Determination of the deflection of a cantilever beam. Solving typical problems on strength of materials. Initial parameters method

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 a force line.

Internal force factors during beam bending.

During plane transverse bending, two internal force factors arise in the beam sections: transverse 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.

Sign rule 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.

Differential relationships have been established between the intensity q of the distributed load, the expressions for the transverse force Q and the bending moment M:

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 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.

The beam is the main element of the supporting structure of the structure. During construction, it is important to calculate the deflection of the beam. In real construction this element 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 be higher limit values. This will allow you to determine optimal material required size, cross-section, flexibility and other indicators.

For the construction of various structures, beams made of durable and durable materials. Such structures may differ in length, shape and cross-section. The most commonly used are wooden and metal structures. For the design deflection scheme great value has an element material. 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 more 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:

Material. Various breeds Trees have different levels of strength, hardness and flexibility.
Cross-sectional shape and other geometric characteristics.
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 stiffness 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:

Simple. When using this method, a magnification factor is applied.
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 for precise definition maximum deflection, the following steps must be performed:

Drawing up a design diagram of the object.
Calculation of the dimensions of the beam and its section.
Calculation maximum load, which acts on the beam.
Determination of the point of application of maximum load.
Additionally, the beam can be tested for strength by maximum bending moment.
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 are talking about a two-story building, then 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. Complex formulas in this case it is not necessary to use. If you use the appropriate coefficients, the calculations can be reduced to simple schemes:

A rod that rests on one rigid and one hinged support and carries a concentrated load.
A rod that rests on a rigid and hinged support and is subject to a distributed load.
Options for loading a cantilever rod that is rigidly fixed.
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. Simple errors like these can 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.

The design process of modern buildings and structures is regulated a huge amount various building codes and regulations. In most cases, standards require certain characteristics to be ensured, for example, deformation or deflection of floor slab beams under static or dynamic load. For example, SNiP No. 2.09.03-85 determines for supports and overpasses the deflection of the beam is no more than 1/150 of the span length. For attic floors this figure is already 1/200, and for interfloor beams it is even less - 1/250. Therefore, one of the mandatory design stages is to perform a beam deflection calculation.

Ways to perform deflection calculations and tests

The reason why SNiPs establish such draconian restrictions is simple and obvious. The smaller the deformation, the greater the margin of strength and flexibility of the structure. For a deflection of less than 0.5%, the load-bearing element, beam or slab still retains elastic properties, which guarantees normal redistribution of forces and maintaining the integrity of the entire structure. As the deflection increases, the building frame bends, resists, but stands; when the permissible value is exceeded, the bonds break, and the structure loses its rigidity and load-bearing capacity like an avalanche.

Use an online software calculator, in which standard conditions are “hardwired”, and nothing more;
Use ready-made reference data for various types and types of beams, for various support load patterns. It is only necessary to correctly identify the type and size of the beam and determine the desired deflection;
Calculate the permissible deflection with your hands and your head; most designers do this, while controlling architectural and construction inspectors prefer the second method of calculation.

For your information! To really understand why it is so important to know the magnitude of the deviation from the initial position, it is worth understanding that measuring the amount of deflection is the only accessible and reliable way to determine the condition of the beam in practice.

By measuring how much the ceiling beam has sagged, you can determine with 99% certainty whether the structure is in disrepair or not.

Method of performing deflection calculations

Before starting the calculation, you will need to remember some dependencies from the theory of strength of materials and draw up a calculation diagram. Depending on how correctly the diagram is executed and the loading conditions are taken into account, the accuracy and correctness of the calculation will depend.

We use the simplest model of a loaded beam shown in the diagram. The simplest analogy of a beam can be a wooden ruler, photo.

In our case, the beam:

Has rectangular section S=b*h, the length of the supporting part is L;
The ruler is loaded with a force Q passing through the center of gravity of the bent plane, as a result of which the ends rotate through a small angle θ, with a deflection relative to the initial horizontal position , equal to f ;
The ends of the beam rest hingedly and freely on fixed supports; accordingly, there is no horizontal component of the reaction, and the ends of the ruler can move in any direction.

To determine the deformation of a body under load, use the formula of the elastic modulus, which is determined by the ratio E = R/Δ, where E is a reference value, R is force, Δ is the amount of deformation of the body.

Calculate moments of inertia and forces

For our case, the dependence will look like this: Δ = Q/(S E) . For a load q distributed along the beam, the formula will look like this: Δ = q h/(S E) .

What follows is the most important point. The above Young diagram shows the deflection of a beam or the deformation of a ruler as if it were crushed under a powerful press. In our case, the beam is bent, which means that at the ends of the ruler, relative to the center of gravity, two bending moments are applied with different sign. The loading diagram for such a beam is given below.

To transform Young's dependence for the bending moment, it is necessary to multiply both sides of the equality by the shoulder L. We obtain Δ*L = Q·L/(b·h·E) .

If we imagine that one of the supports is rigidly fixed, and an equivalent balancing moment of forces M max = q*L*2/8 will be applied to the second, respectively, the magnitude of the beam deformation will be expressed by the dependence Δх = M x/((h/3) b (h/2) E). The quantity b h 2 /6 is called the moment of inertia and is designated W. The result is Δx = M x / (W E) the fundamental formula for calculating a beam for bending W = M / E through the moment of inertia and bending moment.

To accurately calculate the deflection, you will need to know the bending moment and moment of inertia. The value of the first can be calculated, but the specific formula for calculating a beam for deflection will depend on the conditions of contact with the supports on which the beam is located and the method of loading, respectively, for a distributed or concentrated load. The bending moment from a distributed load is calculated using the formula Mmax = q*L 2 /8. The given formulas are valid only for a distributed load. For the case when the pressure on the beam is concentrated at a certain point and often does not coincide with the axis of symmetry, the formula for calculating the deflection must be derived using integral calculus.

The moment of inertia can be thought of as the equivalent of a beam's resistance to bending load. The magnitude of the moment of inertia for a simple rectangular beam can be calculated using the simple formula W=b*h 3 /12, where b and h are the cross-sectional dimensions of the beam.

The formula shows that the same ruler or board of rectangular cross-section can have a completely different moment of inertia and deflection if placed on supports traditional way or put it on edge. No wonder almost all elements rafter system roofs are made not from 100x150 timber, but from 50x150 boards.

Real sections building structures can have a variety of profiles, from square, circle to complex I-beam or channel shapes. At the same time, determining the moment of inertia and the amount of deflection manually, “on paper”, for such cases becomes a non-trivial task for a non-professional builder.

Formulas for practical use

In practice, most often the opposite task is faced - to determine the safety margin of floors or walls for a specific case based on a known deflection value. In the construction business, it is very difficult to assess the safety factor using other, non-destructive methods. Often, based on the magnitude of the deflection, it is necessary to perform a calculation, estimate the safety factor of the building and general condition load-bearing structures. Moreover, based on the measurements taken, it is determined whether the deformation is acceptable, according to the calculation, or whether the building is in emergency condition.

Advice! In the matter of calculation limit state beams in terms of deflection, the requirements of SNiP provide an invaluable service. By setting the deflection limit in a relative value, for example, 1/250, building codes greatly facilitate the determination emergency condition beams or slabs.

For example, if you intend to buy a finished building that has stood for quite a long time on problematic soil, it would be useful to check the condition of the ceiling based on the existing deflection. Knowing everything permissible norm deflection and the length of the beam, one can assess without any calculation how critical the condition of the structure is.

Construction inspection during deflection assessment and assessment bearing capacity overlap goes a more complicated way:

Initially, the geometry of the slab or beam is measured and the deflection value is recorded;
Based on the measured parameters, the assortment of the beam is determined, then the formula for the moment of inertia is selected using the reference book;
The moment of force is determined by the deflection and moment of inertia, after which, knowing the material, you can calculate the actual stresses in a metal, concrete or wooden beam.

The question is why is it so difficult if the deflection can be obtained using the formula for calculation for a simple beam on hinged supports f=5/24*R*L 2 /(E*h) under a distributed force. It is enough to know the span length L, profile height, design resistance R and elastic modulus E for a specific floor material.

Advice! Use in your calculations the existing departmental collections of various design organizations, which contain all the necessary formulas for determining and calculating the maximum loaded state in a condensed form.

Conclusion

Most developers and designers of serious buildings act in a similar way. The program is good, it helps to very quickly calculate the deflection and basic loading parameters of the floor, but it is also important to provide the customer with documentary evidence of the results obtained in the form of specific sequential calculations on paper.

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 to: construct diagrams of shear forces and bending moments, select a beam of circular cross-section with an allowable normal stress kN/cm2 and check the strength of the beam according to tangential stresses with permissible tangential 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 downward, 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.

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 the beam is only acted upon vertical loads, perpendicular to its axis, the horizontal reaction of the fixed hinged support A is equal to 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 visible to us from all actual supporting fastenings and imagine it as if pinched in the section under consideration (that is, we mentally imagine the left edge of a piece of paper as a rigid embedment).

Section 3. 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. We 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.

Straight bend. Plane transverse bending Constructing diagrams of internal force factors for beams Constructing diagrams of Q and M using equations Constructing diagrams of Q and M using characteristic sections (points) Strength calculations for direct bending of beams Principal stresses during bending. A complete check of the strength of beams. The concept of the center of bending. Determination of displacements in beams during bending. Concepts of beam deformation and conditions of their rigidity Differential equation curved axis of a beam Method of direct integration Examples of determining displacements in beams by the method of direct integration Physical meaning of integration constants Method of initial parameters (universal equation of the curved axis of a beam). Examples of determining displacements in a beam using the initial parameters method. Determining displacements using Mohr's method. Rule A.K. Vereshchagin. Calculation of the Mohr integral according to the rule of A.K. Vereshchagina Examples of determining displacements using the Mohr integral Bibliography Direct bending. Flat transverse bend. 1.1. Constructing diagrams of internal force factors for beams Direct bending is a type of deformation in which two internal force factors arise in the cross sections of the rod: a bending moment and a transverse force. In a particular case, the shear force can be zero, then the bending is called pure. In flat transverse bending, all forces are located in one of the main planes of inertia of the rod and perpendicular to its longitudinal axis, and moments are located in the same plane (Fig. 1.1, a, b). Rice. 1.1 The transverse force in an arbitrary cross section of a beam is numerically equal to the algebraic sum of the projections onto the normal to the beam axis of all external forces acting on one side of the section under consideration. Shear force in section m-n beams(Fig. 1.2, a) is considered positive if the resultant of external forces to the left of the section is directed upward, and to the right - downward, and negative - in the opposite case (Fig. 1.2, b). Rice. 1.2 When calculating the shear force in a given section, external forces lying to the left of the section are taken with a plus sign if they are directed upwards, and with a minus sign if they are directed downwards. For the right side of the beam - vice versa. 5 The bending moment in an arbitrary cross section of a beam is numerically equal to the algebraic sum of the moments about the central axis z of the section of all external forces acting on one side of the section under consideration. The bending moment in the m-n section of the beam (Fig. 1.3, a) is considered positive if the resultant moment of external forces to the left of the section is directed clockwise, and to the right - counterclockwise, and negative - in the opposite case (Fig. 1.3, b). Rice. 1.3 When calculating the bending moment in a given section, the moments of external forces lying to the left of the section are considered positive if they are directed clockwise. For the right side of the beam - vice versa. It is convenient to determine the sign of the bending moment by the nature of the deformation of the beam. The bending moment is considered positive if, in the section under consideration, the cut-off part of the beam bends convexly downward, i.e., the lower fibers are stretched. In the opposite case, the bending moment in the section is negative. There are differential relationships between the bending moment M, shear force Q and load intensity q. 1. The first derivative of the shear force along the abscissa of the section is equal to the intensity of the distributed load, i.e. . (1.1) 2. The first derivative of the bending moment along the abscissa of the section is equal to the transverse force, i.e. (1.2) 3. The second derivative with respect to the abscissa of the section is equal to the intensity of the distributed load, i.e. . (1.3) We consider the distributed load directed upward to be positive. From differential dependencies between M, Q, q a number of important conclusions follow: 1. If in the section of the beam: a) the transverse force is positive, then the bending moment increases; b) the shear force is negative, then the bending moment decreases; c) the shear force is zero, then the bending moment has a constant value (pure bending); 6 d) the transverse force passes through zero, changing sign from plus to minus, max M M, in the opposite case M Mmin. 2. If there is no distributed load on the beam section, then the transverse force is constant, and the bending moment changes according to a linear law. 3. If there is a uniformly distributed load on a section of the beam, then the transverse force changes according to a linear law, and the bending moment - according to the law of a square parabola, convexly facing in the direction of the load (in the case of constructing diagram M from the side of stretched fibers). 4. In the section under a concentrated force, diagram Q has a jump (by the magnitude of the force), diagram M has a kink in the direction of the force. 5. In the section where a concentrated moment is applied, the diagram M has a jump equal to the value of this moment. This is not reflected in the Q diagram. When beams are loaded with complex loading, diagrams of transverse forces Q and bending moments M are plotted. Diagram Q(M) is a graph showing the law of change in transverse force (bending moment) along the length of the beam. Based on the analysis of diagrams M and Q, dangerous sections of the beam are determined. Positive ordinates of the Q diagram are laid upward, and negative ordinates are laid down from the base line drawn parallel to the longitudinal axis of the beam. Positive ordinates of the M diagram are laid down, and negative ordinates are laid upward, i.e., the M diagram is constructed from the side of the stretched fibers. The construction of Q and M diagrams for beams should begin with determining the support reactions. For a beam with one clamped end and the other free end, the construction of diagrams Q and M can be started from the free end, without determining the reactions in the embedment. 1.2. The construction of Q and M diagrams using the Beam equations is divided into sections within which the functions for the bending moment and shear force remain constant (do not have discontinuities). The boundaries of the sections are the points of application of concentrated forces, pairs of forces and places of change in the intensity of the distributed load. At each section, an arbitrary section is taken at a distance x from the origin of coordinates, and for this section equations for Q and M are drawn up. Using these equations, diagrams of Q and M are constructed. Example 1.1 Construct diagrams of transverse forces Q and bending moments M for a given beam (Fig. 1.4,a). Solution: 1. Determination of support reactions. We compose equilibrium equations: from which we obtain The reactions of the supports are determined correctly. The beam has four sections Fig. 1.4 loads: CA, AD, DB, BE. 2. Construction of diagram Q. Section CA. In section CA 1, we draw an arbitrary section 1-1 at a distance x1 from the left end of the beam. We define Q as the algebraic sum of all external forces acting to the left of section 1-1: The minus sign is taken because the force acting to the left of the section is directed downward. The expression for Q does not depend on the variable x1. Diagram Q in this section will be depicted as a straight line parallel to the abscissa axis. Section AD. On the section we draw an arbitrary section 2-2 at a distance x2 from the left end of the beam. We define Q2 as the algebraic sum of all external forces acting to the left of section 2-2: 8 The value of Q is constant in the section (does not depend on the variable x2). The Q plot on the section is a straight line parallel to the abscissa axis. Plot DB. On the site we draw an arbitrary section 3-3 at a distance x3 from the right end of the beam. We define Q3 as the algebraic sum of all external forces acting to the right of section 3-3: The resulting expression is the equation of an inclined straight line. Section BE. On the site we draw a section 4-4 at a distance x4 from the right end of the beam. We define Q as the algebraic sum of all external forces acting to the right of section 4-4: 4 Here the plus sign is taken because the resultant load to the right of section 4-4 is directed downward. Based on the obtained values, we construct Q diagrams (Fig. 1.4, b). 3. Construction of diagram M. Section m1. We define the bending moment in section 1-1 as the algebraic sum of the moments of forces acting to the left of section 1-1. – equation of a straight line. Section A 3 We determine the bending moment in section 2-2 as the algebraic sum of the moments of forces acting to the left of section 2-2. – equation of a straight line. Section DB 4 We determine the bending moment in section 3-3 as the algebraic sum of the moments of forces acting to the right of section 3-3. – equation of a quadratic parabola. 9 We find three values at the ends of the section and at the point with coordinate xk, where Section BE 1 We determine the bending moment in section 4-4 as the algebraic sum of the moments of forces acting to the right of section 4-4. – equation of a quadratic parabola, we find three values of M4: Using the obtained values, we construct a diagram of M (Fig. 1.4, c). In sections CA and AD, the Q diagram is limited by straight lines parallel to the abscissa axis, and in sections DB and BE - by inclined straight lines. In sections C, A and B on the Q diagram there are jumps in the magnitude of the corresponding forces, which serves as a check for the correctness of the Q plot. In areas where Q  0, moments increase from left to right. In areas where Q  0, the moments decrease. Under the concentrated forces there are kinks in the direction of the action of the forces. Under the concentrated moment there is a jump in the magnitude of the moment. This indicates the correctness of the construction of diagram M. Example 1.2 Construct diagrams Q and M for a beam on two supports loaded with a distributed load, the intensity of which varies according to a linear law (Fig. 1.5, a). Solution Determination of support reactions. The resultant of the distributed load is equal to the area of the triangle, which is a diagram of the load and is applied at the center of gravity of this triangle. We compile the sums of the moments of all forces relative to points A and B: Constructing diagram Q. Let's draw an arbitrary section at a distance x from the left support. The ordinate of the load diagram corresponding to the section is determined from the similarity of triangles. The resultant of that part of the load that is located to the left of the section. The transverse force in the section is equal. The transverse force changes according to the law of a square parabola. Equating the equation of the transverse force to zero, we find the abscissa of the section in which the diagram Q passes through zero: The Q plot is shown in Fig. 1.5, b. The bending moment in an arbitrary section is equal to The bending moment varies according to the law of a cubic parabola: The bending moment has a maximum value in the section where 0, i.e. at Diagram M is shown in Fig. 1.5, c. 1.3. Constructing diagrams of Q and M from characteristic sections (points) Using differential dependencies between M, Q, q and the conclusions arising from them, it is advisable to construct diagrams of Q and M from characteristic sections (without drawing up equations). Using this method, the values of Q and M are calculated in characteristic sections. The characteristic sections are the boundary sections of sections, as well as sections where a given internal force factor has an extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of the differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct diagrams Q and M for the beam shown in Fig. 1.6, a. Rice. 1.6. Solution: We start constructing the Q and M diagrams from the free end of the beam, while the reactions in the embedment do not need to be determined. The beam has three loading sections: AB, BC, CD. There is no distributed load in sections AB and BC. Shear forces are constant. The Q diagram is limited to straight lines parallel to the x-axis. Bending moments vary linearly. Diagram M is limited by straight lines inclined to the abscissa axis. There is a uniformly distributed load on section CD. Shear forces change according to a linear law, and bending moments - according to the law of a square parabola with convexity in the direction of the action of the distributed load. At the boundary of sections AB and BC, the transverse force changes abruptly. At the boundary of sections BC and CD, the bending moment changes abruptly. 1. Construction of diagram Q. We calculate the values of transverse forces Q in the boundary sections of sections: Based on the calculation results, we construct diagram Q for the beam (Fig. 1, b). From diagram Q it follows that the transverse force on section CD is equal to zero in the section located at a distance qa a q from the beginning of this section. In this section, the bending moment has a maximum value. 2. Constructing diagram M. We calculate the values of bending moments in the boundary sections of sections: At the maximum moment in the section Based on the calculation results, we construct diagram M (Fig. 5.6, c). Example 1.4 Using a given diagram of bending moments (Fig. 1.7, a) for a beam (Fig. 1.7, b), determine the acting loads and construct diagram Q. The circle indicates the vertex of a square parabola. Solution: Let's determine the loads acting on the beam. Section AC is loaded with a uniformly distributed load, since the diagram M in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting clockwise, since in diagram M we have a jump upward by the magnitude of the moment. In the NE section, the beam is not loaded, since the diagram M in this section is limited by an inclined straight line. The reaction of support B is determined from the condition that the bending moment in section C equal to zero, i.e. To determine the intensity of the distributed load, we will compose an expression for the bending moment in section A as the sum of the moments of forces on the right and equate it to zero. Now we will determine the reaction of support A. To do this, we will compose an expression for the bending moments in the section as the sum of the moments of forces on the left Calculation diagram of the beam with load is shown in Fig. 1.7, c. Starting from the left end of the beam, we calculate the values of transverse forces in the boundary sections of the sections: Diagram Q is shown in Fig. 1.7, d. The considered problem can be solved by drawing up functional dependencies for M, Q in each section. Let's choose the origin of coordinates at the left end of the beam. In the AC section, the diagram M is expressed by a square parabola, the equation of which has the form Constants a, b, c are found from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the equation of the parabola, we obtain: The expression for the bending moment will be Differentiating the function M1 , we obtain the dependence for the transverse force. After differentiating the function Q, we obtain an expression for the intensity of the distributed load. In the section NE, the expression for the bending moment is presented in the form of a linear function. To determine the constants a and b, we use the conditions that this straight line passes through two points, the coordinates of which are known. We obtain two equations: ,b from which we have a 20. The equation for the bending moment in the section NE will be After double differentiation of M2, we will find. Using the found values of M and Q, we construct diagrams of bending moments and shear forces for the beam. In addition to the distributed load, concentrated forces are applied to the beam in three sections, where there are jumps on diagram Q and concentrated moments in the section where there is a shock on diagram M. Example 1.5 For a beam (Fig. 1.8, a), determine the rational position of the hinge C, at which the largest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Construct diagrams of Q and M. Solution Determination of support reactions. Even though total number support links is equal to four, the beam is statically determinate. The bending moment in the hinge C is equal to zero, which allows us to create an additional equation: the sum of the moments about the hinge of all external forces acting on one side of this hinge is equal to zero. Let us compile the sum of the moments of all forces to the right of the hinge C. The diagram Q for the beam is limited by an inclined straight line, since q = const. We determine the values of transverse forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation from which the diagram M for the beam is limited by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are written respectively as follows: From the condition of equality of moments we obtain quadratic equation relative to the desired parameter x: Real value x2x 1.029 m. We determine the numerical values of transverse forces and bending moments in characteristic sections of the beam. Figure 1.8, b shows the Q diagram, and in Fig. 1.8, c – diagram M. The problem considered could be solved by dividing the hinged beam into its constituent elements, as shown in Fig. 1.8, d. At the beginning, the reactions of the supports VC and VB are determined. Diagrams of Q and M are constructed for the suspended beam SV from the action of the load applied to it. Then they move to the main beam AC, loading it with an additional force VC, which is the pressure force of the beam CB on the beam AC. After that, diagrams Q and M are built for beam AC. 1.4. Strength calculations for direct bending of beams Strength calculations based on normal and shear stresses. When a beam bends directly in its cross sections, normal and tangential stresses arise (Fig. 1.9). 18 Fig. 1.9 Normal stresses are associated with bending moment, tangential stresses are associated with shear force. With direct pure bend shear stresses are zero. Normal stresses at an arbitrary point in the cross section of a beam are determined by the formula (1.4) where M is the bending moment in a given section; Iz – moment of inertia of the section relative to the neutral axis z; y is the distance from the point where the normal voltage is determined to the neutral z axis. Normal stresses along the height of the section change according to a linear law and reach their greatest value at points farthest from the neutral axis. If the section is symmetrical about the neutral axis (Fig. 1.11), then Fig. 1.11 the greatest tensile and compressive stresses are the same and are determined by the formula,  is the axial moment of resistance of the section during bending. For a rectangular section with width b and height h: (1.7) For a circular section with diameter d: (1.8) For an annular section   – the inner and outer diameters of the ring, respectively. For beams made of plastic materials, the most rational are symmetrical 20 section shapes (I-beam, box-shaped, annular). For beams made of brittle materials that do not equally resist tension and compression, sections that are asymmetrical with respect to the neutral z-axis (T-beam, U-shaped, asymmetrical I-beam) are rational. For beams of constant cross-section made of plastic materials with symmetrical cross-sectional shapes, the strength condition is written as follows: (1.10) where Mmax is the maximum bending moment in modulus; – permissible stress for the material. For beams of constant cross-section made of plastic materials with asymmetrical cross-sectional shapes, the strength condition is written in the following form: (1.11) For beams made of brittle materials with sections that are asymmetrical with respect to the neutral axis, if the diagram M is unambiguous (Fig. 1.12), you need to write two strength conditions - distances from the neutral axis to the most distant points of the stretched and compressed zones, respectively dangerous section; P – permissible stresses for tension and compression, respectively. Fig.1.12. 21 If the diagram of bending moments has sections of different signs (Fig. 1.13), then in addition to checking section 1-1, where Mmax acts, it is necessary to calculate the highest tensile stresses for section 2-2 (with the highest moment of the opposite sign). Rice. 1.13 Along with the main calculation using normal stresses, in a number of cases it is necessary to check the strength of the beam using tangential stresses. Tangential stresses in beams are calculated using the formula of D.I. Zhuravsky (1.13) where Q is the transverse force in the cross section of the beam under consideration; Szотс – static moment relative to the neutral axis of the area of the section part located on one side of a straight line drawn through a given point and parallel to the z axis; b – section width at the level of the point under consideration; Iz is the moment of inertia of the entire section relative to the neutral z axis. In many cases, maximum shear stresses occur at the level of the neutral layer of the beam (rectangle, I-beam, circle). In such cases, the strength condition for tangential stresses is written in the form, (1.14) where Qmax is the largest transverse force in magnitude; – permissible shear stress for the material. For a rectangular section of a beam, the strength condition has the form (1.15) A is the cross-sectional area of the beam. For a circular section, the strength condition is presented in the form (1.16) For an I-section, the strength condition is written as follows: (1.17) where Szo,тmсax is the static moment of the half-section relative to the neutral axis; d – wall thickness of the I-beam. Typically, the cross-sectional dimensions of a beam are determined from the strength condition under normal stresses. Checking the strength of beams by tangential stresses is carried out in mandatory for short beams and beams of any length, if there are concentrated forces of large magnitude near the supports, as well as for wooden, riveted and welded beams. Example 1.6 Check the strength of a box-section beam (Fig. 1.14) using normal and shear stresses, if MPa. Construct diagrams in the dangerous section of the beam. Rice. 1.14 Solution 23 1. Constructing diagrams of Q and M using characteristic sections. Considering the left side of the beam, we obtain The diagram of transverse forces is shown in Fig. 1.14, c. The diagram of bending moments is shown in Fig. 5.14, g. 2. Geometric characteristics of cross section 3. The highest normal stresses in section C, where Mmax acts (modulo): MPa. The maximum normal stresses in the beam are almost equal to the permissible ones. 4. The highest tangential stresses in section C (or A), where max Q acts (modulo): Here is the static moment of the half-section area relative to the neutral axis; b2 cm – section width at the level of the neutral axis. 5. Tangential stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 cm3 is the static moment of the area of the section located above the line passing through point K1; b2 cm – wall thickness at the level of point K1. Diagrams  and  for section C of the beam are shown in Fig. 1.15. Example 1.7 For the beam shown in Fig. 1.16, a, required: 1. Construct diagrams of transverse forces and bending moments along characteristic sections (points). 2. Determine the dimensions of the cross section in the form of a circle, rectangle and I-beam from the condition of strength under normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of beam sections according to tangential stress. Given: Solution: 1. Determine the reactions of the beam supports. Check: 2. Construction of diagrams Q and M. Values of transverse forces in characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, load intensity q = const. Consequently, in these areas the Q diagram is limited to straight lines inclined to the axis. In section DB, the intensity of the distributed load is q = 0, therefore, in this section, the diagram Q is limited to a straight line parallel to the x axis. The Q diagram for the beam is shown in Fig. 1.16, b. Values of bending moments in characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section in which Q = 0: Maximum moment in the second section Diagram M for the beam is shown in Fig. 1.16, c. 2. We create a strength condition based on normal stresses from which we determine the required axial moment of resistance of the section from the expression determined by the required diameter d of a beam of circular cross-section Area of circular cross-section For a beam of rectangular cross-section Required height of the section Area of rectangular cross-section Determine the required number I-beam. Using the tables of GOST 8239-89 we find the nearest higher value axial moment of resistance 597 cm3, which corresponds to I-beam No. 33 with characteristics: A z 9840 cm4. Tolerance check: (underload by 1% of the permissible 5%) the nearest I-beam No. 30 (W 2 cm3) leads to significant overload (more than 5%). We finally accept I-beam No. 33. We compare the areas of the round and rectangular sections with the smallest area A of the I-beam: Of the three sections considered, the most economical is the I-beam section. 3. We calculate the highest normal stresses in the dangerous section 27 of the I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-beam section The diagram of normal stresses in the dangerous section of the beam is shown in Fig. 1.17, b. 5. Determine the highest shear stresses for the selected sections of the beam. a) rectangular section of the beam: b) round section beams: c) I-beam section: Tangential stresses in the wall near the I-beam flange in dangerous section A (right) (at point 2): The diagram of tangential stresses in dangerous sections of the I-beam is shown in Fig. 1.17, c. The maximum tangential stresses in the beam do not exceed the permissible stresses Example 1.8 Determine the permissible load on the beam (Fig. 1.18, a), if 60 MPa, the cross-sectional dimensions are given (Fig. 1.19, a). Construct a diagram of normal stresses in a dangerous section of a beam at an allowable load. Figure 1.18 1. Determination of reactions of beam supports. Due to the symmetry of the system 2. Construction of diagrams Q and M from characteristic sections. Transverse forces in characteristic sections of a beam: Diagram Q for a beam is shown in Fig. 5.18, b. Bending moments in characteristic sections of the beam For the second half of the beam, the ordinates M are along the axes of symmetry. Diagram M for the beam is shown in Fig. 1.18, b. 3. Geometric characteristics of the section (Fig. 1.19). We divide the figure into two simple elements: I-beam - 1 and rectangle - 2. Fig. 1.19 According to the assortment for I-beam No. 20, we have For a rectangle: Static moment of the sectional area relative to the z1 axis Distance from the z1 axis to the center of gravity of the section Moment of inertia of the section relative to the main central axis z of the entire section according to the formulas for the transition to parallel axes 4. Strength condition for normal stresses for dangerous point “a” (Fig. 1.19) in dangerous section I (Fig. 1.18): After substituting numerical data 5. With an allowable load in a dangerous section, the normal stresses at points “a” and “b” will be equal: Diagram of normal stresses for dangerous section 1-1 is shown in Fig. 1.19, b.