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 Russian University friendship of peoples 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 other branches of technology, the most common types of structures are rod ones, in particular beams. Naturally, to determine the real behavior of 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. That's why special meaning acquires simultaneous consideration of plastic deformations and geometric nonlinearity in the calculations of structural systems and, of course, rods.
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. Plastic section occurs at x = x2, in it 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 Russian Academy architecture and building 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.
Calculate bending beam There are several options:
1. Calculation maximum load which she will endure
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle selection of beam section
on two supports loaded with a uniformly distributed load or concentrated force.
To begin with, you will need to find the point (section) at which there will be a maximum moment. This depends on whether the beam is supported or embedded. Below are diagrams of bending moments for the most common schemes.
After finding the bending moment, we must find the moment of resistance Wx of this section using the formula given in the table:
Further, when dividing the maximum bending moment by the moment of resistance in a given section, we get maximum voltage in the beam and we must compare this stress with the stress that our beam of a given material can generally withstand.
For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, A for fragile(cast iron) – tensile strength. We can find the yield strength and tensile strength from the tables below.
Let's look at a couple of examples:
1. [i]You want to check whether an I-beam No. 10 (steel St3sp5) 2 meters long, rigidly embedded in the wall, will support you if you hang on it. Let your mass be 90 kg.
First we need to choose design scheme.
This diagram shows that the maximum moment will be at the seal, and since our I-beam has equal section along the entire length, then the maximum voltage will be in the termination. Let's find it:
P = m * g = 90 * 10 = 900 N = 0.9 kN
M = P * l = 0.9 kN * 2 m = 1.8 kN * m
Using the I-beam assortment table, we find the moment of resistance of I-beam No. 10.
It will be equal to 39.7 cm3. Let's convert to Cubic Meters and we get 0.0000397 m3.
Next, using the formula, we find the maximum stresses that arise in the beam.
b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa
After we have found the maximum stress that occurs in the beam, we can compare it with the maximum permissible stress equal to the limit the fluidity of steel St3sp5 is 245 MPa.
45.34 MPa is correct, which means this I-beam will withstand a mass of 90 kg.
2. [i] Since we have quite a large supply, we will solve the second problem, in which we will find the maximum possible mass that the same I-beam No. 10, 2 meters long, will support.
If we want to find the maximum mass, then we must equate the values of the yield strength and the stress that will arise in the beam (b = 245 MPa = 245,000 kN*m2).
In practice, very often there are cases collaboration rod for bending and tension or compression. This kind of deformation can be caused either by the combined action of longitudinal and transverse forces on the beam, or by longitudinal forces alone.
The first case is shown in Fig. 1. The beam AB is subject to a uniformly distributed load q and longitudinal compressive forces P.
Fig.1.
Let us assume that the deflections of the beam compared to the cross-sectional dimensions can be neglected; then, with a degree of accuracy sufficient for practice, we can assume that even after deformation, the forces P will cause only axial compression of the beam.
Using the method of adding forces, we can find the normal stress at any point of each cross section of the beam as the algebraic sum of the stresses caused by the forces P and the load q.
Compressive stresses from forces P are uniformly distributed over the cross-sectional area F and are the same for all sections
normal stress from bending in a vertical plane in a section with the abscissa x, which is measured, say, from the left end of the beam, are expressed by the formula
Thus, the total stress at a point with coordinate z (counting from the neutral axis) for this section is equal to
Figure 2 shows stress distribution diagrams in the section under consideration from forces P, load q and the total diagram.
The greatest stress in this section will be in the upper fibers, where both types of deformation cause compression; in the lower fibers there can be either compression or tension depending on the numerical values of the stresses and. To create the strength condition, we will find the greatest normal stress.
Fig.2.
Since the stresses from the forces P in all sections are the same and evenly distributed, the fibers that are most stressed from bending will be dangerous. These are the outermost fibers in the cross section with the highest bending moment; for them
Thus, the stresses in the outermost fibers 1 and 2 of the middle section of the beam are expressed by the formula
and the calculated voltage will be equal to
If the forces P were tensile, then the sign of the first term would change, and the lower fibers of the beam would be dangerous.
Denoting the compressive or tensile force with the letter N, we can write general formula to check strength
The described calculation procedure is also applied when inclined forces act on the beam. Such a force can be decomposed into normal to the axis, bending the beam, and longitudinal, compressive or tensile.
beam bending force compression
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 in 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 embedding 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.
2. Construction of diagrams of longitudinal forces Nz
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 external load, applied to the cut-off part of the rod under consideration, causes tension and is 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 laid (on the selected scale) above the axis of the diagram, negative ones - below the axis.
3. Construction of diagrams of torques Mkr.
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).
4. Rules for monitoring diagrams Nz and Mkr.
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.
5. Construction of diagrams of transverse forces Qy and bending moments Mx in beams
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 projections external forces, applied on one side of the section under consideration, on the transverse (vertical) axis.
Sign rule for Qy: Let us agree to consider the transverse force in a 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 the specified form, the Mx diagram always turns out to be constructed from the side of the compressed fibers of the beam.
6. Cantilever beams
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.
Longitudinal-transverse bending is a combination of transverse bending with compression or tension of a 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 cross section with the abscissa (we assume the positive direction of the y-axis is downward, and, therefore, we consider the deflections of the beam 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 .
Full deflection more than the amount deflections that arise under the separate action of transverse load and force S, since in the case of only force S acting 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 calculation using the example of a beam hinged 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):
[minus sign in front right side 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 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, as a rule, includes the minimum moment of inertia of the cross section of the rod, and the expression for the Euler force includes the moment of inertia relative to that of the main axes of inertia of the section that 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.
