Working hypotheses of SOPROMAT
DETERMINATION OF FORCES IN STEP BARS WITH SEVERAL POWER SECTIONS.
ANSWER: Tension or compression of a bar is its type of deformation, in which all external forces are directed along the longitudinal axis, and in the cross sections there is a single internal force factor - the longitudinal force N. Its value is determined using the method of sections: 
, i.e. the longitudinal force in the considered section of the beam is numerically equal to the algebraic sum of all external forces located on one side of the section. We agreed to consider N>0 if it is directed away from the section, i.e. stretches and N<0, если сжимает. Для полного суждения о прочности бруса необходимо построить график изменения продольной силы по длине бруса – эпюру продольной силы (Эп.N).
Stresses and strains in axial tension-compression
In the process of deformation in the cross sections of the beam during axial tension-compression, only normal stresses σ arise, and they are distributed evenly over the cross section.
During tension-compression, the beam experiences only linear deformations.
– absolute longitudinal deformation of the load (elongation)
Relative longitudinal deformation
- absolute transverse deformation (narrowing)
Relative transverse deformation
Relationship between and : - Poisson's ratio.
A linear pattern linking stresses and strains is Hooke's law in axial tension-compression.
Definition of geom. characteristics.
Consider the definition of geom. characteristics, for the most common shaft cross sections.
1.Solid shaft
2.Hollow shaft
3.Thin-walled tubular section
To thin-walled pipes with the ratio 
Determination of movements during bending. Rigidity condition. Differential equation of the bent axis of the beam.
Calculation options for simple statically indeterminate beams
There are several ways to calculate simple beams:
1. Comparison of linear displacements.
ΔВ=ΔВq+ΔBRB=0(1) add. strain equation
The terms in (1) can be found using ready-made tables or universal equations. With regard to a racial subject:
ΔBq=-qe 4 /8EIx; ΔBRB=RBe 3 /3EIx;
ΔB=-qe 4 /8EIx +RBe 3 /3EIx =0 =>RB=3qe/8
2. Comparison of angular displacements.
You can discard the constraint that prevents the rotation of the reference section A and write
ΔA=ΔAq+ΔAMA=0(2)
Also ur-e deformation term means angles of rotation.
3. Compilation of a closed system of ur-I.
3 ur-I statics + unives. ur-e
43. The method of forces for the calculation of complex SNA.
The method in which concentrated moments are taken for the unknown is called the method of forces. It is the most common and used for any elastic systems (beams, frames, flyovers, etc.).
For instance: 
To the three levels of statics for solving this CNN, 3 equations will be added, expressing the equality of 0 displacements in the directions of all discarded connections, i.e. reference section and are not moved by it in horizontal or vertical movements and are not turned over.
X1 ∆1=0
X2 ∆2=0 (1)
X3 ∆3=0
Each equation of system (1) can be written in expanded form:
Δ1=Δ11+Δ12+Δ13+Δ1f=0 (2)
The first character indicates the direction; 2nd voz-e.
Δ1f-displacement of the support section A in the direction of action X, caused by an external load
(2) can be expressed in terms of single movements and the desired unknown (these are the first three terms)
Δ11=δ11-x1 and then the system will be finished. view.
δ11 x1+ δ12 x2+ δ13 x3+ Δ1f=0
δ21 x1+ δ22 x2+ δ23 x3+ Δ2f=0 (3)-coomes system.
δ31 x1+ δ32 x2+ δ33 x3+ Δ3f=0
Canonical ur-I method of forces-KUMS.
The number y-th is equal to the degree of static indeterminacy.
Working hypotheses of SOPROMAT
ANSWER: In contrast to termekh, based on the abs. of a rigid body, in the strength of materials, its own calculation model is adopted - a model of an idealized deformable body. And to simplify the calculations, the following assumptions or hypotheses are accepted: 1) The material of the body has a continuous structure. 2) the material is homogeneous, i.e. properties are the same at all points. 3) the material is isotropic, i.e. properties are the same in all directions. 4) before the application of external forces, there are no initial stresses in the material. 5) when solving real problems, it is advisable to use the principle of superposition, or the principle of independence of the action of forces, i.e. the impact on the structure of a group of forces is equal to the sum of the impacts from each force separately and does not depend on the sequence of application of these forces.
INTERNAL FORCE FACTORS AND THE METHOD OF THEIR DETERMINATION.
ANSWER: Under the action of external forces on the beam, internal forces or internal force factors arise, for the determination of which a single calculation method is adopted in the strength of materials - the method of sections. 1) mentally cut the beam in the section under study into 2 parts I and II. 2) We discard one of the parts. 3) We replace the action of the discarded part II with part I by internal force factors (in the general case there are 6 of them). Q x Q y - transverse forces, N z - longitudinal force, M x M y - bending moments, M z - torque. 4) We balance the remaining part of the beam and, using the equilibrium equations of the termekh, we find the desired force factors.
THE CONCEPT OF STRESSES, DEFORMATIONS AND DISPLACEMENTS.
ANSWER: A measure of the intensity of the action of internal forces in the vicinity of the point of the considered cross section are stresses determined by the ratio of force to unit area [Pa]. If in pop section, select the element DA, to which the force DP will be applied, then DP / DA \u003d p m is the average total stress at the considered point of the cross section. is the total true voltage. The vector is decomposed into and . - normal stress - causes destruction by separation. - shear stress - causes destruction by shear. Displacements and deformations are concepts that characterize changes in the size and shape of the body under study. In this case, displacements are a consequence of deformation.
Under the action of external loads in the section of the structure (rod, beam, etc.), additional forces arise, which are called internal force factors and which are determined by the section method. This is the reaction of the connection of one cut-off part to another, the reaction of the support on the body, the reaction of the flexible connection, etc. The forces of the cut-off part on the considered structural element in relation to it are external forces and are determined by general equations balance.
5. What types of beam deformation determine internal force factors?
Using the method of sections, internal force factors are determined: main vector and the main moment are decomposed into components 
, which define the following types of deformation:
1) Tension (compression) - longitudinal force 
, and all other components are equal to zero.
2) Shear (shear) - shear force 
or 
3) Torsion - torque 
, and all others are equal to zero.
4) Bend - when or 
, or 
, and the rest of the components are equal to zero.
5) Complex resistance - when the combination of any internal efforts is not equal to zero.
What is meant by mechanical stress and what is its dimension?
The stress on a given area is the intensity of internal forces transmitted at a point through a selected area.
The total stress on this area is decomposed into normal and shear stresses, and 
. Stress has the dimension of load intensity, i.e. MPa (kgf/cm2, tf/m2).
1 MPa=106Pa=106N/m2.
Give formulas relating internal force factors to stresses.
Normal and tangential stresses in each cross section of the beam are connected by certain ratios with internal forces acting in this section:
In formulas, the coordinates of the point at which stresses are determined.
What type of deformation is called tension (compression)?
Tension (compression) is such a type of deformation, when in the cross section of the rod under the action of external loads there is only one internal force factor - longitudinal force, and there are no other internal force factors.
The longitudinal force induces normal stresses defined by:
With a uniform distribution over the cross section 
With uneven distribution 
The longitudinal force and stress are positive in tension and negative in compression.
Absolute and relative strain in tension (compression). Poisson's ratio.
If, under the action of a force, a beam in length has changed its longitudinal value by , then this value is called absolute longitudinal deformation (absolute elongation or shortening). In this case, transverse absolute deformation is also observed.
Attitude 
is called the relative longitudinal strain, and the ratio 
- relative transverse deformation.
Attitude 
is called the Poisson's ratio, which characterizes the elastic properties of the material.
Poisson's ratio matters 
. (for steel it is equal to 
)
Formulate Hooke's law in tension (compression).
I form. In the cross sections of the beam under central tension (compression), the normal stresses are equal to the ratio of the longitudinal force to the cross-sectional area:
II form. Relative longitudinal strain is directly proportional to normal stress 
, where 
.
Consider the design scheme of a beam with an arbitrary distributed load (Fig. 2).
Fig.2. Beam bending scheme:
a) calculation model, b) beam fragment
Let's make an equilibrium equation:
Thus, indeed: the first derivative of the internal bending moment with respect to linear coordinate equal to the transverse force in the section.
This well-known property of a function and its first derivative is successfully used to check the correctness of plotting. So, for the design scheme of the cantilever beam (Fig. 1), this connection gives the following verification results:
AND M decreases from 0 before - pl.
AND M x.
Let's consider the second characteristic example of the bending of a two-bearing beam (Fig. 3).
a) calculation scheme, b) model of the first section, c) model of the second section, d) diagram of transverse forces, e) diagram of bending moments
Fig.3. Bending of a two-support beam:
Obviously, the support reactions R A = R B:
- < б) (рис.3 участка первого>
- for the second section (Fig. 3 c) -
Diagrams of internal forces are presented respectively in Fig. 3d and 3 days
Based on differential link Q and M, we get:
- for the first section:
Q > 0 and M increases from zero to .
Q= const and M x
- for the second section:
Q< 0 and M decreases from to zero.
Q = const and M also proportional X, i.e. changes linearly.
Dangerous in this example is the section of the beam in the center of the span:
The third characteristic example is associated with the use of a load distributed along the length of the beam (Fig. 4). Following the methodology adopted earlier, the equality of the support reactions is obvious: , and for the desired section (Fig. 4 b), the expressions for internal forces take the form:
a) calculation scheme, b) cut-off part, c) diagram of transverse forces, d) diagram of internal bending moments
Fig.4 Double support beam with uniformly distributed load:
There is no bending moment on both supports. Nevertheless, the dangerous section of the beam will be the center of the span at . Indeed, based on the properties of the function and the derivative at , the internal bending moment reaches an extremum. To find the original coordinate x 0(Fig. 4 c) in the general case, we equate the expression for the transverse force to zero. As a result, we get
After substituting the bending moment into the expression, we get:
In this way,
It should be noted that the technique of constructing curves during bending is most difficult for listeners to master. You have the opportunity to learn how to “quickly” build diagrams on the simulator tester given in the APPENDIX and solve the positions familiar to you from the problem statement in the final tests on the strength of materials.
Lecture number 5. The concept of stresses and strains
As noted above, the internal forces acting in a certain section from the side of the discarded part of the body can be reduced to the main vector and the main moment. Fix a point M in the section under consideration with a unit normal vector n. In the vicinity of this point, we select a small area F. The main vector of internal forces acting on this site will be denoted by P(Fig. 1 a). With a decrease in the size of the site, respectively
Fig.1. Stress vector composition.
a) total stress vector b) normal and shear stress vector
the main vector and the main moment of internal forces decrease, and the main moment decreases to a greater extent. In the limit at , we obtain
The analogous limit for the principal moment is zero. The vector introduced in this way p n called stress vector at a point. This vector depends not only on the external forces acting on the body and the coordinates of the considered point, but also on the orientation in the space of the site F, characterized by the vector P. The set of all stress vectors at a point M for all possible vector directions P determines the stress state at that point.
In general, the direction of the stress vector p n does not match the direction of the normal vector P. The projection of the vector n onto the direction of the vector n is called the normal stress, and the projection onto the plane passing through the point M and orthogonal to the vector n , - shear stress(Fig. 1 b).
The dimension of stress is equal to the ratio of the dimension of force to the dimension of area. In the international system of SI units, stresses are measured in pascals: 1 Pa \u003d 1 N / m 2.
Under the action of external forces, along with the occurrence of stresses, a change in the volume of the body and its shape occurs, i.e., the body is deformed. In this case, the initial (undeformed) and final (deformed) states of the body are distinguished.
We refer the undeformed body to the Cartesian coordinate system Oxyz(Fig. 2). The position of some point M in this coordinate system is determined by the radius vector r(x, y, z). In the deformed state, the point M will take a new position M / , characterized by the radius vector r" (x, y, z). Vector u=r"-r called vector, displacement points M. Vector projections u on the coordinate axes determine the components of the displacement vector u(x, y, z), v(x, y, z), w(x, y, z), equal differences Cartesian coordinates body points after and before deformation.
A movement in which the relative position of the points of the body does not change is not accompanied by deformations. In this case, the body is said to move as a rigid whole (linear displacement in space or rotation about a certain point). On the other hand, deformation associated with a change in the shape of a body and its volume is impossible without moving its points.
Fig.2. Movement vector composition
Body deformations are characterized by a change relative position body points before and after deformation. Consider, for example, the point M and a point close to it N, the distance between them in the undeformed state along the direction of the vector s will be denoted by (Fig. 2). In the deformed state of the point M and N move to a new position (points M" and N'), the distance between which will be denoted by s". ratio limit
called relative linear deformation at the point M in the direction of the vector s, Fig.3. Considering three mutually perpendicular directions, for example, along the coordinate axes Ooh, ooh and Oz, we obtain three components of relative linear deformations characterizing the change in the volume of the body in the process of deformation. associated with the rotations of the segments
1. Basic concepts in sopromat.
Tasks and methods of sopromat.
All structural elements have strength and rigidity.
The tasks of the strength of materials: the creation of methods for assessing strength.
Sopromat is characterized by approximate methods of calculation.
Calculation schemes and models.
Strength assessment is carried out according to the scheme (model).
Model- a set of basic representations from the main description of the object.
For the same part, several similar schemes can be drawn up. At the same time, for one calculation scheme, one can find various details of material schemes, forms, loading and unloading forces.
reliability models.
material models.
The material is homogeneous, continuous, continuous (you can apply mathematical formulas), isotropic.
Homogeneity of the material - the material is the same throughout the volume.
The calculation model of the material has the properties of elasticity, plasticity and creep.
Elasticity- the property of a material to restore its shape.
Plastic- the property of the body to maintain a changed shape.
Creep- the property of the body to change shape over time (resin).
Form models.
The geometric shape of the bodies is very complex. It is not possible to take into account all forms in the formulas, therefore they are led to 4 schemes:
1. Rod, bar.
2.Plate.
3.Shell.
Varieties of form.
Kernel- the shape of a part in which one size is an order of magnitude larger than the other two.
plate- the shape of a part in which one size is an order of magnitude smaller than the other two.
array- all sizes are different, but differ less than an order of magnitude.
Loading models.
Power is a measure of interaction between two bodies.
Strength is external and internal. External in turn is concentrated, distributed and voluminous.
Concentrated - force applied to small area, which can be considered as a point.
Distributed - a force acting on a significant surface, the size of which must be taken into account.
Volumetric - force distributed over the entire mass of the body.
Force time models.
Distinguish
1. Static
2. Variables
a) Low cycle
b) Multi-cycle (more than 100 thousand changes)
destruction models.
Part destruction- changing its shape into flesh before splitting into parts.
A change in shape and separation into parts will occur when the internal forces exceed the cohesive forces of the individual parts of the material.
To judge strength, internal forces are compared with strength limits. Internal forces are the forces of interatomic interaction arising from the action of external forces.
Consider a body (a), which is in equilibrium under the action of external forces, mentally cut this body into 2 parts by the plane P and consider 1 of them (b). The action of one of them on the other should be replaced by a system of internal forces in the section. Internal forces in sections of body parts are always mutual (the action is equal to the reaction). In sopromat, bodies that are in equilibrium are studied.
To find the resultant (R) and moment (M), we use the equilibrium equations.
We project R and M onto the selected coordinate axes.
The cut off part is in balance
Let's take the coordinate system xyz and decompose and into its component parts.
Then the projections and M on these axes are called internal force factors.
Longitudinal force, - transverse force.
Torque, - bending moments.
To calculate internal forces. Factors needed to solve 6 equilibrium equations.
Stress and strain.
Voltage- intensity of internal forces. factors.
is the total stress at the point.
Point voltage
Tangential and normal stresses.
We decompose the force ΔR into components ΔN - normal and ΔQ - tangential force.
σ is normal and τ is shear stress.
Voltage has the name of force divided by area (N/).
In the SI system, it is expressed in Pascals (Pa).
Relationship between stress and internal force factors.
N-longitudinal force causing bar stress
Shear forces causing shear.
Torque - twisting
Bending moments - curvature of the longitudinal axis.
If a force acts on a body, it means that it is deformed. In the sopromat, all bodies are deformed, but they are extremely small.
Central stretch - compression.
Longitudinal strength.
stretching- type of deformation, in which an internal longitudinal force N arises in the cross section of the rod, while the length increases, and the width decreases.
Under tension conditions, the rod will be under the action of axial forces at the edges (a). The resultant of the system is equal to F.
To determine the longitudinal internal force N, the section method is used.
To determine N in an arbitrary section x of the rod a) consider the equilibrium of the upper cut-off part b). We compose an equilibrium equation, substituting the values we get
The "+" sign indicates that the bar is stretched.
Plot of longitudinal forces.
To judge the strength of the rod, you need to know the longitudinal force at any point.
The graph (epure) of the change in internal forces is on a line drawn parallel to the axis of the rod. Each ordinate of the plot is equal to N.
Plot- some length of the rod, on which there is no change in area or forces.
Let the rod OAB be loaded with forces and have 2 sections OA and AB, sections are selected on them at a distance and from the origin of coordinates. In section longitudinal force
in section
Voltages.
The force N applied at the center of gravity of an arbitrary section of the rod is the resultant of internal forces acting on an infinitely small area dA of the cross section of area A and. Then,
Within the limits of Hooke's law (), the flat cross sections of the rod during deformation are displaced parallel to the initial position, remaining flat (the hypothesis of flat sections), then the norms. the stress at all points of the section is the same, i.e. (Bernoulli's conjecture) and then
When the rod is compressed, the stress has only a different (negative) sign (the normal force is directed into the body of the rod).
Deformation.
A rod of constant cross section with area A under the action of axial tensile forces is extended by an amount, where are the lengths of the rod in the deformed and not deformed state. This length increment is called full or absolute extension.
Relative extension- elongation referred to the original length of the rod ref. linear deformation. ε is measured in%.
Under tension (compression), not only longitudinal, but also transverse deformation of the rod occurs, where a is the transverse dimension.
The ratio of transverse to longitudinal deformation, taken in absolute value, is called Poisson's ratio.
Hooke's law. Rod extension.
There is a linear relationship between stress and small strain, called Hooke's law. For tension (compression) it has the form σ=Еε, where Е is the coefficient of proportionality, elastic modulus.
E is the stress that causes deformation.
Hooke's law for tension (compression) of a rod.
Δl=Fe/EA=λF, where λ is the longitudinal compliance coefficient of the rod.
EA is the tensile stiffness of the rod section.
For a bar with a variable (stepped) section, the elongation is determined by sections (steps) and the results are summarized algebraically:
Material testing diagram.
In calculating the strength of the rod in tension and compression, it is necessary to know the mechanical. Material properties that are revealed during tensile testing of specimens under load. The tensile test makes it possible to judge the behavior of a material under compression, shear, torsion and bending. The graph of the relationship between the tensile force F and the sample elongation Δl is called stretch diagram.
To eliminate the dependence on dimensions, the diagram is rearranged in the coordinates σ - ε.
Characteristics of strength and fluidity.
T.A - proportionality plot (Hooke's conservation law).
Up to t. C - the fluidity of the material.
T. V - max value.
OA - elasticity,
AD - plasticity,
DV - hardening,
VM - local fluidity.
Hooke's law is valid in the OA zone
The value of the elastic limit is close to the limit of proportionality.
The AD zone is the zone of general plasticity. It is characterized by a significant increase in the deformation (length) of the sample without a noticeable increase in load - the yield point (YL). The formation of plastic deformation is caused by a shift in the crystal lattice.
To assess the tension use the characteristic mechan. material properties - yield strength - stress at which a noticeable elongation appears in the material without increasing stress.
Tensile strength.
DV zone - hardening zone; here, the elongation of the sample increases more intensively with increasing load compared to the OA zone. In t. In the stress σ reaches a maximum.
If the sample is loaded at point F, then upon subsequent loading, the material acquires the ability to perceive large loads without residual deformations.
The phenomenon of increasing the elastic properties of a material as a result of preliminary deformation is called work hardening.
The VM zone is called the local fluidity zone. Here, the elongation of the sample occurs with a decrease in force and is accompanied by the formation of a local narrowing - necks. The stress in the cross section of the neck increases. In t. M, the destruction of the sample occurs. The maximum stress on the diagram that the sample can withstand is called the tensile strength (tensile strength).
plasticity and brittleness.
Under plasticity understand the ability of a material to receive large residual deformations without destruction.
fragility- the ability of the material to collapse without the formation of noticeable residual deformations.
Permissible voltages. Design structures.
The tensile strength condition is written in the form, where [σ] is the allowable stress, which is a characteristic of the structural material, which depends on the accepted safety factor n.
n - a value showing how many times the ultimate stress for a given material is greater than the workers [σ]
As a rule, the yield strength (strength) is taken as the ultimate stress.
Shear and twist.
Main questions:
1. The concept of shift
2. Hooke's law in shear
3. Engineering calculations for the shear of the beam material
4. The concept of torsion of a round bar
5. Expressions for shear stresses of angles of twist
6. Strength and stiffness condition
7. Definition of dangerous sections
8. Engineering calculations for torsion.
Internal force factors and deformations. Shear - a type of deformation, when only a shearing force acts in the cross section of the rod, other force factors are absent. Elementary cubes are distorted, stress arises on the side faces.
Shift scheme. Hooke's law. Stress state, at k-m on the faces is highlighted. element, only tangential stress occurs, called pure shear. a-absolute shift, -angle, on which the right angles of the element change, is called relative shift.
Equilibrium equation of the cut-off part, where G is the modulus of elasticity, GA is the shear stiffness of Hooke's shear,
Calculation of structures for shear. Many parts (glued, welded,...) are subject to shear.
Strength condition, - allowable shear stress.
\u003d (0.5 ... 0.6) - for plastic. materials
=(0.7...1.0) - for brittle materials
Torsion.
Torsion is a type of deformation, at which only torque acts.
Internal force factors. To build a diagram, they are divided into sections, dissecting sections at distances x 1, x 2, ... A diagram showing the distribution of cool values. moments along the length of the shaft is called the torque diagram. Sign rule: counterclockwise torque is positive, clockwise is negative.
Plotting torques. Equivalence equation or -right side is similarly considered for all sections.
Conclusion: in any section of the shaft, a torque acts, = the sum of the torques lying on one side of this section. The diagram of torques is a stepped line, which shows the degree of loading of each of the sections of the shaft.
Torsional deformations. During torsion, the generatrices of the cylinder turn into helical lines, round and flat sections retain their shape, the rotation of one section relative to the other occurs at a certain angle of twist, the distance between the cross sections almost does not change. Sections that are flat before twisting remain flat after twisting; the radii of cross sections remain straight during deformation.
Torsion is the result of shifts during mutual rotation of sections.
Beam loading scheme.
Where is the angle of twist per unit length of the rod.
Relates twist angle.
Shear geometry.
Values touch. stresses at the points of the section are proportional to dist. it from the axis of the rod.
Torsional moment.
Torsional stress.
Geometric characteristic is the polar moment of inertia of the section.
Angle of twist per unit rod. is the polar moment of section modulus.
Polar moment of inertia and resistance.
Polar. moment of inertia. , for round section -
Calculation formulas., stiffness condition:
Calculations for strength and stiffness.
Strength condition: .Solid shaft diameter
Angle of twist - determines the rigidity.
The shaft is calculated according to 2 conditions and the larger one is found from the found values.
bend.
Main questions:
1. classification of bends
2. loads and internal force factors
3. construction of load diagrams, the rule of signs
4. normal stresses in pure bending
5. shear stresses in pure bending
6. bending movement
7. differential equation of the elastic line of the beam
8. determination of displacements by direct integration
Classification of bends. Bending is a type of deformation when, under the action of external forces, bending moments occur in the cross section of the rod (beam).
If the bending moment in the section is the only force factor, and there are no transverse and normal forces, it is called clean. If, along with bending moments, transverse forces also act in the cross sections of the rod, the bend is called transverse.
Sometimes several force factors appear in the transverse rod. This is a difficult resistance. The calculations of the rods are based on the principle of independence of the action of forces.
Supports and their reactions. To transfer loads, the rod must be fixed relative to the body with the help of supports - devices that perceive external forces.
There are 3 main types of supports - rigid pinching: 1) sealing - a) excludes axial, angular displacements and perceives axial forces and moment load,
2) hinged-fixed support -b), - allows rotation around the axis and does not perceive the moment,
3) hinged-movable support -c), - does not allow displacement of the rod, only in direction 1 of the axes and transfers the load along this force.
Support reactions. Under the influence of external Loads in the places where the rod is fixed, a support reaction occurs. x is found from the equilibrium conditions. The analysis of internal forces begins after the reaction is determined.
Internal force factors. A rod on 2 supports, loaded with forces F. From the equilibrium condition, we find the support reactions: . Under the influence of external forces and support reactions, the rod b) will be in equilibrium. To determine the internal force factors in the section m1-mi of the section CD of the rod, we mentally cut it into 2 parts, consider the balance of the left c). In order for it to be in balance, we apply unknown internal force factors to point Ci: normal force Nx(xi), cutting, bending moment.
Sign rule. Positive the bending moment bends a horizontally located rod (beam) with a convexity downwards (a), and negative. - bulge up (b).
Positive the transverse force tends to move the left section of the rod up relative to the right or right down relative to the left (a). Negative the transverse force has the opposite direction (b).
Definition of force factors. Shearing force in the section of the rod = the sum of the projections onto the axis of all external forces acting on the mentally cut off part, i.e. . Bending moment in bar section is equal to the sum moments of external forces acting on the cut-off part, taken relative to the center of gravity of the considered section, i.e.
Ur-I statics:, (pure bend). If we make a section m2-m2 on section AC and consider the balance of the left side, then we find that for force factors: (transverse bending)
Schematic of pure bending. Application fields M of the longitudinal force - the arc of a circle, the cross section remains flat, i.e. the hypothesis of flat sections is valid. With pure bending, the fibers on the convex side are stretched, on the concave side they are compressed. There is a layer in which there is no elongation, it is called the neutral layer - the neutral line.
Relationship between stresses and internal factors. Assume that a rod is a set of stretched and compressed elements of rods of length l which are free to lengthen and shorten. Normal stresses are used constant across the width of the section.
The static part of the task. Balance condition between force factors:
Conditions b), c), d) are satisfied identically, conditions a), f), e) have the form: .
fiber deformation., - relative elongation of the layer.
The deformation of some layer depends on its coordinates z, counted from the neutral layer. We use Mr. Hooke: . The ratio is constant for a particular material and a particular case of bending. Therefore stresses are a linear function of the z coordinates. To find the value, you need to know the position of the neutral layer or the radius of curvature.
Normal bending stress.
From equations a), e), f), taking into account k.
From equation a), because then is the static moment of the cross-sectional area. The neutral axis is the central axis. From equation e) we get This is the centrifugal moment of inertia, if it \u003d 0 - the main, central axes. From ur-i e):
where. The calculation formula is obtained by substituting into the last dependence from the formula k.
Calculation formulas.
strength condition:
As follows from the distribution characteristics, the stressed inner layers of the material are underloaded.
Force factors in transverse bending. The sopromat hypotheses extend to transverse bending.
Shear stress formula. We express the forces in terms of normal stress, and the stress in terms of bending moments, taking into account the longitudinal force that causes shear stress, we obtain:
Where A0 is the area of the cut-off part. -static moment of the cut-off part. On the surface in the center = max.
The nature of movement during bending. When bending, there are 2 types of movement: linear and angular.
For small movements.
Curved Axis Equation.
Differential Ur-e of the curved axis of the beam.
Fundamentals of the directed state of the material.
Main questions:
1. types of stress state
3. law of pairing of shear stresses
4. main platforms and main stresses
5. volumetric deformation. Hooke's law
6. specific potential energy
7. plasticity and fracture criteria
8. equivalent voltages
9. strength hypotheses
Types of stress state. The assessment of the strength of a part is a combination of the stress state at the “dangerous” point of the structure with the tensile strength of the material. Such an estimate turns out to be sufficiently accurate for a single-basic stress state (tension, compression).
However, many structural elements operate in a complex stress state. Then the set of stresses at the point of the element are comparable with the mechanical characteristics of its material, that is, an equivalent stress is introduced, i.e. stress in a stretched sample at which the state is equally dangerous with the given one.
Efforts on slopes. The stretched rod is cut by a plane inclined to the cross section at an angle. From the equation of equilibrium follows the resultant of internal forces, in an inclined section, the direction along the axis of the rod and is equal to the external force, I.e.
Let's break it down into components:
Normal
And tangent
The area of the inclined section (A is the area of \u200b\u200bthe normal section).
Normal and shear stresses are uniformly distributed along the inclined section. The maximum normal stresses act in the cross sections of the rod (α=0)
Stresses on inclined mutually perpendicular planes. In inclined sections, normal and shear stresses act simultaneously, which depend on the angle of inclination α. On sites at α=45 and 135 degrees. At α=90, both normal and shear stresses are absent. It is easy to show that the perpendicular section at
Conclusion: 1) in 2 mutually perpendicular planes, the algebraic sum of normal stresses is equal to the normal stress in the cross section
2) tangential stresses are equal to each other in absolute value and proportional in direction (sign) the law of pairing of stresses
Double stretch. Let the element isolated from the body be subjected to normal stresses. Obviously, the direction δ1 and δ2 are the main stresses. Such a stress state is called biaxial or flat. Let us draw an oblique section α, the normal to which forms, with the largest of the normal stresses δ, angle α, counting it as a positive angle counterclockwise.
The area α will be subject to normal and shear stresses. Under the action of only δ1, we obtain
Under the action of δ2:
Stresses under biaxial tension. Under the joint action of δ1 and δ2, it is easy to see:
On site β:
If one voltage takes a maximum, then the second minimum. In this position, the shear stress is zero.
Main sites. Select from the element oblique triangular prism and consider its equilibrium, projecting the forces on the normal and the tangent to the inclined area. ; ;
Investigating the expression for an extremum, one can make sure that the extremum condition for δα coincides with the condition that shear stresses on these areas are equal to zero.
Principal stresses . Normal stresses on these sites are called principal. Principal stresses and the position of principal areas can be found from the first equation. To determine the main areas, we equate the second equation to zero.
The magnitude of stresses on these sites
Volumetric deformation of the material . The volumetric deformation is called the deformation of the element under the action of mutually perpendicular stresses, and it is assumed that δ1> δ2> δ3
Hooke's law is used to determine the strain in the stress of principal stresses. For a linear stress state, the relationship between longitudinal and transverse strain and the principle of independence of the action of forces.
The stress δ1 causes longitudinal and transverse deformation in the directions δ2 and δ3:
Similarly, from the action of δ2 and δ3:
Generalized Hooke's law.
Summing up the deformations of one stress has the form after the transformation of the main deformations:
The ALL equation is a generalized Hooke's law for a volumetric stress state. Changed volume of the element when deforming a single size:
Relative volume change:
Potential energy. Equivalent voltages . The work required to stretch the beam is equal to:
FOR A SINGLE ELEMENT
Generalized formula for volumetric deformation:
To assess the strength, it is necessary to compare the stress at a point in the structure with a complex (flat, bulk) stress state, it is necessary to compare it with the mechanical characteristics of its material, i.e. it is necessary to establish some equivalent stress, which should be created in a stretched sample, so that its stress state is equally dangerous with the given one.
strength hypotheses. 1st hypothesis. A number of strength hypotheses have been developed to assess the hazard of a material under complex stress. Important 4:
1. the greatest normal stresses, that is, δequiv.=δ1, δ2, δ3 are discarded. We compare it with the limiting δequiv.≤δ0(limiting), that is, δ1≤[δ] is in good agreement with the tension of the rod.
2. The theory of the largest linear deformations is destroyed when εmax=ε1≤ε0 after substitution δequiv.=δ1-ν(δ2+δ3)≤[δ] Can be used for brittle materials.
3. Theory of the greatest shear stresses. The material is destroyed if the shear stress reaches the limit. When the beam is deformed from stress δ1, δ2, δ3, the tangential stress is determined by
After replacing the stresses with their values δ1>δ2>δ3, then the greatest shear stress
Well confirmed for ductile materials (steel).
4. energy theory: if the volume change energy does not exceed the limit and the material is strong. From the theory of volume change
For the case of torsion with bending, the form
Complex loading. Oblique bend. Bending with twist .
Main questions :
1. complex loading
2. oblique bend
3. neutral axis
4. determination of displacements and stresses.
5. eccentric stretching
6. determination of stresses in off-center tension
7. torsion bending of shafts
8. definition of effort
9. determination of stresses and calculation of shafts
The concept of complex resistance. Complex resistance is a loaded type, in which 2 or more force factors act in the cross sections of the beam. Lateral bending is also a complex resistance.
In general, there are 6 force factors in the cross section. Previously, methods for calculating stresses and displacements from each of these factors were considered.
Under the action of several factors, using the principle of superposition for a certain total result.
The most common types of compound resistance are oblique bending, eccentric tension, and torsional bending. The most common application and theory of strength for parts.
Scheme of complex loading. The rod is loaded with force F having angles α, β, γ with the coordinate axes, the origin of which is in the center of gravity of the cross section. The Y and Z axes are the main central axes of inertia. The projections of the force F on the coordinate axes are found. Using the method of sections, we establish that the rod works for bending in 2 planes and for axial tension. If the force F is located in the plane of the cross section, then Fx will be absent.
Oblique bend. Oblique bending is understood as such a case of bending, in which the plane of the bending moment does not coincide with any of the main planes of inertia of the rod.
The task of oblique bending is reduced to the simultaneous consideration of two flat (straight) bends, revealing the bending moment in the section by 2 moments, acting in the main planes (pass through the main axes of the section). the stress from force Q is the second order stress from bending.
Scheme of forces in oblique bending.
On fig. a cantilever rod is shown, loaded with a force F, acting perpendicular to its axis and making an angle φ with the main plane xy. Stress at some point B of the cross section at a distance x from the loose end. Moments bending the rod in the vertical and horizontal planes x.
Where Fu and Fz are the vertical and horizontal components of the force.
F,M - component moments in the section
Stresses and the neutral axis in oblique bending. The normal stress at a neutral point with coordinates y and z is determined by the sum of the stresses from the moments Mu and Mz i.e.
The maximum voltage will operate at the points furthest from the neutral line.
The position of the neutral line at an oblique bend is found from the canopy equation δ=0, denoting the coordinates of the neutral line Y0 and Z0 we get
It can be seen that the neutral line is a straight line passing through the origin of coordinates (the center of gravity of the cross section), denoting through α the angle of inclination of the neutral line to the Z axis, we find
Eccentric tension. With an eccentric tension of the rod, the resultant of external forces does not coincide with the axis of the beam, but is shifted relative to the X axis.
In an arbitrary cross section of the rod, internal force factors will act:
YF,XF - COORDINATES OF ANY POINT
MX,MY - bending moments about the axes of the section
Scheme of forces and stresses in the section. To determine the normal stress at an arbitrary point, we find its components from each factor for the case of eccentric tension.
The normal stress is:
THOSE. the stress diagram is a plane.
The position of the point K and the voltage diagram from each factor are shown in fig. To determine the neutral axis, we replace the moments of forces with their value and equate: From here we find the coordinates YK,XK
Stretch bend. In general, both longitudinal and transverse loads can act on a rod.
If we assume a combination of the oblique bend considered above with axial tension or compression, then such loading leads to the appearance of bending moments MX,MY of transverse forces Qz Qy and longitudinal force N in the cross sections of the rod. For example, the following force factors will act in the cross section of the cantilever rod (without taking into account the sign rule).
Stress in the rod. Normal stress causes a tensile force FX , in all cross sections of the rod is equally and evenly distributed over the section. This voltage is determined by the formula: , where A is the cross-sectional area of the rod
Applying the principle of independence of the action of forces, taking into account the previously obtained formula, the normal stress at an arbitrary point С
Then the greatest stress δmax in the cross section:
The transverse condition for allowable stresses in the design cases has the form: δmax≤[δ]
Bending with torsion of shafts.Many structural elements of machines operate under conditions of torsion and bending. For example, gear shafts transmit torque and bending moments from the forces in the meshing of the teeth, belt drive shafts experience the same loads from the difference in belt tension.
To solve the issues of material performance, plots of torque and bending moments are built, and then stresses.
Diagram of the gear shaft. To determine the loads, we build a shaft diagram and load it with bending and torque moments and build diagrams of torque and bending moments.
Stress in cross section. We find dangerous melancholy C and calculate for it , where Z is the distance from the neutral axis ; ρ is the radius vector from the origin of the coordinate axes
Then the equivalent stresses according to the 4th theory of strength
Stability of compressed rods
1) General concepts of system stability
2) Longitudinal bend. Critical Power. Euler formula.
3) Rod End Attachments
4) The limit of applicability of the Euler formula.
5) Practical calculations for compression.
Basic concepts of sustainability. Under the stability of equilibrium is understood the property of the system to maintain its state when it deviates from the initial state of the interaction of external forces. In real conditions, there is a reason why the system deviates from the initial equilibrium. If, after the termination of the action of external forces, the system returns to its original position, then such a position is called stable, if it does not return, it goes into a new state - unstable. The transition from one state to another in this case is a loss of stability. The loss of stability depends on the magnitude of the acting force. The force (or other parameter) is characterized by the transition from a stable state to an unstable one - the critical force. To ensure the operability of the system, it is necessary that the real part is only a part of the critical force.
Stability of compressed rods. Deformable bodies, including rods, are in a stable or unstable state. Other bodies, as well as solid bodies, can be in stable equilibrium. If a thin straight rod is compressed along the geometric axis with a constantly increasing force, then at first it will be straight under the action of compressive stresses, where A - cross-sectional area of the rod. And then, at a certain load Fcr, called critical, the rod bends sharply, the stresses in it increase rapidly, and there is a danger of destruction. This phenomenon is called buckling. If the rod is stretched by a longitudinal force, then it is always in a stable (single) equilibrium position.
Euler's problem. To clarify the conditions under which it becomes possible various states equilibrium, consider an example (the Euler problem) about the compression of a rod. The critical force in this problem will be equal to such an axial force at which the rod can be in a slightly bent state.
With small deflections of the rod, you can use the differential equation of the bent axis in the form. The minus sign on the right side shows that the moment of force tends to increase the negative curvature of the elastic line. The equation can be rewritten as .
Equation solution. , where C1 and From 2- arbitrary constants determined from the boundary conditions: 1) at X = 0 at(0) = 0; 2) at x = l y (1) = 0.
Hence it follows that C 2 = 0; The second condition can be satisfied only if From 1sin kx = 0 . So the equation has two solutions: a) C 1 \u003d 0, b)sin kl = 0 .
At C 1 = C 2 = 0 displacement at are identically equal to zero and the rod retains a rectilinear shape. This case does not satisfy the conditions of the problem, since a bent rod is considered. Therefore, the rod can only bend if kl = np, where P - arbitrary integer.
With a small force F, as long as the value, value and rod will keep a rectilinear shape. As soon as the rod loses its stability and bends.
This power, corresponding n = 1 called the Euler force, or the first critical force. In this case, the rod is bent along a half-wave of a sinusoid.
Bar deflection shapes. The rod is bent along a half wave of a sinusoid . For n=1, with maximum bend С 1 . For n > 1, the elastic line of the rod is represented by a curve including P half wave However, these unstable forms of equilibrium are of no practical importance, since already at n = 1 rod "loses" the bearing capacity.
Influence of rod fixing. The value of Fkr depends on the conditions of fixing the rod, the nature of the loading and the shape of the sections (moment of inertia) of the rod. For example, if a hinged rod is connected to another support in the middle (a), then when it loses stability, it will bend along 2 half-waves (like a two-support rod with a length) and. A rod rigidly fixed at one end and free at the other, loaded end will have the same critical force as a two-support rod with a conditional length of 2 l.
Euler formula.
In the general case, Euler's formula can be represented in the form where m - length reduction factor. The critical load corresponds to the compressive stress
Where l is the coefficient characterizing the reduced flexibility of the rod (taking into account the conditions of its loading and support): , where i- radius of inertia of the section: . Note that the critical force and stresses are determined by the minimum moment of inertia of the section.
Limit stresses. Critical stresses - a characteristic of structures (depends on λ). Curve 1 (Euler hyperbola) is for the elastic state. For very flexible rods (λ>100), buckling occurs at stresses below the yield point, i.e. criterion for the performance of structures. Let the flexibility at limit stresses be proportional, then it depends only on the mechanical characteristics of the material.
Limit of applicability of the Euler formula. For small values of λ<40 стержень теряет работоспособность из-за наступления пластических деформаций, потери устойчивости не происходит и предельное напряжение равно пределу текучести. При средних значениях (40< λ<100) для стержня из стали (Ст3) наблюдается потеря устойчивости стержня, сопровождаемая упругопластическими деформациями (2). Для этого случая нагружения формула Эйлера не справедлива, и критические напряжения вычисляют по эмпирической формуле Ясинского: , основанной на аппроксимации кривой, отрезком прямой. Коэффициент и для сталей марок Ст2, Ст3 и Ст5. В реальных деталях стержневой формы (винтах, стойках и др.) неизбежны отклонения оси стержня от прямолинейного направления и внецентренное приложение сжимающих сил, поэтому потеря устойчивости стержня происходит при напряжениях, меньших критических.
Practical calculations. The calculation of compressed rods is carried out in the same way as for tension rods, but the allowable stresses are taken depending on the flexibility: , where φ is the reduction factor for allowable stresses. The actual stability factor in this case is defined as: .
Fatigue of materials.
1. Cycles of alternating voltages
2. Fatigue of materials
3. Fatigue Curve
4. Endurance limit
5. Influence of design and technological factors on the endurance limit
6. Calculation of strength at variable stresses.
Variable voltages. Most machine parts under operating conditions experience alternating stresses that change cyclically over time (cyclic stresses). They arise in parts from a change in load, as well as in connection with a change in the position of their sections with respect to a constant load (for example, during rotation of a part). Z-ny changes in alternating voltages can be different, but all of them can be represented in the form of the simplest harmonics of a sinusoid and a cosine wave.
The periodic change in stresses over time occurs from the largest value σ max to the smallest σ min and vice versa. Variable stresses can also be tangential.
The number of voltage cycles per second is called load frequency.
Stress cycles
Cycles can be of constant sign (a and v) or alternating, pulsating.
Any cycle can be characterized by average stress σ m = (σ max + σ min)/ 2
and the amplitude of the alternating stress σ a = (σ max - σ min) / 2
The ratio r = σ min / σ max is called the asymmetry coefficient of the cycle.
In a cycle b average voltage \u003d 0, such a cycle is called symmetrical.
(σ max ≥ σ a ≥ σ min; r \u003d -1) If max or min cycle stress \u003d 0, then it is called pulsating or zero (for a cycle v, r = 0)
Fatigue of materials. Variable stresses that appear in machine parts from a change in load or a change in their position relative to a constant load (for example, during rotation) lead to a sudden destruction of the part, although the magnitude of these stresses is significantly lower than the yield strength (permissible stresses). This phenomenon is called fatigue. Fatigue failure begins with the accumulation of damage, the appearance of cracks, gradually developing inwards, which leads to an increase in stresses of the undamaged part.
Endurance limit. The number of cycles until the moment of failure depends on the stress amplitude in a very wide range. The ability of a material to withstand the action of variable loads is called fatigue resistance (there are cases when a part fails at high stresses after several cycles, and at smaller ones it can work indefinitely).
Fatigue resistance is evaluated using the endurance limit determined experimentally for special. machines or stands.( endurance limit- number of cycles before destruction)
Test Method. If a steel sample withstood 10 million cycles, it is believed that it can withstand without destruction a greater number of cycles. 10 7 is the base number.
Fatigue chart. Based on the test results, a fatigue curve is plotted. The highest value of the cycle stress that the sample can withstand before the test base is called the endurance limit. With a symmetrical cycle, the endurance limit is denoted by σ -1, with a pulsating cycle -σ 0, with an asymmetric -σ r. To calculate parts not intended for a long service life, the concept of a limited endurance limit σ rN is introduced, N is a given (less than the base) number of cycles
Approximate values between bending endurance limits and endurance limits for other types of deformation: , .
Fatigue Curve Equation. The relationship between the alternating stress σ max and the number of cycles to failure is quite accurately described by the equation σ max m N=C, where m and c are constants for a given material, temperature and environment. environments ; N σ is the base number of the cycle.
In the logarithmic coordinate. ur-e curve set-ty lgσ max = (I/m) lg N + (1/m) lg C
The tangent of the slope of the straight line β | tgβ | = 1/m
With increasing value m, the slope of the straight line to the axis lg= N decreases and as m→∞ the line becomes horizontal. Usually m=4…10
Ur-e is also valid for the inflection point A of the mouth curve, i.e.
С=σ r m N σ then we get (σ max / σ 1) m = N σ / N, whence N= N σ (σ max / σ 1) m
This dependency is used to determine the resource of work of El-t constructions at a known level of work of variable stresses σ max and values. N σ and σ
Ultimate stress diagram
Endurance limits depend on the cycle asymmetry coefficient. Based on the test results, a diagram of the limiting voltage amplitudes is built. Approximating it, we obtain a linear dependence s a0 = s -1 - sT, where - coefficient, characterizing the sensitivity of the material to the assimilation of the cycle (for tangent or standard stresses), depends on the tensile strength of the material.
Influence of stress concentrators.
Experiments have established that in the zones of sharp changes in constants, an increase in voltage arose - the concentration of stress. Their value σ max = α σ σ n
α σ - conc. coefficient, σ n - nominal value.
Effective coefficient of concentration.
Experiment, but in real.mater. concentration factor Tos =
s -1
/s -1 to, where s -1 to- limit with voltage concentrator.
Conc. coefficient for shafts with fillet.
How<ρ, тем конц.напряж-ий выше.
Scale effect.
The conc. is affected by the size of the part, cat. takes into account the scale factor of the effect.
Pov-ty status.
Learns the quality factor β. coefficient b< 1 характеризует снижение предела выносливости при ухудшении обработки по
сравнению с полировкой. Значения коэффициента b are given in the reference literature.
Conditions of strength at alternating stresses.
Boundary straight line equation for limit stress diagrams
s a0 = s -1 - sT,
to identify the limiting state of the material, we transform to the form
s a0 + s fT,£ s -1 .
If we introduce, as usual, the concept of equivalent alternating voltage
s eq= s a + fs s T,
then the condition for the reliability of the material will be the expression
s eq £ s -1 .
The influence of stress concentration, scale effect and surface state should be attributed, as shown by experimental studies, only to the variable component of the cycle.
With this in mind
s e sq. = s aTos / esbs + fs s T
and the fatigue resistance condition takes the form
s eq = s aTos / esbs + fs s T£s -1.
Under the action of shear stresses, the fatigue resistance condition will be
seq = τ a to τ /e τ b τ + f τ τ t ≤ τ -1
Stocks read at pereem.pryazh-yah.
To assess the reliability of the element, the margin of safety is determined. We accept that during operation, the alternating and constant voltages change proportionally. The margin of safety of a part at a point is the ratio of the limiting value of stresses at a point to the effective equivalent (normal and shear) stresses.
Then, substituting the corresponding stresses from the strength conditions, we obtain
Under the combined action of normal and shear stresses, the margin of safety is found by the formula
The obtained values of safety margins should be compared with their allowable values. Usually accepted P³ 1,5.
Calculation of shells of rotation. Calculations beyond elasticity.
The concept of subtle shells of rotation
Determination of stresses
Strength analysis
Calculation of the strength of the rod in bending and torsion beyond the limit of control.
Thin shell of rotation.
When evaluating strength reliability, a number of common elements of construction are schematized in the form of a thin shell of rotation.
If the load on the shell is axisymmetric, then the definition of stresses in the walls does not cause difficulties. With a wall thickness of not more than 0.1 of the minimum radius of its curvature, it is assumed with an accuracy acceptable for practice that only normal stresses arose in the walls from an external load, which are constant in thickness.
As an example, consider a pressure vessel.
Consider the vessel (a) under pressure. With two transverse and two longitudinal sections, we cut out an infinitely small element from the wall with a length of faces dl (b) In these sections, the effective axial s T and circumferential (ring) σ β stresses, i.e. notch.ele you are in a plane stress state.
Tension-I in the shell.
Axial stresses, caused by axial force:
N=qπR 2 = π Dδ σ m
R is the inner radius of the spherical part, hence D≈ 2R is the average diameter of the cylindrical part of the vessel
δ is the wall thickness.
The area of the annular section A m \u003d π DS, from this equation σ m \u003d N / A m \u003d qD / 4 δ
Circumferential stresses caused by forces dN 0 = δ 0 ⌠ ∂ l, which must balance the force dF R , conditioned. Pressure q, acting on the surface of the element dF R =q dl 2
Equilibrium composition, projecting the forces dN 0 and dF R on the direction of the radius in the middle of the element 2 dN 0 sin (dQ / 2) -dF R \u003d 0
Stress analysis.
Then 2σ 0 δ dl sin(dQ/2)=q dl 2
dl= Rd0 and sin(dQ/2) ≈ dQ/2
We get σ 0 = qD/2 δ.
Comparing with previous σ m voltage in transverse direction (along the ring), sec. 2 times more (in the longitudinal state) than along the pipe (cross section). This situation is taken into account in practice when manufacturing composite tanks. The hem welds are stronger than the transverse seams. In a spherical vessel, the stresses on the faces will be the same.
Scheme of stress in calculations beyond the elastic limit of the material.
Max torsion stress r.m.s. In the outermost fibers and plastic deformation arose first on the contour of the section. Plastic zone at magnification load will develop inside the section. For ideal. An elastic-plastic material transition to the limit state is shown in fig.
On a) showing the distribution of stresses for the elastic state, preserved up to τ max = τ t,
On b) at τ max = τ t, when plastic first appeared. Deform.,
Figure c) shows the limit state.
Limit torque.
The limit value of the moment is easily established from the equilibrium condition.
T cr = A ∫ dT = A ∫ ρ d F t = A ∫ ρ τ t dA = A ∫ ρ2πρ τ t ρ dρ = π τ t ρ 3 /3| a 0 = (πD 3 /12) τ t
Find the ratio of T cr to max values. T m moment in elastic state.
T m = T t W p = τ t W p
We get T cr / T t \u003d 4/3.
Calculation for bending beyond the limit of elasticity.
Experiment. It has been established that with elastic-plastic bending, the s-n of flat sections is preserved. Poet.def.line.depend on cord. y. On fig. A) display. cross section; elastic distribution def. And voltage in height section-I (b and c); elastic-plastic (d) and limit state (e).
Conclusion: Calculation of the limit.torque.Plast.def. zone. spreading inside the section, and when the entire section of the normal stress has reached the yield point, the limit state is reversed and the beam cannot transmit a greater load. The limit moment is found by integrating:
M prev \u003d A ∫ dM \u003d h / 2 ∫ σ t b y d y \u003d σ t b h 2 / 4
Max value of the bending moment for the elastic distribution of stresses, when only in the outermost fiber will be: М τ = σ t b h 2 /6
Ratio M before / M τ = 3/2
Contact voltage.
Contact name voltage in the contact zone of children's machines. in practice, it often becomes necessary to determine the stresses and strains in these zones, both when calculating the contact strength (gear and friction gears), and for evaluating the endurance limit (threaded and press connections ).
Structural contact. problems are solved by methods of the theory of cuprousness, as pr-lo, approximately. Exact solutions have been obtained only for elastic problems. Contact details of a simple form (cylinders, balls, etc.)
To understand the approach principle when solving contact problems, consider the interaction of a cylinder (Hertz problem)
Contact diagram of two cylinders.
Consider. Voltage state 2 long cylinders with || axes, compressed distribution along the length by radial loads P. At a distance E from the square of the passage through the axes of the cylinder, take two points A1 and A2.
If the contact of the cylinders without load occurs along the line || their axes, through t.V, then with a load on the site 2a * l (l is the length of the rod).
Contact homogeneous materials.
If the cylinders are made of material, the cat. E1 = E2 and μ1 = μ2, then
qmax = 0.418√ (pE(R1+R2)/R1R2)
a = 1.52√ (p/E * (R1R2)/(R1+R2))
δ \u003d 2 (1-ν 2) / πE * p * (ln (4R1R2 / a 2) + 0.815)
and depends on p, then the displacement δ is a non-linear function of p, although the material is elastic.
This problem was first solved by Hertz.
Statistically indeterminate systems. The concept of static uncertainty.
It is known from mathematics that the number of ur-ths must be = the number of unknowns. Values, otherwise the system is unsolvable. (statically indeterminate). for the definition of r-tions (or vnutr.str.factorov) is not enough ur-i balance. The number of missing Urs is the degree of uncertainty.
The method of disclosure of static neopr-ty.
Inclusion in the number of ur-th ur-th of the compatibility of the deformation of the rods. The best distribution method of forces, when the Equation of strain compatibility. Vyr-Xia through the acting forces.
For example, the problem of a composite rod.
Let the rod consist of two parts.
From ur-th statics 1 ur-e ∑y =0.
∑y \u003d R K + R B -F \u003d 0.
The problem is not statically solvable
Let's add L-e of compatibility of deformations.
ΔA \u003d - F l 2 / E 2 A 2;
Δx \u003d x l 1 / E 1 A 1 + x l 2 / E 2 A 2;
x \u003d R K \u003d F / (1 + l 1 / l 2 * E 2 A 2 / E 1 A 1);
dynamic loads. Hit .
1) Dynamic loads
2) Accounting for inertia forces
4) Impact loads. Determination of stresses and displacements.
temperature problem. The second group of problems in the statistics of indeterminate systems consists in taking into account the forces from the constraint of thermal deformation. Let us determine the forces and stresses in a fixed rod with a length l, with section A when the temperature changes to, if the coefficient of linear expansion is equal to α, and the modulus of elasticity is E. The statistical side of the problem is reduced to. Let's turn to the geometric side of the problem. The free temperature deformation of the rod is equal to and is realized when the right end of the rod is released.
Solution and analysis. By introducing an unknown force x, we eliminate the thermal expansion, because according to the meaning of the problem and, - does not depend on the size of the rod. When the rod is in compression. In this case, σ reaches large values.
dynamic loads. Dynamic loads occur in structural elements when they move with acceleration. The calculation of internal force factors and stresses is carried out taking into account the forces of inertia and the mechanical properties of materials at high loading rates. The general calculation method is based on the d'Alembert principle, using which a structural element is brought into a state of instantaneous equilibrium by applying inertia forces to it. Next, the method described above determines the strain and stress. Consider the consideration of dynamic loads on the example of oscillatory movements of the rotation of the ring or impact stress.
Scheme of a single-mass oscillatory system. The simplest dynamic system includes a mass attached to a spring. When considering the elastic oscillations of the system, natural and forced oscillations are distinguished. Own (free) vibrations are called which the system makes in the absence of external influences. If at the initial moment we deviate the mass by a value a and leave the system to itself, then natural oscillations will arise and the displacement of the center of mass at time t will be, where a is the oscillation amplitude - the maximum deviation of the center of mass from the equilibrium position; p is the circular frequency of oscillation; .
System Settings. Circular frequency, where λ is the compliance of the spring, mm / N (spring settlement under the action of a force of 1N); m is the weight of the load. The time interval between two identical positions of the system is called the period of oscillation. The reciprocal of the oscillation period is called the oscillation frequency (the number of oscillations in 1s):. Circular frequency representing the number of oscillations in 2π seconds: . Because there are always friction forces in the system, then natural oscillations always damp out. Forced oscillations are called oscillations occurring under the action of an external periodic perturbing force. If an external force is applied to the system, then forced oscillations with frequency ω occur in the system. The deviation of the mass will be, - spring settlement under the statistical action of the amplitude of the perturbing force.
The concept of resonance. Calculation modules of the system. When the frequency ω of the exciting force coincides with the frequency of natural oscillations, resonance sets in, and the oscillation amplitude tends to infinity. In fact, due to friction, the amplitude remains finite, but reaches a larger value. Resonance is a great danger to structural strength and should be avoided. Resonances are most often eliminated by changing the natural frequency of oscillations, less often by changing the frequency of the exciting force. The main task of calculating a structure for vibrations (vibration) is to determine the natural and resonant vibration frequencies. The complexity of the theoretical analysis of oscillations depends on the number of independent coordinates that uniquely determine the positions of points in the system. The mathematical description of such a system can be performed using partial differential equations. In simplified calculations, light parts are considered weightless, but deformable, heavy parts are absolutely rigid bodies - material points. The motion of such a system is described by an ordinary differential equation.
Impact loading of systems. During an impact (sudden loading at a very high speed), the determination of inertial forces is difficult, therefore, for an approximate determination of dynamic stresses and deformations, the law of conservation of energy is used and it is assumed that the striking body does not bounce off the structure after contact, but moves with it (inelastic impact) . Consider the problem of the impact of a load of mass m moving at a speed υ 0 along a rod with axial compliance, where E is the modulus of elasticity of the material, A is the cross-sectional area of the rod.
dynamic factor. The kinetic energy of a dream after touching the rod will accumulate in it in the form of the potential energy of the rod stretched by the value and the change in the potential energy of the load at the same displacement. Energy balance. Where, - elongation of the rod under static loading by the force of gravity of the load; is the dynamism factor.
dynamic stresses. The dynamic coefficient shows how many times the movement of the elastic element increases due to impact compared to the statistical movement. Significantly: the stresses in the body increase proportionally. Dynamic stresses thus depend on kinetic energy.
centrifugal loads. Applying the d'Alembert principle, we determine the stresses in a uniformly rotating ring. This model is used in calculations of transmission belts and other parts. The rotating ring is deformed by centrifugal forces of inertia, uniformly distributed around the circumference (Fig. a). The force of inertia acting on a ring element 1 mm long, where is the mass of the ring element, where ρ is the density of the material; A is the cross-sectional area; ω is the angular velocity of the ring; r is the average radius of the ring.
Stresses in a rotating ring. Using two radial planes, we cut out an infinitesimal element from the ring (Fig.b). The force of inertia acting on an element of length: (q-material mass). This force is balanced by normal forces N 0 in the cross section from circumferential tensile stresses σ 0: . Projecting the forces N 0 onto the line of action of the force, we will compose the equilibrium equations for the element: , where ρ is the density; - circumferential speed of the ring.
When determining the internal force factors, they are considered to be applied in the center of gravity of the section. In reality, the internal forces, being the result of the interaction of body particles, are continuously distributed over the cross section. The intensity of these forces at different points of the section can be different. With an increase in the load on a structural element, internal forces increase and, accordingly, their intensity increases at all points of the section. If at some point the intensity of internal forces reaches a value determined for a given material, a crack occurs at this point, the development of which will lead to the destruction of the element, or unacceptable plastic deformations will occur. Consequently, the strength of structural elements should be judged not by the value of internal force factors, but by their intensity. The measure of the intensity of internal forces is called tension.
In the vicinity of an arbitrary point belonging to the section of some loaded body, we select an elementary area , within which the internal force acts (Fig. 1.6, a).
The average value of the intensity of internal forces on the site, called the average stress, is determined by the formula
Reducing the area , in the limit we obtain the true stress at a given point of the section
The vector quantity is called full voltage at the point. In the international system of units (SI), the unit of voltage is pascal(Pa) is the voltage at which an internal force of 1 N acts on a site of 1 m 2.
Since this unit is very small, a multiple unit of stress is used in the calculations - megapascal (1 MPa \u003d 10 6 Pa).
We decompose the total voltage vector into two components (Fig. 1.6, b).
The projection of the total stress vector onto the normal to a given area is denoted by and is called normal voltage.
Rice. 1.6
The component lying in the section in a given area is denoted by and is called shear stress.
The normal stress directed away from the section is considered positive, directed towards the section - negative.
Normal stresses arise when, under the action of external forces, particles located on both sides of the section tend to move away from one another or approach each other. Shear stresses arise when particles tend to move relative to each other in the section plane.
The shear stress can be decomposed along the coordinate axes into two components and (Fig. 1.6, v). The first index at shows which axis is perpendicular to the section, the second - parallel to which axis the stress acts. If the direction of the shear stress does not matter in the calculations, it is designated without indices.
There is a relationship between the total voltage and its components
An infinite number of sections can be drawn through a point of the body, and for each of them the stresses have their own value. Therefore, when determining stresses, it is necessary to indicate the position of not only the point of the body, but also the section drawn through this point.
The set of stresses for a set of areas passing through a given point forms stress state at this point.
Stresses in cross sections are related to internal force factors by certain dependencies.
Let us take an infinitesimal area in cross section. On this site, in the general case, infinitely small (elementary) internal forces act (Fig. 1.7)
Fig.1.7
The corresponding elementary moments about the coordinate axes , , have the form.
