{ "metadata": { "name": "chapter no.4.ipynb" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4:Stresses in Beams" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.1,Page no.130" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L=5000 #mm #Length of Beam\n", "a=2000 #mm #Length of start of beam to Pt Load\n", "b=3000 #mm #Length of Pt load to end of beam\n", "A=150*250 #m**2 #Area of beam\n", "b=150 #mm #Width of beam\n", "d=250 #mm #Depth of beam\n", "sigma=10#N/mm**2 #stress\n", "l=2000 #m #Load applied from one end\n", "\n", "#Calculations\n", "\n", "#Moment of Inertia\n", "I=1*12**-1*b*d**3 #m**4\n", "\n", "#Distance from N.A to end\n", "y_max=d*2**-1 #m\n", "\n", "#Section Modulus\n", "Z=1*6**-1*b*d**2 #mm**3\n", "\n", "#Moment Carrying Capacity\n", "M=sigma*Z #N-mm\n", "\n", "#Let w be the Intensity of the Load in N/m,then Max moment\n", "#M_max=w*L**2*8**-1 #N-mm\n", "#After substituting values and further simplifying we get\n", "#M_max=w*25*100*8**-1\n", "\n", "#EQuating it to moment carrying capacity,we get max intensity load\n", "w=M*(25*1000)**-1*8*10**-3\n", "\n", "#Part-2\n", "\n", "#Let P be the concentrated load,then max moment occurs under the load and its value\n", "#M1=P*a*b*L**-1 #N-mm\n", "\n", "#Equting it to moment carrying capacity we get\n", "P=M*1200**-1*10**-3 #N\n", "\n", "#Result\n", "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n", "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max Intensity of u.d.l it can carry 5.0 N-mm\n", "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.2,Page no.131" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "D=70 #mm #External Diameter\n", "t=8 #mm #Thickness of pipe\n", "L=2500 #mm #span \n", "sigma=150 #N/mm**2 #stress\n", "\n", "#Calculations\n", "\n", "#Internal Diameter \n", "d=D-2*t #mm\n", "\n", "#M.I Of Pipe\n", "I=pi*64**-1*(D**4-d**4) #mm**4\n", "\n", "y_max=D*2**-1 #mm\n", "Z=I*(y_max)**-1 #mm**3\n", "\n", "#Moment Carrying capacity\n", "M=sigma*Z #N*mm\n", "\n", "#Max moment int the beam occurs at the mid-span and is equal to\n", "#m=P*L*4**-1\n", "\n", "#Equating Max moment to moment carrying capacity we get,\n", "#M=P*2.5*L*4**-1\n", "#After substituting and simplifying we get\n", "P=4*M*(L)**-1*10**-3 #N\n", "\n", "#Result\n", "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max concentrated load that can be applied at the centre of span is 5.22 KN\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.3,Page no.132" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#Flanges Dimension\n", "b1=180 #mm #Width\n", "d1=10 #mm #Thickness\n", "\n", "D=500 #mm #Overall depth\n", "t=8 #mm #Thickness of web\n", "\n", "#Plate Dimensions\n", "b2=240 #mm #Width\n", "t2=12 #mm #Thickness\n", "\n", "sigma=150 #N/mm**2 #Stress\n", "L=3000 #mm #span\n", "\n", "#Calculations\n", "\n", "#Distance of centroid from bottom fibre\n", "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n", "\n", "#M.I of section\n", "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n", "\n", "#Section Modulus\n", "Z=I*(y_bar)**-1 #mm**3\n", "\n", "#Moment or Resistance\n", "M=sigma*Z\n", "\n", "#Let Load on Cantilever be w/m Length \n", "#Max M.I produced\n", "#M_max=w*L**2**-1 \n", "\n", "#Now Equating Moment of resistance to Max moment,we get Max load\n", "#4.5*w=M\n", "#After rearranging and further simplifying we get\n", "w=M*4.5**-1*10**3*10**-9\n", "\n", "#Result\n", "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n", "print\"Load the section can carry is\",round(w,3),\"KN/m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Moment of Resistance is 198770121.83 KN-mm\n", "Load the section can carry is 44.171 KN/m\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.4,Page no.134" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#Flange (Top)\n", "b1=80 #mm #Width \n", "t1=40 #mm #Thickness\n", "\n", "#Flange (Bottom)\n", "b2=160 #mm #width\n", "t2=40 #mm #Thickness\n", "\n", "#web\n", "d=120 #mm #Depth\n", "t3=20 #mm #Thickness\n", "\n", "D=200 #mm #Overall Depth\n", "sigma1=30 #N/mm**2 #Tensile stress\n", "sigma2=90 #N/mm**2 #Compressive stress\n", "L=6000 #mm #Span\n", "\n", "#Calculations\n", "\n", "#Distance of centroid from bottom fibre\n", "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n", "\n", "#Moment of Inertia\n", "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n", "\n", "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n", "\n", "y_t=y_bar #mm\n", "y_c=D-y_bar #mm\n", "\n", "#Moment carrying capacity considering Tensile strength \n", "M1=sigma1*I*y_t**-1*10**-6 #KN-m\n", "\n", "#Moment carrying capacity considering compressive strength \n", "M2=sigma2*I*y_c**-1*10**-6 #KN-m\n", "\n", "#Max Bending moment in simply supported beam 6 m due to u.d.l\n", "#M_max=w*L*10**-3*8**-1\n", "#After simplifying further we get\n", "#M_max=4.5*w\n", "\n", "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n", "w=M1*4.5**-1 #KN/m\n", "\n", "#Result\n", "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max Uniformly Distributed Load is 5.096 KN/m\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.5,Page no.136" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "from scipy.integrate import quad\n", "\n", "#Initilization of Variables\n", "\n", "#Flanges\n", "b=200 #mm #Width\n", "t=25 #mm #Thickness \n", "\n", "D1=500 #mm #Overall Depth\n", "t2=20 #mm #Thickness of web\n", "\n", "d=450 #mm #Depth of web\n", "\n", "#Calculations\n", "\n", "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n", "#Let Bending stress \"sigma_max\"\n", "\n", "#Stress on the element \n", "#sigma=y*(D*2**-1)*sigma_max ..............(1)\n", "\n", "#Area of Element\n", "#A=b*dy .................................(2)\n", "\n", "#Force on Element \n", "#F=y*250**-1*sigma_max*b*dy\n", "\n", "#Let M be the Moment of resistance\n", "#M=y*250**-1*sigma_max*b*dy*y\n", "\n", "#Moment of Resistance of top flange be M1\n", "def integrand(y, b, D):\n", " return b*y**2*D**-1\n", "b=200 \n", "D=250\n", "\n", "X = quad(integrand, 225, 250, args=(b,D))\n", "\n", "Y=2*X[0]\n", "\n", "#M1=Y*sigma\n", "\n", "#Now Moment of Inertia I section is\n", "X=b*D1**3\n", "Y=(b-t2)*d**3\n", "I=(X-Y)*12**-1*10**-8\n", "\n", "#Moment acting on the entire section\n", "#since sigmais the value at y=250\n", "y_max=250\n", "Z=I*10**8*y_max**-1\n", "#M=sigma*Z \n", "#After Simplifying Further we get\n", "#M2=Z*sigma\n", "\n", "#Percentage Moment resisted by Flanges\n", "P1=2258333.3*(2865833.3)**-1*100\n", "\n", "#Percentage Moment resisted by web\n", "P2=100-P1\n", "\n", "#Result\n", "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n", "print\"Percentage Moment resisted by web\",round(P2,2),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Percentage Moment resisted by Flanges 78.8 %\n", "Percentage Moment resisted by web 21.2 %\n" ] } ], "prompt_number": 38 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.6,Page no.137" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#Flanges\n", "b1=200 #mm #Width\n", "t1=10 #mm #Thickness\n", "\n", "#Web\n", "d=380 #mm #Depth \n", "t2=8 #mm #Thickness\n", "\n", "D=400 #mm #Overall Depth\n", "sigma=150 #N/mm**2\n", "\n", "#Calculations\n", "\n", "#Area\n", "A=b1*t1+d*t2+b1*t1 #mm**2\n", "\n", "#Moment of Inertia\n", "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n", "\n", "#Bending Moment\n", "M=sigma*I*(D*2**-1)**-1\n", "\n", "#Square Section\n", "\n", "#Let 'a' be the side\n", "a=A**0.5\n", "\n", "#Moment of Resistance of this section\n", "M1=1*6**-1*a*a**2*sigma\n", "\n", "X=M*M1**-1\n", "\n", "#Rectangular section\n", "#Let 'a' be the side and depth be 2*a\n", "\n", "a=(A*2**-1)**0.5\n", "\n", "#Moment of Rectangular secction\n", "M2=1*6**-1*a*(2*a)**2*sigma\n", "\n", "X2=M*M2**-1\n", "\n", "#Circular section\n", "#A=pi*d1**2*4**-1\n", "\n", "d1=(A*4*pi**-1)**0.5\n", "\n", "#Moment of circular section\n", "M3=pi*32**-1*d1**3*sigma\n", "\n", "X3=M*M3**-1\n", "\n", "#Result\n", "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n", "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n", "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n", "print\"Moment of resistance of circular section\",round(X3,2),\"mm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Moment of resistance of beam section 141536000.0 mm\n", "Moment of resistance of square section 9.58 mm\n", "Moment of resistance of rectangular section 6.78 mm\n", "Moment of resistance of circular section 11.33 mm\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.7,Page no.139" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "F=12 #KN #Force at End of beam\n", "L=2 #m #span\n", "\n", "#Square section \n", "b=d=200 #mm #Width and depth of beam\n", "\n", "#Rectangular section\n", "b1=150 #mm #Width\n", "d1=300 #mm #Depth\n", "\n", "#Calculations\n", "\n", "#Max bending Moment\n", "M=F*L*10**6 #N-mm\n", "\n", "#M=sigma*b*d**2\n", "sigma=M*6*(b*d**2)**-1 #N/mm**2\n", "\n", "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n", "#MAx Moment\n", "#M1=W*L1*4**-1\n", "#After Further simplifying we get\n", "#M1=0.75*10**6 #N-mm\n", "\n", "#The section has a moment of resistance\n", "M1=sigma*1*6**-1*b1*d1**2\n", "\n", "#Equating it to moment of resistance we get max load W\n", "#0.75*10**6*W=M1\n", "#After Further simplifying we get\n", "W=M1*(0.75*10**6)**-1\n", "\n", "#Result\n", "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Minimum Concentrated Load required to brek the beam 54.0 KN\n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.8,Page no.140" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L=3 #m #span\n", "sigma_t=35 #N/mm**2 #Permissible stress in tension\n", "sigma_c=90 #N/mm**2 #Permissible stress in compression\n", "\n", "#Flanges\n", "t=30 #mm #Thickness\n", "d=250 #mm #Depth\n", "\n", "#Web\n", "t2=25 #mm #Thickness\n", "b=600 #mm #Width\n", "\n", "#Calculations\n", "\n", "#Let y_bar be the Distance of N.A from Extreme Fibres\n", "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n", "\n", "#Moment of Inertia\n", "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n", "\n", "#Part-1\n", "\n", "#If web is in Tension\n", "y_t=y_bar #mm\n", "y_c=d-y_bar #mm\n", "\n", "#Moment carrying caryying capacity From consideration of tensile stress\n", "M=sigma_t*I*(y_bar)**-1 #N-mm\n", "\n", "#Moment carrying caryying capacity From consideration of compressive stress\n", "M1=sigma_c*I*(y_c)**-1 #N-mm\n", "\n", "#If w KN/m is u.d.l in beam,Max bending moment\n", "#M=wl**2*8**-1\n", "#After further simplifyng we get\n", "#M=1.125*w*10**6 N-mm\n", "w=M*(1.125*10**6)**-1 #KN\n", "\n", "#Part-2\n", "\n", "#If web is in compression\n", "y_t2=178.299 #mm\n", "y_c2=71.71 #mm \n", "\n", "#Moment carrying caryying capacity From consideration of tensile stress\n", "M2=sigma_t*I*(y_t2)**-1 #N-mm\n", "\n", "#Moment carrying caryying capacity From consideration of compressive stress\n", "M3=sigma_c*I*(y_c2)**-1 #N-mm\n", "\n", "#Moment of resistance is M2\n", "\n", "#Equating it to bending moment we get\n", "#M2=1.125*10**6*w2\n", "#After further simplifyng we get\n", "w2=M2*(1.125*10**6)**-1\n", "\n", "#Result\n", "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n", "print\" :web is in compression\",round(w2,3),\"KN\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n", " :web is in compression 29.446 KN\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.9,Page no.141" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "b1=200 #mm #Width at base\n", "b2=100 #mm #Width at top\n", "\n", "L=8 #m Length\n", "P=500 #N #Load\n", "\n", "#Calculations\n", "\n", "#Consider a section at y metres from top\n", "\n", "#At this section diameter d is\n", "#d=b2+y*L**-1*(b1-b2)\n", "#After Further simplifying we get\n", "#d=b2+12.5*y #mm\n", "\n", "#Moment of Inertia\n", "#I=pi*64**-1*d**4\n", "\n", "#Section Modulus \n", "#Z=pi*32**-1*(b1+12.5*y)**3\n", "\n", "#Moment \n", "#M=5*10**5*y #N-mm\n", "\n", "#Let sigma be the fibre stress at this section then\n", "#M=sigma*Z\n", "#After sub values in above equation and further simplifying we get\n", "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n", "\n", "#For sigma to be Max,d(sigma)*(dy)**-1=0\n", "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n", "#After Further simplifying we get\n", "#b2+12.5*y=37.5*y\n", "#After Further simplifying we get\n", "y=b2*25**-1 #m\n", "\n", "#Stress at this section\n", "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n", "\n", "#Result\n", "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n", "print\"Max stress is\",round(sigma,2),\"N/mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Stress at Extreme Fibre is max 4.0 m\n", "Max stress is 6.04 N/mm**2\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.10,Page no.143" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "H=10 #mm #Height\n", "A1=160*160 #mm**2 #area of square section at bottom\n", "L1=160 #mm #Length of square section at bottom\n", "b1=160 #mm #width of square section at bottom\n", "A2=80*80 #mm**2 #area of square section at top\n", "L2=80 #mm #Length of square section at top\n", "b2=80 #mm #Width of square section at top\n", "P=100 #N #Pull\n", "\n", "#Calculations\n", "\n", "#Consider a section at distance y from top.\n", "#Let the side of square bar be 'a'\n", "#a=L2+y*(H)**-1*(b1-b2)\n", "#After further simplifying we get\n", "#a=L2+8*y\n", "\n", "#Moment of Inertia\n", "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n", "#After further simplifying we get\n", "#I=a**4*12**-1\n", "\n", "#Section Modulus \n", "#Z=a**4*(12*a*(2)**0.5)**-1\n", "#After further simplifying we get\n", "#Z=2**0.5*a**3*(12)**-1 #mm**3\n", "\n", "#Bending moment at this section=100*y N-mm\n", "#M=100*10**3*y #N-mm\n", "\n", "#But\n", "#M=sigma*Z\n", "#After sub values in above equation we get\n", "#sigma=M*Z**-1\n", "#After further simplifying we get\n", "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n", "\n", "#For Max stress df*(dy)**-1=0\n", "#After taking Derivative of above equation we get\n", "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n", "#After further simplifying we get\n", "y=80*16**-1 #m\n", "\n", "#Max stress at this level is\n", "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n", "\n", "#Result\n", "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n", "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max Bending stress is Developed at 5.0 m\n", "Value of Max Bending stress is 2.455 N/mm**2\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.12,Page no.147" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "b=200 #mm #Width of timber \n", "d=400 #mm #Depth of timber\n", "t=6 #mm #Thickness\n", "b2=200 #mm #width of steel plate\n", "t2=20 #mm #Thickness of steel plate\n", "M=40*10**6 #KN-mm #Moment\n", "#Let E_s*E_t**-1=X\n", "X=20 #Ratio of Modulus of steel to timber\n", "\n", "#Calculations\n", "\n", "#let y_bar be the Distance of centroidfrom bottom most fibre\n", "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n", "\n", "#Moment of Inertia\n", "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n", "\n", "#distance of the top fibre from N-A\n", "y_1=d+t-y_bar #mm\n", "\n", "#Distance of the junction of timber and steel From N-A\n", "y_2=y_bar-t #mm\n", "\n", "#Stress in Timber at the top\n", "Y=M*I**-1*y_1 #N/mm**2\n", "\n", "#Stress in the Timber at the junction point\n", "Z=M*I**-1*y_2\n", "\n", "#Coressponding stress in steel at the junction point\n", "Z2=X*Z #N/mm**2 \n", "\n", "#The stress in Extreme steel fibre \n", "Z3=X*M*I**-1*y_bar\n", "\n", "#Result\n", "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Stress in Extreme steel Fibre 69.67 N/mm**2\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.13,Page no.149" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#Timber size\n", "b=150 #mm #Width\n", "b2=120 #mm \n", "d=300 #mm #Depth\n", "\n", "t=6 #mm #Thickness of steel plate\n", "L=6 #m #Span\n", "\n", "#E_s*E_t**-1=20 \n", "#X=E_s*E_t**-1\n", "X=20 \n", "sigma_timber=8 #N/mm**2 #Stress in timber\n", "sigma_steel=150 #N/mm**2 #Stress in steel plate\n", "\n", "#Calculations\n", "\n", "Y\n", "\n", "#Due to synnetry cenroid,the neutral axis is half the depth\n", "I=1*12**-1*\n", "\n", "\n", "#Result\n", "print Z" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "153.0\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.14,Page no.151" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L=6000 #mm #Span of beam\n", "W=20*10**3 #N #Load\n", "sigma=8 #N/mm**2 #Stress\n", "b=200 #mm #Width of section\n", "d=300 #mm #Depth of section\n", "\n", "#Calculations\n", "\n", "#let x be the distance from left side of beam\n", "\n", "#Bending moment\n", "#M=W*2**-1*x #Nmm .......(1)\n", "\n", "#But M=sigma*Z ..........(2)\n", "\n", "#Equating equation 1 and 2 we get\n", "#W*2**-1*x=sigma*Z ............(3)\n", "\n", "#Section Modulus \n", "#Z=1*6*b*d**2 ...............(4)\n", "\n", "#Equating equation 3 and 4 we get\n", "#b*d**2=3*W*x*sigma**-1 .............(5)\n", "\n", "#Beam of uniform strength of constant depth\n", "#b=3*W*x*(sigma*d**2) \n", "\n", "#When x=0\n", "b=0\n", "\n", "#When x=L*2**-1\n", "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n", "\n", "#Beam with constant width of 200 mm\n", "\n", "#We have\n", "#d=(3*W*x*(sigma*d)**-1)**0.5\n", "#thus depth varies as (x)**0.5\n", "\n", "#when x=0\n", "d1=0\n", "\n", "#when x=L*2**-1\n", "d2=(2*W*L*(2*sigma*300)**-1)**0.5 #mm\n", "\n", "#Result\n", "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n", "print\" :\",round(d2,2),\"mm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Cross section of rectangular beam is: 250.0 mm\n", " : 223.61 mm\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.15,Page no.154" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L=800 #mm #Span\n", "n=5 #number of leaves\n", "b=60 #mm #Width\n", "t=10 #mm #thickness\n", "sigma=250 #N/mm**2 #Stress\n", "\n", "#Calculations\n", "\n", "#section Modulus\n", "Z=n*6**-1*b*t**2 #mm**3\n", "\n", "#from the relation\n", "#sigma*Z=M ...................(1)\n", "#M=P*L*4**-1\n", "#sub values of M in equation 1 we get\n", "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n", "\n", "#Length of Leaves\n", "L1=0.2*L #mm\n", "L2=0.4*L #mm\n", "L3=0.6*L #mm\n", "L4=0.8*L #mm\n", "L5=L #mm\n", "\n", "#Result\n", "print\"Max Load it can take is\",round(P,2),\"KN\"\n", "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n", "print\" :L2\",round(L2,2),\"mm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max Load it can take is 6.25 KN\n", "Length of leaves:L1 160.0 mm\n", " :L2 320.0 mm\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.16,Page no.161" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "%matplotlib inline\n", "\n", "#Initilization of Variables\n", "\n", "F=20*10**3 #N #Shear Force\n", "\n", "#Tee section\n", "\n", "#Flange\n", "b=100 #mm #Width\n", "t=12 #mm #Thickness\n", "\n", "#Web\n", "d=88 #mm #Depth\n", "t2=12 #mm #Thicknes\n", "\n", "D=100 #mm #Overall Depth\n", "\n", "#Calculations\n", "\n", "#Distance of C.G from Top Fibre\n", "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n", "\n", "#Moment Of Inertia\n", "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n", "\n", "#shear stress at bottom Flange\n", "\n", "#Area above this level\n", "A=b*t #mm**2\n", "\n", "#C.G of this area from N-A\n", "y2=y-t*2**-1\n", "\n", "#Stress at bottom of flange\n", "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n", "\n", "#sigma2 at same level but in web where width is 12 mm\n", "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n", "\n", "#To find shear stress at N-A\n", "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n", "\n", "sigma3=F*X*(t2*I)**-1 #N/mm**2\n", "\n", "#Shear stress at top and bottom fibre is zero\n", "#sigma4 and sigma5 are top and bottom fibre shear stress\n", "sigma4=sigma5=0\n", "\n", "#Result\n", "print \"The Shear Force and Bending Moment Diagrams are the results\"\n", "\n", "#Plotting the Shear Force Diagram\n", "\n", "X1=[0,t,t,y,D]\n", "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n", "Z1=[0,0,0,0,0]\n", "plt.plot(X1,Y1,X1,Z1)\n", "plt.xlabel(\"Length x in m\")\n", "plt.ylabel(\"Shear Force in kN\")\n", "plt.show()\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Shear Force and Bending Moment Diagrams are the results\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.17,Page no.163" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "%matplotlib inline\n", "\n", "#Initilization of Variables\n", "\n", "F=40*10**3 #N #shear Force\n", "\n", "#I-section\n", "\n", "#Flanges\n", "b=80 #mm #Width of flange\n", "t=20 #mm #Thickness\n", "\n", "#Web\n", "d=200 #mm #Depth\n", "t2=20 #mm #Thickness\n", "\n", "#Flange-2\n", "b2=160 #mm #Width\n", "t3=20 #mm #Thickness\n", "\n", "D=240 #mm #Overall Depth\n", "\n", "#Calculations\n", "\n", "#Distance of N-A from Top Fibre \n", "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n", "\n", "#Moment of Inertia\n", "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n", "\n", "#Shear stress bottom of flange\n", "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n", "\n", "#At same Level but in web\n", "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n", "\n", "#for shear stress at N.A\n", "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n", "sigma3=F*X*(t2*I)**-1 #N/mm**2\n", "\n", "#Shear stress at bottom of web\n", "\n", "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n", "\n", "#Stress at bottom of web\n", "sigma4=F*X*(t2*I)**-1 #N/mm**2\n", "\n", "#Stress at Lower flange\n", "sigma5=F*X*(b2*I)**-1 #N/mm**2\n", "\n", "#Result\n", "print \"The Shear Force Diagram is the result\"\n", "\n", "#Plotting the Shear Force Diagram\n", "\n", "X1=[0,20,20,140,220,220,240]\n", "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n", "Z1=[0,0,0,0,0,0,0]\n", "plt.plot(X1,Y1,X1,Z1)\n", "plt.xlabel(\"Length in mm\")\n", "plt.ylabel(\"Shear Force in N\")\n", "plt.show()\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Shear Force Diagram is the result\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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2vPfee7Ro0cKdZYiIyEXcPmq/YsWKLFy4kD59+pCXl8e4ceMUBCIiNvPIGcgi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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n", "text": [ "" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.18,Page no.164" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "F=30*10**3 #N #Shear Force\n", "\n", "#Channel Section\n", "d=400 #mm #Depth of web \n", "t=10 #mm #THickness of web\n", "t2=15 #mm #Thickness of flange\n", "b=100 #mm #Width of flange\n", "\n", "#Rectangular Welded section\n", "b2=80 #mm #Width\n", "d2=60 #mm #Depth\n", "\n", "#Calculations\n", "\n", "#Distance of Centroid From Top Fibre\n", "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n", "\n", "#Moment Of Inertia of the section about N-A\n", "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n", "\n", "#Shear stress at level of weld\n", "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n", "\n", "#Max Shear Stress occurs at Neutral Axis\n", "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n", "\n", "sigma_max=F*X*((b+t)*I)**-1\n", "\n", "#Result\n", "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n", "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Shear stress in the weld is 3.62 N/mm**2\n", "Max shear stress is 4.48 N/mm**2\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.19,Page no.165" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#Wooden Section\n", "b=300 #mm #Width\n", "d=300 #mm #Depth\n", "\n", "D=100 #mm #Diameter of Bore\n", "F=10*10**3 #N #Shear Force\n", "\n", "#Calculations\n", "\n", "#Moment Of Inertia Of Section\n", "I=1*12**-1*b*d**3-pi*64**-1*D**4\n", "\n", "#Shear stress at crown of circle\n", "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n", "\n", "#Let a*y_bar=X\n", "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n", "\n", "#Shear Stress at Neutral Axis\n", "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n", "\n", "#Result\n", "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n", "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Shearing Stress at Crown of Bore 0.149 N/mm**2\n", "Shear Stress at Neutral Axis 0.246 N/mm**2\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.20,Page no.166" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#flanges\n", "b=200 #mm #width\n", "t1=25 #mm #Thickness\n", "\n", "#web\n", "d=450 #mm #Depth \n", "t2=20 #mm #thickness\n", "\n", "D=500 #mm #Total Depth of section\n", "\n", "#Calculations\n", "\n", "#Moment Of Inertia of the section about N-A\n", "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n", "\n", "#Consider an element in the web at distance y from y from N-A\n", "#Depth of web section=225-y\n", "\n", "#C.G From N-A\n", "#y2=y+(((D*2**-1-t)-y)*2**-1)\n", "\n", "#ay_bar for section at y\n", "#Let ay_bar be X\n", "#X=X1 be of Flange + X2 be of web above y\n", "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n", "#After Sub values and Further simplifying we get\n", "#X=1187500+10*(225**2-y**2)\n", "\n", "#Shear stress at y\n", "#sigma_y=F*(X)*(t2*I)**-1\n", "\n", "#Shear Force resisted by the Element\n", "#F1=F*X*t2*dy*(t2*I)**-1\n", "\n", "#Shear stress resisted by web \n", "#sigma=2*F*I**-1*(X)*dy\n", "\n", "#After Integrating above equation and further simplifying we get\n", "#sigma=0.9578*F\n", "\n", "sigma=0.9578*100\n", "\n", "#Result\n", "print\"Shear Resisted by web\",round(sigma,2),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Shear Resisted by web 95.78 %\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.21,Page no.167" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "#Wooden Beam\n", "\n", "b=150 #mm #width\n", "d=250 #mm #Depth\n", "\n", "L=5000 #mm #span\n", "m=11.2 #N/mm**2 #Max Bending stress\n", "sigma=0.7 #N/mm**2 #Max shear stress\n", "\n", "#Calculations\n", "\n", "#Let 'a' be the distance from left support\n", "#Max shear force\n", "#F=R_A=W*(L-a)*L**-1 \n", "\n", "#Max Moment\n", "#M=W*(L-a)*a*L**-1\n", "\n", "#But M=sigma*Z\n", "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)\n", "\n", "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n", "F=sigma*b*d*1.5**-1\n", "\n", "#W*(L-a)*L**-1=F .....................(2)\n", "\n", "#Dividing Equation 1 nad 2 we get\n", "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n", "\n", "#Sub above value in equation 2 we get\n", "W=(L-a)**-1*L*F*10**-3 #KN \n", "\n", "#Result\n", "print\"Load is\",round(W,2),\"KN\"\n", "print\"Distance from Left support is\",round(a,2),\"mm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Load is 21.87 KN\n", "Distance from Left support is 1000.0 mm\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.22,Page no.168" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L=1000 #mm #span\n", "\n", "#Rectangular Section\n", "\n", "b=200 #mm #width\n", "d=400 #mm #depth\n", "\n", "sigma=1.5 #N/mm**2 #Shear stress\n", "\n", "#Calculations\n", "\n", "#Let AB be the cantilever beam subjected to load W KN at free end\n", "\n", "#MAx shear Force\n", "#F=W*10**3 #KN\n", "\n", "#Since Max shear stress in Rectangular section\n", "#sigma_max=1.5*F*A**-1 \n", "#After sub values and further simplifyng we get\n", "W=1.5*b*d*(1.5*1000)**-1 #KN\n", "\n", "#Moment at fixwed end\n", "M=W*1 #KN-m\n", "y_max=d*2**-1 #mm\n", "\n", "#M.I\n", "I=1*12**-1*b*d**3 #mm**3\n", "\n", "#MAx Stress\n", "sigma_max=M*10**6*I**-1*y_max\n", "\n", "#Result\n", "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Concentrated Load is 15.0 N/mm**2\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4.4.24,Page no.170" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L=4000 #mm #span\n", "\n", "#Rectangular Cross-section\n", "b=100 #mm #Width\n", "d=200 #mm #Thickness\n", "\n", "F_per=10 #N/mm**2 #Max Bending stress\n", "q_max=0.6 #N/mm**2 #Shear stress\n", "\n", "#Calculations\n", "\n", "#If the Load W is in KN/m\n", "\n", "#Max shear Force\n", "#F=w*l*2**-1 #KN\n", "#After substituting values and further simplifying we get\n", "#M=2*w #KN-m\n", "\n", "#Max Load from Consideration of moment\n", "#M=1*6**-1*b*d**2*F_per\n", "#After substituting values and further simplifying we get\n", "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n", "\n", "#Max Load from Consideration of shear stress\n", "#q_max=1.5*F*(b*d)**-1 #N\n", "#After substituting values and further simplifying we get\n", "F=q_max*(1.5)*b*d #N\n", "\n", "#If w is Max Load in KN/m,then\n", "#2*w*1000=8000\n", "#After Rearranging and Further simplifying we get\n", "w2=8000*(2*1000)**-1 #KN/m\n", "\n", "#Result\n", "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }