Longitudinal transverse bending is called a combination of transverse bending with compression or tension of the beam.
When calculating for longitudinal-transverse bending, the calculation of bending moments in the cross sections of the beam is carried out taking into account the deflections of its axis.
Let's consider a beam with hingedly supported ends, loaded with some transverse load and a compressive force 5 acting along the axis of the beam (Fig. 8.13, a). Let us denote the deflection of the beam axis in the cross section with the abscissa (the positive direction of the y axis is taken to be downward, and, therefore, we consider the beam deflections to be positive when they are directed downward). The bending moment M acting in this section is
(23.13)
here the bending moment from the action of the transverse load; - additional bending moment due to force
The total deflection y can be considered to consist of the deflection arising from the action of only the transverse load, and an additional deflection equal to that caused by the force .
The total deflection y is greater than the sum of the deflections that arise under the separate action of the transverse load and the force S, since in the case of the action of only the force S on the beam, its deflections are equal to zero. Thus, in the case of longitudinal-transverse bending, the principle of independent action of forces is not applicable.
When a tensile force S is applied to a beam (Fig. 8.13, b), the bending moment in the section with the abscissa
(24.13)
The tensile force S leads to a decrease in the deflections of the beam, i.e., the total deflections y in this case are less than the deflections caused by the action of only the transverse load.
In the practice of engineering calculations, longitudinal-transverse bending usually means the case of compressive force and transverse load.
With a rigid beam, when the additional bending moments are small compared to the moment, the deflections y differ little from the deflections . In these cases, you can neglect the influence of force S on the magnitude of bending moments and the magnitude of deflections of the beam and carry out its calculation for central compression (or tension) with transverse bending, as described in § 2.9.
For a beam whose rigidity is low, the influence of force S on the magnitude of bending moments and deflections of the beam can be very significant and cannot be neglected in the calculation. In this case, the beam should be designed for longitudinal-transverse bending, meaning by this a calculation for the combined action of bending and compression (or tension), carried out taking into account the influence of the axial load (force S) on the bending deformation of the beam.
Let us consider the method of such a calculation using the example of a beam hingedly supported at the ends, loaded with transverse forces directed in one direction and a compressive force S (Fig. 9.13).
Let's substitute it into an approximate differential equation elastic line (1.13) expression of the bending moment M according to formula (23.13):
[the minus sign in front of the right side of the equation is taken because, unlike formula (1.13), here the downward direction is considered positive for deflections], or
Hence,
To simplify the solution, let us assume that the additional deflection varies along the length of the beam along a sinusoid, i.e. that
This assumption makes it possible to obtain fairly accurate results when a beam is subjected to a transverse load directed in one direction (for example, from top to bottom). Let us replace the deflection in formula (25.13) with the expression
The expression coincides with Euler's formula for the critical force of a compressed rod with hinged ends. Therefore it is designated and called Euler force.
Hence,
It is necessary to distinguish the Euler force from the critical force calculated using the Euler formula. The value can be calculated using Euler’s formula only if the flexibility of the rod is greater than the maximum; the value is substituted into formula (26.13) regardless of the flexibility of the beam. The formula for the critical force usually includes the minimum moment of inertia cross section rod, and the expression of the Euler force includes the moment of inertia relative to that of the main axes of inertia of the section, which is perpendicular to the plane of action of the transverse load.
From formula (26.13) it follows that the ratio between the total deflections of the beam y and the deflections caused by the action of only the transverse load depends on the ratio (the magnitude of the compressive force 5 to the magnitude of the Euler force).
Thus, the ratio is a criterion for the stiffness of the beam during longitudinal-transverse bending; if this ratio is close to zero, then the stiffness of the beam is high, and if it is close to unity, then the stiffness of the beam is small, i.e., the beam is flexible.
In the case when , deflection, i.e. in the absence of force S, deflections are caused only by the action of lateral load.
When the magnitude of the compressive force S approaches the value of the Euler force, the total deflections of the beam increase sharply and can be many times greater than the deflections caused by the action of only the transverse load. In the limiting case at, the deflections y, calculated using formula (26.13), become equal to infinity.
It should be noted that formula (26.13) is not applicable for very large deflections of the beam, since it is based on an approximate expression of curvature. This expression is applicable only for small deflections, and for large ones it should be replaced by the same expression of curvature (65.7). In this case, the deflections at would not be equal to infinity, but would be, although very large, finite.
When a tensile force is applied to the beam, formula (26.13) takes the form.
From this formula it follows that the total deflections are less than the deflections caused by the action of only the transverse load. With a tensile force S numerically equal to the value of the Euler force (i.e., at ), the deflections y are half as large as the deflections
The maximum and minimum normal stresses in the cross section of a beam with hinged ends under longitudinal-transverse bending and compressive force S are equal
Let us consider a two-support beam of I-section with a span. The beam is loaded in the middle with a vertical force P and is compressed by an axial force S = 600 (Fig. 10.13). Beam cross-sectional area moment of inertia, moment of resistance and modulus of elasticity
The transverse ties connecting this beam to the adjacent beams of the structure eliminate the possibility of the beam losing stability in the horizontal plane (i.e., in the plane of least rigidity).
The bending moment and deflection in the middle of the beam, calculated without taking into account the influence of force S, are equal to:
The Euler force is determined from the expression
The deflection in the middle of the beam, calculated taking into account the influence of force S based on formula (26.13),
Let us determine the highest normal (compressive) stresses in the average cross section of the beam using formula (28.13):
from where after conversion
Substituting into expression (29.13) different meanings P (v), we obtain the corresponding voltage values. Graphically, the relationship between, determined by expression (29.13), is characterized by the curve shown in Fig. 11.13.
Let us determine the permissible load P if for the beam material a the required safety factor is therefore the permissible stress for the material
From Fig. 11.23 it follows that stress occurs in the beam under load and stress occurs under load
If we take the load as an allowable load, then the stress safety factor will be equal to the specified value. However, in this case, the beam will have an insignificant load safety factor, since stresses equal to will arise in it already at Rot
Consequently, the load safety factor in this case will be equal to 1.06 (since e. is clearly insufficient.
In order for the beam to have a load safety factor equal to 1.5, the value should be taken as acceptable; the stresses in the beam will be as follows from Fig. 11.13, approximately equal
Above, strength calculations were made based on permissible stresses. This provided the necessary safety margin not only for stresses, but also for loads, since in almost all cases discussed in previous chapters, stresses are directly proportional to the magnitude of the loads.
During longitudinal-transverse bending stress, as follows from Fig. 11.13, are not directly proportional to the load, but change faster than the load (in the case of compressive force S). In this regard, even a slight accidental increase in load above the design one can cause a very large increase in stress and destruction of the structure. Therefore, the calculation of compressed-bent rods for longitudinal-transverse bending should be made not according to the permissible stresses, but according to the permissible load.
By analogy with formula (28.13), let us create a strength condition when calculating longitudinal-transverse bending based on the permissible load.
Compressed-bent rods, in addition to calculations for longitudinal-transverse bending, must also be calculated for stability.
During bending, the cross sections, while remaining flat, rotate relative to each other around certain axes lying in their planes. Beams, axles, shafts and other machine parts and structural elements work for bending. In practice, there are transverse (straight), oblique and clean views bending
Transverse (straight) (Fig. 61, A) called bending when external forces perpendicular to the longitudinal axis of the beam act in a plane passing through the axis of the beam and one of the main central axes of its cross section.
Oblique bending (Fig. 61, b) is a bending when forces act in a plane passing through the axis of the beam, but not passing through any of the main central axes of its cross section.
In the cross sections of beams during bending, two types of internal forces arise - bending moment M and and shear force Q. In the particular case when the shear force is zero and only a bending moment occurs, then pure bending occurs (Fig. 61, c). Pure bending occurs when loaded with a distributed load or under some loadings with concentrated forces, for example, a beam loaded with two symmetrical equal forces.
Rice. 61. Bend: a - transverse (straight) bend; b - oblique bend; c - pure bend
When studying bending deformation, it is mentally imagined that the beam consists of an infinite number of fibers parallel to the longitudinal axis. At pure bend the hypothesis of plane sections is valid: fibers lying on the convex side stretch, lying on the concave side - shrink, and on the boundary between them lies a neutral layer of fibers (longitudinal axis), which only are bent, without changing its length; The longitudinal fibers of the beam do not exert pressure on each other and, therefore, experience only tension and compression.
Internal force factors in beam sections - shear force Q and bending moment M and(Fig. 62) depend on external forces and vary along the length of the beam. The laws of change in shear forces and bending moments are represented by certain equations in which the arguments are the coordinates z cross sections of beams, and functions - Q And M i. To determine internal force factors, we use the section method.
Rice. 62.
Lateral force Q is the resultant of the internal tangential forces in the cross section of the beam. It should be kept in mind that shear force has opposite direction for the left and right parts of the beam, which indicates that the rule of static signs is unsuitable.
Bending moment M and is the resulting moment relative to the neutral axis of the internal normal forces acting in the cross section of the beam. The bending moment, like the shear force, has different direction for the left and right parts of the beam. This indicates that the rule of static signs is unsuitable when determining the bending moment.
Considering the equilibrium of the parts of the beam located to the left and right of the section, it is clear that a bending moment must act in the cross sections M and and shear force Q. Thus, in the case under consideration, at the points of the cross sections there are not only normal stresses corresponding to the bending moment, but also tangent stresses corresponding to the transverse force.
For a visual representation of the distribution of shear forces along the axis of the beam Q and bending moments M and it is convenient to present them in the form of diagrams, the ordinates of which for any abscissa values z give the corresponding values Q And M i. The diagrams are constructed similarly to the construction of diagrams of longitudinal forces (see 4.4) and torques (see 4.6.1.).
Rice. 63. Direction of transverse forces: a - positive; b - negative
Since the rules of static signs are unacceptable for establishing the signs of shear forces and bending moments, we will establish other rules of signs for them, namely:
Rice. 64. Direction of bending moments: a - positive; b - negative
The sign rule for bending moments is related to the nature of the deformation of the beam. The bending moment is considered positive if the beam bends convexly downwards (the stretched fibers are located at the bottom). The bending moment is considered negative if the beam bends convexly upward (the stretched fibers are located at the top).
Using the rules of signs, you should mentally imagine the section of the beam as rigidly clamped, and the connections as discarded and replaced by their reactions. To determine reactions, the rules of static signs are used.
Calculate bending beam There are several options:
1. Calculation maximum load that she can withstand
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle selection of beam section
on two supports loaded with a uniformly distributed load or concentrated force.
To begin with, you will need to find the point (section) at which there will be a maximum moment. This depends on whether the beam is supported or embedded. Below are diagrams of bending moments for the most common schemes.
Further, when dividing the maximum bending moment by the moment of resistance in a given section, we get maximum stress in beam and we must compare this stress with the stress that our beam of a given material can generally withstand.
For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, A for fragile(cast iron) – tensile strength. We can find the yield strength and tensile strength from the tables below.
P = m * g = 90 * 10 = 900 N = 0.9 kN
M = P * l = 0.9 kN * 2 m = 1.8 kN*m
b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa
45.34 MPa is correct, which means this I-beam will withstand a mass of 90 kg.
The whole variety of existing support devices is schematized in the form of a number of basic types of supports, of which
most common: articulated and movablesupport(possible designations for it are presented in Fig. 1, a), hinged-fixed support(Fig. 1, b) and hard pinching, or sealing(Fig. 1, c).
In a hinged-movable support, one support reaction occurs, perpendicular to the support plane. Such a support deprives the support section of one degree of freedom, that is, it prevents displacement in the direction of the support plane, but allows movement in the perpendicular direction and rotation of the support section.
In a hinged-fixed support, vertical and horizontal reactions occur. Here, movements along the directions of the support rods are not possible, but rotation of the support section is allowed.
In a rigid embedment, vertical and horizontal reactions and a support (reactive) moment occur. In this case, the support section cannot shift or rotate. When calculating systems containing a rigid embedment, the resulting support reactions can not be determined, choosing the cut-off part so that the embedment with unknown reactions does not fall into it. When calculating systems on hinged supports, the reactions of the supports must be determined. The static equations used for this depend on the type of system (beam, frame, etc.) and will be given in the relevant sections of this manual.
The longitudinal force in a section is numerically equal to the algebraic sum of the projections of all forces applied on one side of the section under consideration onto the longitudinal axis of the rod.
Rule of signs for Nz: let us agree to consider the longitudinal force in the section positive if the external load applied to the considered cut-off part of the rod causes tension and negative - otherwise.
Example 1.Construct a diagram of longitudinal forces for a rigidly clamped beam(Fig. 2).
Calculation procedure:
1. We outline characteristic sections, numbering them from the free end of the rod to the embedment.
2. Determine the longitudinal force Nz in each characteristic section. In this case, we always consider the cut-off part into which the rigid seal does not fall.
Based on the found values build a diagram Nz. Positive values are plotted (on the selected scale) above the diagram axis, negative values are plotted below the axis.
Torque in the section is numerically equal to the algebraic sum of external moments applied on one side of the section under consideration, relative to the longitudinal Z axis.
Sign rule for microdistrict: let’s agree to count torque in the section is positive if, when looking at the section from the side of the cut-off part under consideration, the external moment is seen directed counterclockwise and negative - otherwise.
Example 2.Construct a diagram of torques for a rigidly clamped rod(Fig. 3, a).
Calculation procedure.
It should be noted that the algorithm and principles for constructing a torque diagram completely coincide with the algorithm and principles constructing a diagram of longitudinal forces.
1. We outline characteristic sections.
2. Determine the torque in each characteristic section.
Based on the found values we build microdistrict diagram(Fig. 3, b).
For diagrams of longitudinal forces and torques are characterized by certain patterns, knowledge of which allows us to evaluate the correctness of the constructions performed.
1. Diagrams Nz and Mkr are always rectilinear.
2. In the area where there is no distributed load, the diagram Nz(Mkr) is a straight line, parallel to the axis, and in the area under a distributed load it is an inclined straight line.
3. Under the point of application of the concentrated force on the diagram Nz there must be a jump in the magnitude of this force, similarly, under the point of application of the concentrated moment on the diagram Mkr there will be a jump in the magnitude of this moment.
A rod that bends is called beam. In sections of beams loaded with vertical loads, as a rule, two internal force factors arise - Qy and bending moment Mx.
Lateral force in the section is numerically equal to the algebraic sum of the projections of external forces applied on one side of the section under consideration onto the transverse (vertical) axis.
Sign rule for Qy: Let us agree to consider the transverse force in the section positive if the external load applied to the cut-off part under consideration tends to rotate this section clockwise and negative otherwise.
Schematically, this sign rule can be represented as
Bending moment Mx in a section is numerically equal to the algebraic sum of the moments of external forces applied on one side of the section under consideration, relative to the x axis passing through this section.
Rule of signs for Mx: let us agree to consider the bending moment in the section positive if the external load applied to the cut-off part under consideration leads to tension in this section of the lower fibers of the beam and negative - otherwise.
Schematically, this sign rule can be represented as:
It should be noted that when using the sign rule for Mx in in the specified form, the Mx diagram always turns out to be constructed from the side of the compressed fibers of the beam.
At plotting Qy and Mx diagrams in cantilever, or rigidly clamped, beams there is no need (as in the previously discussed examples) to calculate the support reactions arising in the rigid embedment, but the cut-off part must be selected so that the embedment does not fall into it.
Example 3.Construct Qy and Mx diagrams(Fig. 4).
Calculation procedure.
1. We outline characteristic sections.
UDC 539.52
ULTIMATE LOAD FOR A RESTRAINTED BEAM LOADED WITH LONGITUDINAL FORCE, UNSYMMETRICALLY DISTRIBUTED LOAD AND SUPPORT MOMENTS
I.A. Monakhov1, Yu.K. Basov2
department construction production Faculty of Civil Engineering Moscow State Mechanical Engineering University st. Pavel Korchagina, 22, Moscow, Russia, 129626
2Department building structures and structures Faculty of Engineering Peoples' Friendship University of Russia st. Ordzhonikidze, 3, Moscow, Russia, 115419
The article develops a method for solving problems of small deflections of beams made of an ideal rigid-plastic material under the action of asymmetrically distributed loads, taking into account preliminary tension-compression. The developed methodology was used to study the stress-strain state of single-span beams, as well as to calculate the ultimate load of beams.
Key words: beam, nonlinearity, analytical.
IN modern construction, shipbuilding, mechanical engineering, chemical industry and in other branches of technology, the most common types of structures are rod ones, in particular beams. Naturally, to determine real behavior rod systems(in particular, beams) and their strength resources, it is necessary to take into account plastic deformations.
Calculation structural systems when taking into account plastic deformations using a model of an ideal rigid-plastic body, it is the simplest, on the one hand, and quite acceptable from the point of view of the requirements of design practice, on the other. If we keep in mind the region of small displacements of structural systems, this is explained by the fact that the bearing capacity (“ultimate load”) of ideal rigid-plastic and elastoplastic systems turns out to be the same.
Additional reserves and stricter assessment bearing capacity structures are revealed by taking into account geometric nonlinearity during their deformation. Currently, taking into account geometric nonlinearity in the calculations of structural systems is a priority task not only from the point of view of the development of calculation theory, but also from the point of view of the practice of designing structures. Acceptability of solutions to problems of structural calculations under conditions of small
displacements is quite uncertain; on the other hand, practical data and properties of deformable systems suggest that large displacements are actually achievable. It is enough to point out the designs of construction, chemical, shipbuilding and mechanical engineering facilities. In addition, the model of a rigid-plastic body means that elastic deformations are neglected, i.e. plastic deformations are much greater than elastic ones. Since deformations correspond to displacements, taking into account large displacements of rigid plastic systems is appropriate.
However, geometrically nonlinear deformation of structures in most cases inevitably leads to the occurrence of plastic deformations. Therefore, the simultaneous consideration of plastic deformations and geometric nonlinearity in the calculations of structural systems and, of course, rods is of particular importance.
This article discusses small deflections. Similar problems were solved in works.
We consider a beam with pinched supports under the action of a step load, edge moments and a previously applied longitudinal force (Fig. 1).
Rice. 1. Beam under distributed load
The equilibrium equation of a beam for large deflections in dimensionless form has the form
d2 t/h d2 w dn
-- + (n ± n)-- + p = ^ - = 0, dx ah ah
x 2w р12 М N,г,
where x ==, w =-, p =--, t =--, n =-, N and M are internal normal
I to 5xЪk b!!bk 25!!bk
force and bending moment, p - transverse uniformly distributed load, W - deflection, x - longitudinal coordinate (origin of coordinates on the left support), 2к - cross-section height, b - cross-section width, 21 - beam span, 5^ - yield strength material. If N is given, then the force N is a consequence of the action p at
available deflections, 11 = = , the line above the letters indicates the dimension of the quantities.
Let's consider the first stage of deformation - “small” deflections. A plastic section occurs at x = x2, in which m = 1 - n2.
Expressions for deflection rates have the form - deflection at x = x2):
(2-x), (x > X2),
The solution to the problem is divided into two cases: x2< 11 и х2 > 11.
Consider the case x2< 11.
For zone 0< х2 < 11 из (1) получаем:
Рх 111 1 Р11 к1р/1 t = + к1 р + р/1 -к1 р/1 -±4- +-^41
x -(1 -n2)±a,
(, 1, r/2 k1 r12L
Рх2 + к1 р + р11 - к1 р11 -+ 1 ^
X2 = k1 +11 - k111 - + ^
Taking into account the appearance of a plastic hinge at x = x2, we obtain:
tx=x = 1 - p2 = - p
(12 k12 L k +/ - k1 - ^ + k "A
k, + /, - k,/, -L +
(/ 2 k/ 2 L k1 + /1 - k1/1 - ^ + M
Considering the case x2 > /1, we obtain:
for zone 0< х < /1 выражение для изгибающих моментов имеет вид
to р-р2 + kar/1+р/1 -к1 р/1 ^ x-(1-П12)±
and for zone 11< х < 2 -
^ р-рЦ + 1^ Л
x -(1 -n-)±a +
(. rg-k1 r1-L
Kx px2 + kh p+
0, and then
I2 12 1 h h x2 = 1 -- + -.
The condition of plasticity implies the equality
where we get the expression for the load:
k1 - 12 + M L2
K1/12 - k2 ¡1
Table 1
k1 = 0 11 = 0.66
table 2
k1 = 0 11 = 1.33
0 6,48 9,72 12,96 16,2 19,44
0,5 3,24 6,48 9,72 12,96 16,2
Table 3
k1 = 0.5 11 = 1.61
0 2,98 4,47 5,96 7,45 8,94
0,5 1,49 2,98 4,47 5,96 7,45
Table 5 k1 = 0.8 11 = 0.94
0 2,24 3,56 4,49 5,61 6,73
0,5 1,12 2,24 3,36 4,49 5,61
0 2,53 3,80 5,06 6,33 7,59
0,5 1,27 2,53 3,80 5,06 6,33
Table 3
k1 = 0.5 11 = 2.0
0 3,56 5,33 7,11 8,89 10,7
0,5 1,78 3,56 5,33 7,11 8,89
Table 6 k1 = 1 11 = 1.33
0 2,0 3,0 4,0 5,0 6,0
0,5 1,0 2,0 3,0 4,0 5,0
Table 7 Table 8
k, = 0.8 /, = 1.65 k, = 0.2 /, = 0.42
0 2,55 3,83 5,15 6,38 7,66
0,5 1,28 2,55 3,83 5,15 6,38
0 7,31 10,9 14,6 18,3 21,9
0,5 3,65 7,31 10,9 14,6 18,3
Setting the load coefficient k1 from 0 to 1, the bending moment a from -1 to 1, the value of the longitudinal force p1 from 0 to 1, the distance /1 from 0 to 2, we obtain the position of the plastic hinge according to formulas (3) and (5), and then we obtain the value of the maximum load using formulas (4) or (6). The numerical results of the calculations are summarized in tables 1-8.
LITERATURE
Basov Yu.K., Monakhov I.A. Analytical solution to the problem of large deflections of a rigid-plastic clamped beam under the action of a local distributed load, supporting moments and longitudinal force. Vestnik RUDN. Series "Engineering Research". - 2012. - No. 3. - P. 120-125.
Savchenko L.V., Monakhov I.A. Large deflections of physically nonlinear round plates // Bulletin of INGECON. Series "Technical Sciences". - Vol. 8(35). - St. Petersburg, 2009. - pp. 132-134.
Galileev S.M., Salikhova E.A. Study of the frequencies of natural vibrations of structural elements made of fiberglass, carbon fiber and graphene // Bulletin of INGECON. Series "Technical Sciences". - Vol. 8. - St. Petersburg, 2011. - P. 102.
Erkhov M.I., Monakhov A.I. Large deflections of a prestressed rigid-plastic beam with hinged supports under a uniformly distributed load and edge moments // Bulletin of the Department of Construction Sciences of the Russian Academy of Architecture and Construction Sciences. - 1999. - Issue. 2. - pp. 151-154. .
THE LITTLE DEFLECTIONS OF THE PREVIOUSLY INTENSE IDEAL PLASTIC BEAMS WITH THE REGIONAL MOMENTS
I.A. Monakhov1, U.K. Basov2
"Department of Building production manufacture Building Faculty Moscow State Machine-building University Pavla Korchagina str., 22, Moskow, Russia,129626
Department of Bulding Structures and Facilities Enqineering Faculty Peoples" Friendship University of Russia Ordzonikidze str., 3, Moskow, Russia, 115419
In the work up the technique of the solution of problems about the little deflections of beams from the ideal hard-plastic material, with various kinds of fastening, for want of action of the asymmetrically distributed loads with allowance for of preliminary stretching-compression is developed. The developed technique is applied for research of the strained-deformed condition of beams, and also for calculation of a deflection of beams with allowance for geometrical nonlinearity.
Key words: beam, analytical, nonlinearity.
