{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 4 : Pure Bending" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.01, Page number 232" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Bending moment = 30.000000 kip.in\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "c=1.25 # Radius(in)\n", "Sy=36 # Stress(ksi)\n", "b=0.8 # Breadth(in)\n", "h=2.5 # Height(in)\n", "\n", "#Calculation \n", "I=(1/12.0)*(b)*(h)**3 # Centroidal moment of inertia(in**4)\n", "M=(I/c)*(Sy) # Bending moment(kip.in)\n", "\n", "# Result\n", "print ('Bending moment = %lf kip.in' %M)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.02, Page number 233" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum tensile stress = 193.397171 MPa\n", "Maximum compressive stress = -142.602829 MPa\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "r=12 # Radius(mm)\n", "p=2.5 # Mean radius(m) \n", "E=70 # Modulus of rigidity(GPa)\n", "n=-1\n", "\n", "#Calculation \n", "Y=(4*r)/(3*(math.pi)) # Ordinate(mm)\n", "c=r-Y # Distance from the neutral axis to the point of crossection(mm) \n", "Em=(c*(10**-3))/p # Maximum absolute value of the strain\n", "Sm=((E*(pow(10,9)))*Em)/(pow(10,6)*(1.0)) # Maximum tensile stress(MPa)\n", "Scomp=(n)*(Y/c)*(Sm) # Maximum compressive stress(MPa) \n", "\n", "# Result\n", "print ('Maximum tensile stress = %lf MPa' %Sm)\n", "print ('Maximum compressive stress = %lf MPa' %Scomp)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.1, Page number 235" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Bending moment M for which factor of safety is 3 = 103.760000 kip.in\n", "Radius of curvature of tube = 110.400000 ft\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "sY=40 # Stress(ksi)\n", "sU=60 # Stress(ksi)\n", "E=(10.6)*(pow(10,6)) # Modulus of rigidity(psi)\n", "FS=3 # Factor of safety\n", "\n", "#Calculation\n", "#Moment of Inertia\n", "E=(10.6)*(pow(10,6)) # Modulus of rigidity(psi)\n", "I=round(((1/12.0)*3.25*pow(5,3))-((1/12)*(2.75)*pow(4.5,3)),2) # Centroidal moment of inertia of a rectangle\n", "#Allowable Stress\n", "sALL=(sU/FS) # Allowable stress(ksi)\n", "#Case(a) Bending Moment\n", "c=(1/2.0)*(5) # Radius(in)\n", "M=((12.97)*(20))/2.5 # Bending moment(kip.in)\n", "#Case(b) Radius of Curvature\n", "p=round((10.6*pow(10,6)*12.97)/(103.8*pow(10,3)),1) # Radius of curvature(in)\n", "p=round((p*0.08333),1) # Converting into feet(ft) \n", "#Alternative Solution.\n", "Em=(sALL/(E*(pow(10,-3))*(1.0))) # Maximum strain(in./in)\n", "p=(c/Em) # Radius of curvature(in)\n", "p=round((p*0.08333),1) # Converting into feet(ft) \n", "\n", "# Result\n", "print ('Bending moment M for which factor of safety is 3 = %lf kip.in' %M)\n", "print ('Radius of curvature of tube = %lf ft' %p)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.2, Page number 236" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum tensile stress = 76.036866 MPa\n", "Maximum compressive stress = -131.336406 MPa\n", "Radius of curvature = 47.740000 ft\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "n=-1\n", "\n", "#Calculation\n", "#Centroid\n", "sumA=3000 # Summing up the area(mm**2) \n", "M=3 # Couple(kN.m) \n", "cA=0.022 # Distance(m) \n", "Y=(114*pow(10,6))/(3000.0) # Distance(mm)\n", "#Centroidal Moment of Inertia\n", "Ix=((1/12.0)*(90)*(pow(20,3)) + (90*20*pow(12,2)) + ((1/12.0)*(30)*(pow(40,3))) + (30*40*pow(18,2)))/(pow(10,12)*(1.0)) # Centroidal moment of inertia(m**4) \n", "#Case(a) Maximum Tensile Stress\n", "sA=((M*cA)/(Ix)*(1.0))/(1000.0) # Maximum tensile stress(MPa) \n", "#Maximum Compressive Stress\n", "sB=n*(3*0.038)/((868*pow(10,-9)*pow(10,3))) # Maximum compressive stress(MPa) \n", "#Case(b) Radius of Curvature\n", "p=((165*868*(pow(10,-9)))/(3))*(pow(10,6)) # Radius of curvature(m)\n", "\n", "\n", "# Result\n", "print ('Maximum tensile stress = %lf MPa' %sA)\n", "print ('Maximum compressive stress = %lf MPa' %sB)\n", "print ('Radius of curvature = %lf ft' %p)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.03, Page number 244" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum stress in brass portion = 11.851852 ksi\n", "Maximum stress in steel portion = 22.909630 ksi\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "Es=29*(pow(10,6)) # Modulus of rigidity(psi)\n", "Eb=15*(pow(10,6))*(1.0) # Modulus of rigidity(psi) \n", "M=40 # Bending moment(kip.in)\n", "h=3 # Height(3)\n", "b=2.25 # Breadth(in) \n", "c=1.5 # Distance(in)\n", "\n", "#Calculation \n", "n=Es/Eb # Ratio\n", "W=0.75*n # width(in)\n", "I=(1/12.0)*(b)*((h)**3) # Moment of inertia of the transformed section(in**4)\n", "Sm=(M*c)/(I) # Maximum stress in the transformed section(ksi)\n", "Sbrass=Sm # Maximum stress in brass portion(ksi)\n", "Ssteel=1.933*(Sbrass) # Maximum stress in steel portion(ksi)\n", "\n", "# Result\n", "print ('Maximum stress in brass portion = %lf ksi' %Sbrass)\n", "print ('Maximum stress in steel portion = %lf ksi' %Ssteel)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.04, Page number 247" ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Smallest allowable width of the groves = 10.400000 mm\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "depth=10 # Depth(mm)\n", "width=60 # Width(mm)\n", "thickness=9 # Thickness(mm)\n", "Smax=150 # Maximum stress(MPa) \n", "M=180 # Bending moment(N.m) \n", "\n", "#Calculation \n", "d=width-(2*depth) # Distance(mm) \n", "c=(1/2.0)*d # Distance(mm) \n", "b=9 # Distance(mm)\n", "I=(1/12.0)*(b*(pow(10,-3)))*((d*(pow(10,3)))**3) # Moment of inertia of the critical cross section(m**4)\n", "Ratio=((M)*(c)*(pow(10,3)))/(I) # Stress(MPa)\n", "k=150/75.0 # Factor \n", "Ratio2=width/(d*1.0) # Ratio\n", "r=0.13*40 # Radius(mm)\n", "wid=2*r # Width(mm)\n", "\n", "# Result\n", "print ('Smallest allowable width of the groves = %lf mm' %wid)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.3, Page number 248" ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum stress in the wood = 4.570000 MPa\n", "Stress in steel = 43.800000 MPa\n" ] } ], "source": [ "import math\n", "\n", "# Variable declaration\n", "Es=200 # Moduluss of rigidity(GPa)\n", "Ew=12.5 # Moduluss of rigidity(GPa) \n", "\n", "#Transformed Section.\n", "n=(Es/Ew) # Ratio\n", "#Neutral Axis\n", "Y=round(((0.160)*(3.2*0.020))/(3.2*0.020+0.470*0.300),2) # Distance(m) \n", "#Centroidal Moment of Inertia\n", "I=round(((1/12)*0.470*(pow(0.3,3)))+(0.470*0.3*(pow(0.05,2)))+((1/12)*(3.2)*(pow(0.020,3)))+(3.2*0.020*(pow(0.160-0.050,2))),5) # Centroidal Moment of Inertia \n", "#Maximum Stress in Wood\n", "sW=((50*(pow(10,3)))*(0.200))/(2.19*pow(10,-3)) # Maximum stress in wood(MPa)\n", "sW=round((sW/(pow(10,6))),2) # Rounding\n", "#Stress in Steel\n", "sS=((16)*(50*(pow(10,3)))*(0.120))/(2.19*(pow(10,-3))) # Stress in steel(MPa) \n", "sS=round((sS/(pow(10,6))),1) # Rounding\n", "\n", "# Result\n", "print ('Maximum stress in the wood = %lf MPa' %sW)\n", "print ('Stress in steel = %lf MPa' %sS)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.4, Page number 249" ] }, { "cell_type": "code", "execution_count": 23, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum stress in concrete = 1.306000 MPa\n", "Stress in steel = 18.520000 MPa\n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve\n", "\n", "# Variable declaration\n", "x=symbols('x') # Variable declartion \n", "d=(5/8.0) # Diameter(in)\n", "Es=(29*(pow(10,6))) # Modulus of elasticity for concrete(psi)\n", "Ec=(3.6*(pow(10,6))) # Modulus of elasticity for stell(psi)\n", "\n", "# Calculation\n", "#Transformed Section\n", "As=round((2)*(math.pi)*(pow(d/2,2)),3) # Cross sectional area(in**2)\n", "n=round(Es/Ec,2) # Ratio\n", "TA=round(n*As,2) # Transformed steel area(in**2)\n", "#Neutral Axis\n", "x=(solve((12*x)*(x/2.0)-(4.95)*(4-x)))\n", "#Moment of inertia\n", "I=(1/3.0)*(12)*((1.450)**3) + 4.95*(4-1.450)**2 # Centroidal moment of inertia(in**4) \n", "#Maximum Stress in Concrete\n", "sC=round(((40)*(1.450))/(44.4),3) # Maximum stress in concrete(ksi) \n", "#Stress in Steel\n", "sS=round(((8.06)*(40)*(2.55))/(44.4),2) # Stress in steel(ksi)\n", "\n", "# Result\n", "print ('Maximum stress in concrete = %lf MPa' %sC)\n", "print ('Stress in steel = %lf MPa' %sS)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.05, Page number 260" ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Thickness of elastic core = 80.000000 mm\n", "Radius of curvature = 33.333333 m\n" ] } ], "source": [ "import math\n", "from sympy import symbols, solve \n", "\n", "#Variable declaration\n", "M=36.8\n", "c=60*(pow(10,-3)) # Half length of rod(mm)\n", "b=50*(pow(10,-3)) # Breadth of rod(mm)\n", "Sy=240 # Stress(MPa) \n", "yY=symbols('yY') # Variable declaration\n", "E=200 # Modulus of elasticity(GPa)\n", "n=-1\n", "\n", "#Calculation \n", "# Case(a)\n", "Rt=(2/3.0)*(b)*(c)**2 # Ratio(m**3)\n", "My=Rt*Sy # Maximum elastic moment(kN.m) \n", "yY=solve(36.8-(3/2.0)*(28.8)*(1-((1/3.0)*(((yY)**2)/(60**2)))),yY) # \n", "# Case(b)\n", "Ey=(Sy*(pow(10,6)))/(E*(pow(10,9)*(1.0)))\n", "p=(yY[0]*(pow(10,-3)))/(Ey)\n", "\n", "# Result\n", "print ('Thickness of elastic core = %lf mm' %(n*(2*yY[0])))\n", "print ('Radius of curvature = %lf m' %(n*p))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.06, Page number 262" ] }, { "cell_type": "code", "execution_count": 27, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Residual stress = 306.666667 MPa\n", "Radius of curvature after unloading = 225 m\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "M=36.8 # Bending moment(kN) \n", "Sy=240 # Yield strength(MPa)\n", "yY=40 # Thickness of elastic core(mm) \n", "n=-1\n", "Sx=n*35.5*(pow(10,6)) # Stress(Pa) \n", "E=200*(pow(10,9))\n", "\n", "#Calculation \n", "# Case(a)\n", "Sml=((36.8)/(120*(pow(10,-6))))/(1000) # Residual stress(MPa) \n", "# Case(b)\n", "Ex=Sx/E # Residual strain \n", "p=(n*(40*(pow(10,-3))))/(Ex) # Radius of Curvature after Unloading(m)\n", "\n", "# Result\n", "print ('Residual stress = %lf MPa' %Sml)\n", "print ('Radius of curvature after unloading = %.lf m' %p)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.5, Page number 263" ] }, { "cell_type": "code", "execution_count": 28, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a) Bending moment = 9525.000000 kip.in\n", "Case(a) Radius of curvature = 4640 in\n", "Case(b) Bending moment = 10226.342000 kip.in\n", "Case(b) Radius of curvature = 338 ft\n" ] } ], "source": [ "import math\n", "\n", "# variable declaration\n", "E=(29*pow(10,6)) # Modulus of elastoplasticity(psi)\n", "sY=50 # Stress(ksi) \n", "\n", "# Calculation\n", "#Case(a) Onset Of Yield\n", "I=round((1/12.0)*(12)*(pow(16,3))-(1/12.0)*(12-0.75)*(pow(14,3)),0) # Centroidal moment of inertia(in**4) \n", "#Bending Moment\n", "sMAX=sY # Stress(ksi)\n", "c=8.0 # Distance(in)\n", "My=(sY*I)/c # Bending moment(kip.in)\n", "#Radius of Curvature \n", "Ey=sY/(E*(1.0)) # Strain\n", "pY=(c/Ey)/(1000.0) # Radius of curvature(in)\n", "#Case(b) Flanges Fully Plastic\n", "R1=50*12*1 # Compressive forces on top(kips) \n", "R4=R1 # Compressive forces on top(kips) \n", "R2=round((1/2.0)*(50)*(7)*(0.75)+0.05,1) # Compressive forces on top half(kips)\n", "R3=R2 # Compressive forces on top half(kips)\n", "#Bending Moment\n", "M=2*((R1*7.5)+(R2*4.67)) # Bending moment(kip.in)\n", "#Radius of Curvature\n", "p=round(((7/0.001724)*0.0833),0) # Radius of curvature(ft) \n", "\n", "# Result\n", "print ('Case(a) Bending moment = %lf kip.in' %My)\n", "print ('Case(a) Radius of curvature = %.lf in' %pY)\n", "print ('Case(b) Bending moment = %lf kip.in' %M)\n", "print ('Case(b) Radius of curvature = %.lf ft' %p)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.6, Page number 264" ] }, { "cell_type": "code", "execution_count": 29, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a) Plastic moment = 44.160000 kN.m\n" ] } ], "source": [ "import math\n", "\n", "\n", "# Variable declaration\n", "sY=240 # Yield strength(MPa)\n", "A1=(0.1*0.02) # Area of cross section(m**2)\n", "A2=(0.02*0.02) # Area of cross section(m**2)\n", "A3=(0.02*0.06) # Area of cross section(m**2)\n", "A4=(0.06*0.02) # Area of cross section(m**2)\n", "\n", "# Calculation\n", "#Neutral Axis\n", "A=(100)*(20) + (80)*(20) + (60)*(20) # Total area(mm**2)\n", "y=(2400-((20)*(100)))/(20) # Distance(mm)\n", "#Plastic Moment\n", "R1=(A1*sY*1000) # Resultant force(kN) \n", "R2=(A2*sY*1000) # Resultant force(kN)\n", "R3=(A3*sY*1000) # Resultant force(kN) \n", "R4=(A4*sY*1000) # Resultant force(kN) \n", "\n", "Mp=(0.030*R1) + (0.010*R2) + (0.030*R3) + (0.070*R4) # Plastic moment(kN.m)\n", "\n", "# Result\n", "print ('Case(a) Plastic moment = %lf kN.m' %Mp)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.7, Page number 265" ] }, { "cell_type": "code", "execution_count": 30, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Case(a) Residual stress = 53.700787 ksi\n", "Case(a) Permanent radius of curvature = 5620.000000 ft\n" ] } ], "source": [ "import math\n", "\n", "# Variable declaration\n", "y=7 # Distance(in)\n", "s=-3.01 # Stress(ksi)\n", "\n", "# Calculation\n", "#Loading\n", "M=10230 # Couple of moment(kip.in)\n", "#Elastic Unloading\n", "sMl=((10230)*(8))/(1524.0) # Maximum stress(ksi)\n", "#Permanent Radius of Curvature\n", "p=round(((7)*(29*pow(10,6))*(pow(10,-3)))/(3.01),-2) # Permanent radius of curvature(in)\n", "p=round((p*0.083333),-1) # Conversion(ft)\n", "\n", "# Result\n", "print ('Case(a) Residual stress = %lf ksi' %sMl)\n", "print ('Case(a) Permanent radius of curvature = %lf ft' %p)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.8, Page number 273" ] }, { "cell_type": "code", "execution_count": 38, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Largest allowable force = 76.972418 kN\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "# Variable declaration\n", "M=symbols('M') # Variable declaration\n", "P=symbols('P') # Variable declaration\n", "\n", "# Calculation\n", "#Properties of Cross Section.\n", "A=(3*pow(10,-3)) # Area of cross section(mm**2)\n", "Y=0.038 # Distance(m)\n", "I=868*(pow(10,-9)) # Moment of inertia(m**4)\n", "d=(0.038-0.010) # Distance(m) \n", "#Force and Couple at C\n", "M=0.028*P # Equivalent force-couple system\n", "s0=(P*(round((1/(3*pow(10,-3))),0))) # Stress compression\n", "s1=(P*round(((0.028)*(0.022))/(868*(pow(10,-9))),0)) # Stress tension \n", "s2=(P*round(((0.028)*(0.038))/(868*pow(10,-9)))) # Stress compression\n", "#Superposition\n", "sA=-s0+s1 # Stress tension\n", "sB=-s0-s2 # Stress compression\n", "#Largest Allowable Force\n", "P1=(30/377.0)*(1000) # The magnitude of P for which the tensile stress at point A is equal to the allowable tensile stress of 30 MPa is found by writing\n", "P2=(120/1559.0)*(1000) # Determine the magnitude of P for which the stress at B is equal to the allowable compressive stress of 120 MPa.\n", "\n", "# Result\n", "print ('Largest allowable force = %lf kN' %P2)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.09, Page number 285" ] }, { "cell_type": "code", "execution_count": 42, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Stress at A = -2.625000 MPa\n", "Stress at B = -1.375000 MPa\n", "Stress at C = 1.625000 MPa\n", "Stress at D = 0.375000 MPa\n", "Neutral axis BG = 36.666667 mm\n", "Neutral axis HA = 70.000000 mm\n" ] } ], "source": [ "import math\n", "\n", "#Variable declaration\n", "F=4.80 # Couple(lb)\n", "l=80 # Length(mm)\n", "b=120 # Breadth(mm)\n", "n=-1\n", "\n", "#Calculation \n", "# Case(a)\n", "Mx=(F)*(40) # Stress in x(N.m)\n", "Mz=(F)*(60-35) # Stress in z(N.m)\n", "A=(0.080)*(0.120) # Area of cross section(m**2)\n", "Ix=(1/12.0)*(0.120)*(0.080**3) # Centroidal moment of inertia(m**4)\n", "Iz=(1/12.0)*(0.080)*(0.120**3) # Centroidal moment of inertia(m**4)\n", "S0=n*(P/A) # Stress(MPa) \n", "S1=((192)*(40))/(5.12*(pow(10,-6))) # Stress in x(MPa)\n", "S2=((120)*(60))/(11.52*(pow(10,-6))) # Stress in y(MPa)\n", "Sa=-0.5-1.5-0.625 # Stress at A(MPa)\n", "Sb=-0.5-1.5+0.625 # Stress at B(MPa)\n", "Sc=-0.5+1.5+0.625 # Stress at C(MPa) \n", "Sd=-0.5+1.5-0.625 # Stress at D(MPa)\n", "\n", "\n", "# Case(b)\n", "BG=((1.375)/(1.625+1.375))*80 # Distance(mm)\n", "HA=((2.625)/(2.625+0.375))*80 # Distance(mm)\n", "\n", "# Result\n", "print ('Stress at A = %lf MPa' %Sa)\n", "print ('Stress at B = %lf MPa' %Sb)\n", "print ('Stress at C = %lf MPa' %Sc)\n", "print ('Stress at D = %lf MPa' %Sd)\n", "print ('Neutral axis BG = %lf mm' %BG)\n", "print ('Neutral axis HA = %lf mm' %HA)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.9, Page number 287" ] }, { "cell_type": "code", "execution_count": 43, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Largest permissible load = 14.251781 kips\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "# Variable declaration\n", "P=symbols('P')\n", "\n", "\n", "#Properties of Cross Section\n", "A=7.46 # Area of cross section(in**2)\n", "Sx=24.7 # Section moduli x(in**3)\n", "Sy=2.91 # Section moduli y(in**3) \n", "#Force and Couple at C\n", "Mx=4.75*P # Moment x()\n", "My=1.5*P # Moment y()\n", "#Normal Stresses\n", "s1=P*round((1/7.46),4) # Normal stresses \n", "s2=P*round((4.75/24.7),4) # Normal stresses\n", "s3=P*round((1.5/2.91),4) # Normal stresses\n", "#Superposition\n", "sA=-s1 + s2 +s3 # Stress at A \n", "sB=-s1+s2-s3 # Stress at B\n", "sD=-s1-s2+s3 # Stress at D\n", "sE=-s1-s2-s3 # Stress at E\n", "\n", "P=12/0.842 # Maximum compressive stress(kips) \n", "\n", "# Result\n", "print ('Largest permissible load = %lf kips' %P)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.10, Page number 288" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Stress at point A = 13.866727 MPa\n", "The angle formed by the neutral axis and the horizontal is = 41.400000 degree\n" ] } ], "source": [ "import math\n", "from sympy import sin,cos\n", "\n", "# Variable declaration \n", "M0=1500 # Couple of magnitude(kN)\n", "yA=50 # Distance() \n", "zA=74\n", "Iy=(3.25*(pow(10,-6))) # Moment of inertia(m**4)\n", "Iz=(4.18*(pow(10,-6))) # Moment of inertia(m**4)\n", "Iyz=(2.87*(pow(10,-6))) # Moment of inertia(m**4) \n", "\n", "# Calculation\n", "# Principal axes\n", "Theta=(80.8)/2.0 # Angle \n", "R=math.sqrt(pow(0.465,2)+pow(2.87,2)) # Radius\n", "R=2.91*(pow(10,-6)) # Converting to meter\n", "Iu=3.72-2.91 # Moment of inertia(m**4)\n", "Iv=3.72+2.91 # Moment of inertia(m**4) \n", "#Loading\n", "Mu=(M0*sin(40.4)) # Applied couple(N.m) \n", "Mv=(M0*cos(40.4)) # Applied couple(N.m)\n", "#Case(a) Stress at A\n", "uA=50*cos(40.4*((2*math.pi)/360.0))+74*sin(40.4*((2*math.pi)/360.0)) # Perpendicular distances(mm)\n", "vA=-50*sin(40.4*((2*math.pi)/360.0))+74*cos(40.4*((2*math.pi)/360.0)) # Perpendicular distances(mm)\n", "sA=((972*0.0239)/(0.810*(pow(10,-6))) - ((1142)*(0.0860))/(6.63*pow(10,-6)))/(pow(10,6)) # Stress at A(MPa)\n", "#Case(b) Neutral Axis\n", "phy=81.8 # Angle neutral axis with the v axis(degree) \n", "B=81.8-40.4 # Angle neutral axis with the horizontal axis(degree)\n", "\n", "# Result\n", "print ('Stress at point A = %lf MPa' %sA)\n", "print ('The angle formed by the neutral axis and the horizontal is = %lf degree' %B)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.10, Page number 298" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The distance e between the centroid and the neutral axis of the cross section = 0.031381 in\n" ] } ], "source": [ "import math\n", "from sympy import integrate,symbols,log\n", "\n", "#Variable declaration\n", "r=6 # mean radius(in)\n", "b=2.5 # Breadth(in) \n", "h=1.5 # Height(in)\n", "n=-1\n", "b=symbols('b') # Variable declaration\n", "h=symbols('h') # Variable declaration\n", "r=symbols('r') # Variable declaration\n", "r1=symbols('r1') # Variable declaration\n", "r2=symbols('r2') # Variable declaration\n", "\n", "\n", "#Calculation \n", "# Case(a)\n", "A=b*h # Area\n", "R=A/(integrate((1/r), (r, r1, r2)))*(b) # Radius\n", "r1=6-((1/2.0)*(1.5)) # Inner radius(in)\n", "r2=6+((1/2.0)*(1.5)) # Outer radius(in)\n", "# Case(b)\n", "R=(1.5)/(log(r2/r1)) # Distance(in)\n", "e=6-R # Distance between the centroid and the neutral axis \n", "\n", "# Result\n", "print ('The distance e between the centroid and the neutral axis of the cross section = %lf in' %e)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4.11, Page number 299" ] }, { "cell_type": "code", "execution_count": 22, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum stress = 7.860974 ksi\n", "Minimum stress = -9.304620 ksi\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "M=8 # Bending moment(kip.in)\n", "A=(2.5)*(1.5) # Area(in**2) \n", "R=5.969 \n", "e=0.0314 # Distance(in)\n", "\n", "#Calculation \n", "# Case(a)\n", "Smax=((8)*(6.75-5.969))/((3.75)*(0.0314)*(6.75)) # Maximum stress(ksi) \n", "Smin=((8)*(5.25-5.969))/((3.75)*(0.0314)*(5.25)) # Minimum stress(ksi)\n", "\n", "# Result\n", "print ('Maximum stress = %lf ksi' %Smax)\n", "print ('Minimum stress = %lf ksi' %Smin)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 4.11, Page number 300" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Largest force that can applied to the component = 8.550000 kN\n" ] } ], "source": [ "import math\n", "from sympy import integrate,symbols,solve\n", "\n", "#Variable declaration\n", "P=symbols('P') # Variable declaration \n", "s=symbols('s') # Variable declaration\n", "A=symbols('A') # Variable declaration\n", "M=symbols('M') # Variable declaration\n", "R=symbols('R') # Variable declaration \n", "e=symbols('e') # Variable declaration\n", "\n", "# Calculation\n", "#Centroid of the Cross Section\n", "r=(120*(pow(10,3)))/(2400.0) # Distance(m) \n", "#Force and Couple at D\n", "M=((50+60)/1000.0)*P # Moment\n", "#Superposition\n", "s=-(P/A) + (M*(r-R))/(A*e*r) # Total stress\n", "#Radius of Neutral Surface\n", "r=symbols('r') # Variable declaration \n", "R=(2400)/((integrate((80/r), (r, 30, 50)))+(integrate((20/r), (r, 50, 90))))# Radius of neutral surface(m) \n", "#Allowable Load\n", "P=solve(-(P/(2.4*pow(10,-3))) + (0.110*P*(0.030-0.04561))/(2.4*(pow(10,-3))*(0.00439)*(0.030)) + (50*pow(10,6)),P) # Allowable load(kN)\n", "P=round((P[0]/1000.0),2) # Rounding off(kN) \n", "\n", "# Result\n", "print ('Largest force that can applied to the component = %lf kN' %P) " ] } ], "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 }