From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- sample_notebooks/bharthkumar/Untitled1.ipynb | 185 --------------------- .../bharthkumar_version_backup/Untitled1.ipynb | 185 +++++++++++++++++++++ 2 files changed, 185 insertions(+), 185 deletions(-) delete mode 100755 sample_notebooks/bharthkumar/Untitled1.ipynb create mode 100755 sample_notebooks/bharthkumar/bharthkumar_version_backup/Untitled1.ipynb (limited to 'sample_notebooks/bharthkumar') diff --git a/sample_notebooks/bharthkumar/Untitled1.ipynb b/sample_notebooks/bharthkumar/Untitled1.ipynb deleted file mode 100755 index c2fe40ea..00000000 --- a/sample_notebooks/bharthkumar/Untitled1.ipynb +++ /dev/null @@ -1,185 +0,0 @@ -{ - "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": {} - } - ] -} \ No newline at end of file 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": {} + } + ] +} \ No newline at end of file -- cgit