{ "metadata": { "name": "", "signature": "sha256:a16e0822e0e7167e54e1f4ffc425aa7247562532992f7dee46fc87221365735a" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 18: Torsion of Beams" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 18.1 Pg.No.527" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#variable declaration\n", "r=100 #radius of cylinder (mm)\n", "l=2*10**3 #length of cylinder (mm)\n", "torque=30*10**6 #torque at mid point (N.mm)\n", "Tmax=15*10**6 \n", "Tau_max=200 #max shear stress (N/mm^2)\n", "G=25000 #shear modulus (N/mm^2)\n", "theta =2 # rotation at mid point (degree)\n", "\n", "#ref equation 18.1 T=2Aq\n", "tmin1=Tmax/(2*math.pi*r**2*Tau_max)\n", "print \"\\nminimum thickness of beam due to stress limited to 200N/mm^2 = %3.1f mm\"%(tmin1)\n", "\n", "z=1*10**3 # twist at mid point (mm)\n", "#equation 18.4 after integration\n", "tmin2=Tmax*200*math.pi*z/(4*(math.pi*r**2)**2*G*theta*math.pi/180)\n", "print \"\\nminimum thickness due to constraint on maximum angle twist = %3.1f mm\"%(tmin2)\n", "t=max(tmin1,tmin2)\n", "print \"\\nthickness which satisfies both conditions = %3.1f mm\"%(t)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "minimum thickness of beam due to stress limited to 200N/mm^2 = 1.2 mm\n", "\n", "minimum thickness due to constraint on maximum angle twist = 2.7 mm\n", "\n", "thickness which satisfies both conditions = 2.7 mm\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 18.3 Pg.No 540" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#reference Fig 18.12\n", "from __future__ import division\n", "import math\n", "from sympy import symbols\n", "from sympy import integrate\n", "\n", "S1,S2,S3=symbols('S1,S2,S3')\n", "\n", "#variable declaration\n", "s1=50 #length of channel (mm)\n", "s2=25 #width of channel (mm)\n", "t1=2.5 #thickness of web(mm)\n", "t2=1.5 #thickness of flange(mm)\n", "T=10*10**3 #torque applied (N.mm)\n", "zeta_s=8.04 \n", "G=25000 #shear modulus (N/mm^2)\n", "\n", "#J=Sum(st^3/3)\n", "J=1/3*(s1*t1**3+2*s2*t2**3)\n", "tau_max=t1*T/J\n", "print \"maximum shear stress = %3.1f N/mm^2\"%(tau_max)\n", "\n", "#in region O2\n", "s1=25\n", "ARf=1/2*zeta_s*s1\n", "ws=-2*ARf*T/(G*J)\n", "print \"warping in O2 region is linear = %3.2f mm\"%(ws)\n", "\n", "#in the wall 21\n", "AR=1/2*zeta_s*s2-0.5*25*s2\n", "w21=-0.03*(zeta_s-s2)\n", "s1=50\n", "tds=2*s2*t2+s1*t1\n", "\n", "A12=25/2*S1\n", "A23=312.5-4.02*S2\n", "A34=111.5+25/2*S3\n", "AR1=1/2*(integrate(A12,(S1,0,25))+integrate(5*(A23),(S2,0,50))+integrate(3*A34,(S3,0,25)))/200\n", "\n", "#equation 18.20\n", "ws=-2*ARf*T/G/J\n", "print \"warping distribution in flange 34 = %1.2f \\n\"%(ws)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "maximum shear stress = 78.9 N/mm^2\n", "warping in O2 region is linear = -0.25 mm\n", "warping distribution in flange 34 = -0.25 \n", "\n" ] } ], "prompt_number": 17 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }