Specific Strength Under Combined Loading

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1. Introduction Solving the problem of increasing strength of structures while simultaneously reducing weight and with minimal consumption of scarce materials leads to the development of methods for evaluating the degree of utilization of the shape and dimensions of the body and the strength of the material. These methods were developed first along the path of assessing the shape and dimensions of the cross-section [1-6] at a constant distribution of material properties across the cross-section. To create a body of equal strength in all its zones and in all directions, it is necessary to evaluate the degree of approximation to an equal-strength state and the correspondence of the stress state in each element of the body to the material strength in the same element. Since in most cases material properties are distributed non-uniformly over the cross-section, either as a result of technological side processes or as a result of their deliberate changes, evaluation of the loadbearing capacity of the body must be made taking into account the variability of mechanical properties across the cross-sections of the body. Reasonable use and creation of variability in material properties plays a similar role to the rational choice of body shape. Recently, methods have been proposed for estimating the load-bearing capacity of bodies taking into account the variability of not only the stress distribution, but also the resistance distribution [2; 7; 8]. In works [7-9] the main conclusion was that from the point of view of increasing the specific strength, it is not the stress distribution itself that is important, but the mutual correspondence of the stress and resistance distributions, and it is necessary to add the analysis of the stress distribution to the analysis of the resistance distribution. 2. Methods Attempts to assess the load-bearing capacity of bodies and individual cross-sections have been made for the simplest loading cases - tension, pure bending, torsion. The bearing capacity of cross-sections is evaluated by specific strength [9-11] using a nondimensional coefficient θC , called the strength utilization factor, which varies from 0 to 1. Specific strength is numerically characterized by coefficient θC . θ =C Fσ ydF = M , (1) Fσc y dFmax y Fmax σav where M is the load-bearing capacity of the cross-section; ymax is the half-height of cross-section; σav is the average reduced resistance of the cross section; F is the area of the cross-section under study. In evaluation of the bearing capacity, the existing bearing capacity must be compared with the maximum possible one and, thus, the unused reserves are identified. The given formula (1) allows to evaluate the degree of utilization of the mechanical properties of the material and the geometry of the body based on the specific strength of the body, i.e. strength per unit cross-sectional area of the body, per unit “arm” and per unit of average strength of the material. The denominator in formula (1) represents the load-bearing capacity of some ideal cross-section, the entire area F of which is concentrated at the end of the “arm” ymax , and the material of which has an average resistance of σav . Based on this, it can be said that the specific strength of a body is the ratio of the bearing capacity of the body to the bearing capacity of an ideal body with the same dimensions. An ideal body (cross-section) has an average resistance, i.e. resistance equal to σav = F1 FσcdF , (2) where σc is the reduced resistance of the material of the cross-section, depending on the coordinates of the cross-section points. The reduced resistance based on the most common theories of strength (first and third) is either the resistance to brittle fracture, or the yield strength, or variable resistance to plastic deformation if hardening is taken into account. Thus, specific strength (similar to efficiency) is a dimensionless number varying from 0 to 1. A specific strength value equal to one (achieved only in rare cases) corresponds to 100% utilization of the given dimensions and strength of the material. Below is an attempt to calculate the specific strength value for combined loading cases for a straight beam. In the general case, the internal forces in the given section of a beam form the resultant vector and the resultant moment [11; 12], which can be decomposed into the following components (Figure 1). The resultant moment is decomposed into two components: perpendicular to the cross-section plane, MT (torque moment), and lying in the section plane, M B (bending moment). The resultant vector - into force P perpendicular to the cross-section (applied at the center of gravity of the cross-section) and the shear force, which is not taken into account. An ideal section can be represented as two concentrated areas Fi spaced from each other at the maximum possible distance h (dimension) perpendicular to the M B vector (Figure 1). This type of cross-section is ideal for bending moment apprehension. For torsion resistance, a thin-walled ring of diameter h with area FK is ideal. It is assumed that steady state is maintained for ideal sections. Both types of cross-sections are able to carry axial load. Thus, the ideal cross-section for a combined load is assumed to consist of the sum of two separate types of ideal cross-sections [13-15]. The load-bearing capacity of the real cross-section is determined by a set of values P M M, B , T , which satisfy the strength conditions (according to the chosen strength theory) with a certain reserved strength. In order to ensure that the specified load acting on the real Figure 1. Two types of ideal cross sections S o u r c e: made by A.K. Kurbanmagomedov cross-section can also be accepted by the ideal cross-section, it is necessary to find the values of the areas Fi and FK of the ideal cross-section. The required areas Fi and FK can be found from the strength conditions for both types of ideal cross-sections. For the part of the ideal cross-section subjected to bending and axial load, the following is obtained: σav = P + 2MB . (3) Fi + FK hFi For the annular part of an ideal cross-section, subjected to torsion and axial load, the following is obtained: ® according to the 1-st theory of strength 2 2 2σ =av P + P +4 2MT , (4) F Fi + K F Fi + K hFK ® according to the 3-rd theory of strength 2 2 σ =av2 P +4 2MT . (5) F Fi + K hFK The presented equations determine the minimum required areas for carrying maximum loads P, M B, MT . With such values of the areas of the ideal cross-section, loads P, M B, MT will be ultimate not only for the given cross-section, but also for the ideal cross-section. The sum of areas Fi +FK gives the total area of the ideal cross-section required to carry the ultimate sum of all types of loads. Moreover, each of the two areas resists with ultimate efficiency only to its own type of load [16-19]. The total area of the ideal crosssection is less than (or equal to) the area of the given cross-section. To assess the degree of utilization of the material properties and the shape of the cross-section, the ratio of the areas can be taken: the minimum area required to withstand the specified combined ultimate load, and the given area that carries the specified combined ultimate load. Then the coefficient evaluating the specific strength will be θC = Fi + FK , (6) F where F Fi, K are the required values of ideal cross-sectional areas calculated using equations (3), (4) or (5). Under the action of only one type of load, coefficient θS according to formula (6) coincides with the strength utilization factor (1) according to the work of N.D. Sobolev and Ya.B. Friedman [7-9]. In fact, substituting the value of the ultimate bending moment from (3) into (1) results in θ =C M = Fi . ymaxFσav F Thus, the ratio of the minimum required area to the given one evaluates the specific strength of the cross-section in the same way as the ratio of the load-bearing capacities. Specific strength in terms of bearing capacity under combined load can be characterized by the ratio of ray segments from the coordinate origin to the real and ideal ultimate load curves: θC = P2 + Mi2 + MT2 , P2 + Mi2 + MK2 where the numerator is the ultimate load of the given cross-section, and the denominator is the ultimate load of the ideal cross-section, the area of which is equal to the area of the given cross-section. However, this method must be abandoned, since adding values with different dimensions can lead to inconsistencies. In formulas (3)-(5), there are maximum loads for the given cross-section, i.e. they satisfy the strength condition according to the maximum normal stress theory 2σ =с (7) or the strength condition according to the maximum shear stress theory 2 σ =с P + MB y + τ4 2 . (8) F Z Here τ is the shear stress at the tangency point of the σc and σ diagrams. The specific strength evaluated using the strength utilization factor (6) shows the degree of utilization of the cross-sectional shape and mechanical properties of the material. As an example of specific strength calculation, some special cases of combined resistance are considered further. 2.1. Pure Bending with Tension 1. Let a beam have rectangular cross-section and constant material resistance. Bending moment M B acts in the principal plane. Using the 3-rd theory of strength according to equation (8) for the given crosssection (rectangle), the following is obtained: Fσ =c P + 6MB . (9) h Substituting the loads from (9) into (3) and then into (5), the strength utilization factor is obtained: θC = P + 2MB = -1 4MB . Fσ σav hF av hFσс In Figure 2, the values of θC are plotted on the lines of ultimate forces and moments [1]. The specific strength of the body equal to 0.33 under bending increases linearly P with the increase of ratio and becomes equal to one M n in the absence of bending. σс 2. Let the resistance increase linearly from on the 2 central axis to the value of σс at the upper and lower Figure 2. The combined effect points of the rectangular cross-section. The average of bending and stretching on the beam S o u r c e : made by A.K. Kurbanmagomedov resistance in this case is equal to σс. Here, when calculating the strength, it is necessary to distinguish between three cases (Figure 3) depending on the points of the cross-section at which the σ and σс diagrams contact. Figure 3. Stress and resistance diagrams for bending and stretching S o u r c e: made by A.K. Kurbanmagomedov From equation (8), the following is obtained: Fσс = +P 6 M , (10) h where for the above three cases the loads are limited as follows: Case І Case 2 Case 3 P=Fσc M <Wσc P=Fσc W =Wσc P<Fσc M >Wσc The strength utilization factor is obtained from (6) by substituting the loads from (10) into (3): θC = 4 P + 8 M . 3 Fσс 3 hFσс In Figure 4, strength utilization factor θC is plotted on the line of ultimate loads P and MB. At a value of P= 6 M the specific strength is the h highest due to the coincidence of σ and σC curves. In tension only, the specific strength decreases to 0.66, and in bending only to 0.45. In Figure 4, strength utilization factor θC is plotted on the line of ultimate loads P and MB. At a value of P= 6 M the specific strength is the h Figure 4. The strength utilization factor under combined highest due to the coincidence of σ and σC curves. bending and stretching of the beam In tension only, the specific strength decreases to S o u r c e: made by A.K. Kurbanmagomedov 0.66, and in bending only to 0.45. 2.2. Tension with Pure Bending and Torsion 3. Let the cross-section be round with constant resistance. For such a cross-section, the following is obtained from (8): 2 2 (Fσc 2 = + P 8 M B + 64 MT . d d Substituting these load values into formulas (3) and (5) and then into (6), the load-bearing capacity coefficient is obtained, which is plotted for the values of the ultimate loads corresponding to the coordinate planes (Figure 5). Figure 5. The combined effect of stretching, torsion and bending on the beam S o u r c e: made by A.K. Kurbanmagomedov 4. Let the same beam have variable resistance of the cross-sectional material. At the center of the σc round cross-section the resistance is equal to . The resistance increases linearly along the radius to the 2 5σc value of σC at the circumference. The average cross-sectional resistance according to (2) is equal to . 6 The point of contact of diagrams σ and σC will be in the center of the circle at P=σc ; n F( σ )2 > + P 8 M B 2 + 64 MT 2 . c F 2 d d Contact of σ and σC will occur on the surface if 2 2 P σc 2 = P F < 2 ; σc F W+MB + 4 MWkTp . If the stress and resistance diagrams contact simultaneously both in the center and on the surface, then 2 2 σc2 = σ2c +MWB +4 MWpT . Based on these cases, Figure 6 shows the surface of ultimate P, MB, MT and the strength utilization factor. The most favorable load ratios are determined by the greatest mutual approach of the σ and σC diagrams, at which the specific strength is the highest [17-20]. Figure 6. Cross-section of regions on coordinate planes S o u r c e: made by A.K. Kurbanmagomedov 2.3. Pure Bending with Torsion 5. Let the cross-section be elliptical with constant resistance. The bending moment acts in the plane of the major axis of the ellipse. When the reduced stress at the end points of the major axis of the ellipse reaches the resistance value, the load-bearing capacity of the section will be exhausted and according to (8) a2 F2σc2 =M MB2 + T2 . 16 In this respect, there are three possible cases: I. II. III. πab2 MT = MB < 4 πab2 MT = MB = 4 πab2 MT < MB > 4 In case 1, σ touches σC at the extreme points of the minor axis of the ellipse (with that a < b); in case 3 - at the extreme points of the major axis of the ellipse (a > b); in case II (a = b) - simultaneously at both of these and other points of the cross-section. Plotting strength utilization factor on the curves of the ultimate MB and MT (Figure 7), it can be observed that the specific strength depends on the ratio of semi-axes of the ellipse and with an increase in b the proportion of torque it varies from 0.25 to 0.5 . a Figure 7. The strength utilization factor in case of combined torsion and bending S o u r c e: made by A.K. Kurbanmagomedov For a particular ratio between bending and torque ultimate moments, the most favorable ratios b of ellipse semi-axes can be found. For example, for case II, equating the derivative of θC in terms of a b to zero, the most favorable cross-sectional shape is obtained at = 0.89. a Let the application of the concept of specific strength to the calculation of the optimal cross-section dimensions be demonstrated using a specific example. Cantilever steel tubular shaft is subjected to static bending moment MB = Pl and torque МT = Pa. The shaft has variable strength in the normal annular cross-section. The outer layer of the shaft with a thickness of σ = 5 mm has a strength K times greater than the core (increasing the strength of the surface layers can be structurally achieved, for example, by freeze casting a stronger material onto the core). The optimal value of the internal diameter d is determined assuming that the remaining parameters are specified and constant, and the violation of strength occurs in the subsurface layer of the shaft. The following notation is adopted: D = 30mm is the outer diameter of the shaft annular section; σsl , σ sc are the resistances (strengths) of the material of the surface layer and the shaft core respectively; C= d ; h= =δ 0.167; K = σsl = =50 1.66. D D σsc 30 From (8), the ultimate value of the force is obtained as πD4(1-c4)σsc P =. 32(D-2δ) l2 +a2 The average resistance according to (2) will be σav = 4h(1-2h)(σ σsl - sc )+σsc. 1-c Substituting P and σav into (3), (5) and (6), the strength utilization factor is obtained as (l + 2a)(1-c4) θC =. 2 1( -2h) 4h(1-h)(k - + -1) (1 c2) l2 +a2 From the condition of maximum specific strength of the cross-section ∂θC =0, the equation for ∂с determining the best value of C is obtained: с4 - 8 1h( -h k)( - +1 2) c2 + =1 0 . For the considered case, this results in C = 0.57, or the internal diameter should be equal to 17 mm. 2.4. Specific Strength of Thick-Walled Pipes The specific strength of a beam under combined loading was discussed above. Let the definition of specific strength be applied to evaluate the extent to which the strength of thick-walled pipes loaded with internal pressure is utilized [21-23]. A thin-walled pipe is assumed to be an ideal body, based on obtaining the maximum load-bearing capacity for a given internal radius of the pipe, considering that the entire thickness of the thick-walled pipe under consideration is concentrated in its internal diameter. The resistance of the material of an ideal pipe, according to definition, is equal to the average resistance of the material of the given thick-walled pipe [23-26]. Then the reference load-bearing capacity of an ideal pipe will be: σav (R-r) Pid = , r where σav = rRσcdρ is average resistance. R r- The specific strength of a thick-walled pipe will be: θC = P = pr ) , Pid σav (R r- where p is the load-bearing capacity of this pipe; R, r are the outer and inner radii of the thick-walled pipe under consideration. The resulting coefficient can be represented as a product of the coefficients of equal strength and shape: θC = P P' =θ θp sh, P P' id where P ' is the maximum internal pressure in the thick-walled pipe, in which the reduced stresses σ throughout the entire thickness of the pipe coincide with the σc resistance diagram. 3. Results and Discussions The coincidence of the diagram of reduced stress from internal pressure with the resistance diagram can be achieved, for example, by continuously increasing the elastic modulus along the thickness of the pipe from the inner to the outer surface. The practical implementation of a continuous change in the elastic modulus according to the desired law is apparently difficult to achieve. The specific strength calculated for some cases of resistance of thick-walled pipes is given in Table. The Table shows that for a jointed pipe and for a pipe, part of the wall of which has crossed into the plastic zone, the strength utilization factor is approximately the same. In the case of a very thick pipe wall, the specific strength drops significantly, which indicates that pipes with very thick walls are unprofitable. By changing the diagram of the yield strength along the thickness of the pipe, it is possible to increase the strength utilization of the pipe yet in the elastic stage to the degree of utilization in case of plastic operation of the pipe with constant resistance. Material Chemical Composition Physical Properties Description and Application Rockwell Hardness Number Cold Bend Angele Cold Finished High Yellow Brass Copper - 65% Zinc - 35% Hard Copper - 64.50-67.50% Zinc - 32.09% Lead - 0.35% Iron - 0.06% Cold Drawn Seamless Tubing All sizes 68.000 40.000 - B75 to B85 90o around radius equal to thickness • Good for flat work. • Suitable for ornamental plate or panel parts. • Used for Squirrel Cage rotor and rings. None Dimensions, Tolerances, Quality and Appearance Cold drawn Seamless tubing high yellow brass Copper - 67.5% Zinc - 32% Lead - 0.5% Light Anneal Copper - 65.00-8.00% Zinc - 0.98% Lead - 0.80% Iron - 0.07% Tin -0.15% Cold Drawn Seamless Tubing All sizes 45.000 18.000 25% - 180o around Pin equal to 1- times • A 7.5 cm long piece of tubing, after being split lengthwise will withstand opening out flat without showing cracks or flaws. • Example of use: Coil spools when ends are to be spun over. None Dimensions, Tolerances, Quality and Appearance Cold drawn Seamless tubing high yellow brass Copper - 67.5% Zinc - 32% Lead - 0.5% Half hard Copper - 65.00-68.00% Zinc - 30.98% Lead - 0.80% Iron - 0.07% Tin - 0.15% Cold Drawn Seamless Tubing All sizes Physical properties on all sizes not available. May be obtained on specific sizes when desired. • Suitable for brass railings. Also coil spools when not spun over. None Dimensions, Tolerances, Quality and Appearance Material specifications S o u r c e: made by A.K. Kurbanmagomedov 4. Conclusion The following conclusion is made based on the obtained results: 1. The paper provides a method for calculating the specific strength for a beam under combined loading, as well as for thick-walled pipes loaded with internal pressure.

About the authors

Arslan K. Kurbanmagomedov

RUDN University

Author for correspondence.
Email: kurbanmagomedov_ak@pfur.ru
ORCID iD: 0000-0001-9158-0378
SPIN-code: 5262-5269

Candidate of Physical and Mathematical Sciences, senior lecturer, Nikolskii Mathematical Institute

Moscow, Russia

Evgeny M. Morozov

National Research Nuclear University MEPhI

Email: evgeny.morozof@gmail.com
ORCID iD: 0000-0002-4824-8481
SPIN-code: 3989-2934

Doctor of Technical Sciences, Professor

Moscow, Russia

References