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diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2.ipynb new file mode 100644 index 00000000..5c92d3ed --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2.ipynb @@ -0,0 +1,327 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f3ccf4fb6d13add26a342446f0908b75d3a6a82c442b340e601708c15ec3ca4f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Axial strains and Deformations in bars "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 page number 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "l_ob = 2000 #mm - length of rod ob\n",
+ "l_bc = 1000 #mm - length of rod bc\n",
+ "l_cd = 1500 #mm - length of rod cd\n",
+ "p_ob = 100 #kN - Force in rods \n",
+ "p_bc = -150 #KN\n",
+ "p_cd = 50 #KN \n",
+ "A_ob = 1000 #mm2 - Area of rod ob\n",
+ "A_bc = 2000 #mm2 - Area of rod bc \n",
+ "A_cd = 1000 #mm2 - Area of rod cd\n",
+ "E = 200.0 #GPA \n",
+ "# the total deflection is algebraic sums of `deflection in each module \n",
+ "e_1 = p_ob*l_ob/(A_ob*E)\n",
+ "e_2 = p_bc*l_bc/(A_bc*E)\n",
+ "e_3 = p_cd*l_cd/(A_cd*E)\n",
+ "#All units are satisfied \n",
+ "e_total = e_1+ e_2 + e_3\n",
+ "print \"The total deflection is :\",round(e_total,3) ,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total deflection is : 1.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 page number 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "p_app = 3 #Kips - applied force \n",
+ "P_A = 2.23 #Kips \n",
+ "p_B = -2.83 #kips - compressive force\n",
+ "l_ab = 6.71 #inch\n",
+ "l_bc = 8.29 #inch\n",
+ "s_ab = 17.8 #ksi - tensile stress\n",
+ "s_bc = -12.9 #ksi - compressive stress\n",
+ "E = 10.6 * pow(10,3) #ksi -youngs modulus \n",
+ "e_ab = s_ab*l_ab/E\n",
+ "\n",
+ "e_bc = s_bc*l_bc/E\n",
+ "x = e_ab/e_bc #the Ratio of cosines of the deflected angles \n",
+ "# t_1 and t_2 be deflected angles \n",
+ "#t_2 = 180-45-26.6-t_1 the sum of angles is 360\n",
+ "#t_1 = 52.2 degress\n",
+ "import math\n",
+ "e = e_ab/math.acos(math.radians(52.2)) #inch\n",
+ "k = p_app/e # kips/in vertical stiffness of the combination\n",
+ "print \"The vertical stiffness of the combination is\",round(k,3),\"kips/inch\"\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.0112677358491\n",
+ "The vertical stiffness of the combination is 113.14 kips/inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 page number 83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "dia = 50 #mm - diameter of aluminium \n",
+ "p = 100 # KN - instant force applid\n",
+ "dia_c = 0.1215 #mm- change in diameter \n",
+ "l_c = 0.219 #mm - change in length\n",
+ "l = 300 #mm - length \n",
+ "strain_dia = dia_c/dia # lateral strain \n",
+ "strain_l = l_c/l #longitudinal strain \n",
+ "po = strain_dia/strain_l # poission ratio \n",
+ "area = 3.14*dia*dia/4 #mm2 area\n",
+ "E = p*l/(area*l_c) #N/mm2 youngs modulus \n",
+ "print \"The lateral strain is:\",strain_dia,\"no units\"\n",
+ "print \"The longitudinal strain is:\",strain_l,\"no units\"\n",
+ "print \"The poissions ratio is:\",po,\"no units\"\n",
+ "print \"Youngs modulus:\",round(E,2),\"N/mm2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lateral strain is: 0.00243 no units\n",
+ "The longitudinal strain is: 0.00073 no units\n",
+ "The poissions ratio is: 3.32876712329 no units\n",
+ "Youngs modulus: 69.8 N/mm2\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 page number 86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "T = 12.9*pow(10,-6) #/F\n",
+ "t = 100.00 # F \n",
+ "e_ab = T*t*l_ab #in-elongation \n",
+ "e_bc = T*t*l_bc #in-elongation\n",
+ "k = e_ab/e_bc # ratio of cosines of deflected angles \n",
+ "# t_1 and t_2 be deflected angles \n",
+ "#t_2 = 180-45-26.6-t_1 the sum of angles is 360\n",
+ "t_1 = 26.6\n",
+ "import math\n",
+ "e = e_ab/math.acos(math.radians(26.6))\n",
+ "print \"The displacement in point B is :\",e ,\"in\"\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The displacement in point B is : 0.00795578950395 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11 page number "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mass = 4 #kg \n",
+ "dist = 1 #mt freely falling distance\n",
+ "l = 1500 #mm length of rod\n",
+ "d = 15 #mm diameter\n",
+ "E = 200 #GPA youngs modulus \n",
+ "k = 4.5 # N/mm stiffness costant\n",
+ "F = mass*9.81# The force applying\n",
+ "Area = 3.14*(d**2)/4 \n",
+ "# Two cases \n",
+ "#youngs modulus \n",
+ "e_y = F*l/(Area*E*pow(10,3))\n",
+ "# stiffness\n",
+ "e_f = F/k \n",
+ "#total\n",
+ "e = e_y +e_f\n",
+ "k = 1+(2/(e*pow(10,-3)))\n",
+ "stress_max_1 = F*(1+pow(k,0.5))/Area\n",
+ "print \"The maximum stress is:\",stress_max_1,\"Mpa\"\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum stress is: 3.59377281766 Mpa\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.12 page number 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "flex_a = 1#f\n",
+ "flex_b = 2#f\n",
+ "#removing lower support and solving FBD\n",
+ "e = -2 -(2+1)#fp\n",
+ "#e_1 = (2+1+1)*R\n",
+ "#e_1 = -e Making the elongations zero since the both ends are fixed\n",
+ "R = e/(2+1+1.0) #p\n",
+ "print \"The reactions at bottom is\",R,\"p\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reactions at bottom is -1.25 p\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.19 page number 113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "l = 30 #in - The length of the rod\n",
+ "p_1 = 80 #kips - The Force on the end\n",
+ "p_2 = 125 #kips - The force on the other end\n",
+ "A_s = 0.5 #in2 - The crossection of the steel rod\n",
+ "A_a = 0.5 #in2 - The crossection of the aluminium \n",
+ "E_a = 10*(10**6) #psi - The youngs modulus of the aluminium \n",
+ "E_s = 30*(10**6) #psi - The youngs modulus of the steel\n",
+ "#Internally stastically indeterminant \n",
+ "p_a = p_1/4 #From solving we get p_s = 3*P_a\n",
+ "#From material properties point of view \n",
+ "#stress_steel = stress_aluminium\n",
+ "e = p_a*l*(10**3)/(A_a*E_a) #The end deflection \n",
+ "print \"The end deflection is\",e,\"in\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The end deflection is 0.12 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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