Calculation of timber with a round cross-section for strength and torsional rigidity. Tension-compression Section stiffness
Axial (central) tension or compression straight beam is caused by external forces, the resultant vector of which coincides with the axis of the beam. When tension or compression occurs in the cross sections of a beam, only longitudinal forces N arise. The longitudinal force N in a certain section is equal to the algebraic sum of the projection onto the axis of the rod of all external forces acting on one side of the section under consideration. According to the rule of signs of longitudinal force N, it is generally accepted that from tensile external loads positive longitudinal forces N arise, and from compressive forces the longitudinal forces N are negative (Fig. 5).
To identify areas of a rod or its section where longitudinal force It has highest value, construct a diagram of longitudinal forces using the section method, discussed in detail in the article:
Analysis of internal force factors in statistically definable systems
I also highly recommend taking a look at the article:
Calculation of statistically determinable timber
If you understand the theory in this article and the tasks in the links, you will become a guru in the topic “Extension-compression” =)
Tensile-compressive stresses.
The longitudinal force N, determined by the section method, is the resultant of the internal forces distributed over the cross section of the rod (Fig. 2, b). Based on the definition of stress, according to expression (1), we can write for the longitudinal force:
where σ is the normal stress at an arbitrary point in the cross section of the rod.
To determine normal stresses at any point on the beam it is necessary to know the law of their distribution over the cross section of the beam. Experimental studies show: if a series of mutually perpendicular lines are applied to the surface of the rod, then after applying an external tensile load the transverse lines do not bend and remain parallel to each other (Fig. 6, a). This phenomenon is spoken of hypothesis flat sections
(Bernoulli's hypothesis): sections that are flat before deformation remain flat after deformation.
Since all the longitudinal fibers of the rod are deformed equally, the stresses in the cross section are the same, and the stress diagram σ along the height of the cross section of the rod looks as shown in Fig. 6, b. It can be seen that the stresses are uniformly distributed over the cross section of the rod, i.e. at all points of the section σ = const. Expression to define voltage values has the form:
Thus, the normal stresses arising in the cross sections of a tensile or compressed beam are equal to the ratio of the longitudinal force to the area of its cross section. Normal stresses are considered to be positive in tension and negative in compression.
Tensile-compressive deformations.
Let us consider the deformations that occur during tension (compression) of the rod (Fig. 6, a). Under the influence of force F, the beam is lengthened by a certain amount Δl called absolute elongation, or absolute longitudinal deformation, which is numerically equal to the difference between the length of the beam after deformation l 1 and its length before deformation l
Absolute relation longitudinal deformation beam Δl to its original length l is called relative elongation, or relative longitudinal deformation:
In tension, the longitudinal strain is positive, and in compression, it is negative. For most construction materials at the stage of elastic deformation, Hooke’s law (4) is satisfied, establishing linear dependence between stresses and strains:
where the modulus of longitudinal elasticity E, also called modulus of elasticity of the first kind is the coefficient of proportionality between stress and strain. It characterizes the stiffness of a material under tension or compression (Table 1).
Table 1
Modulus of longitudinal elasticity for various materials
Absolute transverse deformation of timber equal to the difference in cross-sectional dimensions after and before deformation:
Respectively, relative transverse deformation determined by the formula:
When stretched, the cross-sectional dimensions of the beam decrease, and ε "has negative meaning. Experience has established that, within the limits of Hooke's law, when a beam is stretched, the transverse deformation is directly proportional to the longitudinal one. Attitude transverse deformationε "to the longitudinal strain ε is called the transverse strain coefficient, or Poisson's ratio μ:
It has been experimentally established that at the elastic stage of loading of any material the value μ = const and for various materials the values of Poisson’s ratio range from 0 to 0.5 (Table 2).
table 2
Poisson's ratio.
Absolute elongation of the rodΔl is directly proportional to the longitudinal force N:
This formula can be used to calculate the absolute elongation of a section of a rod with length l, provided that within this section the value of the longitudinal force is constant. In the case when the longitudinal force N changes within a section of the rod, Δl is determined by integration within this section:
The product (EA A) is called section rigidity rod in tension (compression).
Mechanical properties of materials.
Main mechanical properties materials during their deformation are strength, ductility, brittleness, elasticity and hardness.
Strength is the ability of a material to resist external forces without collapsing and without the appearance of residual deformations.
Plasticity is the property of a material to withstand large residual deformations without destruction. Deformations that do not disappear after removal of external loads are called plastic.
Brittleness is the property of a material to collapse with very small residual deformations (for example, cast iron, concrete, glass).
Ideal elasticity– the property of a material (body) to completely restore its shape and size after eliminating the causes that caused the deformation.
Hardness is the property of a material to resist the penetration of other bodies into it.
Consider the tension diagram of a mild steel rod. Let a round rod of length l 0 and initial constant cross-section of area A 0 be statically stretched at both ends by force F.
The rod compression diagram looks like (Fig. 10, a)
where Δl = l - l 0 absolute elongation of the rod; ε = Δl / l 0 - relative longitudinal elongation of the rod; σ = F / A 0 - normal voltage; E - Young's modulus; σ p - limit of proportionality; σ up - elastic limit; σ t - yield strength; σ in - tensile strength (temporary resistance); ε rest - residual deformation after removal of external loads. For materials that do not have a pronounced yield point, a conditional yield strength σ 0.2 is introduced - the stress at which 0.2% of residual deformation is achieved. When the ultimate strength is reached, a local thinning of its diameter (“neck”) occurs in the center of the rod. Further absolute elongation of the rod occurs in the neck zone (local yield zone). When the stress reaches the yield strength σ t, the glossy surface of the rod becomes slightly matte - microcracks (Lüders-Chernov lines) appear on its surface, directed at an angle of 45° to the axis of the rod.
Calculations of strength and rigidity in tension and compression.
The dangerous section in tension and compression is the cross section of the beam in which the maximum normal stress occurs. Allowable stresses are calculated using the formula:
where σ limit is the ultimate stress (σ limit = σ t - for plastic materials and σ limit = σ v - for brittle materials); [n] - safety factor. For plastic materials [n] = = 1.2 ... 2.5; for brittle materials [n] = 2 ... 5, and for wood [n] = 8 ÷ 12.
Calculations of tensile and compressive strength.
The purpose of calculating any structure is to use the results obtained to assess the suitability of this structure for operation under minimum consumption material, which is reflected in the methods of calculation for strength and rigidity.
Strength condition rod when it is stretched (compressed):
At design calculation the dangerous cross-sectional area of the rod is determined:
When determining permissible load the permissible normal force is calculated:
Calculation of rigidity in tension and compression.
Rod performance is determined by its ultimate deformation [l]. The absolute elongation of the rod must satisfy the condition:
Often additional calculations are made for the rigidity of individual sections of the rod.
The stiffness of the section is proportional to the elastic modulus E and the axial moment of inertia Jx, in other words, it is determined by the material, shape and dimensions of the cross section.
The stiffness of the section is proportional to the elastic modulus E and the axial moment of inertia Yx, in other words, it is determined by the material, shape and dimensions of the cross section.
The stiffness of the section is proportional to the modulus of elasticity E and the axial moment of inertia Jx; in other words, it is determined by the material, shape and cross-sectional dimensions.
The stiffness of sections EJx of all frame elements is the same.
The section stiffnesses of all frame elements are the same.
The cross-sectional stiffness of elements without cracks in these cases can be determined by formula (192) as for short-term temperature action, taking vt - 1; cross-sectional stiffness of elements with cracks - according to formulas (207) and (210) as for the case of short-term heating.
The cross-sectional stiffnesses of the frame elements are the same.
Here El is the minimum rigidity of the rod section during bending; G is the length of the rod; P - compressive force; a-coefficient of linear expansion of the material; T - heating temperature (the difference between the operating temperature and the temperature at which movements of the ends of the rod were excluded); EF—stiffness of the rod section under compression; i / I / F is the minimum radius of gyration of the rod section.
If the stiffness of the frame section is constant, the solution is somewhat simplified.
When the stiffness of the sections of a structural element changes continuously along its length, the displacements must be determined by direct (analytical) calculation of the Mohr integral. Such a structure can be calculated approximately by replacing it with a system with elements of step-variable stiffness, after which Vereshchagin’s method can be used to determine the displacements.
Determining the stiffness of sections with ribs by calculation is a complex and, in some cases, impossible task. In this regard, the role of experimental data from testing full-scale structures or models is increasing.
A sharp change in the stiffness of beam sections over a short length causes a significant concentration of stress in welded waist seams in the curvilinear joint zone.
What is the torsional stiffness of a section?
What is the bending stiffness of a section?
What is the torsional stiffness of a section?
What is the bending stiffness of a section?
What is called the cross-sectional stiffness of a rod in shear.
EJ are called the tensile stiffnesses of the bar sections.
The product EF characterizes the stiffness of the section under axial force. Hooke's law (2.3) is valid only in specific area changes in strength. At P Rpc, where Ppc is the force corresponding to the limit of proportionality, the relationship between tensile force and elongation turns out to be nonlinear.
The product EJ characterizes the bending stiffness of the beam section.
Shaft torsion.| Shaft torsional deformation. The product GJр characterizes the torsional rigidity of the shaft section.
If the rigidity of the beam section is constant throughout its entire
Schemes for processing welded parts. a - plane processing. 6 - processing.| Loading of a welded beam with residual stresses. a - beam. b - zones 1 and 2 with high residual tensile stresses. - section of the beam that takes up the load during bending (shown by shading. This reduces the stiffness characteristics of the section EF and EJ. Displacements - deflections, angles of rotation, elongations caused by the load exceed the calculated values.
The product GJP is called the torsional stiffness of the section.
The product G-IP is called the torsional stiffness of the section.
The product G-Ip is called the torsional stiffness of the section.
The product GJp is called the torsional stiffness of the section.
The product ES is called the cross-sectional stiffness of the rod.
The value EA is called the cross-sectional stiffness of the rod in tension and compression.
The product EF is called the cross-sectional stiffness of the rod in tension or compression.
The GJP value is called the torsional stiffness of the shaft section.
The product GJр is called the section stiffness round timber when torsion.
The GJP value is called the torsional stiffness of the section of a round beam.
The loads, lengths and stiffness of beam sections are assumed to be known. In Problem 5.129, establish by how many percent and in what direction the deflection of the midspan of the beam indicated in the figure, determined by the approximate equation of an elastic line, differs from the deflection found exactly by the equation of a circular arc.
The loads, lengths and stiffness of beam sections are assumed to be known.
The product EJZ is usually called the bending stiffness of the section.
The product EA is called the tensile stiffness of the section.
The product EJ2 is usually called the bending stiffness of the section.
The product G 1P is called the torsional stiffness of the section.
Task 3.4.1: The torsional rigidity of the cross section of a round rod is given by the expression...
Possible answers:
1) E.A.; 2) GJP; 3) GA; 4) EJ
Solution: The correct answer is 2).
The relative angle of twist of a rod of circular cross-section is determined by the formula. The smaller, the greater the rigidity of the rod. Therefore the product GJP is called the torsional stiffness of the cross section of the rod.
Task 3.4.2: d loaded as shown in the figure. The maximum value of the relative twist angle is...
The material shear modulus G, moment value M, length l are given.
Possible answers:
1) ; 2) ; 3) ; 4) .
Solution: The correct answer is 1). Let's build a diagram of torques.
When solving the problem, we will use the formula to determine the relative angle of twist of a rod with a circular cross section
in our case we get
Task 3.4.3: From the condition of rigidity at given values and G, the smallest permissible shaft diameter is... Accept.
Possible answers:
1) ; 2) ; 3) ; 4) .
Solution: The correct answer is 1). Since the shaft has a constant diameter, the stiffness condition has the form
Where. Then
Task 3.4.4: Kernel round section diameter d loaded as shown in the figure. Material shear modulus G, length l, moment value M given. The mutual angle of rotation of the extreme sections is equal to...
Possible answers:
1); 2) ; 3) zero; 4) .
Solution: The correct answer is 3). Let us denote the sections where external force pairs are applied B, C,D Accordingly, we will construct a diagram of torques. Section rotation angle D relative to the section B can be expressed as the algebraic sum of the mutual angles of rotation of section C relative to sections B and sections D relative to the section WITH, i.e. . material deformed rod inertia
The mutual angle of rotation of two sections for a rod with a circular cross-section is determined by the formula. In relation to this problem we have
Task 3.4.5: The torsional rigidity condition for a rod of circular cross-section, with a constant diameter along its length, has the form...
Possible answers:
1) ; 2) ; 3) ; 4) .
Solution: The correct answer is 4). The shafts of machines and mechanisms must not only be strong, but also sufficiently rigid. In rigidity calculations, the maximum relative twist angle is limited, which is determined by the formula
Therefore, the rigidity condition for a shaft (rod experiencing torsional deformation) with a constant diameter along its length has the form
where is the permissible relative twist angle.
Task 3.4.6: The loading diagram of the rod is shown in the figure. Length L, torsional rigidity of the cross section of the rod, - permissible angle of rotation of the section WITH given. Based on the maximum stiffness permissible value external load parameter M equals.
1); 2) ; 3) ; 4) .
Solution: The correct answer is 2). The rigidity condition in this case has the form where is the actual angle of rotation of the cross section WITH. We build a torque diagram.
Determine the actual angle of rotation of the section WITH. . We substitute the expression for the actual angle of rotation into the stiffness condition
- 1) oriented; 2) main sites;
- 3) octahedral; 4) secants.
Solution: The correct answer is 2).
When rotating an elementary volume 1, one can find its spatial orientation 2 in which the tangential stresses on its faces disappear and only normal stresses remain (some of them may be equal to zero).
Task 4.1.3: The principal stresses for the stress state shown in the figure are equal to... (The stress values are indicated in MPa).
- 1) y1=150 MPa, y2=50 MPa; 2) y1=0 MPa, y2=50 MPa, y3=150 MPa;
- 3) y1=150 MPa, y2=50 MPa, y3=0 MPa; 4) y1=100 MPa, y2=100 MPa.
Solution: The correct answer is 3). One face of the element is free from shear stress. Therefore, this is the main site, and the normal stress (principal stress) at this site is also zero.
To determine the other two values of the principal stresses, we use the formula
where the positive directions of stress are shown in the figure.
For the given example we have, . After transformations we find, . In accordance with the rule for numbering the principal stresses, we have y1=150 MPa, y2=50 MPa, y3=0 MPa, i.e. plane stress state.
Task 4.1.4: At the studied point of the stressed body on three main sites, the values are determined normal stress: 50MPa, 150MPa, -100MPa. The main stresses in this case are equal...
- 1) y1=150 MPa, y2=50 MPa, y3=-100 MPa;
- 2) y1=150 MPa, y2=-100 MPa, y3=50 MPa;
- 3) y1=50 MPa, y2=-100 MPa, y3=150 MPa;
- 4) y1=-100 MPa, y2=50 MPa, y3=150 MPa;
Solution: The correct answer is 1). The main stresses are assigned indices 1, 2, 3 so that the condition is satisfied.
Task 4.1.5: On the faces of the elementary volume (see figure) the stress values in MPa. Angle between positive axis direction x and the outer normal to the main area, on which the minimum principal stress acts, is equal to ...
1) ; 2) 00; 3) ; 4) .
Solution: The correct answer is 3).
The angle is determined by the formula
Substituting numeric values voltage, we get
We set the negative angle clockwise.
Task 4.1.6: The values of the principal stresses are determined from the solution of the cubic equation. Odds J1, J2, J3 called...
- 1) stress state invariants; 2) elastic constants;
- 3) direction cosines of the normal;
- 4) proportionality coefficients.
Solution: The correct answer is 1). Are the roots of the equation the principal stresses? are determined by the nature of the stress state at the point and do not depend on the choice original system coordinates Consequently, when rotating the coordinate axes system, the coefficients
must remain unchanged.
Calculation of timber with a round cross-section for strength and torsional rigidity
Calculation of timber with a round cross-section for strength and torsional rigidity
The purpose of calculations for strength and torsional rigidity is to determine the cross-sectional dimensions of the beam at which stresses and displacements will not exceed specified values allowed by operating conditions. The strength condition for permissible tangential stresses is generally written in the form This condition means that the highest shear stresses arising in a twisted beam should not exceed the corresponding permissible stresses for the material. The permissible stress during torsion depends on 0 ─ the stress corresponding to the dangerous state of the material, and the accepted safety factor n: ─ yield strength, nt - safety factor for a plastic material; ─ tensile strength, nв - safety factor for brittle material. Due to the fact that it is more difficult to obtain values in torsion experiments than in tension (compression), then, most often, the permissible torsional stresses are taken depending on the permissible tensile stresses for the same material. So for steel [for cast iron. When calculating the strength of twisted beams, three types of problems are possible, differing in the form of using strength conditions: 1) checking stresses (test calculation); 2) selection of section (design calculation); 3) determination of the permissible load. 1. When checking stresses for given loads and dimensions of the beam, the largest tangential stresses occurring in it are determined and compared with those specified according to formula (2.16). If the strength condition is not met, then it is necessary to either increase the cross-sectional dimensions, or reduce the load acting on the beam, or use a material of higher strength. 2. When selecting a section for a given load and a given value of permissible stress, from the strength condition (2.16), the value of the polar moment of resistance of the cross section of the beam is determined. The diameters of the solid round or annular section of the beam are determined by the value of the polar moment of resistance. 3. When determining the permissible load from a given permissible stress and polar moment of resistance WP, based on (3.16), the value of the permissible torque MK is first determined and then, using a torque diagram, a connection is established between K M and external twisting moments. Calculation of timber for strength does not exclude the possibility of deformations that are unacceptable during its operation. twisting of the beam is very dangerous, as it can lead to a violation of the accuracy of processing parts if this beam is a structural element of a processing machine, or torsional vibrations may occur if the beam transmits torsional moments that vary in time, so the beam must also be calculated on its rigidity. The stiffness condition is written in the following form: where ─ the largest relative angle of twist of the beam, determined from expression (2.10) or (2.11). Then the rigidity condition for the shaft will take the form The value of the permissible relative angle of twist is determined by the standards for various elements structures and different types loads vary from 0.15° to 2° per 1 m of beam length. Both in the strength condition and in the rigidity condition, when determining max or max we will use geometric characteristics: WP ─ polar moment of resistance and IP ─ polar moment of inertia. Obviously, these characteristics will be different for a solid round and annular cross sections with the same area of these sections. Through specific calculations, one can be convinced that the polar moments of inertia and the moment of resistance for the annular section are significantly greater than for the irregular circular section, since the annular section does not have areas close to the center. Therefore, a beam with an annular cross-section during torsion is more economical than a beam with a solid circular cross-section, i.e., it requires less material consumption. However, the production of such beams is more difficult and therefore more expensive, and this circumstance must also be taken into account when designing beams operating in torsion. We will illustrate the methodology for calculating timber for strength and torsional rigidity, as well as considerations about cost-effectiveness, with an example. Example 2.2 Compare the weights of two shafts, the transverse dimensions of which are selected for the same torque MK 600 Nm at the same permissible stresses 10 R and 13 Tension along the fibers p] 7 Rp 10 Compression and crushing along the fibers [cm] 10 Rc, Rcm 13 Collapse across the fibers (at a length of at least 10 cm) [cm]90 2.5 Rcm 90 3 Chipping along the fibers during bending [and] 2 Rck 2.4 Chipping along the fibers when cutting 1 Rck 1.2 – 2.4 Chipping across the cuts fibers
The highest shear stresses arising in the twisted beam should not exceed the corresponding permissible stresses:
This requirement is called the strength condition.
The permissible stress during torsion (as well as for other types of deformations) depends on the properties of the material of the beam being calculated and on the accepted safety factor:
In the case of a plastic material, the shear yield strength is taken as the dangerous (ultimate) stress, and in the case of a brittle material, the tensile strength.
Due to the fact that mechanical tests of materials for torsion are carried out much less frequently than for tension, experimentally obtained data on dangerous (ultimate) stresses during torsion are not always available.
Therefore, in most cases, the permissible torsional stresses are taken depending on the permissible tensile stresses for the same material. For example, for steel for cast iron where is the permissible tensile stress of cast iron.
These values of permissible stresses refer to cases of structural elements operating in pure torsion under static loading. Shafts, which are the main objects designed for torsion, in addition to torsion, also experience bending; In addition, the stresses arising in them are variable in time. Therefore, when calculating a shaft only for torsion with a static load without taking into account bending and stress variability, it is necessary to accept reduced values of permissible stresses. Practically, depending on the material and operating conditions, they accept
You should strive to ensure that the material of the beam is used as fully as possible, that is, so that the highest design stresses arising in the beam are equal to the permissible stresses.
The value of tmax in the strength condition (18.6) is the value of the highest shear stress in the dangerous section of the beam in close proximity to its outer surface. A dangerous section of a beam is a section for which absolute value relationships matter most. For a beam of constant cross-section, the most dangerous section is the section in which the torque has the greatest absolute value.
When calculating twisted beams for strength, as well as when calculating other structures, the following three types of problems are possible, differing in the form of using the strength condition (18.6): a) checking stresses (test calculation); b) selection of section (design calculation); c) determination of the permissible load.
When checking stresses for a given load and dimensions of a beam, the largest tangential stresses occurring in it are determined. In this case, in many cases, it is first necessary to construct a diagram, the presence of which makes it easier to determine the dangerous section of the beam. The highest shear stresses in the dangerous section are then compared with the permissible stresses. If condition (18.6) is not satisfied, then it is necessary to change the cross-sectional dimensions of the beam or reduce the load acting on it, or use a material of higher strength. Of course, a slight (about 5%) excess of the maximum design stresses over the permissible ones is not dangerous.
When selecting a section for a given load, the torques in the cross sections of the beam are determined (usually a diagram is drawn), and then using the formula
which is a consequence of formula (8.6) and condition (18.6), the required polar moment of resistance of the cross-section of the beam is determined for each of its sections, in which the cross-section is assumed to be constant.
Here is the value of the largest (in absolute value) torque within each such section.
Based on the magnitude of the polar moment of resistance, the diameter of a solid round is determined using formula (10.6), or using formula (11.6) - outer and internal diameters annular section of the beam.
When determining the permissible load using formula (8.6), based on the known permissible stress and polar moment of resistance W, the value of the permissible torque is determined, then the values of the permissible external loads are established, from the action of which the maximum torque arising in the sections of the beam is equal to the permissible moment.
Calculation of the shaft for strength does not exclude the possibility of deformations that are unacceptable during its operation. Large shaft torsion angles are especially dangerous when they transmit a time-varying torque, since this results in torsional vibrations that are dangerous for its strength. IN technological equipment, For example metal cutting machines, insufficient torsional rigidity of some structural elements (in particular, lead screws of lathes) leads to a violation of the processing accuracy of parts manufactured on this machine. Therefore, in necessary cases shafts are designed not only for strength, but also for rigidity.
The condition for the torsional rigidity of a beam has the form
where is the largest relative angle of twist of the beam, determined by formula (6.6); - permissible relative twist angle accepted for different designs and different types of load equal to from 0.15 to 2° per 1 m of rod length (from 0.0015 to 0.02° per 1 cm of length or from 0.000026 to 0.00035 rad per 1 cm of shaft length).
