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# Graph ploting

Posted 9 years ago
 Hi, I need help with plotting a graph. I have expression and i need to plot a graph, but I do not know how to do it. My expression is: w = 3.36429*10^-9 (-1500 + x) + 0.012242 E^(-0.0428469 x) Cos[0.0428469 x] (0.000818033 + 0.308311 Cos[\[CurlyPhi]]) + (-0.844704 Cos[\[CurlyPhi]] - 1.40784*10^-6 (-1500 + x)^2 Cos[\[CurlyPhi]])/2100 - 0.000524531 E^(-0.0428469 x) (0.00947115 + 3.55308 Cos[\[CurlyPhi]]) (Cos[0.0428469 x] - Sin[0.0428469 x])  This is function of displacement over a cylindrical shell. This probably should be a 3D graph, and i do not know how to do it. Please help.
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Posted 9 years ago
 I add to attachment my code till now. In .nb file you will see picture that is similar to my problem, load distribution under a). Now I will explain problem that I need to solve: long cylindrical shell is loaded with self weight and wind load (similar to picture, except there is no fluid inside) I have to solve displacement (w) and twist (?) along the height of cylinder (x coordinate). Wind load distribution is equal p(?)=-p0cos(?) ?[0,2Pi]. This is why i 2D graph does not show all needed data, for example for x=0 ? changes from [0,2Pi]. I would like to have this solved in Cartesian coordinate system. And if you can show me how to function p(?)=-p0cos(?) i can transform in cartesian coordinate system so i have this load depended on y,z coordinates (like in picture under a)). Thanks Attachments:
Posted 9 years ago
 Hello Benjamin,It would be better if you posted your code by putting it in a separate paragraph, selecting it and using the Code Sample button on the far left above the panel.There are some syntax errors in your expression. You should not use double brackets on the Cos[[CurlyPhi]] expression and you enter the CurlyPhi by using esc j esc. So that part of the expression should be entered as Cos[esc j esc]. The you might define your function this way: f[x_, \[CurlyPhi]_] = 3.36429*10^-9 (-1500 + x) + 0.012242 E^(-0.0428469 x) Cos[ 0.0428469 x] (0.000818033 + 0.308311 Cos[\[CurlyPhi]]) + (-0.844704 Cos[\[CurlyPhi]] - 1.40784*10^-6 (-1500 + x)^2 Cos[\[CurlyPhi]])/2100 - 0.000524531 E^(-0.0428469 x) (0.00947115 + 3.55308 Cos[\[CurlyPhi]]) (Cos[0.0428469 x] - Sin[0.0428469 x]) Here I scale the function up by 10^6 just to have better coordinates on the plot.One possible plot is: Plot3D[10^6 f[x, \[CurlyPhi]], {x, 0, 1}, {\[CurlyPhi], 0, 2 \[Pi]}, AxesLabel -> {"x", "\[CurlyPhi]", "f"}] Another possible plot is: RevolutionPlot3D[{10^6 f[x, \[CurlyPhi]], x}, {x, 0, 1}, {\[CurlyPhi], 0, 2 \[Pi]}, AxesLabel -> {"y", "z", "x"}] Posted 9 years ago
 Are x and Phi the independent variables? Are you picturing cylindrical coordinates, or would cartesian work? Suggestions for formatting posts are at http://community.wolfram.com/groups/-/m/t/270507 .