Jak na matematiku a zlomky?

[jan.busa at tuke.sk]

Dobry den,

mozete pouzit volbu "fleqn", napr.:

\documentclass[11pt,fleqn]{article}

Vztahy medzi
\[

\]

alebo medzi

\begin{displaymath}

\end{displaymath}

alebo

\begin{equation}

\end{equation}

budu odsadzovane zlava.

Vztahy  medzi $$   $$ ostanu centrovane.

Pekny vecer

   Jano Busa
> Dobry den.
>
> Casto pisi matematicke vzorce. Protoze je chci mit na leve strane, tak
> je pisi pomoci "$" a ne "$$". Problem je v tom ze zlomky se pisi dost
> osklive... Jsou male a s okolnim textem se div neprelinaji. Viz ukazka.
> Mozna to delam uplne spatne.. nevite co s tim?
>
> \documentclass[11pt]{article}
> \usepackage[cp1250]{inputenc}
> \usepackage{czech}
> \usepackage{graphicx}
> \usepackage{fancyhdr}
> \usepackage{a4wide}
> \pagestyle{fancy}
> \lhead{\bfseries HW1}
> \rhead{\bfseries  Oldřich Švec}
> \chead{\bfseries{Hydraulics}}
> \begin{document}
> \noindent
> $
> F_{sk} = 192kN\\
> \gamma_f = 1.2\\
> e = 200 mm\\
> S275 => f_y = 275 MPa\\
> $
> *\\$
> F_{sd} =F_{sk}\times \gamma_f= 192 \times 1.2 = 230.4 kN\\
> M_{sd} =F_{sd}\times L \times \gamma_q= 230.4 \times 0.2 \times 1.5 =
> 69.12 kNm\\
> W_{y,pl} =\frac{M_{sd} \times \gamma_{M0}}{\chi_{LT}\times f_y}\\
> => IPE270\\
> W_{pl.y}=484 \times 10^3\\
> W_y= 428.9 \times 10^3\\
> $*\\
> \noindent
> Classification:\\
> 	\indent
> 	Web:\\
> 		\indent\indent
> 		In bending:\\
> 			\indent\indent\indent
> 			$d/t_w = 219.6/6.6 = 33.27<72 \times 0.92 = 66.24 => Class I\\
> 		$\indent\indent
> 		In compression:\\
> 			\indent\indent\indent
> 			$33.27<38 \times 0.92 = 34.96 => Class II\\$
>                  \indent
> 	Flange:\\
>          	\indent\indent
>          	In compression:\\
>                  	\indent\indent\indent
>                          $(b/2-r-t_w/2)/t_f = 4.824<9 \times 0.92 =
>                          8.28 => Class I\\
> $*\\$
> A_{v,z} = 2214 mm^2\\
> V_{pl,Rd} = \frac{A_{v,z} \times f_y}{\gamma_{M0}\times \sqrt 3} =
> \frac{2214 \times 275}{1150 \times \sqrt{3}} = 305.67 kN\\
> 152.835kN < 230.4kN < 305.67 kN\\
> \rho = ({\frac{2\times F_{sd}}{V_{pl,Rd}}-1})^2 = 0.258\\
> W_W = 1/4 \times t_w \times d^2 = 79569.864\\
> M_{v.cr} = \frac{W_{pl}-\rho \times W_W}{\gamma_{M0}} \times f_y =
> \frac{484\times 10^3-0.258 \times 79569.864}{1.15\times 10^6} \times
> 275 = 110.83 > 69.12 kNm\\
> $*\\Weld design:\\
> $a_w = \frac{F_{sd} \times \beta_{w} \times \gamma_{MW} \times
>    \sqrt{3}}{2\times (h_b - 2 \times t_{fb})\times f_u}
> =\frac{305.67\times 0.85 \times 1.5 \times
>    \sqrt{3}}{2\times (270 - 2 \times 10.2})\times 275 = 4.92 mm =>
> 6mm\\
> $
> \\
> \includegraphics[clip]{1.eps}
> \end{document}
>
>