Appendices
I "CONFIRM"
The pro-rector on scientific work
FGBOU IN «National
And villages edovatel s whom Mordo vs whom the state university of
N _ I. Ogaryov the professor
P.V.Senin
THE INTRODUCTION CERTIFICATE
Results of the researches spent by Antoshkinym V.
D. Were used at performance х\д NIR (№ 97-1/17) and on a theme «Working out energoeffektivnyh domes from frame pyramidal panels» in frameworks (ПНР-1) Programs of development of university of 2010-2019 ofThe dean of architectonic-building faculty, academician RAASN. d.t.n. The professor
V.T.Yerofeev
I "CONFIRM"
The pro-rector on scientific work
FGBOU IN «National issledovatel s whom Mordovs whom
The state university of
N.P.Ogaryov, d. So-called., ofessor
P.V.Senin
CERTIFICATE В1ІЕДРЕ ІІ IJA
Antoshkin V. D has developed programs for geometrical and static calculation of mesh domes. It develops offers by calculation and designing of the given coatings which were used on chair of buildings, constructions and highways and on chair of applied mechanics during course and degree designing.
Managing chair
Applied mechanics, the corresponding member
V.D.Tcherkasov
RAASN, d.t.n., the prof.
THE REGIONAL DESIGN-EXPERT CENTER
Society with limited liability «the Regional design-expert centre»
430030, Republic Mordovia, of an island Saransk, Titov's street, d. 34, каб.21
INN/KPP 1326212443/132601001,
OGRN 1091326002460
Banking requisites:
To/account 30101810600000000808;
R/s 40702810100000024269 in Joint-Stock Company AKB "EXPRESS TRAIN-VOLGA", Saratov, Michurin's street, 166/168, the Postal index 410002;
OKATO 89401364000
BIK 046311808
Bodies. A/fax: 8(8342) 23-28-78
Mob.: 8-927-977-07-01 director
À-MA ІІ: 2evReniy@mail.ru
To the chairman
Dissertational council
THE INTRODUCTION CERTIFICATE
According to the contract 97/17 from 01.02.2017, the prisoner between FGBOU IN «the Moscow State University of N.P.Ogaryova» and Open Company «Mordovian regional proektnoekspertnyj the Center» is developed the equipment design dome karkasnotentovogo coatings of summer cafe by span of 8,1 m.
By project working out results of the scientific researches spent by Antoshkinym V. D in a progress of work over the thesis for a doctor's degree were used.
MINOBRNAUKI RUSSIA
FEDERAL STATE BUDGETARY EDUCATIONAL INSTITUTION of HIGHER EDUCATION «NATIONAL RESEARCH MORDOVIAN STATE UNIVERSITY of N.P.OGARYOV» (FGBOU IN «the Moscow State University of N.P.Ogaryov»)
Street Bolshevist, d.
68, Saransk, Republic Mordovia, Russia, 430005,Ph. (8342) 24-37-32, 24-48-88, a fax (8342 47-29-13, E-mail: dep-general@adm.mrsu.ru, http://www.mrsu.ruОКПО 02069964, OGRN 1021300973275, INN/KPP 1326043499/132601001
26.05.2017 № 97-1/17
To the chairman of dissertational council
D 999.94.03
To Dr.Sci.Tech., the professor
Fedorovoj Natalias Vitalevne
Federal state budgetary educational institution of higher education «Southwest state university» 305040, Kursk, street of 50 years of October, 94, a conference hall.
THE INTRODUCTION CERTIFICATE
According to the contract 97-1/17 from 01.02.2017, the prisoner between FGBOU IN «the Moscow State University of N.P.Ogaryova"and"GBPOU RM« Saransk polytechnical technical school »Saransk, in common makes a dome karkasno-awning coating for a building of mobile pavilion.
By working out of the project of pavilion, manufacturing and installation results of the scientific researches spent by Antoshkinym V. D in a progress of work over the thesis for a doctor's degree were used.
The supervisor of studies
Research and design institute
(NIPI ASF FGBOU IN «the Moscow State University of N.P.Ogaryov») academician RAASN the professor, d.t.n.
THE APPENDIX 2.
Listing 2.1:
Listing of operations in program complex AutodeskAutoCad 2015
Stage 1. Definition of axes-borders of sector of sphere. It is necessary to create the circle of individual radius which is passing through the centre of sphere (such circles create «the main lines» on sphere), in plane XOY (it is selected by default) for the further amenity we will designate it «the Circle 1», for this purpose using the button «the Center radius» (prilozh. 2, fig. 2.1)
We create a circle equal 1000, using dynamic input, with the centre in a point [0.0.0]. After that we put a specific cube in position as is shown in
Drawing 2.3 prilozh. 2, we allocate «the Circle 1» and, using function 3D -
Turn
With a base point [0,0,0].
Rice 2 2 Center pjadius
Fig. 2.3. A specific cube
Then we turn «a circle 1» with copying concerning an axis «Y», thereby receiving «the Circle 2». By means of the same actions we create «the Circle 3», copying with turn on 30 degrees
«Circles 2» concerning an axis «Z», and a command "Piece" we spend the pieces limiting sector necessary for us. After performance of the specified actions we necessarily receive 3 circles in different planes as is shown in drawing 2.4 prilozh. 2.
Stage 2. Construction of circles for placing of a hexagonal network.
For formation of listing with certain step of the main lines in this case 15 degrees to us are necessary for setting
«The user system
Co-ordinates »(further PSK) a plane of a circle 3 (prilozh. 2, fig. 2.4), with
The help knopkina panels of problems and to select points 1, 2, 3 (prilozh. 2, fig. 2.4) then to copy «a piece 1» with turn with a base point in the centre of sphere with step of 15 degrees (generally - a step of the main lines). For this purpose it is necessary to press button "Turn" _______________ and in a command line inquiry
«The base point», - to specify the sphere centre. Then it is necessary to press "Copy" then to specify «the Step of the main lines» - in our case of 15 degrees, after that to make the same actions with a following piece and. So further while the subsequent piece will not coincide with a piece 2. It is necessary to copy also a piece one with turn on a corner (15, 30, 45, 60, 75, 90 degrees), Finally it should turn out how it is shown in drawing 2.5 prilozh. 2.1.
For the further constructions it is necessary to construct «a piece 4», for this purpose to us
It is necessary to establish «World system of co-ordinates» to allocate «a piece 1», to press button "Turn" Ξ and in a command line inquiry «a base point» to specify the sphere centre. Then to press "Copy" and after that to specify a corner of turn of 90 degrees. So we receive «a piece 4».sledujushchim a stage we will construct a piece 5, by the same principle, only turn on 90 degrees «a piece 3».tem most we receive a point 8, as is shown in drawing 5, prilozh. 2. After that it is necessary to construct a circle 5, its plane is perpendicular planes of a circle 3 and there is in a plane formed by points [0,0,0], a point 5, a point 6. For construction of a circle 5 it is necessary to set PSK, formed by points a point 1, a point 5, a point 6, iprovesti a circle in this plane with the centre [0,0,0].sleduet in radius to specify a point 5 (prilozh. 2, fig. 2.5).
Then it is necessary to spend «a circle 6» individual radius perpendicularly «circles 2», passing through the centre of the basic sphere and the middle of an arch limited to points 3 and 7 (prilozh. 2, fig. 2.5).
For the given construction it is necessary to establish PSK on points 3,1,4, to establish an objective tie in "Nearest", then a command "Arch" to draw an arch with the centre in a point 1, length to establish points 3, 7 and to select one any point on a circle limited to points 3 and 7. After that we will be set new PSK which will be defined by points: 1, 8, and a point which will be the middle of an arch 3-7 (the arch middle it is necessary to set by means of an objective tie in "Middle")
Thus, we have received points 3 and 7. They are the centres of circles in which further will network hexagons are entered. On the basis of the entered hexagons we will construct triangular networks on sphere. Pieces 3-5 and 5-7 are radiuses of 1st and 2nd circles of a network accordingly. In the beginning we delete all unnecessary lines for the further constructions, and we start construction of the first circle of a network. For this purpose we will spend a piece 1-5 and we will reflect it mirror concerning a piece 1-3 in a plane 3,1,2 and in a plane 3,1,8 or in two any other planes,
To which posess a piece 1-3. Thus we receive three points of a circle. Which lays on a sphere surface, then it is necessary to set new PSK which is defined by three received points after that a command «a Circle, three points» to construct first circle O1сети (fig. 2.16).
By following step it is started construction of 2 circles, sphere laying on a surface, its centre will settle down on crossing of a circle 2 (prilozh. 2.1, fig. 2.5) and the second main line of sphere, i.e. in a point 7. For construction of the given circle it is necessary for us to set new PSK, defined points 1,7,8 and to make mirror copying of a piece 1-5 concerning a piece 1-7. So we will receive a piece 1-8 where the point 8 is the second point of a circle. For circle construction on a surface of sphere it is necessary for us three points, we are set new any PSK so that the piece 1-7 laid in its plane, and we do mirror copying of a piece 1-5 or 1-8. Thus we receive three points for construction of the second circle. Further we define new PSK the turned out points and a command «a Circle, three points» we build second circle О2.
Similarly we build the third circle with the centre in a point of intersection «circles 3» (prilozh. 2.1, fig. 2.5) and the third main line. Further we are set new PSK, the defined following points and mirror we copy a piece to receive the second point of a circle. Thus we receive the third circle which lays on a sphere surface.
The centre of the fourth circle laying on a surface of sphere, is in a point of intersection of circle О2 and the following main line. Construction is made similarly to construction of the second circle of sphere laying on a surface.
The point of the centre of the following circle lays on crossing of the following main line, a perpendicular arch-axis of sector of sphere and a circle of the second circle passing through the centre laying on
Sphere surfaces. And two points of intersection of thirds and the fourth circles of a surface of sphere. Further for construction of sector from a triangular network on sphere it is necessary for us to reflect mirror turned out circles so that circle O1и circle О2остались in the place. We will for this purpose be set new PSK and a command «to Reflect mirror» KZ, we receive circles (fig. of 2.10 chapters 2). Then. Precisely also it is displayed all circles concerning a plane of circle О3. Thus we form sector - a sphere petal. On points of crossings of circles we build the hexagons entered in a circle, thereby receiving structure of a mesh dome (fig. 2.12
Chapters 2).
Stage 3. Formation of a triangular network of an envelopment. By a command "Turn" with copying on 60 degrees it is received full sphere. We delete all auxiliary lines of construction (fig. 2.12) and we add hexagons or their fragments for formation of apertures in a dome coating.
The program 2.1
Co-ordinates
Sector with step | 18 | Degrees through equal radiuses, the big radius R = | 10000 | Mm |
X | Y | Z | X | Y | Z | X | Y | Z | |||
1 | 0 | 0 | 0 | 15 | 4929,386 | 1803,364 | 1488,3 | 29 | 6840,008 | 2535,47 | 3159,99 |
2 | 1016,69 | 0 | 51,8174 | 16 | 4391,555 | 2535,466 | 1381,087 | 30 | 6278,224 | 3212,86 | 2910,47 |
3 | 2107,22 | 0 | 224,541 | 17 | 5776,317 | 0 | 2549,348 | 31 | 8645,845 | 0 | 5024,87 |
4 | 1558,00 | 899,513 | 163,156 | 18 | 6258,656 | 880,4812 | 2250,55 | 32 | 8347,047 | 880,481 | 4563,86 |
5 | 3090,17 | 0 | 489,435 | 19 | 5786,481 | 1756,376 | 2035,593 | 33 | 7964,407 | 1756,38 | 4213,52 |
6 | 2590,69 | 880,481 | 381,629 | 20 | 5183,653 | 2552,624 | 1838,269 | 34 | 7545,672 | 1047,39 | 3521,89 |
7 | 4041,09 | 0 | 852,891 | 21 | 7450,652 | 0 | 3330,084 | 35 | 7089,528 | 3212,86 | 3721,77 |
8 | 3557,62 | 880,481 | 695,804 | 22 | 7058,379 | 867,4721 | 2969,582 | 36 | 6546,537 | 3779,645 | 3453,46 |
9 | 3042,13 | 1756,38 | 637,278 | 23 | 6465,457 | 2153,861 | 2681,615 | 37 | 0 | 0 | 0 |
10 | 5024,87 | 0 | 1354,16 | 24 | 6160,43 | 2552,624 | 2547,939 | 38 | 0 | 0 | 0 |
11 | 4521,50 | 899,513 | 1126,06 | 25 | 5566,079 | 3213,577 | 2338,95 | 39 | 0 | 0 | 0 |
12 | 3988,69 | 1803,36 | 1008,99 | 26 | 8090,17 | 0 | 4122,147 | 40 | 0 | 0 | 0 |
13 | 5877,85 | 0 | 1909,83 | 27 | 7749,45 | 880,4812 | 3741,344 | 41 | 0 | 0 | 0 |
14 | 5436,134 | 880,4812 | 1652,95 | 28 | 7318,385 | 1803,364 | 3428,157 | 42 | 7150,519 | 3495,35 | 3945,86 |
Initial data
о1о5 | 2*oaA2 | o5o13 | o13o26 | o26o43 |
18 | 18 | 18 | 18 | 18 |
о26а31
5,835299
O31
0
o19c12 | 01aA17 | O16=O17 |
5,835299 | 45 | 0 |
o1a2 | o5a3 | o5b6 | 013a10 | o26a21 | o43a37 | o9b6 | o19c15 | o33c28 | o52c45o1o5 |
9 | 5,835299 | 5,835299 | 5,835299 | 5,835299 | 5,835299 | 5,835299 | 5,835299 | 5,835299 | 5,835299 |
o4a2 | o4a3 | o11a7 | o22a17 | o22a17 | o11a10 | o13b14 | o5b6 | O25 | O36 |
6,051625 | 6,051625 | 6,051625 | 6,051625 | 6,051625 | 6,051625 | 5,835299 | 5,835299 | 30 | 30 |
О1 | o1 | х1 | y1 | z1 | O9 | О1А10 | O7 | O4=O5 | O2=O3 |
0 | 0 | 0 | 0 | 0 | 30 | 0 | 0 | 30 | 0 |
o1o9 = arctg (tg o1o5∕cos 30o) = | 33,69007 | ||
x9 = R* sin o1o9 cos O9 | 4803,845 | ||
y9 = R* sin o1o9 sin O9 | 2773,501 | y5 = R* sin o1o5 sin O5 | |
z9 = R * (1 cos o1o9) | 1679,497 | z5 = R* (1-cos o1o5) | |
o1o4 = arctg (tg o1a2/cos 30o) = | 17,19212 | ||
Θ4aA2 = arcsin (sino1o4 sin30 o) = | 8,498781 | ||
x4 = R*sin o1o4 cos O4 | 2559,77 | ||
y4 = R* sin o1o4 sin O4 | 1477,884 | ||
z4 = R* (1-cos o1o4) | 446,81 | ||
o1a2=arctg (tg (o5o9/2)/cos 30o) = | 9,276324 |
о4а2 = arccos (coso1a2coso1o4+sin o1a2sin о1о4 cos 30) =
x2 = R* sin o1a2 cos O2 | 1611,96 | 10,23469 |
y2 = R* sin o1a2 sin O2 | 0 | |
z2 = R* (1-cos o1a2) | 130,7759 | |
o1a3 = o1o5-o5a3 = | 20,72368 | |
x3 = R*sin o1a3 cos O3 | 3538,614 | |
y3 = | 0 | |
z3 = R * (1-cos o1a3) | 647,0211 |
o1b6 = arccos (coso5β6 cos o1o5 + sin о5в6 sin o1o5 cos 60) =
26,49171
Θ1B6 = arcsin (sin05β6 sin60o/sino6) = 18,23747
x6 = R* sin o1b6 cos O1B6 4236,613
y6 = R* sin o1b6 sin O1B6 1395,998
z6 = R * (1 cos o1b6) 1050,011
o1a7 = o1o5+o5a3 = 39,27632
? 7 = R*sin o1a7 cos O7
6330,61
y7 =
08=0lb8 = arccos (coso5B6 cos о1о5 + sin о5в6 sin about 1о5 cos 120) =
O8=O1B8 = arcsin (sin05B6 sin120V s^8) =
x8 = R* sin о1Ь8 cos O1B8 5632,611
13,91982
35,47183
y8 = R* sin о1Ь8 sin O1B8 1395,998
z8 = R * (1 cos о1Ь8)
1855,991
о^10 = o1o13 5а3 =
50,72368
x10 = R*sin o1a10 cos O10 7741,019
y10 =
z10 = R * (1-cos o1a10)
3669,39
А10O1O11=А3O1O4 = arcsin (sin0ta4 sin30o/sino4aА3) = | 56,28036 |
oto11 = arccos (coso11а10 cos о1о10) + sin о11а10 sin о1о10 cos А314) = | 45,62576 |
O11=arcsin (sinO11 sinА10O1O11/s^H ^ | 11,93246 |
x11 = R* sin o1o11 cos O11 6993,42
y11 = R* sin о^11 sin O11 1477,884
z11 = R * (1 cos о^11)
3006,58
О1В6О9=30-О1В6 = | 11,76253 |
O9В6O1=arcsin ^ іпо6sinO1В6O9/sino9в6) = | |
O9C12O1=60+3*Ä9É6Ä1 = | 163,0233 |
o1d2 = arccos (coso19с12 coso1o9 + sino19с12 s ^ 1о9 cos О9С12О1) =
O12=30-O12016=30-arcsin (sino9d2sinO9C12O1/sino12) = 26,01559 42,63783
x12 = R* sin o1d2 cos O12 | 6087,279 |
y12 = R* sin о1с12 sin O12 | 2971,015 |
z12 = R * (1 cos о1с12) | 2643,5 |
o1b14=arccos (coso13в14 co s о1о13 + sin о13в14 sin о1о13 cos О13А10В14) =
O14=arcsin (sino1b14 sin 60/sin о1Ь14) =
x14 = R* sin o1b14 cos O14
8144,009
z14 = R * (1 cos o1b14) 4367,389
o13o19=o5o9 = arcsin (sino1o9 sin30 o) = o1o19 = arccos (coso13c19 cos o1o13) = O19O13O1=arccos ((tgo19o13∕tgo1o19o) = O19O13C15 = 30+2 O9B6O1 =
16,10211 x19 = R* sin o1o19 cos O19
61,28949 y19 = R* sin o1o19 sin O19
80,90256 z19 = R* (1-cos o1o19)
98,68219
O19O1C15 = O19O13C15 - O19O14O1 =
17,77962
18,43495
O19 = arccos ((tg o1o13 ∕ tgo1 o19 o) = o1c15 = arccos (coso1o19cos o19c15 + sin o1o19sin o19c15cos O19O1C15) =
O1C15O19=arcsin (sino19c15 sin O19O1C15∕sino1c15) =
O15 = O19 + O15O19O1 = | 21,99194 |
x15 = R* sin o1c15 cos O15 | 7356,5 |
y15 = R* sin o1c15 sin O15 | 2971,015 |
z15 = R * (1 cos o1c15) | 3912,721 |
3,55699
O1O15O16=30 - O15 | 8,008061 |
o1o16 = arctg (tg o1aA7∕cos 30 o) = | 49,10661 |
o16c15 = arccos (coso1o16 cos o1c15 + sin o1o16 sin o1c15 cos O1O1516) = 7,070335
x16 = R* sin o1o16 cos O16 | 6546,537 |
y16 = R* sin o1o16 sin O16 | 3779,645 |
z16 = R * (1 cos o1o16) | 3453,463 |
17)
o1o25 = arctg (tg o1o13∕cos 30 o) = | 63,43495 | o1a17=o1o13+o13a10 = |
x25 = R*sin o1o25cos O25 | 7745,967 | x17 = R*sin o1o17cos O25 |
y25 = R* sin o1o25 sin O15 | 4472,136 | y17=0 |
z25 = R * (1-cos o1o25) | 5527,864 | z17 = R * (1-cos o1o17) |
o1b18 = arccos (coso1o13 cos o13a10 + sino1 o13 sin o13a10 cos 120) = 64,93404
O18 =arcsin (sino1a2 sin 60/sin o1b18 =
8,865427
x18 = ff*sin o1b18cos O18 | 8949,989 |
y18 = R* sin о1Ь18sin O18 | 1395,998 |
z18 = R * (1-cos o1b18) | 5763,387 |
O1C12O16=30-O12 = 3,98441
О16С12О9 = arcsin ^ іпО1С12О16sino1c12o/sin c^16) = | 22,48131 |
O16C15O1=180-arcsin (sinO1516sin o1c15/sino16s15) = | 116,1091 |
О16 D20O1=О16С15О1+2О16С12О9 =
161,0717
o1d20 = arccos (coso1α16 cos о16 d20 + sin о16 sin о16 d20 cos О16 О20О1) = 55,82923
O20=30-arcsin (sinO16D20O11516sino16d20/sino1d20) = 27,2339
20)
75,67615
D20O1O25 = 2,766103 | o1a21=o1o26-o26a21 = | |
x20 = ff*sin o1d20cos O20 | 7356,5 | x21 = ff*sin o1a21cos O25 |
y20 = R* sin o1d20 sin O20 | 3786,231 | y21=0 |
z20 = R * (1-cos o1d20) | 4383,386 | z21 = R * (1-cos o1a21) |
A21A17B27 = A10O1O11=A3O1O4 = | 56,28036 |
o1o22 = arccos (coso1a21 cos о22а21 + sin o1a21 sin o22a21 cos А21А17О22) =
О22 = arcsin (sin A21A17O22 sino22a21/sino1o22) =
x22 = ff*sin o1o22cos O22
9595,913
7,954047
y22 = R* sin o1o22 sin O22 1340,77
z22 = R * (1-cos o1o22)
7525,976
sin ((o5ol9 + olo5 + olo! 9) ∕2 — olo5) sin (o5ol9 + olo5 + olol9) ∕2
О19О5О1 =
sinBθ5ol9sinolol9
O19C23O1=O19O5O1+O9C12O1 = 179,5848
o1c23 = arccos (coso19с23 cos о^19 + sin о19с23 sin о^19 cos О19С23О1) =
О19С23 = arcsin (sin О19С23О1 sino19s23/sino1c23) =
0,07098
О23=О19-О19С23 = 18,50593
x23 = ff*sin o1c23cos O23 8942,6
y23 = R* sin о^23 sin O23
2993,181
o1o25 = arccos ((tg o1o13 ∕ cos 30 0) =
63,43495 o16aA7 = sin о16 sin30 =
x25 = R*sin o1o25cos O25
7745,967
o1a17=o1o13 + о16аД7 =
y24 = R* sin o1o25 sin O25 4472,136
z25 = R * (1-cos o1o25)
5527,864
o1o29 = arccos (cos o29a17 cos o1a17) =
O29 == arc cos (tg o1a17 ∕ tgo1o29o) = x29 = R*sin o1o29cos O29
y29 = R* sin o1o29 sin O29
z29 = R * (1-cos o1o29)
О29О5О1 = 2arccos
AT
sin ((o5o29 + olo5 + olo29) ∕2 — olo5) sin (o5o29 + olo5 + olo29) ∕2
sino5o29sinolo29
O16C12O9 = arcsin (sinO1216 sino1c12o ∕ sin c12o16) =
o29 d24 = o16c15 = arccos (coso1o16 cos o1c15 + sin o1o16 sin o1c15 cos O16O1C15) =
7,070335
D20O1O25 =
o25o20 = arccos (coso1d20 cos o1o25 + si n o1d20 sin o1o25 cos D20O1O25) =
7,970291
О25О20О1 = arcsin (sin D20O1O25sino1d20/sin d20o25) =
16,73559
o13o25 = arcsin (sino1o25 sin30 0) =
26,56505
О25О13О1 = arc cos (tg o13o25 ∕ tgo25) =
75,52249
02502401 = 2O25O13O1 - О25О20О1 =
134,3094
o1d24 = arccos (coso1o25 d240cos o1o25 + sino1 o25 d24 sin o1o25 cos 02502401) =
69,12926
O1D24O25 = arcsin (sin 02502401 sino25d24/sin o1d24) =
О24 = O25-O1O24O25 =
23,90431
6,095689
x24 = R*sin o1d24cos O24
8542,381
y24 = R* sin o1d24 sin O29
3786,231
z24 = R * (1-cos o1d24)
6437,391
o1o26 =
90
o26β27=o26β32=o33 d34=o26θ21
9,276324
x26 = ff*sin o1o26 cos O26 y26 =
10000
z26 = R * (1-cos o1o26)
10000
o1b27 = arccos (coso1o26 cos о26в27 + sin o1o26 sin o26b27 cos 60 0) =
85,37706
027 = arcsin (sin60 0 sino26β27 ∕ sin o1b27) =
8,051057
x27 = ff*sin o1b27cos O27
9869,224
y27 = R* sin o1b27 sin O27
1395,998
z27 = R * ((1-cos o1b27))
9194,02
o16oA7 = arcsin (sin o16 sin30) =
22,20765
О29С28О16 = 2х О16С12оА7=2 arc cos (tg o16oA7∕tgo16) =
138,5904
О29С28оА17 = О16С12оА7-О16С12О9 =
46,81388
О29С28О1 = О29С28О16-O29 D24O1 =
131,031
о29с28=о16с15 =
0,123401
o1c28 = arccos (coso29с28 cos o1o29 + sin о29с28 sino1o29cos О29С28О1) =
С1О29Є28 = arcsin (sin О29С28О1 sino29s28/sin o1c28) =
80,8309
5,396851
О29 =
22,91134
O28=O29-O2928 =
17,51449
x28 = ff*sin o1c28cos O28
9414,556
y28 = R* sin o1c28 sin O28
2971,015
z28 = R * (1-cos o1c28)
8406,512
С1О29С25=30-О29 = 7,088663
o25o29 = arccos (coso1o29 cos o1o25 + sin o1o29 sin o1o25cos С1О29С25) = 14,32834
О25О1О29=2 arcsin (sino1o13/sin o1o25) = 151,045
o25f30 = o25 d24 =
7,970291 o29f30 = о16с15 = 7,070335
О25О29Б30 =
2arccos
∖
sin ((o29o25 + o25e30 + o29e30) ∕2 - o29e30) si τι (α29o25 + o25e30 + o29e30) ∕2
siftEθ29o2Gsino25e30
Θ25Θ1F30 = Θ25Θ1Θ29 + Θ 25Θ29F30 = | 167,7806 |
o1f30 = arccos (coso1o25 cos o25f30 + sin o1o25 sin o25f30cos Θ25Θ1F30) = 71,2353
O1Θ30O25 = arcsin (sin Θ25Θ1F30sin o25f30 ∕ sino1f30) = 1,776206
Θ30=30 - O1Θ3O25 = 28,22379
x30 = ff*sin o1f30cos O30 | 8342,742 |
y30 = R* sin оИ30 sin O30 | 4477,801 |
z30 = R * (1-cos o1f30) | 6783,176 |
Θ25Θ36F30 = 180 Θ 25Θ1E30 = 12,21943
o1o36=o36 = arc tg (tg o1fA17 ∕ cos30) =
49,10661
o25o36=o1o36-o1o25 =
-14,3283
o36f30 = arccos (coso25o36 cos o25e30 + sin o25o36 sin o25e30cos Θ25Θ36E30) =
22,18094
x36 = ff*sin o1o36cos O36 | 6546,537 | |
y36 = R* sin о^36 sin O36 | 3779,645 | |
z36 = R * (1-cos o1o36) | 3453,463 | |
o1o26=3*o1o3 = | 90 | |
o1a31=o1o26+o26a31 | 99,27632 |
x31 = ff*sin o1a31 cos O31
y31=0
o1b32 = arccos (coso1o26 cos o26a31 + sino1 o26 sin o26a31cos120 o) = | 94,62294 |
Θ32=arcsin (sin Θ26B32Θ1 sino26a31 ∕ sin o1b32) = | 8,051057 |
x32 = ff*sin o1b32cos O32 | 9869,224 |
y32 = R* sin о1Ь32 sin O32
1395,998
z32 = R * (1-cos o1b32)
10805,98
o26o33=o5o9 = 16,10211
o1o33 = arccos (cos o1o26 ∕ cos o5o9) = | 90 | y33 = R* sin о^33 sin O33 |
Θ33 = arcsin (sino26o33 ∕ sin o1o33) = | z33 = R * (1-cos o1o33) | |
16,10211 | ||
Θ29D34O1=O29C28O1+2Θ16C12Θ9 = | 175,9936 |
o1d34 = arccos (coso1o29 cos o29 d34 + sin o1o29 sin o29 d34cos120o) =
79,74398
OW29O34 = arcsin (sino26o33/sino1 o33) = 16,10211
О34=ОЮ29-OW29O34 = | 6,809223 |
x34 = R*sin o1d34cos O34 | 9770,811 |
y34 = R* sin o1d34 sin O34 | 1166,694 |
z34 = R * (1-cos o1d34) | 8219,531 |
a∆17o36=arcsin (sino1o36 sin 30) = | 22,20765 x35 = R*sin o1f35cos O35 |
О36 О29О1 = arc cos (tg a∆17o36 ∕ tgo1o36) = 69,29519 y35 = R* sin o1f35 sin O35
o25o36=o1o36-o1o25 =-14,3283 z35 = R * (1-cos o1f35)
o25 d24 = 7,970291 | О25Е30О36 = О25О36Е30 |
О36Б30О1 = arcsin (sin О25Е30О36 sino25 d24/sin o36∈30) = 4,458487
О36О1Б35 =-2O36O29O1 - О36Е30О1 = | 134,1319 |
o35=arccos (coso1o36 cos o36f35 + sin o1o36 sin o36f35cos О36О1135о) =
O3536=arcsin (sin О36О1О35 sino1o36035/sin o1f35) = 17,26084 65,95289
О35=30-О3536 = 12,73916
О36 Є42О1 = О36О1Б35 + О36Е30О1 = 138,5904
o1g42 = arccos (coso1o36 cos o1o36g 42 + sin o1f36 sin o1o36g 42cos О36 Є42О1о) =
O1G42O36 = arcsin (sin О36 g 42О1 sinoo36 g 42/sin o1 g 42) = 15,75116 66,91067
О42=30-О4236 =
14,24884
X42 = R*sin o1g42cos O4 | 8915,948 |
y42 = R* sin o1g42 sin O42 | 2264,17 |
z42 = R * (1-cos o1 g42) | 6078,342 |
THE APPENDIX 3
Table 3.1 of the Equation of regress of durability of lumber at compression along fibres from duration of a finding in the conditions of the raised moisture content
Table 3.2
Tarirovochnyye wire data
Stages Weighting | Readout on the device for samples of a wire at movings of 5 mm: mm/mpa | Sr znach. | Sredn its sq. otklo nenija | koef - nt varia tsii | Znach. With verojat but - stju 0.95 | |||||
Wuhsi lie, kg | Nap a gabion nija, MPa | 1 | 2 | 2 | 4 | 4 | ||||
50 100 150 200 250 300 350 400 | 26 51 77 102 126 153 177 204 | 11,69 10,51 9,90 8,44 7,36 6,33 5,27 4,24 | 11,67 10,65 9,55 8,47 7,35 6,34 5,23 4,22 | 11,66 10,58 9,46 8.42 7.43 6,39 5,27 4,23 | 11,74 10,57 9,47 8,44 7,39 6,32 5,23 4,36 | 11,68 10,59 9,46 8,49 7,37 6,35 5.28 4.29 | 11,68 | 10,45 | 0,895 | 11,10 |
10,58 | 9,432 | 0,891 | 10,05 | |||||||
9,568 | 8,707 | 0,910 | 9,090 | |||||||
8,452 | 7,554 | 0,894 | 8,029 | |||||||
7,380 | 6,592 | 0,893 | 7,011 | |||||||
6,346 | 5,669 | 0,893 | 6,029 | |||||||
5,256 | 4,707 | 0,896 | 4,993 | |||||||
4,268 | 3,805 | 0,892 | 4,055 |
Table 3.3
Terms of removal of readout on the device for models in Saransk, air temperature/humidity
Sort of skilled materials | Level of pressure in string pieces, N/mm And = 19 mm | ||||||
Models | String pieces, inclined braces | July 2014 24/58 | October 2014 11/55 | July 2015 21/56 | October 2015 11/69 | July 2016 27/58 | October 2016 |
1 | 1-2 | 148,9 | 119,8 | 98,6 | 87,3 | 83,1 | 79,9 |
2-3 | 148,5 | 119,2 | 98,9 | 87,2 | 83,6 | 79,4 | |
3-4 | 148,4 | 118,8 | 98,9 | 86,7 | 83,8 | 79,1 | |
Compare znach. | 148,0 | 119,4 | 98,8 | 87,1 | 83,5 | 79,5 | |
1 | 1 | 127,3 | 97,0 | 86,2 | 78,8 | 75,0 | 68,8 |
2 | 128,2 | 97,5 | 86,7 | 78,0 | 75,3 | 68,8 | |
3 | 128,8 | 96,8 | 86,9 | 77,6 | 75,4 | 69,1 | |
Compare znach. | 128,1 | 97,1 | 86,6 | 78,1 | 75,2 | 68,9 |
Table 3.4
Terms of removal of readout for models in Gelendzhike
Sort of skilled materials | Level of pressure in string pieces, Н/мм2 And = 19 mm 2 | ||||||
Iode Whether | String pieces, inclined braces | July 2014 29/78 | October 2014 14/59 | July 2015 27/68 | October 2015 22/68 | July 2016 28/67 | October 201622/68 |
2 | 7-8 | 170,1 | 136,2 | 119,7 | 106,4 | 77,8 | 68,7 |
8-9 | 169,9 | 136,7 | 119,1 | 106,4 | 77,5 | 68,8 | |
9-10 | 169,8 | 136,3 | 119,1 | 106,8 | 77,4 | 69,3 | |
Compare znach. | 169,7 | 136,3 | 119,3 | 106,5 | 77,6 | 68,9 | |
2 | 7 | 160,2 | 132,5 | 113,9 | 95,1 | 66,7 | 59,9 |
8 | 160,4 | 132,9 | 113,0 | 95,6 | 67,1 | 59,5 | |
9 | 160,9 | 132,6 | 113,1 | 95,8 | 67,5 | 59,6 | |
Compare znach. | 160,5 | 132,7 | 113,3 | 95,5 | 67,1 | 59,7 | |
3 | 13-14 | 139,8 | 119, 7 | 97,7 | 86,5 | 60,9 | 53.4 |
14-15 | 140,5 | 119, 4 | 97,3 | 86,2 | 60,6 | 53.7 | |
15-16 | 141.7 | 119, 0 | 97,0 | 86,0 | 60,1 | 52.9 | |
Compare znach. | 140,7 | 119, 3 | 97,3 | 86,2 | 60,5 | 53.3 | |
3 | 13 | 128,9 | 86,9 | 70,7 | 64,7 | 48,4 | 42,5 |
14 | 128,4 | 86,7 | 70,3 | 65,2 | 48,2 | 42,6 | |
15 | 128,2 | 86,7 | 70,1 | 64,8 | 47,8 | 42,9 | |
Compare znach. | 128,5 | 86,8 | 70,4 | 64,9 | 48,1 | 42,7 |
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