Chapter 1. BENDING OF STRAIGHT BEAMS AND BEAM SYSTEMS

1.1. The main dependencies of the theory of bending of beams

Beamsit is customary to call rods that work in bending under the action of a transverse (normal to the rod axis) load. Beams are the most common elements of ship structures. The beam axis is the geometrical locus of the centers of gravity of its cross-sections in the undeformed state. A beam is called a straight line if the axis is a straight line. The locus of the centers of gravity of the cross-sections of the beam in the bent state is called the elastic line of the beam. The following direction of the coordinate axes is adopted: axis OXaligned with the beam axis, and the axis OY and OZ - with the main central axes of inertia of the cross section (Fig. 1.1).

Beam bending theory is based on the following assumptions.

1. The hypothesis of plane sections is accepted, according to which the cross-sections of the beam, initially flat and normal to the axis of the beam, remain after its bending flat and normal to the elastic line of the beam. Due to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the planes of the beam cross-sections and their rotation relative to the elastic line (Fig.1.2, and).

2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2, b).

3. The beams are considered to be sufficiently rigid, i.e. their deflections are small in comparison with the height of the beams, and the angles of rotation of the sections are small in comparison with the unit (Fig. 1.2, at).

4. Stresses and strains are linearly related; Hooke's law is valid (Fig. 1.2, r).

Figure: 1.2. Bending theory assumptions

We will consider the bending moments and shear forces appearing during bending of the beam in its section as a result of the action of the part of the beam mentally thrown over the section on its remaining part.

The moment of all forces acting in the section relative to one of the main axes is called the bending moment. The bending moment is equal to the sum of the moments of all forces (including support reactions and moments) acting on the discarded part of the beam, relative to the specified axis of the section under consideration.

The projection onto the section plane of the main vector of the forces acting in the section is called the shearing force. It is equal to the sum of the projections on the section plane of all forces (including support reactions) acting on the rejected part of the beam.

We confine ourselves to considering the bending of the beam occurring in the plane XOZ. Such bending will occur when the lateral load acts in a plane parallel to the plane XOZ, and its resultant in each section passes through a point called the center of the section bending. Note that for sections of beams with two axisymmetries, the center of bending coincides with the center of gravity, and for sections with one axis of symmetry, it lies on the axisymmetry, but does not coincide with the center of gravity.

The load of the beams included in the ship's hull can be either distributed (most often evenly distributed along the beam axis, or changing according to a linear law), or applied in the form of concentrated forces and moments.

Let us denote the intensity of the distributed load (the load per unit length of the beam axis) through q(x), external concentrated force - as R , and the external bending moment - as M... Distributed load and concentrated force are positive if their directions of action coincide with the positive direction of the axis OZ(fig. 1.3, and,b). The external bending moment is positive if it is directed clockwise (Figure 1.3, at).

Figure: 1.3. Sign rule for external loads

We denote the deflection of a straight beam during its bending in the plane XOZ across w, and the angle of rotation of the section through θ. Let's accept the rule of signs for bending elements (Fig. 1.4):

1) the deflection is positive if it coincides with the positive direction of the axis OZ(fig. 1.4, and):

2) the angle of rotation of the section is positive if, as a result of bending, the section rotates clockwise (Fig. 1.4, b);

3) bending moments are positive if the beam under their influence bends convex upwards (Fig. 1.4, at);

4) shearing forces are positive if they rotate the selected beam element counterclockwise (Fig. 1.4, r).

Figure: 1.4. Sign rule for bending elements

Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative fiber elongation ε xlocated at zfrom the neutral axis, will be equal to

ε x= −z/ρ ,(1.1)

where ρ - radius of curvature of the beam in the section under consideration.

Figure: 1.5. Beam bending scheme

The neutral axis of the cross section is the locus of points for which the linear deformation in bending is zero. Between curvature and derivatives of w(x) there is a dependence

By virtue of the accepted assumption about the smallness of the angles of rotation for sufficiently rigid beams, the quantitysmall compared to unity, so we can assume that

Substituting 1 / ρ from (1.2) to (1.1), we obtain

Normal bending stresses σ xbased on Hooke's law will be equal

Since it follows from the definition of beams that there is no longitudinal force directed along the axis of the beam, the main vector of normal stresses must vanish, i.e.

where FIs the cross-sectional area of \u200b\u200bthe beam.

From (1.5), we obtain that the static moment of the cross-sectional area of \u200b\u200bthe beam is zero. This means that the neutral axis of the section passes through its center of gravity.

The moment of internal forces acting in the cross section relative to the neutral axis, M ywill be

Considering that the moment of inertia of the cross-sectional area about the neutral axis OY is equal, and substitute this value in (1.6), then we obtain a dependence that expresses the basic differential equation of the beam bending

The moment of internal forces in the section about the axis OZwill be

Since the axes OYand OZby condition are the main central axes of the section, then .

Hence it follows that under the action of a load in a plane parallel to the main plane of bending, the elastic line of the beam will be a flat curve. This bend is called flat... Based on dependencies (1.4) and (1.7), we obtain

Formula (1.8) shows that normal stresses during bending of beams are proportional to the distance from the neutral axis of the beam. Naturally, this follows from the hypothesis of flat sections. In practical calculations, the moment of resistance of the beam section is often used to determine the highest normal stresses

where | z| max is the absolute value of the distance of the most distant fiber from the neutral axis.

Further subscripts y omitted for simplicity.

There is a relationship between the bending moment, the shearing force and the intensity of the transverse load, arising from the equilibrium condition of the element mentally isolated from the beam.

Consider a beam element of length dx (fig. 1.6). It is assumed here that the deformations of the element are negligible.

If the moment acts in the left section of the element Mand cutting force N, then in its right section the corresponding efforts will have increments. Consider only linear increments .

Figure 1.6. Forces acting on a beam element

Equating to zero the projection on the axis OZ of all efforts acting on the element, and the moment of all efforts relative to the neutral axis of the right section, we obtain:

From these equations, up to quantities of higher order of smallness, we obtain

It follows from (1.11) and (1.12) that

Dependencies (1.11) - (1.13) are known as the Zhuravsky – Schwedler theorem. From these dependences it follows that the shear force and bending moment can be determined by integrating the load q:

where N 0 and M 0 - shearing force and bending moment in the section correspondingx \u003dx 0 , which is taken as the origin; ξ,ξ 1 - variables of integration.

Permanent N 0 and M 0 for statically definable beams can be determined from the conditions of their static equilibrium.

If the beam is statically definable, the bending moment in any section can be found from (1.14), and the elastic line is determined by two-fold integration of the differential equation (1.7). However, statically definable beams are extremely rare in ship hull structures. Most of the beams that make up ship structures form statically indeterminate systems many times over. In these cases, equation (1.7) is inconvenient to determine the elastic line, and it is advisable to go to the fourth-order equation.

1.2. Differential Beam Bending Equation

Differentiating equation (1.7) for the general case when the moment of inertia of the section is a function of x, taking into account (1.11) and (1.12) we get:

where primes denote differentiation with respect to x.

For prismatic beams, i.e. beams of constant cross-section, we obtain the following differential equations of bending:

An ordinary inhomogeneous linear differential equation of the fourth order (1.18) can be represented as a set of four differential equations of the first order:

We use further equation (1.18) or the system of equations (1.19) to determine the deflection of the beam (its elastic line) and all unknown bending elements: w(x), θ (x), M(x), N(x).

Integrating (1.18) sequentially 4 times (assuming that the left end of the beam corresponds to the sectionx= x a ), we get:

It is easy to see that the integration constants N a,M a,θ a , w a have a certain physical meaning, namely:

N a- the shearing force at the origin, i.e. at x \u003dx a ;

M a- bending moment at the origin;

θ a - angle of rotation at the origin;

w a - deflection in the same section.

To determine these constants, you can always create four boundary conditions - two for each end of a single-span beam. Naturally, the boundary conditions depend on the structure of the beam ends. The simplest conditions correspond to hinged support on rigid supports or rigid termination.

When the end of the beam is pivotally supported on a rigid support (Fig. 1.7, and) the deflection of the beam and the bending moment are equal to zero:

With a rigid termination on a rigid support (Fig. 1.7, b) the deflection and the angle of rotation of the section are equal to zero:

If the end of the beam (console) is free (Fig. 1.7, at), then in this section the bending moment and the shearing force are equal to zero:

A situation is possible associated with a sliding termination or a termination by symmetry (Fig. 1.7, r). This leads to the following boundary conditions:

Note that the boundary conditions (1.26) concerning deflections and angles of rotation are usually called kinematic, and conditions (1.27) - power.

Figure: 1.7. Types of boundary conditions

Ship structures often have to deal with more complex boundary conditions, which correspond to the support of the beam on elastic supports or elastic termination of the ends.

An elastic support (Fig. 1.8, and) is called a support that has a drawdown proportional to the reaction acting on the support. We will consider the reaction of the elastic support R positive if it acts on the support in the direction of the positive direction of the axis OZ... Then you can write:

w \u003dAR,(1.29)

where A- the coefficient of proportionality, called the coefficient of compliance of the elastic support.

This coefficient is equal to the subsidence of the elastic support under the action of the reaction R \u003d1, i.e. A \u003dw R = 1 .

Elastic supports in ship structures can be beams supporting the beam in question, or pillars and other compression structures.

To determine the compliance coefficient of an elastic support Ait is necessary to load the corresponding structure with a unit force and find the absolute value of subsidence (deflection) at the place of application of the force. Rigid support - a special case of elastic support when A \u003d 0.

Elastic sealing (fig. 1.8, b) is a support structure that prevents free rotation of the section and in which the angle of rotation θ in this section is proportional to the moment, i.e. has a dependency

θ = Â M.(1.30)

Proportionality multiplier  is called the elastic seal compliance coefficient and can be defined as the angle of rotation of the elastic seal at M \u003d 1, i.e.  = θ M \u003d 1 .

A particular case of an elastic seal when  = 0 is a hard termination. In ship structures, elastic fixings are usually beams normal to the considered one and lying in the same plane. For example, beams, etc. can be considered elastically sealed on frames.

Figure: 1.8. Elastic support ( and) and elastic sealing ( b)

If the ends of the beam are long Lsupported on elastic supports (Fig. 1.9), then the reactions of the supports in the end sections are equal to the shear forces, and the boundary conditions can be written:

The minus sign in the first condition (1.31) is accepted because the positive shear force in the left support section corresponds to the reaction acting on the beam from top to bottom, and on the support from bottom to top.

If the ends of the beam are long Lelastically sealed (Fig. 1.9), then for support sections, taking into account the rule of signs for the angles of rotation and bending moments, we can write:

The minus sign in the second condition (1.32) is adopted because with a positive moment in the right support section of the beam, the moment acting on the elastic seal is directed counterclockwise, and the positive angle of rotation in this section is directed clockwise, i.e. the directions of the moment and the angle of rotation do not coincide.

Consideration of the differential equation (1.18) and all boundary conditions shows that they are linear with respect to both the deflections and their derivatives included in them, and the loads acting on the beam. Linearity is a consequence of the assumptions about the validity of Hooke's law and the small deflection of the beam.

Figure: 1.9. A beam, both ends of which are elastically supported and elastically sealed ( and);

forces in elastic supports and elastic fittings corresponding to positive
directions of bending moment and shearing force ( b)

When several loads are applied to a beam, each beam bending element (deflection, angle of rotation, moment and shear force) is the sum of the bending elements from the action of each of the loads separately. This very important position, called the principle of superposition, or the principle of summation of the action of loads, is widely used in practical calculations and, in particular, to disclose the static uncertainty of beams.

1.3. Initial parameter method

The general integral of the differential equation of beam bending can be used to determine the elastic line of a single-span beam in the case when the load of the beam is a continuous function of the coordinate throughout the entire span. If the load contains concentrated forces, moments, or a distributed load acts on a part of the beam length (Fig. 1.10), then expression (1.24) cannot be used directly. In this case, it would be possible, denoting the elastic lines in sections 1, 2 and 3 through w 1 , w 2 , w 3, write down for each of them the integral in the form (1.24) and find all arbitrary constants from the boundary conditions at the ends of the beam and the conjugation conditions at the boundaries of the sections. The conjugation conditions in this case are expressed as follows:

at x \u003d a 1

at x \u003d a 2

at x \u003d a 3

It is easy to see that this way of solving the problem leads to a large number of arbitrary constants equal to 4 nwhere n - the number of sections along the length of the beam.

Figure: 1.10. Beam, in some sections of which loads of different types are applied

It is much more convenient to represent the elastic line of the beam in the form

where the terms behind the double bar are taken into account when x³ a 1, x³ a 2, etc.

Obviously, δ 1 w(x)=w 2 (x)−w 1 (x); δ 2 w(x)=w 3 (x)−w 2 (x); etc.

Differential equations for determining corrections to the elastic line δ iw (x) based on (1.18) and (1.32) can be written in the form

General integral for any correction δ iw (x) to the elastic line can be written in the form (1.24) for x a = a i ... In this case, the parameters N a,M a,θ a , w a have the meaning of a change (jump), respectively: in the shear force, bending moment, angle of rotation and deflection arrow when passing through the section x \u003da i ... This technique is called the method of initial parameters. It can be shown that for the beam shown in Fig. 1.10, the elastic line equation will be

Thus, the method of initial parameters makes it possible, even in the presence of discontinuity in the loads, to write the equation of the elastic line in the form containing only four arbitrary constants N 0 , M 0 , θ 0 , w 0, which are determined from the boundary conditions at the ends of the beam.

Note that for a large number of variants of single-span beams encountered in practice, detailed bending tables have been compiled, which make it easy to find deflections, turning angles and other bending elements.

1.4. Determination of shear stresses in bending beams

The hypothesis of flat sections accepted in the theory of bending of beams leads to the fact that the shear deformation in the section of the beam turns out to be equal to zero, and we do not have the opportunity, using Hooke's law, to determine the shear stresses. However, since, in the general case, shear forces act in the beam sections, the corresponding shear stresses should arise. This contradiction (which is a consequence of the accepted hypothesis of flat sections) can be circumvented by considering the equilibrium conditions. We will assume that when bending a beam composed of thin strips, the shear stresses in the cross section of each of these strips are uniformly distributed over the thickness and directed parallel to the long sides of its contour. This position is practically confirmed by the exact solutions of the theory of elasticity. Consider a beam of an open thin-walled I-profile. In fig. 1.11 shows the positive direction of shear stresses in the flanges and the profile web during bending in the plane of the beam web. We select with a longitudinal section I -I and two cross-sections an element of length dx (fig. 1.12).

We denote the tangential stress in the specified longitudinal section by τ, and the normal forces in the initial cross section by T... The normal forces in the final section will have increments. Consider only linear increments, then.

Figure: 1.12. Longitudinal forces and shear stresses
in the girdle element of the beam

The condition of static equilibrium of the element selected from the beam (equality to zero of the projections of the forces on the axis OX) will be

where; f- the area of \u200b\u200bthe part of the profile, cut off by the line I -I; δ - profile thickness at the section.

From (1.36) it follows:

Since the normal stresses σ x are defined by formula (1.8), then

In this case, we assume that the beam has a constant cross-section along its length. The static moment of a part of the profile (by the clipping line I -I) relative to the neutral axis of the beam section OY is an integral

Then from (1.37) for the absolute value of stresses we obtain:

Naturally, the obtained formula for determining the shear stresses is also valid for any longitudinal section, for example II -II (see fig. 1.11), and the static moment S Ot is calculated for the cut off part of the area of \u200b\u200bthe beam profile relative to the neutral axis without regard to the sign.

Formula (1.38), in the sense of the conclusion drawn, determines the shear stresses in the longitudinal sections of the beam. From the theorem on the pairing of tangential stresses, known from the course of resistance of materials, it follows that the same tangential stresses act at the corresponding points of the cross section of the beam. Naturally, the projection of the main vector of shear stresses on the axis OZ must be equal to the shearing force Nin this section of the beam. Since in the belts beams of this type, as shown in Fig. 1.11, shear stresses are directed along the axis OY, i.e. normal to the plane of action of the load, and are generally balanced, the shear force must be balanced by the shear stresses in the beam web. The distribution of shear stresses along the height of the wall follows the law of variation of the static moment S cut off part of the area relative to the neutral axis (at constant wall thickness δ).

Consider a symmetrical section of an I-beam with a belt area F 1 and wall area ω = hδ (fig. 1.13).

Figure: 1.13. I-beam section

The static moment of the cut-off part of the area for a point located at z from the neutral axis will be

As can be seen from the dependence (1.39), the static moment changes with zaccording to the law of a quadratic parabola. Highest value S from, and, consequently, shear stresses τ , will turn out at the neutral axis, where z \u003d0:

The greatest shear stress in the beam web at the neutral axis

Since the moment of inertia of the section of the considered beam is

then the greatest shear stress will be

Attitude N/ ω is nothing more than the average shear stress in the wall, calculated under the assumption of a uniform stress distribution. Taking, for example, ω \u003d 2 F 1, by formula (1.41) we obtain

Thus, the considered beam has the highest shear stress in the wall at the neutral axis by only 12.5% exceeds the average value of these stresses. It should be noted that for most of the beam profiles used in the ship's hull, the excess of the maximum shear stresses over the average is 10-15%.

If we consider the distribution of shear stresses during bending in the section of the beam shown in Fig. 1.14, then you can see that they form a moment relative to the center of gravity of the section. In the general case, the bending of such a beam in the plane XOZwill be accompanied by twisting.

Beam bending is not accompanied by twisting if the load acts in a plane parallel to XOZpassing through a point called the center of the bend. This point is characterized by the fact that the moment of all tangential forces in the beam section relative to it is zero.

Figure: 1.14. Shear stresses during bending of the channel beam (point AND - bend center)

Designating the distance of the center of the bend AND from the axis of the beam wall through e, we write down the condition of equality to zero of the moment of tangential forces relative to the point AND:

where Q 2 - shear force in the wall equal to the shearing force, i.e. Q 2 =N;

Q 1 =Q 3 - force in the belt, determined on the basis of (1.38) by the dependence

Shear deformation (or shear angle) γ varies along the height of the beam wall in the same way as shear stresses τ , reaching the greatest value at the neutral axis.

As shown, for beams with belts, the change in shear stresses along the height of the wall is very insignificant. This allows us to further consider a certain average shear angle in the beam wall

Shear deformation leads to the fact that the right angle between the plane of the cross-section of the beam and the tangent to the elastic line changes by the value γ Wed A simplified diagram of the shear deformation of a beam element is shown in Fig. 1.15.

Figure: 1.15. Beam element shear deformation diagram

By marking the deflection arrow caused by the shear through w sdv, you can write:

Considering the sign rule for shear force N and the angle of rotation we find

Because the ,

Integrating (1.47), we obtain

Constant a, included in (1.48), defines the displacement of the beam as a rigid body and can be taken equal to any value, since when determining the total arrow of the deflection from bending w exile and shift w SDV

the sum of the integration constants appears w 0 +adetermined from the boundary conditions. Here w 0 - deflection from bending at the origin.

In what follows we put a\u003d 0. Then the final expression for the elastic line caused by the shear takes the form

The flexural and shear components of the elastic line are shown in Fig. 1.16.

Figure: 1.16. Bending ( and) and shear ( b) components of the elastic line of the beam

In the considered case, the angle of rotation of the sections during shear is equal to zero, therefore, taking into account the shear, the angles of rotation of the sections, bending moments and shearing forces are associated only with the derivatives of the elastic line from bending:

The situation is somewhat different in the case of action on the beam of concentrated moments, which, as will be shown below, do not cause deflections from shear, but only lead to an additional rotation of the beam sections.

Consider a beam freely supported on rigid supports, in the left section of which moment acts M... The shearing force in this case will be constant and equal

For the right support section, respectively, we obtain

.(1.52)

Expressions (1.51) and (1.52) can be rewritten as

Expressions in parentheses characterize the relative addition to the angle of rotation of the section caused by shear.

If we consider, for example, a freely supported beam loaded in the middle of its span by the force R (Figure 1.18), then the deflection of the beam under the force will be equal to

The bending deflection can be found in the bending tables of beams. The shear deflection is determined by formula (1.50), taking into account that .

Figure: 1.18. Diagram of a freely supported beam loaded with a concentrated force

As can be seen from formula (1.55), the relative addition to the beam deflection due to shear has the same structure as the relative addition to the angle of rotation, but with a different numerical coefficient.

Let us introduce the notation

where β is a numerical coefficient that depends on the specific problem under consideration, the device of supports and the load of the beam.

Let us analyze the dependence of the coefficient k from various factors.

If we take into account that, instead of (1.56) we get

The moment of inertia of a beam section can always be represented as

,(1.58)

where α is a numerical coefficient depending on the shape and characteristics of the cross section. So, for a beam with an I-profile according to the formula (1.40) at ω \u003d 2 F 1 find I \u003d ωh 2/3, i.e. α \u003d 1/3.

Note that with an increase in the size of the beam flanges, the coefficient α will increase.

Taking into account (1.58), instead of (1.57), we can write:

Thus, the value of the coefficient ksignificantly depends on the ratio of the length of the beam span to its height, on the section shape (through the coefficient α), the arrangement of the supports and the load of the beam (through the coefficient β). The relatively longer the beam ( h /Lsmall), the less the effect of shear deformation. For rolled profile beams related h /Lless than 1/10 ÷ 1/8, the shift correction can be practically ignored.

However, for beams with wide belts, such as, for example, the keel, stringers and flora in the composition of bottom floors, the effect of shear and at the indicated h /Lcan be significant.

It should be noted that shear deformations affect not only the increase in beam deflections, but in some cases also the results of disclosing the static indeterminacy of beams and beam systems.

After determining the initial angle of rotation, the deflection of section A is calculated.

, shown in Fig. 2.3 by a dotted line, is introduced in cases where the deflection is determined in a section that is outside the area of \u200b\u200baction of the distributed load.

The angle of rotation of the section B is calculated by the formula (2.20), in which

2.2.2. Mohr's integral.

Mohr's universal formula for calculating elastic displacements in rod systems is a natural generalization of Castigliano's formula. For linearly elastic rod systems, Castigliano's formula has the form

Δ TO -generalized displacement of the section K,

R K Is the generalized force corresponding to the generalized displacement Δ TO,

U –Function of potential energy.

The potential energy is a quadratic function of the forces and for bending elements is written in the form

(2.22)

In the overwhelming majority of cases, the influence of the transverse force on the potential energy is neglected. A combination of formulas (2.21) and (2.22) gives

(2.23)

The partial derivative corresponds to the function of the bending moment caused by the action of a single generalized force applied in section K in the direction of the desired displacement. Formula (2.23) written in the form

(2.24)

defines a particular form of the universal Mohr formula as applied to the definition of displacements in bending elements.

In practice, the graphical-analytical technique for calculating the Mohr integral is used (Vereshchagin's technique).

- the area of \u200b\u200bthe load diagram (the diagram of the bending moment from the action of a given load);

- the ordinate of the unit diagram (the diagram of the bending moment from the action of a single generalized force), measured under the center of the load diagram.

Calculation of the Mohr integral according to Vereshchagin's formula is called "multiplication" of diagrams in the educational literature.

In a number of cases, when calculating the Mohr integral, it is convenient to use the Simpson formula

(2.26)

where the indices "n", "c", "k" - denote the beginning, middle and end of the section of the multiplied diagrams, respectively.

Example 2.Determine the section deflection AND and the angle of rotation of the section AT beams considered in example 1 (Figure 2.4.a).

Calculate Mohr's integral using Simpson's formula.

To determine the section deflection ANDa freight M p (Fig.2.4.b) and single (Fig.2.4.c) diagrams of bending moments.

Multiplying the load and unit bending moment diagrams by Simpson's formula gives

To determine the angle of rotation of the reference section ATthe second unit diagram of the bending moment is constructed from the action of a unit moment applied in the section AT beams (Figure 2.4.d).

The value of the angle of rotation is determined by multiplying the load and the unit (Figure 2.4.d) diagrams of bending moments.

Note. The minus sign in the answers means that the directions of the actual displacements of the sections ANDand AT will be opposite to the directions of displacements corresponding to unit generalized forces.

2.3. Statically indeterminate beams
(Method of disclosure forces of static indeterminacy)

Statically indeterminate beams contain "redundant" ties (when the redundant ties are removed, the beams become statically determinate). The number of redundant links determines the degree of static indeterminacy of the problem.

A statically definable geometrically invariable beam obtained from a given statically indeterminate one by removing unnecessary constraints is called the main system of the force method.

The algorithm for solving statically indeterminate beams by the force method is considered on the example of a once statically indeterminate beam (Fig. 2.5.a).

The solution of the problem begins with the choice of the main system of the method of forces (Fig. 2.5.b). It should be noted that this is not the only option for choosing the main system (in particular, it is possible to remove internal links by placing a hinge).

The essence of the force method is to negate movements in the direction of the remote connection. Mathematically, this condition is written in the form of the displacement compatibility equation

, (2.27)

δ 11 - movement in the direction of the discarded link caused by the action of a single value of the unknown reaction of the remote link (Fig. 2.5.c)

Δ 1Р - movement in the direction of the discarded connection caused by the action of a given load (Fig. 2.5.d)

Calculation of displacements δ 11, Δ 1Р is made according to the Simpson formula.

The coefficient δ 11 of the canonical equation of the method of forces is determined by multiplying the unit diagram (Fig.2.5.f) by itself

The coefficient Δ 1Р of the canonical equation of the method of forces is calculated by multiplying the unit (Fig. 2.5.f) and cargo (Fig. 2.5. d) diagrams

From the solution of equation (2.27), the reaction X 1 extra connection

This stage of the solution corresponds to the disclosure of the static uncertainty of the problem.

Bending moment plot M x(Fig. 2.5.h) in a statically indeterminate beam is constructed according to the formula

(2.28)

In fig. 2.5.g shows a "corrected" single plot, all ordinates of which are increased in X 1time.

The considered algorithm for solving statically indeterminate problems using the method of forces is also suitable for solving statically indeterminate problems in torsion, with axial action of loads, and also with complex deformation of the bar.

2.4. Stability of compressed rods

For a complete understanding of the operation of a structure, along with strength and stiffness calculations, stability calculations for compressed and compressed-bent elements are required.

Engineering objects, in addition to design loads, can be subject to additional, not provided for in the calculation, small perturbations that can cause non-design deformation in the elements of the object (curvature of the axis of compressed elements, spatial bending of a plane curved element). The result of this additional action depends on the intensity of the loads acting on the structural element. For each element, there is a certain critical value of the load, above which a small random disturbance causes irreversible non-design deformation. This state of the object is dangerous.

Task. For a beam, define displacements incl. AND, AT, FROM, D, select a cross-section of two channels from the strength condition, check the rigidity, show the curved axis of the beam. Material - St3 steel, allowable movement.

1. We define support reactions.

We plot the value of support reactions on design scheme

2. We build plot of moments from a given load - load plot M F .

Because under a uniformly distributed load, the line is a parabolic curve, then an additional point will be required to draw it - we put t. TO in the middle of the load.

We build a plot M F from a given load.

3. We will select section of two channels:

We select 2 channels # 33 cm 3.

Check strength selected section.

Durability is assured.

4. Define moving at the given points. We remove the entire load from the beam. For determining linear displacement (deflections) apply unit force ( F=1 ), and to determine corner displacement - single moment .

Points AND and AT - these are supports, and according to the boundary conditions in the hinged supports no deflection possible, but angular movement is present... At points FROM and D there will be both linear (deflections) and angular (angles of rotation) displacement.

We define angular movement at t. AND ... We apply in AND single moment (fig. b ). We build ep, define the necessary ordinates in it. (fig. at ).

Ep. Ordinates M F- all positive, ep. - also.

We will define displacements mohr's method.

We define moment of inertia I x for the section.

Longitudinal modulus E for St3 E\u003d 2 · 10 5 MPa \u003d 2 · 10 8 kPa... Then:

Angle of rotation φ A turned out positive, it means that the angle of rotation of the section coincides with the direction of the unit moment.

We define angle of rotationφ B. (rice .d, d)

Now let's define the displacement, incl. FROM (linear and angular). We apply a unit force (Fig. e ), we define support reactions and build ep. from a unit force (Fig. f ).

Consider fig. e.

Building ep. :

We define deflection incl. FROM.

To determine the angle of rotation incl. FROM we apply a unit moment (Fig. s ), define the support reactions and plot the diagram of unit moments (Fig. and ).

(sign "— " says that reaction R A directed in the opposite direction... We show this in the design diagram - Fig. s ).

Building ep. ,

Because the m\u003d 1 attached incl. FROM span of the beam, then the moment incl. FROM define both from the left and from the right forces.

We define deflection at point C.

(the "-" sign means that the angle of rotation is opposite to the direction of the unit moment)

Similarly, we define the linear and angular displacement, incl. D .

We define at D . (fig. to ).

Building ep. (fig. l ) :

We define φ D (fig. m ):

Building ep. - (Fig. n ).

We define angle of rotation:

(the angle of rotation is directed in the direction opposite to the unit moment).

Now let's show curved beam axis (elastic line), which became a rectilinear axis under load. For this we will draw initial the position of the axis and on the scale we postpone the calculated displacements (Fig. about ).

Check rigidity beams where f - maximum deflection.

Maximum deflection - rigidity is not provided.

So In this problem, we made sure that the sections selected from the strength condition (in this case, a section of two channels) do not always satisfy the stiffness conditions.

Task.Determine the horizontal displacement of the free end of the frame by Mohr's integral

1. Make up an expression bending moment M F from the current load.

2. Remove all loads from the beam, and at the point where it is necessary to determine the displacement, apply a unit force (if we define a linear displacement) or a unit moment (if we determine an angular displacement) in the direction of the desired displacement. In our task, we apply a horizontal unit force. We compose the expression for the bending moment.

We define moments from a single load F=1

By calculating horizontal movement:

Moving has a positive meaning. This means that it corresponds to the direction of the unit force.

Integral, Mohr's formula. Define the horizontal movement of a point in a curved bar AND... The rigidity is constant over the entire length of the bar.

The axis of the timber is outlined along parabola, the equation of which:

Considering that the timber non-thrust and enough flat (f / ι \u003d 3/15 \u003d 0.2), the influence of longitudinal and transverse forces is neglected... Therefore, to determine the displacement, we will use the formula:

Because hardness EJ constant, then:

Let's compose an expression M 1 for the actual state of the bar ( 1st state) (Fig. and):

We remove all loads from the bar and apply at the point AND horizontal unit force ( 2nd state) (Fig. b). We compose an expression for:

We calculate the required move at point AND :

Sign minus indicates that moving point ANDopposite to the direction of the unit force, i.e. this point is shifted horizontally to the left.

Integral, Mohr's formula. Determine the angle of rotation of the pivot bearing D for a frame with certain support reactions, the stiffnesses of the elements are indicated on the design diagram.

Let's compose an expression M 1, using the system diagram in the 1st state. M 1 - function of the internal bending moment in the force section for a given beam or frame from the action of the given loads of the 1st state.

We release the frame from loads, apply unit moment on support D, we get the system second state.

We compose expressions - this is a function of the internal bending moment on the force section for the auxiliary system of the 2nd state, loaded single effort:Find the desired displacement - the angle of rotation formula (integral):
The value of the rotation angle is positive, which means the direction corresponds to the selected direction of the unit moment.

Integral (Mohr's formula). For the frame, define the horizontal movement of the point C... The stiffnesses of the elements are shown in the figure. Let's call the given system the system the first states. ... We compose for each element expression М₁, taking advantage of scheme of the 1st state of the system:

We remove all loads from the frame and get 2ndframe conditionby applying in the direction of the desired displacement horizontal unit force. We compose the expression of single moments:. We calculate by formula (integral) desired displacement :

Then we get:

Sign minus indicates that the direction of movement is opposite to the direction of the unit force.

For a steel beam, select the dimensions of the cross-section, consisting of two I-beams, based on the strength condition for normal stresses, build diagrams of linear and angular displacements. Given:

Calculation of support reactions and values \u200b\u200bof load diagrams (diagrams of bending moments) will not be presented, we will show without calculations. So, load diagram of moments:

At the same time, on the diagram M, the values \u200b\u200bof bending moments have no signs, fibers are indicated that experience compression... As can be seen from the diagram, in dangerous section: M C \u003d M max \u003d 86.7 kNm.

Let's select a section from two I-beams.Of strength conditions:

According to choose I-beam No. 27а, which one I x 1 \u003d 5500cm 3, h \u003d 27cm. Actual value axial moment of resistance of the entire section W x \u003d 2I x 1 / (h / 2) \u003d 2 5500 / (27/2) \u003d 815 cm 3.

We calculate linear and angular movements beam sections method, applying . The choice of the number of sections required to construct diagrams of linear and angular displacements in a beam depends on the number of sections and the nature of the bending moment diagram. In the considered beam, these include sections A, B, C, D (belong to borders power sections) and sections 1, 2, 3 - in the middle of the sections (the definition of displacements in these sections increases accuracy of plotting).

Section A.As you know, the linear movement of the section in the hinged support y A \u003d 0.

To calculate angular displacement θ a load the auxiliary system with a unit pair of forces - moment equal to one
Equilibrium equations

Solving the equilibrium equations, we get:

Determine the values \u200b\u200bof the moments in the characteristic sections

Section АD:

AT middle section AB value bending moment of load diagram M F equally f \u003d 73.3 · 1- 80 · 1 2/2 \u003d 33.3kNm

We define angular displacement of section Aby :

Angular movement of section A is directed counterclockwise (opposite to the action of a single moment).

Section B

We apply in section B force equal to one, for determining linear displacement, and build a single moment diagram

Equilibrium equations:

From the solution of the equilibrium equations it follows:

Determine the values \u200b\u200bof the moments in characteristic sections:

We define linear displacement y V.

Linear displacement y B \u003d 3.65 × 10 -3 m directed up (opposite to the action of a unit force).

To determine the angular displacement in section B, we apply single moment and build a single moment diagram.

As a result of "multiplication" of the unit diagram and the load diagram, we obtain angular movement:

counterclock-wise.

Section C.

Linear movement:

Angular movement:

Angular movement directed clockwise.

Section D. Linear displacementin this section is equal to zero.

Angular movement:

Angular movement directed clockwise.

Additional sections:

Section 1 (z \u003d 0.5ℓ)

Angular movement:

Angular movement directed counterclock-wise.

Similarly, we construct unit diagrams for section 2 (z \u003d 1.5ℓ) and section 3 (z \u003d 2.5ℓ), we find displacements.

Applying sign rule for linear movements up - plus, down - minus, and for angular displacements counterclockwise - plus, clockwise - minus, build diagrams of linear and angular displacements y and θ.

Determine the maximum deflection and maximum angle of rotation for the beam.

Due to the symmetry of the load support reactions A \u003d B \u003d ql / 2

Curved beam axis differential equation:

We integrate this equation twice. After the first integration, we obtain the equation for the angles of rotation:

(and)

After the second integration, we get the equation of deflections:

(b)

It is necessary to determine the value integration constants - C and D... Let's define them from boundary conditions... In sections A and B, the beam has pivot bearingsmeans deflections in them are equal to zero. Therefore, we have border conditions:

1) z \u003d 0, y \u003d0.

2) z \u003d l, y \u003d0.

We use first boundary condition: z \u003d 0, y = 0.

Then from (b) we have:

The second boundary condition at z \u003d l gives:

from where:

Finally we get.

Rotation angle equation:

Deflection equation:

When the angle of rotation is zero, and the deflection will be maximum:

Sign minus indicates that with the assumed positive direction of the axis up, the deflection will be downward.

The angle of rotation has the greatest value on support sections, for example, at

The minus sign indicates that the angle of rotation at z \u003d 0 directed clockwise.

For the frame, you need to determine the angle of rotation of the section 1 and horizontal movement of the section 2 .

Given: L \u003d 8 m, F \u003d 2 kN, q \u003d 1 kN / m, h \u003d 6 m, moments of inertia I 1 \u003d I, I 2 \u003d 2I

1. Determine support reactions and build a load diagram:

a) Determine support reactions:

The check came together. Vertical reactions are defined correctly. To determine horizontal reactions, use hinge property, namely, to write down the equation of moments relative to the hinge from all forces, located on one side of the frame.

The check came together, so horizontal reactions are defined correctly.

b) We build a cargo plot - a plot from a given load.We will build a cargo plot on stretched fibers.

We break the frame into sections. On each section, we outline sections at the beginning and end of the section, and in sections with a distributed load, an additional section in the middle. In each section, we determine the value of the internal bending moment according to the rule: the bending moment is equal to the algebraic sum of the moments of all external forces located on one side of the section, relative to the center of this section. Sign rule for bending moment: a moment is considered positive if it stretches the lower fibers.

We build cargo plot.

2. Determine the angle of rotation of the section (1)

a) In order to determine the angle of rotation of the specified section, you need sketch the original frame without external load and apply a single moment to the given section.

First, we define the reactions:

The "-" sign means that the cross section is rotated against the direction of the unit moment, i.e. clockwise.

3. Determine the horizontal displacement of the section (2).

a) In order to determine the horizontal displacement in the indicated section, you need to sketch the original frame without external load and apply a unit force in the horizontal direction to the given section.

We define reactions:

We build single moment plot

.

For a beam, determine the linear and angular displacements at points A, B, C, having previously selected the I-beam section from the strength condition.

Given:a\u003d 2 m,b\u003d 4 m, s \u003d 3 m,F\u003d 20 kN, M \u003d 18 kNm,q\u003d 6 kN / m, σ adm\u003d 160 MPa, E \u003d 210 5 MPa

1) We draw a diagram of the beam, we determine the support reactions.In a rigid termination, there is 3 reactions — vertical and horizontal, and reference moment. Since there are no horizontal loads, the corresponding response is zero. In order to find the reactions at point E, we compose equilibrium equations.

∑F y \u003d 0 q7-F + R E \u003d 0

R E \u003d -q7 + F \u003d -67 + 20 \u003d -22kN (the sign indicates that

Find support moment in rigid termination, for which we will solve the equation of moments relative to any selected point.

∑M C: -M E -R E 9-F6-q77 / 2-M \u003d 0

M E \u003d -18-229 + 649/2 \u003d -18-198 + 147 \u003d -69kNm (the sign indicates that the reaction is directed in the opposite direction, we show this in the diagram)

2) We build a load diagram M F - a diagram of moments from a given load.

To construct moment diagrams, we find moments at characteristic points... AT point B define moments from both right and left forcesbecause a moment is applied at this point.

To plot the moment on the line of action of the distributed load (sections AB and BC) we need additional points to plot the curve. Let's define the moments in the middle these sites. These are moments in the middle of sections AB and BC 15.34 kNm and 23.25 kNm... We build cargo plot.

3) To determine linear and angular displacements at a point, it is necessary to apply at this point, in the first case, unit force (F \u003d 1) and build a diagram of moments, in the second case, unit moment (M \u003d 1) and plot the moments. We build diagrams from unit loads for each point - A, B and C.

4) To find the displacements, we use the Simpson formula.

where l i - section length;

EI i - the rigidity of the beam on the site;

M F - values \u200b\u200bof bending moments from the load diagram, respectively at the beginning, in the middle and at the end of the site;

– values \u200b\u200bof bending moments from a single plot, respectively at the beginning, in the middle and at the end of the site.

If the ordinates of the diagrams are located on one side of the beam axis, then the multiplication takes into account the "+" sign, if from different, then the "-" sign.

If the result is obtained with a "-" sign, then the desired displacement in the direction does not coincide with the direction of the corresponding unit force factor.

Consider application of Simpson's formula on the example of determining displacements at point A.

We define deflection, by multiplying the load plot by the plot from the unit force.

Deflection turned out with a "-" sign, means the desired displacement in direction does not coincide with the direction of the unit force (directed upwards).

We define angle of rotationby multiplying the load plot by the plot from a single moment.

The angle of rotation turned out with a "-" sign, means the desired displacement in the direction does not coincide with the direction of the corresponding unit moment (directed counterclockwise).

5) To determine specific displacement values, you need to select a section. We select the I-beam section

where M max - this is maximum moment on the load moment diagram

We select by i-beam No. 30 with W x \u003d 472cm 3 and I x \u003d 7080cm 4

6) Determine displacements at points revealing section stiffness: E - material longitudinal elastic modulus or modulus (2 10 5 MPa),J x - axial moment of inertia of the section

Deflection at point A (up)

Rotation angle (counterclockwise)

First, let's build cargo plot from a given load. Load plot area has a curvilinear outline and is equal to:

Now we remove the load from the beam and apply it at the point where it is necessary to determine the displacement unit force to determine deflection and unit moment for determining the angle of rotation... We build plots from single loads.

Center of gravity of load diagram is at a distance one quarter (see diagram)

The ordinates of the unit diagrams opposite the center of gravity of the load diagram:

Admin in heading.

Bending hypotheses. Neutral layer, radius of curvature, curvature, distribution of deformations and normal stresses along the height of the cross-section of the bar. Shear stresses at plane transverse bending of bars. Calculation of beams for bending strength. Bending movements.

Normal voltages with a pure straight bend. Since normal stresses depend only on bending moments, the derivation of the formula for the calculation can be made in relation to pure bending. Note that the exact dependence for normal stresses in pure bending can be obtained using the methods of the theory of elasticity, but if this problem is solved by methods of resistance of materials, it is necessary to introduce some assumptions.

There are three such hypotheses in bending:

1) the hypothesis of flat sections (Bernoulli's hypothesis) - sections that are flat before deformation remain flat and after deformation, but only rotate about 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, they will be compressed; fibers lying on the neutral axis do not change their length;

2) the hypothesis of the constancy of normal stresses - stresses acting at the same distance y from the neutral axis are constant along the width of the bar;

3) hypothesis about the absence of lateral pressures - adjacent longitudinal fibers do not press against each other.

Figure: 28. Bernoulli's hypothesis

Static plane bend problem... The bending moment in the section is the sum of the moments of all elementary internal normal forces σ.dA arising on the elementary areas of the beam cross-section (Fig. 29), relative to the neutral axis: .

This expression represents the static side of the plane bend problem. But it cannot be used to determine normal stresses, since the law of stress distribution over the section is unknown.

Figure: 29. The static side of the problem

The geometric side of the plane bend problem... Let us select a beam element of length dz with two cross-sections. Under load, the neutral axis is bent (radius of curvature ρ), and the sections rotate about their neutral lines by an angle dθ. The length of the segment of fibers of the neutral layer remains unchanged (Fig. 30, b):

Figure: 30. Geometric side of the problem:
a - beam element; b - curvature of the neutral axis; c - diagram σ.dA; d - diagram ε

Let us determine the length of the fiber segment spaced from the neutral layer at a distance y

dz 1 \u003d (ρ + y) dθ.

The elongation in this case will be

The dependence reflects the geometric aspect of the plane bending problem, from which it can be seen that the deformations of longitudinal fibers vary along the section height according to a linear law.

A set of fibers that do not change their length when the beam is bent is called a neutral layer.

The line along which the cross-section of the beam intersects with the neutral layer of the beam is called the neutral section line.

Physical side of the plane bend problem. Using Hooke's law for axial tension, we obtain

Substituting the value of σ into the expression reflecting the static side of the plane bending problem, we obtain

Substituting the value in the original formula, we get

(13)

This expression reflects the physical side of the plane bending problem, which makes it possible to calculate the normal stresses along the section height.

Although this expression was obtained for the case of pure bending, but as theoretical and experimental studies show, it can be used for plane transverse bending.

Neutral line. The position of the neutral line is determined from the condition of equality to zero of the normal force in the sections of the beam with pure bending

Since M x ≠ 0 and I x ≠ 0, it is necessary that the integral be equal to zero. This integral is the static moment of the section about the neutral axis. Since the static moment of the section is zero only relative to the central axis, therefore, the neutral line in plane bending coincides with the main central axis of inertia of the section.

Shear stresses... The tangential stresses that arise in the sections of the beam during plane transverse bending are determined by the dependence:

(14)

where Q is the shear force in the considered section of the beam; S xo - static moment of the area of \u200b\u200bthe cut-off part of the section relative to the neutral axis of the beam; b is the section width in the layer under consideration; Ix is the moment of inertia of the section about the neutral axis.

Shear stresses are equal to zero in the extreme fibers of the section and are maximum in the fibers of the neutral layer.

Calculation of beams for bending strength. The strength of the beam will be ensured if the following conditions are met:

(15)

The maximum normal bending stresses occur in sections where the maximum bending moment acts, at the points of the section farthest from the neutral axis

The maximum shear stresses arise in the beam sections where the maximum shear force acts

Shear stresses τmax are usually small compared to σmax and are usually not taken into account in calculations. Shear stress check is performed only for short beams.

Bending movements... The calculation of stiffness is understood as the assessment of the elastic compliance of the beam under the action of applied loads and the selection of such cross-sectional dimensions, at which the displacements will not exceed the limits established by the norms.

Bending stiffness condition

The movement of the center of gravity of the section in the direction perpendicular to the axis of the beam is called deflection. The deflection is indicated by the letter W.

The greatest deflection in the span or on the beam console is called the deflection boom and is denoted by the letter ƒ.

Angle q, by which each section rotates with respect to its original position and is the angle of rotation.

The angle of rotation is considered positive when the section is rotated counterclockwise

The angle of rotation of the section is equal to the value of the derivative of the deflection along the Z coordinate in the same section, that is:

Beam elastic line equation

(16)

There are three methods for solving the differential equation of the elastic line of a beam. These are the direct integration method, the Clebsch method and the initial parameter method.

Direct integration method... Having integrated the equation of the elastic line of the beam for the first time, an expression is obtained for determining the angles of rotation:

By integrating the second time, expressions are found to determine the deflections:

The values \u200b\u200bof the integration constants C and D are determined from the initial conditions on the beam supports

Clebsch method... To draw up equations, the following basic conditions must be met:

• the origin of coordinates, for all sections, must be located at the far left end of the beam;
• integrate the differential equation of the elastic line of the beam without opening the brackets;
• when the external concentrated moment M is included in the equation, it must be multiplied by (Z - a), where a is the coordinate of the section in which the moment is applied;
• in the event of a break in the distributed load, it is extended to the end of the beam, and to restore the actual loading conditions, a "compensating" load of the opposite direction is introduced

Initial parameter method

For rotation angles

(17)

For deflections:

(18)

where θ is the angle of rotation of the section; w - deflection; θo - angle of rotation at the origin; w0 - deflection at the origin; dі - distance from the origin to the i-th support of the beam; ai is the distance from the origin of coordinates to the point of application of the concentrated moment Mi; bi is the distance from the origin of coordinates to the point of application of the concentrated force Fi; сi - distance from the origin of coordinates to the beginning of the distributed load section qi; Ri and Мрi - reaction and reactive moment in the beam supports.

Determination of boom deflections for simple cases

Figure: 31. Examples of loads of beams

Mohr calculation of displacements

If knowledge of the equation of the curved line of the bar is not required, but it is necessary to determine only the linear or angular displacements of a separate section, it is most convenient to use the Mohr method. For beams and flat frames, the Mohr integral has the form:

where δ is the desired displacement (linear or angular); M p, M i - analytical expressions of bending moments, respectively, from a given and unit force; EJ x is the stiffness of the beam section in the bending plane. When determining displacements, two states of the system should be considered: 1 - actual state, with an applied external load; 2 - an auxiliary state in which the beam is released from the external load, and a unit force is applied to the section, the displacement of which is determined, if a linear displacement is determined, or a unit moment if an angular displacement is determined (Fig. 32).

Figure: 32. Determination of displacements:
a - actual state; b, c - auxiliary states

Mohr's formula can be obtained, for example. using the principle of possible movements.

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