Can someone assist me in aligning the "=" in the following expressions. In each expression there are two "=" signs. The second row is not aligned. Any suggestions?
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\documentclass[12pt,a4paper]{article}\usepackage[latin1]{inputenc}\usepackage[danish]{babel} % danske overskrifter\usepackage[T1]{fontenc}\usepackage{lmodern}\usepackage{graphicx}\usepackage{mathtools}\usepackage{amsmath}\usepackage{amsfonts}\usepackage{ulem}\usepackage{textcomp}\usepackage{tikz}\usepackage{color}\usepackage{arcs}\usepackage{amsthm}\usepackage[hang,small,bf]{caption}\setlength{\abovecaptionskip}{5pt}\setlength{\captionmargin}{65pt}\○begin{document}\noindent Vindlasten, $\theta=0$, på zonerne $A$, $B$, $C$, $D$ og $E$ fås da til:\begin{alignat*}{2}w_{e,A_{0}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-1,2) \cdot \tfrac{20\text{m}}{5}&=-2,88\tfrac{\text{kN}}{\text{m}} \\w_{e,B_{0}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-0,8) \cdot \tfrac{4\cdot 20\text{m}}{5}&=-7,68\tfrac{\text{kN}}{\text{m}} \\w_{e,C_{0}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-0,5) \cdot (28,8 \text{m}-20\text{m})&=-2,64\tfrac{\text{kN}}{\text{m}} \\w_{e,D_{0}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot 0,71 \cdot 72\text{m}&=30,67\tfrac{\text{kN}}{\text{m}} \\w_{e,E_{0}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-0,33) \cdot 72\text{m}&=-14,26\tfrac{\text{kN}}{\text{m}} \\\end{alignat*}\noindent Formfaktorerne for tilfældet, hvor vinden har en vinkel $\theta=90^{\circ}$ er de samme som for til tilfældet, hvor $\theta=0$. For situationen, hvor vindlasten virker med vinklen $\theta=90^{\circ}$ bliver vindlasten på zonerne $A$, $B$, $C$, $D$ og $E$:\begin{alignat*}{2}w_{e,A_{90}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-1,2) \cdot \tfrac{20\text{m}}{5}&=-2,88\tfrac{\text{kN}}{\text{m}} \\w_{e,B_{90}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-0,8) \cdot \tfrac{4\cdot 20\text{m}}{5}&=-7,68\tfrac{\text{kN}}{\text{m}} \\w_{e,C_{90}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-0,5) \cdot (72 \text{m}-20\text{m})&=-15,6\tfrac{\text{kN}}{\text{m}} \\w_{e,D_{90}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot 0,71 \cdot 28,8\text{m}&=12,27\tfrac{\text{kN}}{\text{m}} \\w_{e,E_{90}}&=0,6\tfrac{\text{kN}}{\text{m}^{2}} \cdot (-0,33) \cdot 28,8\text{m}&=-5,70\tfrac{\text{kN}}{\text{m}} \\\end{alignat*}