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The

strength of

materials

Prof. V.SOLODOV

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Theory of elasticity

Theory

plasticity Mechanics liquids,

gases, plasma

Applied hydromechanics

Theoretical hydromechanics

Hydraulics open channels and

hydraulic works

Acoustics Rheology

Meteorology

Water supply, sewerage

MECHANICS

Continuum mechanics

Climatology Science about the atmosphere, and hydrosphere

Theoretical mechanics Applied

mechanics

Theory of oscillations Technical

mechanics Resistance of materials

Theory machines, mechanisms

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THE SCIENCE OF STRENGTH OF MATERIALS. STUDIED OBJECTS

Resistance of materials is the science of engineering methods of calculating the strength, stiffness and stability of elements of machines and structures.

Strength is understood as the ability of a structure, its parts and parts to withstand a certain load without collapsing.

Rigidity refers to the ability of a structure and its elements to withstand external loads in relation to deformation (changes in shape and size). At given loads, deformations should not exceed a certain value, established in

accordance with the requirements for the structure.

Stability is the ability of a structure or its elements to maintain a certain initial form of elastic equilibrium.

In order for the structure as a whole to meet the requirements of strength, rigidity and stability, it is necessary to give its elements the most rational form.

Knowing the properties of the materials from which they will be made, it is necessary to determine the appropriate dimensions depending on the size and the nature of the acting forces.

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THE SCIENCE OF STRENGTH OF MATERIALS.

STUDIED OBJECTS

Resistance of materials is the science of engineering methods of calculating the strength, stiffness and stability of elements of machines and structures.

During the operation of machines and structures, their elements (rods, beams, plates, bolts, rivets, etc.) to one degree or another participate in the operation of the structure and are subjected to various forces - loads. To ensure normal operation, the structure must meet the necessary conditions for strength, rigidity and stability. Let us consider some of them.

Strength is understood as the ability of a structure, its parts and parts to withstand a certain load without collapsing.

Rigidity refers to the ability of a structure and its elements to withstand external loads in relation to deformation (changes in shape and size). At given loads, deformations should not exceed a certain value, established in accordance with the requirements for the structure.

Stability is the ability of a structure or its elements to maintain a certain initial form of elastic equilibrium.

In order for the structure as a whole to meet the requirements of strength, rigidity and stability, and, therefore, to be reliable in operation, it is necessary to give its elements the most rational form. Knowing the properties of the materials from which they will be made, it is necessary to determine the appropriate dimensions depending on the size and the nature of the acting forces.

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At first view, it may seem that for reliable resistance of structural elements to external load, it is sufficient to increase their dimensions. Indeed, sometimes this leads to the desired results.

However, in cases where the dead weight makes up a significant part of the loads acting on the structure, an increase in the size of its elements, and therefore in weight, will not lead to an increase in strength. An increase in the size of the moving parts of mechanisms and machines leads to an increase in inertial forces, increases the load, and this is undesirable, since it can also lead to destruction.

The resistance of materials solves the indicated strength problems, based on both theoretical and experimental data, which are equally important in this science. In the theoretical part, this science is based on theoretical mechanics and mathematics, and in the experimental part - on physics and materials science.

Resistance of materials is an extremely important general engineering science, necessary for the formation of engineers of any specialty. Without fundamental knowledge in this area, it is impossible to create structures such as various kinds of machines and

mechanisms, civil and industrial structures, bridges, power lines and antennas, hangars, ships, airplanes and helicopters, turbomachines, electrical machines, nuclear power plants, rocket and jet technology, etc.

The increase in size, not caused by the requirements of the reliability of the structure, leads to unnecessary consumption of materials and an increase in its cost. Machines and structures must be built strong and reliable in operation, but at the same time lightweight and cheap.

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Thus, the strength of materials is the most general science about the strength of machines and structures. However, it does not exhaust all questions of the mechanics of deformable bodies.

Other related disciplines are also involved in these issues: structural mechanics of rod systems, the theory of elasticity and the theory of plasticity. A strict boundary cannot be drawn

between these disciplines. The main role in solving strength problems belongs to the resistance of materials.

With all the variety of types of structural elements encountered in structures and machines, they can be reduced to a relatively small number of basic forms.

Bodies with these basic forms are the objects of strength, stiffness and stability calculations.

These include rods, shells, plates, and massive bodies.

A rod or a bar is a body in which one size (length) significantly exceeds two other (transverse) dimensions (Fig. 1, a).

In machines and structures, there are rods both straight-linear (Fig. 1, a) and curvilinear (Fig.

1, b), both prismatic (Fig. 1, a), and variable cross-sections (Fig. 1,c). Examples of straight bars are shafts, axles, beams. Lifting hooks, chain links, and others are examples of curved rods.

Fig.2 Fig.1

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The rods, in which the wall thickness is much less than the overall dimensions of the cross section, are called thin-walled (Fig. 1, d). They are widely used in building structures,

shipbuilding and especially in aircraft construction.

The shell is a body bounded by curved surfaces located at a close distance from each other.

The surface that divides the thickness of the shell into equal parts is called the middle surface. According to the shape of the middle surface, shells are distinguished cylindrical (Fig. 2, a), conical (Fig. 2, b), spherical (Fig. 2, c), etc. The shells include non-flat walls of thin-walled tanks, boilers, domes of buildings, fuselage skins, wings and other parts of aircraft, submarine hulls, etc. If the middle surface is a plane, then the design object is

called a plate (Fig. 2, d). There are plates of round (Fig. 2, e), rectangular (Fig. 2, d) and other outlines. Plates can include flat bottoms and covers of tanks, overlapping of engineering structures, disks of turbomachines, etc.

Bodies that have all three sizes of the same order of magnitude are called massive bodies.

These include the foundations of structures, retaining walls, etc.

In the resistance of materials, problems are usually solved by simple mathematical methods using simplifying hypotheses and using experimental data; In this case, the solutions are brought to calculation formulas suitable for application in engineering practice.

The emergence of the science of the strength of materials is associated with the name of Galileo Galilei (1564-1642), who conducted experiments on the study of strength.

In 1678, the English scientist Robert Hooke (1635-1703) established the law of deformation of elastic bodies, according to which deformation an elastic body is proportional to the force acting on it. This law is fundamental in the theory of strength of materials.

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TYPES OF BAR DEFORMATIONS. THE CONCEPT OF THE DEFORMED STATE OF THE MATERIAL Real bodies can deform, that is, change their shape and size. Deformations of bodies occur as a result of loading them by external forces or changes in temperature. When the body is

deformed, its points, as well as mentally drawn lines or sections, move in a plane or in space relative to their initial position.

When a rigid body is loaded, internal forces arise in it, countering external forces and striving to return the body particles to the position that they occupied before deformation.

Deformations are elastic, that is, they disappear after the cessation of the action of the forces that caused them, and plastic, or residual, which do not disappear.

With an increase in external forces, internal forces also increase, but up to a certain limit, depending on the properties of the material. A moment comes when the body is no longer able to resist a further increase in external forces. Then it collapses. In most cases, certain limitations are set for the magnitude of deformations of structural elements.

The main object considered in the strength of materials is a bar with a straight axis.

In the strength of materials, the following main types bar deformations are studied : tension and compression, shear, torsion and bending. More complex deformations resulting from a combination of several basic ones are also considered.

Tension or compression occurs, for example, when oppositely directed forces are applied to the rod along its axis (Fig. 3). In this case, the sections move along the axis of the rod, which lengthens when stretched, and shortens when compressed. The change in  l to the initial length of the l of rod is called absolute tensile elongation or absolute shortening in

compression.

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The ratio of the absolute elongation (shortening) to the initial length of bar is called the average relative elongation by ɛ=  l/ l .

Tensile or compressive structural elements work in:

truss rods, pumps, tie screws and other details.

A shear or sliding occurs when external forces displace two parallel plane sections of the bar relative to one another at a constant distance between them (Fig. 4).

The amount of displacement  s is called the absolute displacement. The ratio of the absolute shift to the distance a between the shifting planes (the tangent of the γ angle) is called the relative shift. Due to the smallness of the angle γ during elastic deformations, its tangent is taken to be equal to the skew angle of the element in question. Therefore, the relative shift γ=  s/a.

Relative shear is an angular deformation that characterizes the skew of a member. Shear or cut work, for example, with rivets and bolts fastening elements that external forces tend to move one relative to the other.

Torsion occurs when external forces act on the rod, forming a moment relative to the rod axis (Fig. 5). Torsion deformation is accompanied by rotation of the cross-sections of the rod

relative to each other around its axis.

Fig.3

Fig.4

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 The angle of rotation of one section of the rod relative to another, located at a distance I, is called the angle of twist at length I. The ratio of the angle of twist φ to the length I is called the relative angle of twist: θ= φ/ I

 Shafts, spindles of lathes and boring machines and other parts work in torsion.

 Bending deformation (Fig. 6) consists in bending the axis of a straight bar or in a change in the curvature of a curved bar. The outgoing movement of any point of the bar axis is

expressed by a vector, the beginning of which is aligned with the initial position of the point, and the end - with the position of the same point in the deformed bar. In a straight rods, displacements of points directed perpendicular to the initial position of the axis are called deflections and are denoted by the letter w.

 Bending also rotates the bar sections around the axes lying in the section planes. The angles of rotation of the sections relative to their initial positions are designated by the letter θ. For example, axles of railway cars, leaf springs, gear teeth, wheel spokes, floor beams, levers and many other parts work for bending.

Fig.5 Fig. 6

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 To determine the deformation at any point A (Fig. 7) let us draw in an undeformed body a segment of a straight line AB, outgoing from this point in an arbitrary direction and having a length s. After deformation, points A and B will move and occupy positions A1 and B1, respectively, and the distance

s

between them will change by s. The ratio s / s = ɛ is called the average relative linear deformation of the segment AB. Approaching point B to point A, i.e., decreasing the length of the segment s, in the limit we obtain.

The ɛAB value is the relative linear deformation at point A in the direction AB. If it is known that the distance between points A and B increases, then ɛAB is called the relative lengthening, with a

decrease in this distance - the relative shortening.

At the same point A, the relative linear deformations in different directions can be different.

 As a result of the simultaneous action on the body of forces, that cause various types of the indicated basic deformations, a more complex deformation arises. So, often the elements of machines and structures are subjected to the action of forces that

simultaneously cause bending and torsion, bending and stretching or compression, etc.

 The described deformations of the bar give an idea of ​​the change in its shape and

dimensions in general, but they do not say anything about the degree and nature of the deformed state of the material. Research shows that the deformed state of a body is generally uneven and varies from point to point.

0 AB

s

lim s

s

  

Fig. 7

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To fully characterize the deformation at a point, angular deformations are also introduced. If, before the deformation of the body, from point A (Fig. 8) draw two segments AB and AC, forming a right angle, then after moving the points due to deformation of the body, the segments will occupy positions A1B1 and A1C1 and the angle between them will change by the value (BAC —  B1A1C1). Approaching points B and C to point A, in the limit we obtain a change in the initially right angle by the value  1 1 1

0 BAC

s

lim BAC B A C

   

This change in the right angle, expressed in radians, is called the relative angular deformation at point A in the plane where the segments AB and AC lie down. At the same point A, the relative angular deformations in different planes are different. Usually, relative angular deformations are determined in three mutually

perpendicular coordinate planes. Then they are denoted, respectively, through the γxy, γxz, γyz.

The deformed state at a point of the body is completely determined by six components of deformation - three relative linear deformations ɛx, ɛy, ɛz and three relative angular deformations of the γxy, γxz, γyz.

Usually, directions parallel to the axes of the selected rectangular coordinate system are taken as the main ones. Then the relative linear deformations at a point are denoted by ɛX, ɛY, ɛZ, respectively.

Fig. 8

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BASIC HYPOTHESES OF THE SCIENCE OF MATERIAL STRENGTH

 To construct a theory of resistance of materials, some hypotheses are accepted regarding the structure and properties of materials, as well as the nature of deformations. These hypotheses are as follows:

 1. Hypothesis of material continuity. It is assumed that the material completely fills the body shape. The atomic theory of the discrete structure of matter is not taken into account.

 2. Hypothesis of homogeneity and isotropy. The material is assumed to be homogeneous and isotropic, that is, in any volume and in any direction, the properties of the material are considered the same. Although the crystals of which metals are composed are

anisotropic, their chaotic arrangement makes it possible to consider the macro-volumes of metals isotropic.

 In some cases, the assumption of isotropy is unacceptable. For example, anisotropic materials include wood, the properties of which are significantly different along and across the fibers, reinforced materials, etc.

 3. The hypothesis of small deformations. It is assumed that the deformations are small compared to the size of the body. This allows, in most cases, to neglect changes in the location of external forces relative to individual parts of the body and to compose statics equations for an undeformed body. In some cases, this principle has to be deviated from.

Such deviations are discussed separately.

 Small relative deformations are considered as infinitesimal values.

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 4. Hypothesis about the ideal elasticity of the material. All bodies are assumed to be absolutely elastic. Deviations from ideal elasticity, which are always observed when real bodies are loaded, are insignificant and are neglected up to certain deformation limits.

 Most of the problems of resistance of materials are solved in the assumption of a linearly deformable body, that is, one for which Hooke's law is valid, expressing the direct

proportionality between deformations and loads.

 Having accepted the hypotheses about the smallness of deformations and about the linear relationship between deformations and forces, it is possible to apply the principle of superposition (the principle of independence and addition of the action of forces) in solving most problems of material resistance. For example, the efforts in any structural element caused by various factors (several forces, temperature effects) are equal to the sum of the efforts caused by each of these factors, and do not depend on the order of their application. The same is true for deformations.

 The hypotheses listed above, as well as some others, which will be discussed later, allow solving a wide range of problems in calculating strength, stiffness and stability. The

calculation results are in good agreement with the practice data.

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GEOMETRIC CHARACTERISTICS OF FLAT SECTIONS

• The main object studied in the course on resistance of materials is the rod.

• The resistance of a bar to various types of deformation often depends not only on its

material and dimensions, but also on the shape of the axis, the shape of the cross sections and their location. Therefore, in this chapter, abstracting from the physical properties of the object under study, we will consider the main geometric characteristics of its cross- sections, which determine the resistance to various types of deformations. These include cross-sectional areas, static moments and moments of inertia.

• Consider an arbitrary figure (cross-section of the bar) associated with the coordinate axes Oz and Oy (Fig. 9). Select the area

element dF with coordinates z, y. By analogy with the expression for the moment of force about any axis, you can compose an expression for the moment of the area, which is called the static moment. So, the product of the area element dF by the distance y from the Oz axis dSz = ydF called the elementary static

moment of the area element about the axis OZ Fig.9

• Similarly, dSy= zdF is the static moment of the area element relative to the Oy axis.

Summing up such products over the entire area of the figure, we obtain respectively static moments relative to the Z and Y axes:z y

F F

S ydF ; S zdF ;

STATIC MOMENTS OF THE AREA.

CENTER OF GRAVITY OF THE AREA

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Static moments are measured in cubic length units .

Let Zc, Yc must be the coordinates of the center of gravity of the figure. Continuing the analogy with the moments of forces, based on the theorem on the moment of the

resultant, we can write the following expressions: Sz=Fyc; Sy=Fzc; so Zc=Sy/F; Yc=Sz/F From these formulas it follows that the static moments of the area relative to the central axes (axes passing through the center of gravity) are equal to zero.

To calculate the static moments of a complex figure, it is divided into simple parts, for each of which the area F and the coordinates of the center of gravity Zc, and Yc, are known. The static moment of the area of the entire figure relative to a given axis is defined as the sum of the static moments of each part (additive property)

MOMENTS OF INERTIA OF PLANE FIGURES

The axial moment of inertia of the area of a figure is called the integral of the products of elementary areas by the squares of their distances from the axis under consideration.

So, the moments of inertia of an arbitrary figure (Fig. 13) relative to the Z and Y axes

2 2

z y

F F

J y dF ; J z dF

• The polar moment of inertia of the area of a figure relative to a given point (pole O) is called the integral of the products of elementary areas by the squares of their distances from the

pole: p 2 2 2 2 p

2 2

y z

F F

J   dF ;  z y ; J y z dF J J Fig. 13 2.21

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The centrifugal moment of inertia is the integral of the products of the areas of elementary areas by their distance from the Z and Y coordinate axes:

(2.9)

Depending on the position of the axes, the centrifugal moment of inertia can be positive or negative, as well as zero. Indeed, the centrifugal moment of inertia of the area of ​​the figure shown in Fig. 14, a, is positive with respect to the selected system of axes, since the Z

coordinates are positive for all elements. When the axes are rotated around the origin of coordinates by 90 ° (Fig. 14, b), the sign of the centrifugal moment of inertia of the figure is reversed, since in this position the Z coordinates of all elements are positive and the Y

coordinates are negative.

By turning the axes, you can find their position at which the centrifugal moment of inertia is zero. These are the main axes of inertia. Two mutually perpendicular axes, of which at least one is the axis of symmetry of the figure, will always be its main axes of inertia. The main axes passing through the center of gravity of the section are the main central axes.

Moments of inertia are measured in units of length to the fourth power.

zy F

J zydF

Fig.14

b c

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Let us calculate the moments of inertia of the rectangle about the central axes Z, Y, parallel to its sides. To determine the moment of inertia about the Z axis, we select an elementary area in the form of a narrow rectangle parallel to the Z axis. Element width is b, height - dy.

Therefore, dF = bdy 2 0 5 2 3 3

0 5 12 12

. h

z y

F . h

bh b h

J y dF b y dy ; J ;

When calculating the moments of inertia of complex sections, they can be divided into

separate simple parts, the moments of inertia of which are known. From the main property of the integral of the sum it follows that the moment of inertia of a complex figure is equal to the sum of the moments of inertia of its constituent parts.

Let it be required to determine the moment of inertia of a complex figure about the Z-axis (Fig. 22,next slide):

Let's break the figure into simple components I, II and III, as shown in the figure. When calculating the integral, we will successively sum up the products y2dF, covering the areas F1, F2, F3 of simple figures. Then

Obviously, each of the integrals on the right-hand side represents the moment of inertia of the corresponding simple figure.

If there is a hole in the section, it is usually convenient to consider it as part of the figure with negative area. For example, the section shown in Fig. 23, can be divided into two simple parts - a rectangle b X h and a hole of radius r in negative area. Then

2 z

F

J y dF

I I I III

z z z z

J J J J

3 4

12 4

I II

z z

bh r

Jz J J

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MOMENTS OF INERTIA ABOUT PARALLEL AXES

Let the moments of inertia of the figure relative to the central axes Z, Y be known. It is required to determine the moments of inertia about the axes parallel to the central ones (Fig. 24) (Guigens theorem):

The coordinates of any point in the new Z1O1Y1 system can be expressed through the coordinates in the old axes as follows:

z1 = z + b ; y1 = y + a.

Substitute these values into formulas (2.21) and integrate term-wise:. ... Receive (2.25-2.26)

 2

2 2 2 2 2

1 1 2 1

z z y y

F F F F F

J y dF y a dF y dF a dF a ydF J a F ; J J b F

1

zy zy

J J abF

Fig. 24 Fig. 23

Fig. 22

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Therefore: 1) the moment of inertia of the figure relative to any axis is equal to the

moment of inertia about the central axis parallel to the given one, plus the product of the area of ​​the figure by the square of the distance between these axes;

2) the centrifugal moment of inertia relative to any system of rectangular axes is equal to the centrifugal moment relative to the system of central axes parallel to the data, plus the product of the figure's area by the coordinates of its center of gravity in the new axes.

Note that the coordinates a, b included in formula (2.26) should be substituted taking into account their sign.

Formulas (2.25) show that of all the moments of inertia relative to a series of parallel axes, the central moments of inertia will be the smallest.

Let's calculate the moment of inertia of the I-section relative to the central axis Z (Fig. 25).

We divide the section, consisting of two identical shelves b ×and a wall h1 × t into these three simple parts. Then

The moment of inertia of the shelf about the Z axis based on the formula (2.25)

I I I III

z z z z

J J J J

2 3 2

1

1 2 12 2

I III I

z z z

b b h

J J J   F   b

DEPENDENCIES BETWEEN THE MOMENTS OF INERTIA WHEN ROTATING THE COORDINATE AXES

Let the moments of inertia of an arbitrary figure (Fig. 27) are known relative to the coordinate axes Z, Y. Let's rotate the Z and Y axes by an angle a counterclockwise, considering the angle of rotation of the axes in this direction to be positive. Found moments of inertia of the section relative to the rotated axes.

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The coordinates of an arbitrary elementary area in the new Y-axes are expressed in terms of the Z coordinates, for the old system of axes as follows:

1 1

 

 

z z cos y sin y y cos z sin

Substitute these values into expressions (2.31) and integrate term by term:

Taking in view(2.30), finally we can find

 

2 2

1

2 2

1

1 1

2 2

2 1 2

2

z z y zy

y z y zy

z y zy y z

J J cos J sin J sin J J sin J cos J sin

J J cos J J sin

   

   

 

Adding term-by-term formulas (2.34), we find

Thus, when the rectangular axes are rotated, the sum of the moments of inertia does not change and is equal to the polar moment of inertia relative to the origin.

When turning the system of axes at an angle α = 90 °

1 1

z y z y p

J J J J J

1 1 1 1 

z y y z z y zy

J J ; J J ; J J ;

DETERMINATION OF THE DIRECTION OF THE MAIN AXES. MAJOR MOMENTS OF INERTIA Of greatest practical importance are the main central axes, the centrifugal moment of inertia relative to which is zero. We will denote such axes by letters u, v. Consequently, JUV = 0

2 2

1 1 1 1

z y

F F

J y dF ; J z dF ; (2.31)

Fig. 27 Fig. 25

(2.34) (2.30)

(2.30)

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To determine the position of the main central axes of an asymmetric figure, turn an arbitrary initial system of central axes Z, Y (Fig. 28) by some angle α0, at which the centrifugal moment of inertia becomes equal to zero:

JUV = 0=Jzycos2α0 -0.5{Jy-Jz} sin2α0=0 Therefore tan2α0=2Jzy/(Jy-Jz )

The two values of the angle α0 obtained from last formula differ from each other by 90 ° and give the position of the principal axes. It is easy to see that the smaller of these angles does not exceed π/ 4 in absolute value. In what follows, we will only use a smaller angle.

The main axis drawn at this angle (positive or negative) will be denoted by the letter U.

Recall that negative angles α0 are plotted clockwise from the Z-axis.

It is important to note that the main moments of inertia have the property of extremeness.

It is easy to verify this by differentiating the expression for the moment of inertia about an arbitrary axis [see (2.34)].

Considering, that the sum of the moments of inertia relative to two mutually perpendicular axes is a constant value, it can be concluded that the moment of inertia relative to one of the main axes has a maximum value, and relative to the other - minimum.

Note, that the planes drawn through the axis of the bar and the main axes of inertia of its cross-section, are called the main planes.

Fig. 28

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External forces are the forces of interaction between the considered structural element and the bodies associated with it.

If external forces are the result of direct, contact interaction of a given body with other bodies, then they are called surface forces. Surface forces can be continuously

distributed over the entire surface of the body or part of it. For example: steam pressure in the boiler, wind and snow loads, gas pressure in the engine cylinder. The amount of load per unit area is called the load intensity. It is usually denoted by p and is measured in pascals (Pa) or multiples of it (kPa, MPa, GPa). Often the load distributed over the surface (Fig. 36, a) leads to the main plane (Fig. 36, b), as a result of which a load distributed along the line or linear load is obtained. The intensity of such a load (N / m, kN / m, MN / m) is the amount of load per unit length of the line. The intensity can be variable along this length. The nature of the load change is usually shown in the form of a diagram (graph) q(x).

• EXTERNAL AND INTERNAL FORCES. SECTION METHOD. INTERNAL FORCES DIAGRAMS

• Classification of external forces

Fig. 36, Fig. 37,

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In addition, there are loads that can be represented as a concentrated moment (pair). The moments (N • m, kN • m, or MN • m) will usually be depicted in one of two ways shown in Fig. 38, a, b. Sometimes it is convenient to represent the moment as a vector perpendicular to the plane of the action of the pair. We will agree to always consider the moment vector to be right-handed. To distinguish it from the force vector, the line of the moment vector is made wavy (Fig. 38, d) or two arrows are placed (Fig. 38, c).

There are loads that are not the result of contact between two bodies, for example: own weight, inertial forces of a moving body, etc. These forces are applied at each point of the volume occupied by the body, and therefore are called volume or mass forces. The own weight of parts or parts of machines and structures is usually much less than other loads acting on them. Therefore, unless there is a special reservation, we will not take our own weight into account in the entire further presentation.

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Loads distinguish between static and dynamic.

 A load is considered instantaneously applied if it increases from zero to its final value within a very short period of time (fractions of a second). This is the load when igniting a combustible mixture in the cylinder of an internal combustion engine or when

starting off a train.

 For a shock load, it is characteristic that at the moment of its application, the body causing the load has a certain kinetic energy. Such a load is obtained, for example,

when driving piles with a pile driver, in the details of a mechanical forging hammer, etc.

 Many parts of machines (connecting rods, shafts, axles of railway cars, etc.) are

subjects to loads that are continuously and periodically changing over time. Such loads are called re-variable. They are usually associated with cyclically repeating movements of the part. This is a reciprocating movement of the piston rod, vibrations of structural elements, etc.

 Between neighboring particles of a body (crystals, molecules, atoms) there are always certain forces of interaction, otherwise - internal forces. These forces in all cases strive to preserve it as a single whole, counteract any attempt to change the mutual

arrangement of the particles, that is, to deform the body.

d c

b

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• Consequently, the magnitude of the internal forces acting between any two particles in a loaded and unloaded body will be different.

• In the resistance of materials, they do not consider and do not take into account the internal forces acting in the body, which is in its natural (unloaded) state, but only those additional values ​​of internal forces that appear as a result of loading the body are

studied and calculated. Therefore, in what follows, speaking about internal forces, we will mean precisely these additional interaction forces arising as a result of loading.

Internal forces are often called efforts.

 To identify and then calculate the internal forces in the resistance of materials, the method of sections is widely used.

 Consider an arbitrary body loaded with a self-balanced system of forces. In the place of interest to us, mentally dissect it with a certain plane into two parts - A and B (Fig.

40, a). In this case, the section itself will now have two sides: one belonging to part A of the body (left), and the second, belonging to part B (right). At each point of both sides of the section, interaction forces will act (Fig. 40, b). Based on the introduced hypothesis about the continuity of the material, it should be assumed that the

internal forces act at all points of the section and, therefore, represent a distributed load. Depending on the shape of the body and the nature of external loads, the intensity of internal forces at different points can be different.

 External forces, on the contrary, always tend to cause deformation of the body, change the mutual arrangement of particles.

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 It should be emphasized that the internal forces acting over the section belonging to part A of the body, in accordance with Newton's third law, are equal in magnitude and opposite in direction to the internal forces acting over the section belonging to part B of the body (Fig. 40, b).

In other words, the internal forces acting on different parts are reciprocal. Like any system of forces, they can be reduced to one point (usually to the center of gravity of the section), as a result of which on each side of the section we obtain the principal vector and the principal moment of internal forces in the section.

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The cross section method is used to determine the internal forces.

Internal forces obey the 3rd law.

On each side of the section we get 6 force factors: three forces and three moments

The force N gives a longitudinal deformation (tension or compression) - shift of the sides of the section in the direction of the axes.

Mrot gives torsion of the rod, My and Mz - bending of the rod in the main planes.

For the efforts adopted names N - longitudinal force;

Qx, Qy - transverse or cutting forces Mz - torque

Mx , My- bending moments.

INTERNAL FORCES.

SECTION METHOD

N - is equal to the algebraic sum of the projections onto the axis of the rod;

Mz - is equal to the algebraic sum of the moments relative to the axis of the rod.

Fig.40

Fig. 40

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LONGITUDINAL FORCE DIAGRAMS Longitudinal force is positive if it causes

tension, negative when compressed.

To build the plot, we cut the rod into sections between the points of application of

concentrated forces and on sections of the load distributed according to different laws.

For plotting we select the origin at the leftmost point of the rod, the OX axis along the axis of the rod.

Example: (Fig.42) 1st section (0<x<a) left N(x)= P1 = 2kN

right N(x) = P2-P3 = (5-3) kN = 2kN 2nd plot (s)

left N(x) = P1-P2 = (2-5) kN = -3kN right N(x) = -P3 = (-3) kN

if concentrated forces act on the rod, then the lines of the plot are parallel to the axis, the plot consists of rectangles and has jumps in places of external forces

Fig. 42

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TORQUE DIAGRAMS Torsional deformation is common in shafts, for example, when applying moments perpendicular to the axis of the shaft

BEAMS AND THEIR SUPPORTS

a) articulated movable support; Fig. 50a b) pivotally motionless support; Fig. 50b c) pinching. Fig. 50c

The beam in Fig. 51a is simple or single-span or double-supported, distance between

supports - span

Support reactions are found from equilibrium equations;

1) the sum of the projections of all the forces on the beam axis is 0. Where is the reaction HA;

2) The sum of the moments of all external forces with respect to A is 0. We find RB 3) The sum of the moments of all external forces relative to B is 0. We find RA

Fig. 47

In the design scheme, the beam is replaced by the axis. We bring all the loads to the axis of the beam. To calculate the support, we schematize in the form of three main types.

С

The beam (Fig. 52a) is called continuous and С statically indeterminable, since it has 5

unknown support reactions: 3 in support A and one each in supports B and C

18. CALCULATION OF REACTIONS

Fig. 51 Fig. 50

Fig. 52

b)

b)

b)

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BEAMS AND THEIR SUPPORTS

a) articulated movable support; b) pivotally motionless support; c) pinching.

The beam in Fig. 51a is simple or single-span or double-supported, distance between supports - span

To calculate the support, we schematize in the form of three main types.

С)

С)

The beam (Fig. 52a) is called continuous and statically indeterminable, since it has 5 unknown support reactions: 3 in support A and one each in supports B and C

Fig. 51 Fig. 50

Fig. 52

b) b)

b) C)

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TRANSVERSE SHEAR FORCES AND BENDING MOMENTS IN BEAM SECTIONS With a flat bend, the entire load is located in the main

plane of the XY rod, therefore, does not give projections on the axis and moments relative to the X, Y axes.

Sign rules for re-cutting force and bending moment in beam.

1) the force in the cross section is positive, if it’s vector tends to rotate part of the cut clockwise;

2) The moment M in the cross section is positive if it causes compression in the upper fibers and is directed in accordance with fig. 54a;

Negative directions of force and moment are given in Fig. 54b

Compressed zone

Stretched zone

M>0, Q>0 M<0, Q<0

Stretched zone

Compressed zone

Fig.54, a Fig.54, b

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TRANSVERSE SHEAR FORCES AND BENDING MOMENTS IN BEAM SECTIONS With a flat bend, the entire load is located in the main

plane of the XY rod, therefore, does not give projections on the axis and moments relative to the X, Y axes.

1) If an external force tends to rotate the beam relative to the section clockwise, then in the expression for it gives a positive term in this section. For example, the reaction in RA (Fig. 55a) tends to rotate the beam relative to C clockwise, and the

forces P and RB - against it. Transversal force in section C:

QC=RA-P or QC=-RB

2) If the external load creates a moment relative to the section, causing compression of the upper fibers of the beam, then it gives a positive term.

fig. 55

С) b)

Sign for the console. For example, in fig.56 the lower fibers are compressed and M<0.

M>0

M<0

fig. 56 b fig. 56 a

Compressed zone Stretched zone

Stretched zone Compressed

zone

Compressed zone Stretched zone

Stretched zone Compressed

zone

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PLOTTING Q AND M

Evenly distributed intensity load q(x)= const on the console (Fig. 58). The transverse force and bending moment in an arbitrary section K will be calculated as a result of the action of the distributed load located to the left of the section:

Q(x)= -qAK=-q x; M(x)=-qAK*LR= -0.5 qAK2= -0.5 qx2. We also calculate Q and M in an arbitrary section X on the right, only the force P acts. Therefore, Q(x)=P, M(x)=-P*KB=-P(l-x)

Fig. 58 Fig. 57 We also calculate M in an arbitrary section X on the right,

only the force P acts.

The concentrated force at the free end of the console (Fig.

57). The beam consists of one section. The origin of

coordinates is selected at the extreme left point A. The axis OX is directed to the right.

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PLOTTING Q AND M

Evenly distributed intensity load q(x)=const on the the entire length of the span doubly-supported beam. Resultant is

equal ql and applied in the middle of beam (Fig. 59).

ΣMB=RAl-0.5ql=0; ΣMA=Rbl-0.5ql=0; →RA=RA=0.5ql;

Transversal force and bending moment:

are the result of acting force left-wise of section K.

Q(x)=RA-qx=𝑞𝑙2 − 𝑞𝑥; M(x)=RAx-qx𝑥2= 𝑞𝑙2x- 𝑞𝑥

2

2

Q(0)= 𝑞𝑙2; Q(l)=-0.5ql; M(0)=0=M(l); M(2𝑙)= 𝑞𝑙

2

8 ;

Concentrated force applied to the double-beam beam.

First, support reactions should be found (fig. 60).

fig. 60

0 0

B A A B A B

Pb Pa

M R l Pb ; M R l Pa ; R ; R

l l

 

AC: Q(x)=RA= 𝑃𝑏𝑙 , M(x)=Rax= 𝑃𝑏𝑙 x, {0, 𝑃𝑏𝑙 a}

CB: Q(x)=-RB=-𝑃𝑎𝑙 , M(x)=RB K2B= 𝑃𝑎𝑙 (l-x), {0, 𝑃𝑏𝑙 a}

M(x=a); Mc(2𝑙)= -𝑃𝑎𝑙 (𝑙 − 𝑎) =𝑃𝑎𝑏𝑙 ; M(x=l) → MB=0

fig. 59

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PLOTTING Q AND M Evenly distributed intensity load q(x)= const on the entire length of the span doubly-supported beam

• The concentrated moment in the span of a two support beam. First, support reactions should be found (fig. 61).

 Concentrated moments on the supports of a two-supported beam. Find the

reactions of the supports.( Fig 63).

0 0

0 0

B A A

A B B

M R l M M R ;

M R l M M R ;

 

then for an arbitrary section at a distance X from the left support:

A 0

Q( x )R; M ( x )Mconst

fig. 61

fig. 63

1 0 1 0

B A A B

M R lM ; M R lM ;

A 1 B 1

M M

R ; R

l l

 

AC: Q(x)=-RA=- 𝑀𝑙1 , M(x)=-RAx= -𝑀𝑙1𝑥

CB: Q(x)=-RB=- 𝑀𝑙1 , M(x)=-RB K2B=-𝑀𝑙1 (l-x),

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DIFFERENTIAL DEPENDENCIES IN BENDING. FEATURES OF DIAGRAMS for Q and M

dQ q( x )

dx dM

Q( x )

dx dM

Q( x ) m

dx

2 2

d M q( x ) dx

 

0

   

Y Q q dx Q dQ   0

oz 2

M  M Qdx qdx dx M dM

1)In areas without distributed load, the diagrams Q are limited by straight lines parallel to the base, and the diagrams M are inclined by straight lines (Fig. 66).

2)In areas of evenly distributed load, the diagrams are limited by oblique straight lines parallel to the base, and the diagrams М by quadratic parabolas (Fig. 66). Since the M plot is built on compressed fibers, the convexity of the parabola is opposite to the direction of the load (Fig. 67a,b).

3)In sections where Q= 0, the tangent to the plot is parallel to the base of the plot (Fig. 66,67).

4)In sections Q>0, the moment M increases, i.e. from left to right, the positive ordinates of the M plot increase and the negative ordinates decrease (Fig. 65, 66, AC, BE sections). In sections Q<0, moment M decreases (Fig. 65, 66, sections CE, DB).

Fig. 64

Fig. 65

Fig. 66

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b) on the diagram M there will be fractures (Fig. 68) , and the tip of the fracture is directed against the action of forces

DIFFERENTIAL DEPENDENCIES IN BENDING. FEATURES OF DIAGRAMS for Q and M

7. If a concentrated moment is applied to the beam at the end of the console or in the end support, then in this section the bending moment is equal to the external moment (Fig.

71, sections B and C). If the beam is not loaded with an

external moment in the end hinged support or at the end of the console, then M = 0 in them (Fig. 65 and 66 of section A and E).

The diagram Q is a diagram of the derivative of the diagram M. This means that the diagrams Q are proportional to the

tangent of the tangent to the diagram M. Fig. 71 5)In sections where concentrated forces are applied to the beam

a) The diagram Q will show jumps by the magnitude, and in direction of applied forces, (in Fig. 65 and 66, the jumps are marked by thick lines with arrows)

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8. Diagram Q is a diagram of the derivative of diagram M. Therefore, the ordinates of

diagram Q are proportional to the tangent of the angle of inclination of the tangent to the diagram M.

To substantiate the listed properties of diagrams, consider the following. If there is no distributed load, then 𝒅𝑸𝒅𝒙 = q = 0.

Integrating, we get:

Q (x) = C1 = const.

Consequently,

𝒅𝑴

𝒅𝒙 = Q = C1,

from where M(x)=C1x+C2

q𝑎2 + +

- -

Fig.67 Fig.68

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Note, the appearance of jumps on the diagram Q is associated with the introduction of the conventional concept of a concentrated force.

Fig.69 Fig.70 Fig.71

6. In sections where concentrated moments are applied to the beam, jumps by their size will be on the diagram M (there will be no changes on the diagram Q). The direction of the jump depends on the direction of the external moment.

7. If a concentrated moment is applied to the beam at the end of the console or in the end support, then in this section the bending moment is equal to the external moment (Fig. 71, sections B and C). If the beam is not loaded with an external moment in the end hinged support or at the end of the console, then M = 0 in them (Fig. 65 and 66 of section A and E).

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Note, the appearance of jumps on the diagram Q is associated with the introduction of the conventional concept of a concentrated force (fig.72).

Fig.73 Fig.72

In some cases it is possible to construct diagrams without drawing up the Q and M

expressions for arbitrary sections of the sections. It is enough to calculate the values ​​of Q and M in the characteristic sections. For these cases, the following order of plotting can be recommended:

1. Find support reactions (you may not find them for the console).

2. Along the jumps and slopes, walking along the beam, be sure to build a diagram Q from left to right (you do not need to make any notes for this).

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EXAMPLE 6. Let us construct diagrams Q and M for a simple beam loaded with a distributed load changing according to a linear law (Fig. 73).

We define support reactions. The result is distributed throughout! load is equal to ql / 2 and passes through the center of gravity of the load diagram, which is l/3 from the right support.

therefore

ΣMB= Ral -𝑞𝑙2 3𝑙= 0; → RA=𝑞𝑙6 ΣMA = RBI - 𝑞𝑙2 2𝑙3= 0. → RB=𝑞𝑙3

We calculate the transverse force and bending moment in an arbitrary section K as a result of the action of forces located to the left of the section K, the reaction RA and the resultant of distributed load (1/2)q (x) x. From the similarity of triangles

q {x) = q𝒙𝒍

4. Calculate the values ​​of M in the characteristic sections and construct a diagram of M using the found ordinates. In this case, one should be guided by the general properties of the

diagrams, and for the cantilever parts of the beams it is advisable to use the diagrams known for them (Figs. 57 and 58).

3. Find the characteristic sections of the beam. The characteristic sections are those in which concentrated forces and moments are applied, the distributed load begins or ends, as well as those in which Q vanishes.

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Example 6. 5. Construct diagrams Q and M for a beam loaded with a distributed load, the intensity of which varies linearly (Fig. 6. 7, a).

Solution. 1) Determination of support reactions.

It is easy to see that the resultant of the distributed load (denote this resultant Po) is equal to P0=(qmaxl)/2

In other words, it is half the resultant of an equally distributed load with intensity qmax. The force Po is applied at the center of gravity of the triangle, which is a diagram of the load, i.e., applied at a distance l/3 from right support.

Summing up the moments of all forces relative to the centers of the hinges (supports), we have

2) Plotting transverse forces.

Let's make an arbitrary section at a distance x from the left support; the ordinate of the load diagram corresponding to this section is determined from the similarity of triangles q(x)/x=qmax/l, therefore q(x)=qmaxx/l

0 0

2 2

0 0

3 3 3

max

A B B

q l m ; R lP l  R P

0 0

0 0 1

3 3 6

max

B A A

q l m ; R lP l  R P

Fig. 6. 7 The resultant of that part of the distributed load, which is

located to the left of the section, Po(x)=12q(x).x =𝑞𝑚𝑎𝑥𝑥

2

2𝑙

b

c

References

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