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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 8 : Mechanical Testing"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 8.1 pageno : 195"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "b = 225.;\t\t\t#in mm\n",
      "h = 10. \t\t\t#in mm\n",
      "l = 1100.;\t\t\t#in mm\n",
      "f1 = 250.;\t\t\t#in N\n",
      "f2 = 350;\t\t\t#in N at which glass breaks\n",
      "\n",
      "# Calculations\n",
      "m = f1*l/4.;\t\t\t#in N-mm\n",
      "f = f1/2.;  \t\t\t#in N\n",
      "a = (6*m)/(b*h**2);\t\t\t#in N/mm**2\n",
      "t = (3*f)/(2*b*h);\t\t\t#in N/sqmm\n",
      "r = f2*l/4;\t\t\t        #in N-mm\n",
      "i = (b*h**3)/12;\t\t\t#in mm**4\n",
      "y = h/2;\t        \t\t#in mm\n",
      "mr = r*y/i;\t\t        \t#in n/sqmm\n",
      "\n",
      "# Results\n",
      "print \"Flexural Strength (in N/sqmm)  =  %.2f N/mm**2\"%a\n",
      "print \"Shear Strength (in N/sqmm)  =  %3f N/mm**2\"%t\n",
      "print \"Modulous of Rupture (in N/sqmm)  =  %.2f N/mm**2\"%mr\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Flexural Strength (in N/sqmm)  =  18.33 N/mm**2\n",
        "Shear Strength (in N/sqmm)  =  0.083333 N/mm**2\n",
        "Modulous of Rupture (in N/sqmm)  =  25.67 N/mm**2\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 8.2 pageno : 201"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "d = 5.;     \t\t\t#in mm\n",
      "\n",
      "# Calculations\n",
      "id = 32.5/10;\t\t\t#indentation diameter in mm\n",
      "p = 30*d**2;\t\t\t#load for steel specimen in kgf\n",
      "bhn = p/((3.14*d/2)*(d-math.sqrt(d**2-id**2)));\t\t\t#in kgf/sqmm\n",
      "\n",
      "# Results\n",
      "print \"Load P for steel specimen (in kgf)  =  %.f kgf\"%p\n",
      "print \"BRINELL HARDNESS NUMBER of the steel specimen  =  %.1f\"%bhn\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Load P for steel specimen (in kgf)  =  750 kgf\n",
        "BRINELL HARDNESS NUMBER of the steel specimen  =  79.6\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 8.3 pageno : 209"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "l = 0.1;\t\t\t#frictinal and windage losses in kgf-m\n",
      "dr = 5.9;\t\t\t#dial reading in kgf-m\n",
      "w = 19.33;\t\t\t#weight of hammer in kgf-m\n",
      "t = 10.;\t\t\t#in mm\n",
      "ui = 30.;\t\t\t#in kgf-m\n",
      "a = 160.;\t\t\t#angle in degrees\n",
      "r = 0.8;\t\t\t#swing radius in m\n",
      "\n",
      "\n",
      "# Calculations\n",
      "u = dr-l;\t        \t\t#in kgf-m\n",
      "d = t/5;\t\t        \t#depth of V-notch in mm\n",
      "te = t-d;\t\t    \t    #effective thickness in mm\n",
      "ve = 75.*10*te; \t\t\t#effective volume in cu. mm\n",
      "vem = ve*10.**-9;\t\t\t#in cu. m\n",
      "mr = u/vem;\t        \t\t#in kgf/sqm\n",
      "ae = t*te;      \t\t\t#effective area of cross section in sqmm\n",
      "aem = ae*10**-6;\t\t\t#in sqm\n",
      "is_ = u/aem;\t\t        \t#in kg/m\n",
      "uf = ui-u;\t\t\t#in kgf-m\n",
      "hf = uf/w;\t\t\t#in m\n",
      "B = math.degrees(math.acos(1-(uf/(w*r))))\n",
      "\n",
      "# Results\n",
      "print \"Rupture Energy (in kgf-m)  =  %.1f kgf-m\"%u\n",
      "print \"Modulous Of Rupture (in kgf/sqm)  =  %.1e kgf/m**2\"%mr\n",
      "print \"Notch Imapct Strength (in kg/m)  =  %.2e kgm\"%is_\n",
      "print \"Height risen by Hammer (in m)  =  %.2f m\"%hf\n",
      "print \"Angle after Breaking the specimen (in degress)  =  %.1f degrees\"%(B)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rupture Energy (in kgf-m)  =  5.8 kgf-m\n",
        "Modulous Of Rupture (in kgf/sqm)  =  9.7e+05 kgf/m**2\n",
        "Notch Imapct Strength (in kg/m)  =  7.25e+04 kgm\n",
        "Height risen by Hammer (in m)  =  1.25 m\n",
        "Angle after Breaking the specimen (in degress)  =  124.4 degrees\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 8.4 pageno : 211"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "a_m = 70.;  \t    \t\t#mean stress in Mpa\n",
      "a_r = 210.;\t    \t    \t#stress amplitude in Mpa\n",
      "\n",
      "# Calculations\n",
      "a_max = ((2*a_m)+a_r)/2;\t\t\t#maximum stress in MPa\n",
      "a_min = 2*a_m-a_max;\t    \t\t#Minimum stress in MPa\n",
      "s = a_min/a_max;\t\t\t        #stress ratio\n",
      "sr = a_max-a_min;       \t\t\t#stress range in MPa\n",
      "\n",
      "# Results\n",
      "print \"Maximum Stress Level (in MPa)  =  \",a_max\n",
      "print \"Minimum Stress Level (in MPa)  =  \",a_min\n",
      "print \"Stress Ratio  =  \",s\n",
      "print \"Stress Range (in MPa)  =  \",sr\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum Stress Level (in MPa)  =   175.0\n",
        "Minimum Stress Level (in MPa)  =   -35.0\n",
        "Stress Ratio  =   -0.2\n",
        "Stress Range (in MPa)  =   210.0\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 8.5 pageno : 212"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "p_min = 20.;\t\t\t#in kN\n",
      "p_max = 50.;\t\t\t#in kN\n",
      "l = 500.;   \t\t\t#in mm\n",
      "d = 60.;\t    \t\t#in mm\n",
      "a_u = 650.;\t\t    \t#in MPa\n",
      "a_y = 520.;\t\t    #in MPa\n",
      "fos = 1.8;\t\t\t    #factor of safety\n",
      "\n",
      "# Calculations\n",
      "m_max = p_max*l/4;\t\t\t#maximum bending moment in kN mm\n",
      "m_min = p_min*l/4;\t\t\t#minimum bending moment in kN mm\n",
      "m_m = (m_max+m_min)/2;\t\t\t#mean bending moment in kN mm\n",
      "m_a = (m_max-m_min)/2;\t\t\t#alternating bending moment in kN mm\n",
      "z = 3.14*d**3/32;\n",
      "a_m = (m_m/z)*1000;\t\t\t#mean bending stress in MPa\n",
      "a_a = (m_a/z)*1000;\t\t\t#alternating bending stress in MPa\n",
      "a_e1 = a_a/((1/fos)-(a_m/a_u)**2*fos);\t\t\t#in MPa\n",
      "a_e2 = a_a/((1/fos)-(a_m/a_u));\t\t\t#in MPa\n",
      "a_e3 = a_a/((1/fos)-(a_m/a_y));\t\t\t#in MPa\n",
      "\n",
      "# Results\n",
      "print \"ENDURANCE STRESS FROM Gerbers Parabolic Function (in MPa)  =  %.2f MPa\"%a_e1\n",
      "print \"ENDURANCE STRESS FROM Goodman Straight Line Relation (in MPa)  =  %.2f MPa\"%a_e2\n",
      "print \"ENDURANCE STRESS FROM Soderberg Straight Line Relation (in MPa)  =  %.2f MPa\"%a_e3\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "ENDURANCE STRESS FROM Gerbers Parabolic Function (in MPa)  =  236.52 MPa\n",
        "ENDURANCE STRESS FROM Goodman Straight Line Relation (in MPa)  =  371.71 MPa\n",
        "ENDURANCE STRESS FROM Soderberg Straight Line Relation (in MPa)  =  557.78 MPa\n"
       ]
      }
     ],
     "prompt_number": 11
    }
   ],
   "metadata": {}
  }
 ]
}