<<

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 of

The 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

<< |
A source: Antoshkin Vasily Dmitrievich. is constructive-TECHNOLOGICAL DECISIONS of PRECAST SPHERICAL SHELLS. The dissertation on competition of a scientific degree of a Dr.Sci.Tech. Saransk - 2017. 2017

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