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diff --git a/sample_notebooks/SalilKapur/IntroductionConcept_of.ipynb b/sample_notebooks/SalilKapur/IntroductionConcept_of.ipynb new file mode 100755 index 00000000..050d69c7 --- /dev/null +++ b/sample_notebooks/SalilKapur/IntroductionConcept_of.ipynb @@ -0,0 +1,282 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Introduction—Concept of Stress\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 1.1, Page number 18 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Shearing Stress in Pin A = 6790.6 psi\n", + "Case(b): Shearing Stress in Pin C = 7639 psi\n", + "Case(c): Largest Normal Stress in Link ABC = 2286 psi\n", + "Case(d): Average Shearing Stress at B = 171 psi\n", + "Case(e): Bearing Stress in Link at C = 6000 psi\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Fac = 750 #Force on rod AC(lb)\n", + "D = 0.375 #Diameter at the upper junction of rod ABC(in)\n", + "\n", + "\n", + "#Calculation \n", + "#Case(a)\n", + "A=(1/4)*((math.pi)*pow(D,2)) #Area at the upper junction of rod ABC(in^2) \n", + "tA=(Fac/A) #Shearing Stress in Pin A(psi) \n", + "#Case(b) \n", + "Ab=(1/4)*((math.pi)*pow(0.25,2)) #Area at the lower junction of rod ABC(in^2)\n", + "tC=(((1/2)*Fac)/Ab) #Shearing Stress in Pin C(psi)\n", + "#Case(c)\n", + "Anet=(3/8)*(1.25-0.375) #Area of cross section at A(in^2)\n", + "sA=(Fac/Anet) #Largest Normal Stress in Link ABC(psi)\n", + "#Case(d)\n", + "F1=750/2 #Force on each side(lb)\n", + "Ad=(1.25*1.75) #Area at junction B(in^2)\n", + "tB=(F1/Ad) #Average Shearing Stress at B\n", + "#Case(e)\n", + "Ae=0.25*0.25 #Area at point C(in^2)\n", + "sB=(F1/Ae) #Bearing Stress in Link at C\n", + "\n", + "\n", + "#Result\n", + "print('Case(a): Shearing Stress in Pin A = %.1f psi' %tA)\n", + "print('Case(b): Shearing Stress in Pin C = %.f psi' %tC)\n", + "print('Case(c): Largest Normal Stress in Link ABC = %.f psi' %sA)\n", + "print('Case(d): Average Shearing Stress at B = %.f psi' %tB)\n", + "print('Case(e): Bearing Stress in Link at C = %.f psi' %sB)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 1.2, Page number 19" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Diameter of the bolt = 28 mm\n", + "Case(a): Dimension b at Each End of the Bar = 62 mm\n", + "Case(a): Dimension h of the Bar = 34.300000 mm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "P = 120 #Maximum allowable tension force \n", + "s = 175 #Maximum allowable stress\n", + "t = 100 #Maximum allowable stress\n", + "Sb = 350 #Maximum allowable stress\n", + "\n", + "\n", + "#Calculation\n", + "#Case(a)\n", + "F1=P/2 #Current(A)\n", + "d=math.sqrt(((P/2)*1000)/((22/(4*7))*(100000000))) #Diameter of bolt(m)\n", + "d=d*1000 #Diameter of bolt(mm)\n", + "d=round(d,0) #Rounding of the value of diameter of bolt(mm)\n", + "Ad=(0.020*0.028) #Area of cross section of plate \n", + "tb=((P*1000)/Ad)/(1000000) #Stress between between the 20-mm-thick plate and the 28-mm-diameter bolt\n", + "tb=round(tb,0) #Rounding of the above calculated stress to check if it is less than 350\n", + "a=(P/2)/((0.02)*(175)) #Dimension of cross section of ring \n", + "a=round(a,2) #Rounding dimension of cross section of ring to two decimal places\n", + "b=28 + (2*(a)) #Dimension b at Each End of the Bar\n", + "b=round(b,2) #Rounding the dimension b to two decimal places \n", + "h=(P)/((0.020)*(175)) #Dimension h of the Bar\n", + "h=round(h,1) #Rounding dimension h of bar to 1 decimal place\n", + "\n", + "\n", + "#Result\n", + "print ('Case(a): Diameter of the bolt = %.f mm' %d)\n", + "print ('Case(a): Dimension b at Each End of the Bar = %.f mm' %b)\n", + "print ('Case(a): Dimension h of the Bar = %f mm' %h)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 1.3, Page number 34" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Diameter of the bolt = 16.730000 mm\n", + "Case(a): Dimension b at Each End of the Bar = 22.000000 mm\n", + "Case(a): Dimension h of the Bar = 6.000000 mm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Su = 600 #ultimate normal stress(MPa) \n", + "FS = 3.3 #Factor of safety with respect to failure\n", + "tU=350 #Ultimate shearing stress(MPa)\n", + "Cx=40 #X Component of reaction at C(kN)\n", + "Cy=65 #Y Component of reaction at C(kN)\n", + "Smax=300 #Allowable bearing stress of the steel \n", + "\n", + "#Calculation\n", + "C=math.sqrt((math.pow(40,2))+(math.pow(65,2)))\n", + "\n", + "#Case(a)\n", + "P=(15*0.6 + 50*0.3)/(0.6) #Allowable bearing stress of the steel(MPa)\n", + "Sall=(Su/FS) #Allowable Stress(MPa)\n", + "Sall=round(Sall,1) #Rounding Allowable stress to 1 decimal place(MPa)\n", + "Areqa=(P/(Sall*(1000))) #Cross Sectional area(m^2)\n", + "Areqa=round(Areqa,5) #Rounding cross sectional area to 5 decimal places(m^2)\n", + "dAB=math.sqrt(((Areqa)*(4))/(22/7)) #Diameter of AB(m)\n", + "dAB=dAB*1000 #Diameter of AB(mm)\n", + "dAB=round(dAB,2) #Rounding Diameter of AB(mm)\n", + "\n", + "#Case(b)\n", + "tALL=tU/FS #Stress(MPa)\n", + "tALL=round(tALL,1) #Rounding of Stress\n", + "AreqC=((C/2)/tALL) #Cross sectional area(m^2)\n", + "AreqC=AreqC*1000 \n", + "AreqC=round(AreqC,0) #Rounding the cross sectional area\n", + "dC=math.sqrt((4*AreqC)/(22/7)) #Diameter at point C\n", + "dC=round((dC+1),0) #Rounding of the diameter at C\n", + "\n", + "#Case(c)\n", + "\n", + "Areq=((C/2)/Smax) \n", + "Areq=Areq*1000 #Cross sectional area(mm^2)\n", + "t=(Areq/22) #Thickness of the bracket\n", + "t=round(t,0)\n", + "\n", + "#Result\n", + "print ('Case(a): Diameter of the bolt = % f mm' %dAB)\n", + "print ('Case(a): Dimension b at Each End of the Bar = % f mm' %dC)\n", + "print ('Case(a): Dimension h of the Bar = % f mm' %t)" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Example 1.4, Page number 35" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Control Rod = 5.263672 kips\n", + "Case(b): Bolt at B = 5.156250 kips\n", + "Case(c): Bolt at D = 6.865179 kips\n", + "Case(d): Bolt at C = 5.238095 kips\n", + "Summary. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the smallest value, namely:= 5.156250 kips\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "tU=40 #ultimate tensile stress\n", + "sU=60 #ultimate shearing stress\n", + "FS=3 #Mimnimum factor of safety\n", + "dA=(7/16) #Diameter of bolt at A(in)\n", + "dB=3/8 #Diameter of bolt at B(in) \n", + "dD=3/8 #Diameter of bolt at D(in)\n", + "dC=1/2 #Diameter of bolt at C(in)\n", + "\n", + "\n", + "#Calculation\n", + "Sall=(sU/FS) #Total tensile stress(kips)\n", + "B=Sall*((1/4)*(22/7)*(pow((7/16),2))) #Allowable force in the control rod(kips)\n", + "C1=1.75*(B) #Control Rod(kips)\n", + "tall=(tU/FS) #Total shearing stress\n", + "B=2*(tall*(1/4)*(22/7)*(3/8)*(3/8)) #Allowable magnitude of the force B exerted on the bolt\n", + "C2=1.75*B #Bolt at B(kips) \n", + "D=B #Bolt at D. Since this bolt is the same as bolt B, the allowable force is same(kips) \n", + "C3=2.33*D #Bolt at D(kips)\n", + "C4=2*(tall*(1/4)*(22/7)*(1/2)*(1/2)) #Bolt at C(kips) \n", + "list1=[C1,C2,C3,C4] #Adding all the maximum allowable forces on C(kips) \n", + "\n", + "\n", + "#Result\n", + "print ('Case(a): Control Rod = % f kips' %C1)\n", + "print ('Case(b): Bolt at B = % f kips' %C2)\n", + "print ('Case(c): Bolt at D = % f kips' %C3)\n", + "print ('Case(d): Bolt at C = % f kips' %C4)\n", + "print ('Summary. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the smallest value, namely:= % f kips' %min(list1));" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |