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@Eric2Steve
Created April 29, 2013 23:07
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A file my professor gave me
{
"metadata": {
"name": "New Concrete Beam"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "code",
"collapsed": false,
"input": [
"import Concrete_Beam as Beam\n",
"from IPython.core.display import Latex"
],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": 1
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"dir(Beam)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "pyout",
"prompt_number": 2,
"text": [
"['A_s_min',\n",
" 'Beta1',\n",
" 'Design_Analysis',\n",
" 'Math',\n",
" 'NA',\n",
" 'Phi',\n",
" 'Required_Width',\n",
" 'Yield_Strain',\n",
" '__builtins__',\n",
" '__doc__',\n",
" '__file__',\n",
" '__name__',\n",
" '__package__',\n",
" 'bar_options',\n",
" 'bar_value',\n",
" 'ceil',\n",
" 'double_Mn',\n",
" 'numpy',\n",
" 'round',\n",
" 'simply_analize']"
]
}
],
"prompt_number": 2
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"f_c = 3000. #psi\n",
"f_y = 60000. #psi\n",
"d = 24. #in\n",
"b = 12. #in\n",
"e_t = 0.009 #in/in\n",
"(output, latex) = Beam.Design_Analysis(f_c, f_y, d, b, e_t)\n",
"Latex(latex)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"As_min was checked.\n",
"\n"
]
},
{
"latex": [
"$$\\beta_1 = 0.85 \\text{ since }f_c \\leq 4,000 \\text{ psi} $$$$\\text{The distance down to the neutral axis } c = \\frac{0.003 d}\n",
" {\\varepsilon_t +0.003} = 6.00\\text{\"}$$ $ $$$\\text{The height of the whitney stress block }a = \\beta_{1} c\n",
" = 5.10\\text{\"}$$$$\\text{The required tension steel area assuming yielding is }\n",
" A_{s} = \\frac{a b (0.85 f_{c})}{f_{y}} =2.60\\text{ in}^2$$$$\\text{Since }3 \\sqrt{f_{c}} \\leq 200 \\text{ then } A_{s, min}\n",
" =\\frac{200 b d}{f_{y}} = 0.96\\text{ in}^{2} $$$$\\text{Note that }A_{s} \\geq A_{s, min}$$$$\\rho = \\frac{A_{s}}{b d}= 0.0090$$ $ $$$Z = d - \\frac{a}{2} =21.45\\text{\"}$$ $ $$$M_{n} = A_{s} f_{y} Z = 279.0\\text{ kip ft}$$ $ $$$\\phi = 0.90 \\text{ since }\\varepsilon_t \\geq 0.005$$$ $$$M_{u} = \\phi M_{n} = 251.1\\text{ kip ft}$$$$\\text{The beam's self weight is not accounted for and is equal to \n",
" 0.300} \\frac{\\text{kip}}{\\text{ft}}$$$$\\bar{k} = \\frac{M_{n}}{b d^{2}} = 0.4843\\text{ ksi}$$"
],
"output_type": "pyout",
"prompt_number": 3,
"text": [
"<IPython.core.display.Latex at 0x2c86350>"
]
}
],
"prompt_number": 3
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
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