Flat transverse bending beams Internal bending forces. Differential dependencies of internal forces. Rules for checking diagrams of internal bending forces. Normal and shear stresses during bending. Strength calculation based on normal and tangential stresses.
Bending is a type of loading in which the rod is loaded with moments in planes passing through the longitudinal axis of the rod.
A rod that bends is called a beam (or timber). In the future, we will consider rectilinear beams, the cross section of which has at least one axis of symmetry.
The resistance of materials is divided into flat, oblique and complex bending.
Plane bending is a bending in which all the forces bending the beam lie in one of the planes of symmetry of the beam (in one of the main planes).
The main planes of inertia of a beam are the planes passing through the main axes cross sections and the geometric axis of the beam (x-axis).
Oblique bending is a bending in which the loads act in one plane that does not coincide with the main planes of inertia.
Complex bending is a bending in which loads act in different (arbitrary) planes.
Let's consider two typical cases of bending: in the first, the cantilever beam is bent by a concentrated moment M o ; in the second - concentrated force F.
Using the method of mental sections and composing equilibrium equations for the cut off parts of the beam, we determine the internal forces in both cases:
The remaining equilibrium equations are obviously identically equal to zero.
Thus, in the general case of plane bending in the section of a beam, out of six internal forces, two arise - bending moment M z and shear force Q y (or when bending relative to another main axis - bending moment M y and shear force Q z).
Moreover, in accordance with the two loading cases considered, flat bend can be divided into pure and transverse.
Pure bending is a flat bending in which only one out of six internal forces occurs in the sections of the rod - a bending moment (see the first case).
Transverse bend– bending, in which in the sections of the rod, in addition to the internal bending moment, a transverse force also arises (see the second case).
Strictly speaking, to simple types resistance only applies pure bend; transverse bending is conventionally classified as a simple type of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected when calculating strength.
When determining internal efforts, we will adhere to next rule signs:
1) the transverse force Q y is considered positive if it tends to rotate the beam element in question clockwise;
2) bending moment M z is considered positive if, when bending a beam element, the upper fibers of the element are compressed and the lower fibers are stretched (umbrella rule).
Thus, the solution to the problem of determining the internal forces during bending will be built according to the following plan: 1) at the first stage, considering the equilibrium conditions of the structure as a whole, we determine, if necessary, the unknown reactions of the supports (note that for a cantilever beam the reactions in the embedment can be and not found if we consider the beam from the free end); 2) at the second stage, we select characteristic sections of the beam, taking as the boundaries of the sections the points of application of forces, points of change in the shape or size of the beam, points of fastening of the beam; 3) at the third stage, we determine the internal forces in the sections of the beam, considering the conditions of equilibrium of the beam elements in each section.
Let us establish some relationships between internal forces and external bending loads, as well as characteristics diagrams Q and M, knowledge of which will facilitate the construction of diagrams and allow you to control their correctness. For convenience of notation, we will denote: M ≡ M z, Q ≡ Q y.
Let us select a small element dx in a section of a beam with an arbitrary load in a place where there are no concentrated forces and moments. Since the entire beam is in equilibrium, element dx will also be in equilibrium under the action of shear forces, bending moments and external load applied to it. Since Q and M generally change along the axis of the beam, then in the sections of the element dx there will be shear forces Q and Q +dQ, as well as bending moments M and M +dM. From the equilibrium condition of the selected element we obtain
∑ F y = 0 Q + q dx − (Q + dQ) = 0;
∑ M 0 = 0 M + Q dx + q dx dx 2 − (M + dM ) = 0.
From the second equation, neglecting the term q dx (dx /2) as an infinitesimal quantity of the second order, we find
Relations (10.1), (10.2) and (10.3) are called differential dependencies of D.I. Zhuravsky during bending.
Analysis of the above differential dependencies during bending allows us to establish some features (rules) for constructing diagrams of bending moments and transverse forces:
a – in areas where there is no distributed load q, diagrams Q are limited to straight lines parallel to the base, and diagrams M are limited to inclined straight lines;
b – in areas where a distributed load q is applied to the beam, diagrams Q are limited by inclined straight lines, and diagrams M are limited by quadratic parabolas. Moreover, if we construct diagram M “on a stretched fiber,” then the convexity of the pa-
the work will be directed in the direction of action q, and the extremum will be located in the section where the diagram Q intersects the base line;
c – in sections where a concentrated force is applied to the beam, on the diagram Q there will be jumps by the magnitude and in the direction of this force, and on the diagram M there will be kinks, the tip directed in the direction of action of this force; d – in sections where a concentrated moment is applied to the beam on the epi-
there will be no changes in re Q, and on the diagram M there will be jumps by the value of this moment; d – in areas where Q >0, the moment M increases, and in areas where Q<0, момент М убывает (см. рисунки а–г).
Let us consider the case of pure plane bending of a beam and derive a formula for determining normal stresses for this case. Note that in the theory of elasticity it is possible to obtain an exact dependence for normal stresses during pure bending, but if this problem is solved by methods of resistance of materials, it is necessary to introduce some assumptions.
a – hypothesis of plane sections (Bernoulli hypothesis)
– sections that are flat before deformation remain flat after deformation, but only rotate relative to a certain line, which is called the neutral axis of the beam section. In this case, the fibers of the beam lying on one side of the neutral axis will stretch, and on the other, compress; fibers lying on the neutral axis do not change their length;
b – hypothesis about the constancy of normal stresses
niy – stresses acting at the same distance y from the neutral axis are constant across the width of the beam;
c – hypothesis about the absence of lateral pressures – co-
The gray longitudinal fibers do not press on each other.
Straight bend. Plane transverse bending Constructing diagrams of internal force factors for beams Constructing diagrams of Q and M using equations Constructing diagrams of Q and M using characteristic sections (points) Strength calculations for direct bending of beams Principal stresses during bending. A complete check of the strength of beams. The concept of the center of bending. Determination of displacements in beams during bending. Concepts of deformation of beams and conditions for their rigidity Differential equation of the curved axis of a beam Method of direct integration Examples of determining displacements in beams by the method of direct integration Physical meaning of integration constants Method of initial parameters (universal equation of the curved axis of a beam). Examples of determining displacements in a beam using the initial parameters method. Determining displacements using Mohr's method. Rule A.K. Vereshchagin. Calculation of the Mohr integral according to the rule of A.K. Vereshchagina Examples of determining displacements using the Mohr integral Bibliography 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. The transverse force in the m-n section of the beam (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 section m-n 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 with respect to 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 to be 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: 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: 8 The value of Q is constant in 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: 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. Plot 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. Section A 3 We determine 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. Section DB 4 We determine 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 Section BE 1 We determine 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 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. Rice. 1.6. Solution: We start constructing the Q and M diagrams from the free end of the beam, while the reactions in the embedment do not need to 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 on 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. Constructing diagram M. We calculate the values of bending moments in the boundary sections of sections: At 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. To do this, we will compose an expression for bending moments in the section as the sum of the moments of forces on the left. 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: ,b from which we have a 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. Despite the fact that the total number of support links is 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 written respectively as follows: From the condition of equality of moments, we obtain a quadratic equation for the desired parameter x: Real value x2x 1.029 m. 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). 18 Fig. 1.9 Normal stresses are associated with bending moment, tangential 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, is the axial moment of resistance of the 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 – 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 section shapes, the strength condition is written in the following form: (1. 11) 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), it is necessary to write down two strength conditions - the distance from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; P – permissible stresses for 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 magnitude; – permissible shear stress for the material. For a rectangular section of a beam, the strength condition has the form (1.15) A is the 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) using normal and shear stresses, if 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): MPa. The maximum normal stresses in the beam are almost equal to the permissible ones. 4. The highest tangential stresses in section C (or A), where max Q acts (modulo): Here is the static moment of the half-section area relative to the neutral axis; b2 cm – section width at the level of the neutral axis. 5. Tangential stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 cm3 is the static moment of the area of the section located above the line passing through point K1; b2 cm – wall thickness at the level of point K1. Diagrams and for section C of the 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. Given: Solution: 1. Determine the reactions of the beam supports. Check: 2. Construction of diagrams Q and M. Values of transverse forces in characteristic sections of the beam 25 Fig. 1.16 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 597 cm3, which corresponds to I-beam No. 33 with the characteristics: A z 9840 cm4. Tolerance check: (underload by 1% of the permissible 5%) the nearest I-beam No. 30 (W 2 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 60 MPa, 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 2. Construction of diagrams Q and M 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 an allowable load in a dangerous section, the normal stresses at points “a” and “b” will be equal: Diagram of normal stresses for dangerous section 1-1 is shown in Fig. 1.19, b.
Bend is called deformation in which the axis of the rod and all its fibers, i.e. longitudinal lines parallel to the axis of the rod, are bent under the influence of external forces. The simplest case of bending occurs when external forces lie in a plane passing through the central axis of the rod and do not produce projections onto this axis. This type of bending is called transverse bending. There are flat bends and oblique bends.
Flat bend- such a case when the curved axis of the rod is located in the same plane in which external forces act.
Oblique (complex) bend– a case of bending when the bent axis of the rod does not lie in the plane of action of external forces.
A bending rod is usually called beam.
During flat transverse bending of beams in a section with the coordinate system y0x, two internal forces can arise - transverse force Q y and bending moment M x; in what follows we introduce the notation for them Q And M. If there is no transverse force in a section or section of a beam (Q = 0), and the bending moment is not zero or M is const, then such a bend is usually called clean.
Lateral force in any section of the beam is numerically equal to the algebraic sum of the projections onto the axis of all forces (including support reactions) located on one side (either) of the drawn section.
Bending moment in a beam section is numerically equal to the algebraic sum of the moments of all forces (including support reactions) located on one side (any) of the drawn section relative to the center of gravity of this section, more precisely, relative to the axis passing perpendicular to the drawing plane through the center of gravity of the drawn section.
Force Q is resultant distributed over the cross-section of internal shear stress, A moment M– sum of moments around the central axis of section X internal normal stress.
There is a differential relationship between internal forces
which is used in constructing and checking Q and M diagrams.
Since some of the fibers of the beam are stretched, and some are compressed, and the transition from tension to compression occurs smoothly, without jumps, in the middle part of the beam there is a layer whose fibers only bend, but do not experience either tension or compression. This layer is called neutral layer. The line along which the neutral layer intersects the cross section of the beam is called neutral line th or neutral axis sections. Neutral lines are strung on the axis of the beam.
Lines drawn on the side surface of the beam perpendicular to the axis remain flat when bending. These experimental data make it possible to base the conclusions of the formulas on the hypothesis of plane sections. According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and turn out to be perpendicular to the curved axis of the beam when it is bent. The cross section of the beam is distorted when bending. Due to transverse deformation, the cross-sectional dimensions in the compressed zone of the beam increase, and in the tensile zone they are compressed.
Assumptions for deriving formulas. Normal voltages
1) The hypothesis of plane sections is fulfilled.
2) Longitudinal fibers do not press on each other and, therefore, under the influence of normal stresses, linear tension or compression operates.
3) Deformations of fibers do not depend on their position along the cross-sectional width. Consequently, normal stresses, changing along the height of the section, remain the same along the width.
4) The beam has at least one plane of symmetry, and all external forces lie in this plane.
5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same.
6) The relationship between the dimensions of the beam is such that it operates under plane bending conditions without warping or twisting.
In case of pure bending of a beam, only normal stress, determined by the formula:
where y is the coordinate of an arbitrary section point, measured from the neutral line - the main central axis x.
Normal bending stresses along the height of the section are distributed over linear law. At the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero.
The nature of normal stress diagrams for symmetrical sections relative to the neutral line
The nature of normal stress diagrams for sections that do not have symmetry with respect to the neutral line
Dangerous points are the points furthest from the neutral line.
Let's choose some section
For any point of the section, let's call it a point TO, the beam strength condition for normal stresses has the form:
, where n.o. - This neutral axis
This axial section modulus relative to the neutral axis. Its dimension is cm 3, m 3. The moment of resistance characterizes the influence of the shape and dimensions of the cross-section on the magnitude of the stresses.
Normal stress strength condition:
The normal stress is equal to the ratio of the maximum bending moment to the axial moment of resistance of the section relative to the neutral axis.
If the material does not equally resist tension and compression, then two strength conditions must be used: for the tensile zone with the permissible tensile stress; for a compression zone with permissible compressive stress.
During transverse bending, the beams on the platforms in its cross-section act as normal, so tangents voltage.
Bend is the type of loading of a beam in which a moment is applied to it lying in a plane passing through the longitudinal axis. Bending moments occur in the cross sections of the beam. When bending, deformation occurs in which the axis of a straight beam bends or the curvature of a curved beam changes.
A beam that bends is called beam . A structure consisting of several bendable rods, most often connected to each other at an angle of 90°, is called frame .
The bend is called flat or straight , if the load plane passes through the main central axis of inertia of the section (Fig. 6.1).
Fig.6.1
When plane transverse bending occurs in a beam, two types of internal forces arise: transverse force Q and bending moment M. In a frame with flat transverse bending, three forces arise: longitudinal N, transverse Q forces and bending moment M.
If the bending moment is the only internal force factor, then such bending is called clean (Fig. 6.2). When there is a shear force, bending is called transverse . Strictly speaking, simple types of resistance include only pure bending; transverse bending is conventionally classified as a simple type of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected when calculating strength.
22.Flat transverse bend. Differential dependencies between internal forces and external load. There are differential relationships between the bending moment, shear force and the intensity of the distributed load, based on the Zhuravsky theorem, named after the Russian bridge engineer D.I. Zhuravsky (1821-1891).
This theorem is formulated as follows:
The transverse force is equal to the first derivative of the bending moment along the abscissa of the beam section.
Let's discard the right side of the beam and replace its action on the left side with a transverse force and a bending moment. For ease of calculation, let’s cover the discarded right side of the beam with a piece of paper, aligning the left edge of the sheet with the section 1 under consideration.
The transverse force in section 1 of the beam is equal to the algebraic sum of all external forces that are visible after closure
We see only the reaction of the support directed downwards. Thus, the shear force is:
kN.
We took the “minus” sign because the force rotates the part of the beam visible to us relative to the first section counterclockwise (or because it is in the same direction as the direction of the transverse force according to the sign rule)
The bending moment in section 1 of the beam is equal to the algebraic sum of the moments of all the forces that we see after closing the discarded part of the beam, relative to the section 1 under consideration.
We see two forces: the reaction of the support and the moment M. However, the force has a shoulder that is practically equal to zero. Therefore, the bending moment is equal to: 
kNm.
Here we took the “plus” sign because the external moment M bends the part of the beam visible to us with a convex downward. (or because it is opposite to the direction of the bending moment according to the sign rule)
Determination of shear forces and bending moments - section 2
Unlike the first section, the reaction force now has a shoulder equal to a.
shear force:
kN;
bending moment:
Determination of shear forces and bending moments - section 3
shear force:
bending moment:
Determination of shear forces and bending moments - section 4
Now it's more convenient cover the left side of the beam with a sheet.
shear force:
bending moment:
Determination of shear forces and bending moments - section 5
shear force:
bending moment:
Determination of shear forces and bending moments - section 1
shear force and bending moment:
.
Using the found values, we construct a diagram of transverse forces (Fig. 7.7, b) and bending moments (Fig. 7.7, c).
CONTROL OF THE CORRECTNESS OF CONSTRUCTION OF DIAGRAMS
Let's make sure that diagrams are constructed correctly based on external features, using the rules for constructing diagrams.
Checking the shear force diagram
We are convinced: under unloaded areas the diagram of transverse forces runs parallel to the axis of the beam, and under a distributed load q - along a downward inclined straight line. On the diagram of the longitudinal force there are three jumps: under the reaction - down by 15 kN, under the force P - down by 20 kN and under the reaction - up by 75 kN.
Checking the bending moment diagram
In the diagram of bending moments we see kinks under the concentrated force P and under the support reactions. The fracture angles are directed towards these forces. Under a distributed load q, the diagram of bending moments changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6 on the diagram of the bending moment there is an extremum, since the diagram of the transverse force in this place passes through the zero value.
Bending deformation is characterized by the loss of straightness or original shape by the beam line (its axis) when an external load is applied. In this case, unlike shear deformation, the beam line changes its shape smoothly.
It is easy to see that resistance to bending is affected not only by the cross-sectional area of the beam (beam, rod, etc.), but also by the geometric shape of this section.
Since the bending of a body (beam, timber, etc.) is carried out relative to any axis, the resistance to bending is affected by the value of the axial moment of inertia of the body’s section relative to this axis.
For comparison, during torsional deformation, the section of the body is subject to twisting relative to the pole (point), therefore, the resistance to torsion is influenced by the polar moment of inertia of this section.
Many structural elements can bend - axles, shafts, beams, gear teeth, levers, rods, etc.
In the strength of materials, several types of bends are considered:
- depending on the nature of the external load applied to the beam, there are pure bend And transverse bending;
- depending on the location of the plane of action of the bending load relative to the axis of the beam - straight bend And oblique bend.
Pure bending is a type of deformation in which only a bending moment occurs in any cross section of the beam ( rice. 2).
Pure bending deformation will, for example, occur if two pairs of forces equal in magnitude and opposite in sign are applied to a straight beam in a plane passing through the axis. Then in each section of the beam only bending moments will act.
If bending occurs as a result of applying a transverse force to the beam ( rice. 3), then such a bend is called transverse. In this case, in each section of the beam, both a transverse force and a bending moment act (except for the section to which an external load is applied).
If the beam has at least one axis of symmetry, and the plane of action of the loads coincides with it, then direct bending occurs, but if this condition is not met, then oblique bending occurs.
When studying bending deformation, we will mentally imagine that the beam (timber) consists of an innumerable number of longitudinal fibers parallel to the axis.
To visualize the deformation of a straight bend, we will conduct an experiment with a rubber bar on which a grid of longitudinal and transverse lines is applied. 
Having subjected such a beam to straight bending, one can notice that ( rice. 1):
The transverse lines will remain straight during deformation, but will turn at an angle to each other;
- the sections of the beam will expand in the transverse direction on the concave side and narrow on the convex side;
- longitudinal straight lines will bend.
From this experience we can conclude that:
For pure bending, the hypothesis of plane sections is valid;
- fibers lying on the convex side are stretched, on the concave side they are compressed, and on the border between them there is a neutral layer of fibers that only bend without changing their length.
Assuming that the hypothesis that there is no pressure on the fibers is valid, it can be argued that with pure bending in the cross section of the beam, only normal tensile and compressive stresses arise, unevenly distributed over the cross section.
The line of intersection of the neutral layer with the cross-sectional plane is called neutral axis. It is obvious that on the neutral axis the normal stresses are zero.
As is known from theoretical mechanics, the support reactions of beams are determined by composing and solving static equilibrium equations for the entire beam. When solving problems of resistance of materials, and determining the internal force factors in the beams, we took into account the reactions of the connections along with the external loads acting on the beams.
To determine the internal force factors, we will use the section method, and we will depict the beam with only one line - the axis to which active and reactive forces are applied (loads and reaction reactions).
Let's consider two cases:
1. Two pairs of forces of equal and opposite sign are applied to a beam.
Considering the equilibrium of the part of the beam located to the left or right of section 1-1 (Fig. 2), we see that in all cross sections only a bending moment M and equal to the external moment occurs. Thus, this is a case of pure bending.
The bending moment is the resulting moment about the neutral axis of the internal normal forces acting in the cross section of the beam.
Let us note that the bending moment has a different direction for the left and right parts of the beam. This indicates the unsuitability of the static sign rule when determining the sign of the bending moment.
2. Active and reactive forces (loads and reaction reactions) perpendicular to the axis are applied to the beam (rice. 3). Considering the equilibrium of the parts of the beam located on the left and right, we see that the bending moment M must act in the cross sections And
and shear force Q.
It follows from this that in the case under consideration, at the points of the cross sections there are not only normal stresses corresponding to the bending moment, but also tangent stresses corresponding to the transverse force.
The transverse force is the resultant of the internal tangential forces in the cross section of the beam.
Let us pay attention to the fact that the transverse force has the opposite direction for the left and right parts of the beam, which indicates that the rule of static signs is unsuitable when determining the sign of the transverse force.
Bending, in which a bending moment and shear force act in the cross section of the beam, is called transverse.
For a beam that is in water equilibrium under the action of a plane system of forces, the algebraic sum of the moments of all active and reactive forces relative to any point is equal to zero; therefore, the sum of the moments of external forces acting on the beam to the left of the section is numerically equal to the sum of the moments of all external forces acting on the beam to the right of the section.
Thus, the bending moment in the beam section is numerically equal to the algebraic sum of the moments relative to the center of gravity of the section of all external forces acting on the beam to the right or left of the section.
For a beam in equilibrium under the action of a plane system of forces perpendicular to the axis (i.e., a system of parallel forces), the algebraic sum of all external forces is equal to zero; therefore, the sum of external forces acting on the beam to the left of the section is numerically equal to the algebraic sum of the forces acting on the beam to the right of the section.
Thus, the transverse force in the beam section is numerically equal to the algebraic sum of all external forces acting to the right or left of the section.
Since the rules of static signs are unacceptable for establishing the signs of bending moment and shear force, we will establish other sign rules for them, namely: If an external load tends to bend the beam with its convexity downward, then the bending moment in the section is considered positive, and vice versa, if the external load tends to bend beam with a convex upwards, then the bending moment in the section is considered negative ( Fig 4,a).
If the sum of external forces lying on the left side of the section gives a resultant directed upward, then the transverse force in the section is considered positive; if the resultant is directed downward, then the transverse force in the section is considered negative; for the part of the beam located to the right of the section, the signs of the shear force will be opposite ( rice. 4,b). Using these rules, you should mentally imagine the section of the beam as rigidly clamped, and the connections as discarded and replaced by reactions.
Let us note once again that to determine the reactions of bonds, the rules of signs of statics are used, and to determine the signs of bending moment and transverse force, the rules of signs of resistance of materials are used.
The rule of signs for bending moments is sometimes called the “rule of rain”, meaning that in the case of a convexity downward, a funnel is formed in which rainwater is retained (the sign is positive), and vice versa - if under the influence of loads the beam bends in an arc upward, there is no water on it delayed (the sign of the bending moments is negative).
Materials from the "Bending" section:
