What is cross section and transverse camber. Bend. Moment of inertia of a rectangular section
Classification of types of rod bending
Bend This type of deformation is called in which bending moments occur in the cross sections of the rod. A rod that bends is usually called beam. If bending moments are the only internal force factors in the cross sections, then the rod experiences clean bend. If bending moments occur together with transverse forces, then such bending is called transverse.
Beams, axles, shafts and other structural parts work for bending.
Let's introduce some concepts. The plane passing through one of the main central axes of the section and the geometric axis of the rod is called main plane. The plane in which external loads act, causing the beam to bend, is called force plane. The line of intersection of the force plane with the cross-sectional plane of the rod is called power line. Depending on the relative position of the force and main planes of the beam, straight or oblique bending is distinguished. If the force plane coincides with one of the main planes, then the rod experiences straight bend(Fig. 5.1, A), if it does not match - oblique(Fig. 5.1, b).
Rice. 5.1. Rod bend: A- straight; b- oblique
From a geometric point of view, the bending of the rod is accompanied by a change in the curvature of the rod axis. The initially straight axis of the rod becomes curved when it is bent. At straight bend the curved axis of the rod lies in the force plane, while in the case of an oblique rod it lies in a plane different from the force plane.
Observing the bending of a rubber rod, you can notice that part of its longitudinal fibers are stretched, and the other part is compressed. Obviously, between the stretched and compressed fibers of the rod there is a layer of fibers that experience neither tension nor compression - the so-called neutral layer. The line of intersection of the neutral layer of the rod with the plane of its cross section is called neutral section line.
As a rule, loads acting on a beam can be classified into one of three types: concentrated forces R, concentrated moments M distributed loads of intensity ts(Fig. 5.2). Part I of the beam located between the supports is called in flight, part II of the beam located on one side of the support - console.
Straight transverse bend occurs when all loads are applied perpendicular to the axis of the rod, lie in the same plane and, in addition, the plane of their action coincides with one of the main central axes of inertia of the section. Straight transverse bending refers to simple view resistance is flat stress state, i.e. two principal stresses are non-zero. With this type of deformation, internal forces arise: shear force and bending moment. A special case of direct transverse bending is pure bend, with such resistance there are load areas within which the transverse force becomes zero and the bending moment is non-zero. In the cross sections of the rods during direct transverse bending, normal and tangential stresses arise. Stresses are a function of internal force, in this case normal stresses are a function of bending moment, and tangential stresses are a function of shear force. For direct transverse bending, several hypotheses are introduced:
1) The cross sections of the beam, flat before deformation, remain flat and orthogonal to the neutral layer after deformation (hypothesis of plane sections or J. Bernoulli’s hypothesis). This hypothesis is satisfied under pure bending and is violated when shear forces, shear stresses, and angular deformation occur.
2) There is no mutual pressure between the longitudinal layers (hypothesis of non-pressure of fibers). From this hypothesis it follows that longitudinal fibers experience uniaxial tension or compression, therefore, with pure bending, Hooke's law is valid.
A rod undergoing bending is called beam. When bending, one part of the fibers stretches, the other part contracts. The layer of fibers located between the stretched and compressed fibers is called neutral layer, it passes through the center of gravity of the sections. The line of its intersection with the cross section of the beam is called neutral axis. Based on the introduced hypotheses at pure bending a formula was obtained for determining normal stresses, which is also used for direct transverse bending. The normal stress can be found using the linear relationship (1), in which the ratio of the bending moment to the axial moment of inertia ( 
) in a particular section is a constant value, and the distance ( y) along the ordinate axis from the center of gravity of the section to the point at which the stress is determined varies from 0 to
.
.
(1)
To determine the shear stress during bending in 1856. Russian engineer and bridge builder D.I. Zhuravsky became addicted
.
(2)
The shear stress in a particular section does not depend on the ratio of the transverse force to the axial moment of inertia ( 
), because this value does not change within one section, but depends on the ratio of the static moment of the area of the cut-off part to the width of the section at the level of the cut-off part ( 
).
When straight transverse bending occurs movements: deflections (v
) and rotation angles (Θ
)
. To determine them, use the equations of the initial parameters method (3), which are obtained by integrating the differential equation of the curved axis of the beam ( 
).
Here v 0 , Θ 0 ,M 0 , Q 0 – initial parameters, x – distance from the origin to the section in which the displacement is determined , a– the distance from the origin of coordinates to the place of application or the beginning of the load.
Strength and stiffness calculations are made using the strength and stiffness conditions. Using these conditions, you can solve verification problems (check the fulfillment of a condition), determine the size of the cross section, or select the permissible value of the load parameter. There are several strength conditions, some of which are given below. Normal stress strength condition has the form:
,
(4)
Here 
–
moment of resistance of the section relative to the z axis, R – design resistance based on normal stresses.
Strength condition for tangential stresses looks like:
,
(5)
here the notations are the same as in Zhuravsky’s formula, and R s – calculated shear resistance or calculated resistance to tangential stresses.
Strength condition according to the third strength hypothesis or the hypothesis of the greatest tangential stresses can be written in the following form:
.
(6)
Severity conditions can be written for deflections (v ) And rotation angles (Θ ) :
where the displacement values in square brackets are valid.
Example of completing individual task No. 4 (term 2-8 weeks)
During transverse bending in the cross section of a beam (beam), in addition to the bending moment, a transverse force also acts. If the transverse bending is straight, then the bending moment acts in a plane coinciding with one of the main planes of the beam.
The transverse force in this case is usually parallel to the plane of action of the bending moment and, as shown below (see § 12.7), passes through a certain point in the cross section, called the center of bending. The position of the bending center depends on the shape and dimensions of the cross-section of the beam. For a cross section that has two axes of symmetry, the center of bending coincides with the center of gravity of the section.
Experimental and theoretical studies show that the formulas obtained for the case of straight pure bending are also applicable for straight transverse bending.
The transverse force acting in a section of a beam is related to the shear stresses arising in this section, the dependence
where is the component of the shear stress in the cross section of the beam, parallel to the y-axis and the force
The quantity represents the elementary tangential force (parallel to the force Q) acting on the elementary area of the cross section of the beam.
Let's look at some cross section timber (Fig. 37.7). Tangential stresses at points near the section contour are directed tangentially to the contour. Indeed, if the tangential stress had a component directed along the normal to the contour, then, according to the law of pairing of tangential stresses, the same stress would arise on the side surface of the beam, which is impossible, since the side surface is stress-free.
The shear stress at each point of the section can be decomposed into two components: .
Let's consider the definition of the components. The definition of components is discussed in § 12.7 only for some types of cross sections.
It is assumed that the components of tangential stresses across the entire width of the section in the direction parallel to the axis are the same (Fig. 37.7), i.e., that the value changes only along the height of the section.
To determine the vertical components of tangential stresses, we select element 1-2-3-4 from a beam of constant cross-section, symmetrical about the y-axis, with two cross sections drawn at distances from the left end of the beam, and one section parallel to the neutral layer, spaced from it (Fig. 38.7).
In the cross section of the beam with the abscissa there is a bending moment M, and with the abscissa there is a bending moment M. In accordance with this, the normal stresses a and acting along areas 1-2 and 3-4 of the selected element are determined by the expressions [see. formula (17.7)]
Diagrams of normal stresses acting on sites 1-2 and 3-4 at positive value M, shown in Fig. 39.7. Tangential stresses also act on these same areas, also shown in Fig. 39.7. The magnitude of these stresses varies along the height of the section.
Let us denote the magnitude of the shear stress at the lower points of areas 1-2 and 3-4 (at level ). According to the law of pairing of tangential stresses, it follows that tangential stresses of the same magnitude act along the lower area 1-4 of the selected element. Normal stresses along this area are considered equal to zero, since in the theory of bending it is assumed that the longitudinal fibers of the beam do not exert pressure on each other.
Platform 1-2 or 3-4 (Fig. 39.7 and 40.7), i.e. the part of the cross section located above the level (above platform 1-4), is called the cut-off part of the cross section. Let's denote its area
Let's create an equilibrium equation for element 1-2-3-4 in the form of the sum of the projections of all forces applied to it onto the axis of the beam:
Here is the resultant of the elementary forces arising along the area of 1-2 elements; - the resultant of the elementary forces arising on the site of 3-4 elements; - resultant of elementary tangential forces arising along the area of 1-4 elements; - width of the cross section of the beam at level y
Let us substitute expressions using formulas (26.7) into equation (27.7):
But based on Zhuravsky’s theorem [formula (6.7)]
The integral represents the static moment of area about the neutral axis of the beam cross section.
Hence,
According to the law of pairing of tangential stresses, the stresses at points of the cross section of the beam, spaced at a distance from the neutral axis, are equal (by absolute value) i.e.
Thus, the values of tangential stresses in the cross sections of the beam and in sections of its planes parallel to the neutral layer are determined by the formula
Here Q is the shear force in the cross section of the beam under consideration; - static moment (relative to the neutral axis) of the cut-off part of the cross section located on one side of the level at which the shear stresses are determined; J is the moment of inertia of the entire cross section relative to the neutral axis; - the width of the cross section of the beam at the level at which shear stresses are determined.
Expression (28.7) is called the Zhuravsky formula.
Tangential stresses are determined using formula (28.7) in the following order:
1) a cross section of the beam is drawn;
2) for this cross section, the values of the transverse force Q and the value J of the moment of inertia of the section relative to the main central axis coinciding with the neutral axis are determined;
3) in the cross section at the level for which the tangential stresses are determined, a straight line is drawn parallel to the neutral axis, cutting off part of the section; the length of the segment of this straight line, enclosed inside the contour of the cross section, is the width included in the denominator of formula (28.7);
4) the static moment S of the cut-off (located on one side of the straight line specified in paragraph 3) part of the section relative to the neutral axis is calculated;
5) formula (28.7) determines the absolute value of the shear stress. The sign of the tangential stresses in the cross section of the beam coincides with the sign of the transverse force acting in this section. The sign of the tangential stresses in areas parallel to the neutral layer is opposite to the sign of the transverse force.
Let us determine, as an example, the tangential stresses in the rectangular cross section of the beam shown in Fig. 41.7, a. The transverse force in this section acts parallel to the y-axis and is equal to
Moment of inertia of the cross section about the axis
To determine the shear stress at a certain point C, we draw a straight line 1-1 through this point, parallel to the axis (Fig. 41.7, a).
Let us determine the static moment S of the part of the section cut off by straight line 1-1 relative to the axis. Both the part of the section located above straight line 1-1 (shaded in Fig. 41.7, a) and the part located below this straight line can be taken as cut off.
For the top
Let us substitute the values of Q, S, J and b into formula (28.7):
From this expression it follows that shear stresses vary along the height of the cross section according to the law of a square parabola. At voltage The highest voltages are present at the points of the neutral axis, i.e. at
where is the cross-sectional area.
Thus, in case rectangular section the greatest tangential stress is 1.5 times greater than its average value, equal to The diagram of tangential stresses, showing their change along the height of the beam section, is shown in Fig. 41.7, b.
To check the resulting expression [see formula (29.7)] we substitute it into equality (25.7):
The resulting identity indicates the correctness of expression (29.7).
Parabolic diagram of tangential stresses shown in Fig. 41.7, b, is a consequence of the fact that with a rectangular section, the static moment of the cut-off part of the section changes with the change in the position of straight line 1-1 (see Fig. 41.7, a) according to the law of a square parabola.
For sections of any other shape, the nature of the change in tangential stresses along the height of the section depends on the law by which the ratio changes; if in certain sections of the section height the width b is constant, then the stresses in these sections change according to the law of change in the static moment
At the points of the cross section of the beam that are farthest from the neutral axis, the tangential stresses are equal to zero, since when determining the stresses at these points, the value of the static moment of the cut-off part of the section, equal to zero, is substituted into formula (28.7).
Value 5 reaches a maximum for points located on the neutral axis, however, shear stresses for sections with variable width b may not be maximum on the neutral axis. So, for example, the diagram of tangential stresses for the section shown in Fig. 42.7, and has the form shown in Fig. 42.7, b.
Tangential stresses arising during transverse bending in planes parallel to the neutral layer characterize the interaction forces between the individual layers of the beam; these forces tend to move adjacent layers relative to each other in the longitudinal direction.
If there is not sufficient connection between the individual layers of the beam, then such a shift will occur. For example, boards laid on top of each other (Fig. 43.7, a) will resist external load, like a whole beam (Fig. 43.7, b), until the forces along the planes of contact of the boards exceed the friction forces between them. When the friction forces are surpassed, the boards will move one over the other, as shown in Fig. 43.7, c. In this case, the deflections of the boards will increase sharply.
Tangential stresses acting in the cross sections of the beam and in sections parallel to the neutral layer cause shear deformations, as a result of which the right angles between these sections are distorted, i.e., they cease to be straight. The greatest distortions of angles occur at those points of the cross section at which the greatest tangential stresses act; There are no angular distortions at the upper and lower edges of the beam, since the tangential stresses there are zero.
As a result of shear deformations, the cross sections of the beam bend during transverse bending. However, this does not significantly affect the deformation of the longitudinal fibers, and therefore the distribution of normal stresses in the cross sections of the beam.
Let us now consider the distribution of shear stresses in thin-walled beams with cross sections symmetrical with respect to the y-axis, in the direction of which the transverse force Q acts, for example, in an I-section beam shown in Fig. 44.7, a.
To do this, using the Zhuravsky formula (28.7), we determine the tangential stresses at some characteristic points of the cross section of the beam.
At the top point 1 (Fig. 44.7, a) there are shear stresses since the entire cross-sectional area is located below this point, and therefore the static moment 5 relative to the axis (part of the cross-sectional area located above point 1) is zero.
At point 2, located directly above the line passing through the bottom edge top shelf I-beam, tangential stresses calculated using formula (28.7),
Between points 1 and 2, the stresses [determined by formula (28.7)] change along a square parabola, as for a rectangular section. In the I-beam wall at point 3, located directly below point 2, shear stresses
Since the width b of the I-beam flange is significantly greater than the thickness d vertical wall, then the diagram of tangential stresses (Fig. 44.7, b) has a sharp jump in the level corresponding to the lower edge of the upper flange. Below point 3, the tangential stresses in the I-beam wall change according to the law of a square parabola, as for a rectangle. The highest shear stresses occur at the level of the neutral axis:
The diagram of tangential stresses, constructed from the obtained values of and , is shown in Fig. 44.7, b; it is symmetrical about the ordinate.
According to this diagram, at points located at the inner edges of the flanges (for example, at points 4 in Fig. 44.7, a), tangential stresses act perpendicular to the section contour. But, as already noted, such stresses cannot arise near the section contour. Consequently, the assumption of uniform distribution of tangential stresses along the width b of the cross section, which is the basis for the derivation of formula (28.7), is not applicable to the flanges of an I-beam; it is not applicable to some elements of other thin-walled beams.
The tangential stresses in the flanges of the I-beam cannot be determined by methods of resistance of materials. These stresses are very small compared to the stresses in the wall of the I-beam. Therefore, they are not taken into account and the tangential stress diagram is built only for the I-beam wall, as shown in Fig. 44.7, c.
In some cases, for example, when calculating composite beams, the value T of the tangential forces acting in sections of the beam parallel to the neutral layer and per unit length is determined. We find this value by multiplying the voltage value by the section width b:
Let's substitute the value using formula (28.7):
As in § 17, we assume that the cross section of the rod has two axes of symmetry, one of which lies in the bending plane.
In the case of transverse bending of a rod, tangential stresses arise in its cross section, and when the rod is deformed, it does not remain flat, as in the case of pure bending. However, for a beam of solid cross-section, the influence of tangential stresses during transverse bending can be neglected and it can be approximately assumed that, just as in the case of pure bending, the cross-section of the rod remains flat during its deformation. Then the formulas for stress and curvature derived in § 17 remain approximately valid. They are accurate for the special case of a constant shear force along the length of the rod 1102).
Unlike pure bending, in transverse bending the bending moment and curvature do not remain constant along the length of the rod. The main task in the case of transverse bending is to determine the deflections. To determine small deflections, you can use the known approximate dependence of the curvature of a bent rod on deflection 11021. Based on this dependence, the curvature of a bent rod x c and deflection V e, resulting from the creep of the material, are related by the relation x c = = dV
Substituting curvature into this relation according to formula (4.16), we establish that
Integrating the last equation makes it possible to obtain the deflection resulting from the creep of the beam material.
Analyzing the above solution to the problem of creep of a bent rod, we can conclude that it is completely equivalent to the solution to the problem of bending a rod made of a material for which the tension-compression diagrams can be approximated power function. Therefore, the determination of deflections arising due to creep, in the case under consideration, can also be made using the Mohr integral to determine the movement of rods made of material that does not obey Hooke’s law)
