diff options
Diffstat (limited to 'Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb')
| -rwxr-xr-x | Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb | 450 |
1 files changed, 450 insertions, 0 deletions
diff --git a/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb new file mode 100755 index 00000000..d8fd7d88 --- /dev/null +++ b/Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb @@ -0,0 +1,450 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d43d7deae359298d302c004afe7a99ad4d721fd85b5d0a2361afaf05440c9100" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 06:Deflection of Beam" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.1, Page No:196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "wo=400 #loading in lb/ft\n", + "E=29*10**6 #Modulus of elasticity in psi\n", + "I=285 #Moment of inertia in in^4\n", + "S=45.6 #Sectional Modulus in in^3\n", + "L=8 #Span in ft\n", + "\n", + "#Calculations\n", + "#Part 1\n", + "#Part1 is theoretical in nature hence not coded\n", + "\n", + "#Part 2\n", + "delta_max=((wo*12**-1)*(L*12)**4)/(8*E*I) #maximum deflection in inches\n", + "M_max=(wo*12**-1)*(L*12)**2 #Maximum moment\n", + "sigma_max=M_max/(2*S) #Maximum bending stress in psi\n", + "\n", + "#Result\n", + "print M_max\n", + "print \"The maximum deflection is\",round(delta_max,4),\"in\"\n", + "print \"The maximum Bending Stress is\",round(sigma_max),\"psi\"\n", + "\n", + "#Answer in the textbook for sigma_max is incorrect" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "307200.0\n", + "The maximum deflection is 0.0428 in\n", + "The maximum Bending Stress is 3368.0 psi\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.3, Page No:198" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "P=300 #Point Load in N\n", + "R_a=100 #Reaction at A in N\n", + "R_c=200 #Reaction at C in N\n", + "E=12 #Youngs Modulus in GPa\n", + "L1=2 #Length of the load from A in m\n", + "L2=1 #Length of the load from C in m\n", + "b=0.04 #Width of the CS of the beam in m\n", + "h=0.08 #Depth of the CS of the beam in m\n", + "\n", + "#Claculations\n", + "#Moment of inertia \n", + "I=b*h**3*12**-1 #Moment of Inertia in m^4\n", + "#Flexural Rigidity\n", + "FR=E*10**9*I #FLexural rigidity in N.m^2\n", + "\n", + "#Moments in terms of x are\n", + "#Given\n", + "#After the variable Calculations we get\n", + "C1=-400/3 #Constant\n", + "C3=C1 #Constant\n", + "C2=0 #Constant\n", + "C4=0 #Constant\n", + "\n", + "#to get max displacement x we have\n", + "x=(6.510/2.441)**0.5 #Length at which displacement is maximum in m\n", + "v=(0.8138*x**3-6.510*x) #Displacement in mm\n", + "\n", + "#Largest slope\n", + "theta=(2.441*(L1+L2)**2-(7.324*(L1+L2-L1)**2)-6.150)*10**-3#Angle in radians\n", + "\n", + "#Result \n", + "print \"The maximum displacement is\",round(-v,2),\"mm downwards\"\n", + "print \"The maximum angle is\",round(theta*180*pi**-1,3),\"degrees anticlockwise\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum displacement is 7.09 mm downwards\n", + "The maximum angle is 0.487 degrees anticlockwise\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.4, Page No:200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "#The computation is mostly variable based hence values will be directly declared \n", + "C1=19.20*10**3 #lb.ft^2\n", + "C2=-131.6*10**3 #lb.ft^2\n", + "C3=14.7*10**3 #lb.ft^2\n", + "C4=-112.7*10**3 #lb.ft^2\n", + "EI=10**7 #Flexural Rigidity in psi\n", + "\n", + "#Calculations\n", + "v=-(C2*12**3)/(EI*40) #Displacement in inches\n", + "\n", + "#Result\n", + "print \"The maximum displacement is\",round(v,3),\"in downwards\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum displacement is 0.569 in downwards\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.6, Page No:210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "L1=3 #Length in m\n", + "L2=1 #Length in m\n", + "L3=8 #Length in m\n", + "L4=4 #Length in m\n", + "L5=6 #Length in m\n", + "\n", + "#Calculations\n", + "#Deflection midway\n", + "EIv=250*3**-1*L1**3-(50*3**-1*(L1-L2)**4)-(3925*3**-1*L1) #Deflection in N.m^3\n", + "\n", + "#Deflection at E\n", + "EIv_E=250*3**-1*L3**3-(50*3**-1*(L3-L2)**4)+(50*3**-1*(L3-L4)**4)+(650*3**-1*(L3-L5)**3)-(3925*3**-1*L3) #Deflection in N.m^3\n", + "\n", + "#Result\n", + "print \"The deflection at midspan is\",round(-EIv),\"N.m^3 downwards\"\n", + "print \"The deflection at point E is\",round(-EIv_E),\"N.m^3 downwards\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The deflection at midspan is 1942.0 N.m^3 downwards\n", + "The deflection at point E is 1817.0 N.m^3 downwards\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.8, Page No:223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "x1=16*3**-1 #Centroid of the triangle in ft\n", + "x2=3 #Centroid of the lower parabola in ft\n", + "x3=6 #Centroid of the rectangle in ft\n", + "x4=20*3**-1 #Centroid of the triangle in ft\n", + "#Moment values\n", + "M1=4800 #Moment in lb.ft\n", + "M2=14400 #Moment in lb.ft\n", + "\n", + "#Calcualtions\n", + "P=((3**-1*4*M1*x2)+(4*M1*x3)+(0.5*4*M1*2*x4))*(x1*8*8*0.5)**-1 #Force P in lb\n", + "\n", + "#Result\n", + "print \"The magnitude of force P is\",round(P,1),\"lb\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The magnitude of force P is 1537.5 lb\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.9, Page No:225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "P=300 #Force in N\n", + "L1=1 #Length in m\n", + "L2=2 #Length in m\n", + "R_a=100 #Reaction at A in N\n", + "R_c=200 #Reaction at C in N\n", + "EI=20.48*10**3 #Flexural Rigidity in N.m^2\n", + "\n", + "#Calculations\n", + "#Part 1\n", + "tC_A=(0.5*(L1+L2)*P*L1-(0.5*L1*P*(L1+L2)**-1))*EI**-1 #First Moment in m\n", + "theta_A=tC_A/(L1+L2) #Angle in radians \n", + "\n", + "#Part 2\n", + "tD_A=0.5*L1*R_a*(L1+L2)**-1*EI**-1 #First Moment in m\n", + "delta_D=(theta_A*L1-tD_A) #Displacement in m \n", + "\n", + "#Result\n", + "print \"The angle in part 1 is\",round(theta_A*180*pi**-1,3),\"Degrees\"\n", + "print \"The displacement in part 2 is\",round(delta_D*1000,2),\"mm downward\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angle in part 1 is 0.373 Degrees\n", + "The displacement in part 2 is 5.7 mm downward\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.10, Page No:227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "P1=150 #Load in lb\n", + "P2=30 #Load in lb\n", + "R_A=78 #Reaction at A in lb\n", + "R_C=102 #Reaction at C in lb\n", + "L1=4 #Length in ft\n", + "L2=6 #Length in ft\n", + "M1=780 #Moment in lb.ft\n", + "M2=900 #Moment in lb.ft\n", + "M3=120 #Moment in lb.ft\n", + "\n", + "#Calculations\n", + "EI_AC=0.5*(L1+L2)*M1*(2*3**-1)*(L1+L2)-(0.5*L2*M2*(L1+(2*3**-1)*L2)) #Deflection in lb.ft^3\n", + "EI_thetaC=EI_AC/(L1+L2) #Deflection in lb.ft^2\n", + "\n", + "EI_DC=-0.5*L1*M3*2*3**-1*L1 #Deflection in lb.ft^3\n", + "EI_deltaD=EI_thetaC*L1-(-EI_DC) #Deflection in lb.ft^2\n", + "\n", + "#Result\n", + "print \"The deflection is\",round(EI_deltaD),\"lb.ft^2 upwards\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The deflection is 1120.0 lb.ft^2 upwards\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.11, Page No:234" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "P1=80 #Load in lb\n", + "P2=100 #Load in lb\n", + "b1=3 #Distance of load from end in ft\n", + "b2=2 #Distance of load from end in ft\n", + "L=9 #Span of the beam in ft\n", + "\n", + "#Calcualtions\n", + "EI_delta1=(P1*b1*48**-1)*(3*L**2-4*b1**2) #Deflection in lb.ft^3\n", + "EI_delta2=(P2*b2*48**-1)*(3*L**2-4*b2**2) #Deflection in lb.ft^3\n", + "EI_delta=EI_delta1+EI_delta2 #Deflection at modspan in lb.ft^3\n", + "\n", + "#Result\n", + "print \"The deflection at midspan is\",round(EI_delta),\" lb.ft^3 downward\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The deflection at midspan is 1981.0 lb.ft^3 downward\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6.12, Page no:234" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable Decleration\n", + "wo=600 #Load in N/m\n", + "L=6 #Span of the beam in m\n", + "b=2 #Distance of the load from end in m\n", + "a=1 #Distance of the load from end in m\n", + "\n", + "#Calulations\n", + "EI_delta1=wo*384**-1*(5*L**4-12*L**2*b**2+8*b**4) #Deflection in N.m^3\n", + "EI_delta2=wo*96**-1*a**2*(3*L**2-2*a**2) #Deflection in N.m^3\n", + "\n", + "EI_delta=EI_delta1-EI_delta2 #Total Delfection at midspan in N.m^3\n", + "\n", + "#Result\n", + "print \"The total Deflection at midpsan is\",round(EI_delta),\"N.m^3 downwards\"\n", + "\n", + "#NOTE:The answer varies due to decimal point accuracy" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total Deflection at midpsan is 6963.0 N.m^3 downwards\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file |