Beam loaded with longitudinal force. Differential relationships between longitudinal force, load, deformation. Construction of diagrams of longitudinal forces Nz
Bending moment, shear force, longitudinal force- internal forces arising from external loads (bending, transverse external load,tension-compression).
Diagrams- graphs of changes in internal forces along the longitudinal axis of the rod, plotted on a certain scale.
Ordinate on the diagram shows the value of the internal force at a given point on the section axis.
17.Bending moment. Rules (order) for constructing a diagram of bending moments.
Bending moment- internal force arising from the action of an external load (bending, eccentric compression-tension).
The procedure for constructing a diagram of bending moments:
1. Determination of the support reactions of a given structure.
2. Determination of sections of a given structure within which the bending moment will change according to the same law.
3. Make a section of this structure in the vicinity of the point that separates the sections.
4. Discard one of the parts of the structure, divided in half.
5. Find the moment that will balance the action on one of the remaining parts of the structure of all external loads and coupling reactions.
6.Place the value of this moment, taking into account the sign and the selected scale, on the diagram.
Question No. 18. Lateral force. Constructing a diagram shear forces using the bending moment diagram.
Lateral forceQ– internal force arising in the rod under the influence of external load (bending, lateral load). The transverse force is directed perpendicular to the axis of the rod.
The diagram of transverse forces Q is constructed based on the following differential relationship: , i.e. The first derivative of the bending moment along the longitudinal coordinate is equal to the transverse force.
The sign of the shear force is determined based on the following position:
If the neutral axis of the structure on the moment diagram rotates clockwise to the diagram axis, then the shear force diagram has a plus sign, if counterclockwise, it has a minus sign.
Depending on diagram M, diagram Q can take one form or another:
1. if the diagram of moments has the form of a rectangle, then the diagram of transverse forces is equal to zero.
2. If the moment diagram is a triangle, then the shear force diagram is a rectangle.
3. If the diagram of moments has the form of a square parabola, then the diagram of transverse forces has a triangle and is constructed according to the following principle
Question number 19. Longitudinal force. A method for constructing a diagram of longitudinal forces using a diagram of transverse forces. Rule of signs.
Weaving force N is the internal force arising due to central and eccentric tension-compression. The longitudinal force is directed along the axis of the rod.
In order to construct a diagram of longitudinal forces you need:
1.Cut the node of this design. If we are dealing with a one-dimensional structure, then make a section on the section of this structure that interests us.
2.Remove from the diagram Q the values of the forces acting in the immediate vicinity of the cut node.
3. Give direction to the vectors of transverse forces, based on the sign of the given transverse force on diagram Q according to the following rules: if the transverse force has a plus sign on diagram Q, then it must be directed so that it rotates this unit clockwise, if the shear force has a minus sign, counterclockwise. If an external force is applied to a node, then it must be left and the node must be considered together with it.
4. Balance the assembly using longitudinal forces N.
5. Sign rule for N: if the longitudinal force is directed towards the section, then it has a minus sign (works in compression). If the longitudinal force is directed away from the section, it has a plus sign (works in tension).
Question No. 20. Rules used to check the correctness of constructing diagrams of internal forcesM, Q, N.
1. In the section where a concentrated force F is applied, the diagram Q will have a jump equal to the value of this force and directed in the same direction (when constructing the diagram from left to right), and the diagram M will have a fracture directed in the direction of the force F .
2. In the section where a concentrated bending moment is applied on the diagram M, there will be a jump equal to the value of the moment M; there will be no changes on the Q diagram. In this case, the direction of the jump will be downwards (when constructing a diagram from left to right) if the concentrated moment acts clockwise, and upwards if counterclockwise.
3. If in a section where there is a uniformly distributed load, the shear force in one of the sections is zero (Q=M"=0), then the bending moment in this section takes on an extreme value M extra - maximum or minimum (here tangent to the diagram M horizontal).
4. To check the correctness of constructing the diagram M, you can use the method of cutting out nodes. In this case, the moment applied in the node must be left when cutting the node.
The correctness of constructing the Q and M diagrams can be checked by duplicating the method of cutting out nodes using the section method and vice versa.
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 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 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 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 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 constructing 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.
Calculate bending beam There are several options:
1. Calculation of the maximum load that it will 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.
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 select a 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 yield strength of steel St3sp5 - 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).
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.
Keywords: 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. 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 ± u)-- + 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 on the left support), 2k - height cross section, b - cross-section width, 21 - beam span, 5^ - yield strength of the 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 natural vibration frequencies 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.
In practice, very often there are cases of joint work of a rod in 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 only cause axial compression of the beam.
Using the method of adding forces, we can find normal voltage at any point of each cross section of the beam as the algebraic sum of stresses caused by forces P and load q.
Compressive stresses from forces P are uniformly distributed over the cross-sectional area F and are the same for all sections
normal bending stresses in 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
