{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 10: Columns" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 10.01, Page number 638" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Size of cross section if the column is to safetly support 100 kN = 0.100000 psi \n", "Case(b): Size of cross section if the column is to safetly support 200 kN = 129 psi \n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "L=2 # Length(m)\n", "E=13*(pow(10,9)) # Modulus of elasticity(GPa) \n", "Sall=12 # Stress(MPa)\n", "FS=2.5 # Factor of safety(2.5)\n", "Ld1=100 # Load force(kN) \n", "Ld2=200 # Load force(kN) \n", "\n", "\n", "#Calculation \n", "#(a) For the 100-kN Load\n", "Pcr=FS*Ld1*(1000.0) # Pressure(kN) \n", "I=(Pcr*(L**2))/(((math.pi)**2)*E) # Moment of inertia(m**4) \n", "a1=round((I*12)**(1/4.0),1) # Side of square(mm)\n", "S=(100)/((0.1)**2) # Normal stress in column(MPa) \n", "\n", "#(b) For the 200-kN Load \n", "Pcr=FS*(Ld2)*(1000.0) # Pressure(kN)\n", "I=(Pcr*(L**2))/(((math.pi)**2)*E) # Moment of inertia(m**4)\n", "a=((I)*12)**(1/4.0) # Side of square(mm)\n", "S=(200/(0.11695)**2) # Normal stress(MPa)\n", "A=(200/12.0)*(pow(10,-3)) # Area of cross section(m**2)\n", "a2=(A)**(1/2.0)*(1000) # Side of square(mm)\n", "\n", "#Result\n", "print('Case(a): Size of cross section if the column is to safetly support 100 kN = %lf psi '%a1)\n", "print('Case(b): Size of cross section if the column is to safetly support 200 kN = %.lf psi '%a2)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 10.1, Page number 643" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Ratio for most efficient design = 0.350000 \n", "Case(b): Design for given data breadth = 1.619708 in \n", "Case(b): Design for given data length = 0.566898 in \n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "L=20 # Length of aluminium column(in)\n", "E=10.1*(pow(10,6)) # Modulus of elasticity(psi)\n", "P=5 # Stress(kips) \n", "L=symbols('L') # Variable declaration\n", "Le=symbols('Le') # Variable declaration \n", "Le=0.7*L # Effective length of column\n", "a=symbols('a') # Variable declaration \n", "b=symbols('b') # Variable declaration \n", "Scr=symbols('Scr') # Variable declaration\n", "ry=(b)/(math.sqrt(12)) # Radius of gyration \n", "FS=2.5 # Factor of safety\n", "\n", "#Calculation \n", "#Buckling in xy Plane\n", "Ix=(1/12.0)*(a**3)*b # Moment of inertia\n", "A=a*b # Area of cross section\n", "rz=(((1/12.0)*b*(a**3))/(a*b))**(1/2.0) # Radius of gyration\n", "ESRxy=(Le)/(rz) # Effective slenderness ratio \n", "\n", "\n", "#Buckling in xz Plane \n", "ESRxz=(Le)/(ry) # Effective slenderness ratio\n", "\n", "# Case(a) Most Efficient Design\n", "Ans=((0.7*L)*(math.sqrt(12.0)))/((2.0*L)*(math.sqrt(12.0))) # Most efficient design \n", "\n", "# Case(b) Design for Given Data.\n", "Pcr=(FS)*(P) # Pressure(kips)\n", "a=0.35*b # Distance \n", "A=a*b # Area\n", "Scr=(Pcr/A) # Stress\n", "b=(((12500)*((138.6)**2))/(0.35*10.1*(pow(10.0,6))*((math.pi)**2.0)))**(1/4.0) # Distance(in)\n", "\n", "\n", "#Result\n", "print('Case(a): Ratio for most efficient design = %lf '%Ans)\n", "print('Case(b): Design for given data breadth = %lf in '%b)\n", "print('Case(b): Design for given data length = %lf in '%(0.35*b))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 10.2, Page number 654 " ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a): Allowable load = 31.056698 kips\n", "Case(a): Allowable stress = 8.773078 ksi \n", "Case(b): The horizontal deflection of the top of the column = 0.939000 in \n", "Case(b): Maximum normal stress in the column = 21.981585 ksi \n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "E=(29*(pow(10,6))) # Modulus of elasticity(psi)\n", "FS=2 # Factor of safety \n", "A=3.54 # Area of cross section(in**2) \n", "I=8.00 # Moment of inertia(in**4)\n", "r=1.50 # Radius(in)\n", "c=2.00 # Distance(in) \n", "Lab=8\n", "\n", "#Calculation \n", "# Effective Length\n", "Le=2*(Lab) # Effective length(in)\n", "# Critical Load\n", "Pcr=((((math.pi)**2)*E*(8.0))/(192.0)**2)/(1000.0) # Critical load(kips)\n", "\n", "#Case(a) Allowable Load and Stress\n", "Pall=Pcr/FS # Allowable load(kips)\n", "S=Pall/A # Allowable Stress(ksi)\n", "\n", "#Case(b) Eccentric Load\n", "ym=(0.75)*(2.252-1) # Distance(in)\n", "Sm=(31.1/3.54)*(1+(0.667)*(2.252)) # Distance(in)\n", "\n", "#Result\n", "print('Case(a): Allowable load = %lf kips'%Pall)\n", "print('Case(a): Allowable stress = %lf ksi '%S)\n", "print('Case(b): The horizontal deflection of the top of the column = %lf in '%ym)\n", "print('Case(b): Maximum normal stress in the column = %lf ksi '%Sm)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 10.02, Page number 664" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The longest unsupported length L = 2.318480 m\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "A=1460*(pow(10,-6)) # Area of cross section(m**2)\n", "rx=41.7 # Radius of gyration(mm)\n", "ry=14.6 # Radius of gyration(mm)\n", "P=60*(pow(10,3)) # Force(N)\n", "E=200*(pow(10,9)) # Modulus of elasticity(GPa)\n", "L=symbols('L') # Variable Declaration\n", "r=symbols('r') # Variable Declaration\n", "Scr=symbols('Scr')\n", "\n", "#Calculation \n", "Sall=(P/A) # Stress of all(Pa)\n", "Scr=0.8778*((((math.pi)**2)*(E))/(L/r)**2) # Critical stress\n", "Sall=Scr/1.67 # Stress of all(Pa)\n", "R1=((1.037*(pow(10,12)))/(1.41*(pow(10,6))))**(1/2) # Slenderness ratio\n", "R1l=(4.71)*((200*(pow(10,9)))/(250*(pow(10,6))))**(1/2) # Slenderness ratio\n", "L=158.8*(14.6*(pow(10,-3))) # Choosing the smaller of the two radii of gyration\n", "\n", "#Result\n", "print('The longest unsupported length L = %lf m'%L)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 10.3, Page number 668 " ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Effective centric load P if the effective length of the column is 24 = 81.649348 kips\n", "Effective centric load P if bracing is provided to prevent the movement of the midpoint C in the xz plane = 187.639873 ksi\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "Sy=36 # Stress(ksi)\n", "E=(29*(pow(10,6))) # Modulus of elasticity(psi)\n", "A=11.5 # Area(in**2) \n", "FS=2 # Factor of safety\n", "\n", "\n", "#Calculation \n", "ratio=(4.71)*(E/(36*(pow(10,3)))) # Value of the slenderness ratio\n", "\n", "#Case(a) Effective Length\n", "Sr=(24*12)/(1.98) # Value of the slenderness ratio\n", "Scr=((0.877)*((math.pi)**2)*(29*(pow(10,3))))/(145.5)**2 # Value of the slenderness ratio\n", "Sall=(Scr/1.67) # Allowable stress(ksi)\n", "Pall1=Sall*A # Pressure(kips)\n", "#Case(b) Bracing at Midpoint C \n", "#xz Plane\n", "Elxz=(144)/(1.98) # Slenderness ratio\n", "#yz Plane\n", "Elyz=(288)/(4.27) # Slenderness ratio\n", "\n", "Se=(((math.pi)**2)*(E))/(72.7)**2 # Stress(ksi)\n", "Scr=(0.658)**(36/54.1)*(36) # Stress(ksi) \n", "\n", "Sall=(Scr)/(1.67) # Allowable load(ksi)\n", "Pall2=Sall*A # Force(ksi) \n", "\n", "#Result\n", "print('Effective centric load P if the effective length of the column is 24 = %lf kips'%Pall1)\n", "print('Effective centric load P if bracing is provided to prevent the movement of the midpoint C in the xz plane = %lf ksi'%Pall2)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 10.4, Page number 669 " ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The smallest diameter rod that can be used to support the centric load P=60 kN if L=750 mm = 0.036628 \n", "The smallest diameter rod that can be used to support the centric load P=60 kN if L=300 mm = 0.023900 \n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "P=60 # Centric load(kN)\n", "c=symbols('c') # Variable declaration\n", "I=((math.pi)/4.0)*(c**4) # Moment of inertia\n", "A=(math.pi)*(c**2) # Area of cross section \n", "\n", "#Calculation \n", "r=(I/A)**(1/2.0) # Radius\n", "\n", "#Case(a) Length of 750 mm\n", "c=((60*(pow(10,3))*((0.75)**2)*4)/((math.pi)*(382.0)*(pow(10.0,9))))**(1/4.0)\n", "Sr=(750)/(18.31/2.0)\n", "d1=2*c\n", "\n", "#Case(b) Length of 300 mm\n", "c=11.95 # Radius(mm)\n", "\n", "Sr=((300)*(2))/(11.95) # Slenderness ratio\n", "d2=(2*c)/(1000.0) # Required diamter(mm)\n", "\n", "#Result\n", "print('The smallest diameter rod that can be used to support the centric load P=60 kN if L=750 mm = %lf '%d1)\n", "print('The smallest diameter rod that can be used to support the centric load P=60 kN if L=300 mm = %lf '%d2)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 10.04, Page number 676" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The maximum load that can be safely applied is = 23.278244 kips\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "L=28 # Effective length(in)\n", "P=symbols('P')\n", "\n", "#Calculation \n", "A=2**2 # Area of cross ection(in**2)\n", "I=(1/12.0)*(2**4) # Moment of inertia(in**4)\n", "r=(I/A)**(1/2.0) # Radius(in)\n", "R1=(28/0.5774) # Ratio\n", "Sall=(30.9-(0.229*48.5)) # Stress for the aluminum column(ksi) \n", "P=solve((P/4.0)+((P*0.8*1)/(1.333))-19.79,P) # Maximum load that can be safely applied(kips)\n", "\n", "#Result\n", "print('The maximum load that can be safely applied is = %lf kips'%P[0])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 10.05, Page number 677" ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The maximum load that can be safely applied is = 26.568262 kips\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "Sallcentric=19.79 # Stress all centric(ksi) \n", "Sallbending=24 # Stress all bending(ksi)\n", "P=symbols('P')\n", "\n", "#Calculation \n", "P=solve((P/4.0)/(19.79) + ((P*0.8*1)/(1.333*24))-1,P)\n", "\n", "#Result\n", "print('The maximum load that can be safely applied is = %lf kips'%P[0])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 10.5, Page number 678" ] }, { "cell_type": "code", "execution_count": 28, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The largest allowable load P is thus = 327.340184 kN\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "E=200 # Modulus of elasticity(GPa)\n", "Sy=250 # Stress(MPa)\n", "P=symbols('P') # Variable declaration\n", "\n", "#Calculation \n", "Sall=162.2/1.67 # Allowable stress(MPa)\n", "\n", "P=solve(P*((1/(9.42*(pow(10,-3))))+(0.200/(1.050*(pow(10,-3)))))-97.1,P) # Largest load P(kN)\n", "\n", "#Result\n", "print('The largest allowable load P is thus = %lf kN'%(P[0]*1000))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 10.6, Page number 678" ] }, { "cell_type": "code", "execution_count": 29, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The largest allowable load P is thus = 423.169834 kN\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "E=200 # Modulus of elasticity(GPa)\n", "Sy=150 # Stress(MPa)\n", "P=symbols('P') # Variable declaration\n", "\n", "\n", "#Calculation \n", "P=solve(P*((1/(97.1*9.42*(pow(10,3))))+(0.200/(150*1.050*(pow(10,3)))))-1,P) # Largest allowable load(kN)\n", "\n", "#Result\n", "print('The largest allowable load P is thus = %lf kN'%(P[0]/1000)) " ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## SAMPLE PROBLEM 10.7, Page number 679" ] }, { "cell_type": "code", "execution_count": 30, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Selection of Shape. The shape to be used is W8*48\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "A=symbols('A') # Variable declaration \n", "E=29*(pow(10,6)) # Modulus of elasticity(psi) \n", "Sy=36 # Stress(ksi) \n", "SR=133.7 # Slenderness ratio\n", "Scr=22.5 # Slenderness ratio\n", "\n", "#Calculation \n", "A=solve(22-(85/A)-((425*4)/(A*(pow(3.5,2)))),A) # Area of cross section(in**2) \n", "\n", "#Trial 1: W8*35\n", "#Allowable Bending Stress\n", "Sall=22 # Allowable bending stress(ksi) \n", "#Allowable Concentric Stress\n", "Ratio1=(8.25/13.46)+(13.62/22.0) # Allowable concentric stress(ksi) \n", "#Since 1.232 > 1.000, the requirement expressed by the interaction formula is not satisfied; we must select a larger trial shape.\n", "\n", "#Trial 2: W8*48\n", "Ratio2=(6.03/13.76)+(9.82/22.0) # Allowable concentric stress(ksi)\n", "#The W8*48 shape is satisfactory but may be unnecessarily large\n", "\n", "#Trial 3: W8*40\n", "#Following again the same procedure, we find that the interaction formula is not satisfied.\n", "\n", "print ('Selection of Shape. The shape to be used is W8*48')" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.1" } }, "nbformat": 4, "nbformat_minor": 0 }