Proracun
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Idejno rešenje dispozicije baterije ćelija silosa: kapacitet 9000 t žita oblik ćelija kvadratni temeljenje na šipovima „Franki“ Ø520, nosivosti 1100 kN
Predpostavljena sopstvena težina ćelija silosa, prema težini skladištenja materijala u odnosu 45% konstrukcija 55% skladišteni materijal. 9000 : g = 55 : 45 45 ⋅ 9000 g≈ = 7363 usvojeno : g = 7500t 55 Iz usvojenog oblika ćelije Slika 1 imamo: V ' = 4.22 = 17.64 m3 / m
90000 = 12000 m3 7.5
3
12000 = 28.35m 24 ⋅ 17.64
30
30
pretp : n = 24 → hpot =
( težina žita 7.5 kN/m ; 1t = 10kN )
450 420
V pot =
450 420
Slika 1
G + P = 75000 + 90000 = 165000 kN 165000 = 150 kom n= 1100 raspored šipova i širine naglavnih greda date su na Slika 5, pa je prema tome usvojeni broj „Franki“ Ø520 šipova 182 komada.
5.5
30 40
Mašinske prostorije
4.5
Potreban broj šipova: S dop = 1100 kN
Slika 2
1
Novak Novaković 14/02
450
450
450
450
450
450
172.5
225
225
112.5
120
330
120
300 150 300 150 300 150 225
225
225
330
120
330
225
225
225
225
225
225
225
60
120
330
120
330
120
330
120
112.5
300
225
180
60
60
135
60
450
270
135
112.5
2370
150 300
180
450
270
180
450
270
180
450
270
180
450
60
135
60
172.5
3045
Slika 3
Otporni moment grupa šipova, Slika 4: x1 = 14.62m ( n = 20 )
450
450
450
60
172.5
1a
2a
3a
4a
450
x3 = 10.12m
1
3
4
5
6
7
270
x4 = 7.88m
2
180
60
x2 = 12.38m
x5 = 5.62m x6 = 3.38m
Slika 4
x7 = 1.12m x1a = 13.5m
( n = 12 )
x2 a = 9.0m x3a = 4.5m
2
Novak Novaković 14/02
W = 20 ⋅ (14.622 + 12.382 + 10.122 + 7.882 + 5.622 + 3.382 + 1.122 ) = 11515.62 m 2 Wa = 12 ⋅ (13.52 + 92 + 4.52 ) = 3402 m 2 W = 11515.62 + 3402 = 14917.62 m 2
Proračun naprezanja šipova Pun silos V = 90000 + 75000 = 165000 kN 165000 S= = 906.59 kN < S dop = 1100 kN 182 Za najnepovoljniji način punjenja, Slika 6 2 P = ⋅ 90000 = 60000 kN 3 V = 75000 + 60000 = 135000 kN
Tp
Tg 4.5
Slika 6
Pri zemljotresu, seizmička zona a=0.11g seizmička sila deluje na polovini visine silosa, Slika 7 G = 75000 kN
Mašinske prostorije
P = 90000 kN ( pun silos ) 2 ⋅ 1100 = 1450 kN 1.5 0.11g S= ( 75000 + 90000 ) = 18150 kN g M = 18150 ⋅ 20 = 363000 kN sile u najopterećenijim šipovina 165000 363000 + ⋅ 14.62 = 1262.25 kN < Sdop , se = 1450 kN S max = 182 14917.62 165000 363000 − ⋅ 14.62 = 550.8 kN > 0 S min = 14917.62 182
4.5
M = 60000 ⋅ 4.5 = 270000 kNm 135000 270000 S max = + ⋅ 14.62 = 1006.3 kN < Sdop = 1100kN 182 14917.62 135000 270000 S min = − ⋅ 14.62 = 472.22 kN > 0 ( nema zatezanja u šipu ) 182 14917.62
S dop , se =
3
5.5
20
30 40
S
Slika 7
Novak Novaković 14/02
Proračun opterećenja za zid silosa (sa korakom od 1.0 m) μ = 0.3 λ = 0.6 Za izračunavanje maksimalnih projektnih opterećenja, uzimaju se sledeće kombinacije za μ i λ. ph :1.15λ 0.9 μ pv : 0.9λ
0.9 μ
pw :1.15λ
1.15μ
p
v
h
5.5
za kvadratni silos R = a / 2 = 4.2 / 2 = 2.1m R z0 = λ '⋅ μ ' Cz = 1 − e
p
pw
30 40
karakteriskitke materijala za proračun sila γ = 8.5 kN / m3
Slika 8
⎛ z ⎞ ⎜− ⎟ ⎝ z0 ⎠
γ ⋅R ⋅ Cz ( z ) ⋅ C μ' γ ⋅R ⋅ Cz ( z ) ⋅ C pve ( z ) = λ '⋅ μ ' pwe ( z ) = γ ⋅ R ⋅ C z ( z ) ⋅ C phe ( z ) =
koeficient nadpritiska C, uvećava opterećenje pri pražnjenju i iznosi 1.35 za ovaj silos (h/d>1.5)
4
Novak Novaković 14/02
Tabelarni prikaz opterećenja sa korakom od 1m. λ= 0.6 λ’= μ= 0.3 μ’= γ= 8.5 kN/m3 C= 1.35 z0= 11.272 z 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
phf phe Cz 0.000 0.00 0.00 0.085 5.61 7.58 0.163 10.75 14.51 0.234 15.45 20.86 0.299 19.75 26.66 0.358 23.68 31.97 0.413 27.29 36.84 0.463 30.58 41.29 0.508 33.60 45.36 0.550 36.36 49.08 0.588 38.88 52.49 0.623 41.20 55.61 0.655 43.31 58.47 0.684 45.25 61.08 0.711 47.02 63.47 0.736 48.64 65.66 0.758 50.12 67.66 0.779 51.48 69.50 0.797 52.72 71.17 0.815 53.86 72.71 0.830 54.90 74.11 0.845 55.85 75.40 0.858 56.72 76.57 0.870 57.52 77.65 0.881 58.25 78.63 0.891 58.92 79.54 0.900 59.53 80.36 0.909 60.09 81.12 0.917 60.60 81.81 0.924 61.06 82.44 0.930 61.49 83.02
ph0=61.49 kN/m2
λ= 0.6 λ’= μ= 0.3 μ’= γ= 8.5 kN/m3 C= 1.35 z0= 14.403
0.69 0.27
z
pvf
Cz 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
0.000 0.067 0.130 0.188 0.242 0.293 0.341 0.385 0.426 0.465 0.501 0.534 0.565 0.594 0.622 0.647 0.671 0.693 0.713 0.733 0.751 0.767 0.783 0.797 0.811 0.824 0.836 0.847 0.857 0.866 0.875
5
0.00 8.21 15.87 23.02 29.69 35.91 41.71 47.12 52.18 56.89 61.28 65.38 69.21 72.78 76.11 79.22 82.12 84.82 87.34 89.70 91.89 93.94 95.85 97.63 99.30 100.85 102.29 103.64 104.90 106.08 107.18
λ= μ= γ= C= z0=
0.54 0.27
pve 0.00 11.09 21.43 31.08 40.08 48.48 56.31 63.62 70.44 76.80 82.73 88.27 93.43 98.25 102.75 106.94 110.86 114.51 117.91 121.09 124.05 126.82 129.40 131.80 134.05 136.14 138.10 139.92 141.62 143.21 144.69
z
0.6 λ’= 0.3 μ’= 8.5 kN/m3 1.35 8.821 pwf
Cz 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
0.000 0.107 0.203 0.288 0.365 0.433 0.493 0.548 0.596 0.639 0.678 0.713 0.743 0.771 0.795 0.817 0.837 0.854 0.870 0.884 0.896 0.907 0.917 0.926 0.934 0.941 0.948 0.953 0.958 0.963 0.967
0.00 1.91 3.62 5.15 6.51 7.72 8.81 9.78 10.64 11.41 12.10 12.72 13.27 13.76 14.20 14.59 14.94 15.25 15.53 15.78 16.00 16.20 16.38 16.53 16.67 16.80 16.91 17.01 17.10 17.18 17.25
0.69 0.345
pwe 0.00 2.58 4.89 6.95 8.78 10.43 11.89 13.20 14.37 15.41 16.34 17.17 17.91 18.58 19.17 19.70 20.17 20.59 20.97 21.30 21.60 21.87 22.11 22.32 22.51 22.68 22.83 22.97 23.09 23.20 23.29
Novak Novaković 14/02
Rasppodela horizzontalnog op pterećenja po p zidu silosa prikazana na Slikaa 9, nsjoptreećeniji preseek je najnižaa tačka silosa, kod spojja ćelijee sa levkom m. Račuunski potrebbna armaturra u najopterrećenijem ppreseku h=00m (sa grafika 30m) ma opterećenja, zajednoo sa dodatniim opterećeenjem Shem p = 83.02 8 kN / m 2
30
25
20
pdod = 0.2 ⋅ 83.022 = 16.60 kN N / m2 15
deo na n koji deluuje dodatno opterećenjee je 0.8 ⋅
A 177.64 = 0.84 m = 0.8 ⋅ 166.8 u 10
183
5
0
84 450
83.02
0
16.60
20
40 4
60
80
100
183
Slik ka 9
Uticaaji: samo o od ph
ph+dopuunsko
6
Novak Novakovićć 14/02
Proračun potrebene armature za preseke u uglovima 83.02 ⋅ 4.5 = 186.79 kN Z= 2 M max = 147.84 kNm Z u = 1.8 ⋅ 186.79 = 336.22 kN M u = 1.8 ⋅ 140.10 = 252.18 kNm ⎛ 0.30 ⎞ − 0.045 ⎟ = 216.88 kNm M au = 252.18 − 336.22 ⋅ ⎜ ⎝ 2 ⎠ RA400 / 500 MB 40 25.5 = 2.765 → μ = 14.152% 216.88 2.55 100 ⋅ 25.5 2.55 336.22 ⋅ + Aa = 14.152 ⋅ = 26.90 cm 2 40 100 40
k=
usvojeno : R∅ 22 / 20 + R∅16 / 20 Aa , stv =
381 201 + = 19.05 + 10.05 = 29.10 cm 2 20 20
Proračun potrebne armature za sredine zidova ćelija
83.02 ⋅ 4.5 = 186.79 kN 2 M max = 76.53 kNm Z=
Z u = 1.8 ⋅ 186.79 = 336.22 kN M u = 1.8 ⋅ 76.53 = 137.54 kNm ⎛ 0.30 ⎞ − 0.045 ⎟ = 102.45 kNm M au = 137.54 − 336.22 ⋅ ⎜ ⎝ 2 ⎠ RA400 / 500 MB 40 25.5
= 4.023 → μ = 6.591% 102.45 2.55 100 ⋅ 25.5 2.55 336.22 ⋅ + Aa = 6.591 ⋅ = 18.89 cm 2 100 40 40 3.80 ⋅ 100 ea ( R∅ 22 ) = = 20.11cm → usvojeno : R∅ 22 / 20 18.89 k=
7
Novak Novaković 14/02
Proračun opterećenja za levak prema JUS.ISO.11697 ph 0 = 61.49 kN / m 2
α = 30o cos α = 0.866 sin α = 0.5 λ = 0.9 ⋅ 0.6 = 0.54 μ = 0.9 ⋅ 0.3 = 0.27 ⎛1 ⎞ ⎛ 1 ⎞ pn1 = 1.5 ⋅ ph 0 ⋅ ⎜ ⋅ cos 2 α + sin 2 α ⎟ = 1.5 ⋅ 61.49 ⋅ ⎜ ⋅ 0.8662 + 0.52 ⎟ = 151.163 kN / m 2 ⎝λ ⎠ ⎝ 0.54 ⎠ 1.5 1.5 pn 2 = ⋅ ph 0 ⋅ cos 2 α = ⋅ 61.49 ⋅ 0.8662 = 128.1 kN / m 2 λ 0.54 3 ⋅ 17.64 8.5 ⋅ 0.54 3⋅ A γ ⋅λ pn 3 = ⋅ ⋅ sin 2 α = ⋅ ⋅ 0.52 = 7.33 kN / m 2 u 16.8 μ 0.27 pt = μ ⋅ ( pn1 + pn 2 + pn 3 ) = 0.27 ⋅ (151.16 + 128.1 + 7.33) = 85.98 kN / m 2 ps = 2 ⋅ ph 0 = 2 ⋅ 61.49 = 122.98 kN / m 2
8
Novak Novaković 14/02
Predpostavljena sopstvena težina ćelija silosa, prema težini skladištenja materijala u odnosu 45% konstrukcija 55% skladišteni materijal. 9000 : g = 55 : 45 45 ⋅ 9000 g≈ = 7363 usvojeno : g = 7500t 55 Iz usvojenog oblika ćelije Slika 1 imamo: V ' = 4.22 = 17.64 m3 / m
90000 = 12000 m3 7.5
3
12000 = 28.35m 24 ⋅ 17.64
30
30
pretp : n = 24 → hpot =
( težina žita 7.5 kN/m ; 1t = 10kN )
450 420
V pot =
450 420
Slika 1
G + P = 75000 + 90000 = 165000 kN 165000 = 150 kom n= 1100 raspored šipova i širine naglavnih greda date su na Slika 5, pa je prema tome usvojeni broj „Franki“ Ø520 šipova 182 komada.
5.5
30 40
Mašinske prostorije
4.5
Potreban broj šipova: S dop = 1100 kN
Slika 2
1
Novak Novaković 14/02
450
450
450
450
450
450
172.5
225
225
112.5
120
330
120
300 150 300 150 300 150 225
225
225
330
120
330
225
225
225
225
225
225
225
60
120
330
120
330
120
330
120
112.5
300
225
180
60
60
135
60
450
270
135
112.5
2370
150 300
180
450
270
180
450
270
180
450
270
180
450
60
135
60
172.5
3045
Slika 3
Otporni moment grupa šipova, Slika 4: x1 = 14.62m ( n = 20 )
450
450
450
60
172.5
1a
2a
3a
4a
450
x3 = 10.12m
1
3
4
5
6
7
270
x4 = 7.88m
2
180
60
x2 = 12.38m
x5 = 5.62m x6 = 3.38m
Slika 4
x7 = 1.12m x1a = 13.5m
( n = 12 )
x2 a = 9.0m x3a = 4.5m
2
Novak Novaković 14/02
W = 20 ⋅ (14.622 + 12.382 + 10.122 + 7.882 + 5.622 + 3.382 + 1.122 ) = 11515.62 m 2 Wa = 12 ⋅ (13.52 + 92 + 4.52 ) = 3402 m 2 W = 11515.62 + 3402 = 14917.62 m 2
Proračun naprezanja šipova Pun silos V = 90000 + 75000 = 165000 kN 165000 S= = 906.59 kN < S dop = 1100 kN 182 Za najnepovoljniji način punjenja, Slika 6 2 P = ⋅ 90000 = 60000 kN 3 V = 75000 + 60000 = 135000 kN
Tp
Tg 4.5
Slika 6
Pri zemljotresu, seizmička zona a=0.11g seizmička sila deluje na polovini visine silosa, Slika 7 G = 75000 kN
Mašinske prostorije
P = 90000 kN ( pun silos ) 2 ⋅ 1100 = 1450 kN 1.5 0.11g S= ( 75000 + 90000 ) = 18150 kN g M = 18150 ⋅ 20 = 363000 kN sile u najopterećenijim šipovina 165000 363000 + ⋅ 14.62 = 1262.25 kN < Sdop , se = 1450 kN S max = 182 14917.62 165000 363000 − ⋅ 14.62 = 550.8 kN > 0 S min = 14917.62 182
4.5
M = 60000 ⋅ 4.5 = 270000 kNm 135000 270000 S max = + ⋅ 14.62 = 1006.3 kN < Sdop = 1100kN 182 14917.62 135000 270000 S min = − ⋅ 14.62 = 472.22 kN > 0 ( nema zatezanja u šipu ) 182 14917.62
S dop , se =
3
5.5
20
30 40
S
Slika 7
Novak Novaković 14/02
Proračun opterećenja za zid silosa (sa korakom od 1.0 m) μ = 0.3 λ = 0.6 Za izračunavanje maksimalnih projektnih opterećenja, uzimaju se sledeće kombinacije za μ i λ. ph :1.15λ 0.9 μ pv : 0.9λ
0.9 μ
pw :1.15λ
1.15μ
p
v
h
5.5
za kvadratni silos R = a / 2 = 4.2 / 2 = 2.1m R z0 = λ '⋅ μ ' Cz = 1 − e
p
pw
30 40
karakteriskitke materijala za proračun sila γ = 8.5 kN / m3
Slika 8
⎛ z ⎞ ⎜− ⎟ ⎝ z0 ⎠
γ ⋅R ⋅ Cz ( z ) ⋅ C μ' γ ⋅R ⋅ Cz ( z ) ⋅ C pve ( z ) = λ '⋅ μ ' pwe ( z ) = γ ⋅ R ⋅ C z ( z ) ⋅ C phe ( z ) =
koeficient nadpritiska C, uvećava opterećenje pri pražnjenju i iznosi 1.35 za ovaj silos (h/d>1.5)
4
Novak Novaković 14/02
Tabelarni prikaz opterećenja sa korakom od 1m. λ= 0.6 λ’= μ= 0.3 μ’= γ= 8.5 kN/m3 C= 1.35 z0= 11.272 z 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
phf phe Cz 0.000 0.00 0.00 0.085 5.61 7.58 0.163 10.75 14.51 0.234 15.45 20.86 0.299 19.75 26.66 0.358 23.68 31.97 0.413 27.29 36.84 0.463 30.58 41.29 0.508 33.60 45.36 0.550 36.36 49.08 0.588 38.88 52.49 0.623 41.20 55.61 0.655 43.31 58.47 0.684 45.25 61.08 0.711 47.02 63.47 0.736 48.64 65.66 0.758 50.12 67.66 0.779 51.48 69.50 0.797 52.72 71.17 0.815 53.86 72.71 0.830 54.90 74.11 0.845 55.85 75.40 0.858 56.72 76.57 0.870 57.52 77.65 0.881 58.25 78.63 0.891 58.92 79.54 0.900 59.53 80.36 0.909 60.09 81.12 0.917 60.60 81.81 0.924 61.06 82.44 0.930 61.49 83.02
ph0=61.49 kN/m2
λ= 0.6 λ’= μ= 0.3 μ’= γ= 8.5 kN/m3 C= 1.35 z0= 14.403
0.69 0.27
z
pvf
Cz 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
0.000 0.067 0.130 0.188 0.242 0.293 0.341 0.385 0.426 0.465 0.501 0.534 0.565 0.594 0.622 0.647 0.671 0.693 0.713 0.733 0.751 0.767 0.783 0.797 0.811 0.824 0.836 0.847 0.857 0.866 0.875
5
0.00 8.21 15.87 23.02 29.69 35.91 41.71 47.12 52.18 56.89 61.28 65.38 69.21 72.78 76.11 79.22 82.12 84.82 87.34 89.70 91.89 93.94 95.85 97.63 99.30 100.85 102.29 103.64 104.90 106.08 107.18
λ= μ= γ= C= z0=
0.54 0.27
pve 0.00 11.09 21.43 31.08 40.08 48.48 56.31 63.62 70.44 76.80 82.73 88.27 93.43 98.25 102.75 106.94 110.86 114.51 117.91 121.09 124.05 126.82 129.40 131.80 134.05 136.14 138.10 139.92 141.62 143.21 144.69
z
0.6 λ’= 0.3 μ’= 8.5 kN/m3 1.35 8.821 pwf
Cz 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
0.000 0.107 0.203 0.288 0.365 0.433 0.493 0.548 0.596 0.639 0.678 0.713 0.743 0.771 0.795 0.817 0.837 0.854 0.870 0.884 0.896 0.907 0.917 0.926 0.934 0.941 0.948 0.953 0.958 0.963 0.967
0.00 1.91 3.62 5.15 6.51 7.72 8.81 9.78 10.64 11.41 12.10 12.72 13.27 13.76 14.20 14.59 14.94 15.25 15.53 15.78 16.00 16.20 16.38 16.53 16.67 16.80 16.91 17.01 17.10 17.18 17.25
0.69 0.345
pwe 0.00 2.58 4.89 6.95 8.78 10.43 11.89 13.20 14.37 15.41 16.34 17.17 17.91 18.58 19.17 19.70 20.17 20.59 20.97 21.30 21.60 21.87 22.11 22.32 22.51 22.68 22.83 22.97 23.09 23.20 23.29
Novak Novaković 14/02
Rasppodela horizzontalnog op pterećenja po p zidu silosa prikazana na Slikaa 9, nsjoptreećeniji preseek je najnižaa tačka silosa, kod spojja ćelijee sa levkom m. Račuunski potrebbna armaturra u najopterrećenijem ppreseku h=00m (sa grafika 30m) ma opterećenja, zajednoo sa dodatniim opterećeenjem Shem p = 83.02 8 kN / m 2
30
25
20
pdod = 0.2 ⋅ 83.022 = 16.60 kN N / m2 15
deo na n koji deluuje dodatno opterećenjee je 0.8 ⋅
A 177.64 = 0.84 m = 0.8 ⋅ 166.8 u 10
183
5
0
84 450
83.02
0
16.60
20
40 4
60
80
100
183
Slik ka 9
Uticaaji: samo o od ph
ph+dopuunsko
6
Novak Novakovićć 14/02
Proračun potrebene armature za preseke u uglovima 83.02 ⋅ 4.5 = 186.79 kN Z= 2 M max = 147.84 kNm Z u = 1.8 ⋅ 186.79 = 336.22 kN M u = 1.8 ⋅ 140.10 = 252.18 kNm ⎛ 0.30 ⎞ − 0.045 ⎟ = 216.88 kNm M au = 252.18 − 336.22 ⋅ ⎜ ⎝ 2 ⎠ RA400 / 500 MB 40 25.5 = 2.765 → μ = 14.152% 216.88 2.55 100 ⋅ 25.5 2.55 336.22 ⋅ + Aa = 14.152 ⋅ = 26.90 cm 2 40 100 40
k=
usvojeno : R∅ 22 / 20 + R∅16 / 20 Aa , stv =
381 201 + = 19.05 + 10.05 = 29.10 cm 2 20 20
Proračun potrebne armature za sredine zidova ćelija
83.02 ⋅ 4.5 = 186.79 kN 2 M max = 76.53 kNm Z=
Z u = 1.8 ⋅ 186.79 = 336.22 kN M u = 1.8 ⋅ 76.53 = 137.54 kNm ⎛ 0.30 ⎞ − 0.045 ⎟ = 102.45 kNm M au = 137.54 − 336.22 ⋅ ⎜ ⎝ 2 ⎠ RA400 / 500 MB 40 25.5
= 4.023 → μ = 6.591% 102.45 2.55 100 ⋅ 25.5 2.55 336.22 ⋅ + Aa = 6.591 ⋅ = 18.89 cm 2 100 40 40 3.80 ⋅ 100 ea ( R∅ 22 ) = = 20.11cm → usvojeno : R∅ 22 / 20 18.89 k=
7
Novak Novaković 14/02
Proračun opterećenja za levak prema JUS.ISO.11697 ph 0 = 61.49 kN / m 2
α = 30o cos α = 0.866 sin α = 0.5 λ = 0.9 ⋅ 0.6 = 0.54 μ = 0.9 ⋅ 0.3 = 0.27 ⎛1 ⎞ ⎛ 1 ⎞ pn1 = 1.5 ⋅ ph 0 ⋅ ⎜ ⋅ cos 2 α + sin 2 α ⎟ = 1.5 ⋅ 61.49 ⋅ ⎜ ⋅ 0.8662 + 0.52 ⎟ = 151.163 kN / m 2 ⎝λ ⎠ ⎝ 0.54 ⎠ 1.5 1.5 pn 2 = ⋅ ph 0 ⋅ cos 2 α = ⋅ 61.49 ⋅ 0.8662 = 128.1 kN / m 2 λ 0.54 3 ⋅ 17.64 8.5 ⋅ 0.54 3⋅ A γ ⋅λ pn 3 = ⋅ ⋅ sin 2 α = ⋅ ⋅ 0.52 = 7.33 kN / m 2 u 16.8 μ 0.27 pt = μ ⋅ ( pn1 + pn 2 + pn 3 ) = 0.27 ⋅ (151.16 + 128.1 + 7.33) = 85.98 kN / m 2 ps = 2 ⋅ ph 0 = 2 ⋅ 61.49 = 122.98 kN / m 2
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Novak Novaković 14/02
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