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| author | hardythe1 | 2014-07-28 14:01:58 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-28 14:01:58 +0530 |
| commit | 73b0bffc1781c6cf1eec459813767508da507c33 (patch) | |
| tree | 81c5dc85e69261df791d26c88dca60dbf765970b /Mechanics_of_Materials/Chapter8.ipynb | |
| parent | 5571c8f5bda1334edf8ef9d8c69928d46cc9f163 (diff) | |
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diff --git a/Mechanics_of_Materials/Chapter8.ipynb b/Mechanics_of_Materials/Chapter8.ipynb new file mode 100755 index 00000000..366c18a1 --- /dev/null +++ b/Mechanics_of_Materials/Chapter8.ipynb @@ -0,0 +1,412 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8:Combined Loadings"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page No 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "di=4*12 #inch, diameter\n",
+ "ri=di/2.0 #Radius\n",
+ "t=0.5 #inch, thickness\n",
+ "sigma=20.0 #Ksi, stress\n",
+ "\n",
+ "#Calculation\n",
+ "#Cylindrical Pressure Vessel\n",
+ "p1=(t*sigma)/ri #sigma = pr/t\n",
+ "#Spherical Vessel\n",
+ "p2=(2*t*sigma)/(ri) #sigma = pr/2t\n",
+ "\n",
+ "#Display\n",
+ "print\"The maximum internal pressure the cylindrical pressure vessel can sustainis\",round(p1*1000,0),\"psi\"\n",
+ "print\"The maximum internal pressure a spherical pressure vessel can sustain is\",round(p2*1000,0),\"psi\"\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum internal pressure the cylindrical pressure vessel can sustainis 417.0 psi\n",
+ "The maximum internal pressure a spherical pressure vessel can sustain is 833.0 psi\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page No 414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine stress at point B and C\n",
+ "\n",
+ "#Given\n",
+ "P = 15000.0 #N, force,\n",
+ "a = 40.0 #mm, length\n",
+ "b = 100.0 #mm, breadth\n",
+ "\n",
+ "#CAlculation\n",
+ "#Normal Force\n",
+ "A = a*b #Area\n",
+ "sigma = P/A\n",
+ "#Bending Moment\n",
+ "I = (a*b**3)/12.0 #I = (1/12)*bh**3\n",
+ "M = P*(b/2.0) \n",
+ "c = b/2.0\n",
+ "sigma_max =(M*c)/I\n",
+ "\n",
+ "#Superposition\n",
+ "x = ((sigma_max-sigma)*b)/((sigma_max+sigma)+(sigma_max-sigma))\n",
+ "sigma_b = (sigma_max-sigma)\n",
+ "sigma_c = (sigma_max + sigma)\n",
+ "\n",
+ "#Display\n",
+ "print\"The state of stress at B is(tensile)\",sigma_b,\"psi\"\n",
+ "print\"The state of stress at C is (compressive)\",sigma_c,\"psi\"\n",
+ "\n",
+ "\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The state of stress at B is(tensile) 7.5 psi\n",
+ "The state of stress at C is (compressive) 15.0 psi\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page No 415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the stress\n",
+ "\n",
+ "#Given:\n",
+ "ri =24 #inch, radius\n",
+ "t = 0.5 #inch\n",
+ "ro = ri+t\n",
+ "sp_wt_water = 62.4 #lb/ft**3\n",
+ "sp_wt_steel = 490 #lb/ft**3\n",
+ "l_a = 3 #m depth of point A from the top\n",
+ "\n",
+ "#Internal Loadings:\n",
+ "import math\n",
+ "v = (math.pi*l_a)*((ro/12.0)**2 - (ri/12.0)**2)\n",
+ "W_st = sp_wt_steel*v\n",
+ "p = sp_wt_water*l_a #lb/ft**2,Pascal's Law\n",
+ "p_=p*0.0069 #psi\n",
+ "#Circumferential Stress:\n",
+ "sigma1 = (p_*ri)/t\n",
+ "#Longitudinal Stress:\n",
+ "A_st = (math.pi)*(ro**2 - ri**2)\n",
+ "sigma2 = W_st/A_st\n",
+ "\n",
+ "#Display:\n",
+ "print\"The state of stress at A (Circumferential)\",round(sigma1,0),\"KPa\"\n",
+ "print\"The state of stress at A (Longitudinal) \",round(sigma2,1),\"KPa\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The state of stress at A (Circumferential) 62.0 KPa\n",
+ "The state of stress at A (Longitudinal) 10.2 KPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page No 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the state of stress\n",
+ "\n",
+ "#Given\n",
+ "y_c = 125/1000.0 #m, length\n",
+ "x_c = 1.5 #m\n",
+ "y_b = 1.5 #m\n",
+ "x_b = 6.0 #m\n",
+ "udl = 50.0 #kN/m, force per unit length\n",
+ "l_udl = 2.5 #m\n",
+ "l = 250/1000.0 #m\n",
+ "width = 50/1000.0 #m \n",
+ "\n",
+ "\n",
+ "#Internal Loadings:\n",
+ "N = 16.45 #kN\n",
+ "V = 21.93 #kN\n",
+ "M = 32.89 #kNm\n",
+ "\n",
+ "#Stress Components:\n",
+ "#Normal Force:\n",
+ "A = l*width\n",
+ "sigma1 = N/(A*1000)\n",
+ "#Shear Force:\n",
+ "tou_c = 0\n",
+ "#Bending Moment:\n",
+ "c = y_c\n",
+ "I = (1/12.0)*(width*l**3)\n",
+ "sigma2 = (M*c)/(I*1000)\n",
+ "#Superposition:\n",
+ "sigmaC = sigma1+sigma2\n",
+ "\n",
+ "#Display:\n",
+ "print\"The stress due to normal force at C \",round(sigma1,2),\"MPa\"\n",
+ "print\"The stress due to shear force at C \",tou_c,\"MPa\"\n",
+ "print\"The stress due to bending moment at C \",round(sigma2,1),\"MPa\"\n",
+ "print\"The resultant stress at C \",round(sigmaC,1),\"MPa\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress due to normal force at C 1.32 MPa\n",
+ "The stress due to shear force at C 0 MPa\n",
+ "The stress due to bending moment at C 63.1 MPa\n",
+ "The resultant stress at C 64.5 MPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page No 418"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given:\n",
+ "r = 0.75*10 #mm, radius\n",
+ "f_x =40000 #N, force along x\n",
+ "f_y =800 #N force along y\n",
+ "l1 = 0.8 #mm\n",
+ "l2 = 0.4 #mm\n",
+ "\n",
+ "#Stress Components:\n",
+ "#Normal Force:\n",
+ "A1 =l1*l2\n",
+ "sigma1 = f_x/A1 #stress = P/A\n",
+ "\n",
+ "#Bending Moment:\n",
+ "M_y1 = 8000 #N\n",
+ "c1 = l2/2.0\n",
+ "I1 = (1/12.0)*(l1*l2**3)\n",
+ "sigma_A1 = (M_y1*c1)/I1 \n",
+ "M_y2 = 16000 #N\n",
+ "c2 = l2\n",
+ "I2 = (1/12.0)*(l2*l1**3)\n",
+ "sigma_A2 = (M_y2*c2)/I2 \n",
+ "\n",
+ "#Resultant:\n",
+ "res_normal= -sigma1-sigma_A1-sigma_A2\n",
+ "\n",
+ "#Display:\n",
+ "\n",
+ "print\"The stress due to normal force at A \",sigma1/1000,\"KPa\"\n",
+ "print\"The stress due to bending moment 8KN at A \",sigma_A1/1000,\"KPa\"\n",
+ "print\"The stress due to bending moment 16KN at A \",sigma_A2/1000,\"KPa\"\n",
+ "print\"The resultant normal stress component at A \",res_normal/1000,\"KPa\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress due to normal force at A 125.0 KPa\n",
+ "The stress due to bending moment 8KN at A 375.0 KPa\n",
+ "The stress due to bending moment 16KN at A 375.0 KPa\n",
+ "The resultant normal stress component at A -875.0 KPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 Page No 420"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate Stress at A\n",
+ "\n",
+ "#Given:\n",
+ "P = 500 #lb, load\n",
+ "r=0.75 #inch, radius\n",
+ "#Stress Components:\n",
+ "\n",
+ "#Normal Force:\n",
+ "import math\n",
+ "A = math.pi*r**2\n",
+ "sigma = P/A\n",
+ "\n",
+ "#Bendng Moments:\n",
+ "M_x =7000 #lb\n",
+ "cy = r\n",
+ "Ix = (1/4.0)*math.pi*(r**4) \n",
+ "sigma_max_1 = (M_x*cy)/Ix \n",
+ "\n",
+ "M_y = P*l_bc/2.0\n",
+ "cx = l_bc/2.0\n",
+ "Iy = (1/12.0)*(l_ab*l_bc**3) #I = (1/12)*(bh**3)\n",
+ "sigma_max_2 = (M_y*cx)/Iy #sigma = My/I\n",
+ "#Superposition\n",
+ "sigmaf=round(sigma/1000,3)+round(sigma_max_1/1000,1)\n",
+ "\n",
+ "#Display:\n",
+ "print\"The normal stress at corner A \",round(sigma/1000,3),\"ksi\"\n",
+ "print\"The normal stress at point A for Bending Moment \",round(sigma_max_1/1000,1),\"ksi\"\n",
+ "print\"The normal stress at point A for Superimposition \",round(sigmaf,1),\"ksi\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The normal stress at corner A 0.283 ksi\n",
+ "The normal stress at point A for Bending Moment 21.1 ksi\n",
+ "The normal stress at point A for Superimposition 21.4 ksi\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page No 421"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the state of stress at point A\n",
+ "\n",
+ "#Given\n",
+ "r=0.75 #radius,inch\n",
+ "V=800 #Forca, lb\n",
+ "\n",
+ "#Calculation\n",
+ "#shear force\n",
+ "import math\n",
+ "Q=(4*r/(3*math.pi))*(0.5*(math.pi*r**2))\n",
+ "Ix=(1/4.0)*math.pi*(r**4) \n",
+ "tau=V*Q/(Ix*2*r)\n",
+ "#Since point A is on neutral axis\n",
+ "sigmaA=0\n",
+ "T=11200 #lb inch, force \n",
+ "Iy=(1/2.0)*math.pi*(r**4) \n",
+ "sigma_a=T*r/Iy\n",
+ "#Superimposition\n",
+ "sigmayzA=tau+sigma_a\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The stress for shear stress distribution is\",round(tau/1000,3),\"ksi\"\n",
+ "print\"The stress for Bending moment is\",sigmaA,\"ksi\"\n",
+ "print\"The stress for torque\",round(sigma_a/1000,2),\"ksi\"\n",
+ "print\"The stress for Superimposition \",round(sigmayzA/1000,1),\"ksi\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress for shear stress distribution is 0.604 ksi\n",
+ "The stress for Bending moment is 0 ksi\n",
+ "The stress for torque 16.9 ksi\n",
+ "The stress for Superimposition 17.5 ksi\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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