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diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb deleted file mode 100755 index 5f5f30b9..00000000 --- a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb +++ /dev/null @@ -1,622 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:c5e7c024d1eb3a7426e4241fc719f61d59003d11703e75e2542d5874bd3a6f9f" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 02:Strain" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Examples No:2.2.1, Page No:36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "#Axial Forces in lb in member AB, BC and CD\n", - "P_AB=2000 \n", - "P_BC=2000\n", - "P_CD=4000\n", - "#Other Variables\n", - "E=29*10**6 #Modulus of Elasticity in psi\n", - "#Length of each member in inches\n", - "L_AB=5*12\n", - "L_BC=4*12\n", - "L_CD=4*12\n", - "#Diameter of each member in inches\n", - "D_AB=0.5\n", - "D_BC=0.75\n", - "D_CD=0.75\n", - "\n", - "#Calculation\n", - "#Area Calculation of each member in square inches\n", - "A_AB=(pi*D_AB**2)/4\n", - "A_BC=(pi*D_BC**2)/4\n", - "A_CD=(pi*D_CD**2)/4\n", - "\n", - "#Using relation delta=(PL/AE) to compute strain\n", - "#As stress in Member CD is compression\n", - "delta=(E**-1)*((P_AB*L_AB*A_AB**-1)+(P_BC*L_BC*A_BC**-1)-(P_CD*L_CD*A_CD**-1))\n", - "\n", - "#Result\n", - "print \"The elongation in the total structure is\",round(delta,5),\"in\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The elongation in the total structure is 0.01358 in\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.2, Page No:36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "from scipy.integrate import quad\n", - "\n", - "#Variable Decleration\n", - "E=200*10**9 #Modulus of elasticity in Pa\n", - "P=10**5 #Force acting in N\n", - "\n", - "#Calculations\n", - "#Using quad integration\n", - "#Area has been defined as a quadratic equation to integrate\n", - "def integrand(x, a, b):\n", - " return 1/(a * x + b)\n", - "a = 160\n", - "b = 800\n", - "I = quad(integrand, 0, 10, args=(a,b))\n", - "#Using delta=(P/E)*I where I is the integrand\n", - "delta=(P*E**-1)*10**6*I[0]\n", - "\n", - "#Result\n", - "print \"The elongation in the member is\",round(delta*1000,2),\"mm\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The elongation in the member is 3.43 mm\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.3, Page No:37" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decelration\n", - "A_AC=0.25 #Cross Sectional Area in square inch\n", - "Load=2000 #Load at point C in lb\n", - "E=29*10**6 #Modulus of elasticity in psi\n", - "theta=(pi*40)/180 #Angle in radians\n", - "L_BC=8 #Length in ft\n", - "\n", - "#Calculations\n", - "#Using sum of forces \n", - "P_AC=Load/sin(theta) #Force in cable AC in lb\n", - "L_AC=(L_BC*12)/cos(theta) #Length of cable AC in in\n", - "\n", - "delta_AC=(P_AC*L_AC)/(E*A_AC) #elongation in inches\n", - "\n", - "delta_C=delta_AC/sin(theta) #displacement of point C in inches\n", - "\n", - "#Result\n", - "print \"The displacement of point C is\",round(delta_C,4),\"in\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The displacement of point C is 0.0837 in\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.4, Page No:46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "d=0.05 #Diameter of the rod in mm\n", - "P=8000 #Load on the bar in N\n", - "E=40*10**6 #Modulus of elasticity in Pa\n", - "v=0.45 #Poisson Ratio\n", - "L=300 #Length of the rod in mm\n", - "\n", - "#Calculation\n", - "A=((pi*d**2)/4) #Area of the bar in mm^2\n", - "sigma_x=-P/A #Axial Stress in the bar in Pa\n", - "#As contact pressure resists the force\n", - "p=(v*sigma_x)/(1-v)\n", - "#Using Axial Strain formula\n", - "e_x=(sigma_x-(v*2*p))/E\n", - "#Corresponding change in length\n", - "delta=e_x*L #contraction in mm\n", - "#Without constrains of the wall\n", - "delta_w=(-P*(L*10**-3))/(E*A) #Elongation in m\n", - "\n", - "#Result\n", - "print \"The elongation in the bar is\",-round(delta,2),\"mm contraction\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The elongation in the bar is 8.06 mm contraction\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.5, Page No:47" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "E=500 #Modulus of elasticity in psi\n", - "v=0.48 #Poisson ratio\n", - "V=600 #Force in lb\n", - "w=5 #Width of the plate in inches\n", - "l=9 #Length of the plate in inches\n", - "t=1.75 #Thickness of the rubber layer in inches\n", - "\n", - "#Calculations\n", - "tau=V*(w*l)**-1 #Shear stress in rubber in psi\n", - "G=E/(2*(1+v)) #Bulk modulus in psi\n", - "gamma=tau/G #Shear Modulus \n", - "disp=t*gamma #Diplacement in inches\n", - "\n", - "#Result\n", - "print \"The displacement of the rubber layer is\",round(disp,4),\"in\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The displacement of the rubber layer is 0.1381 in\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.6, Page No:52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "P=10**6 #Force on the member in N\n", - "Es=200 #Modulus of elasticity of steel in GPa\n", - "Ec=14 #Modulus of elasticity concrete in GPa\n", - "As=900*10**-6 #Area of steel in m^2\n", - "Ac=0.3**2 #Area of concrete block in m^2\n", - "\n", - "#Calculation\n", - "#Cross Sectional Areas\n", - "Ast=4*As #Cross Sectional Area in m^2 of Steel\n", - "Act=Ac-Ast #Cross Sectional Area of Concrete in m^2\n", - "\n", - "#Applying equilibrium to the structure\n", - "#Using the ratio of stress and modulii of elasticity we obtain the following eq\n", - "sigma_ct=P/(((Es*Ec**-1)*Ast)+Act) #Stress in Concrete in Pa\n", - "sigma_st=sigma_ct*Es*Ec**-1 #Stress in Steel in Pa\n", - "\n", - "#Result\n", - "print \"The stress in steel and concrete is as follows\",round(sigma_st*10**-6,1),\"MPa and\",round(sigma_ct*10**-6,3),\"Mpa respectively\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The stress in steel and concrete is as follows 103.6 MPa and 7.255 Mpa respectively\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.7, Page No:52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "#Say the ratio of stress in steel to concrete is R\n", - "R=14.286 \n", - "sigma_co=6*10**6 #Stress in concrete in Pa\n", - "Ast=3.6*10**-3 #Area of steel in m^2\n", - "Aco=86.4*10**-3 #Area of Concrete in m^2\n", - "\n", - "#Calculation\n", - "sigma_st=R*sigma_co #Stress in steel in Pa\n", - "#Here stress is below the allowable hence safe\n", - "P=sigma_st*Ast+sigma_co*Aco #Allowable force in N\n", - "\n", - "#Result\n", - "print \"The maximum allowable force is\",round(P*10**-3),\"kN\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The maximum allowable force is 827.0 kN\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.8, Page No:53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#NOTE:The NOtation has been changed to ease coding\n", - "#Variable Decleration\n", - "d=0.005 #difference in length in inch\n", - "L=10 #Length in inch\n", - "#Area of copper and aluminium in sq.in\n", - "Ac=2 #Area of copper \n", - "Aa=3 #Area of aluminium \n", - "#Modulus of elasticity of copper and aluminium in psi\n", - "Ec=17000000 #Copper\n", - "Ea=10**7 #Aluminium\n", - "#Allowable Stress in psi\n", - "Sc=20*10**3 #Copper\n", - "Sa=10*10**3 #Aluminium\n", - "\n", - "#Calculation\n", - "#Equilibrium is Pc+Pa=P\n", - "#Hookes Law is delta_c=delta_a+0.005\n", - "#Simplfying the solution we have constants we can directly compute\n", - "A=d*Ec*(L+d)**-1\n", - "B=Ec*Ea**-1\n", - "C=L*B*(L+d)**-1\n", - "sigma_a=(Sc-A)*C**-1\n", - "\n", - "#Using equilibrium equation\n", - "P=Sc*Ac+sigma_a*Aa #Safe load in lb\n", - "\n", - "#Result\n", - "print \"The safe load on the structure is\",round(P),\"lb\"\n", - "#NOTE:Answer in the textbook has been rounded off and hence the discrepancy" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The safe load on the structure is 60312.0 lb\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.9, Page No:54" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "import numpy as np\n", - "\n", - "#Variable Decleration\n", - "P=50*10**3 #Load applied in N\n", - "x1=0.6 #Length in m\n", - "x2=1.6 #Length in m\n", - "L1=1 #Length of steel cable in m\n", - "L2=2 #Length of bronze cable in m\n", - "L=2.4 #Length in m\n", - "#Area in m^2\n", - "Ast=600*10**-6 #Steel\n", - "Abr=300*10**-6 #Bronze\n", - "#Modulus of elasticity in GPa\n", - "Est=200 #Steel\n", - "Ebr=83 #Bronze\n", - "\n", - "#Calculations\n", - "#Applying the equilibrium and Hookes law we solve by matrix method\n", - "a=np.array([[x1,x2],[1,-((x1*Est*Ast*L2)/(x2*Ebr*Abr))]])\n", - "b=np.array([L*P,0])\n", - "y=np.linalg.solve(a,b)\n", - "\n", - "#Stresses in Pa\n", - "sigma_st=y[0]*Ast**-1 #Stress in steel\n", - "sigma_br=y[1]/Abr #Stress in bronze\n", - "\n", - "#Result\n", - "print \"The stresses in steel and bronze are as follows\"\n", - "print round(sigma_st*10**-6,1),\"MPa and\",round(sigma_br*10**-6,1),\"MPa respectively\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The stresses in steel and bronze are as follows\n", - "191.8 MPa and 106.1 MPa respectively\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.10, Page No:62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "L=2.5 #Length in m\n", - "A=1200 #Cross sectional Area in mm^2\n", - "delta_T=40 #Temperature drop in degree C\n", - "delta=0.5*10**-3 #Movement of the walls in mm\n", - "alpha=11.7*10**-6 #Coefficient of thermal expansion in /degreeC\n", - "E=200*10**9 #Modulus of elasticity in Pa\n", - "\n", - "#Calculation\n", - "#Part(1)\n", - "sigma_1=alpha*delta_T*E #Stress in the rod in Pa\n", - "\n", - "#Part(2)\n", - "#Using Hookes Law\n", - "sigma_2=E*((alpha*delta_T)-(delta*L**-1)) #Stress in the rod in Pa\n", - "\n", - "print \"The Stress in part 1 in the rod is\",round(sigma_1*10**-6,1),\"MPa\"\n", - "print \"The Stress in part 2 in the rod is\",round(sigma_2*10**-6,1),\"MPa\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Stress in part 1 in the rod is 93.6 MPa\n", - "The Stress in part 2 in the rod is 53.6 MPa\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.11, Page No:63" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "delta=100 #Increase in the temperature in degreeF\n", - "Load=12000 #Load on the beam in lb\n", - "#Length in inch\n", - "Ls=2*12 #Steel\n", - "Lb=3*12 #Bronze\n", - "#Area in sq.in\n", - "As=0.75 #Steel\n", - "Ab=1.5 #Bronze\n", - "#Modulus of elasticity in psi\n", - "Es=29*10**6 #Steel\n", - "Eb=12*10**6 #Bronze\n", - "#Coefficient of thermal expansion in /degree C\n", - "alpha_s=6.5*10**-6 #Steel\n", - "alpha_b=10**-5 #Bronze\n", - "\n", - "#Calculations\n", - "#Applying the Hookes Law and equilibrium we get two equations\n", - "a=np.array([[Ls*(Es*As)**-1,-Lb*(Eb*Ab)**-1],[2,1]])\n", - "b=np.array([(alpha_b*delta*Lb-alpha_s*delta*Ls),Load])\n", - "y=np.linalg.solve(a,b)\n", - "\n", - "#Stresses\n", - "sigma_st=y[0]*As**-1 #Stress in steel in psi (T)\n", - "sigma_br=y[1]*Ab**-1 #Stress in bronze in psi (C)\n", - "\n", - "#Result\n", - "print \"The Stress in steel and bronze are as follows\"\n", - "print sigma_st,\"psi (T) and\", -sigma_br,\"psi (C)\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Stress in steel and bronze are as follows\n", - "11600.0 psi (T) and 3600.0 psi (C)\n" - ] - } - ], - "prompt_number": 58 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2.12, Page No:64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\n", - "#Variable Decleration\n", - "P=6000 #Force in lb\n", - "Est=29*10**6 #Modulus of elasticity of steel in psi\n", - "L1=24 #Length in inches\n", - "L2=36 #Length in inches\n", - "alpha_1=6.5*10**-6 #coefficient of thermal expansion in /degree F of steel\n", - "alpha_2=10**-5 #coefficient of thermal expansion in /degree F of bronze\n", - "As=0.75 #Area os steel in sq.in\n", - "\n", - "#Calculations\n", - "delta_T=((P*L1)/(Est*As))/(alpha_2*L2-alpha_1*L1) #Change in temperature in degree F\n", - "\n", - "print \"The change in the Temperature is\",round(delta_T,1),\"F\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The change in the Temperature is 32.5 F\n" - ] - } - ], - "prompt_number": 60 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
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