3) We write down the equation of bending moment from a unit force.

4) Determine movements

Example 2.

Determine the movement of a point TO beams according to Vereshchagin's method.

Solution.

1) We are building a cargo diagram.

2) We apply a unit force at point K.

3) We build a single diagram.

4) Determine the deflection

Example 3.

Determine the angles of rotation on the supports A And IN

Solution.

We construct diagrams from a given load and from individual moments applied in sections A And IN(see picture). We determine the required displacements using Mohr integrals

,

, which we calculate using Vereshchagin’s rule.

Finding the plot parameters

C 1 = 2/3, C 2 = 1/3,

and then the rotation angles on the supports A And IN

Example 4.

Determine the angle of rotation of the section WITH for a given beam (see figure).

Solution.

Determining support reactions R A =R B ,

, , R A = R B = qa.

We construct diagrams of the bending moment from a given load and from a single moment applied in the section WITH, where the rotation angle is sought. We calculate the Mohr integral using Vereshchagin's rule. Finding the plot parameters

C 2 = -C 1 = -1/4,

and along them the desired movement

Example 5.

Determine the deflection in the section WITH for a given beam (see figure).

Solution.

Diagram M F(Fig. b)

Support reactions:

BE: , ,

, R B + R E = F, R E = 0;

AB: , R A = R IN = F; , .

We calculate moments at characteristic points, M B = 0, M C = Fa and build a diagram of the bending moment from a given load.

Diagram(Fig. c).

In cross section WITH, where the deflection is sought, we apply a unit force and construct a bending moment diagram from it, first calculating the support reactions BE - , , = 2/3; , , = 1/3, and then moments at characteristic points , , .

2. Determination of the desired deflection. Let's use Vereshchagin's rule and first calculate the parameters of the diagrams and:

,

Section deflection WITH

Example 6.

Determine the deflection in the section WITH for a given beam (see figure).

Solution.

WITH. Using Vereshchagin’s rule, we calculate the parameters of the diagrams ,

and find the desired deflection

Example 7.

Determine the deflection in the section WITH for a given beam (see figure).

Solution.

1. Constructing diagrams of bending moments.

Support reactions:

, , R A = 2qa,

, R A + R D = 3qa, R D = qa.

We construct diagrams of bending moments from a given load and from a unit force applied at a point WITH.

2. Determination of movements. To calculate the Mohr integral, we use Simpson’s formula, sequentially applying it to each of the three sections into which the beam is divided.

PlotAB :

PlotSun :

PlotWITH D :

Required movement

Example 8.

Determine the section deflection A and section rotation angle E for a given beam (Fig. A).

Solution.

1. Constructing diagrams of bending moments.

Diagram M F(rice. V). Having determined the support reactions

, , R B = 19qa/8,

, R D = 13qa/8, we build diagrams of transverse force Q and bending moment M F from a given load.

Diagram(Fig. d). In cross section A, where the deflection is sought, we apply a unit force and construct a diagram of the bending moment from it.

Diagram(Fig. e). This diagram is constructed from a single moment applied in the section E, where the rotation angle is sought.

2. Determination of movements. Section deflection A we find using Vereshchagin's rule. Epure M F at the sites Sun And CD We break it down into simple parts (Fig. d). We present the necessary calculations in the form of a table.

We get.

The minus sign in the result means that the point A does not move downwards, as the unit force was directed, but upwards.

Section rotation angle E we find in two ways: by Vereshchagin’s rule and by Simpson’s formula.

According to Vereshchagin’s rule, multiplying diagrams M F and, by analogy with the previous one, we obtain

,

To find the angle of rotation using Simpson's formula, we calculate the preliminary bending moments in the middle of the sections:

The required displacement, increased by EI x once,

Example 9.

Determine at what value of the coefficient k section deflection WITH will be equal to zero. When the value is found k construct a diagram of the bending moment and depict an approximate view of the elastic line of the beam (see figure).

Solution.

We construct diagrams of bending moments from a given load and from a unit force applied in the section WITH, where the deflection is sought.

According to the conditions of the problem V C= 0. On the other hand, . Integral on the plot AB we calculate using Simpson’s formula, and in the section Sun– according to Vereshchagin’s rule.

We find in advance

Moving a section WITH ,

From here , .

When the value is found k determine the value of the support reaction at the point A: , , , from which we find the position of the extremum point on the diagram M according to the condition .

Based on the moment values ​​at characteristic points

We build a diagram of the bending moment (Fig. d).

Example 10.

IN cantilever beam shown in the figure.

Solution.

M from the action of an external concentrated force F: M IN = 0, M A = –F 2l(linear plot).

According to the conditions of the problem, it is necessary to determine the vertical displacement at IN points IN cantilever beam, therefore we build a unit diagram of the action of a vertical unit force F i = 1 applied at the point IN.

Considering that the cantilever beam consists of two sections with different bending rigidities, diagrams and M We multiply using Vereshchagin’s rule by sections separately. Diagrams M and multiply the first section using the formula , and the diagrams of the second section - as the area of ​​the diagram M second section Fl 2 / 2 to ordinate 2 l/3 diagrams of the second section under the center of gravity of the triangular diagram M the same area.

In this case the formula gives:

Example 11.

Determine the vertical displacement of a point IN single-span beam shown in the figure. The beam has a constant bending rigidity along its entire length. EI.

Solution.

We build a diagram of bending moments M from the action of an external distributed load: M A = 0; M D = 0;

Apply at the point IN unit vertical force F i = 1 and build a diagram (see figure):

where R a = 2/3;

Where R d = 1/3, so M a = 0; M d = 0; .

Let's divide the beam in question into 3 sections. Multiplying diagrams of the 1st and 3rd sections does not cause difficulties, since we multiply triangular diagrams. In order to apply Vereshchagin’s rule to the 2nd section, let’s split the diagram M 2nd section into two components of the diagram: rectangular and parabolic with area (see table).

Center of gravity of the parabolic part of the diagram M lies in the middle of the 2nd section.

So the formula using Vereshchagin's rule gives:

Example 12.

Determine the maximum deflection in a two-support beam loaded with a uniformly distributed load of intensity q(see picture).

Solution.

Finding bending moments:

From a given load

From a unit force applied at a point WITH where the deflection is sought.

We calculate the required maximum deflection that occurs in the middle section of the beam

Example 13.

Determine the deflection at a point IN beam shown in the figure.

Solution.

We construct diagrams of bending moments from a given load and a unit force applied at a point IN. To multiply these diagrams, the beam must be divided into three sections, since a single diagram is limited to three different straight lines.

The operation of multiplying diagrams in the second and third sections is carried out simply. Difficulties arise when calculating the area and coordinates of the center of gravity of the main diagram in the first section. In such cases, constructing layered diagrams greatly simplifies the solution of the problem. In this case, it is convenient to take one of the sections conditionally as stationary and construct diagrams for each of the loads, approaching this section from the right and left. It is advisable to take the section at the fracture site as a stationary one in the diagram of unit loads.

A layered diagram in which the section is taken as the stationary one IN, is shown in the figure. Having calculated the areas of the component parts of the layered diagram and the corresponding ordinates of the unit diagram, we obtain

Example 14.

Determine the displacements at points 1 and 2 of the beam (Fig. a).

Solution.

Here are the diagrams M And Q for beams at A=2 m; q=10 kN/m; WITH=1,5A; M=0,5qa 2 ; R=0,8qa; M 0 =M; =200 MPa (Fig. b And V).

Let us determine the vertical displacement of the center of the section where the concentrated moment is applied. To do this, consider a beam in a state under the influence of only a concentrated force applied at point 1 perpendicular to the axis of the beam (in the direction of the desired displacement) (Fig. d).

Let's calculate the support reactions by composing three equilibrium equations

Examination

The reactions were found correctly.

To construct a diagram, consider three sections (Fig. d).

1 plot

2nd section

Section 3

Using these data, we construct a diagram (Fig. e) from the side of the stretched fibers.

Let us determine by Mohr's formula using Vereshchagin's rule. In this case, a curved diagram in the area between the supports can be represented as the addition of three diagrams. Arrow

The minus sign means that point 1 moves up (in the opposite direction).

Let us determine the vertical displacement of point 2, where the concentrated force is applied. To do this, consider a beam in a state under the influence of only a concentrated force applied at point 2 perpendicular to the axis of the beam (in the direction of the desired displacement) (Fig. e).

The diagram is constructed similarly to the previous one.

Point 2 moves up.

Let us determine the angle of rotation of the section where the concentrated moment is applied.

Straight bend. Flat transverse bend 1.1. Construction of diagrams of internal force factors for beams 1.2. Construction of diagrams Q and M using equations 1.3. Construction of Q and M diagrams using characteristic sections (points) 1.4. Strength calculations for direct bending of beams 1.5. Main stresses during bending. Full check of beam strength 1.6. The concept of the center of bending 1.7. Determination of displacements in beams during bending. Concepts of beam deformation and conditions of their rigidity 1.8. Differential equation of the curved axis of a beam 1.9. Direct integration method 1.10. Examples of determining displacements in beams using the direct integration method 1.11. Physical meaning of integration constants 1.12. Method of initial parameters (universal equation of the curved axis of a beam) 1.13. Examples of determining displacements in a beam using the initial parameters method 1.14. Determination of displacements using Mohr's method. Rule A.K. Vereshchagina 1.15. Calculation of the Mohr integral according to the rule of A.K. Vereshchagina 1.16. Examples of determining displacements using the Mohr integral Bibliography 4 1. Direct bending. Flat transverse bend. 1.1. Constructing diagrams of internal force factors for beams Direct bending is a type of deformation in which two internal force factors arise in the cross sections of the rod: a bending moment and a transverse force. In a particular case, the shear force can be zero, then the bending is called pure. In flat transverse bending, all forces are located in one of the main planes of inertia of the rod and perpendicular to its longitudinal axis, and moments are located in the same plane (Fig. 1.1, a, b). Rice. 1.1 The transverse force in an arbitrary cross section of a beam is numerically equal to the algebraic sum of the projections onto the normal to the beam axis of all external forces acting on one side of the section under consideration. Lateral force in cross section m-n beams (Fig. 1.2, a) is considered positive if the resultant of external forces to the left of the section is directed upward, and to the right - downward, and negative - in the opposite case (Fig. 1.2, b). Rice. 1.2 When calculating the transverse force in a given section, external forces lying to the left of the section are taken with a plus sign if they are directed upwards, and with a minus sign if they are directed downwards. For the right side of the beam - vice versa. 5 The bending moment in an arbitrary cross section of a beam is numerically equal to the algebraic sum of the moments about the central axis z of the section of all external forces acting on one side of the section under consideration. The bending moment in the m-n section of the beam (Fig. 1.3, a) is considered positive if the resultant moment of external forces to the left of the section is directed clockwise, and to the right - counterclockwise, and negative - in the opposite case (Fig. 1.3, b). Rice. 1.3 When calculating the bending moment in a given section, the moments of external forces lying to the left of the section are considered positive if they are directed clockwise. For the right side of the beam - vice versa. It is convenient to determine the sign of the bending moment by the nature of the deformation of the beam. The bending moment is considered positive if, in the section under consideration, the cut-off part of the beam bends convexly downward, i.e., the lower fibers are stretched. In the opposite case, the bending moment in the section is negative. There are differential relationships between the bending moment M, shear force Q and load intensity q. 1. The first derivative of the shear force along the abscissa of the section is equal to the intensity of the distributed load, i.e. . (1.1) 2. The first derivative of the bending moment along the abscissa of the section is equal to the transverse force, i.e. (1.2) 3. The second derivative along the abscissa of the section is equal to the intensity of the distributed load, i.e. (1.3) We consider the distributed load directed upward positive. A number of important conclusions follow from the differential relationships between M, Q, q: 1. If on the beam section: a) the transverse force is positive, then the bending moment increases; b) the shear force is negative, then the bending moment decreases; c) the transverse force is zero, then the bending moment has a constant value (pure bending); 6 d) the transverse force passes through zero, changing sign from plus to minus, max M M, in the opposite case M Mmin. 2. If there is no distributed load on the beam section, then the transverse force is constant, and the bending moment changes according to a linear law. 3. If there is a uniformly distributed load on a section of the beam, then the transverse force changes according to a linear law, and the bending moment - according to the law of a square parabola, convexly facing in the direction of the load (in the case of constructing diagram M from the side of stretched fibers). 4. In the section under a concentrated force, diagram Q has a jump (by the magnitude of the force), diagram M has a kink in the direction of the force. 5. In the section where a concentrated moment is applied, the diagram M has a jump equal to the value of this moment. This is not reflected in the Q diagram. When beams are loaded with complex loading, diagrams of transverse forces Q and bending moments M are plotted. Diagram Q(M) is a graph showing the law of change in transverse force (bending moment) along the length of the beam. Based on the analysis of diagrams M and Q, dangerous sections of the beam are determined. Positive ordinates of the Q diagram are laid upward, and negative ordinates are laid down from the base line drawn parallel to the longitudinal axis of the beam. Positive ordinates of the M diagram are laid down, and negative ordinates are laid upward, i.e., the M diagram is constructed from the side of the stretched fibers. The construction of Q and M diagrams for beams should begin with determining the support reactions. For a beam with one clamped end and the other free end, the construction of diagrams Q and M can be started from the free end, without determining the reactions in the embedment. 1.2. The construction of Q and M diagrams using the Beam equations is divided into sections within which the functions for the bending moment and shear force remain constant (do not have discontinuities). The boundaries of the sections are the points of application of concentrated forces, pairs of forces and places of change in the intensity of the distributed load. At each section, an arbitrary section is taken at a distance x from the origin of coordinates, and for this section equations for Q and M are drawn up. Using these equations, diagrams of Q and M are constructed. Example 1.1 Construct diagrams of transverse forces Q and bending moments M for a given beam (Fig. 1.4,a). Solution: 1. Determination of support reactions. We compose equilibrium equations: from which we obtain The reactions of the supports are determined correctly. The beam has four sections Fig. 1.4 loads: CA, AD, DB, BE. 2. Construction of diagram Q. Section CA. In section CA 1, we draw an arbitrary section 1-1 at a distance x1 from the left end of the beam. We define Q as the algebraic sum of all external forces acting to the left of section 1-1: 1 Q 3 0 kN. The minus sign is taken because the force acting to the left of the section is directed downward. The expression for Q does not depend on the variable x1. Diagram Q in this section will be depicted as a straight line parallel to the abscissa axis. Section AD. On the section we draw an arbitrary section 2-2 at a distance x2 from the left end of the beam. We define Q2 as the algebraic sum of all external forces acting to the left of section 2-2: The value of Q is constant over the section (does not depend on the variable x2). The Q plot on the section is a straight line parallel to the abscissa axis. Plot DB. On the site we draw an arbitrary section 3-3 at a distance x3 from the right end of the beam. We define Q3 as the algebraic sum of all external forces acting to the right of section 3-3: . The resulting expression is the equation of an inclined straight line. Section BE. On the site we draw a section 4-4 at a distance x4 from the right end of the beam. We define Q as the algebraic sum of all external forces acting to the right of section 4-4: Here the plus sign is taken because the resultant load to the right of section 4-4 is directed downward. Based on the obtained values, we construct Q diagrams (Fig. 1.4, b). 3. Construction of diagram M. Section CA m1. We define the bending moment in section 1-1 as the algebraic sum of the moments of forces acting to the left of section 1-1. – equation of a straight line. Plot. 3We define the bending moment in section 2-2 as the algebraic sum of the moments of forces acting to the left of section 2-2. – equation of a straight line. Plot. 4We define the bending moment in section 3-3 as the algebraic sum of the moments of forces acting to the right of section 3-3. – equation of a quadratic parabola. 9 We find three values ​​at the ends of the section and at the point with coordinate xk, where since here we have kNm. Plot. 1We define the bending moment in section 4-4 as the algebraic sum of the moments of forces acting to the right of section 4-4. – equation of a quadratic parabola, we find three values ​​of M4: Using the obtained values, we construct a diagram of M (Fig. 1.4, c). In sections CA and AD, the Q diagram is limited by straight lines parallel to the abscissa axis, and in sections DB and BE - by inclined straight lines. In sections C, A and B on the Q diagram there are jumps in the magnitude of the corresponding forces, which serves as a check for the correctness of the Q plot. In sections where Q 0, the moments increase from left to right. In areas where Q 0, the moments decrease. Under the concentrated forces there are kinks in the direction of the action of the forces. Under the concentrated moment there is a jump in the magnitude of the moment. This indicates the correctness of the construction of diagram M. Example 1.2 Construct diagrams Q and M for a beam on two supports loaded with a distributed load, the intensity of which varies according to a linear law (Fig. 1.5, a). Solution Determination of support reactions. The resultant of the distributed load is equal to the area of ​​the triangle, which is a diagram of the load and is applied at the center of gravity of this triangle. We compile the sums of the moments of all forces relative to points A and B: Constructing diagram Q. Let's draw an arbitrary section at a distance x from the left support. The ordinate of the load diagram corresponding to the section is determined from the similarity of triangles The resultant of that part of the load that is located to the left of the section The transverse force in the section is equal The transverse force changes according to the law of a square parabola Equating the equation of the transverse force to zero, we find the abscissa of the section in which the diagram Q passes through zero: The Q plot is shown in Fig. 1.5, b. The bending moment in an arbitrary section is equal to. The bending moment varies according to the law of a cubic parabola: The bending moment has a maximum value in the section where Q 0, i.e. at Diagram M is shown in Fig. 1.5, c. 1.3. Constructing diagrams of Q and M from characteristic sections (points) Using differential dependencies between M, Q, q and the conclusions arising from them, it is advisable to construct diagrams of Q and M from characteristic sections (without drawing up equations). Using this method, the values ​​of Q and M are calculated in characteristic sections. The characteristic sections are the boundary sections of sections, as well as sections where a given internal force factor has an extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of the differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct diagrams Q and M for the beam shown in Fig. 1.6, a. We start constructing the Q and M diagrams from the free end of the beam, while the reactions in the embedment need not be determined. The beam has three loading sections: AB, BC, CD. There is no distributed load in sections AB and BC. Shear forces are constant. The Q diagram is limited to straight lines parallel to the x-axis. Bending moments vary linearly. Diagram M is limited by straight lines inclined to the abscissa axis. There is a uniformly distributed load on section CD. Transverse forces vary according to a linear law, and bending moments - according to the law of a square parabola with convexity in the direction of the distributed load. At the boundary of sections AB and BC, the transverse force changes abruptly. At the boundary of sections BC and CD, the bending moment changes abruptly. 1. Construction of diagram Q. We calculate the values ​​of transverse forces Q in the boundary sections of sections: Based on the calculation results, we construct diagram Q for the beam (Fig. 1, b). From diagram Q it follows that the transverse force in section CD is equal to zero in the section located at a distance qa a q  from the beginning of this section. In this section, the bending moment has its maximum value. 2. Construction of diagram M. We calculate the values ​​of bending moments in the boundary sections of sections: At Kx3, the maximum moment in the section. Based on the calculation results, we construct diagram M (Fig. 5.6, c). Example 1.4 Using a given diagram of bending moments (Fig. 1.7, a) for a beam (Fig. 1.7, b), determine the acting loads and construct diagram Q. The circle indicates the vertex of a square parabola. Solution: Let's determine the loads acting on the beam. Section AC is loaded with a uniformly distributed load, since the diagram M in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting clockwise, since in diagram M we have a jump upward by the magnitude of the moment. In the NE section, the beam is not loaded, since the M diagram in this section is limited by an inclined straight line. The reaction of support B is determined from the condition that the bending moment in section C is equal to zero, i.e. To determine the intensity of the distributed load, we create an expression for the bending moment in section A as the sum of the moments of forces on the right and equate it to zero. Now we determine the reaction of support A. For this Let's create an expression for bending moments in the section as the sum of the moments of forces on the left, from where Fig. 1.7 Check The design diagram of the beam with a load is shown in Fig. 1.7, c. Starting from the left end of the beam, we calculate the values ​​of transverse forces in the boundary sections of the sections: Diagram Q is shown in Fig. 1.7, d. The considered problem can be solved by drawing up functional dependencies for M, Q in each section. Let's choose the origin of coordinates at the left end of the beam. In the AC section, the diagram M is expressed by a square parabola, the equation of which has the form Constants a, b, c are found from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the equation of the parabola, we obtain: The expression for the bending moment will be Differentiating the function M1 , we obtain the dependence for the transverse force. After differentiating the function Q, we obtain an expression for the intensity of the distributed load. In the section NE, the expression for the bending moment is presented in the form of a linear function. To determine the constants a and b, we use the conditions that this straight line passes through two points, the coordinates of which are known. We obtain two equations: from which we have a 10, b  20. The equation for the bending moment in the section NE will be After double differentiation of M2, we will find. Using the found values ​​of M and Q, we construct diagrams of bending moments and shear forces for the beam. In addition to the distributed load, concentrated forces are applied to the beam in three sections, where there are jumps on diagram Q and concentrated moments in the section where there is a shock on diagram M. Example 1.5 For a beam (Fig. 1.8, a), determine the rational position of the hinge C, at which the largest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Construct diagrams of Q and M. Solution Determination of support reactions. Although total number support links is equal to four, the beam is statically determinate. The bending moment in the hinge C is zero, which allows us to create an additional equation: the sum of the moments about the hinge of all external forces acting on one side of this hinge is equal to zero. Let us compile the sum of the moments of all forces to the right of the hinge C. The diagram Q for the beam is limited by an inclined straight line, since q = const. We determine the values ​​of transverse forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation from which the diagram M for the beam is limited by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are respectively written as follows: From the condition of equality of moments, we obtain a quadratic equation for the desired parameter x: Real value. We determine the numerical values ​​of transverse forces and bending moments in characteristic sections of the beam. Figure 1.8, b shows the diagram Q, and in Fig. 1.8, c – diagram M. The problem considered could be solved by dividing the hinged beam into its constituent elements, as shown in Fig. 1.8, d. At the beginning, the reactions of the supports VC and VB are determined. Diagrams of Q and M are constructed for the suspended beam SV from the action of the load applied to it. Then they move to the main beam AC, loading it with an additional force VC, which is the pressure force of the beam CB on the beam AC. After that, diagrams Q and M are built for beam AC. 1.4. Strength calculations for direct bending of beams Strength calculations based on normal and shear stresses. When a beam bends directly in its cross sections, normal and tangential stresses arise (Fig. 1.9). Normal stresses are associated with bending moment, shear stresses are associated with shear force. In straight pure bending, the shear stresses are zero. Normal stresses at an arbitrary point in the cross section of a beam are determined by the formula (1.4) where M is the bending moment in a given section; Iz – moment of inertia of the section relative to the neutral axis z; y is the distance from the point where the normal voltage is determined to the neutral z axis. Normal stresses along the height of the section change according to a linear law and reach their greatest value at points farthest from the neutral axis. If the section is symmetrical about the neutral axis (Fig. 1.11), then Fig. 1.11 the greatest tensile and compressive stresses are the same and are determined by the formula - axial moment of resistance of a section during bending. For a rectangular section with width b and height h: (1.7) For a circular section with diameter d: (1. 8) For an annular section (1.9) where d0 and d are the inner and outer diameters of the ring, respectively. For beams made of plastic materials, the most rational are symmetrical 20 section shapes (I-beam, box-shaped, annular). For beams made of brittle materials that do not equally resist tension and compression, sections that are asymmetrical with respect to the neutral z-axis (T-beam, U-shaped, asymmetrical I-beam) are rational. For beams of constant cross-section made of plastic materials with symmetrical cross-sectional shapes, the strength condition is written as follows: (1.10) where Mmax is the maximum bending moment in modulus; – permissible stress for the material. For beams of constant cross-section made of plastic materials with asymmetrical cross-sectional shapes, the strength condition is written in the following form: For beams made of brittle materials with sections that are asymmetrical with respect to the neutral axis, if the diagram M is unambiguous (Fig. 1.12), two strength conditions must be written where yP,max , yC,max – distances from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; – permissible stresses in tension and compression, respectively. Fig.1.12. 21 If the diagram of bending moments has sections of different signs (Fig. 1.13), then in addition to checking section 1-1, where Mmax acts, it is necessary to calculate the highest tensile stresses for section 2-2 (with the highest moment of the opposite sign). Rice. 1.13 Along with the main calculation using normal stresses, in some cases it is necessary to check the strength of the beam using tangential stresses. Tangential stresses in beams are calculated using the formula of D.I. Zhuravsky (1.13) where Q is the transverse force in the cross section of the beam under consideration; Szотс – static moment relative to the neutral axis of the area of ​​the section part located on one side of a straight line drawn through a given point and parallel to the z axis; b – section width at the level of the point under consideration; Iz is the moment of inertia of the entire section relative to the neutral z axis. In many cases, maximum shear stresses occur at the level of the neutral layer of the beam (rectangle, I-beam, circle). In such cases, the strength condition for tangential stresses is written in the form, (1.14) where Qmax is the largest transverse force in absolute value; – permissible shear stress for the material. For a rectangular section of a beam, the strength condition has the form 22 (1.15) A – cross-sectional area of ​​the beam. For a circular section, the strength condition is presented in the form (1.16) For an I-section, the strength condition is written as follows: (1.17) where Szo,тmсax is the static moment of the half-section relative to the neutral axis; d – wall thickness of the I-beam. Typically, the cross-sectional dimensions of a beam are determined from the strength condition under normal stresses. Checking the strength of beams by shear stress is mandatory for short beams and beams of any length if there are concentrated forces of large magnitude near the supports, as well as for wooden, riveted and welded beams. Example 1.6 Check the strength of a box-section beam (Fig. 1.14) by normal and tangential stresses, if 0 MPa. Construct diagrams in the dangerous section of the beam. Rice. 1.14 Solution 23 1. Constructing diagrams of Q and M using characteristic sections. Considering the left side of the beam, we obtain The diagram of transverse forces is shown in Fig. 1.14, c. . The diagram of bending moments is shown in Fig. 5.14, g. 2. Geometric characteristics of cross section 3. The highest normal stresses in section C, where Mmax acts (modulo): The maximum normal stresses in the beam are almost equal to the permissible ones. 4. The greatest tangential stresses in section C (or A), where the static moment of the half-section area relative to the neutral axis acts; b2 cm – section width at the level of the neutral axis. 5. Tangential stresses at a point (in the wall) in section C: Here is the static moment of the area of ​​the part of the section located above the line passing through point K1; b2 cm – wall thickness at point K1. Diagrams for section C of beam are shown in Fig. 1.15. Example 1.7 For the beam shown in Fig. 1.16, a, required: 1. Construct diagrams of transverse forces and bending moments along characteristic sections (points). 2. Determine the dimensions of the cross section in the form of a circle, rectangle and I-beam from the condition of strength under normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of beam sections according to tangential stress. Solution: 1. Determine the reactions of the beam supports from where Checking: 2. Constructing diagrams Q and M. Values ​​of transverse forces in characteristic sections of the beam In sections CA and AD, load intensity q = const. Consequently, in these areas the Q diagram is limited to straight lines inclined to the axis. In section DB, the intensity of the distributed load is q = 0, therefore, in this section, the diagram Q is limited to a straight line parallel to the x axis. The Q diagram for the beam is shown in Fig. 1.16, b. Values ​​of bending moments in characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section in which Q = 0: Maximum moment in the second section Diagram M for the beam is shown in Fig. 1.16, c. 2. We create a strength condition based on normal stresses, from which we determine the required axial moment of resistance of the section from the expression determined by the required diameter d of a beam of a circular section. Area of ​​a circular section. For a beam of a rectangular section. Required height of the section. Area of ​​a rectangular section. Determine the required number of the I-beam. Using the tables of GOST 8239-89, we find the nearest higher value of the axial moment of resistance which corresponds to I-beam No. 33 with the characteristics: Tolerance check: (underload by 1% of the permissible 5%) the nearest I-beam No. 30 (W  472 cm3) leads to significant overload ( more than 5%). We finally accept I-beam No. 33. We compare the areas of the round and rectangular sections with the smallest area A of the I-beam: Of the three sections considered, the most economical is the I-beam section. 3. We calculate the highest normal stresses in the dangerous section 27 of the I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-beam section The diagram of normal stresses in the dangerous section of the beam is shown in Fig. 1.17, b. 5. Determine the highest shear stresses for the selected sections of the beam. a) rectangular section of the beam: b) round section of the beam: c) I-beam section: Tangential stresses in the wall near the flange of the I-beam in dangerous section A (right) (at point 2): The diagram of tangential stresses in dangerous sections of the I-beam is shown in Fig. 1.17, c. The maximum tangential stresses in the beam do not exceed the permissible stresses. Example 1.8 Determine the permissible load on the beam (Fig. 1.18, a) if the cross-sectional dimensions are given (Fig. 1.19, a). Construct a diagram of normal stresses in a dangerous section of a beam at an allowable load. Figure 1.18 1. Determination of reactions of beam supports. Due to the symmetry of the system VVB A8qa . 29 2. Construction of Q and M diagrams using characteristic sections. Transverse forces in characteristic sections of a beam: Diagram Q for a beam is shown in Fig. 5.18, b. Bending moments in characteristic sections of the beam For the second half of the beam, the ordinates M are along the axes of symmetry. Diagram M for the beam is shown in Fig. 1.18, b. 3. Geometric characteristics of the section (Fig. 1.19). We divide the figure into two simple elements: I-beam - 1 and rectangle - 2. Fig. 1.19 According to the assortment for I-beam No. 20, we have For a rectangle: Static moment of the sectional area relative to the z1 axis Distance from the z1 axis to the center of gravity of the section Moment of inertia of the section relative to the main central axis z of the entire section according to the formulas for the transition to parallel axes 4. Strength condition for normal stresses for dangerous point “a” (Fig. 1.19) in dangerous section I (Fig. 1.18): After substituting numerical data 5. With permissible load q in dangerous section, normal stresses at points “a” and “ b" will be equal to: The normal stress diagram for dangerous section 1-1 is shown in Fig. 1.19, b. Example 1.9 Determine the required dimensions of the cross-section of a cast-iron beam (Fig. 1.20), having previously selected a rational location of the section. Make Decision 1. Determine the reactions of the beam supports. 2. Construction of Q and M diagrams. The diagrams are presented in Fig. 1.20, in, g. The largest (in absolute value) bending moment occurs in section “b”. In this section, the stretched fibers are located at the top. Most of the material should be located in the tension zone. Therefore, it is rational to position the beam section as shown in Fig. 1.20, b. 3. Determination of the position of the center of gravity of the section (by analogy with the previous example): 4. Determination of the moment of inertia of the section relative to the neutral axis: 5. Determination of the required dimensions of the beam section from the condition of strength under normal stresses. Let us denote by y, respectively, the distances from the neutral axis to the most distant points in the tension and compression zones (for section B): then the points of the tension zone that are most distant from the neutral axis are dangerous. We create a strength condition for point m in section B: or after substituting numerical values. In this case, the stresses at point n, the most distant from the neutral axis in the compressed zone (in section B), will be MPa. Diagram M is ambiguous. It is necessary to check the strength of the beam in section C. Here is the moment, but the lower fibers are stretched. The dangerous point will be point n: In this case, the stresses at point m will be From the calculations we finally accept The diagram of normal stresses for the dangerous section C is shown in Fig. 1.21. Rice. 1.21 1.5. Main stresses during bending. Full check of the strength of beams Above, examples of calculating beams for strength using normal and shear stresses are discussed. In the vast majority of cases, this calculation is sufficient. However, in thin-walled beams of I-beam, T-beam, channel and box sections, significant shear stresses arise at the junction of the wall and the flange. This occurs in cases where a significant shear force is applied to the beam and there are sections in which M and Q are simultaneously large. One of these sections will be dangerous and is checked 34 by principal stresses using one of the strength theories. Checking the strength of beams using normal, tangential and principal stresses is called a complete check of the strength of beams. This calculation is discussed below. The main thing is to calculate the beam using normal stresses. The strength condition for beams, the material of which equally resists tension and compression, has the form where Mmax ─ maximum bending moment (modulo), taken from the diagram M, Wz ─ axial moment of resistance of the section relative to the neutral axis of the beam; [ ]─ permissible normal stress for the material. From the strength condition (1), the required cross-sectional dimensions of the beam are determined. The selected dimensions of the beam section are checked by shear stresses. The strength condition for tangential stresses has the form (formula of D.I. Zhuravsky): where Qmax ─ maximum transverse force taken from the diagram Q; Szots.─ static moment (relative to the neutral axis) of the cut-off part of the cross section located on one side of the level at which shear stresses are determined; I z ─ moment of inertia of the entire cross section relative to the neutral axis; b─ the width of the beam section at the level where the shear stresses are determined; ─ permissible tangential stress of the material during bending. The normal stress strength test refers to the point furthest from the neutral axis in the section where Mmax acts. The shear stress strength test refers to a point located on the neutral axis in the section where Qmax acts. In beams with a thin-walled cross-section (I-beam, etc.), a point located in the wall in a section where M and Q are both large can be dangerous. In this case, the strength is checked using principal stresses. The main and extreme tangential stresses are determined by analytical dependencies obtained from the theory of the plane stress state of bodies: The angle of inclination of the main areas is determined by the formula (1.22) Having the values ​​of the main stresses, strength conditions are drawn up according to one or another strength theory. For example, According to the third theory of the highest tangential stresses, we have After substituting the values ​​of the principal stresses, we finally obtain (1.23) According to the fourth energy theory of strength, the strength condition has the form (1.24) From formulas (1.6) and (1.7) it is clear that the design stress Eq depends on. Consequently, the material element of the beam for which it will be large at the same time is subject to verification. This is carried out in the following cases: 1) the bending moment and shear force reach highest value in the same section; 2) the width of the beam changes sharply near the edges of the section (I-beam, etc.). If the specified conditions do not hold, then it is necessary to consider several sections in which the highest values ​​of eq. Example 1.10 A welded beam of I-beam cross-section with a span of l = 5 m, simply supported at the ends, is loaded with a uniformly distributed load of intensity q and a concentrated force P 5qa applied at a distance a = 1 m from the right support (Fig. 1.22). Determine the permissible load on the beam from the strength condition for normal stresses and check for tangential and principal stresses according to 36 4th (energy) strength theory. Construct diagrams in a dangerous section using principal stresses and examine the stressed state of an element selected in the wall near the flange in the indicated section. Allowable tensile and compressive stress: bending 160 MPa; and shear 100 MPa. Rice. 1.22 Solution 1. Determination of reactions of beam supports: 2. Construction of diagrams M and Q using characteristic sections (points): 3. Calculation of geometric characteristics of the beam section. a) axial moment of inertia of the section relative to the neutral axis z: 37 b) Axial moment of resistance relative to the neutral axis z: 4. Determination of the permissible load on the beam from the condition of strength by normal stresses: Permissible load on the beam 5. Checking the strength of the beam by tangential stresses using the formula D.I. Zhuravsky Static moment of the half-section of an I-beam relative to the neutral axis z: Section width at point level 3: Maximum transverse force Maximum shear stresses in the beam 6. Checking the strength of the beam by principal stresses. Dangerous in terms of principal stresses is the section D, in which M and Q are both large, and the dangerous points in this section are points 2 and 4, where  and  are both large (Fig. 1.23). For points 2 and 4, we check the strength by principal stresses, using the 4th theory of strength where (2) and (2)─ normal and shear stresses at point 2(4), respectively (Fig. 1.2). Rice. 1.23 distance from the neutral axis to point 2. where Sz is the static moment of the flange relative to the neutral axis z. cm ─ section width along a line passing through point 3. Equivalent stresses according to the 4th theory of strength at point 2 of section D: The strength condition according to the 4th theory of strength is satisfied. 7. Construction of diagrams of normal, tangential, principal and extreme tangential stresses in dangerous section D (based on principal stresses). a) calculate the stresses at points (1-5) of section D using the appropriate formulas. Point 2 (in the wall) Previously, the values ​​of normal and shear stresses at point 2 were calculated. We find the main and extreme shear stresses at the same point 2: Point 3. Normal and shear stresses at point 3: Main and extreme shear stresses at point 3: The voltages at points 4 and 5 are found similarly. Based on the data obtained, we construct diagrams, max. 8. The stressed state of the element selected in the vicinity of point 2 in section D is shown in Fig. 1.24, angle of inclination of the main platforms 1.6. The concept of the center of bending As mentioned above, the tangential stresses in the cross sections of thin-walled rods during bending (for example, an I-beam or a channel) are determined by the formula In Fig. 194 shows diagrams of tangential stresses in an I-section. Using the technique described in paragraph 63, you can construct diagram 41 also for the channel. Let's consider the case when the channel is embedded in a wall, and at the other end it is loaded with a force P applied at the center of gravity of the section. Rice. 1.25 The general view of the τ diagram in any section is shown in Fig. 1.25, a. Tangential stresses τу arise in the vertical wall. As a result of the action of stresses τу, a total shear force T2 arises (Fig. 1.25, b). If we neglect the tangential stresses τу in the flanges, then we can write the approximate equality In horizontal flanges, tangential stresses τх arise, which are directed horizontally. The greatest shear stress in the flange τx max is equal to Here S1OTS is the static moment of the flange area relative to the Ox axis: Consequently, the Total shear force in the flange will be determined as the area of ​​the shear stress diagram multiplied by the thickness of the flange. Exactly the same shear force acts on the lower flange as on top, but it is directed in the opposite direction. Two forces T1 form a pair with a moment (1.25) Thus, due to the tangential stresses τу and τх, three internal tangential forces arise, which are shown in Fig. 1.25, b. From this figure it is clear that the forces T1 and T2 tend to rotate the channel section relative to the center of gravity in the same direction. Rice. 1.25 Consequently, an internal torque appears in the channel section, directed clockwise. So, when a channel beam is bent by a force applied at the center of gravity of the section, the beam simultaneously twists. Three tangential forces can be reduced to a principal vector and a principal moment. The magnitude of the main moment depends on the position of the point to which the forces are brought. It turns out that it is possible to choose a point A relative to which the main moment is equal to zero. This point is called the center of the bend. Equating the moment of tangential forces to zero: we obtain Taking into account expression (1.25), we will finally find the distance from the axis of the vertical wall to the center of bending: If an external force is applied not at the center of gravity of the section, but at the center of bending, then it will create the same moment relative to the center of gravity as create internal tangential forces, but only of the opposite sign. With such a load (Fig. 1.25, c), the channel will not twist, but will only bend. That is why point A is called the center of the bend. A detailed description of the calculation of thin-walled rods is given in Chapter. XIII. 1.7. Determination of displacements in beams during bending. Concepts of deformation of beams and conditions for their rigidity Under the influence of an external load, the beam is deformed and its axis is bent. The curve into which the axis of the beam turns after applying a load is called an elastic line, provided that the stresses of the beam do not exceed the proportionality limit. Depending on the direction of the load, the location of the diagrams, the elastic line can have a convexity upward (Fig. 1.26, a), downward (Fig. 1.26, b) or a combination (Fig. 1.26, c). In this case, the centers of gravity of the cross sections move respectively either up or down, and the sections themselves rotate relative to the neutral axis, remaining perpendicular to the curved axis of the beam (Fig. 1.26, a). Strictly speaking, the centers of gravity of the cross sections also move in the direction of the longitudinal axis of the beam. However, due to the smallness of these movements for beams, they are neglected, i.e., it is assumed that the center of gravity of the section moves perpendicular to the axis of the beam. Let us denote this movement by y, and in the future we will understand by it the deflection of the beam (see Fig. 1.26). The deflection of a beam in a given section is the movement of the center of gravity of the section in a direction perpendicular to the axis of the beam. Rice. 1.26 Deflections in various sections of a beam depend on the position of the sections and are a variable value. So, for a beam (Fig. 1.26, a) at point B the deflection will have a maximum value, and at point D it will be zero. As already noted, along with the movement of the center of gravity of the section, the sections rotate relative to the neutral axis of the section. The angle by which the section is rotated relative to its original position is called the angle of rotation of the section. We will denote the angle of rotation by (Fig. 1.26, a). Since when a beam is bent, the cross section always remains perpendicular to its curved axis, the angle of rotation can be represented as the angle between the tangent to the curved axis at a given point and the original axis of the beam (Fig. 1.26, a) or perpendicular to the original and curved axes of the beam at the point in question. The angle of rotation of the section for beams is also a variable value. For example, for a beam (Fig. 1.26, b), it has a maximum value in the hinged supports, and a minimum value of 0 for the section in which the deflection has a maximum value. For a cantilever beam (Fig. 1.26, a) the maximum angle of rotation will be at its free end, i.e. at point B. To ensure normal operation of beams, it is not enough that they satisfy the strength condition. It is also necessary that the beams have sufficient rigidity, that is, that the maximum deflection and angle of rotation do not exceed the permissible values ​​determined by the operating conditions of the beams. This situation is called the condition of beam rigidity during bending. In a short mathematical form of notation, the stiffness conditions have the form: where [y] and, accordingly, the permissible deflection and angle of rotation. 45 The permissible deflection is usually specified as part of the distance between the supports of the beam (span length l), i.e. where m is a coefficient depending on the value and operating conditions of the system in which this beam is used. In each branch of mechanical engineering, this value is determined by design standards and varies widely. As follows: - for crane beams m = 400 - 700; - For railway bridges m = 1000; - for spindles of lathes m= 1000-2000. The permissible angles of rotation for beams usually do not exceed 0.001 rad. The left side of equations (1.26) includes the maximum deflection ymax and rotation angle max, which are determined by calculation based on known methods: analytical, graphical and graphic-analytical, some of which are discussed below. 1.8. Differential equation for the curved axis of a beam Under the influence of external forces, the axis of the beam is bent (see Fig. 1.26, a). Then the equation of the curved axis of the beam can be written in the form and the angle of rotation  for any section will be equal to the angle of inclination of the tangent to the curved axis at a given point. The tangent of this angle is numerically equal to the derivative of the deflection along the abscissa of the current section x, i.e. Since the deflections of the beam are small compared to its length l (see above), we can assume that the angle of rotation (1.27) When deriving the normal stress formula during bending, it was found that the following relationship exists between the curvature of the neutral layer and the bending moment: This formula shows that the curvature changes along the length of the beam according to the same law according to which the value Mz changes. If a beam of constant cross-section experiences pure bending (Fig. 5.27), in which the moment along the length does not change, its curvature is: Therefore, for such a beam, the radius of curvature is also a constant value and the beam in this case will bend along a circular arc. However, in the general case, it is not possible to directly apply the law of change in curvature to determine deflections. To solve the problem analytically, we use the well-known expression for curvature from mathematics. (1.29) Substituting (1.28) into (1.29), we obtain the exact differential equation for the curved axis of the beam: . (1.30) Equation (1.30) is nonlinear, and its integration is associated with great difficulties. Considering that deflections and rotation angles for real beams used in mechanical engineering, construction, etc. are small, then the value can be neglected. Taking into account this, and also the fact that for the right coordinate system the bending moment and curvature have the same sign (Fig. 1.26), then for the right coordinate system the minus sign in equation (1.26) can be omitted. Then the approximate differential equation will have the form 1.9. Direct integration method This method is based on the integration of equation (1.31) and allows us to obtain the equation of the elastic axis of the beam in the form of deflections y f (x) and the equation of rotation angles. Having integrated equation (1.31) for the first time, we obtain the equation of rotation angles (1. 32) where C is the integration constant. Integrating a second time, we obtain the deflection equation where D is the second constant of integration. The constants C and D are determined from the boundary conditions of the support of the beam and the boundary conditions of its sections. So for a beam (Fig. 1.26, a), at the place of embedding (x l), the deflection and angle of rotation of the section are equal to zero, and for a beam (see Fig. 1.26, b) the deflection y and deflection yD 0, at x .l For hinged supported beam with consoles (Fig. 1.28), when the origin of coordinates is aligned with the end of the left support and the choice of the right coordinate system, the boundary conditions have the form Taking into account the boundary conditions, the integration constants are determined. After substituting the integration constants into the equations of rotation angles (1.32) and deflections (1.33), the rotation angles and deflections of a given section are calculated. 1.10. Examples of determining displacements in beams using the direct integration method Example 1.11 Determine the maximum deflection and angle of rotation for a cantilever beam (Fig. 1.26, a). Solution The origin of coordinates is aligned with the left end of the beam. The bending moment in an arbitrary section at a distance x from the left end of the beam is calculated using the formula Taking into account the moment, the approximate differential equation has the form Integrating for the first time, we have (1.34) Integrating for the second time Boundary conditions Taking into account the second condition, from which Similarly, from the first condition we will have Taking into account of the found integration constants C and D, the equation for the angles of rotation and deflection will have the form: When (see Fig. 1.26, a) the angle of rotation and deflection have maximum values: A positive value of the angle  indicates that the section when bending the beam rotates in the direction opposite to the movement clockwise. A negative y value indicates that the center of gravity of the section is moving downward. 1.11. Physical meaning of the integration constants If we turn to equations (1.32), (1.33) and (1.34), (1.35), the examples considered above, then it is easy to notice that for x 0 it follows from them. Thus, we can conclude that the integration constants C and D represent the product of the beam stiffness, respectively, by the angle: rotation 0 and deflection y0 at the origin. Dependencies (1.36) and (1.37) always turn out to be valid for beams that have one loading section, if we calculate the bending moment from the forces located between the section and the origin. The same remains valid for beams with any number of loading sections, if special techniques for integrating the differential equation of the curved axis of the beam are used, which will be discussed below. 1.12. Method of initial parameters (universal equation of the curved axis of a beam) When determining deflections and rotation angles by the method of direct integration, it is necessary to find two integration constants C and D, even in cases where the beam has one loading section. In practice, beams are used that have several loading areas. In these cases, the bending moment law will be different in different loading areas. Then the differential equation of the curved axis will need to be compiled for each section of the beam and for each of them its constants of integration C and D must be found. Obviously, if a beam has n loading sections, then the number of integration constants will be equal to twice the number of sections. To determine them, you will need to solve 2 equations. This task is time consuming. To solve problems that have more than one loading area, the method of initial parameters, which is a development of the method of direct integration, has become widespread. It turns out that by observing certain conditions and techniques for composing and integrating equations over sections, it is possible to reduce the number of integration constants, regardless of the number of loading sections, to two, representing the deflection and rotation angle at the origin. Let us consider the essence of this method using the example of a cantilever beam (Fig. 1.28), loaded with an arbitrary load, but creating a positive moment in any section of the beam. Let a beam of constant cross-section be given, and the cross-section has an axis of symmetry coinciding with the y-axis, and the entire load is located in one plane passing through this axis. Let us set the task of establishing the dependencies that determine the angle of rotation and deflection of an arbitrary section of a beam. Rice. 1.29 When solving problems, we agree: 1. The origin of coordinates will be associated with the left end of the beam, and it is common to all sections. 2. The bending moment in an arbitrary section will always be calculated for the section of the beam located to the left of the section, i.e., between the origin and the section. 3. We will integrate the differential equation of the curved axis in all sections without opening the brackets of some expressions containing brackets. So, for example, integration of an expression of the form P x(b) is carried out without opening the parentheses, namely according to the following formula. Integration according to this formula differs from integration with preliminary opening of the brackets only in the value of an arbitrary constant. 4. When composing an expression for the bending moment in an arbitrary section caused by an external concentrated moment M, we will add the factor (x)a0 1. Adhering to these rules, we will compose and integrate an approximate differential equation for each of the five sections of the beam indicated in Fig. 1.28 in Roman numerals. The approximate differential equation for the indicated sections has the same form: (1.38) but for each section the bending moment has its own law of change. The bending moments for the sections have the form: Substituting the expressions for the bending moment into equation (1.38), for each of the sections after integration we obtain two equations: the equation of rotation angles and the equation of deflections, which will include their two integration constants Ci and Di. Since the beam has five sections, there will be ten such integration constants. However, taking into account that the curved axis of the beam is a continuous and elastic line, then at the boundaries of adjacent sections the deflection and the angle of rotation have the same values, i.e., etc. Because of this, from a comparison of the equations for the angles of rotation and deflections of neighboring sections, we obtain that the constants of integration Thus, instead of ten constants of integration, to solve the problem posed, it is necessary to determine only two constants of integration C and D. From the consideration of the integral equations of the first section it follows that at x 0: i.e. they represent the same dependencies (1.36) and (1.37). The initial parameters 0 and y0 о are determined from the boundary conditions that were discussed in the previous section. Analyzing the obtained expressions for the angles of rotation and deflections y, we see that the most general form equations corresponds to the fifth section. Taking into account the integration constants, these equations have the form: The first of these equations represents the equation of rotation angles, and the second represents the equation of deflections. Since more than one concentrated force can act on a beam, a moment or a beam can have more than one section with a distributed load, then for the general case, equations (1.38), (1.39) will be written in the form: Equations (1.41), (1.42) are called universal equations the curved axis of the beam. The first of these equations is the equation of rotation angles, and the second is the equation of deflections. Using these equations, it is possible to determine the deflections and angles of rotation of sections for any statically determinate beams whose stiffness along their length is constant EI  const. In equations (1.41), (1.42): M, P, q, qx ─ external load located between the origin of coordinates and the section in which the displacements (angle of rotation and deflection) are determined; a, b, c, d ─ distances from the origin of coordinates to the points of application, respectively, of the moment M, concentrated force P, the beginning of a uniformly distributed load and the beginning of an unevenly distributed load. It is necessary to pay attention: 53 1. In the opposite direction of the external load, which is accepted when deriving universal equations, in front of the corresponding term of the equations the sign changes to the opposite, i.e., to minus. 2. The last two terms of equations (1.41), (1.42) are valid only if the distributed load does not terminate before the section in which the deflection and angle of rotation are determined. If the load does not reach this section, then it must be continued to this section and at the same time added on the extended section the same distributed load, but of opposite sign, this idea is explained in Fig. 1.30. The dotted line shows the added distributed load on the extended section. Rice. 1.30 When determining rotation angles  and deflections y, the origin of coordinates should be placed at the left end of the beam, directing the y axis upward and the x axis to the right. The compiled equation for rotation angles and deflections includes only those forces that are located to the left of the section, i.e. on the section of the beam between the origin of coordinates and the section in which the deflection and angle of rotation are determined (including the forces acting in the section coinciding with the origin of coordinates). 1.13. Examples of determining displacements in a beam using the initial parameters method Example 1.12 For a beam (Fig. 1.31), clamped at the left end and loaded with a concentrated force P, determine the angle of rotation and deflection at the point of application of the force, as well as the free end (section D). Beam stiffness Fig. 1.31 Solution of the static equilibrium equation: 1) Note that the reactive torque is directed counterclockwise, so it will enter the equation of the curved axis with a minus sign. 2. Combine the origin of coordinates with point B and set the initial parameters. In pinching ()B there is no deflection and no rotation angle, i.e. 0 0. We write down the equation of rotation angles and deflections for an arbitrary section of the second section, i.e. located at a distance x from the origin of coordinates Taking into account the reactive forces, as well as the equality to zero of the initial parameters, these equations have the form For x l we have the angle of rotation and deflection of the section C, respectively 55 For section D, x1l 12(1)2 Example 1.13 Determine the maximum deflection and angle rotation on the right support of the beam, loaded in the middle of the span with a concentrated force (Fig. 1.32). Solution 1. Determine the support reactions From the static equations we have B 2. Place the origin of coordinates at the left end of the beam (point B). Rice. 1.32 3. Set the initial parameters. Deflection at the origin By0, since the support does not allow vertical movement. It should be noted that if the support were spring-loaded, then the deflection at the origin would be equal to the deformation of the spring. The angle of rotation at the origin of coordinates is not equal to zero, i.e. 4. Determine the angle of rotation at the origin of coordinates 0. To do this, we use the condition that at x l the deflection is equal to zero yD 0: 3 Since the beam is symmetrical relative to the load P, the angle of rotation on the right support is equal to the angle of rotation on the left support. 2 BD 16z Pl EI . The maximum deflection will be in the middle of the beam at x. Therefore, Example 1.14 Determine the deflection in the middle of the span and at the right end of the beam (Fig. 1.33), if the beam is made of I-beam No. 10 (moment of inertia Iz 198 scm4), loaded with a distributed load q 2.N/m, concentrated by a moment M force. P kkNN Fig. 1.33 Solution 1. Determining the support reactions From where: Checking the correctness of determining the reactions 2. Combine the origin of coordinates with point B and set the initial parameters. From Fig. 1.33 it follows that at the origin of coordinates the deflection y0 0 and the angle of rotation. 57 3. Determine the initial parameters y0 and 0. To do this, we use the boundary conditions that when: To implement the boundary conditions, we create an equation for the curved axis. for two sections: section BC 0 mm1: When writing this equation, it was taken into account that the distributed load was interrupted at point C, therefore, according to what was said above, it was continued and a compensating load of the same magnitude, but in the opposite direction, was introduced in the continued section. Taking into account the boundary conditions (point 3) and the load, equations (1.43) and (1.44) have the form: From the joint solution of these equations we have 4. We determine the deflection in sections K and E. For section K at x 2 mm we have 1.14. Determination of displacements using Mohr's method Rule A.K. Vereshchagin's Mohr method is general method determination of displacements in linearly deformable rod systems. The determination of displacements (linear, angular) in the design sections is made using the Mohr formula (integral), which is easy to obtain based on the theorem on the reciprocity of work (Betti’s theorem) and the theorem on the reciprocity of displacements (Maxwell’s theorem). Let, for example, be given a flat elastic system in the form of a beam (Fig. 1.34), loaded with a flat balanced arbitrary load. We will call the given state of the system cargo and denote it by the letter P. Under the influence of an external load, deformation will occur, and displacements will occur at point K, in particular, in the direction perpendicular to the axis - deflection cr. Let us introduce a new (auxiliary) state of the same system, but loaded at point K in the direction of the desired displacement (cr) by a unit dimensionless force (Fig. 1.34). We will denote such a state of the system by the letter i, and will call it a single state. 59 Fig. 1.34 Based on Betti’s theorem, the possible work of the forces of the cargo state pi A and the forces of a single state pi A are equal to (1.45) The possible work of the forces of the cargo state, expressed in terms of internal forces, is determined by the formula and the forces of a single state - by the formula (1.47) Taking into account (1.46 ), (1.47) from (1.45) we have (1.48) where M p , Qp, Np ─ respectively the bending moment, transverse and longitudinal forces arising in the system from the external load; Mi, Qi, Ni ─ respectively, bending moment, transverse and longitudinal forces arising in the system from a unit load applied in the direction of the determined displacement; k ─ coefficient taking into account the unevenness of tangential stresses across the section; I ─ axial moment of inertia relative to the main central axis; A─ cross-sectional area of ​​the rod in the area; 60 E, G ─ elastic moduli of the material. The uneven distribution of tangential stresses in a section depends on the shape of the section. For rectangular and triangular sections k 1.2, circular section k 1.11, circular annular section k 2. Formula (1.48) allows you to determine the displacement at any point of a flat elastic system. When determining the deflection in the section (K), we apply a unit force (dimensionless) at this point. In the case of determining the angle of rotation of the section at point K, it is necessary to apply a unit dimensionless moment

Analytical determination of displacements in beams

Example 1

The task

For the beam shown in Fig. 4.20, A, you need to find the deflection in the section WITH, angle of rotation in section IN analytically and check the stiffness condition if the permissible deflection is equal to l/100. The beam is made of wood and has a cross-section of three logs with a radius of 12 cm. (For the selection of the cross-section of this beam, see Section 4.1.2, example 1.)

Solution

To determine the displacements of the beam in an analytical way, we will compose the differential equation of the curved axis (4.16), using the Clebsch rules for recording the expression for the bending moment. In the problem under consideration, it is more rational to choose the origin of coordinates on the right (in the close). The distributed load, which does not reach the left end of the beam, will be extended to the section WITH(Fig. 4.20, V). The expression for the bending moment will look like this:

.

Let's substitute this expression into the differential equation (4.16) and integrate it twice:

;

;

.

To determine constants WITH And D Let's write down the boundary conditions: in the embedment (in the section A, where the origin of coordinates is located), the angle of rotation and deflection of the beam are equal to zero, that is

AND .

Substituting these conditions into the expressions for the angle of rotation and deflection in the first section, we find that

Now you can define the specified movements. To determine the angle of rotation in a section IN Let’s substitute into the expression for the angle of rotation in the first section (only up to the line numbered I) the value:

According to the sign rule, the negative sign of the rotation angle for the selected origin X on the right means that the section is rotated clockwise.

In cross section WITH, where you want to find the deflection, coordinate X is equal to , and this section is located in the third section of the beam, so we substitute X= 4 m in the expression for deflections, using terms in all three sections:

kN m 3.

The minus sign for the found deflection indicates that the section WITH moves up. Let us show the found displacements on the curved axis of the beam. To draw the axis of the beam after deformation, we construct a diagram of bending moments (Fig. 4.20, b). Positive sign of the diagram M in a section shows that the beam in this section bends convexly downward, with a negative sign M the curved axis convexes upward. In addition, the deformed axis of the beam must satisfy the conditions of fastening: in our case, at the right end the beam is rigidly pinched, and, as already noted when writing the boundary conditions, the deflection and angle of rotation in the pinching must be equal to zero. In Fig. 4.20, G the axis of the beam under consideration is shown after deformation, satisfying these conditions. The curved axis shows the found deflections in section WITH and section rotation angle IN taking into account their signs.

In conclusion, let’s calculate the deflection of the beam in centimeters, the angle of rotation in radians, and check the rigidity condition. Let's find the rigidity EI the considered wooden beam of three logs with a radius of 12 cm. Moment of inertia of the cross section

cm 4.

Modulus of elasticity of wood E= 10 4 MPa = 10 3 kN / cm 2. Then

Beam deflection in section WITH

cm,

and the angle of rotation of the section IN

glad.

Obviously (see Fig. 4.20, G), that the found deflection of the beam in the section WITH is maximum, therefore, to check the rigidity condition, we compare it with the permissible deflection. For a beam length m, the permissible deflection according to the condition cm. Thus, the maximum deflection is cm is less than permissible, and the rigidity condition is satisfied.

Example 2

The task

In a beam with two consoles, shown in Fig. 4.21, A we need to find the angle of rotation of the section A and section deflection D using an analytical method. The cross section of the beam is I-beam No. 24.

Solution

Let us select the origin of the coordinate X at the left end of the beam at the point A and write down the expression for the bending moment in all sections, taking into account Clebsch’s rules:

Let's substitute this expression into the differential equation of the curved axis (4.16) and integrate it twice:

Let's find arbitrary constants WITH And D from the boundary conditions. At points IN And WITH where the supports are located, deflections are not possible. That's why

We obtained a system of two equations with two unknowns WITH And D. Solving this system, we find WITH= 40 kN m 2, D= – 40 kN m 3. Let us analyze the result using the geometric meaning of arbitrary constants WITH And D. In Fig. 4.21, V the curved axis of the beam is shown, corresponding to the diagram of bending moments and fastening conditions. Dot A, located at the origin, moves upward, and therefore we should expect that will have a negative sign in accordance with the sign rule. Section at a point A rotates clockwise, so the constant must be positive. Received marks WITH And D do not contradict the analysis performed.

4.1. Differential equation of the curved axis of a beam and its integration.

When bending, the axis of the beam is bent, and the cross sections move translationally and rotate around neutral axes, while remaining normal to the curved longitudinal axis (Fig. 8.22). The deformed (curved) longitudinal axis of the beam is called an elastic line, and the translational displacements of the sections are equal to the displacements y= y(x) their centers of gravity of the sections are the deflections of the beam.

Between deflections y(x) and angles of rotation of sections θ (x) there is a certain dependence. From Fig. 8.22 it can be seen that the angle of rotation of the section θ equal to angle φ the inclination of the tangent to the elastic line ( θ And φ - angles with mutually perpendicular sides). But according to the geometric meaning of the first derivative y / = tgθ . Hence, tgθ =tgφ =y / .

Within the limits of elastic deformations, beam deflections are usually significantly less than the section height h, and the rotation angles θ do not exceed 0.1 – 0.15 rad. In this case, the relationship between deflections and rotation angles is simplified and takes the form θ =y / .

Let us now determine the shape of the elastic line. Influence of shearing forces Q the deflection of beams is, as a rule, insignificant. Therefore, with sufficient accuracy we can assume that during transverse bending, the curvature of the elastic line depends only on the magnitude of the bending moment Mz and rigidity EIz(see equation (8.8)):

Equating the right-hand sides of (8.26) and (8.27) and taking into account that the sign rules for Mz And y// were accepted independently of each other, we get

The choice of sign on the right side of (8.29) is determined by the direction of the coordinate axis y, since the sign of the second derivative depends on this direction y// . If the axis is directed upward, then, as can be seen from Fig. 8.23, signs y// And Mz coincide, and the plus sign must be left on the right side. If the axis is directed downward, then the signs y// And Mz are opposite, and this forces us to select the minus sign on the right side.

Note that equation (8.29) is valid only within the limits of applicability of Hooke’s law and only in those cases when the plane of action of the bending moment Mz contains one of the main axes of inertia of the section.

Integrating (8.29), we first find the angles of rotation of the sections

The integration constants are determined from the boundary conditions. In sections with different analytical expressions for bending moments, the differential equations of the elastic line are also different. Integrating these equations at n plots gives 2 n arbitrary constants. To determine them, the conditions for equality of deflections and rotation angles at the junction of two adjacent sections of the beam are added to the boundary conditions on the supports.

2013_2014 academic year II semester Lecture No. 2.6 page 12

Deformation of beams during bending. Differential equation for the curved axis of a beam. Initial parameters method. Universal equation of an elastic line.

6. Deformation of beams during plane bending

6.1. Basic concepts and definitions

Let us consider the deformation of a beam during plane bending. The axis of the beam under the influence of load is bent in the plane of action of forces (plane x 0y), while the cross sections are rotated and shifted by a certain amount. The curved axis of a beam during bending is called curved axis or elastic line.

We will describe the deformation of beams during bending by two parameters:

deflection(y) – displacement of the center of gravity of the beam section in a direction perpendicular to

rice. 6.1 to its axis.

Don't confuse deflection y with coordinate y beam section points!

The greatest deflection of the beam is called the deflection arrow ( f= y max);

2) section rotation angle() – the angle by which the section rotates relative to its original position (or the angle between the tangent to the elastic line and the original axis of the beam).

In general, the amount of deflection of a beam at a given point is a function of the coordinate z and can be written as the following equation:

Then the angle between the tangent to the curved axis of the beam and the axis x will be determined from the following expression:

.

Due to the smallness of the angles and displacements, we can assume that

the angle of rotation of the section is the first derivative of the deflection of the beam along the abscissa of the section.

6.2. Differential equation of the curved axis of a beam

Based on the physical nature of the bending phenomenon, we can assert that the curved axis of a continuous beam must be a continuous and smooth (without kinks) curve. In this case, the deformation of a particular section of the beam is determined by the curvature of its elastic line, that is, the curvature of the beam axis.

Previously, we obtained a formula for determining the curvature of a beam (1/ρ) during bending

.

On the other hand, from the course higher mathematics It is known that the equation of curvature of a plane curve is as follows:

.

Equating the right-hand sides of these expressions, we obtain a differential equation for the curved axis of the beam, which is called the exact equation for the curved axis of the beam

In the coordinate system of deflections z0 y when the axis y is directed upward, the sign of the moment determines the sign of the second derivative of y By z.

Integrating this equation obviously presents some difficulties. Therefore, it is usually written in a simplified form, neglecting the value in parentheses compared to unity.

Then differential equation of the elastic line of a beam we will consider it in the form:

(6.1)

We find the solution to the differential equation (6.1) by integrating both of its parts over the variable z:

(6.2)

(6.3)

Constants of integration C 1 , D 1 is found from the boundary conditions - the conditions for securing the beam, and for each section of the beam its own constants will be determined.

Let us consider the procedure for solving these equations using a specific example.

D ano:

Cantilever beam length l loaded with shear force F. Beam material ( E), the shape and dimensions of its cross-section ( I x) we also consider known.

ABOUT limit law of change of angle of rotation ( z) and deflection y(z) beams along its length and their values ​​in characteristic sections.

Solution

a) determine the reactions in the sealing

b) using the method of sections, we determine the internal bending moment:

c) determine the angle of rotation of the beam sections

Constant C 1 we find from the conditions of fastening, namely, in a rigid embedment the angle of rotation is equal to zero, then


(0) = 0  C 1 =0.

Let us find the angle of rotation of the free end of the beam ( z = l) :

The minus sign indicates that the section has rotated clockwise.

d) determine the deflections of the beam:

Constant D 1 we find from the fastening conditions, namely, in a rigid embedment the deflection is equal to zero, then

y(0) = 0 + D 1  D 1 = 0

Let us find the deflection of the free end of the beam ( x= l)

.

The minus sign indicates that the cross section has moved downwards.