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| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d (patch) | |
| tree | 22789c9dbe468dae6697dcd12d8e97de4bcf94a2 /Mechanics_of_Materials_by_R._C._Hibbeler/Chapter13.ipynb | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
| download | Python-Textbook-Companions-6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d.tar.gz Python-Textbook-Companions-6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d.tar.bz2 Python-Textbook-Companions-6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d.zip | |
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diff --git a/Mechanics_of_Materials_by_R._C._Hibbeler/Chapter13.ipynb b/Mechanics_of_Materials_by_R._C._Hibbeler/Chapter13.ipynb new file mode 100755 index 00000000..2e2e42c7 --- /dev/null +++ b/Mechanics_of_Materials_by_R._C._Hibbeler/Chapter13.ipynb @@ -0,0 +1,636 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13:Buckling of Columns"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page NO 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l = 12 #ft, length\n",
+ "E = 29*10**3 #GPa, stress\n",
+ "ro = 75 #mm, outside radius\n",
+ "ri = 70 #mm, inside radius\n",
+ "sigma_y = 250 #MPa, stress\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Ix=110\n",
+ "Iy=37\n",
+ "A = 9.13\n",
+ "Pcr = (math.pi**2*(E*10**3)*Iy)/((l*12)**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "sigma_cr = (Pcr*1000)/A\n",
+ "p=36*A\n",
+ " \n",
+ "print\"The maximum allowable axial load that the column can support = \",round(p,0),\"kip\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum allowable axial load that the column can support = 329.0 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.2 Page NO 668"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E = 29*10**3 #GPa, stress\n",
+ "lx= 144 #inch, length\n",
+ "ly=100.8 #inch\n",
+ "A =4.43 #inch**2, area\n",
+ "sigma_y = 60 #ksi, stress\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Ix=29\n",
+ "Iy=9.32\n",
+ "\n",
+ "Pcrx = ((math.pi**2)*E*Ix)/(lx**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "Pcry = ((math.pi**2)*E*Iy)/(ly**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "sigma_cr = (Pcr*1000)/A\n",
+ "sigmacr = Pcry/A #in kN\n",
+ "if sigmacr<sigma_y:\n",
+ " print\"Buckling will occue before the material yield. So Psr=\",round(Pcry,0),\"kip\"\n",
+ "else:\n",
+ " print\"n\"\n",
+ " \n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Buckling will occue before the material yield. So Psr= 263.0 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.3 Page No 669"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E = 70 #GPa\n",
+ "Ix = 61.3*10**-6 #Moment of inertia along x-axis\n",
+ "Iy = 23.2*10**-6 ##Moment of inertia along y-axis\n",
+ "l = 5\n",
+ "KLx = 2*l #m\n",
+ "KLy = 0.7*(l) #m\n",
+ "FS = 3 #Factor of safety\n",
+ "sigma_y = 215 #MPa\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Pcrx = (math.pi**2*E*10**6*Ix)/(KLx**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "Pcry = (math.pi**2*E*10**6*Iy)/(KLy**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "Pcr = min(Pcrx,Pcry)\n",
+ "A = 7.5*10**-3 #mm**2\n",
+ "P_allow = Pcr/FS\n",
+ "sigma_cr = (Pcr*10**-3)/A\n",
+ "\n",
+ "\n",
+ "if(sigma_cr<sigma_y):\n",
+ " print\"The largest allowable load that the column can support = \",round(P_allow,0),\"kN\"\n",
+ "else:\n",
+ " print\"n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The largest allowable load that the column can support = 141.0 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page NO 683"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "z1 = 4*1000 #mm, length\n",
+ "e = 200 #mm, elongation\n",
+ "KLy = 100.8 #inch**2\n",
+ "Iy = 49.1 #inch**4\n",
+ "E = 29*10**3 #ksi, stress\n",
+ "sigma_y =421.2 #MPa\n",
+ "\n",
+ "#Calculation\n",
+ "#y-y Axis Buckling\n",
+ "import math\n",
+ "Pcry = (math.pi**2*E*10**6*Iy)/(KLy**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "Pcry = Pcry/1000\n",
+ "#x-x Axis Yielding\n",
+ "Kx= 2\n",
+ "KLx = 288 #inch\n",
+ "A=11.7\n",
+ "c = (8.25)/2.0\n",
+ "rx = 3.53 #inch\n",
+ "\n",
+ "#Solved by applying the Secant Formula and then finding Px by trial and error\n",
+ "trial_Px = 88.4 #kN\n",
+ "A = 7850#mm**2\n",
+ "sigma = (trial_Px*1000)/(A)\n",
+ "\n",
+ "if(Pcry>trial_Px and sigma<sigma_y):\n",
+ " print'The maximum eccentric load that the column can support = ',trial_Px\n",
+ " print'Failure will occur about the x-x axis.'\n",
+ "\n",
+ "else:\n",
+ " print\"n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum eccentric load that the column can support = 88.4\n",
+ "Failure will occur about the x-x axis.\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5 Page NO 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d = 30 #mm, diameter\n",
+ "r = d/2 #mm, radius\n",
+ "L = 600 #mm\n",
+ "sigma_pl = 150#MPa, stress\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "I = (math.pi/4)*(r**4)\n",
+ "A = math.pi*r**2\n",
+ "r_gyr = sqrt(I/A)\n",
+ "K = 1\n",
+ "sl_ratio = (K*L)/(r_gyr)\n",
+ "flag1 = 0\n",
+ "\n",
+ "#Assuming the critical stress is elastic\n",
+ "E = 150/0.001\n",
+ "sigma_cr1 = (math.pi**2*E)/(sl_ratio**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ "\n",
+ "\n",
+ "if(sigma_cr1 > sigma_pl):\n",
+ " Et = (270 - 150)/(0.002 - 0.001)\n",
+ " sigma_cr2 = (math.pi**2*Et)/(sl_ratio**2) #Pcr = (math.pi**2*EI)/(l**2)\n",
+ " \n",
+ "if(sigma_cr2>150 and sigma_cr2<270): \n",
+ " Pcr = sigma_cr2*A\n",
+ " Pcr = Pcr/1000.0 #in kN\n",
+ " print'The critical load when used as a pin supported column = ',round(Pcr,0),\"kN\"\n",
+ " \n",
+ "else:\n",
+ " print\"\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical load when used as a pin supported column = 131.0 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.6 page No 696"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "l=16 #16 ft\n",
+ "A=29.4 #in**2, area\n",
+ "rx=4.60 #in\n",
+ "ry=2.65 #in\n",
+ "k=1\n",
+ "E=29*10**3 #Stress\n",
+ "sigmay=36.0 #ksi\n",
+ "#calculation\n",
+ "import math\n",
+ "x1=k*l*12/ry\n",
+ "x=math.sqrt(2*math.pi**2*E/sigmay)\n",
+ "a=(1-(x1**2/(2.0*x**2)))*((sigmay))\n",
+ "b=(5/3.0)+((3/8.0)*(x1)/(x))-((x1**3)/(8.0*(x**3)))\n",
+ "sigmaallow=a/b\n",
+ "P=sigmaallow*A\n",
+ "\n",
+ "#result\n",
+ "print\" The largest load is\",round(P,0),\"kip\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The largest load is 476.0 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.7 Page NO 697"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P = 18 #kip, load\n",
+ "E = 29*10**3 #ksi, stress\n",
+ "sigma_y = 50 #ksi, shear stress\n",
+ "l = 15 #ft, length\n",
+ "k =0.5 #shape factor\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "I_by_d = (1/4.0)*(math.pi)*(d/2.0)**4\n",
+ "A_by_d = (1/4.0)*(math.pi)*d**2\n",
+ "r_by_d = math.sqrt(I_by_d/A_by_d)\n",
+ "sl_ratio_c = math.sqrt((2*math.pi**2*E)/(sigma_y))\n",
+ "\n",
+ "a1=math.sqrt(2*(math.pi)**2*E/(sigma_y))\n",
+ "\n",
+ "d_=((18*4*16*23*(k**2)*(l**2)*12**2)/(12*math.pi**3*E))**(1/4.0)\n",
+ "print \"The smallest diameter is \",round(d_,2),\"inch. So use d=2.25 inch\"\n",
+ "d=2.25\n",
+ "a1=k*l*12/(d/4.0)\n",
+ "if a1<200:\n",
+ " print\"Use of equation is appropriate\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The smallest diameter is 2.11 inch. So use d=2.25 inch\n",
+ "Use of equation is appropriate\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page NO 698"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L = 30 #inch\n",
+ "P = 12 #kip\n",
+ "sigma =28.0 #ksi\n",
+ "K = 1\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "b2 = (P)/(2*sigma)\n",
+ "b_ = math.sqrt(b2)\n",
+ "A = 2*b_*b_\n",
+ "Iy = (1/12.0)*(2*b_*b_**3)\n",
+ "ry = sqrt(Iy/A)\n",
+ "sl_ratio = (K*L)/(ry)\n",
+ "if(sl_ratio>12):\n",
+ " b4 = (P*103.9**2)/(2*54000) #Eqn 13.26\n",
+ " b = b4**(1/4.0)\n",
+ " \n",
+ " sl_ratio_ = (2598.1)/(b)\n",
+ " w = 2*b\n",
+ "else:\n",
+ " print\"j\"\n",
+ " \n",
+ "if(sl_ratio>55):\n",
+ " print'The thickness of the bar = ',round(b,2),\"inch\"\n",
+ " \n",
+ "else:\n",
+ " print\"h\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of the bar = 1.05 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9 Page NO 699"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P = 5.0 #kip. load\n",
+ "y1 = 5.5 #inch, length\n",
+ "x1 = 1.5 #inch\n",
+ "A = (x1*y1) #area\n",
+ "d = 1.5 #inch\n",
+ "K = 1\n",
+ "\n",
+ "#Eqn 13.29\n",
+ "L2 = (540*A*d**2)/(P)\n",
+ "L = sqrt(L2)\n",
+ "KL_d = (K*L)/(d)\n",
+ "\n",
+ "if(KL_d>26 and KL_d<=50):\n",
+ " print'The greatest allowable length L as specified by the NFPA = ',round(L,1),\"inch\"\n",
+ "else:\n",
+ " print\"j\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The greatest allowable length L as specified by the NFPA = 44.8 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.10 Page NO 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#the given dimansion are\n",
+ "L = 80.0 #inch\n",
+ "K = 2.0\n",
+ "l = 4.0 #inch\n",
+ "b = 2.0 #inch\n",
+ "e = 1 #inch\n",
+ "c = 2 #inch\n",
+ "\n",
+ "#Calculations\n",
+ "I1 = (1/12.0)*(l*b**3)\n",
+ "A = l*b\n",
+ "r = sqrt(I1/A)\n",
+ "sl_ratio = (K*L)/(r)\n",
+ "\n",
+ "#Eqn 13.26\n",
+ "sigma_allow = (54000)/(sl_ratio**2)\n",
+ "I2 = (1/12.0)*(b*l**3)\n",
+ "coefficient = (1/A) + (e*c)/I2\n",
+ "\n",
+ "sigma_max = sigma_allow\n",
+ "P = sigma_max/coefficient\n",
+ "\n",
+ "#Display\n",
+ "print'The load that can be supported if the column is fixed at its base ',P,\"kip\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The load that can be supported if the column is fixed at its base 2.25 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page No 706"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "import math\n",
+ "sigmaB_allow = 22 #ksi, allowable stress\n",
+ "E = 29*10**3 #ksi, stress\n",
+ "sigma_y = 36 #ksi, shear stress\n",
+ "K= 1 #shape factor\n",
+ "A = 5.87 #inch**2, area\n",
+ "Ix = 41.4 #inch**4, moment of inertia\n",
+ "ry = 1.5 #inch\n",
+ "d = 6.2 #inch\n",
+ "c= d/2.0 \n",
+ "e = 30 #inch\n",
+ "L = 15 #ft\n",
+ "\n",
+ "sl_ratio = (K*L*12)/(ry)\n",
+ "sl_ratio_c = math.sqrt((2*math.pi**2*E)/(sigma_y))\n",
+ "\n",
+ "\n",
+ "\n",
+ "if(sl_ratio<sl_ratio_c):\n",
+ " num = (1 - (sl_ratio**2/(2*sl_ratio_c**2)))*sigma_y\n",
+ " denom1 = (5/3.0) + ((3/8.0)*sl_ratio/sl_ratio_c)\n",
+ " denom2 = (sl_ratio**3)/(8*sl_ratio_c**3)\n",
+ " sigmaA_allow = num/(denom1 - denom2)\n",
+ " \n",
+ " coeffP = 1/(sigmaA_allow*A) + (e*c)/(Ix*sigmaB_allow)\n",
+ " P = 1/coeffP\n",
+ " \n",
+ " sigA = (P/A)/(sigmaA_allow)\n",
+ "else:\n",
+ " print\"k\"\n",
+ " \n",
+ " \n",
+ "if(sigA < 0.15):\n",
+ " print'The maximum allowable value of eccentric load = ',round(P,2),\"kN\"\n",
+ "else:\n",
+ " print\"h\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum allowable value of eccentric load = 8.43 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12 Page NO 707"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K = 2 #shape factor\n",
+ "d= 3.0 #inch, diameter\n",
+ "L = 60 #inch, length\n",
+ "e = 4 #inch\n",
+ "c = d\n",
+ "l = 3.0 #inch\n",
+ "b =6.0 #inch\n",
+ "A = l*b #inch**2\n",
+ "\n",
+ "#Calculations\n",
+ "sl_ratio = (K*L)/(d)\n",
+ "\n",
+ "if(sl_ratio>26 and sl_ratio<50):\n",
+ " sigma_allow = (540)/(sl_ratio**2)\n",
+ " sigma_max = sigma_allow\n",
+ " \n",
+ " I = (1/12.0)*(l*b**3)\n",
+ " coeffP = (1/A) + (e*c)/(I)\n",
+ " P = sigma_max/coeffP\n",
+ " print'The eccentric load that can be supported = ',round(P,2),\"kip\"\n",
+ "else:\n",
+ " print\"no\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The eccentric load that can be supported = 1.22 kip\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |