For a cantilever beam loaded with a distributed load of intensity kN/m and a concentrated moment of kN m (Fig. 3.12), it is required: construct diagrams of shear forces and bending moments, select a round beam cross section at permissible normal stress kN/cm2 and check the strength of the beam by shear stress at permissible shear stress kN/cm2. Beam dimensions m; m; m.

Calculation scheme for the problem of direct transverse bending

Rice. 3.12

Solution of the problem "straight transverse bending"

Determining support reactions

The horizontal reaction in the embedment is zero, since external loads in the z-axis direction do not act on the beam.

We choose the directions of the remaining reactive forces arising in the embedment: we will direct the vertical reaction, for example, downward, and the moment – ​​clockwise. Their values ​​are determined from the static equations:

When composing these equations, we consider the moment to be positive when rotating counterclockwise, and the projection of the force to be positive if its direction coincides with the positive direction of the y-axis.

From the first equation we find the moment at the seal:

From the second equation - vertical reaction:

Received by us positive values for the moment and vertical reaction in the embedment indicate that we guessed their directions.

In accordance with the nature of the fastening and loading of the beam, we divide its length into two sections. Along the boundaries of each of these sections we will outline four cross sections (see Fig. 3.12), in which we will use the method of sections (ROZU) to calculate the values ​​of shearing forces and bending moments.

Section 1. Let's mentally discard the right side of the beam. Let's replace its action on the remaining left side with a cutting force and a bending moment. For the convenience of calculating their values, 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 under consideration.

Let us recall that the shear force arising in any cross section must balance all external forces (active and reactive) that act on the part of the beam being considered (that is, visible) by us. Therefore, the shearing force must be equal to the algebraic sum of all the forces that we see.

Let us also present the rule of signs for the shearing force: an external force acting on the part of the beam under consideration and tending to “rotate” this part relative to the section in a clockwise direction causes a positive shearing force in the section. Such an external force is included in the algebraic sum for the definition with a plus sign.

In our case, we see only the reaction of the support, which rotates the part of the beam visible to us relative to the first section (relative to the edge of the piece of paper) counterclockwise. That's why

kN.

The bending moment in any section must balance the moment created by the external forces visible to us relative to the section in question. Consequently, it is equal to the algebraic sum of the moments of all forces that act on the part of the beam we are considering, relative to the section under consideration (in other words, relative to the edge of the piece of paper). In this case, the external load, bending the part of the beam under consideration with its convexity downwards, causes a positive bending moment in the section. And the moment created by such a load is included in the algebraic sum for determination with a “plus” sign.

We see two efforts: reaction and closing moment. However, the force's leverage relative to section 1 is zero. That's why

kNm.

We took the “plus” sign because the reactive moment bends the part of the beam visible to us with a convex downward.

Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now, unlike the first section, the force has a shoulder: m. Therefore

kN; kNm.

Section 3. Closing the right side of the beam, we find

kN;

Section 4. Cover the left side of the beam with a sheet. Then

kNm.

kNm.

.

Using the found values, we construct diagrams of shearing forces (Fig. 3.12, b) and bending moments (Fig. 3.12, c).

Under unloaded areas, the diagram of shearing forces goes parallel to the axis of the beam, and under a distributed load q - along an inclined straight line upward. Under the support reaction in the diagram there is a jump down by the value of this reaction, that is, by 40 kN.

In the diagram of bending moments we see a break under the support reaction. The bend angle is directed towards the support reaction. Under a distributed load q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6 on the diagram there is an extremum, since the diagram of the shearing force in this place passes through the zero value.

Determine the required cross-sectional diameter of the beam

The normal stress strength condition has the form:

,

where is the moment of resistance of the beam during bending. For a beam of circular cross-section it is equal to:

.

The largest absolute value of the bending moment occurs in the third section of the beam: kN cm

Then the required beam diameter is determined by the formula

cm.

We accept mm. Then

kN/cm2 kN/cm2.

"Overvoltage" is

,

what is allowed.

We check the strength of the beam by the highest shear stresses

The greatest tangential stresses arising in the cross section of a beam of circular cross-section are calculated by the formula

,

where is the cross-sectional area.

According to the diagram, the largest algebraic value of the shearing force is equal to kN. Then

kN/cm2 kN/cm2,

that is, the strength condition for tangential stresses is also satisfied, and with a large margin.

An example of solving the problem "straight transverse bending" No. 2

Condition of an example problem on straight transverse bending

For a simply supported beam loaded with a distributed load of intensity kN/m, concentrated force kN and concentrated moment kN m (Fig. 3.13), it is necessary to construct diagrams of shear forces and bending moments and select a beam of I-beam cross-section with an allowable normal stress kN/cm2 and permissible tangential stress kN/cm2. Beam span m.

An example of a straight bending problem - calculation diagram

Rice. 3.13

Solution of an example problem on straight bending

Determining support reactions

For a given simply supported beam, it is necessary to find three support reactions: , and . Since only vertical loads perpendicular to its axis act on the beam, the horizontal reaction of the fixed hinged support A is zero: .

The directions of vertical reactions are chosen arbitrarily. Let us direct, for example, both vertical reactions upward. To calculate their values, let’s create two static equations:

Let us recall that the resultant of the linear load , uniformly distributed over a section of length l, is equal to , that is, equal to the area of ​​the diagram of this load and it is applied at the center of gravity of this diagram, that is, in the middle of the length.

;

kN.

Let's check: .

Recall that forces whose direction coincides with the positive direction of the y-axis are projected (projected) onto this axis with a plus sign:

that is true.

We construct diagrams of shearing forces and bending moments

We divide the length of the beam into separate sections. The boundaries of these sections are the points of application of concentrated forces (active and/or reactive), as well as points corresponding to the beginning and end of the distributed load. There are three such sections in our problem. Along the boundaries of these sections, we will outline six cross sections, in which we will calculate the values ​​of shear forces and bending moments (Fig. 3.13, a).

Section 1. Let's mentally discard the right side of the beam. For the convenience of calculating the shearing force and bending moment arising in this section, we will cover the part of the beam we discarded with a piece of paper, aligning the left edge of the sheet of paper with the section itself.

The shearing force in the beam section is equal to the algebraic sum of all external forces (active and reactive) that we see. In this case, we see the reaction of the support and the linear load q distributed over an infinitesimal length. The resultant linear load is zero. That's why

kN.

The plus sign is taken because the force rotates the part of the beam visible to us relative to the first section (the edge of a piece of paper) clockwise.

The bending moment in the beam section is equal to the algebraic sum of the moments of all the forces that we see relative to the section under consideration (that is, relative to the edge of the piece of paper). We see the support reaction and linear load q distributed over an infinitesimal length. However, the force has a leverage of zero. The resultant linear load is also zero. That's why

Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now we see the reaction and load q acting on a section of length . The resultant linear load is equal to . It is attached in the middle of a section of length . That's why

Let us recall that when determining the sign of the bending moment, we mentally free the part of the beam we see from all the actual supporting fastenings and imagine it as if pinched in the section under consideration (that is, we mentally imagine the left edge of the piece of paper as a rigid embedment).

Section 3. Let's close the right side. We get

Section 4. Cover the right side of the beam with a sheet. Then

Now, to check the correctness of the calculations, let’s cover the left side of the beam with a piece of paper. We see the concentrated force P, the reaction of the right support and the linear load q distributed over an infinitesimal length. The resultant linear load is zero. That's why

kNm.

That is, everything is correct.

Section 5. As before, close the left side of the beam. Will have

kN;

kNm.

Section 6. Let's close the left side of the beam again. We get

kN;

Using the found values, we construct diagrams of shearing forces (Fig. 3.13, b) and bending moments (Fig. 3.13, c).

We make sure that under the unloaded area the diagram of shearing forces runs parallel to the axis of the beam, and under a distributed load q - along a straight line sloping downwards. There are three jumps in the diagram: under the reaction - up by 37.5 kN, under the reaction - up by 132.5 kN and under the force P - down by 50 kN.

In the diagram of bending moments we see breaks under the concentrated force P and under the support reactions. The fracture angles are directed towards these forces. Under a distributed load of intensity q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. Under the concentrated moment there is a jump of 60 kN m, that is, by the magnitude of the moment itself. In section 7 on the diagram there is an extremum, since the diagram of the shearing force for this section passes through the zero value (). Let us determine the distance from section 7 to the left support.

The beam is the main element load-bearing structure structures. During construction, it is important to calculate the deflection of the beam. IN real construction This element is affected by wind force, loading and vibration. However, when performing calculations, it is customary to take into account only the transverse load or the applied load, which is equivalent to the transverse one.

Beams in the house

When calculating, the beam is perceived as a rigidly fixed rod, which is installed on two supports. If it is installed on three or more supports, calculating its deflection is more complex, and it is almost impossible to do it yourself. The main load is calculated as the sum of forces that act in the direction of the perpendicular section of the structure. A design diagram is required to determine the maximum deformation, which should not exceed the limit values. This will allow you to determine optimal material required size, cross-section, flexibility and other indicators.

Beams made of strong and durable materials are used for the construction of various structures. Such structures may differ in length, shape and cross-section. The most commonly used are wooden and metal constructions. For the deflection calculation scheme, the material of the element is of great importance. The specifics of calculating the deflection of a beam in this case will depend on the homogeneity and structure of its material.

Wooden

For the construction of private houses, cottages and other individual construction, wooden beams are most often used. Wooden structures, working in bending, can be used for ceilings and floors.

Wooden floors

To calculate the maximum deflection, consider:

1. Material. Different types of wood have different strength, hardness and flexibility.
2. Cross-sectional shape and other geometric characteristics.
3. Various types of load on the material.

The permissible deflection of the beam takes into account the maximum real deflection, as well as possible additional operational loads.

Coniferous wood structures

Steel

Metal beams have a complex or even composite cross-section and are most often made from several types of metal. When calculating such structures, it is necessary to take into account not only their rigidity, but also the strength of the connections.

Steel floors

Metal structures are made by connecting several types of rolled metal, using the following types of connections:

• electric welding;
• rivets;
• bolts, screws and other types of threaded connections.

Steel beams are most often used for multi-story buildings and other types of construction where high structural strength is required. In this case, when using high-quality connections, a uniformly distributed load on the beam is guaranteed.

To calculate the beam for deflection, this video can help:

Beam strength and rigidity

To ensure the strength, durability and safety of the structure, it is necessary to calculate the deflection value of the beams at the design stage of the structure. Therefore, it is extremely important to know the maximum deflection of the beam, the formula of which will help draw a conclusion about the likelihood of using a certain building structure.

Using a calculation scheme of rigidity allows you to determine the maximum changes in the geometry of the part. Calculating a structure using experimental formulas is not always effective. It is recommended to use additional coefficients to add the necessary safety margin. Not leaving an additional margin of safety is one of the main construction mistakes, which leads to the impossibility of using the building or even serious consequences.

There are two main methods for calculating strength and stiffness:

1. Simple. When using this method, a magnification factor is applied.
2. Accurate. This method includes the use of not only safety factors, but also additional calculations of the boundary state.

The last method is the most accurate and reliable, because it helps determine exactly what load the beam can withstand.

Calculation of beams for deflection

Stiffness calculation

To calculate the bending strength of a beam, the formula is used:

M is the maximum moment that occurs in the beam;

W n,min – moment of resistance of the section, which is a tabular value or is determined separately for each type of profile.

R y is the design resistance of steel in bending. Depends on the type of steel.

γ c is the operating condition coefficient, which is a tabular value.

Calculating the stiffness or deflection of a beam is quite simple, so even an inexperienced builder can perform the calculations. However, to accurately determine the maximum deflection, you must perform the following steps:

1. Drawing up a design diagram of the object.
2. Calculation of the dimensions of the beam and its section.
3. Calculation of the maximum load that acts on the beam.
4. Determination of the point of application of maximum load.
5. Additionally, the beam can be tested for strength by maximum bending moment.
6. Calculation of the stiffness value or maximum deflection of a beam.

To create a calculation scheme, you will need the following data:

• beam dimensions, length of consoles and span between them;
• cross-sectional size and shape;
• features of the load on the structure and its exact application;
• material and its properties.

If a two-support beam is being calculated, then one support is considered rigid, and the second is considered hinged.

Calculation of moments of inertia and section resistance

To perform stiffness calculations, you will need the moment of inertia of the section (J) and the moment of resistance (W). To calculate the moment of resistance of a section, it is best to use the formula:

An important characteristic when determining the moment of inertia and resistance of a section is the orientation of the section in the cut plane. As the moment of inertia increases, the stiffness index also increases.

Determination of maximum load and deflection

To accurately determine the deflection of a beam, it is best to use this formula:

q is a uniformly distributed load;

E – elastic modulus, which is a tabular value;

l – length;

I – moment of inertia of the section.

To calculate the maximum load, static and periodic loads must be taken into account. For example, if we're talking about about a two-story building, then on wooden beam there will be a constant load from its weight, equipment, and people.

Features of deflection calculations

Deflection calculations are required for any floors. It is extremely important to accurately calculate this indicator under significant external loads. In this case, it is not necessary to use complex formulas. If you use the appropriate coefficients, the calculations can be reduced to simple schemes:

1. A rod that rests on one rigid and one hinged support and carries a concentrated load.
2. A rod that rests on a rigid and hinged support and is subject to a distributed load.
3. Options for loading a cantilever rod that is rigidly fixed.
4. The effect of a complex load on a structure.

Using this method for calculating deflection allows you to ignore the material. Therefore, the calculations are not affected by the values ​​of its main characteristics.

Example of calculating deflection

To understand the process of calculating the stiffness of a beam and its maximum deflection, you can use a simple calculation example. This calculation is carried out for a beam with the following characteristics:

• material of manufacture – wood;
• density is 600 kg/m3;
• length is 4 m;
• the cross-section of the material is 150*200 mm;
• the mass of the covering elements is 60 kg/m²;
• the maximum load of the structure is 249 kg/m;
• the elasticity of the material is 100,000 kgf/m²;
• J is equal to 10 kg*m².

To calculate the maximum permissible load the weight of the beam, floors and supports is taken into account. It is also recommended to take into account the weight of furniture, appliances, decoration, people and other heavy things, which will also have an impact on the structure. For the calculation you will need the following data:

• weight of one meter of beam;
• weight m2 of floor;
• the distance that is left between the beams;

To simplify the calculation this example, we can take the mass of the floor as 60 kg/m², the load on each floor as 250 kg/m², the load on the partitions as 75 kg/m², and the weight of a meter of beam as 18 kg. With a distance between beams of 60 cm, the coefficient k will be equal to 0.6.

If you plug all these values ​​into the formula, you get:

q = (60 + 250 + 75) * 0.6 + 18 = 249 kg/m.

To calculate the bending moment, use the formula f = (5 / 384) * [(qn * L4) / (E * J)] £ [¦].

Substituting the data into it, we get f = (5 / 384) * [(qn * L4) / (E * J)] = (5 / 384) * [(249 * 44) / (100,000 * 10)] = 0 .13020833 * [(249 * 256) / (100,000 * 10)] = 0.13020833 * (6,3744 / 10,000,000) = 0.13020833 * 0.0000063744 = 0.00083 m = 0.83 cm.

This is precisely the indicator of deflection when a maximum load is applied to the beam. These calculations show that when a maximum load is applied to it, it will bend by 0.83 cm. If this indicator is less than 1, then its use at the specified loads is allowed.

The use of such calculations is in a universal way calculating the rigidity of the structure and the amount of their deflection. It is quite easy to calculate these values ​​yourself. It is enough to know the necessary formulas and also calculate the values. Some data needs to be taken in a table. When performing calculations, it is extremely important to pay attention to units of measurement. If the value in the formula is in meters, then it needs to be converted into this form. Such simple mistakes may render calculations useless. To calculate the rigidity and maximum deflection of a beam, it is enough to know the basic characteristics and dimensions of the material. This data should be plugged into a few simple formulas.

Bending is a type of deformation in which the longitudinal axis of the beam is bent. Straight beams that bend are called beams. Direct bending is a bend in which the external forces acting on the beam lie in one plane (force plane) passing through the longitudinal axis of the beam and the main central axis of inertia of the cross section.

The bend is called pure, if only one bending moment occurs in any cross section of the beam.

Bending, in which a bending moment and a transverse force simultaneously act in the cross section of a beam, is called transverse. The line of intersection of the force plane and the cross-sectional plane is called the force line.

Internal force factors during beam bending.

When flat transverse bending in beam sections, two internal force factors arise: shear force Q and bending moment M. To determine them, the method of sections is used (see lecture 1). The transverse force Q in the beam section is equal to the algebraic sum of the projections onto the section plane of all external forces acting on one side of the section under consideration.

Rule of signs for shear forces Q:

The bending moment M in a beam section is equal to the algebraic sum of the moments relative to the center of gravity of this section of all external forces acting on one side of the section under consideration.

Sign rule for bending moments M:

Zhuravsky's differential dependencies.

Between the intensity q of the distributed load, the expressions for the shear force Q and the bending moment M are established differential dependencies:

Based on these dependencies, the following general patterns of diagrams of transverse forces Q and bending moments M can be identified:

Features of diagrams of internal force factors during bending.

1. In the section of the beam where there is no distributed load, the Q diagram is presented straight line , parallel to the base of the diagram, and diagram M - an inclined straight line (Fig. a).

2. In the section where a concentrated force is applied, Q should be on the diagram leap , equal to the value of this force, and on the diagram M - breaking point (Fig. a).

3. In the section where a concentrated moment is applied, the value of Q does not change, and the diagram M has leap , equal to the value of this moment (Fig. 26, b).

4. In a section of a beam with a distributed load of intensity q, the diagram Q changes according to a linear law, and the diagram M changes according to a parabolic law, and the convexity of the parabola is directed towards the direction of the distributed load (Fig. c, d).

5. If, within a characteristic section, the diagram Q intersects the base of the diagram, then in the section where Q = 0, the bending moment has an extreme value M max or M min (Fig. d).

Normal bending stresses.

Determined by the formula:

The moment of resistance of a section to bending is the quantity:

Dangerous cross section during bending, the cross section of the beam in which the maximum normal stress occurs is called.

Shear stresses during straight bending.

Determined by Zhuravsky's formula for shear stresses at straight bend beams:

where S ots is the static moment of the transverse area of ​​the cut-off layer of longitudinal fibers relative to the neutral line.

Calculations of bending strength.

1. At verification calculation The maximum design stress is determined and compared with the permissible stress:

2. At design calculation the selection of the beam section is made from the condition:

3. When determining the permissible load, the permissible bending moment is determined from the condition:

Bending movements.

Under the influence of bending load, the axis of the beam bends. In this case, tension of the fibers is observed on the convex part and compression on the concave part of the beam. In addition, there is a vertical movement of the centers of gravity of the cross sections and their rotation relative to the neutral axis. To characterize bending deformation, the following concepts are used:

Beam deflection Y- movement of the center of gravity of the cross section of the beam in the direction perpendicular to its axis.

Deflection is considered positive if the center of gravity moves upward. The amount of deflection varies along the length of the beam, i.e. y = y(z)

Section rotation angle- angle θ through which each section rotates relative to its original position. The rotation angle is considered positive when the section is rotated counterclockwise. The magnitude of the rotation angle varies along the length of the beam, being a function of θ = θ (z).

The most common methods for determining displacements is the method Mora And Vereshchagin's rule.

Mohr's method.

The procedure for determining displacements using Mohr's method:

1. An “auxiliary system” is built and loaded with a unit load at the point where the displacement is required to be determined. If linear displacement is determined, then a unit force is applied in its direction; when angular displacements are determined, a unit moment is applied.

2. For each section of the system, expressions for bending moments M f from the applied load and M 1 from the unit load are written down.

3. Over all sections of the system, Mohr’s integrals are calculated and summed, resulting in the desired displacement:

4. If the calculated displacement has a positive sign, this means that its direction coincides with the direction of the unit force. A negative sign indicates that the actual displacement is opposite to the direction of the unit force.

Vereshchagin's rule.

For the case when the diagram of bending moments from a given load has an arbitrary outline, and from a unit load – a rectilinear outline, it is convenient to use the graphic-analytical method, or Vereshchagin’s rule.

where A f is the area of ​​the diagram of the bending moment M f from a given load; y c – ordinate of the diagram from a unit load under the center of gravity of the diagram M f; EI x is the section stiffness of the beam section. Calculations using this formula are made in sections, in each of which the straight-line diagram should be without fractures. The value (A f *y c) is considered positive if both diagrams are located on the same side of the beam, negative if they are located on different sides. A positive result of multiplying diagrams means that the direction of movement coincides with the direction of a unit force (or moment). A complex diagram M f should be divided into simple figures (the so-called “plot stratification” is used), for each of which it is easy to determine the ordinate of the center of gravity. In this case, the area of ​​each figure is multiplied by the ordinate under its center of gravity.

Chapter 1. BENDING OF RIGHT LINEAR BEAMS AND BEAM SYSTEMS

1.1. Basic dependencies of the theory of beam bending

Beams It is customary to call rods that bend under the action of a transverse (normal to the axis of the rod) load. Beams are the most common elements of ship structures. The axis of a beam is the geometric location of the centers of gravity of its cross sections in an undeformed state. A beam is called straight if its axis is a straight line. The geometric location of the centers of gravity of the cross sections of a beam in a bent state is called the elastic line of the beam. The following direction of the coordinate axes is accepted: axis OX aligned with the axis of the beam, and the axis OY And OZ– with the main central axes of inertia of the cross section (Fig. 1.1).

The theory of beam bending is based on the following assumptions.

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

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

3. Beams are considered sufficiently rigid, i.e. their deflections are small compared to the height of the beams, and the angles of rotation of the sections are small compared to unity (Fig. 1.2, V).

4. Stresses and strains are related by a linear relationship, i.e. Hooke's law is valid (Fig. 1.2, G).

Rice. 1.2. Assumptions of beam bending theory

We will consider the bending moments and shearing forces that appear during bending of a beam in its cross-section as a result of the action of a part of the beam that is mentally thrown along the cross-section onto its remaining part.

The moment of all forces acting in a 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 rejected 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 forces acting in the section is called shear force. It is equal to the sum of projections onto the cross-sectional plane of all forces (including support reactions) acting on the rejected part of the beam.

Let us limit 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 bending of the section. Note that for sections of beams that have two axes of symmetry, the center of bending coincides with the center of gravity, and for sections that have one axis of symmetry, it lies on the axis of symmetry, but does not coincide with the center of gravity.

The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying 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) by q(x), external concentrated force – as R, and the external bending moment is as M. Distributed load and concentrated force are positive if the directions of their action coincide with the positive direction of the axis OZ(Fig. 1.3, A,b). The external bending moment is positive if it is directed clockwise (Fig. 1.3, V).

Rice. 1.3. Sign rule for external loads

Let us denote the deflection of a straight beam when it is bent in a plane XOZ through w, and the angle of rotation of the section is through θ. Let us accept the rule of signs for bending elements (Fig. 1.4):

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

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 bends convex upward under their influence (Fig. 1.4, V);

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

Rice. 1.4. Sign rule for bending elements

Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε x, separated by z from the neutral axis, it will be equal

ε x= −z/ρ ,(1.1)

Where ρ – radius of curvature of the beam in the section under consideration.

Rice. 1.5. Beam bending diagram

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

Due to the accepted assumption that the rotation angles are small for sufficiently rigid beams, the valuesmall compared to unity, therefore we can assume that

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

Normal bending stress σ x based 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 F– cross-sectional area of ​​the beam.

From (1.5) we obtain that the static moment of the cross-sectional area of ​​the beam is equal to 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 y will

If we take into account that the moment of inertia of the cross-sectional area relative to the neutral axis OY is equal to , and substitute this value into (1.6), we obtain a dependence that expresses the basic differential equation for beam bending

Moment of internal forces in the section relative to the axis OZ will

Since the axes OY And OZ by condition are the main central axes of the section, then .

It follows that when a load is applied in a plane parallel to the main bending plane, 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 stress when bending beams, they are proportional to the distance from the neutral axis of the beam. Naturally, this follows from the hypothesis of plane sections. In practical calculations, the moment of resistance of the beam section is often used to determine the highest normal stresses

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

In what follows, subscripts y omitted for simplicity.

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

Consider a beam element with length dx (Fig. 1.6). Here it is assumed that the deformations of the element are negligible.

If a moment acts in the left section of the element M and cutting force N, then in its right section the corresponding forces will have increments. Let's consider only linear increments .

Fig.1.6. Forces acting on a beam element

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

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

From (1.11) and (1.12) it follows that

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

Where N 0 and M 0 – shear force and bending moment in the section corresponding tox =x 0 , which is taken as the starting point; ξ,ξ 1 – integration variables.

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

If the beam is statically determinate, the bending moment at any section can be found using (1.14), and the elastic line is determined by integrating the differential equation (1.7) twice. However, statically definable beams are extremely rare in ship hull structures. Most of the beams that make up ship structures form multiple statically indeterminate systems. In these cases, equation (1.7) is inconvenient for determining the elastic line, and it is advisable to move on to a fourth-order equation.

1.2. Differential equation for bending beams

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 obtain:

where the primes indicate differentiation with respect to x.

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

The 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 the following equation (1.18) or 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) 4 times successively (assuming that the left end of the beam corresponds to the sectionx= xa ), we get:

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

N a– shearing force at the beginning of the count, i.e. at x =xa ;

M a– bending moment at the beginning of the reference;

θ a – angle of rotation at the beginning of the count;

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 arrangement of the ends of the beam. The simplest conditions correspond to hinged support on rigid supports or rigid embedding.

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

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

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

A possible situation is associated with sliding embedding or symmetry embedding (Fig. 1.7, G). This leads to the following boundary conditions:

Note that the boundary conditions (1.26) concerning deflections and rotation angles are usually called kinematic, and conditions (1.27) – by force.

Rice. 1.7. Types of boundary conditions

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

Elastic support (Fig. 1.8, A) is 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 we can write:

w =AR,(1.29)

Where A– 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= 1, i.e. A=w R = 1 .

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

To determine the compliance coefficient of an elastic support A it is necessary to load the corresponding structure with a unit force and find the absolute value of the subsidence (deflection) at the point of application of the force. Rigid support – special case elastic support at A= 0.

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

θ = Â M.(1.30)

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

A special case of elastic sealing with  = 0 is hard termination. In ship structures, elastic embeddings are usually beams normal to the one under consideration and lying in the same plane. For example, beams, etc. can be considered elastically embedded on frames.

Rice. 1.8. Elastic support ( A) and elastic seal ( b)

If the ends of the beam are long L are supported on elastic supports (Fig. 1.9), then the reactions of the supports in the end sections are equal to the shearing 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 rotation angles and bending moments, we can write:

The minus sign in the second condition (1.32) is accepted because with a positive moment in the right supporting 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 included in them and their derivatives, and the loads acting on the beam. Linearity is a consequence of the assumptions about the validity of Hooke's law and the smallness of beam deflections.

Rice. 1.9. A beam, both ends of which are elastically supported and elastically embedded ( A);

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

When several loads are applied to a beam, each bending element of the beam (deflection, rotation angle, moment and shear force) is the sum of the bending elements due to the action of each load 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 reveal the static indetermination of beams.

1.3. Initial parameters method

The general integral of the differential equation for beam bending can be used to determine the elastic line of a single-span beam in the case where the beam load is a continuous function of the coordinate throughout the entire span. If the load contains concentrated forces, moments, or a distributed load acts on part of the length of the beam (Fig. 1.10), then expression (1.24) cannot be used directly. In this case, it would be possible to designate elastic lines in sections 1, 2 and 3 through w 1 , w 2 , w 3, write out the integral for each of them 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 pairing conditions in the case under consideration are expressed as follows:

at x=a 1

at x=a 2

at x=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 n, Where n– number of sections along the length of the beam.

Rice. 1.10. Beam with loads applied in certain areas different types

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

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

It is obvious that δ 1 w(x)=w 2 (x)−w 1 (x); δ2 w(x)=w 3 (x)−w 2 (x); etc.

Differential equations to determine 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) with xa = a i . In this case, the parameters Na,Ma,θ a , w a have the meaning of changes (jumps) respectively: in shearing force, bending moment, angle of rotation and deflection arrow when passing through the section x =a i . This technique is called the initial parameters method. It can be shown that for the beam shown in Fig. 1.10, the equation of the elastic line 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 a 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 that make it easy to find deflections, rotation angles and other bending elements.

1.4. Determination of shear stresses during bending of beams

The hypothesis of flat sections adopted in the theory of beam bending leads to the fact that the shear deformation in the beam section is equal to zero, and we are unable to determine shear stresses using Hooke’s law. However, since in general case When shearing forces act in the sections of the beam, corresponding tangential stresses should arise. This contradiction (which is a consequence of the accepted hypothesis of plane sections) can be circumvented by considering equilibrium conditions. We will assume that when a beam composed of thin strips is bent, the tangential stresses in the cross section of each of these strips are uniformly distributed throughout the thickness and directed parallel to the long sides of its contour. This position is practically confirmed by exact solutions of the theory of elasticity. Let's consider a beam of an open thin-walled I-beam. In Fig. Figure 1.11 shows the positive direction of tangential stresses in the flanges and the profile wall during bending in the plane of the beam wall. Let us highlight with a longitudinal section I -I and two cross-sections of an element length dx (Fig. 1.12).

Let us denote the tangential stress in the indicated longitudinal section by τ, and the normal forces in the initial cross section by T. Normal forces in the final section will have increments. Let's consider only linear increments, then .

Rice. 1.12. Longitudinal forces and shear stresses
in the beam flange element

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

Where ; f– area of ​​the profile part cut off by the line I –I; δ – profile thickness at the section.

From (1.36) it follows:

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

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

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

Naturally, the resulting formula for determining shear stresses is also valid for any longitudinal section, for example II –II(see Fig. 1.11), and static moment S ots is calculated for the cut-off part of the beam profile area relative to the neutral axis without taking into account the sign.

Formula (1.38), in the sense of the derivation, determines the tangential stresses in the longitudinal sections of the beam. From the theorem on the pairing of tangential stresses, known from the course on strength 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 tangential stresses onto the axis OZ must be equal to the shear force N in a given section of the beam. Since in the corbels of beams of this type, as shown in Fig. 1.11, tangential 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 tangential stresses along the height of the wall follows the law of change in static moment S ots of the cut-off part of the area relative to the neutral axis (with constant thickness walls δ ).

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

Rice. 1.13. Section of an I-beam

Static moment of the cut-off part of the area for a point located at z from the neutral axis, there will be

As can be seen from dependence (1.39), the static moment varies with z according to the law of quadratic parabola. Highest value S ots , and therefore tangential stresses τ , will be obtained at the neutral axis, where z = 0:

The highest shear stress in the beam wall at the neutral axis

Since the moment of inertia of the section of the beam in question is equal to

then the maximum shear stress will be

Attitude N/ω is nothing more than the average shear stress in the wall, calculated assuming a uniform stress distribution. Taking for example ω = 2 F 1 , according to formula (1.41) we get

Thus, the beam under consideration has the greatest tangential stress in the wall at the neutral axis by only 12.5% exceeds the average value of these voltages. It should be noted that for most beam profiles used in ship hulls, the maximum shear stresses exceed the average ones by 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 XOZ will be accompanied by twisting.

Bending of the beam is not accompanied by twisting if the load acts in a plane parallel to XOZ passing 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 section of the beam relative to it is equal to zero.

Rice. 1.14. Tangential stresses during channel beam bending (point A – center of bend)

Indicating the distance of the center of the bend A from the axis of the beam wall through e, we write down the condition for the moment of tangential forces to be equal to zero relative to the point A:

Where Q 2 – tangential force in the wall, equal to the shearing force, i.e. Q 2 =N;

Q 1 =Q 3 – force in the belt, determined based on (1.38) by the dependence

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

As has been shown, for beams with chords, the change in tangential 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 cross-sectional plane of the beam and the tangent to the elastic line changes by the amount γ Wed A simplified diagram of the shear deformation of a beam element is shown in Fig. 1.15.

Rice. 1.15. Beam element shear deformation diagram

Having indicated the arrow of deflection caused by shear through w sdv, we can write:

Taking into account the rule of signs for cutting force N and find the angle of rotation

Because the ,

Integrating (1.47), we obtain

Constant a, included in (1.48), determines the displacement of the beam as solid and can be taken equal to any value, since when determining the total arrow of deflection from bending w bending and shear w SDV

the sum of the integration constants will appear w 0 +a, determined from the boundary conditions. Here w 0 – deflection from bending at the origin.

Let us put in the future a=0. Then the final expression for the elastic line caused by the shear will take the form

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

Rice. 1.16. Bend ( A) and shear ( b) components of the elastic line of the beam

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

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

Let us consider a beam freely supported on rigid supports, in the left section of which moment is valid M. The shearing force in this case will be constant and equal

For the right reference section, we respectively obtain

.(1.52)

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

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

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

The bending deflection can be found from beam bending tables. The shear deflection is determined by formula (1.50), taking into account the fact that .

Rice. 1.18. Diagram of a simply 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 depending on the specific task under consideration, the design of supports and the load of the beam.

Let's analyze the dependence of the coefficient k from various factors.

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

The moment of inertia of a beam section can always be represented in the form

,(1.58)

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

Note that as the size of the beam flanges increases, the coefficient α will increase.

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

Thus, the value of the coefficient k significantly depends on the ratio of the span of the beam to its height, on the shape of the section (through the coefficient α), the arrangement of supports and the load of the beam (through the coefficient β). The relatively longer the beam ( h/L small), the smaller the influence of shear deformation. For beams rolled profile related h/L less than 1/10÷1/8, the shift correction can practically not be taken into account.

However, for beams with wide flanges, such as, for example, keels, stringers and floras in the composition of bottom floors, the influence of shear and at the indicated h/L may turn out to be significant.

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

Calculate bending beam There are several options:
1. Calculation of the maximum load that it will withstand
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle selection of beam section on two supports loaded with a uniformly distributed load or concentrated force.
To begin with, you will need to find the point (section) at which there will be a maximum moment. This depends on whether the beam is supported or embedded. Below are diagrams of bending moments for the most common schemes.

After finding the bending moment, we must find the moment of resistance Wx of this section using the formula given in the table:

Further, when dividing the maximum bending moment by the moment of resistance in a given section, we get maximum stress in beam and we must compare this stress with the stress that our beam of a given material can generally withstand.

For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, A for fragile(cast iron) – tensile strength. We can find the yield strength and tensile strength from the tables below.

Let's look at a couple of examples:
1. [i]You want to check whether an I-beam No. 10 (steel St3sp5) 2 meters long, rigidly embedded in the wall, will support you if you hang on it. Let your mass be 90 kg.
First, we need to select a design scheme.

This diagram shows that the maximum moment will be at the seal, and since our I-beam has equal section along the entire length, then the maximum voltage will be in the termination. Let's find it:

P = m * g = 90 * 10 = 900 N = 0.9 kN

M = P * l = 0.9 kN * 2 m = 1.8 kN * m

Using the I-beam assortment table, we find the moment of resistance of I-beam No. 10.

It will be equal to 39.7 cm3. Let's convert it to cubic meters and get 0.0000397 m3.
Next, using the formula, we find the maximum stresses that arise in the beam.

b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa

After we have found the maximum stress that occurs in the beam, we can compare it with the maximum permissible voltage equal to the limit the fluidity of steel St3sp5 is 245 MPa.

45.34 MPa is correct, which means this I-beam will withstand a mass of 90 kg.

2. [i] Since we have quite a large supply, we will solve the second problem, in which we will find the maximum possible mass that the same I-beam No. 10, 2 meters long, will support.
If we want to find the maximum mass, then we must equate the values ​​of the yield strength and the stress that will arise in the beam (b = 245 MPa = 245,000 kN*m2).