From 567954f0f9089a76901da46619e67bcaa8eaf0d4 Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Fri, 17 Jun 2016 23:30:26 +0600 Subject: Added(A)/Deleted(D) following books A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap10_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap11_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap12_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap13_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap16_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap17_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap18_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap19_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap20_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap21_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap22_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap23_1.ipynb A 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A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/Chap9_1.ipynb A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/screenshots/Ch28Capacitance_1.png A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/screenshots/Ch28OpFreq_1.png A A_Textbook_of_Electronic_Circuits_by_R._S._Sedha/screenshots/Ch28VoltageGain_1.png A Electronics_Circuits_and_Systems_by_Y._N._Bapat/README.txt A Engineering_Mechanics_of_Solids_by_Popov_E_P/Chapter1_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_5.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_6.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3_7.ipynb A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/1_4.PNG A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/2_4.PNG A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/3_4.PNG M Fluid_Flow_For_The_Practicing_Chemical_Engineer_by_J._P._Abulencia_And_L._Theodore/Chapter-27.ipynb M Fluid_Flow_For_The_Practicing_Chemical_Engineer_by_J._P._Abulencia_And_L._Theodore/Chapter-27_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter10_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9_1.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen1_1.png A Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen2_1.png A Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen3_1.png A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter10_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter11_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter13_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter14_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter15_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter17_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter18_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter19_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter1_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter21_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter22_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter23_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter25_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter26_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter3_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter4_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter5_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter6_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter7_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter9_1.ipynb A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/screenshots/Ch14PathOfAscent_1.png A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/screenshots/Ch14RandomNumberGenerator_1.png A Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/screenshots/Ch26EulerMethod_1.png A "sample_notebooks/GundlaKeerthi vani/J.B.Gupta_Chapter_6_(1).ipynb" --- .../Chapter1_6.ipynb | 393 ++++++++++ .../chapter10_6.ipynb | 781 +++++++++++++++++++ .../chapter11_6.ipynb | 376 +++++++++ .../chapter12_6.ipynb | 411 ++++++++++ .../chapter13_5.ipynb | 215 +++++ .../chapter2_6.ipynb | 325 ++++++++ .../chapter4_6.ipynb | 439 +++++++++++ .../chapter5_6.ipynb | 865 +++++++++++++++++++++ .../chapter6_6.ipynb | 570 ++++++++++++++ .../chapter7_6.ipynb | 257 ++++++ .../chapter8_6.ipynb | 345 ++++++++ .../chapter9_6.ipynb | 229 ++++++ .../charpter_3_7.ipynb | 288 +++++++ .../screenshots/1_4.PNG | Bin 0 -> 55244 bytes .../screenshots/2_4.PNG | Bin 0 -> 43946 bytes .../screenshots/3_4.PNG | Bin 0 -> 55584 bytes 16 files changed, 5494 insertions(+) create mode 100644 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+ }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The bearing stress at C is 0.875 MPA\n", + "The maximum normal stress in BD bolt is: 62.0 MPA\n", + "The tensile strss at shank of the bolt is: 40.0 MPA\n" + ] + } + ], + "source": [ + "#Given\n", + "import math\n", + "d_bolt = 20.0 #mm,diameter,This is not the minimum area\n", + "d_bolt_min = 16.0 #mm This is at the roots of the thread \n", + "#This yealds maximum stress \n", + "A_crossection = (math.pi)*(d_bolt**2)/4 #mm*2\n", + "A_crossection_min = (math.pi)*(d_bolt_min**2)/4 #mm*2 ,This is minimum area which yeilds maximum stress\n", + "load = 10.0 #KN\n", + "BC = 1.0 #m\n", + "CF = 2.5 #m\n", + "contact_area = 200*200 # mm*2 , The contact area at c\n", + "\n", + "#caliculations \n", + "#Balancing forces in the x direction:\n", + "# Balncing the moments about C and B:\n", + "Fx = 0 \n", + "R_cy = load*(BC+CF) #KN , Reaction at C in y-direction\n", + "R_by = load*(CF) #KN , Reaction at B in y-direction\n", + "#Because of 2 bolts\n", + "stress_max = (R_by/(2*A_crossection_min))*(10**3) # MPA,maximum stess records at minimum area\n", + "stress_shank = (R_by/(2*A_crossection))*(10**3) # MPA\n", + "Bearing_stress_c = (R_cy/contact_area)*(10**3) #MPA, Bearing stress at C\n", + "\n", + "print\"The bearing stress at C is \",(Bearing_stress_c) ,\"MPA\"\n", + "print\"The maximum normal stress in BD bolt is: \",round(stress_max),\"MPA\"\n", + "print\"The tensile strss at shank of the bolt is: \",round(stress_shank),\"MPA\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2 page number 26" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total weight of pier is 25.0 KN\n", + "The stress at 1 m above is 28.75 MPA\n" + ] + } + ], + "source": [ + "#Given \n", + "load_distributed = 20 #KN/m*2, This is the load distributed over the pier\n", + "H = 2 # m, Total height \n", + "h = 1 #m , point of investigation \n", + "base = 1.5 #m The length of crossection in side veiw \n", + "top = 0.5 #m ,The length where load is distributed on top\n", + "base_inv = 1 #m , the length at the point of investigation \n", + "area = 0.5*1 #m ,The length at a-a crossection \n", + "density_conc = 25 #KN/m*2\n", + "#caliculation of total weight \n", + "\n", + "v_total = ((top+base)/2)*top*H #m*2 ,The total volume \n", + "w_total = v_total* density_conc #KN , The total weight\n", + "R_top = (top**2)*load_distributed #KN , THe reaction force due to load distribution \n", + "reaction_net = w_total + R_top\n", + "\n", + "#caliculation of State of stress at 1m \n", + "v_inv = ((top+base_inv)/2)*top*h #m*2 ,The total volume from 1m to top\n", + "w_inv = v_inv*density_conc #KN , The total weight from 1m to top\n", + "reaction_net = w_inv + R_top #KN\n", + "Stress = reaction_net/area #KN/m*2\n", + "print\"The total weight of pier is\",w_total,\"KN\"\n", + "print\"The stress at 1 m above is\",Stress,\"MPA\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3 page number 27" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tensile stress in main bar AB: 17.89 Ksi\n", + "Tensile stress in clevis of main bar AB: 11.18 Ksi\n", + "Comprensive stress in main bar BC: 12.93 Ksi\n", + "Bearing stress in pin at C: 18.86 Ksi\n", + "torsion stress in pin at C: -25.62 Ksi\n" + ] + } + ], + "source": [ + "#Given\n", + "from math import pow\n", + "d_pins = 0.375 #inch\n", + "load = 3 #Kips\n", + "AB_x = 6 #inch,X-component\n", + "AB_y = 3 #inch,Y-component \n", + "BC_y = 6 #inch,Y-component\n", + "BC_x = 6 #inch,X-component\n", + "area_AB = 0.25*0.5 #inch*2 \n", + "area_net = 0.20*2*(0.875-0.375) #inch*2 \n", + "area_BC = 0.875*0.25 #inch*2 \n", + "area_pin = d_pins*2*0.20 #inch*2 \n", + "area_pin_crossection = 3.14*((d_pins/2)**2)\n", + "#caliculations\n", + "\n", + "slope = AB_y/ AB_x #For AB\n", + "slope = BC_y/ BC_x #For BC\n", + "\n", + "#momentum at point C:\n", + "F_A_x = (load*AB_x )/(BC_y + AB_y ) #Kips, F_A_x X-component of F_A\n", + "\n", + "#momentum at point A:\n", + "F_C_x = -(load*BC_x)/(BC_y + AB_y ) #Kips, F_C_x X-component of F_c\n", + "\n", + "#X,Y components of F_A\n", + "F_A= (pow(5,0.5)/2)*F_A_x #Kips\n", + "F_A_y = 0.5*F_A_x #Kips\n", + "\n", + "#X,Y components of F_C \n", + "F_C= pow(2,0.5)*F_C_x #Kips\n", + "F_C_y = F_C_x #Kips\n", + "\n", + "T_stress_AB = F_A/area_AB #Ksi , Tensile stress in main bar AB\n", + "stress_clevis = F_A/area_net #Ksi ,Tensile stress in clevis of main bar AB\n", + "c_strees_BC = F_C/area_BC #Ksi , Comprensive stress in main bar BC\n", + "B_stress_pin = F_C/area_pin #Ksi , Bearing stress in pin at C\n", + "To_stress_pin = F_C/area_pin_crossection #Ksi , torsion stress in pin at C\n", + "\n", + "print\"Tensile stress in main bar AB:\",round(T_stress_AB,2),\"Ksi\"\n", + "print\"Tensile stress in clevis of main bar AB:\",round(stress_clevis,2),\"Ksi\"\n", + "print\"Comprensive stress in main bar BC:\",round(-c_strees_BC,2),\"Ksi\"\n", + "print\"Bearing stress in pin at C:\",round(-B_stress_pin,2),\"Ksi\"\n", + "print\"torsion stress in pin at C:\",round(To_stress_pin,2),\"Ksi\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4 page number 38" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The factor 2.5 is less than assumed factor 2.7 so this can be considered\n" + ] + } + ], + "source": [ + "#Given\n", + "strength_steel = 120 #Ksi\n", + "factor = 2.5\n", + "F_C = 2.23 #Ksi\n", + "\n", + "#caliculations\n", + "\n", + "stress_allow = strength_steel/factor #Ksi\n", + "A_net = F_C/strength_steel #in*2 , \n", + "#lets adopt 0.20x0.25 in*2 and check wether we are correct or not? \n", + "\n", + "A_net_assumption = 0.25*0.20 #in*2 , this is assumed area which is near to A_net\n", + "stress = 2.23/A_net_assumption #Ksi\n", + "factor_assumed = strength_steel/stress \n", + "\n", + "if factor_assumed > factor :\n", + " print \"The factor\",factor,\"is less than assumed factor\",round(factor_assumed,1),\"so this can be considered\"\n", + "else:\n", + " print \"The assumed factor\",factor, \"is more than assumed factor\",factor_assumed,\"factor_assumed\"\n", + " \n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6 page number 35" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required size of rod is: 49.35 m*2\n" + ] + } + ], + "source": [ + "#Given\n", + "mass = 5 #Kg\n", + "frequency = 10 #Hz\n", + "stress_allow = 200 #MPa\n", + "R = 0.5 #m\n", + "\n", + "#caliculations \n", + "from math import pi\n", + "w = 2*pi*frequency #rad/sec\n", + "a = (w**2)*R #m*2/sec\n", + "F = mass*a #N\n", + "A_req = F/stress_allow #m*2 , The required area for aloowing stress\n", + "print\"The required size of rod is:\",round(A_req,2),\"m*2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7 page number 45" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the allowable area for live load 1.0 is 0.273 in*2\n", + "the allowable area for live load 15 is 0.909 in*2\n", + "the crossection area for live load 1.0 is 0.235 in*2\n", + "the crossection area for live load 15 is 0.926 in*2\n" + ] + } + ], + "source": [ + "#Given\n", + "D_n = 5.0 #kips, dead load\n", + "L_n_1 = 1.0 #kips ,live load 1\n", + "L_n_2 = 15 #kips ,live load 2\n", + "stress_allow = 22 #ksi\n", + "phi = 0.9 #probalistic coefficients\n", + "y_stress = 36 #ksi,Yeild strength\n", + "#According to AISR \n", + "\n", + "#a\n", + "p_1 = D_n + L_n_1 #kips since the total load is sum of dead load and live load\n", + "p_2 = D_n + L_n_2 #kips, For second live load\n", + "\n", + "Area_1 = p_1/stress_allow #in*2 ,the allowable area for the allowed stress\n", + "Area_2 = p_2/stress_allow #in*2\n", + "print \"the allowable area for live load\",L_n_1,\"is\",round(Area_1,3),\"in*2\"\n", + "print \"the allowable area for live load\",L_n_2,\"is\",round(Area_2,3),\"in*2\"\n", + "\n", + "#b\n", + "#area_crossection= (1.2*D_n +1.6L_n)/(phi*y_stress)\n", + "\n", + "area_crossection_1= (1.2*D_n +1.6*L_n_1)/(phi*y_stress) #in*2,crossection area for first live load\n", + "area_crossection_2= (1.2*D_n +1.6*L_n_2)/(phi*y_stress) #in*2,crossection area for second live load\n", + "print \"the crossection area for live load\",L_n_1,\"is\",round(area_crossection_1,3),\"in*2\"\n", + "print \"the crossection area for live load\",L_n_2,\"is\",round(area_crossection_2,3),\"in*2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.8 page number 51" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of the Weld 1: 2.54 in\n", + "Length of the Weld 2: 4.65 in\n" + ] + } + ], + "source": [ + "#Given\n", + "A_angle = 2 #in*2 \n", + "stress_allow = 20 #ksi, The maximum alowable stress\n", + "F = stress_allow*A_angle #K, The maximum force\n", + "AD = 3 #in, from the figure\n", + "DC = 1.06 #in, from the figure\n", + "strength_AWS = 5.56 # kips/in,Allowable strength according to AWS\n", + "\n", + "#caliculations \n", + "#momentum at point \"d\" is equal to 0\n", + "R_1 = (F*DC)/AD #k,Resultant force developed by the weld\n", + "R_2 = (F*(AD-DC))/AD #k,Resultant force developed by the weld\n", + "\n", + "l_1 = R_1/strength_AWS #in,Length of the Weld 1\n", + "l_2 = R_2/strength_AWS #in,Length of the Weld 2\n", + " \n", + "print \"Length of the Weld 1:\",round(l_1,2),\"in\"\n", + "print \"Length of the Weld 2:\",round(l_2,2),\"in\" \n", + " \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_6.ipynb new file mode 100644 index 00000000..c57f065e --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_6.ipynb @@ -0,0 +1,781 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10:Deflections of beams " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1 page number 501" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum bending stress developed in the saw 300.0 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia = 400 #mm - The diameter of a pulley\n", + "E = 200 #Gpa - Youngs modulus\n", + "t = 0.6 #mm - The thickness of band\n", + "c = t/2 #mm - The maximum stress is seen \n", + "#Caliculations\n", + "\n", + "stress_max = E*c*(10**3)/(dia/2) #Mpa - The maximum stress on the crossection occurs at the ends\n", + "print \"The maximum bending stress developed in the saw \",stress_max,\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3 page number 512" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The maximum displacement in y direction is -0.0130208333333 W(l**4)/EI \n", + "a) The maximum deflection occured at 0.5 L\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b) The above graph is bending moment graph\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b)The above graph is beam displacement graph\n", + "b)The maximum occures in the middle from the above graph \n" + ] + } + ], + "source": [ + "#Given\n", + "import numpy\n", + "l_ab = 1.0 #L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "#The precision is very less while caliculating through this equation because the least count in X direction is 0.1\n", + "print \"a) The maximum displacement in y direction is\",min(Y),\"W(l**4)/EI \"\n", + "print \"a) The maximum deflection occured at\",l_1[Y.index(min(Y))],\"L\"\n", + "\n", + "#Part - B\n", + "#Graphs\n", + "import numpy as np\n", + "values = M_1\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "%matplotlib inline\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b) The above graph is bending moment graph\"\n", + "values = Y \n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b)The above graph is beam displacement graph\"\n", + "print \"b)The maximum occures in the middle from the above graph \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5 page number 517" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction at A is 0.375 WL\n", + "The reaction at B is 0.625 WL\n", + "The reaction at C is 0.375 WL\n" + ] + } + ], + "source": [ + "#Given \n", + "#because of symmetry the problem can be solved by considering first half\n", + "#Given\n", + "import numpy\n", + "\n", + "l_ab = 1.0 #L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "\n", + "\n", + "# M_1_intg2[10] = 0, the displacement at the end of rod is 0 since its rigid \n", + "R_A = (F_D*(l_1[10]**4)/24.0 + F_D*(l_ab**3)*l_1[10]/48.0)/((l_1[10]**3)/6.0)\n", + "R_C = R_A #WL - symmetry\n", + "R_B = 1-R_A # WL - F_Y = 0, the equilibrium in Y direction\n", + "print \"The reaction at A is\",R_A ,\"WL\"\n", + "print \"The reaction at B is\",R_B ,\"WL\"\n", + "print \"The reaction at C is\",R_C ,\"WL\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.7 page number 521 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false, + "scrolled": true + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "%matplotlib inline\n", + "import numpy as np\n", + "# This problem is divided into two parts\n", + "#Part _1\n", + "#Given\n", + "\n", + "l = 1.0 #l - The length of the beam\n", + "p = 1.0 #W - The total load applied\n", + "#since it is triangular distribution \n", + "l_com = 0.66*l #l - The distance of force of action from one end\n", + "#F_Y = 0\n", + "#R_A + R_B = p\n", + "#M_a = 0 Implies that R_B = 2*R_A\n", + "R_A = p/3.0\n", + "R_B = 2.0*p/3\n", + "\n", + "#Taking Many sections \n", + "\n", + "#Section 1----1\n", + "l = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = p*(l[i]**2) - p/3.0\n", + " M[i] = p*(l[i]**3)/(3.0)- p*l[i]/3.0\n", + "\n", + "v[10] = R_B #again concluded Because the value is tearing of \n", + "\n", + "\n", + "#Graph\n", + "values = M\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "values = v\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "\n", + "\n", + "#part B\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_6.ipynb new file mode 100644 index 00000000..411b3c0a --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_6.ipynb @@ -0,0 +1,376 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11:Stability of Equilibrium: columns " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2 page number 589" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The length after which the beam starts buckling is 1539.0 mm\n" + ] + } + ], + "source": [ + "#Given \n", + "import math\n", + "h = 60 #mm - the length of the crossection \n", + "b = 100 #mm - the width of hte crossection \n", + "E = 200 #Gpa - The youngs modulus\n", + "stress_cr = 250 #Mpa - The proportionality limit\n", + "#Caliculations \n", + "\n", + "I = b*(h**3)/12 #mm3 The momentof inertia of the crossection\n", + "A = h*b #mm2 - The area of teh crossection \n", + "#From Eulier formula\n", + "r_min = pow((I/A),0.5) #mm - The radius of the gyration \n", + "#(l/r)**2= (pi**2)*E/stress_cr #From Eulier formula\n", + "l = (((math.pi**2)*E*(10**3)/stress_cr)**0.5)*r_min #mm - the length after which the beam starts buckling\n", + "print \"The length after which the beam starts buckling is \",round(l,0),\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.6 page number 613" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a)The following approch is solvable\n", + "a) The allowable stress in this case is 18.9 Kips\n", + "b) The following approch is solvable\n", + "b) The allowable stress in this case is 11.59 Kips\n" + ] + } + ], + "source": [ + "#Given\n", + "import math\n", + "L = 15 #ft - The length of the each rod\n", + "A = 46.7 #in2 - The length of the crossection \n", + "r_min = 4 #in - The radius of gyration\n", + "stress_yp = 36 #Ksi - the yielding point stress\n", + "E = 29*(10**3) #ksi - The youngs modulus\n", + "C_c = ((2*(math.pi**2)*E/stress_yp)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation \n", + "#According to AISC formulas \n", + "if C_s 1\n", + "The following approch is solvable\n", + "The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b 0.9 <1\n" + ] + } + ], + "source": [ + "import math\n", + "P = 200.0 #K The force on the beam \n", + "L = 15 #ft - The length of the rod\n", + "F_y = 50.0 #Ksi \n", + "F_a = F_y/(5.0/3) #Ksi -AISC MANUAL ,allowable axial stress if axial force is alone\n", + "F_b = F_a #Allowable compressive bending stress\n", + "M_1 = 600.0 #k-in - The moment acting on the ends of the rod\n", + "M_2 = 800.0 #k-in - the moment acting on the other end of teh rod\n", + "B_x = 0.264 #in - Extracted from AISC manual \n", + "E = 29*(10**3) \n", + "A = P/F_a + M_2*B_x/F_b #in2- The minimum area \n", + "print \"The minimum area is \",round(A,2),\"in2\"\n", + "#we will select W10x49 section \n", + "A_s = 14.4 #in2 - The area of the section \n", + "r_min = 2.54 #in The minimum radius \n", + "r_x = 4.35 #in \n", + "f_a = P/A_s #Ksi- The computed axial stress\n", + "f_b = M_2*B_x/A_s #Computed bending stess\n", + "C_c = ((2*(math.pi**2)*E/F_y)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation\n", + "if C_s 1:\n", + " print \"The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",round(k,3),\">1\"\n", + "else:\n", + " print \"The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",k,\"<1\"\n", + " \n", + "#trail - 2\n", + "#Lets take W10 x 60\n", + "A_s = 17.6 #in2 - The area of the section \n", + "r_min = 2.57 #in The minimum radius \n", + "r_x = 4.39 #in \n", + "f_a = P/A_s #Ksi- The computed axial stress\n", + "f_b = M_2*B_x/A_s #Computed bending stess\n", + "C_c = ((2*(math.pi**2)*E/F_y)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation\n", + "if C_s 1:\n", + " print \"The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",round(k,3),\">1\"\n", + "else:\n", + " print \"The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",round(k,2),\"<1\"\n", + " \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_6.ipynb new file mode 100644 index 00000000..71380130 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_6.ipynb @@ -0,0 +1,411 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Chapter 12:Energy and Virtual-work Methods" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.1 page number 645 " + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The deflection is downwards 0.044 in\n", + "b) The deflection is upwards 0.104 in\n", + "c) The deflection is downwards 0.039 in\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading\n", + "F_ab = 2500 #lb\n", + "F_bc = -2500 #lb\n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "#Part_a\n", + "e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) #in the deflection\n", + "if e_a<0:\n", + " print \"a) The deflection is downwards\",round(-e_a,3),\"in\"\n", + "else:\n", + " print \"a) The deflection is upwards\",round(e_a,3),\"in\"\n", + "#part-b\n", + "x = 0.125 #Shortening of member Ab\n", + "e_b = p_ab*(-x) + p_bc*0 #in - in\n", + "if e_b<0:\n", + " print \"b) The deflection is downwards\",round(-e_b,3),\"in\"\n", + "else:\n", + " print \"b) The deflection is upwards\",round(e_b,3),\"in\"\n", + "#Part-c\n", + "S = 6.5*(10**-6) #Thermal specific heat\n", + "T = 120 #F - The cahnge in temperature\n", + "e_t = -S*T*l_ab #in - The change in length of member\n", + "e_c = p_bc*e_t #in the deflection\n", + "if e_c<0:\n", + " print \"c) The deflection is downwards\",round(-e_c,3),\"in\"\n", + "else:\n", + " print \"c) The deflection is upwards\",round(e_c,3),\"in\"\n", + "\n", + "\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3 page number 648" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The point C deflects 0.019 mt down\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading\n", + "#Two parts \n", + "#Part -1 \n", + "p_ab = 5 #KN The recorded virtual loading\n", + "p_bc = -4 #KN The recorded virtual loading\n", + "F_ab = 10 #KN\n", + "F_bc = -8 #KN\n", + "l_ab = 2.5 #mt - The length of the rod\n", + "l_bc = 2 #mt - The length of the rod\n", + "A_ab = 5*(10**-4) #mt2 the areaof ab\n", + "A_bc = 5*(10**-3) #mt2 the areaof bc\n", + "E = 70 #Gpa The youngs modulus of the material\n", + "e_a =(p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E))*(10**-6) #KN-m\n", + "#Part -2 due to flexure\n", + "I = 60*10**6 #mm4 - the moment of inertia \n", + "#After solving the integration \n", + "e_b = 0.01525 #KN-m\n", + "#Total\n", + "e = (e_a+e_b)*1 #m\n", + "print \"The point C deflects\",round(e,3),\"mt down\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5 page number 651" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction force at D is 1578.98 lb\n", + "The deflection of nodal point B 0.0211 in\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading Without f_d\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading\n", + "F_ab = 2500 #lb\n", + "F_bc = -2500 #lb\n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "#Part_a\n", + "e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) #lb-in the deflection\n", + "#With f_d\n", + "p_bd = 1 #lb The recorded virtual loading \n", + "F_bd = 1 #lb\n", + "l_bd = 40 #in - The length of the rod\n", + "A_bd = 0.1 #in2 the areaof ab\n", + "e_a_1 =p_ab*p_ab*l_ab/(A_ab*E) + p_bc*p_bc*l_bc/(A_bc*E) +p_bd*p_bd*l_bd/(A_bd*E) #lb-in the deflection\n", + "#Since the produced defelection should compensate the other one\n", + "x_d = e_a/e_a_1\n", + "print \"The reaction force at D is\",round(-x_d,2),\"lb\"\n", + "\n", + "#Part - B\n", + "e_b = -x_d*l_bd/(A_bd*E ) #in - The deflection of nodal point B\n", + "print\"The deflection of nodal point B\",round(e_b,4),\"in\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.6 page number 655" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction at A 2.5 k\n", + "The reaction at B -2.5 k\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading\n", + "import numpy as np\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading \n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "K_1 = A_ab*E/l_ab #k/in - Stiffness\n", + "K_2 = A_bc*E/l_bc #k/in - Stiffness\n", + "#soving for e_1 and e_2 gives a liner euations to solve\n", + "# 128*e_1 + 24*e_2 = 0\n", + "#24*e_1 + 72*e_2 = -3\n", + "#Solving for e_1,e_2\n", + "a = np.array([[128,24], [24,72]])\n", + "b = np.array([0,-3])\n", + "x = np.linalg.solve(a, b)\n", + "e_1 = x[0] #in\n", + "e_2 = x[1] #in\n", + "u_1 = 0.8*e_1 - 0.6*e_2 #Taking each components\n", + "F_1 = K_1*u_1*(10**-3) #k The reaction at A Force = stiffness x dislacement \n", + "u_2 = 0.8*e_1 + 0.6*e_2 #Taking each components\n", + "F_2 = K_2*u_2*(10**-3) #k The reaction at B Force\n", + "print \"The reaction at A \",F_1,\"k\"\n", + "print \"The reaction at B \",F_2,\"k\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.7 page number 655" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction at A 1.18 k\n", + "The reaction at B -1.18 k\n", + "The reaction at D -1.58 k\n" + ] + } + ], + "source": [ + "#Virtual loading\n", + "import numpy as np\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading \n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "K_1 = A_ab*E/l_ab #k/in - Stiffness\n", + "K_2 = A_bc*E/l_bc #k/in - Stiffness\n", + "p_bd = 1 #lb The recorded virtual loading \n", + "F_bd = 1 #lb\n", + "l_bd = 40 #in - The length of the rod\n", + "A_bd = 0.1 #in2 the areaof ab\n", + "K_3 = A_ab*E/l_ab #k/in - Stiffness\n", + "#soving for e_1 and e_2 gives a liner euations to solve\n", + "# 128*e_1 + 24*e_2 = 0\n", + "#24*e_1 + 72*e_2 = -3\n", + "#Solving for e_1,e_2\n", + "a = np.array([[128,24], [24,147]])\n", + "b = np.array([0,-3])\n", + "x = np.linalg.solve(a, b)\n", + "e_1 = x[0] #in\n", + "e_2 = x[1] #in\n", + "u_1 = 0.8*e_1 - 0.6*e_2 #Taking each components\n", + "F_1 = K_1*u_1*(10**-3) #k The reaction at A Force = stiffness x dislacement \n", + "u_2 = 0.8*e_1 + 0.6*e_2 #Taking each components\n", + "F_2 = K_2*u_2*(10**-3) #k The reaction at B Force\n", + "u_3 = e_2 #Taking each components\n", + "F_3 = K_3*u_3*(10**-3) #k The reaction at D Force\n", + "print \"The reaction at A \",round(F_1,2),\"k\"\n", + "print \"The reaction at B \",round(F_2,2),\"k\"\n", + "print \"The reaction at D \",round(F_3,2),\"k\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.8 page number 659" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b) The vertical component of the nodal force is [ 12.] \n", + "b) The vertical component of the nodal force is [ 24.] \n", + "a) The components of displacement of point B are 18.75 L/AE and 16.67 L/AE\n" + ] + } + ], + "source": [ + "#Given\n", + "#First we will solve part B\n", + "import numpy as np\n", + "u_1 =5 #L/AE, elastic elongation\n", + "u_2 =25 #L/AE,elastic elongation\n", + "f_1 = u_1#, Units got neutralized , Constitutive relation for elastic bars\n", + "f_2 = u_2# Units got neutralized\n", + "#u_1 = 0.8*e_1 - 0.6*e_2\n", + "#u_2 = 0.8*e_1 + 0.6*e_2\n", + "#u = A*e Matric multiplication \n", + "A = np.array([[0.8,-0.6],[0.8,0.6]]) #The matrix form of A\n", + "F = np.array([[f_1],[f_2]])\n", + "P = np.dot((A.T),F) #Nodal forces matrix\n", + "print \"b) The vertical component of the nodal force is\",P[1],\"\"\n", + "print \"b) The vertical component of the nodal force is\",P[0],\"\"\n", + "#Part A\n", + "#F_1 = (5/8.0)*P_1 - (5/6.0)*p_2 , From statics\n", + "#F_1 = (5/8.0)*P_1 + (5/6.0)*p_2\n", + "#F = BP ,Matric multiplication \n", + "B = np.array([[(5/8.0),-(5/6.0)],[(5/8.0),(5/6.0)]]) #The matrix form of A\n", + "U = np.array([[u_1],[u_2]])\n", + "e = P = np.dot((B.T),U) #L/AE, Nodal forces matrix\n", + "print \"a) The components of displacement of point B are\",round(e[0],2),\"L/AE and\",round(e[1],2),\"L/AE\" \n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.10 page number 667" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The deflection is 0.018 in\n" + ] + } + ], + "source": [ + "#Given\n", + "A_1 = 0.125 #in2 , The area of the crossection of AB\n", + "A_2 = 0.219 #in2 , The area of the crossection of BC\n", + "l_1 = 3*(5**0.5) #in , The length of AB\n", + "l_2 = 6*(2**0.5) #in , The length of BC\n", + "p = 3 #k , Force acting on the system \n", + "E = 10.6*(10**3) #Ksi - youngs modulus of the material\n", + "p_1 = (5**0.5)*p/3 #P, The component of p on AB\n", + "p_2 = -2*(2**0.5)*p/3 #P, The component of p on AB\n", + "\n", + "e = p_1*l_1*p_1/(p*E*A_1) + p_2*l_2*p_2/(p*E*A_2) #in, By virtual deflection method \n", + "print \"The deflection is\",round(e,3),\"in\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_5.ipynb new file mode 100644 index 00000000..4d2cad2f --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_5.ipynb @@ -0,0 +1,215 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: Statically Indeterminate Problems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2 page number 693" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The maximum displacement in y direction is -0.208333333333 W(l**4)/EI \n", + "a) The maximum deflection occured at 1.0 L\n", + "The reaction at the mid of the bar 1.25 WL\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b)The above graph is beam displacement graph\n", + "b)The minimum occures in the middle from the above graph \n" + ] + } + ], + "source": [ + "#Given \n", + "#First we will solve without the reaction at middle\n", + "#Given\n", + "import numpy\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "l_ab = 1.0 #2L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "Y_min = 16*min(Y)\n", + "print \"a) The maximum displacement in y direction is\",16*min(Y),\"W(l**4)/EI \"\n", + "print \"a) The maximum deflection occured at\",2*l_1[Y.index(min(Y))],\"L\"\n", + "f_bb = 2**3/48.0 #l**3/EI - flexibility coefficient\n", + "Reac = - Y_min/f_bb #WL , The reaction at the mid of the bar\n", + "print \"The reaction at the mid of the bar\",Reac ,\"WL\"\n", + "\n", + "#Graphs \n", + "Y.extend(Y) #Because of symmetry\n", + "values = Y \n", + "y = np.array(values)\n", + "t = np.linspace(0,1,22)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b)The above graph is beam displacement graph\"\n", + "print \"b)The minimum occures in the middle from the above graph \"\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3 page number 694 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reactive moment at A i.e M_A -0.0714285714286 WL**2\n", + "The reactive force at A i.e R_A -1.14285714286 WL\n" + ] + } + ], + "source": [ + "#Given \n", + "#First we will solve without the reaction at middle\n", + "#Given\n", + "import numpy as np\n", + "l_ab = 1.0 #2L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "e_1 = 16*min(Y) #WL4/EI - The maximum defection \n", + "e_2 = - F_D*((2*l_ab)**3)/24.0 #WL3/EI - The maximum angle\n", + "#Caliculating for momentum and force\n", + "f_ab = ((2*l_ab)**2)/16.0 #L2/EI \n", + "f_bb = ((2*l_ab)**3)/48.0 #L3/EI \n", + "f_aa = 2*l_ab/3.0 #L/EI\n", + "f_ba = ((l_ab)**2)/4.0 #L2/EI\n", + "#F*X = e - Matrix multiplication \n", + "#Solving for X\n", + "a = np.array([[f_aa,f_ba], [f_ba,f_bb]])\n", + "b = np.array([e_2,e_1])\n", + "x = np.linalg.solve(a, b)\n", + "print \"The reactive moment at A i.e M_A\",x[0],\"WL**2\"\n", + "print \"The reactive force at A i.e R_A\",x[1],\"WL\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_6.ipynb new file mode 100644 index 00000000..0c496644 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_6.ipynb @@ -0,0 +1,325 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2:Axial strains and Deformations in bars " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1 page number 77" + ] + }, + { + "cell_type": "code", + "execution_count": 77, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total deflection is : 1.0 mm\n" + ] + } + ], + "source": [ + "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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4 page number 80" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.0112677358491\n", + "The vertical stiffness of the combination is 113.14 kips/inch\n" + ] + } + ], + "source": [ + "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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6 page number 83" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "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" + ] + } + ], + "source": [ + "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\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7 page number 86" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The displacement in point B is : 0.00795578950395 in\n" + ] + } + ], + "source": [ + "T = 12.9*pow(10,-6) #/F\n", + "t = 100.00 # F \n", + "l_ab = 6.71 #inch\n", + "l_bc = 8.29 #inch\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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.11 page number " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress is: 3.59377281766 Mpa\n" + ] + } + ], + "source": [ + "mass = 4 #kg \n", + "dist = 1 #mt freely falling distance\n", + "l = 1500 #mm length of rod\n", + "d = 15 #mm diameter\n", + "l_ab = 6.71 #inch\n", + "l_bc = 8.29 #inch\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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.12 page number 103" + ] + }, + { + "cell_type": "code", + "execution_count": 75, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reactions at bottom is -1.25 p\n" + ] + } + ], + "source": [ + "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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.19 page number 113" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The end deflection is 0.12 in\n" + ] + } + ], + "source": [ + "#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" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_6.ipynb new file mode 100644 index 00000000..f687074e --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_6.ipynb @@ -0,0 +1,439 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4:Torsion" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 page number 183" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The maximum shear due to torsion is 152.87 Mpa\n", + "stress tensor matrix [[ 0. 152.9 0. ]\n", + " [ 152.9 0. 0. ]\n", + " [ 0. 0. 0. ]]\n" + ] + } + ], + "source": [ + "#Given\n", + "dia = 10 #diameter of shaft(A-C)\n", + "c = dia/2 #mm - Radius\n", + "T = 30 #N/mm -Torque in the shaft \n", + "#Caliculations\n", + "\n", + "J = 3.14*(dia**4)/32 #mm4\n", + "shear_T = T*c*pow(10,3)/J # The torsion shear in the shaft AC\n", + "import numpy as np \n", + "print \"The maximum shear due to torsion is \",round(shear_T,2),\"Mpa\"\n", + "arr_T = np.zeros((3,3))\n", + "arr_T[0][1]=round(shear_T,1) #arranging the elements in array\n", + "arr_T[1][0]=round(shear_T,1)\n", + "print \"stress tensor matrix\",(arr_T),\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 page number 184" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum shear due to torsion is 43.15 Mpa\n", + "The minimum shear due to torsion is 34.52 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia_out = 20 #mm- outer diameter of shaft\n", + "dia_in = 16 #mm- inner diameter of shaft \n", + "c_out = dia_out/2 #mm - outer Radius of shaft \n", + "c_in = dia_in/2 #mm - inner radius of shaft \n", + "T = 40 #N/mm -Torque in the shaft \n", + "#caliculations\n", + "\n", + "J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4\n", + "shear_T_max = T*c_out*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "shear_T_min = T*c_in*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "print \"The maximum shear due to torsion is \",round(shear_T_max,2),\"Mpa\"\n", + "print \"The minimum shear due to torsion is \",round(shear_T_min,2),\"Mpa\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4 page number 187" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Diameter of the shaft used is 15.26 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "hp = 10 # horse power of motor \n", + "f = 30 # given \n", + "shear_T = 55 #Mpa - The maximum shearing in the shaft \n", + "#caliculations\n", + "\n", + "T = 119*hp/f # N.m The torsion in the shaft \n", + "#j/c=T/shear_T=K\n", + "k = T*pow(10,3)/shear_T #mm3\n", + "#c3=2K/3.14\n", + "c = pow((2*k/3),0.33) #mm - The radius of the shaft \n", + "diamter = 2*c #mm - The diameter of the shaft\n", + "print \"The Diameter of the shaft used is\",round(diamter,2),\"mm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5 page number 188" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Diameter of the shaft1 is 6.87 mm\n", + "The Diameter of the shaft2 is 0.702590481015 mm\n" + ] + } + ], + "source": [ + "#Given \n", + "hp = 200 #Horse power\n", + "stress_sh = 10000 #psi- shear stress\n", + "rpm_1 = 20.0 # The rpm at which this shaft1 operates \n", + "rpm_2 = 20000.0 # The rpm at which this shaft2 operates\n", + "T_1= hp*63000.0/rpm_1 #in-lb Torsion due to rpm1\n", + "T_2= hp*63000/rpm_2 #in-lb Torsion due to rpm1\n", + "#caliculations \n", + "\n", + "#j/c=T/shear_T=K\n", + "k_1= T_1/stress_sh #mm3\n", + "#c3=2K/3.14\n", + "c_1= pow((2*k_1/3),0.33) #mm - The radius of the shaft \n", + "diamter_1 = 2*c_1 #mm - The diameter of the shaft\n", + "print \"The Diameter of the shaft1 is\",round(diamter_1,2),\"mm\"\n", + "\n", + "#j/c=T/shear_T=K\n", + "k_2= T_2/stress_sh #mm3\n", + "#c3=2K/3.14\n", + "c_2= pow((2*k_2/3),0.33) #mm - The radius of the shaft \n", + "diamter_2 = 2*c_2 #mm - The diameter of the shaft\n", + "print \"The Diameter of the shaft2 is\",diamter_2,\"mm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 page number 193" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum angle rotated is 0.0232628450106 radians \n" + ] + } + ], + "source": [ + "#Given \n", + "T_ab = 0 #N.m - torsion in AB \n", + "T_bc = 150 #N.m - torsion in BC\n", + "T_cd = 150 #N.m - torsion in CD\n", + "T_de = 1150 #N.m - torsion in DE\n", + "l_ab = 250 #mm - length of AB\n", + "l_bc = 200 #mm - length of BC\n", + "l_cd = 300 #mm - length of cd \n", + "l_de = 500.0 #mm - length of de\n", + "d_1 = 25 #mm - outer diameter \n", + "d_2 = 50 #mm - inner diameter\n", + "G = 80 #Gpa -shear modulus\n", + "#Caliculations \n", + "\n", + "J_ab = 3.14*(d_1**4)/32 #mm4\n", + "J_bc = 3.14*(d_1**4)/32 #mm4\n", + "J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) # adding the maximum radians roteted in each module\n", + "print \"The maximum angle rotated is \",rad,\"radians \" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9 Pagenumber 196" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Torsion at rigid end A is -141.72 N-m\n", + "The Torsion at rigid end B is 1291.72 N-m\n" + ] + } + ], + "source": [ + "#given \n", + "#its a statistally indeterminant \n", + "#we will take of one of the support \n", + "#Given \n", + "T_ab = 0 #N.m - torsion in AB \n", + "T_bc = 150 #N.m - torsion in BC\n", + "T_cd = 150 #N.m - torsion in CD\n", + "T_de = 1150 #N.m - torsion in DE\n", + "l_ab = 250 #mm - length of AB\n", + "l_bc = 200 #mm - length of BC\n", + "l_cd = 300 #mm - length of cd \n", + "l_de = 500.0#mm - length of de\n", + "d_1 = 25 #mm - outer diameter \n", + "d_2 = 50 #mm - inner diameter\n", + "#Caliculations \n", + "\n", + "J_ab = 3.14*(d_1**4)/32 #mm4\n", + "J_bc = 3.14*(d_1**4)/32 #mm4\n", + "J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "G = 80 #Gpa -shear modulus\n", + "rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) \n", + "#now lets consider T_A then the torsion is only T_A\n", + "# T_A*(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) +rad = 0\n", + "# since there will be no displacement \n", + "T_A =-rad/(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) #Torsion at A\n", + "T_B = 1150 - T_A #n-m F_X = 0 torsion at B\n", + "print \"The Torsion at rigid end A is\",round(T_A,2),\"N-m\"\n", + "print \"The Torsion at rigid end B is\",round(T_B,2),\"N-m\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.12 Pagenumber 202" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The allowable torsion on the 8 bolt combination 27129600.0 N-m\n" + ] + } + ], + "source": [ + "#Given\n", + "dai_bc = 240 #mm- daimeter of '8'bolt circle \n", + "dia = dai_bc/8 #Diameter of each bolt\n", + "A = 0.25*(dia**2)*3.14 # Area of a bolt\n", + "S_allow = 40 #Mpa - The maximum allowable allowable shear stress \n", + "P_max = (S_allow)*A #N - The maximum allowable force \n", + "D = 120.0 #mm - the distance from central axis \n", + "T_allow =P_max*D*8 #N-m The allowable torsion on the 8 bolt combination \n", + "print \"The allowable torsion on the 8 bolt combination\",T_allow ,\"N-m\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.15 page number 211" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Equivalent Torsion constant is 1.97 in4\n" + ] + } + ], + "source": [ + "#Given \n", + "#AISC MANUALS\n", + "#approximated by three narrow tubes \n", + "#J = Bbt^3\n", + "B = 0.33 # constant mentiones in AISC\n", + "#three rods \n", + "\n", + "#rod_1\n", + "t_1 = 0.605 #inch - Thickness \n", + "b = 12.0 #inches - width \n", + "J_1 = B*b*(t_1**3) #in4 - Torsion constant \n", + "\n", + "#rod_2\n", + "t_2 = 0.605 #inch - Thickness \n", + "b = 12 #inches - width \n", + "J_2 = B*b*(t_2**3) #in4 - Torsion constant \n", + "\n", + "#rod_3\n", + "t_3 = 0.390 #inch - Thickness \n", + "b = 10.91 #inches - width \n", + "J_3 = B*b*(t_3**3) #in4 - Torsion constant \n", + "\n", + "#Equivalent\n", + "J_eq = J_1+J_2+J_3 #in4 - Torsion constant \n", + "print \"the Equivalent Torsion constant is \",round(J_eq,2), \"in4\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.16 page number 214" + ] + }, + { + "cell_type": "code", + "execution_count": 58, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum shear due to torsion is 345.23 Mpa\n", + "The minimum shear due to torsion is 276.18 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia_out = 10 #mm- outer diameter of shaft\n", + "dia_in = 8 #mm- inner diameter of shaft \n", + "c_out = dia_out/2 #mm - outer Radius of shaft \n", + "c_in = dia_in/2 #mm - inner radius of shaft \n", + "T = 40 #N/mm -Torque in the shaft \n", + "#caliculations\n", + "\n", + "J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4\n", + "shear_T_max = T*c_out*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "shear_T_min = T*c_in*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "print \"The maximum shear due to torsion is \",round(shear_T_max,2),\"Mpa\"\n", + "print \"The minimum shear due to torsion is \",round(shear_T_min,2),\"Mpa\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_6.ipynb new file mode 100644 index 00000000..d6f0670f --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_6.ipynb @@ -0,0 +1,865 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Chapter 5:Axial force, Shear and Bending moment " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example5.2 pagenumber 231" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The X,Y components of reaction force at A is 0 , -410.0 N\n", + "The X,Y components of reaction force at B is 0 , 670.0 N\n" + ] + } + ], + "source": [ + "#Given \n", + "L_ab = 0.4 #mt The total length of the rod\n", + "M = 200 #N_m - the moment acting on rod\n", + "l_1 = 0.1 #mt -moment acting point the distance from 'a'\n", + "R_1 = 100 #N - The Force acting \n", + "l_2 = 0.2 #mt -R_1 acting point the distance from 'a'\n", + "R_2 = 160 #N The Force acting \n", + "l_3 = 0.3 #mt -R_2 acting point the distance from 'a'\n", + "#caliculations\n", + "\n", + "#F_X = 0 forces in x directions \n", + "R_A_X = 0 # since there are no forces in X-direction \n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "\n", + "# M + R_1*l_2 + R_2*l_3 = R_B*L_ab *the moment for a force is FxL\n", + "R_B_Y = (M + R_1*l_2 + R_2*l_3)/L_ab\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "# R_A_Y*L_ab + M - R_1*l_2 - R_2*0.1 = 0\n", + "\n", + "R_A_Y = -(M - R_1*l_2 - R_2*0.1)/L_ab\n", + " \n", + "print \"The X,Y components of reaction force at A is \",R_A_X,\",\",R_A_Y,\"N\"\n", + "print \"The X,Y components of reaction force at B is \",R_B_X,\",\",R_B_Y,\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2 page number 233" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The X,Y components of reaction force at A is 0 , -9.0 N\n", + "The X,Y components of reaction force at B is 0 , 6.0 N\n" + ] + } + ], + "source": [ + "#Given \n", + "P_Max = 10 #N - the maximum distribution in a triangular distribution\n", + "L = 3 #mt the total length of force distribution \n", + "L_X = 5 #mt - the horizantal length of the rod\n", + "#caliculations \n", + "\n", + "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n", + "L_com = 2*L /3 #mt - the resultant force acting as a result of distribution acting position \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = 0 # since there are no forces in X-direction\n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "#F_y*L_com - R_B_Y*L_X = 0\n", + "R_B_Y = F_y*L_com/L_X\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "#- R_A_Y*L_X = F_y*(L_X-L )\n", + "\n", + "R_A_Y = - F_y*L/L_X \n", + "print \"The X,Y components of reaction force at A is \",R_A_X,\",\",R_A_Y,\"N\"\n", + "print \"The X,Y components of reaction force at B is \",R_B_X,\",\",R_B_Y,\"N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3 page number 233 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The X,Y components and resultant of reaction force at A is 4 , 3 , 5.0 N\n", + "The X,Y components and resultant of reaction force at B is 1 , 1 , 1.41 N\n" + ] + } + ], + "source": [ + "#given\n", + "F = 5 #K - force acting on the system\n", + "tan = (4/3) # the Tan of the angle of force with x axis\n", + "l_ab = 12 #inch - the total length of ab \n", + "l = 3 # inch - Distance from 'a'\n", + "#caliculation\n", + "F_X = 4 #K\n", + "F_Y = 3 #k\n", + "\n", + "#M_A = 0 momentum at point a is zero\n", + "# F_X*l- R_B_Y*l_ab = 0 \n", + "R_B_Y = F_X*l/l_ab\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "# R_A_Y*l_ab - F_X*(l_ab - l)\n", + "R_A_Y = F_X*(l_ab - l)/l_ab\n", + " \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = F_Y + R_B_Y \n", + "R_B_X = R_B_Y # since the angle is 45 degrees\n", + "\n", + "#resultants \n", + "R_A = pow(R_A_X**2 + R_A_Y**2,0.5)\n", + "R_B = pow(R_B_X**2 + R_B_Y**2,0.5)\n", + "\n", + "print \"The X,Y components and resultant of reaction force at A is \",R_A_X,\",\",R_A_Y,\",\",R_A,\"N\"\n", + "print \"The X,Y components and resultant of reaction force at B is \",R_B_X,\",\",R_B_Y,\",\",round(R_B,2),\"N\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4 page number 239" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n", + "The force and moment in section b--b are 6.0 KN , 6.0 KN-m\n" + ] + } + ], + "source": [ + "#Given \n", + "P_Max = 10 #N - the maximum distribution in a triangular distribution\n", + "L = 3 #mt the total length of force distribution \n", + "L_X = 5 #mt - the horizantal length of the rod\n", + "#caliculations \n", + "\n", + "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n", + "L_com = 2*L /3 #mt - the resultant force acting as a result of distribution acting position \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = 0 # since there are no forces in X-direction\n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "#F_y*L_com - R_B_Y*L_X = 0\n", + "R_B_Y = F_y*L_com/L_X\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "#- R_A_Y*L_X = F_y*(L_X-L )\n", + "\n", + "R_A_Y = - F_y*L/L_X\n", + "\n", + "#For a---a section \n", + "l_a = 2 #mt - a---a section from a \n", + "l_com_a = 2*l_a/3\n", + "v_a = R_A_Y + 0.5*l_a*(10.0*2/3) #*(10*2/3) because the maximum moves\n", + "\n", + "M_a = (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a \n", + "\n", + "#For b---b section \n", + "\n", + "v_b = F_y + R_A_Y #equilabrium conditions\n", + "M_b = (F_y + R_A_Y)*(-1)\n", + "\n", + "print \"The force and moment in section a--a are\",round(v_a,2),\"KN ,\",M_a,\"KN-m\"\n", + "print \"The force and moment in section b--b are\",v_b,\"KN ,\",M_b,\"KN-m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5 page number 241" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given \n", + "#Lets divide the section into two sections \n", + "l_ac = 10 # ft -The total length of the rod\n", + "R = 5 #k - The applies force at c\n", + "tan = 4/3 # The tan of the angle of the force \n", + "l_ab = 5 #ft - The distance of applied force from A\n", + "R_y = 4 #k,downwards X- component of the force\n", + "R_X = 3 #k Y- component of the force , lets consider only Y direction since we are concentrating on the Shears \n", + "\n", + "#F_Y = 0 forces in Y directions\n", + "#R_A +R_B = R_y\n", + "#M_c = 0 making the moment zero at point c \n", + "#Caliculations \n", + "# R_A= R_B*(l_ac - l_ab)/(l_ab) so R_A = R_B\n", + "\n", + "R_A = R_y/2 #F_Y = 0\n", + "R_B = R_y/2\n", + "#considering section x--x\n", + "l_x = 2 #ft - length of section from A\n", + "v_x = R_A #k ,F_X = 0 \n", + "M_x = R_A*l_x #k-ft M_c = 0\n", + "\n", + "#considering section at midpoint t--t\n", + "l_t = 2 #ft - length of section from A\n", + "v_t = 0 #k ,F_X = 0 \n", + "M_t = (R_A)*l_t #k-ft M_c = 0\n", + "\n", + "##considering section y---y\n", + "l_y = 2 #ft - length of section from B\n", + "v_y = - R_B #k ,F_X = 0 \n", + "M_y = R_B*l_y #k-ft M_c = 0\n", + "\n", + "#Graph\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,2,4.9999999999999,5,5.00000000000000001,7,10] # For graph precision \n", + "\n", + "V = [R_A,v_x,v_x,v_t,v_y,v_y,-R_B];\t\t\t#Shear matrix\n", + "M = [0,M_x,M_t,M_t,M_t,M_y,0];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t#Shear diagram\n", + "plot(X,M,'r');\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6 pagenumber 243 " + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "l = 1 #L - Length of the cantilever \n", + "F_app = ((2**0.5))*2 #p - force applies \n", + "tan = 1 # The angle of force applied\n", + "F_app_x = F_app/((2**0.5)) #p The horizantal component of the force , neglected \n", + "F_app_y = F_app/((2**0.5)) #p The Vertical component of the force \n", + "#F_Y = 0 \n", + "R_A = 1 #p\n", + "\n", + "#Considering section 1-----1\n", + "l_1 = 0.5 # The length of the section from one end\n", + "v_1 = R_A #F_Y = 0\n", + "M_1 = -R_A*l_1 #MAking moment at section 1 = 0\n", + "\n", + "#considering end of cantilever\n", + "l_2 = 1 # The length of the section from one end\n", + "v_2 = R_A #F_Y = 0\n", + "M_2 = -R_A*l_2#MAking moment at section 1 = 0\n", + "\n", + "#Graph\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,0.5,1] # For graph precision \n", + "\n", + "V = [R_A,v_1,v_2 ];\t\t\t#Shear matrix\n", + "M = [0,M_1,M_2];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t #Shear diagram\n", + "plot(X,M,'r');\t\t\t #Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7 page number 243" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "l_ab = 3 #L - The total length lets say '3L'\n", + "R_1 = 1 #p - The force applied at b\n", + "R_2 = 1 #p - The force applied at c\n", + "l_ab = 1 #L\n", + "l_bc = 1 #L \n", + "\n", + "#Logical step \n", + "#Since the system is in symmetry we can avoid moment M = 0 caliculations\n", + "\n", + "#F_Y = 0 \n", + "R_A = (R_1 + R_2)/2\n", + "R_B = (R_1 + R_2)/2\n", + "\n", + "#Lets take '3' sections \n", + "#Considering section 1-----1 at 0.5L\n", + "l_1 = 0.5 #L - distance of the section from the A\n", + "v_1 = R_A #F_Y = 0 \n", + "M_1 = R_A*l_1 #MAking moment at section 1 = 0\n", + "\n", + "#Considering section 2-----2 at 1L\n", + "l_2 = 1 #L - distance of the section from the A\n", + "v_2 = R_A #F_Y = 0 \n", + "M_2 = R_A*l_2 #MAking moment at section 2 = 0\n", + "\n", + "#Considering section 3-----3 at 1.5L\n", + "l_3 = 1.5 #L - distance of the section from the A\n", + "v_3 = 0 #F_Y = 0 \n", + "M_3 = R_A*l_2 #MAking moment at section 2 = 0 and symmetry \n", + "\n", + "#GRAPH\n", + "#Since the symmetry exists the graphs are also symmetry\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,0.5,1,1.000000001,1.5,1.999999999999,2,2.5,3] # For graph precision \n", + "\n", + "V = [R_A,v_1,v_2,v_3,v_3,v_3,-v_2,-v_1,-R_B];\t\t\t#Shear matrix\n", + "M = [0,M_1,M_2,1,1,1,M_2,M_1,0];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t#Shear diagram\n", + "plot(X,M);\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8 page nmber 244" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.5\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given \n", + "import numpy as np\n", + "l_ab = 1.0 #L - The length of the beam\n", + "F_D = 1.0 #W - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "\n", + "#considering many sections \n", + "\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2 #M = 0 in the section\n", + "print R_A\n", + "#Graphs\n", + "import numpy as np\n", + "values = [0.5,0,-0.5]\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,3)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "\n", + "import numpy as np\n", + "values = M_1\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.9 page number 245" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n" + ] + } + ], + "source": [ + "#Given \n", + "P_Max = 10 #N - the maximum distribution in a triangular distribution\n", + "L = 3 #mt the total length of force distribution \n", + "L_X = 5 #mt - the horizantal length of the rod\n", + "#caliculations \n", + "\n", + "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n", + "L_com = 2*L /3 #mt - the resultant force acting as a result of distribution acting position \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = 0 # since there are no forces in X-direction\n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "#F_y*L_com - R_B_Y*L_X = 0\n", + "R_B_Y = F_y*L_com/L_X\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "#- R_A_Y*L_X = F_y*(L_X-L )\n", + "\n", + "R_A_Y = - F_y*L/L_X\n", + "\n", + "#caliculating for some random value\n", + "#For a---a section \n", + "l_a = 2 #mt - a---a section from a \n", + "l_com_a = 2*l_a/3\n", + "v_a = R_A_Y + 0.5*l_a*(10.0*2/3) #*(10*2/3) because the maximum moves\n", + "\n", + "M_a = (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a\n", + "\n", + "print \"The force and moment in section a--a are\",round(v_a,2),\"KN ,\",M_a,\"KN-m\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.13 page number 254" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "l_ab = 4 #L - The total length lets say '3L'\n", + "R_1 = 1 #p - The force applied at b\n", + "R_2 = 1 #p - The force applied at c\n", + "l_ab = 1 #L\n", + "l_bc = 3 #L \n", + "\n", + "#Logical step \n", + "#Since the system is in symmetry we can avoid moment M = 0 caliculations\n", + "\n", + "#F_Y = 0 \n", + "R_A = (R_1 + R_2)/2\n", + "R_B = (R_1 + R_2)/2\n", + "\n", + "#Lets take '3' sections \n", + "#Considering section 1-----1 at 0.5L\n", + "l_1 = 0.5 #L - distance of the section from the A\n", + "v_1 = R_A #F_Y = 0 \n", + "M_1 = R_A*l_1 #MAking moment at section 1 = 0\n", + "\n", + "#Considering section 2-----2 at 1L\n", + "l_2 = 1 #L - distance of the section from the A\n", + "v_2 = R_A #F_Y = 0 \n", + "M_2 = R_A*l_2 #MAking moment at section 2 = 0\n", + "\n", + "#Considering section 3-----3 at 1.5L\n", + "l_3 = 3 #L - distance of the section from the A\n", + "v_3 = 0 #F_Y = 0 \n", + "M_3 = R_A*l_2 #MAking moment at section 2 = 0 and symmetry \n", + "\n", + "#GRAPH\n", + "#Since the symmetry exists the graphs are also symmetry\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,0.5,1,1.0000001,2,2.9999999999,3,3.5,4] # For graph precision \n", + "\n", + "V = [R_A,v_1,v_2,v_3,v_3,v_3,-v_2,-v_1,-R_B];\t\t\t#Shear matrix\n", + "M = [0,M_1,M_2,M_3,M_3,M_3,M_2,M_1,0];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t#Shear diagram\n", + "plot(X,M);\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.14 page number 255" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "l = 1.0 #l - The length of the beam\n", + "p = 1.0 #W - The total load applied\n", + "#since it is triangular distribution \n", + "l_com = 0.66*l#l - The distance of force of action from one end\n", + "#F_Y = 0\n", + "#R_A + R_B = p\n", + "#M_a = 0 Implies that R_B = 2*R_A\n", + "R_A = p/3.0\n", + "R_B = 2.0*p/3\n", + "\n", + "#Taking Many sections \n", + "\n", + "#Section 1----1\n", + "l = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = p*(l[i]**2) - p/3.0\n", + " M[i] = p*(l[i]**3)/(3.0)- p*l[i]/3.0\n", + "\n", + "v[10] = R_B #again concluded Because the value is tearing of \n", + "\n", + "\n", + "#Graph\n", + "values = M\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "%matplotlib inline\n", + "values = v\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.16 page number 259" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.166666666667\n", + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "l = 6.0 #a - length of the rod\n", + "F = 1.0 #p - force applies in x direction \n", + "d = 1.0 #a \n", + "M = 1.0 #pa - torque applies on the rod\n", + "l_ab = 4.0 #a application of torque point from A\n", + "#M = 0 implies that\n", + "R_A = F/6.0 #p - The reaction at A\n", + "R_B = - R_A #F_Y = 0\n", + "\n", + "#Caliculations \n", + "\n", + "#Taking sections \n", + "#Section 1---1\n", + "l_1 = 1 #a - the length of the section \n", + "M_1 = - R_A*l_1 #M = 0\n", + "\n", + "#Section 2---2\n", + "l_2 = 4 #a - the length of the section \n", + "M_2 = - R_A*l_2 #M = 0\n", + "\n", + "l_4 = 2 #a - the length of the section \n", + "M_4 = 1/3.0 #pa #M = 0 '-M' because there is moment couple in between\n", + "\n", + "\n", + "#Section 3---3\n", + "l_3 = 1 #a - the length of the section \n", + "M_3 = 1/6.0#pa M = 0 '-M' because there is moment couple in between\n", + "print R_A\n", + "\n", + "#GRAPH\n", + "#Since the symmetry exists the graphs are also symmetry\n", + "%matplotlib inline\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,1,4,4.00001,5,6] # For graph precision \n", + "M = [0,M_1,M_2,M_4,M_3,0];\t\t\t#Bending moment matrix\n", + "plot(X,M);\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_6.ipynb new file mode 100644 index 00000000..6b74082c --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_6.ipynb @@ -0,0 +1,570 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6:Pure Bending and Bending with Axial force " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 page number 293" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The moment of inertia of total system is 655000.0 mm4\n" + ] + } + ], + "source": [ + "#Given \n", + "#Entire area - hallow area\n", + "l_e = 60.0 #mm - length of the entire area\n", + "b_e = 40 #mm - width of the entire area\n", + "l_h = 30 #mm - length of the hallow area\n", + "b_h = 20 #mm - width of the hallow area\n", + "A_e = l_e*b_e #mm2 - The entire area\n", + "A_h = -l_h*b_h #mm2 - The hallow area '-' because its hallow\n", + "A_re = A_e + A_h #mm2 resultant area\n", + "y_e = l_e/2 # mm com from bottom \n", + "y_h = 20+l_h/2 #mm com from bottom \n", + "y_com = (A_e*y_e + A_h*y_h)/A_re \n", + "#moment of inertia caliculatins - bh3/12 +ad2\n", + "I_e = b_e*(l_e**3)/12 + A_e*((y_e-y_com)**2) #Parallel axis theorm\n", + "I_h = b_h*(l_h**3)/12 - A_h*((y_h-y_com)**2) #Parallel axis theorm\n", + "I_total = I_e - I_h\n", + "print \"The moment of inertia of total system is \",I_total,\"mm4\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 page number 295" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress at 2 mt is 4.81 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 400 #mm - Length \n", + "b = 300 #mm - breath \n", + "F = 20 #KN _ the force applied on the beam \n", + "F_d = 0.75 #KN-m - The force distribution \n", + "d = 2 #mt - the point of interest from the free end\n", + "#caliculations \n", + "#From moment diagram\n", + "M = F*d - F_d*d*1\n", + "I = b*(l**3)/12 #mm4 - Bending moment diagram \n", + "c = l/2 # the stress max at this C\n", + "S = I/c #The maximum shear stress \n", + "shear_max = M*(10**6)/S #MPA - the maximum stress \n", + "print \"The maximum stress at 2 mt is\",round(shear_max,2),\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 pagr number 297" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum tensile stress 16.0 Ksi\n", + "The maximum compressive stress 21.6 Ksi\n" + ] + } + ], + "source": [ + "#Given \n", + "#We will divide this into three parts\n", + "F = 8 #k - force applied\n", + "d = 16 #inch -distance\n", + "l_1 = 1 #in \n", + "l_2 = 3 #in \n", + "b_1 = 4 #in \n", + "b_2 = 1 #in\n", + "A_1 = l_1* b_1 #in2 - area of part_1\n", + "y_1 = 0.5 #in com distance from ab\n", + "A_2 =l_2*b_2 #in2 - area of part_1\n", + "y_2 = 2.5 #in com distance from ab\n", + "A_3 = l_2*b_2 #in2 - area of part_1\n", + "y_3 = 2.5 #in com distance from ab\n", + "\n", + "y_net = (A_1*y_1 +A_2*y_2 + A_3*y_3)/(A_1+A_2+A_3) #in - The com of the whole system\n", + "c_max = (4-y_net) #in - The maximum distace from com to end\n", + "c_min = y_net #in - the minimum distance from com to end\n", + "I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n", + "I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n", + "I_3 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n", + "I_net = I_1 + I_2 + I_3 #in4 - the total moment of inertia\n", + "M_c = F*d*c_max \n", + "stress_cmax = M_c/I_net #Ksi - The maximum compressive stress\n", + "\n", + "M_t= F*d*c_min \n", + "stress_tmax = M_t/I_net #Ksi - The maximum tensile stress\n", + "print \"The maximum tensile stress\",stress_tmax ,\"Ksi\"\n", + "print \"The maximum compressive stress\",round(stress_cmax,1) ,\"Ksi\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 page number 303" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in steel 11.49 Mpa\n", + "The maximum stress in wood 97.09 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "#Given \n", + "#We will divide this into two parts\n", + "E_w = 10.0 #Gpa - Youngs modulus of wood\n", + "E_s = 200.0 #Gpa - Youngs modulus of steel\n", + "M = 30.0 #K.N-m _ applied bending moment \n", + "n = E_s/E_w \n", + "l_1 = 250 #mm \n", + "l_2 = 10 #mm\n", + "b_1 = 150.0 #mm\n", + "b_2 = 150.0*n #mm\n", + "A_1 = l_1* b_1 #mm2 - area of part_1\n", + "y_1 = 125.0 #mm com distance from top\n", + "A_2 =l_2*b_2 #mm2 - area of part_1\n", + "y_2 = 255.0 #mm com distance from top\n", + "y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system from top\n", + "I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n", + "I_2 = b_2*(l_2**3)/12.0 + A_2*((y_2-y_net)**2)\n", + "I_net = I_1 + I_2 #mm4 - the total moment of inertia\n", + "c_s= y_net # The maximum distance in steel \n", + "stress_steel = M*(10.0**6)*c_s/I_net #Mpa - The maximum stress in steel \n", + "\n", + "c_w= l_1+l_2-y_net # The maximum distance in wood \n", + "stress_wood = n*M*(10.0**6)*c_w/I_net #MPa - The maximum stress in wood \n", + "\n", + "print \"The maximum stress in steel \",round(stress_steel,2) ,\"Mpa\"\n", + "print \"The maximum stress in wood\",round(stress_wood,2) ,\"Mpa\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9 page number 305" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in concrete 834.07 psi\n", + "The stress in steel 17427.61 psi\n" + ] + } + ], + "source": [ + "#Given \n", + "M = 50000 #ft-lb , positive bending moment applied\n", + "N = 9 # number of steel bars \n", + "n = 15 # The ratio of steel to concrete \n", + "A_s = 30 #in2 area of steel in concrete\n", + "#(10*y)*(y/2) = 30*(20-y)\n", + "#y**2 + 6*y -120\n", + "#solving quadractic equation \n", + "import math\n", + "\n", + "a = 1\n", + "b = 6\n", + "c = -120\n", + "# calculate the discriminant\n", + "d = (b**2) - (4*a*c)\n", + "\n", + "# find two solutions\n", + "sol1 = (-b-math.sqrt(d))/(2*a)\n", + "sol2 = (-b+math.sqrt(d))/(2*a)\n", + "y = sol2 # Nuetral axis is found\n", + "l_1 = y #in- the concrete below nuetral axis is not considered\n", + "b_1 = 10 #in - width\n", + "A_1 = l_1* b_1 #in2 - area of concrete\n", + "y_1 = y/2 #in com of the concrete \n", + "y_2 = 20-y #in com of the transformed steel \n", + "I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y)**2) #in4 parallel axis theorm\n", + "I_2 = A_s*((y_2)**2) #in4 first part is neglected\n", + "I_net = I_1 + I_2 #in4 - the total moment of inertia\n", + "c_c= y #in The maximum distance in concrete \n", + "stress_concrete = M*12*c_c/I_net #psi - The maximum stress in concrete \n", + "c_s= 20- y \n", + "stress_steel =n*M*12*c_s/I_net #psi - The maximum stress in concrete \n", + "print \"The maximum stress in concrete \",round(stress_concrete,2) ,\"psi\"\n", + "print \"The stress in steel\",round(stress_steel,2) ,\"psi\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 6.10 page number 309" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress upward in straight case is 99.984 Mpa\n", + "The maximum stress downward in straight case is -99.984 Mpa\n", + "The maximum stress upward in curved case is 107.093207632 Mpa\n", + "The maximum stress downward in curved case is -93.6813516989 Mpa\n", + "The maximum stress upward in curved case2 is 128.733538525 Mpa\n", + "The maximum stress downward in curved case2 is -81.0307692623 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 50.0 #mm - the length of the beam \n", + "b = 50.0 #mm - the width of the beam\n", + "M = 2083 #Nm\n", + "A = l*b #mm2 - The area \n", + "#straight beam \n", + "I = b*(l**3)/12.0 #mm4 - The moment of inertia of the beam\n", + "c_1= l/2 # the distance where the stress is maximum \n", + "c_2 = -l/2 # the distance where the stress is maximum \n", + "s_1 = I/c_1\n", + "s_2 = I/c_2\n", + "stress_max_1 = M*(10**3)/s_1 #Mpa - the maximum strss recorded in the crossection\n", + "stress_max_2 = M*(10**3)/s_2 #Mpa - the maximum strss recorded in the crossection \n", + "print \"The maximum stress upward in straight case is\",stress_max_1,\"Mpa\"\n", + "print \"The maximum stress downward in straight case is\",stress_max_2,\"Mpa\"\n", + "\n", + "#curved beam \n", + "import math\n", + "r = 250.0 #mm Radius of beam curved \n", + "r_0 = r - l/2 # inner radius \n", + "r_1 = r + l/2 # outer radius\n", + "R = l/(math.log(r_1/r_0)) #mm \n", + "e = r - R \n", + "stressr_max_1 = M*(10**3)*(R-r_0)/(r_0*A*e)\n", + "stressr_max_2 = M*(10**3)*(R-r_1)/(r_1*A*e)\n", + "print \"The maximum stress upward in curved case is\",stressr_max_1,\"Mpa\"\n", + "print \"The maximum stress downward in curved case is\",stressr_max_2,\"Mpa\"\n", + "\n", + "#curved beam _2 \n", + "import math\n", + "r = 75.0 #mm Radius of beam curved \n", + "r_0 = r - l/2 # inner radius \n", + "r_1 = r + l/2 # outer radius\n", + "R = l/(math.log(r_1/r_0)) #mm \n", + "e = r - R \n", + "stressr_max_1 = M*(10**3)*(R-r_0)/(r_0*A*e)\n", + "stressr_max_2 = M*(10**3)*(R-r_1)/(r_1*A*e)\n", + "print \"The maximum stress upward in curved case2 is\",stressr_max_1,\"Mpa\"\n", + "print \"The maximum stress downward in curved case2 is\",stressr_max_2,\"Mpa\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page number 6.14 page number 318" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The plastic moment of the system is 120784.313725 ft-lb\n" + ] + } + ], + "source": [ + "#given \n", + "#from example 6.9\n", + "St_ul = 2500 #psi - ultimate strength\n", + "st_yl = 40000 #psi _ yielding strength \n", + "b = 10 #in - width from example \n", + "A = 2 #in2 The area of the steel\n", + "d = 20 \n", + "t_ul = st_yl*A #ultimate capasity\n", + "y = t_ul/(St_ul*b*0.85) #in 0.85 because its customary\n", + "M_ul = t_ul*(d-y/2)/12 #ft-lb Plastic moment \n", + "print \"The plastic moment of the system is \",M_ul,\"ft-lb\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.15 page number 231" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angle at which nuetral axis locates is 0.0226547191205 radians\n" + ] + } + ], + "source": [ + "#Given \n", + "#From example 5.8 \n", + "W = 4.0 #N/m - The force distribution \n", + "L = 3 # m - The length of the force applied\n", + "M = W*L/8.0 # KN.m The moment due to force distribution\n", + "o = 30 # the angle of force applid to horizantal\n", + "l = 150.0 #mm length of the crossection \n", + "b = 100.0 #mm - width of the crossection \n", + "import math \n", + "M_z = M*(math.cos(3.14/6))\n", + "M_y = M*(math.sin(math.pi/6))\n", + "I_z = b*(l**3)/12.0\n", + "I_y = l*(b**3)/12.0\n", + "#tanb = I_z /I_y *tan30\n", + "b = math.atan(math.radians(I_z*math.tan(3.14/6.0)/I_y ))\n", + "print \"The angle at which nuetral axis locates is\",b,\"radians\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.16 pagenumber 323" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum tensile stress -76.1 Mpa\n", + "The maximum compressive stress 67.73 Mpa\n" + ] + } + ], + "source": [ + "import math\n", + "M = 10 #KN.m - The moment applied\n", + "I_max = 23.95*(10**6) #mm4 - I_z The moment of inertia\n", + "I_min = 2.53*(10**6) #mm4 - I_y The moment of inertia\n", + "o = 14.34 # degress the principle axis rotated\n", + "#Coponents of M in Y,Z direction \n", + "M_z = M*(10**6)*math.cos(math.radians(o))\n", + "M_y = M*(10**6)*math.sin(math.radians(o))\n", + "#tanb = I_z /I_y *tan14.34\n", + "b = math.atan((I_max*math.tan(math.radians(o))/I_min ))\n", + "B = math.degrees(b) \n", + "y_p = 122.9 # mm - principle axis Y cordinate\n", + "z_p = -26.95 #mm - principle axis z cordinate\n", + "stress_B = - M_z*y_p/I_max + M_y*z_p/I_min #Mpa - Maximum tensile stress\n", + "y_f = -65.97 # mm - principle axis Y cordinate\n", + "z_f = 41.93 #mm - principle axis z cordinate\n", + "stress_f = - M_z*y_f/I_max + M_y*z_f/I_min #Mpa - Maximum compressive stress\n", + "print \"The maximum tensile stress\",round(stress_B,2) ,\"Mpa\"\n", + "print \"The maximum compressive stress\",round(stress_f,2),\"Mpa\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.18 page number 328" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in the beam 3.332 Mpa \n" + ] + } + ], + "source": [ + "l = 50 #mm - The length of the beam \n", + "b = 50 #mm - The width of the beam \n", + "A = l*b #mm2 - The area of the beam \n", + "p = 8.33 #KN - The force applied on the beam \n", + "stress_max = p*(10**3)/A #Mpa After cutting section A--b\n", + "print \"The maximum stress in the beam\",stress_max ,\"Mpa \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.24 page number 339" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "By sketching the line with angle 53.2 degrees The farthest point associated with B and F\n" + ] + } + ], + "source": [ + "import math\n", + "M = 10 #KN.m - The moment applied\n", + "I_max = 23.95*(10**6) #mm4 - I_z The moment of inertia\n", + "I_min = 2.53*(10**6) #mm4 - I_y The moment of inertia\n", + "o = 14.34 # degress the principle axis rotated\n", + "#Coponents of M in Y,Z direction \n", + "M_z = M*(10**6)*math.cos(math.radians(o))\n", + "M_y = M*(10**6)*math.sin(math.radians(o))\n", + "#tanb = I_z /I_y *tan14.34\n", + "b = math.atan((I_max*math.tan(math.radians(o))/I_min ))\n", + "B = math.degrees(b) \n", + "y_p = 122.9 # mm - principle axis Y cordinate\n", + "z_p = -26.95 #mm - principle axis z cordinate\n", + "stress_B = - M_z*y_p/I_max + M_y*z_p/I_min #Mpa - Maximum tensile stress\n", + "y_f = -65.97 # mm - principle axis Y cordinate\n", + "z_f = 41.93 #mm - principle axis z cordinate\n", + "stress_f = - M_z*y_f/I_max + M_y*z_f/I_min #Mpa - Maximum compressive stress\n", + "#location of nuetral axis To show these stresses are max and minimum \n", + "#tanB = MzI_z + MzI_yz/MyI_y +M_YI_yz\n", + "I_z = 22.64 *(10**6) #mm4 moment of inertia in Z direction\n", + "I_y = 3.84 *(10**6) #mm4 moment of inertia in Y direction\n", + "I_yz =5.14 *(10**6) #mm4 moment of inertia in YZ direction \n", + "M_y = M #KN.m bending moment in Y dorection \n", + "M_z = M #KN.m bending moment in Y dorection \n", + "B = math.atan(( M_z*I_yz)/(M_z*I_y )) #radians location on neutral axis\n", + "beta = math.degrees(B)\n", + "print \"By sketching the line with angle\",round(beta,1),\"degrees The farthest point associated with B and F\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_6.ipynb new file mode 100644 index 00000000..cec54f64 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_6.ipynb @@ -0,0 +1,257 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 7:Shear stress in Beams and Related Problems " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1 page number 365" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimal space between the nails 42.0 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "shear_v = 3000 #N - Transmitted vetical shear \n", + "shear_al = 700 #N - The maximum allowable \n", + "#We will divide this into two parts\n", + "l_1 = 50.0 #mm \n", + "l_2 = 200.0 #mm \n", + "b_1 = 200.0 #mm \n", + "b_2 = 50.0 #mm\n", + "A_1 = l_1* b_1 #mm2 - area of part_1\n", + "y_1 = 25.0 #mm com distance \n", + "A_2 =l_2*b_2 #mm2 - area of part_1\n", + "y_2 = 150.0 #in com distance \n", + "y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system\n", + "c_max = (4-y_net) #mm - The maximum distace from com to end\n", + "c_min = y_net #mm - the minimum distance from com to end\n", + "I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n", + "I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n", + "I_net = I_1 + I_2 #mm4 - the total moment of inertia\n", + "Q = A_1*(-y_1+y_net) #mm3\n", + "q = shear_v*Q/I_net #N/mm - Shear flow\n", + "d = shear_al/q # The space between the nails \n", + "print \"The minimal space between the nails \",round(d,0) ,\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2 pagenumber 365" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimal space between the nails 123.0 mm\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 6 #m -length of the beam \n", + "p = 3 #KN-m _ the load applied\n", + "R_a = l*p/2 #KN -The reaction at a, Since the system is symmetry \n", + "R_b = l*p/2 #KN -The reaction at b \n", + "l_s = 10 #mm - The length of the screw \n", + "shear_al = 2 #KN - The maximum load the screw can take \n", + "I = 2.36*(10**9) #mm2 The moment of inertia of the whole system\n", + "#We will divide this into two parts\n", + "l_1 = 50.0 #mm \n", + "l_2 = 50.0 #mm \n", + "b_1 = 100.0 #mm \n", + "b_2 = 200.0 #mm\n", + "A_1 = l_1* b_1 #in2 - area of part_1\n", + "y_1 = 200.0 #mm com distance \n", + "A_2 =l_2*b_2 #mm2 - area of part_1\n", + "y_2 = 225.0 #in com distance\n", + "Q = 2*A_1*y_1 + A_2*y_2 # mm3 For the whole system\n", + "q = R_a*Q*(10**3)/I #N/mm The shear flow \n", + "d = shear_al*(10**3)/q #mm The space between the nails\n", + "print \"The minimal space between the nails \",round(d,0),\"mm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6 page number 376" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The shear centre from outside vertical face is 1.825 in\n" + ] + } + ], + "source": [ + "#Given\n", + "#we will divide this into two equal parts and other part\n", + "l = 10.0 # in - The height \n", + "t = 0.1 # in - The width\n", + "b = 5.0 #mm- The width of the above part \n", + "A = t* b #in2 - area of part\n", + "y_net = l/2 # The com of the system \n", + "y_1 = l # The position of teh com of part_2\n", + "I_1 = t*(l**3)/12 #in4 The moment of inertia of part 1\n", + "I_2 = 2*A*((y_1-y_net)**2) #in4 The moment of inertia of part 2 \n", + "I = I_1 + I_2 #in4 The moment of inertia \n", + "e = (b**2)*(l**2)*t/(4*I) #in the formula of channels\n", + "l_sc = e - t/2 #in- The shear centre \n", + "print \"The shear centre from outside vertical face is \",l_sc ,\"in\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8 page number 387" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The direct maximum stress 4.25 Mpa\n", + "The torsion maximum stress 101.91 Mpa\n", + "The total stress 106.16 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia = 10.0 #mm - The diameter of the cylinder \n", + "c = dia/2 #mm - the radius of the cylinder \n", + "A = 3.14*(c**2) #mm2 The area of the crossection \n", + "y = 4*c/(3*3.14) #mm The com of cylinder \n", + "I = 3.14*(c**4)/4 #mm4 - The moment of inertia of the cylinder\n", + "j = 3.14*(dia**4)/32 #mm4\n", + "T = 20.0 #N.m - The torque \n", + "V = 250.0 #N - The shear \n", + "M = 25.0 #N-m The bending moment \n", + "Q = A*y/2 #mm\n", + "stress_dmax = 4*V/(3*A) #V*Q/(I*d) #Mpa The direct maximum stress\n", + "stress_tmax = T*c*(10**3)/j #-Mpa The torsion maximum stress\n", + "stress_total = stress_dmax + stress_tmax #Mpa The total stress\n", + "print \"The direct maximum stress\",round(stress_dmax,2),\"Mpa\"\n", + "print \"The torsion maximum stress\",round(stress_tmax,2),\"Mpa\"\n", + "print \"The total stress\",round(stress_total,2),\"Mpa\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9 page number 393" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in the system 4.84 Mpa\n" + ] + } + ], + "source": [ + "#Given\n", + "dia = 15 #mm - The diameter of the rod\n", + "h = 0.5 #mt - The freely falling height \n", + "A = 3.14*(dia**2)/4 #mm2 The area of the crossection\n", + "E = 200 #Gpa -Youngs modulus\n", + "L = 750 #mm - The total length of the rod\n", + "G = 80 #gpa - Shear modulus \n", + "N = 10 #number of live coils\n", + "d = 5 #mm the diameter of live coil \n", + "m = 3 # the mass of freely falling body\n", + "H = 500 #mm -from mass to spring \n", + "F= m*9.81 #Kg the force due to that mass\n", + "p = 3 #KN-m _ the load applied\n", + "#e = e_rod + e_spr\n", + "#e_rod\n", + "e_rod = p*L*(10**-3)/(A*E) #mm The elongation due to freely falling body\n", + "#e_spr\n", + "e_spr = 64*F*(dia**3)*N*(10**-3)/(G*(d**4)) #mm The elongation due to spring\n", + "e = e_rod + e_spr #mm The total elongation \n", + "p_dyn =F*(1+pow((1+(2*H/e)),0.5))\n", + "Stress_max = p_dyn/A #MPa - The maximum stress in the system \n", + "print \"The maximum stress in the system \",round(Stress_max,2),\"Mpa\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_6.ipynb new file mode 100644 index 00000000..4066d756 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_6.ipynb @@ -0,0 +1,345 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8:Transformation of stress and strain and Yield and Fracture criteria " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1 page number 405 " + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The stress action in normal direction on AB 1.29 Mpa\n", + "The stress action in tangential direction on AB 2.12 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "import math \n", + "from math import radians\n", + "o = 22.5 #degrees , The angle of infetisimal wedge \n", + "A = 1 #mm2 The area of the element \n", + "A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB\n", + "A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC\n", + "S_1 = 3 #MN The stresses applying on the element \n", + "S_2 = 2 #MN\n", + "S_3 = 2 #MN\n", + "S_4 = 1 #MN \n", + "F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas \n", + "F_2 = S_2*A_ab\n", + "F_3 = S_3*A_bc\n", + "F_4 = S_4*A_bc\n", + "#sum of F_N = 0 equilibrim in normal direction \n", + "N = (F_1-F_3)*(math.cos(radians(o))) + (F_4 - F_2)*(math.sin(radians(o)))\n", + "\n", + "#sum of F_s = 0 equilibrim in tangential direction \n", + "\n", + "S = (F_2-F_4)*(math.cos(radians(o))) + (F_1 - F_3)*(math.sin(radians(o)))\n", + "\n", + "Stress_Normal = N/A #Mpa - The stress action in normal direction on AB\n", + "Stress_tan = S/A #Mpa - The stress action in tangential direction on AB\n", + "print \"The stress action in normal direction on AB\",round(Stress_Normal,2),\"Mpa\"\n", + "print \"The stress action in tangential direction on AB\",round(Stress_tan,2),\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2 page number 413" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The stress action in normal direction on AB 4.12 Mpa\n", + "a) The stress action in tangential direction on AB 0.71 Mpa\n", + "b) The principle stress 4.2 Mpa tension\n", + "b) The principle stress -0.06 Mpa compression\n", + "b) The principle plane angles are 32.0 , 122.0 degrees\n", + "c) The maximum shear is -2.24 Mpa\n", + "a) [ 4.2 -0.1 0. ] Mpa\n", + "b) [ 2. -2.24 -2.24 2. ] Mpa\n" + ] + } + ], + "source": [ + "#Given\n", + "o = -22.5 #degrees , The angle of infetisimal wedge \n", + "A = 1 #mm2 The area of the element \n", + "import math \n", + "from math import radians\n", + "from numpy import array\n", + "A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB\n", + "A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC\n", + "S_1 = 3.0 #MN The stresses applying on the element \n", + "S_2 = 2.0 #MN\n", + "S_3 = 2.0 #MN\n", + "S_4 = 1.0 #MN\n", + "#Caliculations \n", + "\n", + "F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas \n", + "F_2 = S_2*A_ab\n", + "F_3 = S_3*A_bc\n", + "F_4 = S_4*A_bc\n", + "#sum of F_N = 0 equilibrim in normal direction \n", + "N = (F_1-F_3)*(math.cos(radians(o))) + (F_4 - F_2)*(math.sin(radians(o)))\n", + "\n", + "#sum of F_s = 0 equilibrim in tangential direction \n", + "\n", + "S = (F_2-F_4)*(math.cos(radians(o))) + (F_1 - F_3)*(math.sin(radians(o)))\n", + "\n", + "Stress_Normal = N/A #Mpa - The stress action in normal direction on AB\n", + "Stress_tan = S/A #Mpa - The stress action in tangential direction on AB\n", + "print \"a) The stress action in normal direction on AB\",round(Stress_Normal,2),\"Mpa\"\n", + "print \"a) The stress action in tangential direction on AB\",round(Stress_tan,2),\"Mpa\"\n", + "\n", + "#Part- b\n", + "\n", + "S_max = (S_4+S_1)/2 + (((((S_4-S_1)/2)**2) + S_3**2)**0.5) #Mpa - The maximum stress\n", + "S_min = (S_4+S_1)/2.0 - (((((S_4-S_1/2))**2) + S_3**2)**0.5) #Mpa - The minumum stress\n", + "k = 0.5*math.atan(S_3/((S_1-S_4)/2)) #radians The angle of principle axis\n", + "k_1 = math.degrees(k)\n", + "k_2 = k_1+90 #The principle plane angles\n", + "print \"b) The principle stress \",round(S_max,1),\"Mpa tension\"\n", + "print \"b) The principle stress \",round(S_min,2),\"Mpa compression\"\n", + "print \"b) The principle plane angles are\",round(k_1,0),\",\",round(k_2,0),\"degrees\"\n", + "\n", + "#part-c\n", + "#The maximum shear stress case\n", + "t_xy = (((((S_4-S_1)/2)**2) + S_3**2)**0.5) #Mpa - The maximum shear stress case\n", + "K = 0.5*math.atan((-(S_1-S_4)/(2*S_3))) #radians The angle of principle axis\n", + "K_0 = math.degrees(K)\n", + "if K_0<0:\n", + " K_1 = K_0+90\n", + "else:\n", + " K_1 = K_0\n", + "K_2 = K_1+90 #PRinciple plain angles\n", + "T_xy = -((S_1-S_4)/2)*(math.sin(radians(2*K_1))) + ((S_4+S_1)/2)*(math.cos(radians(2*K_1))) # Shear stress\n", + "print \"c) The maximum shear is \",round(T_xy,2),\"Mpa\" \n", + "S_mat_a = array([round(S_max,1),round(S_min,1),0]) #MPa maximum stress matrix\n", + "S_mat_b = array([(S_4+S_1)/2,round(T_xy,2),round(T_xy,2),(S_4+S_1)/2]) #MPa maximum stress matrix at maximum shear\n", + "print \"a)\",S_mat_a,\"Mpa\"\n", + "print \"b)\",S_mat_b,\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3 page number 421" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The principle stresses are 6.0 Mpa -4.0 Mpa\n", + "The maximum shear stress 5.0 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "import math \n", + "from math import radians \n", + "S_x = -2 #Mpa _ the noraml stress in x direction\n", + "S_y = 4 #Mpa _ the noraml stress in Y direction\n", + "c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle \n", + "point_x = -2 #The x coordinate of a point on mohr circle\n", + "point_y = 4 #The y coordinate of a point on mohr circle\n", + "Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle\n", + "S_1 = Radius +1#MPa The principle stress\n", + "S_2 = -Radius +1 #Mpa The principle stress\n", + "S_xy_max = Radius #Mpa The maximum shear stress\n", + "print \"The principle stresses are\",S_1 ,\"Mpa\",S_2,\"Mpa\"\n", + "print \"The maximum shear stress\",S_xy_max,\"Mpa\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4 page number 423" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The normal stress on the 221/2 plane 4.82 Mpa\n", + "The tangential stress on the 221/2 plane 1.43 Mpa\n" + ] + } + ], + "source": [ + "#Given\n", + "import math \n", + "S_x = 3.0 #Mpa _ the noraml stress in x direction\n", + "S_y = 1.0 #Mpa _ the noraml stress in Y direction\n", + "c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle \n", + "point_x = 1 #The x coordinate of a point on mohr circle\n", + "point_y = 3 #The y coordinate of a point on mohr circle\n", + "#Caliculations \n", + "\n", + "Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle\n", + "#22.5 degrees line is drawn \n", + "o = 22.5 #degrees \n", + "a = 71.5 - 2*o #Degrees, from diagram \n", + "stress_n = c + Radius*math.sin(math.degrees(o)) #Mpa The normal stress on the plane \n", + "stress_t = Radius*math.cos(math.degrees(o)) #Mpa The tangential stress on the plane\n", + "print \"The normal stress on the 221/2 plane \",round(stress_n,2),\"Mpa\"\n", + "print \"The tangential stress on the 221/2 plane \",round(stress_t,2),\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.7 page number 437" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The principle strains are 400 um/m -600 um/m\n", + "The angle of principle plane 18.43 degrees\n" + ] + } + ], + "source": [ + "import math\n", + "e_x = -500 #10-6 m/m The contraction in X direction\n", + "e_y = 300 #10-6 m/m The contraction in Y direction\n", + "e_xy = -600 #10-6 m/m discorted angle\n", + "centre = (e_x + e_y)/2 #10-6 m/m \n", + "point_x = -500 #The x coordinate of a point on mohr circle\n", + "point_y = 300 #The y coordinate of a point on mohr circle\n", + "Radius = 500 #10-6 m/m - from mohr circle\n", + "e_1 = Radius +centre #MPa The principle strain\n", + "e_2 = -Radius +centre #Mpa The principle strain\n", + "k = math.atan(300.0/900) # from geometry\n", + "k_1 = math.degrees(k)\n", + "print \"The principle strains are\",e_1,\"um/m\",e_2,\"um/m\"\n", + "print \"The angle of principle plane\",round(k_1,2) ,\"degrees\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.8 page number 441" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The principle stresses are 48.35 Mpa -105.49 MPa\n" + ] + } + ], + "source": [ + "#Given\n", + "e_0 = -500 #10-6 m/m \n", + "e_45 = 200 #10-6 m/m \n", + "e_90 = 300 #10-6 m/m\n", + "E = 200 #Gpa - youngs modulus of steel \n", + "v = 0.3 # poissions ratio \n", + "#Caliculations \n", + "\n", + "e_xy = 2*e_45 - (e_0 +e_90 ) #10-6 m/m from equation 8-40 in text\n", + "# from example 8.7\n", + "e_x = -500 #10-6 m/m The contraction in X direction\n", + "e_y = 300 #10-6 m/m The contraction in Y direction\n", + "e_xy = -600 #10-6 m/m discorted angle\n", + "centre = (e_x + e_y)/2 #10-6 m/m \n", + "point_x = -500 #The x coordinate of a point on mohr circle\n", + "point_y = 300 #The y coordinate of a point on mohr circle\n", + "Radius = 500 #10-6 m/m - from mohr circle\n", + "e_1 = Radius +centre #MPa The principle strain\n", + "e_2 = -Radius +centre #Mpa The principle strain\n", + "\n", + "stress_1 = E*(10**-3)*(e_1+v*e_2)/(1-v**2) #Mpa the stress in this direction \n", + "stress_2 = E*(10**-3)*(e_2+v*e_1)/(1-v**2) #Mpa the stress in this direction \n", + "print\"The principle stresses are \",round(stress_1,2),\"Mpa\",round(stress_2,2),\"MPa\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_6.ipynb new file mode 100644 index 00000000..34d5fcb7 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_6.ipynb @@ -0,0 +1,229 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9:Elastic stress analysis and design" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4 pagenumber 465" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a)The principle stresses are 16.67 MPa, -16.67 Mpa\n", + "b)The stresses on inclines plane 11.11 Mpa noraml, -7.06 Mpa shear \n" + ] + } + ], + "source": [ + "#Given \n", + "import math \n", + "b = 40.0 #mm - The width of the beam crossection\n", + "h = 300.0 #mm - The length of the beam crossection \n", + "V = 40.0 #KN - The shear stress in teh crossection\n", + "M = 10.0 #KN-m - The bending moment on K----K crossection \n", + "c = h/2 #mm -The position at which maximum stress occurs on the crossection\n", + "I = b*(h**3)/12 #mmm4 - the moment of inertia \n", + "#Caliculations \n", + "\n", + "stress_max_1 = M*c*(10**6)/I #The maximum stress occurs at the end\n", + "stress_max_2 = -M*c*(10**6)/I #The maximum stress occurs at the end\n", + "y = 140 #mm The point of interest, the distance of element from com\n", + "n = y/(c) # The ratio of the distances from nuetral axis to the elements\n", + "stress_L_1 = n*stress_max_1 #The normal stress on elements L--L\n", + "stress_L_2 = -n*stress_max_1 #The normal stress on elements L--L\n", + "x = 10 #mm The length of the element\n", + "A = b*x #mm3 The area of the element \n", + "y_1 = y+x/2 # the com of element from com of whole system\n", + "stress_xy = V*A*y_1*(10**3)/(I*b) #Mpa - The shear stress on the element \n", + "#stresses acting in plane 30 degrees \n", + "o = 60 #degrees - the plane angle\n", + "stress_theta = stress_L_1/2 + stress_L_1*(math.cos(math.radians(o)))/2 - stress_xy*(math.sin(math.radians(o))) #Mpa by direct application of equations\n", + "stress_shear = -stress_L_1*(math.sin(math.radians(o)))/2 - stress_xy*(math.cos(math.radians(o))) #Mpa Shear stress\n", + " \n", + "print \"a)The principle stresses are \",round(stress_max_1,2),\"MPa,\",round(stress_max_2,2),\"Mpa\"\n", + "print \"b)The stresses on inclines plane \",round(stress_theta,2),\"Mpa noraml, \",round(stress_shear,2),\"Mpa shear \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5 page number 476" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The stress developed 0.4 is in allowable ranges for 30077.85 mm2 area\n", + "The minimum area is 5714.28571429 mm2\n" + ] + } + ], + "source": [ + "#Given\n", + "M = 10 #KN-m moment\n", + "v = 8.0 #KN - shear Stress \n", + "stress_allow = 8 #MPa - The maximum allowable stress\n", + "shear_allow_per = 1.4 #Mpa - The allowable stress perpendicular to grain\n", + "stress_allow_shear = 0.7 #MPa - The maximum allowable shear stress\n", + "#Caliculations \n", + "\n", + "S = M*(10**6)/stress_allow #mm3 \n", + "#lets arbitarly assume h = 2b\n", + "#S = b*(h**2)/6\n", + "h = pow(12*S,0.333) #The depth of the beam\n", + "b = h/2 #mm The width of the beam\n", + "A = h*b #mm2 The area of the crossection , assumption\n", + "stress_shear = 3*v*(10**3)/(2*A) #Mpa The strear stress \n", + "if stress_shear