A manual for the calculation of wooden structures. Examples of calculations of wooden structures of forest engineering structures. Calculation of wooden floors
For all building materials there are areas of rational and effective use. This also applies to wood, which is a local building material in many areas of our country. In some areas, wood is available in abundance (in so-called forest-surplus areas).
Our country is the first in the world in terms of the number of forest areas (Brazil ranks 2nd, Canada ranks 3rd, and the USA ranks 4th), which occupy almost half of Russia’s territory – approximately 12.3 million km 2 . The main part of Russian forests (about ¾ of the part) is located in the regions of Siberia, the Far East, and in the northern regions of the European part of the country. The predominant species are conifers: 37% of the forests are larch, 19% - pine, 20% - spruce and fir, 8% - cedar. Deciduous trees occupy about ¼ of our forest area. The most common species is birch, occupying about 1/6 of the total forest area.
Wood reserves in our forests amount to about 80 billion m3. About 280 million m3 are harvested annually. industrial wood (i.e. suitable for the manufacture of structures and products). However, this amount does not exhaust the natural annual growth of wood in remote areas of Siberia and the Far East.
History of creation wooden buildings and structures dates back to ancient times. First structural form The buildings were made of logs, rectangular in plan. The area and volume of structures under construction gradually increased, and the functional purpose of the premises expanded. Log houses began to be built polygonal in plan with the presence of internal walls, ensuring the immutability of structures and the stability of external walls.
The presence of huge forest reserves on the territory of Russia was the basis for the centuries-old use of wood as a building material for the construction of buildings and structures for residential, commercial, religious and other purposes. To this day, unique structures made by architects in the form of a log house more than 250 years ago have been preserved. An example of such construction are the existing churches in Kizhi on Lake Onega, buildings in Malye Karely in the Arkhangelsk region (Fig. 1).
The first engineering structures of mankind - pile buildings, bridges and dams - were also made of wood. Since the end of the 17th century, when it became possible to saw logs into beams and boards, wooden construction reached new stage. More economical and lighter sections of wood made it possible to create effective rod systems that could span significant spans, which gave impetus to the development of architecture and bridge construction. Most a shining example the use of wood as rafter structures is the design of the Admiralty spire (Fig. 2), carried out according to the design of I.K. Korobova and saved by A.D. Zakharov during the reconstruction of the tower at the beginning of the 19th century, trusses for covering the Manege in Moscow with a span of 48 m, built in 1817 by A.A. Betancourt (Fig. 3).
Fig. 1 – Wooden churches in Kizhi on Lake Onega
Fig. 2 – Admiralty building in St. Petersburg
Fig. 3 – Installation of Manege covering trusses in Moscow
Many years of experience in building construction for various purposes allowed us to determine rational areas of application of wooden structures:
1. Visual and public buildings, athletic facilities, exhibition pavilions, markets and others with a span of 18 to 100 m (see example in Fig. 4).
2. Coatings of civil, industrial and agricultural buildings. It is advisable to use plank and cobblestone trusses with assembly at the construction site (the effectiveness of application is determined by lightness, strength and favorable conditions for combating shortcomings).
3. Buildings with a chemically aggressive environment. First of all, warehouse buildings with a span of up to 45 m for reloading and storing mineral fertilizers.
4. Low-rise wooden housing construction.
5. Industrial agricultural buildings.
6. Unheated buildings for production and auxiliary purposes of industrial enterprises.
7. Unheated buildings and sheds for storage and processing of agricultural products.
8. Prefabricated buildings of complete supply of small spans for remote areas of the Far North.
9. Engineering structures - power transmission line supports (with voltage up to 35 kV), triangulation and radio-transparent masts and towers, light-weight bridges, pedestrian bridges.
Fig. 4 – Diagram of the frame of the indoor track and field athletics arena at the Meteor sports complex in Zhukovsky with load-bearing laminated board arches
It is not advisable to use wooden structures in places where measures to protect wood from fire and alternating moisture (and therefore rotting) are difficult:
Hot shops;
Industrial building with large crane loads;
Premises with high operational humidity (except baths).
Despite the centuries-old use of wood as building structures, the search for new technical solutions continues. Over the past 20 years, the development of rigid connections of laminated wood elements (by analogy with embedded parts of reinforced concrete structures) has been underway, which has made it possible to open a new direction of prefabricated laminated wood structures. In construction practice in Russia and abroad, it has been implemented a large number of long-span buildings and structures made of prefabricated laminated timber structures. The combination of glued-laminated beams with linear reinforcement of laminated timber elements is a further step in the development of laminated timber structures for very long span buildings.
Progressive forms of industrial wood structures:
1. Monolithic laminated board and glued plywood structures in the form of beams, arches, frames and combined systems.
2. Metal-wood trusses with laminated board top chord.
3. Circular mesh spatial designs from standard solid and glued jambs.
Unlike wood, plastics began to be used in structures from the middle of the last century, after the emergence industrial production synthetic materials.
The main structural construction plastics include:
High-strength fiberglass;
Transparent less durable fiberglass;
Plexiglas;
Viniplast;
Styrofoam;
Air- and waterproof fabrics and films;
Wood plastics.
Plastic structures are used mainly in the form wall panels, covering slabs, translucent enclosing elements of various shapes and many individual designs produced in small batches.
The most durable fiberglass plastics, the calculated compressive and tensile strength of which reaches 100 MPa, are used to make elements of load-bearing building structures. However, this application is only possible with a technical and economic feasibility study. Transparent fiberglass is used as translucent elements of building envelopes. Transparent parts of fencing are made from especially transparent plexiglass and transparent vinyl plastic, allowing all parts of the solar spectrum to pass through. Ultralight foam plastics are used in the middle layers of lightweight enclosing coatings and walls.
A special class of plastic structures are membranes (strong, thin air- and waterproof fabrics), which are used in the form of pneumatic and awning structures. The material in them works in tension and there is no danger of loss of stability.
CHAPTER 1. WOOD AND PLASTICS - BUILDING MATERIALS
1.1 ADVANTAGES AND DISADVANTAGES OF WOOD
The main advantages of wood include:
Light weight. Wood has an average density of 550 kg/m3 and is 14 times lighter than steel, 4.5 times lighter than concrete, which makes it possible to significantly reduce material costs for transportation, construction of foundations, and do without heavy lifting mechanisms during the construction of buildings and structures.
Strength. One of the indicators of the effectiveness of using structures made of various materials is the specific strength of the material, which is expressed by the ratio of the density of the material to its volumetric weight. For laminated wood this ratio is 3.66×10 -4 1/m, for carbon steel 3.7×10 -4 1/m, for concrete class 22.5 ÷ 1.85×10 -4 1/m. This confirms the feasibility of using laminated wood structures along with steel in long-span buildings, where self-weight is critical.
Deformability and viscosity. Of all the traditional building materials, only wood reacts to a lesser extent to uneven settlement of foundation foundations. The viscous nature of wood destruction (with the exception of chipping) allows the redistribution of forces in the elements, which does not cause instantaneous failure of structures.
Temperature expansion. The coefficient of linear expansion of wood varies along the grain and at an angle to it. Along the fibers, the value of this coefficient is 7-10 times less than across the fibers, and 2-3 times less than for steel. This fact makes it possible to ignore the influence of temperature and does not require dividing the building into temperature blocks.
Thermal conductivity. The low thermal conductivity of wood, due to its structure, is the basis for its widespread use in the walls of enclosing structures. The thermal conductivity coefficient of wood is 6 times lower than that of ceramic bricks, 2 times lower than that of expanded clay concrete, gas-foam concrete with a density of 800 kg/m 3 and is equivalent to gas-foam concrete with a density of 300 kg/m 3, i.e. density is almost half that of wood.
Chemical resistance wood. Wood can be used without additional protection or protected by painting or surface impregnation in a chemically aggressive environment. Wooden structures are used in the construction of warehouses for chemically aggressive bulk materials such as potassium and sodium salts, mineral fertilizers that destroy concrete and steel. Most organic acids do not attack wood at normal temperatures.
Self-renewability of wood. The main advantage of wood compared to other structural materials is the constant renewal of its reserves. The production of other structural materials (steel, concrete, plastic, etc.) requires large amounts of energy and consumes a large amount of raw materials, the reserves of which are constantly running out.
Ease of processing. Wood is easily processed by simple manual or electric tool. The deformability of wood allows structures made from it to be given various rectilinear and curvilinear shapes. Production of small span structures from solid wood can be mastered practically at logging stations, at any base of the construction industry, which is impossible for the production of metal or reinforced concrete structures.
Wood, like other materials, has disadvantages:
Heterogeneity, anisotropy of wood and defects. The heterogeneity of wood is manifested in the difference in the structure and properties of the annual layers formed during the growth of the tree, depending on environmental conditions (climatic conditions).
The heterogeneity of wood affects the variability of strength indicators, which complicates obtaining reliable calculated characteristics of wood.
Wood is a body with three axes of anisotropy along the main structural directions - along and across the fibers in the tangential and radial directions. Significant differences in the strength of wood when forces are applied along and across the fibers significantly complicate the design of wooden structures and, first of all, nodal connections, which often leads to an irrational increase in the cross-sections of the connected elements.
The main defects include knots, cracks and cross-layers. The presence of a knot changes the direction of the wood fibers or interrupts them, which significantly affects the strength, especially when stretching, because uneven loading of all fibers across the cross-section occurs.
Dependence of physical and mechanical properties of wood on humidity. Wood has the ability to absorb moisture due to its hygroscopicity. Its physical and mechanical properties also largely depend on the amount of moisture in wood. Density of freshly cut coniferous wood (except larch) and soft hardwood(aspen, poplar, alder, linden) is 850 kg/m3. As moisture is removed, the density decreases. At 15-25% humidity the density is assumed to be 600 kg/m3, and at 6-12% humidity the density is assumed to be 500 kg/m3. Larch has a density of 800 kg/m 3 and 650 kg/m 3, respectively, with humidity in the range of 15-25% and 6-12%, respectively. Wood for construction is distinguished:
Raw with humidity above 25%;
Semi-dry with humidity 12-25%;
Air-dry with humidity 6-12%.
Creep of wood. Under short-term exposure to a load, wood works almost elastically, but under long-term exposure to a constant load, deformations increase over time. Even at low stress levels, creep can continue for years.
Biodestruction of wood. Directly related to wood moisture content. When humidity is more than 18%, as well as in the presence of oxygen and positive temperature, conditions arise for the life of wood-decaying fungi. Wood is also destroyed by the activity of insects that damage unbarked wood in the forest, in warehouses, cutting areas and destroy debarked wood during its processing and during use in structures.
Spread of fire occurs as a result of the combination of wood carbon with oxygen. Combustion begins at approximately 250 °C. And if the wood quickly burns from the outside, then due to its low thermal conductivity and the appearance of a thick charring layer, which prevents the flow of oxygen, the further process slows down greatly. Therefore, wooden structures with a massive cross-section have greater fire resistance compared to unprotected metal structures.
1.2 WOOD STRUCTURE AND PHYSICAL PROPERTIES
In the cross section of a coniferous wood trunk (pine, spruce), several characteristic layers can be seen (Fig. 1.1).
Outer layer consists of bark - 1 and bast - 2 . Under the phloem is a thin layer of cambium. The purpose of the bast in a growing tree is to carry the nutrient organic substances formed in the leaves down the trunk.
In cross-section, the main part is occupied by sapwood and core. The sapwood consists of young cells, the core consists entirely of dead cells. In trees of all species, at an early age, the wood consists only of sapwood, and only over time does the death of living cells occur, usually accompanied by darkening.
During spring, when a lot of sap appears in the trunk, the cambium develops great activity, depositing a significant number of large cells in the inner part. In summer, as the amount of nutrient juices decreases, the activity of the cambium slows down, and fewer cells and smaller sizes are deposited. IN winter time the vital activity of the cambium subsides and the growth of the tree stops. The deposition of spring and summer parts of wood, which occurs periodically from year to year, is the cause of the formation of annual layers (rings). The growth layer consists of a light layer of wood (earlywood) facing towards the pith and a darker, denser layer of summerwood facing towards the bark (latewood).
Mechanical function in wood, they are performed primarily by prosenchymal cells - tracheids, which are mainly located vertically. The joining of tracheids in the longitudinal direction occurs during the growth process. With their pointed ends, they grow between each other and into other anatomical elements, the so-called “parenchyma cells”, which have the same dimensions in all three axial directions. These cells are part of the “core rays”, which penetrate several annual layers in a perpendicular direction.
Tracheids make up 90% of the total volume of wood, and 1 cm 3 of them contains approximately 420,000 pieces. The tracheids of the early part of the annual layer have thin walls (2-3 µm) and large internal cavities, while the tracheids of the late part of the annual layer have thicker walls (5-7 µm) and smaller cavities. The length of the tracheids is 2-5 mm, the cross-sectional size is 50-60 times less than the length.
For a more complete picture of the structure of wood, three sections of the trunk are considered: transverse, radial and tangential (Fig. 1.2).
Deciduous wood has a slightly different structure from coniferous wood. The spiral direction of the cell walls of hardwood wood leads to large warping and cracking of lumber during drying, and deterioration of nailability. The presence of these shortcomings and low resistance to decay limits the use of hardwood for wooden structures. Higher strength characteristics of hardwood are realized by using them for manufacturing connecting elements(dowels, dowels, linings), as well as supporting antiseptic parts.
Physical properties wood
Density. Since moisture makes up a significant part of the mass of wood, the density value is established at a certain humidity. With increasing humidity, the density increases and, therefore, for calculations when determining permanent loads, the average indicators presented in the standards are used.
For structures operated in conditions where equilibrium humidity does not exceed 12% (heated and unheated rooms with relative humidity up to 75%), the density of pine and spruce is 500 kg/m 3, and larch 650 kg/m 3.
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For structures used outdoors or indoors with high humidity more than 75%, the density of pine and spruce is 600 kg/m 3, and larch 800 kg/m 3.
Thermal conductivity of wood depends on density, humidity and direction of fibers. At equal density and humidity, thermal conductivity across the fibers is 2.5-3 times less than along the fibers. The coefficient of thermal conductivity across the fibers at a standard humidity of 12% is more than 2 times lower than at a humidity of 30%. These indicators are explained by the tubular structure of wood fibers.
Temperature expansion. The coefficient of linear expansion across the grain is proportional to the density of the wood, and is 7 to 10 times greater than the coefficient of expansion along the grain. This is explained by the fact that when heated, wood loses moisture and changes its volume.
In design practice, thermal deformations are practically not considered, since the coefficient of linear expansion along the fibers is insignificant.
1.3 MECHANICAL PROPERTIES OF WOOD
Features of wood.
Size: px
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Transcript
1 Federal agency of Education Government agency higher professional education Ukhta State Technical University Examples of calculations of wooden forest structures engineering structures Tutorial in the discipline "Forest Engineering Structures" Ukhta 008
2 UDC 634* 383 (075) Ch90 Chuprakov, A.M. Examples of calculation of wooden structures of forest engineering structures [Text]: textbook. manual for the discipline “Forest engineering structures” / A.M. Chuprakov. Ukhta: USTU, village: ill. ISBN The textbook is intended for students of the specialty “Forestry Engineering”. The textbook contains examples of the calculation of load-bearing elements and structures made of wood, which consistently outline the application of the basic design provisions to the solution practical problems. At the beginning of each paragraph, brief information is provided to explain and justify the calculation methods used. Toolkit reviewed and approved by the department of “Technologies and Logging Machines”, protocol 14 dated December 7, 007 and proposed for publication. Recommended for publication by the Editorial and Publishing Council of Ukhta State Technical University. Reviewers: V.N. Pantileenko, Ph.D., professor, head. Department of Industrial and Civil Engineering; E.A. Chernyshov, General Director of Severny Les Company LLC. Ukhta State Technical University, 008 Chuprakov A.M., 008 ISBN
3 INTRODUCTION This manual has a primarily educational and methodological goal of teaching students to apply the theoretical information presented in the course “Forest Engineering Structures” and the ability to apply SNiP to solve practical problems. The calculation examples in each section are preceded by brief information to explain and justify the calculation methods and design techniques used. This publication is intended as a guide when conducting practical classes during the study of engineering structures made of wood, when performing calculations coursework, as well as when developing the constructive part of diploma projects. Target this manual fill the gap in the calculation of elements of wooden structures, the ability to apply SNiP for the design of wooden structures in connection with the exclusion of the discipline “Fundamentals of Construction” from the curriculum in the specialty “Forestry Engineering”. It is necessary to design wooden structures in strict accordance with SNiPII.5.80 “Wooden structures. Design standards" and SNiPII.6.74 "Loads and impacts. Design standards". At the end of the manual, auxiliary and reference data necessary for structural calculations are provided in the form of appendices. 3
4 CHAPTER 1 CALCULATION OF ELEMENTS OF WOODEN STRUCTURES Wooden structures are calculated based on two limit states: bearing capacity(strength or stability) and by deformation (by deflection). When calculating using the first limit state, you need to know design resistance, and according to the second, the elastic modulus of wood. The main calculated resistances of pine and spruce wood in structures protected from moisture and heat are given in. The calculated resistances of wood of other species are obtained by multiplying the main calculated resistances by the transition coefficients given in. Unfavorable operating conditions of structures are taken into account by introducing coefficients for reducing design resistances, the values of which are given in [1, table. 10]. When determining the deformations of structures under normal operating conditions, the modulus of elasticity of wood, regardless of the species of the latter, is taken equal to E = kgf/cm. Under unfavorable operating conditions, correction factors are introduced according to. The moisture content of wood used for the manufacture of wooden structures should be no more than 15% for glued structures, no more than 0% for non-glued structures of industrial, public, residential and warehouse buildings and no more than 5% for livestock buildings, outdoor structures and inventory structures temporary buildings and structures. Here and further in the text, numbers in square brackets indicate the serial numbers of the list of references given at the end of the book. 4
5 1. CENTRALLY EXTENSION ELEMENTS Central extension elements are calculated using the formula where N is the design axial force; ** net area of the cross-section under consideration; N R, (1.1) p 5 NT; N T b r o s l b gross cross-sectional area; osl weakening cross-sectional area; R p is the calculated tensile strength of wood along the fibers, Appendix 4. When determining the area of the LT, all weakening located in a section 0 cm long are taken as if combined in one section. Example 1.1. Check the strength of the wooden hanger of the rafters, weakened by two notches h bp = 3.5 cm, side cuts h st = 1 cm and a bolt hole d = 1.6 cm (Fig. 1.1). Calculated tensile force N = 7700 kgf, log diameter D = 16 cm. Solution. Gross cross-sectional area of the rod D 4 = 01 cm. Segment area at cutting depth h bp = 3.5 cm (Appendix 1), 1 = 3.5 cm. Segment area at cutting depth h st = 1 cm = 5.4 cm Since between the weakening by the notches and the weakening of the hole Fig. 1. Tensile element Here and in all subsequent formulas, unless a reservation is made, force factors are expressed in kgf, and geometric characteristics in cm
6 for bolt distance 8 cm< 0 см, то условно считаем эти ослабления совмещенными в одном сечении. Площадь ослабления отверстием для болта осл = d (D h ст) = 1,6 (1,6 1) =,4 см. Площадь сечения стержня нетто за вычетом всех ослаблений нт = бр осл = 01 3,5 5,4,4 = 103 см. Напряжение растяжения по формуле (1.1) кгс/см ЦЕНТРАЛЬНОСЖАТЫЕ ЭЛЕМЕНТЫ Центральносжатые деревянные стержни в расчетном отношении можно разделить на три группы: стержни малой гибкости (λ < 30), стержни средней гибкости (λ = 30 70) и стержни большой гибкости (λ >70). Low-flexibility rods are calculated only for strength using the formula N R. (1.) c High-flexibility rods are calculated only for stability using the HT formula N r a s h R s. (1.3) Rods of medium flexibility with weakening must be calculated for both strength according to formula (1.) and stability according to formula (1.3). The calculated area (calculation) of the rod for calculating stability in the absence of weakening and with weakening that does not extend to its edges (Fig. a), if the area of weakening does not exceed 0.5 br, is taken equal to 6
7 calculated = 6p, where 6p is the gross cross-sectional area; for weakening that does not extend to the edges, if the weakening area exceeds 0.5 6p, the calculation is taken equal to 4/3 NT; with symmetrical weakening extending to the edges (Fig. b), calculation = NT. The longitudinal bending coefficient is determined depending on the calculated flexibility of the element using the formulas: with element flexibility λ 70 1 a 100 ; (1.4) with element flexibility λ > 70 Fig. Weakening of compressed elements: a) not extending to the edge; b) facing edge A, (1.5) where: coefficient a = 0.8 for wood and a = 1 for plywood; coefficient A = 3000 for wood and A = 500 for plywood. The coefficient values calculated using these formulas are given in the Appendix. The flexibility λ of solid rods is determined by the formula l 0, (1.6) where l 0 is the design length of the element. To determine the design length of straight elements loaded with longitudinal forces at the ends, the coefficient μ 0 should be taken equal: with hinged ends, as well as with hinged connections at intermediate points of element 1 (Fig. 3.1); r 7
8 with one hinged and the other pinched ends 0.8 (Fig. 3.); with one pinched and the other free loaded end (Fig. 3.3); with both ends pinched 0.65 (Fig. 3.4). r radius of inertia of the element’s section. Rice. 3 Schemes for fastening the ends of the rods The radius of inertia r in the general case is determined by the formula r J br, (1.7) br where J br and 6p the moment of inertia and the gross cross-sectional area of the element. For a rectangular section with side dimensions b and h r x = 0.9 h; r y = 0.9 b. For a circular cross section (1.7a) r D 0.5 D. (1.7b) 4 8
9 The design flexibility of compressed elements should not exceed the following limit values: for the main compressed elements of the chords, support braces and support posts of trusses, columns 10; for secondary compressed elements, intermediate posts and truss braces, etc. 150; for link elements 00. The selection of sections of centrally compressed flexible rods is carried out in the following order: a) they are set by the flexibility of the rod (for the main elements λ =; for secondary elements λ =) and find the corresponding value of the coefficient; b) determine the required radius of gyration and set a smaller cross-sectional size; c) determine the required area and set the second cross-sectional size; d) check the accepted cross section using formula (1.3). Compressed elements made of logs while maintaining their conicity are calculated using a section in the middle of the length of the rod. The diameter of the log in the design section is determined by the formula D calculated = D 0 +0.008 x, (1.8) where D 0 is the diameter of the log at the thin end; x is the distance from the thin end to the section under consideration. Example 1. Check the strength and stability of a compressed rod weakened in the middle of the length by two holes for bolts d = 16 mm (Fig. 4, a). Rod cross-section b x h = 13 x 18 cm, length l =.5 m, ends are hinged. Design load N = kgf. Solution. Estimated free length of the rod l 0 = l =.5 m. Minimum radius of gyration of the section r = 0.9 b = 0.9 13 = 3.76 cm. 9
10 Fig. 4. Centrally compressed elements The greatest flexibility, 7 6 Therefore, the rod must be designed for both strength and stability. Net area of the rod nt = br osl = .6 13 = 19.4 cm. Compressive stress according to formula (1.) k g / s m. 1 9. 4 10
11 Buckling coefficient according to formula (1.4) 6 6, 6 1 0, 8 0, The weakening area is from the gross area of \u200b\u200babout sl br 1, 8 5% Therefore, the calculated area in this case calc = br = = 34 cm. Stress when calculating stability according to formula (1.3) to g s / s m R c 0, Example 1.3. Select the cross-section of the wooden block rack (Fig. 4, b) with the following data: design compressive force N = kgf; stand length l = 3.4 m, the ends are hinged. Solution. We set the flexibility of the rack to λ = 80. The coefficient corresponding to this flexibility is = 0.48 (Appendix). Find the required minimum radius of gyration (at λ = 80) l l 1 l cm; 0 0 r tr l, 5 cm 80 and the required cross-sectional area of the rack (at φ = 0.48) tr N cm R 0, c Then the required cross-sectional width of the beam according to formula (1.7a) b tr rtr 4, 5 1 4, 7 cm. 0, 9 0, 9 In accordance with the assortment of lumber, we accept b = 15 cm. The required height of the beam section. eleven
12 h tr tr 7 1 8.1 cm b 15 Take h = 18 cm; = = 70 cm. Flexibility of the rod of the accepted cross-section Stress l, 5 y r 0, m and n; u = 0.5. N k g s / s m 0, Example 1.4. A wooden post with a round cross-section, while maintaining a natural slope, carries a load N = (Fig. 4, c). The ends of the stand are hinged. Determine the diameter of the rack if its height is l = 4 m. Solution. We set the flexibility λ = 80 and find the coefficient corresponding to this flexibility = 0.48 (Appendix). We determine the required radius of gyration and the corresponding cross-section diameter: r tr l 400 r 0 tr 5 cm; D " 0 cm tr 80 0.5 We determine the required area and the corresponding cross-section diameter: hence tr N cm R 0, D "" tr Average required diameter c; tr 4 tr, 9 cm 3.1 4 D tr D " D " 1 9. 4 5 cm. D; 4. 1
13 We take the diameter of the log at the thin end D 0 = 18 cm. Then the diameter in the design section located in the middle of the length of the element is determined by formula (1.8): D = , = 19.6 cm; D 3, 6 30 cm. 4 4 Checking the accepted cross-section, 5 1 9, 6 ; 0, 4 6 ; k g s / s m 0, BENDING ELEMENTS Elements of wooden structures that work in bending (beams) are calculated for strength and deflection. Strength calculations are carried out using the formula M R, (1.9) u W where M is the bending moment from the design load; W HT the net moment of resistance of the section under consideration; R u is the calculated bending resistance of wood. The deflections of bending elements are calculated from the action of standard loads. The deflection values should not exceed the following values: for beams between floors 1 / 50 l; for attic floor beams, purlins and rafters 1 / 00 l; for lathing and flooring 1/150 l, where l is the design span of the beam. The values of bending moments and deflections of beams are calculated using general formulas structural mechanics. For a beam on two supports loaded with a uniformly distributed load, the moment and relative deflection are calculated using the formulas: HT 13
14 ql 8 M; (1.10) f 5 q l l H 3. (1.11) 384EJ The design span is taken equal to the distance between the centers of the beam supports. If the beam support width is unknown in preliminary calculations, then the clear span l 0 increased by 5% is taken as the design span of the beam, i.e. l = 1.05 l 0. When calculating elements made of solid logs or logs sawn by one , two or four edges, take into account their natural run (taper). With a uniformly distributed load, the calculation is carried out along the section in the middle of the span. Example 1.5. Design and calculate the attic floor using wooden beams located at B = 1 m from one another. The width of the room (clear span) l 0 = 5 m. Solution. We accept this floor design (Fig. 5, a). Skull bars are nailed to the wooden beams l, resting on the walls of the building, on which are laid rolling boards 3, consisting of a solid plank flooring and four bars hemmed to it (Fig. 5, b). Dry gypsum plaster 4, coated on the inside with bitumen, is nailed to the bevel bars from below. On top of the board flooring, a vapor barrier 5 is first laid in the form of a cm thick layer of impregnated clay, and then insulation 6 is expanded perlite, vermiculite or other fireproof backfill materials, prepared from local raw materials and having a density (volumetric mass) γ = kg/m 3. Thickness layer of insulation 1 cm. A protective lime-sand crust 7 cm thick is placed on top of the insulation. Calculate loads. We determine the loads per 1 m of flooring (Table 1.1). 14
15 Fig. 5. To the calculation of attic floor beams Table 1.1 Elements and calculation of loads Lime-sand crust, 0, Insulation, 0.1 350 Clay lubricant, 0, Rolling boards (flooring + 50% on bars), 0.5 Dry plaster with bitumen, 0, 5 Payload Total... Standard load, kgf/m g, Load factor 1, 1, 1, 1.1 1.1 1.4 Design load, kgf/m 38.4 50.4 38.4 15.6 17, We do not take into account the own weight of the beams, since the loads from all other floor elements listed in the table were assumed to be distributed over the entire area without excluding the areas occupied by the beams. 15
16 Calculation of floor beams. When placing beams every 1 m, the linear load on the beam is: standard q H = 11 1 = 11 kgf/m; calculated q=65 1=65 kgf/m. Design span of the beam l = 1.05 l 0 = 1.05 5 = 5.5 m. Bending moment according to formula (1.10) M k gf / m. 8 Required moment of resistance of the beam W tr M cm. R and 130 Given the section width b = 10 cm, find h tr 6W tr, 6 cm. b 10 We take a beam with a cross section bxh = 10 x cm with W = 807 cm 3 and J = 8873 cm 4. Relative deflection according to formula (1.11) f l 3 5, Calculation of the shield roll forward. We calculate the panel deck for two loading cases: a) permanent and temporary load; b) assembly centered design load P = 10 kgf. In the first case, we calculate the flooring for a strip 1 m wide. Load per 1 linear line. m of design strip: q H = 11 kgf/m; q = 65 kgf/m. Design span of the flooring a 4 l B b cm. H Here B is the distance between the axes of the beams; b beam section width; and the cross-sectional width of the cranial block.. 16
17 Bending moment M 6 5 0.8 6 4.5 k gf / m. 8 The thickness of the flooring boards is taken equal to δ = 19 mm. The moments of resistance and inertia of the design strip of the flooring are equal to: W Bending stress J, cm; , cm, k g s / s m. 6 0, Relative deflection fl 3 5, Significant reserves of strength and rigidity of the flooring make it possible to use grade III semi-edged boards for its production. When the thickness of the flooring is reduced to 16 mm, its deflection will be more than the maximum. If there are distribution bars hemmed from below, the concentrated load is assumed to be distributed over a deck width of 0.5 m. We consider the load to be applied in the middle of the deck span. Bending moment M Pl H k g s / s m. 4 4 Moment of resistance of the design strip. W 5 0 1.1 cm. 6 17
18 Bending stress, g s / s m, 3 0.1 where 1 is a coefficient taking into account the short duration of action installation load. 4. TENSION-BENDING AND COMPRESSION-BENDING ELEMENTS Tension-bending and compression-bending elements are subject to the simultaneous action of axial forces and a bending moment resulting from transverse bending of the rod or eccentric application of longitudinal forces. Tensile bending rods are calculated using the formula N M R p R. (1.1) p W R H T H T and Compression bending rods in the bending plane are calculated using the formula N M R c R W R H T H T u c, (1.13) where the coefficient taking into account the additional moment from the longitudinal force during deformation of the rod, determined by the formula 1 N 3100 R with br. Compressed bending rods with lower cross-sectional rigidity in the plane perpendicular to the bending must be checked in this plane for general stability without taking into account the bending moment according to formula (1.3). 18
19 Example 1.6. Check the strength of a beam with a cross section of 13 x 18 cm (Fig. 6), stretched by a force N = kgf and bent by a concentrated load P = 380 kgf, applied in the middle of the span l = 3 m. The cross section of the rod in this place is weakened by two holes for bolts d = 16 mm. Rice. 6. Tensile bending element Solution. Maximum bending moment M Pl k g s / m. 4 4 Net cross-sectional area nt = b (h d) = 13 (18 1.6) = 19.4 cm Moment of inertia of weakened section bh J b d a cm HT 1 1 Moment of resistance W HT J 5750 HT see 0.5 h 9 19
20 Stress according to formula (1.1), k g s / s m. 1 9, Example 1.7. Check the strength and stability of the compressed-bending rod, hinged at the ends (Fig. 7). Section dimensions b x h = 13 x 18 cm, rod length l = 4 m. Design compressive force N = 6500 kgf, design concentrated force applied in the middle of the rod length, P = 400 kgf. Rice. 7. Compressed bending elements Solution. Let's check the strength of the rod in the bending plane. Design bending moment from transverse load M Pl k g s / m. 4 4 Section area = = 34 cm. Sectional moment of resistance W x = bh /6 = 70 cm 3. 0
21 Radius of inertia of the section relative to the X axis r к = 0.9 h = 0.9 18 = 5, cm Flexibility of the rod x 5, Coefficient according to formula (1.14), Stress according to formula (1.13) k g s / s m 3 4 0, Let's check the stability of the rod in a plane perpendicular to the bend. Radius of inertia of the section relative to the Y axis r y = 0.9 b = 0.9 13 = 3.76 cm. Flexibility of the rod relative to the Y axis y 3.7 6 Buckling coefficient (as applied) φ = 0.76. Stress according to formula (1.3) k g s / s m 0,
22 CHAPTER CALCULATION OF CONNECTIONS OF ELEMENTS OF WOODEN STRUCTURES 5. JOINTS ON NOTCHES Elements on notches are connected mainly in the form of frontal notches with one tooth (Fig. 8). Frontal notches are designed for crushing and spalling based on the condition that the design force acting on the connection does not exceed the design load-bearing capacity of the latter. Rice. 8. Frontal cut
23 Calculation of frontal notches for crushing is carried out according to the basic work plane crushing, located perpendicular to the axis of the adjacent compressed element, to the total force acting in this element. The calculated load-bearing capacity of the connection from the crushing condition is determined by the formula T R cm cm cm, (.1) where is the crushing area; R cm cm calculated resistance of wood to crushing at an angle to the direction of the fibers, determined by the formula R cm R cm R cm sin R cm 90. (.) The depth of notches in the support nodes of rod structures should be no more than 1 3 h, and in intermediate nodes not more than 1 4 h, where h is the cross-sectional size of the element in the cutting direction. The design load-bearing capacity of a connection based on the shearing condition is determined by the formula where is the shearing area; sk av, (.3) s k s k s k T R av R calculated average chipping resistance of wood over the sk cleaving area. The length of the shearing area l sk in frontal cuts must be at least 1.5 h. The average calculated chipping resistance over the shearing area with a platform length of no more than h and ten insertion depths in joints made of pine and spruce is taken equal to avg 1 /. R k gf s m For length l ck more than h, the calculated shear resistance is reduced and is taken according to Table 1. 3
24 sr l sk h Table.1,4,6,8 3 3, 3.33 R, k gf / s msk 1 11.4 10.9 10.4 10 9.5 9. 9 For intermediate values of the ratio l sk / h the values of the calculated resistances are determined by interpolation. Example.1. Check the load-bearing capacity of the truss support unit, solved by a frontal notch with one tooth (Fig. 8, a). Section of beams b x h = 15 x 0 cm; angle between belts " "(s in 0, 3 7 1; c o s 0, 9 8); cutting depth h = 5.5 cm; length of the shearing platform l ск = 10 h рр = 55 cm; calculated compressive force in the upper belt N c = 8900 kgf. Solution. Calculated resistance of wood to crushing at an angle according to the formula (.) Crushing area 130 R / 130 k gf s m cm, cm bhv 1 5 5. 5 8 8. 8 cm c o s 0. 9 8 Load-bearing capacity of the connection from the condition of bearing strength according to the formula (.1) T 8 8, N to gs. cm Design force acting on the shearing area, T N N c o s to gf. Shearing area p c c c c k l b cm c.. 4
25 Calculated average chipping resistance of wood at the ratio l sk / h = 55/0 =.75 avg 1 0.1 / (see Table 1). R k gf s m Load-bearing capacity of the connection from the condition of chipping strength according to formula (.3) T sk, k gf. Example.. Calculate the frontal notch of the support unit of a triangular truss truss (Fig. 8, b). The truss chords are made of logs with a design diameter at the node D = cm. The angle between the chords is a = 6 30" (sin a = 0.446; cos a = 0.895). The design compressive force in the upper chord is N c = kgf. Solution. Design resistance of wood crushing at a given angle cm / (Appendix 4). cm cm Using Appendix 1, we find that with D = cm, the nearest area seg = 93.9 cm corresponds to the cutting depth h bp = 6.5 cm. We accept h bp = 6.5 cm, which is less than the maximum cutting depth, which in this case, taking into account the necessary undercutting of the log of the lower belt to a depth of h CT = cm is 1 D h st h h 6, 6 7 cm wr Length of the cutting chord (width of the shearing plane) at h wr = 6.5 cm b = 0.1 cm (Appendix 15
26 Required length of the shearing plane at av R = 1 kgf/cm: sk l sk N c o s , c 3 7.1 cm av br 0.1 1 sk We accept l sk = 38 cm, which is more than 1.5 h = 1.5 () = 30 cm. Since the length of the shearing plane turned out to be less than h = () = 40 cm, cp, then the accepted value R = 1 kgf/cm corresponds to the standards. sk We arrange the support beam from plates with a diameter of cm. For the support cushion we take the same plate with a top edge of cm, which will provide a support width b 1 = 1.6 cm (Appendix 1). Bearing stress over the area of contact between the sub-beam and the support cushion N c sin, 4 k gf / s m 1. 6 cm where 4 kgf / cm is the calculated bearing resistance R CM90 across the fibers in the supporting planes of the structures.., 6. CONNECTIONS ON CYLINDRICAL DOGS Estimated load-bearing capacity the ability for one cut of a cylindrical dowel in joints of elements made of pine and spruce when the forces are directed along the fibers of the elements is determined by the formulas: according to the bending of the dowel T and = 180 d + a, but not more than 50 d; by collapse of the middle element with thickness T c = 50 cd; according to the collapse of the outermost element with thickness a T a = 80 ad. (.4a) (.4b) (.4c) The number of dowels n H that must be placed in the connection to transmit force N is found from expression 6
27 n H N, (.5) where T n is the smaller of the three values of the load-bearing capacity of the dowel, calculated using formulas (.4); p s number of dowel cuts. The calculated load-bearing capacity of the dowel T n can also be determined using Appendix 5. The distance between the axes of the dowels must be at least: along the fibers s 1 = 7 d; across the fibers s = 3.5 d and from the edge of the element s 3 = 3 d. The calculated load-bearing capacity of a cylindrical dowel T n when the force is directed at an angle a to the fibers of the elements is determined as the smaller of the three according to the formulas: H nt (1 8 0), but not more than T k d a c H T c = k α 50 cd; T a = k α 80 cd. k 50d ; (.6a) (.6b) (.6c) Angle α and degrees Table. Coefficient k a for steel dowels with a diameter in mm 1, 1.4 1.6 1.8, 0.95 0.95 0.9 0.9 0.9 0.9 0.75 0.75 0.7 0.675 0, 65 0.65 0.7 0.65 0.6 0.575 0.55 0.55 Note. The values of the coefficient ka for intermediate angles are determined by interpolation. Example.3. The joint of the lower stretched belt of the truss truss (Fig. 9, a) is made using plank overlays connected to the belt with dowels made of round steel. The belt is made of logs with a diameter at the joint of 19 cm. To ensure a tight fit of the overlays, the logs are hewn on both sides by 3 cm to a thickness of c = 13 cm. The overlays are made from boards with a cross section a x h = 6 x 18 cm. Design tensile force N = kgf. Calculate the connection. 7
28 Fig. 9. Connections on steel cylindrical dowels Solution. The diameter of the dowels is set approximately equal to (0.0.5) a, where a is the thickness of the lining. We accept d = 1.6 cm. We determine the calculated load-bearing capacity of the dowel per section using formulas (.4): H , ; T k gs k gs T c T a , k gs; , to Ms. 8
29 The smallest calculated load-bearing capacity Tn = 533 kgf. Double-cut dowels. Required number of dowels according to formula (.5): n H , 9 pcs We accept 1 dowels, of which 4 are bolts on each side of the joint. We place the dowels in two longitudinal rows. Distance between dowels along the fibers: s 1 = 7 d 7 1, 6 = 11, cm (assuming 1 cm). The distance from the axis of the dowels to the edge of the overlays is s 3 = 3 d 3 1, 6 = 4.8 cm (assuming 5 cm). The distance between the dowels across the fibers is s h s = 8 cm > 3.5 d = 5.6 cm. 3 Net cross-sectional area of the belt minus side stitches and weakening by holes for dowels. D 8 4 8, 8 1,. seg d c cm HT 4 Weakened cross-sectional area of the linings HT () 6 (1 8 1, 6) 1 7 7, 6. a h d cm Tensile stress in the linings N, k gf / s m. HT 1 7 7, 6 Example.4. In the crossbar of inclined rafters (Fig. 9, b) a tensile force of N = 500 kgf occurs. The crossbar is made of two plates with a diameter Dpl = 18 cm. The plates cover a rafter leg made of logs D = cm on both sides and are attached to it with two bolts d = 18 mm, working as double-cut dowels. Grinding depth 9
30 of the rafter leg at the junction of the crossbar h "ST = 3 cm. For a tight fit of the bolt washers, the plates are hewn to a depth of h St = cm. The angle between the direction of the crossbar and the rafter leg is a = 30. Check the strength of the connection. Solution. Load-bearing capacity of a steel cylindrical dowel per cut with the direction of the force at an angle to the fibers is determined by the formulas (.6): H 0, 9 (, 8 7) , ; 9 coefficient k a, determined from the table.; c = D h st = 3 = 16 cm thickness of the middle element; a = 0.5 D pl h st = 0, = 7 cm thickness of the outer element. The smallest load-bearing capacity of the dowel T n = 647 kgf. Full load-bearing capacity of the connection p n p s T n = == 588 > 500 kgf. The distance from the axis of the dowel to the end of the crossbar is taken s 1 = 13 cm > 7 1, 8 = 1.6 cm. The distance between the axes of the dowels across to the axis of the crossbar we take s = 6 cm and across to the axis of the rafter leg. So, let's summarize: "s = 9 cm. The ability of a material to resist external force influences is called mechanical properties. TO mechanical properties wood include: strength, elasticity, ductility and hardness. The strength of wood is characterized by its ability to resist external forces (loads). thirty
31 Forces that resist external influences (loads) are called internal forces or stresses. Thus, in the sections of wooden structures, compressive, tensile, bending, shearing (crushing) or chipping stresses arise. The considered methods for calculating wooden structures are focused on typical types of structures studied in the discipline “Forest Engineering Structures”. . It is necessary to design wooden structures in strict accordance with SNiP and GOST. 31
32 Applications 3
33 Diameter in cm Indicators B B B B B B B B B B B B B B B B B B 4.8 1.6 5 1.68 5.3 1.75 5.37 1.8 5.57 1.87 5.76 1.93 5.91 1.98 6.08, 04 6.5.09 6.4.14 6.55, 6.7.4 6.85.3 Dimensions of chords b in cm and areas in cm of segments Cutting depth 0.5 1 1.5.5 3 3.5 4 4.5 5 7.34 7.14.39 7.7.45 7.41.49 7.55.5 7.67.57 6.6 4.5 6.9 4.7 7, 4.88 7.47 5.06 7.8 5.4 8 5.4 8, 5.56 7.94 8.18 8.3 8.65 8.67 8.85 9.0 9, 9.3 9.51 9.6 9.83 9.9 10.1 8.5 5.7 10, 10.4 8.7 5.87 8.9 6 9, 6.17 9.4 6.31 9.6 6.44 9.8 6.58 10.5 10.7 8.91 1.4 9.39 1.9 9.8 13.6 9.75 17, 10, 17.8 10.7 18.6 10, 14 11 ,1 19.7 10.6 14.5 10.4.1 10.9 3, 11.5 4, 11.6 0 1.5 6.1 10.3 15.4 11.7 15.9 10, 8 11 1.3 16.8 11.1 11.3 11.4 11.5 11.6 11.8 10 6.71 1.1 1, 10, 6.85 10.4 6.96 10.6 7 ,1 10.8 7.3 1.4 1.4 1.8.1 1 16.3 13.6 1.6 17.1.9 17.6 11.9 1 13.6 18.4 1.4 1.5 1.6 1.7 13.6 3.3 10.9 7.5 11.5 8.8 1.1 30.1 1 5.1 1.7 31.4 13.4 7.9 13 .8 8.8 14.3 9.6 14.7 30.4 14 3.9 15.1 31.1 14.3 4.4 15.5 31.9 13.7 5 15.9 3.6 13 ,8 18.8 14.1 19.1 14.4 19.5 1.7 19.9 13.1 13, 15 5.5 16, 33.4 13, 3.5 13.7 33.7 14, 34.8 14.7 35.9 15, 36.9 15.6 37.9 15.1 38.9 16.5 39.9 16.9 40.9 17.3 41.8 15.3 6 16, 7 4.6 15.7 6.6 16 1.7 16.3 7.6 15 0.4 16.6 8.7 18.1 43.6 17.3 35.4 17.7 36.1 18, 5 44.4 18.9 45.8 19.3 46.3 11.4 1.4 40.7 1.7 36.6 13.3 37.8 13.9 39.3 14.4 40.5 43 .7 13.1 4.8 13.8 44.7 14.4 46.6 49.7 16.51.4 16.7 5.9 16.54, 17.7 55.9 17.4 48.4 17.9 49.5 18.3 50.7 18.8 51.8 19.5.9 18.57.4 18.7 58.8 19.60.1 19.7 61.4 0.1 6, 7 Appendix 1 14.1 51.5 14.8 53.7 15.5 55.7 16.1 57.7 16.7 59.6 17.3 61.4 17.9 63, 18.4 64.6 19.5 68.3 0 69.9 0.5 71.6 54 0.6 64 1.4 74.4 58.1 1 65.5 1.9 76 1.4 66.5.4 77.4 33
34 34 End adj. 1 in round sections For various depths insert h VR in cm 5.5 6 6.5 7 7.5 8 8.5 9 9.9 63.6 16.6 65.3 17, 68.1 17.7 76.8 17.9 70, 18 ,3 79.3 18.7 88.5 18.5 7.6 19.4 91.19.1 74.3 19.6 84 0.1 93.9 0.6 76.3 0.86, 0, 7 96.5 1, 107 1, 78, 0.8 88.4 1.3 99 1.8 110, 11.6 13 0.7 80.1 1.4 90.5 1.9 101.4 113, 9 14 3, 81.9 1.9 9.7.7 84.5 94.7 3, 130 4.6 14 5.4 167, 85.4 3 96.7 3, 10 4, 171.7 87, 1 3.5 98.7 4, 111 4.8 13 5, 188 3, 88.9 19 8.3 06
35 35 Flexibility λ Appendix Value of coefficient φ Coefficient φ .99 0.99 0.988 0.986 0.984 0.98 0.98 0.977 0.974 0.968 0.965 0.961 0.958 0.954 0.95 0.946 0.94 0.937 0.98 0.93 0.918 0.913 0.907 0.891 0.884 0.87 0.866 0.859 0.85 0.845 0.838 0.831 0.84 0.810 0.8 0.79 0.784 0.776 0.768 0.758 0.749 0.74 0.731 0.71 0J0 0.69 0.68 0 .67 0.66 0.65 0.641 0.63 0.608 0.597 0.585 0.574 0.56 0.55 0.535 0.53 0.508 0.484 0.473 0.461 0.45 0.439 0.49 0.419 0.409 0.4 0.383 0.374 0.3 66 0.358 0.351 0.344 0.336 0.33 0.33 0.31 0.304 0.98 0.9 0.87 0.81 0.76 0.71 0.66 0.61
36 36 End adj. Flexibility λ Coefficient φ .56 0.5 0.47 0.43 0.39 0.34 0.3 0.6 0, 0.16 0.1 0.08 0.05 0.0 0.198 0.195 0.19 0.189 0.183 0.181 0.178 0.175 0.173 0.17 0.168 0.165 0.163 0.158 0.156 0.154 0.15 0.15 0.147 0.145 0.144 0.14 0.138 0.136 0.134 0.13 0.13 0.19 0.17 0.16 0.14 0.11 0.1 0.118 0.117 0.115 0.114 0.11 0.111 0.11 0.107 G, 106 0.105 0.104 0.10 0.101 0.1 0.099 0.098 0.096 0.095 0.094 0.093 0, 09 0.091 0.09 0.089 0.086 0.085 0.084 0.083 0.08 0.081 0.081 0.08 0.079 0.078
37 Appendix 3 Calculated data Height h=k 1 D 1 0.5 Sectional area =k D 0.785 0.393 Distance from the neutral axis to the outermost fibers: z 1 =k 3 D z =k 4 D 0.5 0.5 0.1 0.9 Moment of inertia: J x =k 5 D 4 J y =k 6 D 4 0.0491 0.0491 0.0069 0.045 Moment of resistance: W x =k 7 D 3 W y =k 8 D 3 0.098 0.098 0.038 0.0491 Maximum radius of gyration r min =k 9 D 0.5 0.13 37
38 End adj.971 0.933 0.943 0.866 0.393 0.779 0.763 0.773 0.740 0.5 0.475 0.447 0.471 0.433 0.5 0.496 0.486 0.471 0.433 0.04 5 0.0476 0.441 0.461 0.0395 0.0069 0.0491 0.0488 0.490 0.0485 0 .0491 0.0960 0.0908 0.0978 0.091 0.038 0.0981 0.0976 0.0980 0.097 0.13 0.47 0.41 0.44 0.031 38
39 Design characteristics of materials Appendix 4 Stress state and characteristics of elements Designation Design resistance MPa leniya, for kgf/cm graded wood Bending, compression and crushing of fibers: a) elements of rectangular cross-section (except for those specified in subparagraphs “b” and “c”) with height up to 50 cm b) elements of a rectangular section with a width of over 11 to 13 cm with a section height of over 11 to 50 cm c) elements of a rectangular section with a width of over 13 cm with a section height of over 13 to 50 cm d) elements made of round timber without inserts in the design section . Tension along the fibers: a) non-glued elements b) glued elements 3. Compression and crushing over the entire area across the fibers 4. Local crushing across the fibers: a) in the supporting parts of structures, frontal and nodal junctions of elements b) under washers at crushing angles of 90 to Chipping along the fibers: a) when bending non-glued elements b) when bending glued elements c) in frontal cuttings for maximum stress R and, R c, R cm R and, R c, R cm R and, R c, R cm R i, R c, R cm R p R p R c.90, R cm.90 R cm.90 R cm.90 R ck R ck R ck.8 18 1.6 16.6 16 1.5 15.6 16 1.5 15.1 1 39
40 Stress state and characteristics of elements Design characteristics of materials Designation End adj. 4 Calculated resistance MPa leniya, for kgf/cm graded wood 1 3 g) local in adhesive joints for maximum stress 6. Shearing across the grain: a) in joints of non-glued elements b) in joints of glued elements 7. Tension across the fibers of elements made of laminated wood R ck R ck.90 R ck.90 R p.90.7 7 0.35 3.5.1 1 0.8 8 0.7 7 0.3 3.1 1 0.6 6 0.6 6 0.35 3.5 NOTE: 1. The design resistance of wood to crushing at an angle to the direction of the fibers is determined by formula R cm R cm 3 1 (1) s in R R cm 90. The calculated resistance of wood to chipping at an angle to the direction of the fibers is determined by the formula R cm sk. R sk 3 1 (1) sin R R sk.90 sk.. 40
41 Bibliography 1. SNiP II Wooden structures. Design standards.. SNiP IIB. 36. Steel structures. Design standards. 3. SNiP II6.74. Loads and impacts. Design standards. 4. Ivanin, I.Ya. Examples of design and calculation of wooden structures [Text] / I.Ya. Ivanin. M.: Gosstroyizdat, Shishkin, V.E. Structures made of wood and plastic [Text] / V.E. Shishkin. M.: Stroyizdat, Forest engineering structures [Text]: guidelines to the implementation of a wooden bridge project for students of the specialty “Forestry Engineering” / A.M. Chuprakov. Ukhta: USTU,
42 Contents Introduction... 3 Chapter 1 Calculation of elements of wooden structures Centrally tensile elements... 5 Centrally compressed elements Bendable elements Tensile-bending and compression-bending elements Chapter Calculation of connections of elements of wooden structures... 5 Connections on notches... 6 Connections on cylindrical dowels.. 6 Applications... 3 Bibliography
43 Educational publication Chuprakov A.M. Examples of calculation of wooden structures of forest engineering structures Textbook Editor I.A. Bezrodnykh Corrector O.V. Moisenia Technical editor L.P. Korovkin Plan 008, position 57. Signed for printing. Computer typesetting. Times New Roman typeface. Format 60x84 1/16. Offset paper. Screen printing. Conditional oven l.,5. Uch. ed. l., 3. Circulation 150 copies. Order 17. Ukhta State Technical University, Ukhta, st. Pervomaiskaya, 13 Department of operational printing of USTU, Ukhta, st. Oktyabrskaya, 13.
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Calculation of wooden floors
Calculating a wooden floor is one of the easiest tasks, and not only because wood is one of the lightest building materials. Why this is so, we will find out very soon. But I’ll tell you right away, if you are interested in the classic calculation, according to the requirements regulatory documents, then you here .
When building or repairing a wooden house, using metal, and even more so reinforced concrete floor beams is somehow out of the question. If the house is wooden, then it is logical to make the floor beams wooden. It’s just that you can’t tell by eye what kind of timber can be used for floor beams and what kind of span should be made between the beams. To answer these questions, you need to know exactly the distance between the supporting walls and at least approximately the load on the floor.
It is clear that the distances between the walls are different, and the load on the floor can also be very different. It’s one thing to calculate the floor if there is a non-residential attic on top, and a completely different thing to calculate the floor for the room in which partitions will be built in the future. cast iron bathtub, bronze toilet and much more.
Calculation of wooden structures should be done:
- on load-bearing capacity (strength, stability) for all structures;
- on deformations for structures in which the magnitude of deformations may limit the possibility of their operation.
Calculation of bearing capacity should be carried out under the influence of design loads.
Calculation of deformations should be carried out under the influence of standard loads.
Deformations (deflections) of bending elements should not exceed the values given in table. 37.
Table 37. Limit deformations (deflections) of bending elements
Note. If there is plaster, the deflection of the floor elements is only from payload should not be more than 1/350 of the span.
Centrally stretched elements
Calculation of centrally stretched elements is carried out according to the formula:
where N is the calculated longitudinal force,
mр - coefficient of operating conditions of the element in tension, accepted: for elements that do not have weakening in the design section, mр = 1.0; for elements with weakening, mр = 0.8;
Rp is the calculated tensile strength of wood along the grain,
Fnt is the net area of the cross-section under consideration: when determining Fnt, weakenings located in a section 20 cm long are taken to be combined in one section. Centrally compressed elements. Calculation of centrally compressed elements is carried out according to the formulas: for strength
for sustainability
where mс is the coefficient of operating conditions of compression elements, taken equal to unity,
Rc is the calculated resistance of wood to compression along the grain,
The buckling coefficient, determined from the graph (Fig. 4),
Fnt - net cross-sectional area of the element, Fcalc - calculated cross-sectional area for stability calculations accepted:
1) in the absence of weakening: Fcalc=Fbr;
2) for weakening that does not extend to the edge - Fcalc = Fbr, if the area of the weakening does not exceed 25% of Fbr and Fcalc = 4/3Fnt, if their area exceeds 25% of Fbr;
3) with symmetrical weakening facing the edge: Fcalc=Fnt
Flexibility? solid elements is determined by the formula:
Note. For asymmetrical weakening extending to the ribs, the elements are calculated as eccentrically compressed.
Figure 4. Graph of buckling coefficients where Io is the estimated length of the element,
r - radius of inertia of the element’s section, determined by the formula:
l6p and F6p are the moment of inertia and gross cross-sectional area of the element.
The estimated length of the element l0 is determined by multiplying its actual length by the coefficient:
with both hinged ends - 1.0; with one end pinched and the other freely loaded - 2.0;
with one end pinched and the other hinged - 0.8;
with both ends pinched - 0.65.
Bendable elements
Calculation of bending elements for strength is carried out according to the formula:
where M is the design bending moment;
mi - coefficient of operating conditions of the element for bending; Ri is the design bending resistance of wood,
Wnt is the net moment of resistance of the cross section under consideration.
The coefficient of operating conditions for bending elements mi is accepted: for boards, bars and beams with cross-sectional dimensions of less than 15 cm and glued elements of rectangular cross-section mi = 1.0; for beams with side dimensions of 15 cm or more, with the ratio of the height of the element’s section to its width h/b? 3.5 - mi = 1.15
Calculation of solid cross-section elements for strength during oblique bending is carried out according to the formula:
where Mx, My are the components of the design bending moment, respectively, for the main axes x and y
mi - coefficient of operating conditions of the element for bending;
Wx, Wy are the net moments of resistance of the cross section under consideration for the x and y axes. Eccentrically extended and extracentrically compressed elements. Calculation of eccentrically stretched elements is carried out according to the formula:
Calculation of eccentrically compressed elements is carried out according to the formula:
where? is a coefficient (valid in the range from 1 to 0), taking into account the additional moment from the longitudinal force N during deformation of the element, determined by the formula;
At low bending stresses M/Wbr, not exceeding 10% of the stress
stress N/Fbr, eccentrically compressed elements are calculated on
stability according to formula N where Q is the calculated shear force; mck=1 - coefficient of operating conditions of a solid element for chipping during bending; Rck is the calculated resistance of wood to chipping along the grain; Ibr is the gross moment of inertia of the section under consideration; Sbr is the gross static moment of the shifted part of the section relative to the neutral axis; b - section width. Vladimir Fedorovich Ivanov The book outlines the basics of design, calculation, manufacturing and installation, rules of operation and reinforcement of structures made of wood and using plastics; measures to protect them from decay, fire and other harmful effects are indicated; The physical and mechanical properties of wood and structural plastics are considered. Introduction (3) SECTION ONE Chapter 1. Raw material base of wood and its importance for use in the national economy (16) Chapter 2. Structure of wood, its physical and mechanical properties (20) Chapter 3. Mechanical properties of wood (27) SECTION TWO Chapter 4. Protection of wooden structures from fire (41) Chapter 5. Protection of wooden structures from rotting (43) SECTION THREE Chapter 6. Calculation of wooden structures using the limit state method (50) Chapter 7. Calculation of elements of wooden structures of solid section (56) Chapter 8. Solid beams (69) SECTION FOUR Chapter 9. General data 72 Chapter 10. Connections on notches and keys (76) Chapter 11. Dowel connections (87) Chapter 12. Connections on stretched working links (95) Chapter 13. Adhesive joints (97) SECTION FIVE Chapter 14. Calculation of composite elements based on elastic-yielding bonds (101) Chapter 15. Calculation of composite elements on elastic-yielding bonds using the approximate method SNiP II-B.4-62 (103) SECTION SIX Chapter 16. Types of continuous systems of wooden structures (110) Chapter 17. Structures of wooden beams of composite section (113) Chapter 18. Spacer systems for solid timber structures (129) SECTION SEVEN Chapter 19. Main types of through wooden structures (141) Chapter 20. Combined timber construction systems (149) Chapter 21. Beam trusses made of logs and beams (154) Chapter 22. Metal-wood trusses with a glued top chord and segmental trusses on nails (161) SECTION EIGHT Chapter 24. Ensuring spatial rigidity during operation and installation (173) SECTION NINE Chapter 25. Basic types of spatial wooden structures (180) Chapter 26. Circular reticulate vaults (185) Chapter 27. Wooden shell vaults and folds (193) Chapter 28. Wooden domes (196) SECTION TEN Chapter 29. Towers (206) Chapter 30. Silos, tanks and bunkers (213) Chapter 31. Masts (215) Chapter 32. General information about wooden bridges (218) Chapter 33. Scaffolding, scaffolding and circles for the construction of buildings and engineering structures (230) SECTION ELEVEN Chapter 34. Timber industry (236) Chapter 35. Sawmilling (246) Chapter 36. Drying wood (249) Chapter 37 Basics of organizing the manufacture of wooden structures (251) Chapter 38. Operation, repair and strengthening of wooden structures (257) SECTION TWELVE Chapter 39. Plastics as a structural building material (261) Chapter 40. Features of the calculation of structural elements using plastics (286) Chapter 41. Layered structures (304) Chapter 42. Pneumatic structures (315) SECTION THIRTEEN Chapter 43. Prospects for the development and application of structures made of wood and plastics (324) Applications (330) Don’t forget about the topic: “Your scans, our processing and translation into DJVU.” 
Structures made of wood and plastics
(textbook for universities)
1966
The book is intended for students of construction universities and faculties as a textbook
WOOD AS A CONSTRUCTION MATERIAL
§ 1. Raw material base of wood (-)
§ 2. Wood as a building material and its use in construction (17)
§ 3. The structure of wood and its properties (-)
§ 4. Moisture in wood and its effect on physical and mechanical properties (23)
§ 5. Chemical effects on wood (25)
§ 6. Physical properties of wood (26)
§ 7. Anisotropy of wood and general characteristics of its mechanical properties (-)
§ 8. The influence of the structure and some basic defects of wood on its mechanical properties (29)
§ 9. Long-term resistance of wood (31)
§ 10. Work of wood in tension, compression, transverse bending, crushing and chipping (33)
§ 11. Selection of timber during the construction of load-bearing wooden structures (39)
PROTECTION OF WOODEN STRUCTURES FROM FIRE, BIOLOGICAL DEATH AND IMPACT OF CHEMICAL REAGENTS
§ 12. Fire resistance of building structure elements (-)
§ 13. Measures to protect wooden structures from fire (-)
§ 14. General information (-)
§ 15. Wood-destroying fungi and conditions for their development (-)
§ 16. Constructive prevention to combat rotting of elements of wooden structures (44)
§ 17. Protection of wooden structures from the effects of chemical reagents 47
§ 18. Chemical measures to protect wood from decay (antiseptic treatment) (-)
§ 19. Damage to wood by insects and measures to combat them (49)
CALCULATION AND DESIGN OF ELEMENTS OF WOODEN STRUCTURES
§ 20. Initial provisions for the calculation of elements of wooden structures (-)
§ 21. Data for the calculation of wooden structures using the limit state method (52)
§ 22. Central stretch (-)
§ 23. Central compression (57)
§ 24. Transverse bending (62)
§ 25. Oblique bend (65)
§ 26. Compressed-bent elements (66)
§ 27. Stretched-curved elements (68)
§ 28. Single-span beams of solid section (-)
§ 29. Beams of solid section, reinforced with sub-beams (-)
§ 30. Cantilever-beam and continuous purlin systems (70)
CONNECTIONS OF STRUCTURE ELEMENTS
§ 31. Classification of connections (connections) (-)
§ 32. General instructions for calculating connections of elements of wooden structures (74)
§ 33. Frontal cuts (-)
§ 34. Simple, double and three-lobe stops (80)
§ 35. Connections with keys (82)
§ 36. Prismatic transverse, longitudinal and inclined keys (84)
§ 37. Metal keys and washers (86)
§ 38. General information (-)
§ 39. Main features of pin connections (89)
§ 40. Calculation of dowel connections based on the limit state (90)
§ 41. Bolts (-)
§ 42. Clamps, staples, nails, screws, screws and screws (96)
§ 43. Types of adhesives (-)
§ 44. Bonding technology (98)
§ 45. Constructions of glued joints and cleestal washers (99)
COMPONENT ELEMENTS OF WOODEN STRUCTURES ON ELASTIC-COMPLIABLE LINK
§ 46. General information (-)
§ 47. Transverse bending of constituent elements (-)
§ 48. Central compression of constituent elements (105)
§ 49. Eccentric compression of composite elements (107)
§ 50. Examples of calculation of composite elements (108)
FLAT SOLID WOODEN STRUCTURES
§ 51. General information (-)
§ 52. Composite beams of the Derevyagin system (-)
§ 53. Design and calculation of laminated beams (117)
§ 54. Design and calculation of glued plywood beams (121)
§ 55. Manufacturing of laminated beams (123)
§ 56. Design and calculation of I-beams with a double plank cross wall on nails (124)
§ 57. Three-hinged arches from beams of the Derevyagin system (-)
§ 58. Circular arch systems (131)
§ 59. Arched structures of an I-profile with a double cross wall on nail connections (132)
§ 60. Glued arches (134)
§ 61. Solid frame structures (138)
§ 62. Manufacturing of arched and frame structures and their installation (139)
FLAT THROUGH-THROUGH WOODEN STRUCTURES
§ 63. General information (-)
§ 64. Fundamentals of designing structures of through trusses (145)
§ 65. Truss beams (-)
§ 66. Suspended and braced systems of wooden structures (152)
§ 67. Log and cobblestone trusses on frontal cuts (-)
§ 68. Metal-wood trusses TsNIISK (156)
§ 69. Metal-wood trusses with an upper chord made of Derevyagin beams (160)
§ 70. Metal-wood trusses with a rectangular glued top chord (-)
§ 71. Metal-wood segment trusses with glued top chord (162)
§ 72. Segmental trusses made of bars and boards on nails (165)
Chapter 23. Arch and frame through structures. Lattice racks (-)
§ 73. Three-hinged arches from segmental, crescent-shaped and polygonal beam trusses (-)
§ 74. Frame through wooden structures and lattice racks (169)
SPATIAL FIXING OF FLAT WOODEN STRUCTURES
§ 75. Measures to ensure spatial rigidity of flat wooden structures (-)
§ 76. Work of flat wooden structures during installation (176)
SPATIAL WOODEN STRUCTURES
§ 77. General provisions (-)
§ 78. Vault systems (-)
§ 79. Metal-free circular-mesh vault of the system of S. I. Peselnik (188)
§ 80. Circular-reticulate vault of the Zollbau system (-)
§ 81. Basic principles of construction of circle-mesh vaults (189)
§ 82. Calculation of circular-mesh vaults (-)
§ 83. General concepts of the cross and closed vault of the circle-mesh system (191)
§ 84. General information (-)
§ 85. Domes of the radial system (-)
§ 86. Domes of a circle-mesh design (200)
§ 87. Thin-walled and ribbed spherical domes and methods for their calculation (202)
WOODEN STRUCTURES AND SPECIAL PURPOSE STRUCTURES
§ 88. General information (-)
§ 89. Towers with lattice and mesh shaft construction (-)
§ 90. Towers with solid shafts (212)
§ 91. Design and principles of calculation (-)
§ 92. Guyed masts (-)
§ 93. Bridges and overpasses (-)
§ 94. Roadway for road bridges and its connection with the embankment (219)
§ 95. Supports of wooden bridges of the beam system (221)
§ 96. Wooden beam bridges of solid section (224)
§ 97. Strut systems for wooden bridges (-)
§ 98. Arched systems of wooden bridges (225)
§ 99. Span structures of wooden bridges of through systems (226)
§ 100. General concepts about forests and circles (-)
§ 101. Schemes and designs of scaffolding (231)
PRODUCTION OF WOODEN STRUCTURES AND PARTS FOR CONSTRUCTION
§ 102. Logging and woodworking industry (-)
§ 103. Basic technological processes of mechanical woodworking (237)
§ 104. Sawmill frames (239)
§ 105. Circular saws (-)
§ 106. Band saw machines (240)
§ 107. Planing machines (242)
§ 108. Milling and tenoning machines (-)
§ 109. Drilling machines (244)
§ 110. Slotting machines (-)
§ 111. Grinding machines (245)
§ 112. Lathes and other equipment (-)
§ 113. Electrified portable tools (-)
§ 114. General information (-)
§ 115. Natural drying of wood (-)
§ 116. Artificial drying of wood and types of drying chambers (-)
§ 117. Construction shop (-)
§ 118. Workshop for the production of laminated wood and structures made from it (252)
§ 119. Production of plywood and some other types of treated wood (254)
§ 120. Safety precautions and labor protection in the manufacture of wooden structures and building parts (256)
§ 121. Basic rules for the operation of wooden structures (-)
§ 122. Repair and strengthening of wooden structures (-)
BUILDING STRUCTURES AND PRODUCTS USING PLASTICS
§ 123. General information about plastics and their components (-)
§ 124. Brief information on methods for processing polymers into building materials and products (265)
§ 125. Basic requirements for plastics used in building structures (268)
§ 126. Fiberglass plastics (269)
§ 127. Wood-laminated plastics (chipboard) (276)
§ 128. Fiberboards (FPV) (273)
§ 129. Particle boards (PDS) (-)
§ 130. Organic glass (polymethyl methacrylate) (280)
§ 131. Hard vinyl plastic (VN) (281)
§ 132. Foam plastics (282)
§ 133. Honeycombs and mipores (283)
§ 134. Heat-, sound- and waterproofing materials obtained from plastics and used in building structures (284)
§ 135. Features of some physical and mechanical properties of structural plastics (285)
§ 136. Central tension and compression (-)
§ 137. Transverse bending of plastic elements (289)
§ 138. Tensile-curved and compressed-curved elements made of plastics (295)
§ 139. Data for the calculation of building structures using plastics (-)
§ 140. Connection of structural elements made of plastics (299)
§ 141. Synthetic adhesives for gluing different materials (301)
§ 142. Schemes and design solutions of layered structures (-)
§ 143. Calculation method for three-layer slab panels (310)
§ 144. Some examples of the use of laminated panels in buildings for various purposes (312)
§ 145. Plastic pipelines (314)
§ 146. General information and classification of pneumatic structures (-)
§ 147. Fundamentals of calculation of pneumatic structures (318)
§ 148. Examples of pneumatic structures in structures for various purposes (320)
USING WOOD AND PLASTICS IN THE STRUCTURES OF THE FUTURE
§ 149. General information (-)
§ 150. Prospects for the use of wood in structures (326)
§ 151. Prospects for the use of plastics in structures (328)
Literature (346)
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