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diff --git a/Mechanics_of_Materials/Chapter3.ipynb b/Mechanics_of_Materials/Chapter3.ipynb new file mode 100755 index 00000000..355e0ff6 --- /dev/null +++ b/Mechanics_of_Materials/Chapter3.ipynb @@ -0,0 +1,334 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3: Mechanics Properties of Materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "offset = 0.2 #\n",
+ "a_x = 0.0016 #in/in\n",
+ "a_y = 50 #ksi\n",
+ "\n",
+ "#Calculations\n",
+ "#Modulus of Elasticity\n",
+ "E = a_y/(a_x*10**3) #E is the slope in GPa.\n",
+ "#Yield Strength\n",
+ "sigma_ys = 68 #ksi, Graphically, for a strain of 0.0016 mm/mm\n",
+ "#Ultimate Stress\n",
+ "sigma_u = 108 #ksi B is the peak of stress strain graph.\n",
+ "#Fracture Stress\n",
+ "ep_f = 0.23 #in/in\n",
+ "sigma_f = 90 #ksi from the graph.\n",
+ "\n",
+ "#Display\n",
+ "print\"The Modulus of Elasticity is \",E,\"ksi\"\n",
+ "print\"The Yield Strength from the graph \",sigma_ys,\"ksi\"\n",
+ "print\"The Ultimate Stress from the graph is \",sigma_u,\"ksi\"\n",
+ "print\"The Fracture Stress from the graph is \",sigma_f,\"ksi\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Modulus of Elasticity is 31.25 ksi\n",
+ "The Yield Strength from the graph 68 ksi\n",
+ "The Ultimate Stress from the graph is 108 ksi\n",
+ "The Fracture Stress from the graph is 90 ksi\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page No 95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "stress_b = 600 #MPa, stress\n",
+ "strain_b = 0.023 #mm/mm, strain\n",
+ "stress_a = 450 #Mpa, stress\n",
+ "strain_a = 0.006 #mm/mm, strain\n",
+ "\n",
+ "#Calculations\n",
+ "#Permanent Strain\n",
+ "E = stress_a/strain_a\n",
+ "strain_cd = stress_b/E #The recovered elastic strain\n",
+ "perm_strain = strain_b - strain_cd #mm/mm\n",
+ "\n",
+ "#Modulus of Resilience\n",
+ "ur_initial = (0.5*stress_a*strain_a)\n",
+ "ur_final = (0.5*stress_b*strain_cd) \n",
+ "\n",
+ "#Display\n",
+ "print\"The Permanent Strain is =\",perm_strain,\"mm/mm\"\n",
+ "print\"The Initial Modulus of Resilience is = \",ur_initial,\"MJ/mm**3\"\n",
+ "print\"The Final Modulus of Resilience is = \",ur_final,\"MJ/mm**3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Permanent Strain is = 0.015 mm/mm\n",
+ "The Initial Modulus of Resilience is = 1.35 MJ/mm**3\n",
+ "The Final Modulus of Resilience is = 2.4 MJ/mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page No 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Given\n",
+ "p = 10000 #N, load\n",
+ "E_al = 70*(10**3) #MPa,pressure\n",
+ "#The given dimension are\n",
+ "l_ab = 600.0 #mm\n",
+ "d_ab = 20.0 #mm\n",
+ "l_bc = 400.0 #mm\n",
+ "d_bc = 15.0 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "a_ab = (math.pi/4.0)*(d_ab**2) # Area of AB\n",
+ "a_bc = (math.pi/4.0)*(d_bc**2)\n",
+ "stress_ab = p/a_ab\n",
+ "stress_bc = p/a_bc\n",
+ "e_ab = stress_ab/E_al \n",
+ "e_bc = 0.045 #mm/mm . From the graph for stress_bc\n",
+ "elongation = (l_ab*e_ab)+ (l_bc*e_bc)\n",
+ "strain_rec = stress_bc/E_al \n",
+ "e_og = e_bc-strain_rec\n",
+ "rod_elong = e_og*l_bc\n",
+ "\n",
+ "#Display\n",
+ "print\"The elongation of the rod when load is applied is\",round(elongation,1),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The elongation of the rod when load is applied is 18.3 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page No 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P = 80 \t#kN load\n",
+ "#Dimensions are\n",
+ "l_z = 1.5\t #m\n",
+ "l_y = 0.05\t#m\n",
+ "l_x = 0.1 \t#m\n",
+ "\n",
+ "#Calculations\n",
+ "Aa= l_x*l_y\n",
+ "normal_stress_z = (P*(10**3))/Aa\n",
+ "Est = 200 #GPa - from the tables.\n",
+ "strain_z = (normal_stress_z)/(Est*(10**9)) \n",
+ "axial_elong = strain_z*l_z #elongation in the y direction\n",
+ "nu_st = 0.32 #Poisson's Ratio - from the tables.\n",
+ "strain_x = -(nu_st)*(strain_z) #strain in the x direction.\n",
+ "strain_y = strain_x\n",
+ "#Elongations\n",
+ "delta_x = strain_x*l_x\n",
+ "delta_y = strain_y*l_y\n",
+ "\n",
+ "#Display\n",
+ "print\"The change in the length (z direction) =\",axial_elong*10**6,\"micrometer\"\n",
+ "print\"The change in the cross section (x direction)=\",delta_x*10**6,\"micrometer\"\n",
+ "print\"The change in the cross section (y direction)=\",delta_y*10**6,\"micrometer\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in the length (z direction) = 120.0 micrometer\n",
+ "The change in the cross section (x direction)= -2.56 micrometer\n",
+ "The change in the cross section (y direction)= -1.28 micrometer\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page No 105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#Refer to the graph of shear stress-strain of titanium alloy.\n",
+ "x_A = 0.008 #rad - x co-ordinate of A\n",
+ "y_A = 52\t #MPa - y co-ordinate of A\n",
+ "height = 2 \t#mm\n",
+ "l = 3.0 \t\t #mm\n",
+ "b = 4.0 \t #mm\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "#Shear Modulus\n",
+ "G = y_A/x_A\n",
+ "#Proportional Limit\n",
+ "tou_pl = 52 #ksi Point A\n",
+ "#Ultimate Stresss\n",
+ "tou_u = 73 #ksi - Max shear stress at B\n",
+ "#Maximum Elastic Displacement\n",
+ "tanA= x_A # tan theta is approximated as theta.\n",
+ "d = tanA*height\n",
+ "#Shear Force\n",
+ "A = l*b\n",
+ "V = tou_pl*A\n",
+ "\n",
+ "#Display\n",
+ "print\"The Shear Modulus = \",G,\"ksi\"\n",
+ "print\"The Proportional Limit = \",tou_pl,\"ksi\"\n",
+ "print\"The Ultimate Shear Stress = \",tou_u,\"ksi\"\n",
+ "print\"The Maximum Elastic Displacement = \",d,\"inch\"\n",
+ "print\"The Shear Force = \",V,\"kip\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Modulus = 6500.0 ksi\n",
+ "The Proportional Limit = 52 ksi\n",
+ "The Ultimate Shear Stress = 73 ksi\n",
+ "The Maximum Elastic Displacement = 0.016 inch\n",
+ "The Shear Force = 624.0 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page No 106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d_o = 0.025 \t\t#m, outside diameter\n",
+ "l_o =0.25 \t\t #m\n",
+ "F =165 \t\t\t #kN, load\n",
+ "delta = 1.2 \t\t#mm\n",
+ "G_al = 26 \t\t #GPa\n",
+ "sigma_y = 440 \t\t#MPa\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "#Modulus of Elasticity\n",
+ "A = (math.pi/4)*(d_o**2)\n",
+ "avg_normal_stress = (F*10**3)/A\n",
+ "avg_normal_strain = delta/l_o\n",
+ "E_al = avg_normal_stress/ avg_normal_strain\n",
+ "E_al = E_al/10**6\n",
+ "#Contraction of Diameter\n",
+ "nu = (E_al/(2*G_al))-1\n",
+ "strain_lat = nu*(avg_normal_strain) \n",
+ "d_contraction = strain_lat* d_o \n",
+ "\n",
+ "#Display\n",
+ "print\"The Modulus of Elasticity is = \",round(E_al,0),\"GPa\"\n",
+ "print\"The contraction in diameter due to the force = \",round(d_contraction,4),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "70.0281749604\n",
+ "The Modulus of Elasticity is = 70.0 GPa\n",
+ "The contraction in diameter due to the force = 0.0416 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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