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diff --git a/sample_notebooks/bharthkumar/bharthkumar_version_backup/Untitled1.ipynb b/sample_notebooks/bharthkumar/bharthkumar_version_backup/Untitled1.ipynb new file mode 100755 index 00000000..c2fe40ea --- /dev/null +++ b/sample_notebooks/bharthkumar/bharthkumar_version_backup/Untitled1.ipynb @@ -0,0 +1,185 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2206f2855e4232dc4600c4e414262a1e7f06df22f4a8f22ba905ba92f7813175"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Introduction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##calculate the\n",
+ "## initialization of variables\n",
+ "import math\n",
+ "## part (a)\n",
+ "a=700. ## M Pa from figure 1.8\n",
+ "b=100. ## M Pafrom figure 1.8\n",
+ "m=1/6. ## from figure 1.8\n",
+ "Y=450. ## M Pa from figure 1.9\n",
+ "##calculations\n",
+ "sigma_u=a+m*b\n",
+ "## results\n",
+ "print('\\n part (a) \\n')\n",
+ "print\"%s %.2f %s\"%(' The ultimate strength is sigma = ',sigma_u,' M Pa')\n",
+ "print\"%s %.2f %s\"%('\\n and the yield strength is Y = ',Y,'M Pa')\n",
+ "\n",
+ "## part (b)\n",
+ "c1=62. ## from figure 1.8\n",
+ "d1=0.025 ## from figure 1.8\n",
+ "c2=27. ## from figure 1.10a\n",
+ "d2=0.04 ## from figure 1.10a\n",
+ "## calculations\n",
+ "U_f1=c1*b*d1*10**6\n",
+ "U_f2=c2*b*d2*10**6\n",
+ "## results\n",
+ "print('\\n part (b)')\n",
+ "print\"%s %.2e %s\"%('\\n The modulus of toughness for alloy steel is Uf = ',U_f1,' N/m^2')\n",
+ "print\"%s %.2e %s\"%('\\n and structural steel is Uf = ',U_f2,' N/m^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " part (a) \n",
+ "\n",
+ " The ultimate strength is sigma = 716.67 M Pa\n",
+ "\n",
+ " and the yield strength is Y = 450.00 M Pa\n",
+ "\n",
+ " part (b)\n",
+ "\n",
+ " The modulus of toughness for alloy steel is Uf = 1.55e+08 N/m^2\n",
+ "\n",
+ " and structural steel is Uf = 1.08e+08 N/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##calculate the permanet strain\n",
+ "## initialization of variables\n",
+ "import math\n",
+ "sigma=500. ## Stress M Pa\n",
+ "eps=0.0073 ## Strain\n",
+ "sigma_A=343. ## M Pa from figure 1.9\n",
+ "eps_A=0.00172 ## from figure 1.9\n",
+ "## part (a)\n",
+ "E=sigma_A/eps_A\n",
+ "\n",
+ "## part (B)\n",
+ "eps_e=sigma/E\n",
+ "eps_p=eps-eps_e\n",
+ "## results\n",
+ "print(' part (a) \\n')\n",
+ "print\"%s %.2f %s\"%(' The modulus of elasticity of the rod is E = ',E/1000,' G Pa')\n",
+ "print('\\n part (b)')\n",
+ "print\"%s %.4f %s\"%('\\n the permanent strain is = ',eps_p,'')\n",
+ "print\"%s %.4f %s\"%('\\n and the strain recovered is =',eps_e,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " part (a) \n",
+ "\n",
+ " The modulus of elasticity of the rod is E = 199.42 G Pa\n",
+ "\n",
+ " part (b)\n",
+ "\n",
+ " the permanent strain is = 0.0048 \n",
+ "\n",
+ " and the strain recovered is = 0.0025 \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##calculate the diameter\n",
+ "## initialization of variables\n",
+ "import math\n",
+ "D=25. ## kN\n",
+ "L=60. ## kN\n",
+ "W=30. ##kN\n",
+ "Y=250. ## M Pa\n",
+ "safety=5./3. ## AISC, 1989\n",
+ "## calculations\n",
+ "Q=(D+L+W)*10**3. ## converted to N\n",
+ "A=safety*Q/Y\n",
+ "r=math.sqrt(A/math.pi)+0.5 ## additional 0.5 mm is for extra safety\n",
+ "d=1.8*r ## diameter\n",
+ "## results\n",
+ "print('Part (a) \\n ')\n",
+ "print\"%s %.2f %s %.2f %s \"%('A rod of ',d,' mm'and ' in diameter, with a cross sectional area of ',math.pi*(d**2./4.),' mm^2, is adequate')\n",
+ "## The diameter is correct as given in the textbook. Area doesn't match due to rounding off error and partly because it's a design problem.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part (a) \n",
+ " \n",
+ "A rod of 29.02 in diameter, with a cross sectional area of 661.39 mm^2, is adequate \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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