OBLICZENIA – PŁYTA
Schemat statyczny
Rozkład obciążeń
g – zebrany ciężar działający na płytę zgodnie z projektem wstępnym
p – obciążenie użytkowe
q’ = 0,5p + g
q’’ = 0,5p
obciążenia stropu nad parterem – wartości obliczeniowe:
$$q^{'} = g + 0,5p = 6,134*1,35 + 0,5*1,5*1,9 = 11,13\ \frac{\text{kN}}{m}$$
$$q^{''} = 0,5p = 0,5*1,5*1,9 = 1,43\frac{\text{kN}}{m}$$
$$q = q^{''} + q^{'} = 11,52 + 4,5 = 12,56\frac{\text{kN}}{m}$$
Wyznaczenie momentów zginających
Współczynniki do obliczenia momentów przęsłowych i podporowych dobrano na podstawie tablic zawartych w „Konstrukcje żelbetowe” tom II, Kobiak, Stachurski.
Momenty przęsłowe
Płyta – schemat 2’
$\frac{l_{y'}}{l_{x'}} = \frac{6,5}{6,4} = 1,02$ φx = 0, 0344 (na kierunku globalnego y)
φy = 0, 0260 (na kierunku globalnego x)
φ1x = 0, 0380
φ1y = 0, 0352
$$M_{2'x,\max} = \left( \varphi_{2'y}*q^{'} + \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,026*11,13 + 0,038*1,43 \right)*{6,5}^{2} = 14,52\frac{\text{kNm}}{m}$$
$$M_{2'x,\min} = \left( \varphi_{2^{'}y}*q^{'} - \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,026*11,13 - 0,038*1,43 \right)*{6,5}^{2}\ = 9,93\frac{\text{kNm}}{m}$$
$$M_{2'y,\max} = \left( \varphi_{2'x}*q^{'} + \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0344*11,13 + 0,0352*1,43 \right)*{6,4}^{2} = 17,74\frac{\text{kNm}}{m}$$
$$M_{2'y,\min} = \left( \varphi_{2'x}*q^{'} - \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0344*11,13 - 0,0352*1,43 \right)*{6,4}^{2} = 13,62\frac{\text{kNm}}{m}$$
Płyta – schemat 3
$\frac{l_{y}}{l_{x}} = \frac{6,2}{6,4} = 0,97$ φx = 0, 0257 φ1x = 0, 0343
φy = 0, 0199 φ1y = 0, 0391
$$M_{3x,\max} = \left( \varphi_{3x}*q^{'} + \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0257*11,13 + 0,0343*1,43 \right)*{6,4}^{2} = 13,73\frac{\text{kNm}}{m}$$
$$M_{3x,\min} = \left( \varphi_{3x}*q^{'} - \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0257*11,13 - 0,0343*1,43 \right)*{6,4}^{2}\ = 9,71\frac{\text{kNm}}{m}$$
$$M_{3y,\max} = \left( \varphi_{3y}*q^{'} + \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0199*11,13 + 0,0391*1,43 \right)*{6,2}^{2} = 10,66\frac{\text{kNm}}{m}$$
$$M_{3y,\min} = \left( \varphi_{3y}*q^{'} - \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0199*11,13 - 0,0391*1,43 \right)*{6,2}^{2} = 6,36\frac{\text{kNm}}{m}$$
Płyta – schemat 4
Wariant 1
$\frac{l_{y}}{l_{x}} = \frac{6,2}{6,5} = 0,95$ φx = 0, 0242 φ1x = 0, 0328
φy = 0, 0298 φ1y = 0, 0408
$$M_{4x,1,\max} = \left( \varphi_{4x}*q^{'} + \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0242*11,13 + 0,0328*1,43 \right)*{6,5}^{2} = 13,36\frac{\text{kNm}}{m}$$
$$M_{4x,1,\min} = \left( \varphi_{4x}*q^{'} - \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0242*11,13 - 0,0328*1,43 \right)*{6,5}^{2}\ = 9,40\frac{\text{kNm}}{m}$$
$$M_{4y,1\max} = \left( \varphi_{4y}*q^{'} + \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0298*11,13 + 0,0408*1,43 \right)*{6,2}^{2} = 14,99\frac{\text{kNm}}{m}$$
$$M_{4y,1,\min} = \left( \varphi_{4y}*q^{'} - \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0298*11,13 - 0,0408*1,43 \right)*{6,2}^{2} = 10,51\frac{\text{kNm}}{m}$$
Wariant 2
$\frac{l_{y}}{l_{x}} = \frac{6,4}{5,7} = 1,12$ φx = 0, 0332 φ1x = 0, 0454
φy = 0, 0212 φ1y = 0, 0290
$$M_{4x,2,\max} = \left( \varphi_{4x}*q^{'} + \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0332*11,13 + 0,0454*1,43 \right)*{5,7}^{2} = 14,11\frac{\text{kNm}}{m}$$
$$M_{4x,2,\min} = \left( \varphi_{4x}*q^{'} - \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0332*11,13 - 0,0454*1,43 \right)*{5,7}^{2}\ = 9,90\frac{\text{kNm}}{m}$$
$$M_{4y,2\max} = \left( \varphi_{4y}*q^{'} + \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0212*11,13 + 0,029*1,43 \right)*{6,4}^{2} = 11,36\frac{\text{kNm}}{m}$$
$$M_{4y,2,\min} = \left( \varphi_{4y}*q^{'} - \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0212*11,13 - 0,029*1,43 \right)*{6,4}^{2} = 7,97\frac{\text{kNm}}{m}$$
Wariant 3
$\frac{l_{y}}{l_{x}} = \frac{6,2}{5,7} = 1,10$ φx = 0, 0322 φ1x = 0, 0440
φy = 0, 0220 φ1y = 0, 0300
$$M_{4x,3,\max} = \left( \varphi_{4x}*q^{'} + \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0322*11,13 + 0,044*1,43 \right)*{5,7}^{2} = 13,69\frac{\text{kNm}}{m}$$
$$M_{4x,3,\min} = \left( \varphi_{4x}*q^{'} - \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0322*11,13 - 0,044*1,43 \right)*{5,7}^{2}\ = 9,60\frac{\text{kNm}}{m}$$
$$M_{4y,3,\max} = \left( \varphi_{4y}*q^{'} + \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,022*11,13 + 0,03*1,43 \right)*{6,2}^{2} = 11,06\frac{\text{kNm}}{m}$$
$$M_{4y,3,\min} = \left( \varphi_{4y}*q^{'} - \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,022*11,13 - 0,03*1,43 \right)*{6,2}^{2}\ \ \ = 7,76\frac{\text{kNm}}{m}$$
Płyta – schemat 5
$\frac{l_{y}}{l_{x}} = \frac{6,2}{5,7} = 1,10$ φx = 0, 0257 φ1x = 0, 0440
φy = 0, 0153 φ1y = 0, 0300
$$M_{5x,\max} = \left( \varphi_{3x}*q^{'} + \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0257*11,13 + 0,044*1,43 \right)*{5,7}^{2} = 11,34\frac{\text{kNm}}{m}$$
$$M_{5x,\min} = \left( \varphi_{3x}*q^{'} - \varphi_{1x}*q^{''} \right)*l_{x}^{2} = \left( 0,0257*11,13 - 0,044*1,43 \right)*{5,7}^{2}\ = 7,25\frac{\text{kNm}}{m}$$
$$M_{5y,\max} = \left( \varphi_{3y}*q^{'} + \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0153*11,13 + 0,03*1,43 \right)*{6,2}^{2}\ = 8,19\frac{\text{kNm}}{m}$$
$$M_{5y,\min} = \left( \varphi_{3y}*q^{'} - \varphi_{1y}*q^{''} \right)*l_{y}^{2} = \left( 0,0153*11,13 - 0,03*1,43 \right)*{6,2}^{2}\ \ = 4,90\frac{\text{kNm}}{m}$$
Momenty podporowe w osiach podpór
współczynniki χ:
schemat 2’: χ = 0, 729 (na kierunku globalnego y)
schemat 3: χ = 0, 813
schemat 4.1: χ = 0, 448
schemat 4.2: χ = 0, 610
schemat 4.3: χ = 0, 594
schemat 5: χ = 0, 745
Moment podporowy a
$$M_{a}^{x} = 0,5\left( M_{4,1}^{x} + M_{3}^{x} \right) = - 0,5q\left( \frac{\chi_{4,1}}{8}l_{x,4,1}^{2} + \frac{\chi_{3}}{12}l_{x,3}^{2} \right) = - 0,5*12,56\left( \frac{0,448}{8}{6,5}^{2} + \frac{0,813}{12}{6,4}^{2} \right) = - 32,29\ \frac{\text{kNm}}{m}\ $$
Moment podporowy b
$$M_{b}^{x} = 0,5\left( M_{3}^{x} + M_{5}^{x} \right) = - 0,5q\left( \frac{\chi_{3}}{12}l_{x,3}^{2} + \frac{\chi_{5}}{12}l_{x,5}^{2} \right) = - 0,5*12,56\left( \frac{0,813}{12}{6,4}^{2} + \frac{0,745}{12}{5,7}^{2} \right)\ \ = - 30,09\ \frac{\text{kNm}}{m}\ $$
Moment podporowy c
$$M_{c}^{x} = 0,5\left( M_{4,2}^{x} + M_{4,2}^{x} \right) = - 0,5q\left( \frac{\chi_{4,2}}{8}l_{x,4,2}^{2} + \frac{\chi_{4,2}}{8}l_{x,4,2}^{2} \right) = - 0,5*12,56\left( \frac{0,61}{8}{5,7}^{2} + \frac{0,61}{8}{5,7}^{2} \right)\ \ = - 31,12\ \frac{\text{kNm}}{m}$$
Moment podporowy d
$$M_{d}^{x} = 0,5\left( M_{5}^{x} + M_{4,3}^{x} \right) = - 0,5q\left( \frac{\chi_{5}}{12}l_{x,5}^{2} + \frac{\chi_{4,3}}{8}l_{x,4,3}^{2} \right) = - 0,5*12,56\left( \frac{0,745}{12}{5,7}^{2} + \frac{0,594}{12}{5,7}^{2} \right)\ \ = - 27,82\ \frac{\text{kNm}}{m}$$
Moment podporowy e
$$M_{e}^{y} = 0,5\left( M_{4,1}^{y} + M_{2'}^{y} \right) = - 0,5q\left( \frac{{1 - \chi}_{4,1}}{8}l_{y,4,1}^{2} + \frac{\chi_{2'}}{8}l_{y,2'}^{2} \right) = - 0,5*12,56\left( \frac{1 - 0,448}{8}{6,2}^{2} + \frac{0,729}{8}{6,4}^{2} \right)\ \ = - 40,10\ \frac{\text{kNm}}{m}$$
Moment podporowy f
$$M_{f}^{y} = 0,5\left( M_{5}^{y} + M_{4,2}^{y} \right) = - 0,5q\left( \frac{{1 - \chi}_{5}}{8}l_{y,5}^{2} + \frac{{1 - \chi}_{4,2}}{8}l_{y,4,2}^{2} \right) = - 0,5*12,56\left( \frac{1 - 0,745}{8}{6,2}^{2} + \frac{1 - 0,61}{8}{6,4}^{2} \right)\ \ = - 20,23\ \frac{\text{kNm}}{m}$$
Moment podporowy g
$$M_{g}^{y} = 0,5\left( M_{4,3}^{y} + M_{4,2}^{y} \right) = - 0,5q\left( \frac{{1 - \chi}_{4,3}}{8}l_{y,4,3}^{2} + \frac{{1 - \chi}_{4,2}}{8}l_{y,4,2}^{2} \right) = - 0,5*12,56\left( \frac{1 - 0,594}{8}{6,2}^{2} + \frac{1 - 0,61}{8}{6,4}^{2} \right)\ \ = - 24,79\ \frac{\text{kNm}}{m}$$
Momenty krawędziowe (podporowe na krawędziach)
współczynniki χ: patrz pkt. 3.2
b=0,25m - szerokość żebra w obu kierunkach
qx = χ * q
qy = (1 − χ)*q
Uwaga: Obliczono tylko większe wartości momentów (mniejsza wartość χ )
$${q'}_{x4,1} = \chi_{4,1}q = 0,448*12,56 = 5,63\ \frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{a}\rbrack = M_{a} + 0,25{q'}_{x4,1}l_{x}b = - 32,29 + 0,25*5,63*6,4*0,25 = - 30,04\ \frac{\text{kNm}}{m}$$
$${q'}_{x5} = \chi_{5}q = 0,745*12,56 = 9,36\ \frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{b}\rbrack = M_{b} + 0,25{q'}_{x5}l_{x}b = - 30,09 + 0,25*9,36*5,7*0,25 = - 26,76\ \frac{\text{kNm}}{m}$$
$${q'}_{x4,2} = \chi_{4,2}q = 0,61*12,56 = 7,66\ \frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{c}\rbrack = M_{c} + 0,25{q'}_{x4,2}l_{x}b = - 31,12 + 0,25*7,66*5,7*0,25 = - 27,36\ \frac{\text{kNm}}{m}$$
$${q'}_{x4,3} = \chi_{4,3}q = 0,594*12,56 = 7,46\frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{d}\rbrack = M_{d} + 0,25{q'}_{x4,2}l_{x}b = - 27,82 + 0,25*7,46*5,7*0,25 = - 25,16\ \frac{\text{kNm}}{m}$$
$${q'}_{y4,1} = \chi_{4,1}q = 0,448*12,56 = 5,63\frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{e}\rbrack = M_{e} + 0,25{q'}_{x4,1}l_{y}b = - 40,10 + 0,25*5,63*6,2*0,25 = - 37,92\ \frac{\text{kNm}}{m}$$
$${q'}_{y4,2} = \chi_{4,2}q = 0,61*12,56 = 7,66\frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{f}\rbrack = M_{f} + 0,25{q'}_{x4,2}l_{y}b = - 20,23 + 0,25*7,66*6,4*0,25 = - 17,17\ \frac{\text{kNm}}{m}$$
$${q'}_{y4,3} = \chi_{4,3}q = 0,594*12,56 = 7,46\frac{\text{kN}}{m^{2}}$$
$${\lbrack M}_{g}\rbrack = M_{g} + 0,25{q'}_{x4,3}l_{y}b = - 24,79 + 0,25*7,46*6,2*0,25 = - 21,90\frac{\text{kNm}}{m}$$
Momenty częściowego zamocowania na krawędziach zewnętrznych
indeksy momentów zgodnie ze schematem z punktu 1.
do obliczeń wykorzystano tylko maksymalne wartości momentów przęsłowych
$$M_{1} = M_{13} = - 0,15*M_{2',x} = - 0,15*14,52 = - 2,178\ \frac{\text{kNm}}{m}$$
$$M_{2} = - 0,15*M_{4,x,1} = - 0,15*13,36 = - 2,004\ \frac{\text{kNm}}{m}$$
$$M_{3} = - 0,15*M_{4,y,1} = - 0,15*14,99 = - 2,249\ \frac{\text{kNm}}{m}$$
$$M_{4} = M_{12} = - 0,15*M_{3,y} = - 0,15*10,66 = - 1,599\ \frac{\text{kNm}}{m}$$
$$M_{5} = - 0,15*M_{5,y} = - 0,15*8,19 = - 1,229\ \frac{\text{kNm}}{m}$$
$$M_{6} = - 0,15*M_{4,y,3} = - 0,15*11,06 = - 1,659\ \frac{\text{kNm}}{m}$$
$$M_{7} = - 0,15*M_{4,x,3} = - 0,15*13,69 = - 2,054\ \frac{\text{kNm}}{m}$$
$$M_{8} = M_{11} = - 0,15*M_{4,x,2} = - 0,15*14,11 = - 2,117\ \frac{\text{kNm}}{m}$$
$$M_{9} = M_{10} = - 0,15*M_{4,y,2} = - 0,15*11,36 = - 1,704\ \frac{\text{kNm}}{m}$$
$$M_{14} = - 0,15*M_{2',y} = - 0,15*17,74 = - 2,661\frac{\text{kNm}}{m}$$
Wymiarowanie zbrojenia
Dane materiałowe
Beton C30/37
fck = 30 MPa
$$f_{\text{cd}} = \frac{30\ \text{MPa}}{1.5} = 20\ MPa$$
fctm = 2, 9 MPa
Ecm = 32 GPa
Stal A-IIIN RB500W:
fyk = 500 MPa
$$f_{\text{yd}} = \frac{500}{1,15} = 434,783\ \text{MPa}$$
Es = 200 GPa
$$\xi_{\text{eff},\lim} = 0,8\frac{\varepsilon_{cu3}}{\varepsilon_{cu3} + \frac{f_{\text{yd}}}{E_{s}}} = 0,8\frac{0,0035}{0,0035 + \frac{434,783}{200000}} = 0,493$$
Otulenie zbrojenia
cmin = max{cmin, b; cmin, dur + cdur; γ − cdur, st − cdur, add, 10mm}
cmin, b – min. otulenie ze względu na przyczepność; ⌀ pręta zbrojeniowego, wstępnie przyjmuję pręty zbrojeniowe 10mm
cmin, b = 10 mm
cmin, dur = 10 mm
cdur; γ = 0
cdur, st = 0
cdur, add = 0
cdev = 10 mm
cmin = max{ ⌀=10mm;10+0−0−0, 10mm} = 10mm
cnom = cmin + cdev
cnom = 10 + 10 = 20 mm
Wysokość użyteczna przekroju
dx = hf − c − 0, 5 ⌀=20 − 2 − 0, 5 = 17, 5cm
dy = dx − ⌀ = 17, 5 − 1 = 16, 5cm
wys. użyteczna dla max momentów podporowych na żebrze:
$${d'}_{x} = d_{x} + \frac{b_{z}}{6} = 17,5 + \frac{25}{6} = 21,667cm$$
$${d'}_{y} = d_{y} + \frac{b_{z}}{6} = 16,5 + \frac{25}{6} = 20,667cm$$
Minimalny przekrój zbrojenia podłużnego na pasmo 1m
$A_{s1x,min} = max\left\{ \begin{matrix} 0,26 \times \frac{f_{\text{ctm}}}{f_{\text{yk}}} \times b \times d_{x} \\ 0,0013 \times b \times d_{x} \\ \end{matrix} \right.\ = max\left\{ \begin{matrix} 0,26 \times \frac{2,9}{500} \times 100 \times 17,5 = 2,64\ cm^{2} \\ 0,0013 \times 100 \times 17,5 = 2,28\ cm^{2} \\ \end{matrix} = 2,64\ cm^{2} \right.\ $
$$A_{s1y,min} = max\left\{ \begin{matrix}
0,26 \times \frac{f_{\text{ctm}}}{f_{\text{yk}}} \times b \times d_{y} \\
0,0013 \times b \times d_{y} \\
\end{matrix} \right.\ = max\left\{ \begin{matrix}
0,26 \times \frac{2,9}{500} \times 100 \times 16,5 = 2,49\ cm^{2} \\
0,0013 \times 100 \times 16,5 = 2,15\ cm^{2} \\
\end{matrix} = 2,49\ cm^{2} \right.\ $$
$$A_{s1x',min} = max\left\{ \begin{matrix}
0,26 \times \frac{f_{\text{ctm}}}{f_{\text{yk}}} \times b \times {d'}_{x} \\
0,0013 \times b \times {d'}_{x} \\
\end{matrix} \right.\ = max\left\{ \begin{matrix}
0,26 \times \frac{2,9}{500} \times 100 \times 21,667 = 3,27cm^{2} \\
0,0013 \times 100 \times 21,667 = 2,82cm^{2} \\
\end{matrix} = 3,27\ cm^{2} \right.\ $$
$$A_{s1x',min} = max\left\{ \begin{matrix}
0,26 \times \frac{f_{\text{ctm}}}{f_{\text{yk}}} \times b \times {d'}_{y} \\
0,0013 \times b \times {d'}_{y} \\
\end{matrix} \right.\ = max\left\{ \begin{matrix}
0,26 \times \frac{2,9}{500} \times 100 \times 20,667 = 3,12\ cm^{2} \\
0,0013 \times 100 \times 20,667 = 2,69\ cm^{2} \\
\end{matrix} = 3,12\ cm^{2} \right.\ $$
Rozstaw maksymalny
Zbrojenie główne: Smax = 2 * hf = 40cm ≤ 25 cm
- momenty przęsłowe, podporowe
Zbrojenie drugorzędne: Smax = 3 * hf = 60cm ≤ 40 cm
- momenty uśrednione, zbrojenie rozdzielcze
Wyznaczenie ilości potrzebnego zbrojenia
Przykładowe obliczenia – moment przęsłowy na kierunku x w płycie 2’:
$$\mu_{\text{eff}} = \frac{\left| M_{\text{ED}} \right|}{b \times d^{2} \times f_{\text{cd}}} = \frac{14,52\ \lbrack\frac{\text{kNm}}{m}\rbrack}{1,0\lbrack m\rbrack*0,175^{2}\lbrack m\rbrack*20*10^{3}\lbrack kPa\rbrack} = 0,024\ \lbrack - \rbrack$$
$$\xi_{\text{eff}} = 1 - \sqrt{1 - 2 \times \mu_{\text{eff}}} = 1 - \sqrt{1 - 2 \times 0,024} = 0,024\ \leq \ \xi_{eff,lim} = 0,493\ \lbrack - \rbrack$$
$$A_{s1} = \xi_{\text{eff}} \times b \times d \times \frac{f_{\text{cd}}}{f_{\text{yd}}} = 0,024*100*17,5*\frac{20}{434,783} = 1,93\ \text{cm}^{2}\ \geq A_{s1x,min} = \ 2,64\ cm^{2}$$
As1 = 2, 64 cm2, przyjęto pręty 8mm w rozstawie co 12,5cm/m
$$\text{\ A}_{s1x,prov} = \frac{b}{s}*\varnothing*\frac{\pi}{4} = \frac{100}{12,5}*{0,8}^{2}*\frac{\pi}{4} = 4,02\text{cm}^{2}$$
$$\rho_{L} = \frac{A_{s1,prov}}{b*d}*100\% = \frac{2,65\ }{100*0,175}*100\% = 0,23\ \%$$
Zestawienie zbrojenia
Zbrojenie teoretyczne | Zbrojenie przyjęte | |
---|---|---|
Msd [kNm/m] | d[cm] | meff [-] |
M2'x | 14,52 | 17,5 |
M2'y | 17,74 | 16,5 |
M3x | 13,73 | 17,5 |
M3y | 10,66 | 16,5 |
M4,1x | 13,36 | 17,5 |
M4,1y | 14,99 | 16,5 |
M4,2x | 14,11 | 17,5 |
M4,2y | 11,36 | 16,5 |
M4,3x | 13,69 | 17,5 |
M4,3y | 11,06 | 16,5 |
M5x | 11,34 | 17,5 |
M5y | 8,19 | 16,5 |
Ma | 32,29 | 21,7 |
[Ma] | 30,04 | 17,5 |
Mb | 30,09 | 21,7 |
[Mb] | 26,76 | 17,5 |
Mc | 31,12 | 21,7 |
[Mc] | 27,36 | 17,5 |
Md | 27,82 | 21,7 |
[Md] | 25,16 | 17,5 |
Me | 40,10 | 20,7 |
[Me] | 37,92 | 16,5 |
Mf | 20,23 | 20,7 |
[Mf] | 17,17 | 16,5 |
Mg | 24,79 | 20,7 |
[Mg] | 21,90 | 16,5 |
M1=M13 | 2,18 | 17,5 |
M2 | 2,00 | 17,5 |
M3 | 2,25 | 16,5 |
M4=M12 | 1,60 | 16,5 |
M5 | 1,23 | 16,5 |
M6 | 1,66 | 16,5 |
M7 | 2,05 | 17,5 |
M8=M11 | 2,12 | 17,5 |
M9 | 1,70 | 16,5 |
M10 | 1,70 | 16,5 |
M14 | 2,661 | 16,5 |