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For Crushing Glass,
Mini pressure applied in the cylinder (P) = 6.86x105N/m2       
Diameter of the cylinder (D) = 32 mm 
Diameter of the piston rod (d) = 17 mm               
Stroke length = 100 mm      
Area of cylinder (A) = (3.14/4*(D2)
                 = (0.785x.0322)
               A= 8.0384 x 10-4 m2
Force during forward stroke,
F = {π/4x D^2 x P}
F = (6.86 x 105 N /m2) (3.14/4*(0.0322)
F = 551.43 N 
       1kg = 9.81 N 
    Load capacity during forward stroke W= F / a
                                      W =    551.43 / 9.81
                                      W =     56.21 KG  
Force during return stroke,
F = {π/4x (D^2 – d^2) x P}
F = {π/4x (0.032^2 – 0.017^2) x 6.86x105}
 F = 395.99 N
Load capacity during return stroke W= F / a
                                      W =    395.99 / 9.81
                                      W =     40.36 KG 
For  Plastic bottles,
Mini pressure applied in the cylinder (P) = 5.39x105N/m2       
Diameter of the cylinder (D) = 32 mm 
Diameter of the piston rod (d) = 17 mm               
Stroke length = 100 mm      
Area of cylinder (A) = (3.14/4*(D2)
                 = (0.785x.0322)
               A= 8.0384 x 10-4 m2
Force during forward stroke,
F = {π/4x D^2 x P}
F = (5.39 x 105 N /m2) (3.14/4*(0.0322)
F = 433.26 N 
       1kg = 9.81 N 
    Load capacity during forward stroke W= F / a
                                      W =    433.26 / 9.81
                                      W =     44.16 KG  
Force during return stroke,
F = {π/4x (D^2 – d^2) x P}
F = {π/4x (0.032^2 – 0.017^2) x 5.39x105}
 F = 311.13 N
Load capacity during return stroke W= F / a
                                      W =    311.13/ 9.81
                                      W =     31.71 KG  

For Crushing aluminium cans,
Mini pressure applied in the cylinder (P) = 4.41x105N/m2       
Diameter of the cylinder (D) = 32 mm 
Diameter of the piston rod (d) = 17 mm               
Stroke length = 100 mm      
Area of cylinder (A) = (3.14/4*(D2)
                 = (0.785x.0322)
               A= 8.0384 x 10-4 m2
Force during forward stroke,
F = {π/4x D^2 x P}
F = (4.41 x 105 N /m2) (3.14/4*(0.0322)
F = 354.49 N 
       1kg = 9.81 N 
    Load capacity during forward stroke W= F / a
                                      W =    354.49 / 9.81
                                      W =     36.135 KG  
Force during return stroke,
F = {π/4x (D^2 – d^2) x P}
F = {π/4x (0.032^2 – 0.017^2) x 4.41x105}
 F = 254.56 N
Load capacity during return stroke W= F / a
                                      W =    254.56 / 9.81
                                      W =     25.94 KG  
2
  • 1
    what is your question? Apr 11, 2016 at 20:07
  • \raggedright or the verbatim environment are options you might consider. The flalign environment from amsmath could prove useful as well. Apr 11, 2016 at 20:33

1 Answer 1

2

I think you should set up a two-column table, with the textual descriptions on the left and the formulas and numerical examples on the right. Do also take care to typeset the scientific units correctly.

The first few rows might look as follows:

enter image description here

\documentclass{report}
\usepackage{amsmath,longtable,booktabs,array,siunitx}
\sisetup{per-mode=symbol}
\begin{document}

\begin{longtable}{@{}l >{$}l<{$} @{}}
\toprule
\endhead
\bottomrule
\endfoot
For Crushing Glass,\\
Mini pressure applied in the cylinder & P = 6.86 \cdot 105\, \si{\newton\per\meter\squared} \\
Diameter of the cylinder &D = \SI{32}{\milli\meter}\\ 
Diameter of the piston rod & d = \SI{17}{\milli\meter}\\               
Stroke length & \SI{100}{\milli\meter}   \\    
Area of cylinder & 
\!\begin{aligned}[t] 
A &= \pi/4\cdot D^2\\
  &= 0.785\times .0322\\
  &= \SI{8.0384e-4}{\meter\squared}
\end{aligned} \\
Force during forward stroke &
\!\begin{aligned}[t]
F &= \pi /4\cdot D^2 \cdot P\\
  &= 6.86 \cdot 105\,\si{\newton\per\meter\squared} \quad(3.14/4\cdot 0.0322)\\
 &= \SI{551.43}{\newton} \quad (\SI{1}{\kilo\gram} =\SI{9.81}{\newton})
\end{aligned} \\ 
Load capacity during forward stroke & W=\dots \\
\end{longtable}
\end{document} 

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