{ "metadata": { "name": "chapter 17 som.ipynb" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 17:Welded Joints" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.1,Page no.379" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "b=12 #cm #width of steel plates\n", "t=1 #cm #thickness of steel plates\n", "sigma=75 #MPa #stress\n", "\n", "#Calculations\n", "\n", "#The maximum Load which the plate can carry\n", "P=b*t*10**-6*sigma*10**6 #N \n", "\n", "#Length of weld for static loading\n", "\n", "#size of weld is equal to thickness of plate\n", "S=t #cm\n", "\n", "#P=2**0.5*l*S*sigma\n", "\n", "#After substituting values and simplifying above equation, we get, \n", "l=((P)*(2**0.5*S*sigma)**-1) #cm\n", "\n", "#add 1.25 to allow start and stop of weld run\n", "L=l+1.25 #cm \n", "\n", "#Length of weld for Dynamic loading\n", "\n", "#The stress concentration factor for transverse fillet weld is 1.5\n", "\n", "sigma_2=sigma*1.5**-1 #MPa #Permissible tensile stress\n", "\n", "#P=2**0.5*l_2*S*sigma_2 \n", "\n", "#After substituting values and simplifying above equation, we get,\n", "l_2=((P)*(2**0.5*S*sigma_2)**-1) #cm\n", "\n", "#add 1.25 cm\n", "l_3=l_2+1.25 #cm \n", "\n", "#Result\n", "print\"Length of weld for static loading\",round(L,2),\"cm\"\n", "print\"Length of weld for Dynamic loading\",round(l_3,2),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Length of weld for static loading 9.74 cm\n", "Length of weld for Dynamic loading 13.98 cm\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.2,Page no.380" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "d=6 #cm #diameter of rod\n", "L=40 #cm #Length of steel plate\n", "P=12 #KN #Load\n", "sigma=180 #MPa #Allowable stress\n", "\n", "#Calculations\n", "\n", "A=pi*4**-1*d**2 #cm**2 #Area of rod\n", "M=P*10**3*L #Ncm\n", "\n", "F=M*A**-1 #N/cm #Force per unit cm of weld at top and bottom\n", "\n", "V_s=P*10**3*(pi*d)**-1 #N/cm #vertical shear\n", "\n", "R=(F**2+V_s**2)**0.5 #N/cm #resultant Load\n", "\n", "S=R*(sigma)**-1*10**-2 #cm #size of weld\n", "\n", "#Result\n", "print\"size of weld is\",round(S,2),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "size of weld is 0.94 cm\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.3,Page no.380" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "b=12 #cm #width of plate\n", "S=t=1 #cm #thickness of plate\n", "P=50 #KN #load\n", "sigma_s=60 #MPa #shear stress\n", "\n", "#Calculations (part-1)\n", "\n", "#Under static Loading\n", "\n", "#P=2**0.5*l*S*sigma_s\n", "\n", "l=((P*10**3)*(2**0.5*S*sigma_s*10**-4*10**6)**-1) #cm \n", "\n", "#add 1.25 cm to start and stop weld run\n", "\n", "L=l+1.25 #cm #length of weld\n", "\n", "#Calculations (part-2)\n", "\n", "#Under Fatigue load\n", "\n", "#stress concentration factor for parallel fillet weld is 2.7\n", "\n", "sigma_s_2=sigma_s*2.7**-1 #MPa #permissible shear stress\n", "\n", "#P=2**0.5*l_2*S*sigma_s_2\n", "\n", "l_2=((P*10**3)*(2**0.5*S*sigma_s_2*10**-4*10**6)**-1) #cm\n", "\n", "#add 1.25 cm \n", "\n", "l_3=l_2+1.25 #cm #length of weld\n", "\n", "#Result\n", "print\"Length of weld Under static Loading is\",round(L,3),\"cm\"\n", "print\"Length of weld Under Ftigue Loading is\",round(l_3,3),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Length of weld Under static Loading is 7.143 cm\n", "Length of weld Under Ftigue Loading is 17.16 cm\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.4,Page no.381" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "sigma_t=100 #MPa #tensile stress\n", "P=170 #KN #Load\n", "\n", "#Calculations\n", "\n", "#For equal stress in the welds A and B, the load shared by the fillet welds will be proportional to size of weld\n", "\n", "#t_a=0.7*s #Effective throat thickness of weld A in upper plate\n", "#s=size of weld \n", "\n", "#t_b=1.05*s #Effective throat thickness of weld B in lower plate\n", "\n", "#For weld A\n", "#P_1=l*t*sigma_t \n", "\n", "#After substituting values and simplifying above equation, we get,\n", "#P_1=84000*s #N (equation 1)\n", "\n", "#P_2=l*t_2*sigma_t\n", "\n", "#After substituting values and simplifying above equation, we get,\n", "#P_2=126000*s #N (equation 2)\n", "\n", "#After adding equation 1 and 2, we get,\n", "#P=210000*s (equation 3)\n", "\n", "#Now equating total forces of the fillets to load on the plates\n", "s=P*10**3*210000**-1 #cm\n", "\n", "#Result\n", "print\"size of end fillet is\",round(s,2),\"cm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "size of end fillet is 0.81 cm\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.5,Page no.381" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "L_1=30 #cm #length of longitudinal weld\n", "L_2=16 #cm #length of transverse weld\n", "#t=0.7*s #Effective thickness of weld \n", "sigma_t_1=100 #MPa #working stress for transverse welds\n", "sigma_t_2=85 #MPa #working stress for longitudinal welds\n", "P=150 #KN #load\n", "\n", "#Calculations\n", "\n", "#For transverse weld\n", "#P_1=L_1*t*10**-4*sigma_t_1*10**6 \n", "\n", "#After substituting values and simplifying above equation, we get,\n", "#P_1=112000*s #N\n", "\n", "#For longitudinal weld\n", "#P_2=L_2*t*10**-4*sigma_t_2*10**6\n", "\n", "#Total force of resistance of weld\n", "#P=P_1+P_2 #N\n", "\n", "#after adding we get,\n", "#P=290500*s #N\n", "\n", "#Now equating total forces of resistance to pull of the joint\n", "s=P*10**3*290500**-1 #cm\n", "\n", "#Result \n", "print\"size of weld is\",round(s,3),\"cm\"\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "size of weld is 0.516 cm\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.6,Page no.382" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "P=200 #KN #Load carried by the angle \n", "S=0.6 #mm #size of weld\n", "b=4.46 #cm #Distance of centre of gravity of the angle from the top shorter leg\n", "a=10.54 #cm #Distance of centre of gravity of the angle from the top edge of the angle\n", "sigma_s=102.5 #MPa #shear stress\n", "#l_1=Length of the top weld\n", "#l_2=length of the bottom weld\n", "#L=l_1+l_2 #cm #total length weld\n", "\n", "#Using the relation\n", "#P=L*0.7*S*sigma_s\n", "\n", "#After substituting values and simplifying we get\n", "L=(P*10**3)*(0.7*S*sigma_s*10**-4*10**6)**-1 #cm (equation 1)\n", "\n", "#Using the relation\n", "l_1=(L*b)*(a+b)**-1 #cm\n", "\n", "#substituting this value in equation 1 we have,\n", "l_2=L-l_1 #cm \n", "\n", "#Result\n", "print\"Distance of centre of gravity of the angle from the top edge of the angle\",round(l_2,2),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Distance of centre of gravity of the angle from the top edge of the angle 32.64 cm\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Problem no.17.7,Page no.383" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Initilization of Variables\n", "\n", "P=12 #KN #Load\n", "sigma_s=75 #N/mm**2 #shear stress\n", "e=12 #cm\n", "r_1=2.5 #cm\n", "\n", "#Calculations\n", "\n", "#A=(2*S*l)*(2)**0.5\n", "#sigma_s=P*A**-1 #MPa #shear stress\n", "\n", "#After substituting values and simplifying we get\n", "#sigma_s=16.97*S**-1 #MPa\n", "\n", "#I_g=S*l*(3*b**2+l**2)*(6)**-1 #cm**4 #Polar moment of Inertia of weld\n", "\n", "#After substituting values and simplifying we get\n", "#I_g=180.833*S #cm**4\n", "r_2=((8*2**-1)**2)+((5*2**-1)**2)**0.5 #cm #max radius of weld\n", "\n", "#sigma_s_2=P*e*r_2*I_g**-1 #MPa #shear stress due to bending moment\n", "\n", "cos_theta=r_1*r_2**-1\n", "\n", "#Now using the relation\n", "#sigma_s=(sigma_s_1**2+sigma_s_2**2+2sigma_s_1*sigma_s_2*cos_theta\n", "\n", "S=(2363.8958*5625**-1)**0.5 #cm #size of the weld\n", "\n", "#Result\n", "print\"size of the weld\",round(S,3),\"cm\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "size of the weld 0.648 cm\n" ] } ], "prompt_number": 11 } ], "metadata": {} } ] }