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diff --git a/Mechanics_of_Materials/chapter6.ipynb b/Mechanics_of_Materials/chapter6.ipynb new file mode 100644 index 00000000..344830c9 --- /dev/null +++ b/Mechanics_of_Materials/chapter6.ipynb @@ -0,0 +1,444 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Stresses in Beams Advanced Topics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1, page no. 400" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "largest tensile and compressive stresses in the wood & the max. and min. tensile stresses in the steel\n", + "\"\"\"\n", + "\n", + "import math \n", + "\n", + "#initialisation\n", + "# 4*6 inch wood beam dimension\n", + "# 4*0.5 inch steel beam dimension\n", + "M = 60.0 # Moment in k-in\n", + "E1 = 1500. # in Ksi\n", + "E2 = 30000.0 # in Ksi\n", + "h1 = 5.031 # Distance between top surface and neutral axis of the beam in inch by solving 1500*(h1-3)*24 + 30000*(h1-6.25)*2 = 0\n", + "\n", + "#calculation\n", + "h2 = 6.5 - h1 \n", + "I1 = (1.0/12.0)*(4*6**3) + (4*6)*(h1-3)**2 # Momeny of inertia of the wooden cross section\n", + "I2 = (1.0/12.0)*(4*0.5**3) + (4*0.5)*(h2-0.25)**2 # Momeny of inertia of the steel cross section\n", + "I = I1 + I2 # Moment of inertia of whole cross section\n", + "\n", + "# Material 1\n", + "s1a = -(M*h1*E1)/((E1*I1)+(E2*I2)) # Maximum compressive stress in ksi where y = h1\n", + "s1c = -(M*(-(h2-0.5))*E1)/((E1*I1)+(E2*I2)) # Maximum tensile stress in ksi where y = -(h2-0.5)\n", + "print \"Maximum compressive stress in wood is\", round(s1a,3)*1000, \"psi\"\n", + "print \"Maximum tensile stress in wood is\", round(s1c,3)*1000, \"psi\"\n", + "\n", + "# Material 2\n", + "s2a = -(M*(-h2)*E2)/((E1*I1)+(E2*I2)) # Maximum tensile stress in ksi where y = -h2\n", + "s2c = -(M*(-(h2-0.5))*E2)/((E1*I1)+(E2*I2)) # Minimum tensile stress in ksi where y = -(h2-0.5)\n", + "print \"Maximum tensile stress in steel is\", round(s2a,3)*1000, \"psi\"\n", + "print \"Minimum tensile stress in steel is\", round(s2c,3)*1000, \"psi\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum compressive stress in wood is -1305.0 psi\n", + "Maximum tensile stress in wood is 251.0 psi\n", + "Maximum tensile stress in steel is 7622.0 psi\n", + "Minimum tensile stress in steel is 5028.0 psi\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, page no. 402" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "maximum tensile and compressive stresses in the faces and the core using: general theory for composite beams and \n", + "approximate theory for sandwich beams\n", + "\"\"\"\n", + "\n", + "import math \n", + "\n", + "#initialisation\n", + "\n", + "M = 3000 # moment in N-m\n", + "t = 0.005 # thickness of alluminiun in m\n", + "E1 = 72e09 # Modulus of elasticity of alluminium in Pa\n", + "E2 = 800e06 # Modulus of elasticity of Plastic core in Pa\n", + "b = 0.2 # Width of cross section in m\n", + "h = 0.160 # Height of cross section in m\n", + "hc = 0.150 # Height of Plastic core cross section in m\n", + "\n", + "#calculation\n", + "I1 = (b/12.0)*(h**3 - hc**3) # Moment of inertia of alluminium cross section\n", + "I2 = (b/12.0)*(hc**3) # Moment of inertia of Plastic core cross section\n", + "f = (E1*I1) + (E2*I2) # Flexural rigidity of the cross section\n", + "s1_max = (M*(h/2.0)*E1)/f \n", + "s1c = -s1_max # Maximum compressive stress in alluminium core in Pa\n", + "s1t = s1_max # Maximum tensile stress in alluminium core in Pa\n", + "print \"Maximum compressive stress on alluminium face by the general theory for composite beams is\", s1c, \"Pa\"\n", + "print \"Maximum tensile stress on alluminium face by the general theory for composite beams is\", s1t, \"Pa\"\n", + "s2_max = (M*(hc/2.0)*E2)/f \n", + "s2c = -s2_max # Maximum compressive stress in Plastic core in Pa\n", + "s2t = s2_max # Maximum tensile stress in Plastic core in Pa\n", + "print \"Maximum compressive stress in plastic core by the general theory for composite beams is\", s2c, \"Pa\"\n", + "print \"Maximum tensile stress in plastic core by the general theory for composite beams is\", s2t, \"Pa\"\n", + "\n", + "# Part (b) : Calculation from approximate theory of sandwitch\n", + "s1_max1 = (M*h)/(2*I1) \n", + "s1c1 = -s1_max1 # Maximum compressive stress in alluminium core in Pa\n", + "s1t1 = s1_max1 # Maximum tensile stress in alluminium core in Pa\n", + "print \"Maximum compressive stress on alluminium core by approximate theory of sandwitch is\", s1c1, \"Pa\"\n", + "print \"Maximum tensile stress on alluminium core by approximate theory of sandwitch is\", s1t1, \"Pa\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum compressive stress on alluminium face by the general theory for composite beams is -18984838.497 Pa\n", + "Maximum tensile stress on alluminium face by the general theory for composite beams is 18984838.497 Pa\n", + "Maximum compressive stress in plastic core by the general theory for composite beams is -197758.734344 Pa\n", + "Maximum tensile stress in plastic core by the general theory for composite beams is 197758.734344 Pa\n", + "Maximum compressive stress on alluminium core by approximate theory of sandwitch is -19972260.749 Pa\n", + "Maximum tensile stress on alluminium core by approximate theory of sandwitch is 19972260.749 Pa\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3, page no. 407" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "calculate largest tensile & compressive stresses in the wood\n", + "also, the maximum and minimum tensile stresses in the steel\n", + "\"\"\"\n", + "\n", + "import math \n", + "\n", + "#initialisation\n", + "# 4*6 inch wood beam dimension\n", + "# 4*0.5 inch steel beam dimension\n", + "M = 60.0 # Moment in k-in\n", + "E1 = 1500.0 # in Ksi\n", + "E2 = 30000.0 # in Ksi\n", + "b = 4.0 # width of crosssection in inch\n", + "\n", + "#calculation\n", + "# Transformed Section\n", + "n = E2/E1 # Modular ratio\n", + "b1 = n*4 # Increased width of transformed cross section\n", + "\n", + "# Neutral axis\n", + "h1 = ((3*4*6)+(80*0.5*6.25))/((4*6)+(80*0.5)) # Dismath.tance between top surface and neutral axis of the beam in inch\n", + "h2 = 6.5 - h1 # in inch\n", + "\n", + "# Moment of inertia\n", + "It = (1.0/12.0)*(4*6**3) + (4*6)*(h1-3)**2 + (1.0/12.0)*(80*0.5**3) + (80*0.5)*(h2-0.25)**2 # Moment of inertia of transformed cross section\n", + "\n", + "# Material 1\n", + "s1a = -(M*h1)/It # Maximum tensile stress in ksi where y = h1\n", + "s1c = -(M*(-(h2-0.5)))/It # Maximum compressive stress in ksi where y = -(h2-0.5)\n", + "print \"Maximum tensile stress in wood is\", s1a*1000, \"psi\"\n", + "print \"Maximum compressive stress in wood is\", s1c*1000, \"psi\"\n", + "\n", + "# Material 2\n", + "s2a = -(M*(-h2)*n)/It # Maximum tensile stress in ksi where y = -h2\n", + "s2c = -(M*(-(h2-0.5)*n))/It # Minimum tensile stress in ksi where y = -(h2-0.5)\n", + "print \"Maximum tensile stress in steel\", s2a*1000, \"psi\"\n", + "print \"Minimum tensile stress in steel\", s2c*1000, \"psi\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum tensile stress in wood is -1305.28781191 psi\n", + "Maximum compressive stress in wood is 251.328709125 psi\n", + "Maximum tensile stress in steel 7620.9350509 psi\n", + "Minimum tensile stress in steel 5026.57418251 psi\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4,page no. 412" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "maximum tensile and compressive stresses in the beam\n", + "\"\"\"\n", + "\n", + "import math\n", + "import numpy\n", + "\n", + "#initialisation\n", + "\n", + "q = 3000.0 # Uniform load intensity in N/m\n", + "a = 26.57 # tilt of the beam in degree\n", + "b = 0.1 # width of the beam\n", + "h = 0.15 # height of the beam\n", + "L = 1.6 # Span of the beam\n", + "\n", + "#calculation\n", + "qy = q*math.cos(math.radians(a)) # Component of q in y direction\n", + "qz = q*math.sin(math.radians(a)) # Component of q in z direction\n", + "My = (qz*L**2.0)/8.0 # Maximum bending moment in y direction\n", + "Mz = (qy*L**2.0)/8.0 # Maximum bending moment in z direction\n", + "Iy = (h*b**3.0)/12.0 # Moment of inertia along y\n", + "Iz = (b*h**3.0)/12.0 # Moment of inertia alon z\n", + "s = ((3*q*L**2)/(4*b*h))*((math.sin(math.radians(a))/b)+(math.cos(math.radians(a))/h))\n", + "sc = -s # Maximum compressive stress\n", + "st = s # Maximum tensile stress\n", + "print \"Maximum compressive stress in the beam is\", sc, \"Pa\"\n", + "print \"Maximum tensile stress in the beam is\", st, \"Pa\"\n", + "\n", + "# Neutral axis\n", + "l = (h/b)**2\n", + "t = math.sin(math.radians(a)/math.cos(math.radians(a)))\n", + "j = l*(math.sin(math.radians(a)/math.cos(math.radians(a))))\n", + "be = math.degrees((numpy.arctan((j)))) # Inclination of Neutral axis to z axis\n", + "print \"Inclination of Neutral axis to z axis is\", round(be,2), \"degree\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum compressive stress in the beam is -4007231.57248 Pa\n", + "Maximum tensile stress in the beam is 4007231.57248 Pa\n", + "Inclination of Neutral axis to z axis is 48.11 degree\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, page no. 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "maximum bending stresses in the beam for various conditions\n", + "\"\"\"\n", + "\n", + "import math\n", + "import numpy \n", + "\n", + "#initialisation\n", + "L = 12.0 # Length of the beam in ft\n", + "P = 10.0 # Load in k acting in vertical direction\n", + "\n", + "#Part (a)\n", + "h = 24.0 # Height of beam in inch\n", + "Iz = 2100 # Moment of inertia along z axis in in4\n", + "Iy = 42.2 # Moment of inertia along y axis in in4\n", + "\n", + "#calculation\n", + "s_max = (P*(h/2.0)*L*12)/Iz # Maximum stress in Ksi\n", + "print \"Maximum tensile stress in the beam at the top of the beam\", round(s_max*1000), \"psi\"\n", + "print \"Maximum compressive stress in the beam at the bottom of the beam\", round(-s_max*1000), \"psi\"\n", + "\n", + "#Part (b)\n", + "a = 1 # Angle between y axis and the load\n", + "My = -(P*math.sin(math.radians(a)))*L*12 # Moment along y-axis in K-in\n", + "Mz = -(P*math.cos(math.radians(a)))*L*12 # Moment along z-axis in K-in\n", + "ba = math.radians(numpy.arctan(((My*Iz)/(Mz*Iy)))) # Orientation of neutral axis\n", + "z = -3.5\n", + "y = 12.0 # Coordinates of the point A and B where maximum stress occur\n", + "s = ((My*z)/Iy)-((Mz*y)/Iz) # Stress in Ksi\n", + "sa = s # Tensile stress at A\n", + "sb = -s # Compressive stress in B\n", + "print \"The tensile stress at A is\", round(sa*1000), \"psi\"\n", + "print \"The compressive stress at B is\", round(sb*1000), \"psi\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum tensile stress in the beam at the top of the beam 8229.0 psi\n", + "Maximum compressive stress in the beam at the bottom of the beam -8229.0 psi\n", + "The tensile stress at A is 10312.0 psi\n", + "The compressive stress at B is -10312.0 psi\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, page no. 420" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "the bending stresses at points A and B\n", + "\"\"\"\n", + "\n", + "import math \n", + "import numpy\n", + "\n", + "#initialisation\n", + "M = 15 # Bending moment in k-in\n", + "t = 10 # Angle between line of action of moment and z-axis\n", + "\n", + "# Properties of cross section\n", + "c = 0.634 # Location of centroid on the axis of symmetry\n", + "Iy = 2.28 # Moment of inertia in y-direction in in4\n", + "Iz = 67.4 # Moment of inertia in z-direction in in4\n", + "ya = 5\n", + "za = -2.6+0.634 # Coordinates of point A\n", + "yb = -5\n", + "zb = 0.634 # Coordinates of point B\n", + "My = M*math.sin(math.radians(t)) # Moment along y-axis\n", + "Mz = M*math.cos(math.radians(t)) # Moment along z-axis\n", + "sa = ((My*za)/Iy)-((Mz*ya)/Iz) # Bending stress at point A in ksi\n", + "sb = ((My*zb)/Iy)-((Mz*yb)/Iz) # Bending stress at point B in ksi\n", + "print \"The bending stress at point A is\", round(sa*1000), \"psi\"\n", + "print \"The bending stress at point B is\", round(sb*1000), \"psi\"\n", + "\n", + "# Neutral axis\n", + "j = (Iz/Iy)*(math.sin(math.radians(t)/math.cos(math.radians(t))))\n", + "be = numpy.degrees(numpy.arctan((j))) # Inclination of neutral axis to z-axis in degree\n", + "print \"Inclination of neutral axis to z-axis is\", round(be,1), \"degree\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The bending stress at point A is -3342.0 psi\n", + "The bending stress at point B is 1820.0 psi\n", + "Inclination of neutral axis to z-axis is 79.1 degree\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9, page no. 448" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\"\"\"\n", + "calculate magnitude of the moment M\n", + "\"\"\"\n", + "\n", + "import math \n", + "\n", + "#initialization\n", + "b = 5 # in inch\n", + "b1 = 4 # in inch\n", + "h = 9 # in inch\n", + "h1 = 7.5 # in inch\n", + "sy = 33 # stress along y axis in ksi\n", + "\n", + "#Calculations\n", + "M = (sy/12.0)*((3*b*h**2)-(b+(2*b1))*(h1**2)) # Bending moment acting in k-in\n", + "\n", + "#Result\n", + "print \"the magnitude of the moment M is\", round(M), \"k-in\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the magnitude of the moment M is 1330.0 k-in\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
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