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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Ch-10, Mechanical tests and factors affecting mechanical properties"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.1 : page no-298"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "#given\n",
      "#initial gauge length of the specimen\n",
      "l0=50*10**-3   #m\n",
      "#initia l gauge diameter of the specimen\n",
      "d0=12*10**-3  #m\n",
      "#extended gauge length of fracture\n",
      "lf=58*10**-3   #m\n",
      "#reduced gauge diameter\n",
      "df=7*10**-3  #m\n",
      "#initial an final cross sectional areas are\n",
      "A_i=3.14/4*d0**2   #m**2\n",
      "A_f=3.14/4*df**2   #m**2\n",
      "#various applied loads are in k N\n",
      "P1=0\n",
      "P2=5\n",
      "P3=10\n",
      "P4=15\n",
      "P5=20\n",
      "P6=25\n",
      "P7=30\n",
      "P8=32\n",
      "P9=33\n",
      "P10=32\n",
      "P11=31\n",
      "P12=35\n",
      "P13=40\n",
      "P16=130\n",
      "#corresponding to these load we have recorded elongation as\n",
      "delta1=0\n",
      "delta2=0.011\n",
      "delta3=0.022\n",
      "delta4=0.035\n",
      "delta5=0.048\n",
      "delta6=0.059\n",
      "delta7=0.073\n",
      "delta8=0.088\n",
      "delta9=0.100\n",
      "delta10=0.125\n",
      "delta11=0.150\n",
      "delta12=0.230\n",
      "delta13=0.400\n",
      "delta16=8.000\n",
      "#stress and strain corresponding to these loads and elongations are\n",
      "sigma1=P1/A_i\n",
      "strain1=delta1/l0\n",
      "sigma2=P2/A_i\n",
      "strain2=delta2/l0\n",
      "sigma3=P3/A_i\n",
      "strain3=delta3/l0\n",
      "sigma4=P4/A_i\n",
      "strain4=delta4/l0\n",
      "sigma5=P5/A_i\n",
      "strain5=delta5/l0\n",
      "sigma6=P6/A_i\n",
      "strain6=delta6/l0\n",
      "sigma7=P7/A_i\n",
      "strain7=delta7/l0\n",
      "sigma8=P8/A_i\n",
      "strain8=delta8/l0\n",
      "sigma9=P9/A_i\n",
      "strain9=delta9/l0\n",
      "sigma10=P10/A_i\n",
      "strain10=delta10/l0\n",
      "sigma11=P11/A_i\n",
      "strain11=delta11/l0\n",
      "sigma12=P12/A_i\n",
      "strain12=delta12/l0\n",
      "sigma13=P13/A_i\n",
      "strain13=delta13/l0\n",
      "sigma16=P16/A_i\n",
      "strain16=delta16/l0\n",
      "#part(a)\n",
      "#modulus of elasticity\n",
      "E=(sigma4-sigma1)/(strain4-strain1)/10**3   #G Pa\n",
      "#part(b)\n",
      "#ultimate stress (maximum)\n",
      "ultimate_sigma=P16/A_i*10**3/10**6   #M N/m**3\n",
      "#part (c)\n",
      "#upper yield point at C (shown in fig 10.3)\n",
      "u_yield=P9/A_i*1000/10**6   #M N/m**2\n",
      "#lower yield point at D (shown in fig 10.3)\n",
      "l_yield=P11/A_i*1000/10**6   #M N/m**2\n",
      "#part(d)\n",
      "#percentage reduction in area\n",
      "percent_A=(d0**2-df**2)/d0**2*100  #%\n",
      "#part (e)\n",
      "#percentage elongation\n",
      "percent_l=(lf-l0)/l0*100  #%\n",
      "#part(f)\n",
      "#apparent breaking stress\n",
      "app_breaking_stress=P16*1000/A_i/10**6   #M N/m**2\n",
      "#actual breaking stress\n",
      "actual_breaking_stress=P16*1000/A_f/10**6   #M N/m**2\n",
      "print \"\"\"the modulus of elasticity is %0.3f G Pa\n",
      "the ultimate (maximum) stress is %0.3f M n/m**2 \n",
      "upper yield point is %0.3f M N/m**2 \n",
      "lower yield point is %0.3f M n/m**2\n",
      "percentage reduction in area is %0.3f \n",
      "percentage elongation in length %0.3f\n",
      "apparent breaking stress is %0.3f M n/mm**2\n",
      "actual breaking point is %0.3f M n/m**2\"\"\"%(E,ultimate_sigma,u_yield,l_yield,percent_A,percent_l,app_breaking_stress,actual_breaking_stress)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the modulus of elasticity is 189.566 G Pa\n",
        "the ultimate (maximum) stress is 1150.035 M n/m**2 \n",
        "upper yield point is 291.932 M N/m**2 \n",
        "lower yield point is 274.239 M n/m**2\n",
        "percentage reduction in area is 65.972 \n",
        "percentage elongation in length 16.000\n",
        "apparent breaking stress is 1150.035 M n/mm**2\n",
        "actual breaking point is 3379.696 M n/m**2\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.2 : page no-303"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#initial length of the specimen\n",
      "h0=24.02*10**-3   #m\n",
      "#initial gauge diameter of the specimen\n",
      "d0=18.74*10**-3  #m\n",
      "#final length of specimen\n",
      "hf=18.70*10**-3   #m\n",
      "#final diameter\n",
      "df=21.54*10**-3  #m\n",
      "#initial an final cross sectional areas are\n",
      "A_i=3.14/4*d0**2   #m**2\n",
      "A_f=3.14/4*df**2   #m**2\n",
      "#various applied loads are in k N\n",
      "P1=0\n",
      "P2=5\n",
      "P3=10\n",
      "P4=15\n",
      "P5=20\n",
      "P6=25\n",
      "P7=30\n",
      "P8=35\n",
      "P9=40\n",
      "P10=45\n",
      "P11=50\n",
      "P12=55\n",
      "P13=60\n",
      "P14=65\n",
      "P15=70\n",
      "P16=75\n",
      "P17=80\n",
      "P18=85\n",
      "P19=131\n",
      "#corresponding to these load we have recorded contraction as\n",
      "delta1=0\n",
      "delta2=0.004\n",
      "delta3=0.008\n",
      "delta4=0.012\n",
      "delta5=0.015\n",
      "delta6=0.017\n",
      "delta7=0.020\n",
      "delta8=0.023\n",
      "delta9=0.025\n",
      "delta10=0.028\n",
      "delta11=0.032\n",
      "delta12=0.036\n",
      "delta13=0.040\n",
      "delta14=0.044\n",
      "delta15=0.049\n",
      "delta16=0.054\n",
      "delta17=0.061\n",
      "delta18=0.069\n",
      "#stress and strain corresponding to these loads and elongations are\n",
      "sigma1=P1/A_i\n",
      "strain1=delta1/h0\n",
      "sigma2=P2/A_i\n",
      "strain2=delta2/h0\n",
      "sigma3=P3/A_i\n",
      "strain3=delta3/h0\n",
      "sigma4=P4/A_i\n",
      "strain4=delta4/h0\n",
      "sigma5=P5/A_i\n",
      "strain5=delta5/h0\n",
      "sigma6=P6/A_i\n",
      "strain6=delta6/h0\n",
      "sigma7=P7/A_i\n",
      "strain7=delta7/h0\n",
      "sigma8=P8/A_i\n",
      "strain8=delta8/h0\n",
      "sigma9=P9/A_i\n",
      "strain9=delta9/h0\n",
      "sigma10=P10/A_i\n",
      "strain10=delta10/h0\n",
      "sigma11=P11/A_i\n",
      "strain11=delta11/h0\n",
      "sigma12=P12/A_i\n",
      "strain12=delta12/h0\n",
      "sigma13=P13/A_i\n",
      "strain13=delta13/h0\n",
      "sigma14=P14/A_i\n",
      "strain14=delta14/h0\n",
      "sigma15=P15/A_i\n",
      "strain15=delta15/h0\n",
      "sigma16=P16/A_i\n",
      "strain16=delta16/h0\n",
      "sigma17=P17/A_i\n",
      "strain17=delta17/h0\n",
      "sigma18=P18/A_i\n",
      "strain18=delta18/h0\n",
      "#part(a)\n",
      "#modulus of elasticity\n",
      "E=(sigma13-sigma1)/(strain13-strain1)/10**3   #G Pa\n",
      "#part(b)\n",
      "# yield stress at D (shown in fig 10.6)\n",
      "Yield=P15/A_i*1000/10**6   #M N/m**2\n",
      "#part (c)\n",
      "#ultimate stress (maximum)\n",
      "ultimate_sigma=P19/A_i*10**3/10**6   #M N/m**3\n",
      "#part (d)\n",
      "#percentage contraction\n",
      "percent_l=(h0-hf)/h0*100  #%\n",
      "#part(e)\n",
      "#percentage increase in area\n",
      "percent_A=(df**2-d0**2)/d0**2*100  #%\n",
      "#part(f)\n",
      "#apparent breaking stress\n",
      "app_breaking_stress=P19*1000/A_i/10**6   #M N/m**2\n",
      "#actual breaking stress\n",
      "actual_breaking_stress=P19*1000/A_f/10**6   #M N/m**2\n",
      "print \"\"\"The modulus of elasticity is %0.3f G Pa \n",
      "Yield stress is %0.3f M n/m**2\n",
      "The ultimate (maximum) stress is %0.3f M n/m**2 \n",
      "Percentage contraction in length %0.3f\n",
      "Percentage increase in area is %0.3f \n",
      "Apparent breaking stress is %0.3f M n/mm**2\n",
      "Actual breaking point is %0.3f M n/m**2\"\"\"%(E,Yield,ultimate_sigma,percent_l,percent_A,app_breaking_stress,actual_breaking_stress)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The modulus of elasticity is 130.694 G Pa \n",
        "Yield stress is 253.915 M n/m**2\n",
        "The ultimate (maximum) stress is 475.185 M n/m**2 \n",
        "Percentage contraction in length 22.148\n",
        "Percentage increase in area is 32.115 \n",
        "Apparent breaking stress is 475.185 M n/mm**2\n",
        "Actual breaking point is 359.675 M n/m**2\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.3 page no-307"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#length of glass piece\n",
      "l=1.1*10**3  #mm\n",
      "#width of glass piece\n",
      "b=225   #mm\n",
      "#height or thicness of plate\n",
      "h=10  #mm\n",
      "#load\n",
      "P=250  #N\n",
      "#for a simply supported beam subjected to concentrated load in the middle  of its span,\n",
      "M=P*l/4  #N mm\n",
      "#and force\n",
      "F=P/2  #N\n",
      "#part(a)\n",
      "#flexure strength\n",
      "sigma=6*M/b/h**2  #N/mm**2\n",
      "#part (b)\n",
      "#shear strength\n",
      "tau=3*F/2/b/h  #N/mm**2\n",
      "#part (c)\n",
      "P1=350  #N\n",
      "M1=P1*l/4\n",
      "#ineria\n",
      "I=b*h**3/12  #mm**4\n",
      "y=h/2  #mm\n",
      "#the modulus of rupture  is given by\n",
      "sigmar=M1*y/I\n",
      "print \"The flexture strength, shear strength and modulus of rupture are :\\n%0.3f N/mm**2, %0.3f N/mm**2 and %0.3f N/mm**2 resp\"%(sigma, tau, sigmar)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The flexture strength, shear strength and modulus of rupture are :\n",
        "18.333 N/mm**2, 0.083 N/mm**2 and 25.667 N/mm**2 resp\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.4 page no-313"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi, sqrt\n",
      "#given\n",
      "#diameter of ball\n",
      "D=0.5*10   #mm\n",
      "#indentation diameter\n",
      "d=32.5/10  #mm (diveided by 10 because it is 10 times magnified)\n",
      "#from table- 10.3 of book , the load for steel specimen is\n",
      "P=30*D**2  #kg f\n",
      "#hardness\n",
      "BHN=P/((pi)*D/2*(D-sqrt(D**2-d**2))) \n",
      "print \"The hardness is %0.3f\" %(BHN)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The hardness is 79.556\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.5 page no-321"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sin, cos, acos\n",
      "#given\n",
      "#dimension of steel specimen\n",
      "l=75  #mm\n",
      "b=10  #mm\n",
      "t=10  #mm\n",
      "#depth of V-notch is t/5\n",
      "#in the absence of specimen, frictional and windage loss\n",
      "L1=0.1  #kg f.m\n",
      "#in the presence ofspecimen, which is placed on support breaks\n",
      "L2=5.9  #kg f m\n",
      "#rupture energy\n",
      "U=L2-L1  #kg f m\n",
      "#since the depth of V-notch is t/5\n",
      "#so t/5=2\n",
      "te=t-2  #mm\n",
      "#volume of specimen\n",
      "Ve=l*b*te*10**-9  #m**3\n",
      "#modulus of rupture\n",
      "Ur=U/Ve  #kg f/m**2\n",
      "#effective area of cross section\n",
      "Ae=b*te*10**-6  #m**2\n",
      "#notch impact strength\n",
      "Is=U/Ae  #kg/m\n",
      "#given that \n",
      "Ui=30   #kg f.m\n",
      "alpha=160*(pi)/180  #radians\n",
      "#swing diameter\n",
      "D=1600  #mm\n",
      "R=D/2*10**-3  #m\n",
      "#weight of hammer\n",
      "#as we know that\n",
      "#Ut=W*R(1-cos(aplha))\n",
      "#so\n",
      "W=Ui/R/(1-cos(alpha))  #kg f.m\n",
      "#capacity of izod impact testing machine\n",
      "L3=30  #kg f.m\n",
      "#so Uf will be\n",
      "Uf=L3-L2   #kbg f.m\n",
      "#we know that energy after rupture\n",
      "#Uf=W*R(1-cos(beta))\n",
      "bet=acos(1-Uf/W/R)  #radins\n",
      "#beta in degrees\n",
      "Beta=bet*180/(pi)  #degrees\n",
      "#also we know that Uf=W*hf\n",
      "#so hf will be\n",
      "hf=Uf/W  #m\n",
      "print \"\"\"rupture energy is %0.3f kg f.m \n",
      "modulus of rupture %0.3f kg f/m**2 \n",
      "notch impact strength %0.3f kg/m \n",
      "angle of hammer after striking %0.3f degrees\n",
      "height risen by hammer after breaking %0.3f m\"\"\"%(U,Ur,Is,Beta,hf)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "rupture energy is 5.800 kg f.m \n",
        "modulus of rupture 966666.667 kg f/m**2 \n",
        "notch impact strength 72500.000 kg/m \n",
        "angle of hammer after striking 123.933 degrees\n",
        "height risen by hammer after breaking 1.247 m\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.6 page no-323"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#diameter  of circular section of bean\n",
      "d=60  #mm\n",
      "#length of circular section of beam\n",
      "l=500  #mm\n",
      "#maximum and minimum load \n",
      "Pmin= 20  #kN\n",
      "Pmax= 50  #kN\n",
      "#ultimate strength\n",
      "sigmau=650  #MPa\n",
      "#yoeld strength\n",
      "sigmay=520  #MPa\n",
      "#factor of safety\n",
      "fos=1.8\n",
      "#maximum bending moment\n",
      "Mmax=Pmax*l/4  #kN mm\n",
      "#minimum bending moment\n",
      "Mmin=Pmin*l/4  #kN mm\n",
      "#mean bending moment\n",
      "Mm=(Mmax+Mmin)/2  #kN mm\n",
      "#alerting (variable) bending moment\n",
      "Ma=(Mmax-Mmin)/2  #kN mm\n",
      "#section modulus of beam\n",
      "Z=(pi)*d**3/32  #mm**3\n",
      "#mean bending stress\n",
      "sigmam=Mm/Z*1000  #MPa\n",
      "#variable bending stress\n",
      "sigmaa=Ma/Z*1000    #MPa\n",
      "#endurance stress from\n",
      "#(i) gerber's parabolic relation\n",
      "sigmae1=sigmaa/(1/fos-(sigmam/sigmau)**2*fos)  #MPa\n",
      "#(ii) goodman's straight line relation\n",
      "sigmae2=sigmaa/(1/fos-sigmam/sigmau)  #MPa\n",
      "#(iii) soderberg's straight line realtion\n",
      "sigmae3=sigmaa/(1/fos-sigmam/sigmay)  #MPa\n",
      "print \"\"\"Endurance strength of the material are :\n",
      "Gerbers parabolic fomula %0.3f MPa\n",
      "goodmans straight line formula %0.3f MPa\n",
      "sodergerbs straight line relation %0.3f MPa \"\"\"%(sigmae1,sigmae2,sigmae3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Endurance strength of the material are :\n",
        "Gerbers parabolic fomula 236.280 MPa\n",
        "goodmans straight line formula 371.272 MPa\n",
        "sodergerbs straight line relation 556.791 MPa \n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.7 page no-329"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#yeild strength of polycrystalline material increases from sigmay1 to sigmay2\n",
      "sigmay1=118  #MPa\n",
      "sigmay2=207  #MPa\n",
      "#decreasing grain diameter from d1 to d2\n",
      "d1=0.253*10**-3   #m\n",
      "d2=0.0224*10**-3  #m\n",
      "#to find the yield strngth at \n",
      "d=0.095*10**-3  #m\n",
      "#as we know that according to hall and petch equation,\n",
      "#sigmay=sigma0+C/sqrt(d)\n",
      "#putting sigmay1,sigmay2,d1 and d2.. we get 2 equations\n",
      "#sigma0+C/sqrt(d1)=sigmay1  -------(1)\n",
      "#and sigma0+C/sqrt(d2)=sigmay2    -----(2)\n",
      "#solving equation 1 and 2 we get\n",
      "sigma0=80.3  #MPa\n",
      "#and\n",
      "C=0.1896  #MN/m**(3/2)\n",
      "#so the yield stress for the grain size\n",
      "d=0.095*10**-3  #m\n",
      "sigma=sigma0+C/sqrt(d)  #MPa\n",
      "print \"The yield stress for a grainof size of 0.095 mm is %0.3f MPa\"%(sigma)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The yield stress for a grainof size of 0.095 mm is 99.753 MPa\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.8 page no-330"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#ASTM number\n",
      "n=12\n",
      "#as we know that\n",
      "N=2**(n-1) \n",
      "#1 square inch=645 mm**2\n",
      "#so grain diameter for ASTM number 12 will be\n",
      "d=1/sqrt((N/645)*10**4)  #mm\n",
      "print \"The grain diameter of ASTM number 12 is %0.4f mm\"%(d)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The grain diameter of ASTM number 12 is 0.0056 mm\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.9 page no-330"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "#ASTM number of grain\n",
      "n=5\n",
      "#as we know that\n",
      "N=2**(n-1) #grains/inche**2 at magnification 100*\n",
      "# as lineal and areal magnifications are related as *100=10,000 areal\n",
      "N1=N/0.01/0.01  #grains/inch**2 at 1*\n",
      "#average area of one grain\n",
      "A=2.54*2.54/N1  #cm**2\n",
      "#now N1 grains/ inch**2 of surface is = sqrt(160,000)=400 grain/inch of length and this is equal to =(400)**3=6.4*10**7  grains/m**3 of volume\n",
      "#surface area of eaxh cubic surface\n",
      "S=(1/400)**2  #inch**2\n",
      "#there are 6 surfaces in a cubic grain\n",
      "#so total surface area of each grain\n",
      "TS=6*S  #inch**2\n",
      "#each boundary is composed of two grain surfaces, therefore , total boundary in the volume is\n",
      "TotS=1/2*TS*(400)**3  #inch**2 boundary per cubic of steel\n",
      "#total suface area in cm**2\n",
      "TotalS=TotS/(2.54)  #cm**2 boundary per cubic cm of steel\n",
      "print \"total boundary in the volume is %0.3f cm**2 per cm**3 of steel\"%(TotalS)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "total boundary in the volume is 472.441 cm**2 per cm**3 of steel\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "example-10.16 page no-332"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "#given\n",
      "#ASTM number\n",
      "n=4\n",
      "#as we know that\n",
      "N=2**(n-1)  #per inch**2 at a magnification of 100\n",
      "#let r be the radius of grain\n",
      "#so\n",
      "#N*A=1/100 inch**2  where A=(pi)*r**2\n",
      "#so\n",
      "r=sqrt(1/100/N/(pi))   #inch\n",
      "#radius of grain in mm\n",
      "R=r*25.4  #mm\n",
      "print \"the radius of grain is %0.3f mm\"%(R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the radius of grain is 0.507 mm\n"
       ]
      }
     ],
     "prompt_number": 24
    }
   ],
   "metadata": {}
  }
 ]
}