path: root/MECHANICS_OF_SOLIDS_by_S.S._Bhavikatti/Chapter11.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter11-PRINCIPAL STRESSES AND STRAINS"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.2 page number 352"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a) P= 14726.22 N\n",
      "(b) P= -44178.65 N compressive\n",
      "Material fails because of maximum shear and not by axial compression.\n",
      "P= 24544.0 N\n",
      "The plane of qmax is at 45° to the plane of px. \n"
     ]
    }
   ],
   "source": [
    "from math import sqrt,pi\n",
    "\n",
    "#A material has strength in tension, compression and shear as 30N/mm2, 90 N/mm2 and 25 N/mm2, respectively. If a specimen of diameter 25 mm is tested in tension and compression identity the failure surfaces and loads. \n",
    "\n",
    "#variable declaration\n",
    "\n",
    "#In tension: Let axial direction be x direction. Since it is uniaxial loading, py  = 0, q = 0 and only px exists.when the material is subjected to full tensile stress, px = 30 N/mm^2.\n",
    "\n",
    "pt=float(30)\n",
    "pc=float(90)\n",
    "ps=float(25)\n",
    "\n",
    "d=float(25)\n",
    "px=float(30)         #N/mm^2\n",
    "py=0\n",
    "q=0\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "qmax=(px-py)/2\n",
    "\n",
    "#Hence failure criteria is normal stress p1\n",
    "\n",
    "A=pi*pow(d,2)/4\n",
    "\n",
    "#Corresponding load P is obtained by\n",
    "p=p1\n",
    "P=p1*A\n",
    "\n",
    "print \"(a) P=\",round(P,2),\"N\"\n",
    "\n",
    "#In case of compression test,\n",
    "\n",
    "px=-pc\n",
    "py=q=0\n",
    "\n",
    "P=-px*A\n",
    "\n",
    "print \"(b) P=\",round((-P),2),\"N compressive\"\n",
    "\n",
    "#at this stage\n",
    "\n",
    "qmax=sqrt((pow((px-py)/2,2))+pow(q,2))\n",
    "\n",
    "print \"Material fails because of maximum shear and not by axial compression.\"\n",
    "qmax=25\n",
    "px=2*qmax\n",
    "\n",
    "P=px*A\n",
    "print\"P=\",round(P),\"N\"\n",
    "print \"The plane of qmax is at 45° to the plane of px. \"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.3 page number 354"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "pn= 30.0 N/mm^2\n",
      "pt= 86.6 N/mm^2\n",
      "p= 91.65 N/mm^2\n",
      "alpha= 19.1 °\n"
     ]
    }
   ],
   "source": [
    "#The direct stresses at a point in the strained material are 120 N/mm2 compressive and 80 N/mm2 tensile. There is no shear stress.\n",
    "\n",
    "from math import sqrt,cos,sin,atan,pi\n",
    "#variable declaration\n",
    "\n",
    "#The plane AC makes 30° (anticlockwise) to the plane of px (y-axis). Hence theta= 30°. px = 80 N/mm^2 py = – 120 N/mm^2 ,q = 0\n",
    "\n",
    "px=float(80)\n",
    "py=float(-120)\n",
    "q=float(0)\n",
    "theta=30\n",
    "pn=((px+py)/2)+((px-py)/2)*cos(2*theta*pi/180)+q*sin(2*theta*pi/180)\n",
    "\n",
    "print\"pn=\",round(pn),\"N/mm^2\"\n",
    "\n",
    "pt=((px-py)/2)*sin(2*theta*pi/180)-q*cos(2*theta*pi/180)\n",
    "\n",
    "print\"pt=\",round(pt,1),\"N/mm^2\"\n",
    "p=sqrt(pow(pn,2)+pow(pt,2))\n",
    "\n",
    "print\"p=\",round(p,2),\"N/mm^2\"\n",
    "\n",
    "alpha=atan(pn/pt)*180/pi\n",
    "\n",
    "print \"alpha=\", round(alpha,1),\"°\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.4 page number 355"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "theta= 37.98 ° and  127.98 °\n",
      "p1= 278.08 N/mm^2\n",
      "p2= 71.92 N/mm^2\n",
      "qmax= 103.08 N/mm^2\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt,cos,sin,atan,pi\n",
    "#variable declaration\n",
    "#Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then\n",
    "\n",
    "px=float(200)                    #N/mm^2\n",
    "py=float(150)                    #N/mm^2\n",
    "q=float(100)                     #N/mm^2\n",
    "\n",
    "theta1=(atan((2*q)/(px-py))*180)/(pi*2)                        \n",
    "theta2=90+theta1\n",
    "print\"theta=\",round(theta1,2),\"°\" \" and \",round(theta2,2),\"°\"\n",
    "\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p1=\",round(p1,2),\"N/mm^2\"\n",
    "\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p2=\",round(p2,2),\"N/mm^2\"\n",
    "\n",
    "qmax=sqrt((pow((px-py)/2,2))+pow(q,2))\n",
    "\n",
    "print\"qmax=\",round(qmax,2),\"N/mm^2\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.5 page number 356"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "p1= 82.8 N/mm^2\n",
      "p2= -62.8 N/mm^2\n",
      "qmax= 72.8 N/mm^2\n",
      "theta= 7.97 ° and  97.97 °\n",
      "theta'= 37.03 ° and= 52.97 °\n",
      "answer in book is wrong\n"
     ]
    }
   ],
   "source": [
    "\n",
    "from math import sqrt,cos,sin,atan,pi\n",
    "#variable declaration\n",
    "#Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then\n",
    "\n",
    "px=float(80)                    #N/mm^2\n",
    "py=float(-60)                    #N/mm^2\n",
    "q=float(20)                     #N/mm^2\n",
    "\n",
    "\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p1=\",round(p1,2),\"N/mm^2\"\n",
    "\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p2=\",round(p2,2),\"N/mm^2\"\n",
    "\n",
    "qmax=sqrt((pow((px-py)/2,2))+pow(q,2))\n",
    "\n",
    "print\"qmax=\",round(qmax,2),\"N/mm^2\"\n",
    "\n",
    "#let theta be the inclination of principal stress to the plane of px.\n",
    "\n",
    "\n",
    "theta1=(atan((2*q)/(px-py))*180)/(pi*2)                        \n",
    "theta2=90+theta1\n",
    "print\"theta=\",round(theta1,2),\"°\" \" and \",round(theta2,2),\"°\"\n",
    "\n",
    "#Planes of maximum shear make 45° to the above planes.\n",
    "theta11=45-theta1\n",
    "theta22=theta2-45\n",
    "print\"theta'=\",round(theta11,2),\"°\",\"and=\",round(theta22,2),\"°\"\n",
    "\n",
    "print\"answer in book is wrong\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.6 page number 357"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "p1= -35.96 N/mm^2\n",
      "p2= -139.04 N/mm^2\n",
      "qmax= 51.54 N/mm^2\n",
      "theta= 37.98 ° and  127.98 °\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt,cos,sin,atan,pi\n",
    "#variable declaration\n",
    "#Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then\n",
    "\n",
    "px=float(-100)                    #N/mm^2\n",
    "py=float(-75)                    #N/mm^2\n",
    "q=float(-50)                     #N/mm^2\n",
    "\n",
    "\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p1=\",round(p1,2),\"N/mm^2\"\n",
    "\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p2=\",round(p2,2),\"N/mm^2\"\n",
    "\n",
    "qmax=sqrt((pow((px-py)/2,2))+pow(q,2))\n",
    "\n",
    "print\"qmax=\",round(qmax,2),\"N/mm^2\"\n",
    "\n",
    "#let theta be the inclination of principal stress to the plane of px.\n",
    "\n",
    "\n",
    "theta1=(atan((2*q)/(px-py))*180)/(pi*2)                        \n",
    "theta2=90+theta1\n",
    "print\"theta=\",round(theta1,2),\"°\" \" and \",round(theta2,2),\"°\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.7 page number 358"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(i) p1= 131.07 N/mm^2\n",
      "p2= -81.07 N/mm^2\n",
      "(ii) qmax= 106.07 N/mm^2\n",
      "theta= -22.5 ° clockwise\n",
      "theta2= 22.5 °\n",
      "p= 108.97 N/mm^2\n",
      "phi= 13.3 °\n",
      "mitake in book answer is wrong\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt,cos,sin,atan,pi\n",
    "#variable declaration\n",
    "#Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then\n",
    "\n",
    "px=float(-50)                    #N/mm^2\n",
    "py=float(100)                    #N/mm^2\n",
    "q=float(75)                     #N/mm^2\n",
    "\n",
    "\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"(i) p1=\",round(p1,2),\"N/mm^2\"\n",
    "\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "\n",
    "print \"p2=\",round(p2,2),\"N/mm^2\"\n",
    "\n",
    "qmax=sqrt((pow((px-py)/2,2))+pow(q,2))\n",
    "\n",
    "print\"(ii) qmax=\",round(qmax,2),\"N/mm^2\"\n",
    "\n",
    "#let theta be the inclination of principal stress to the plane of px.\n",
    "\n",
    "\n",
    "theta1=(atan((2*q)/(px-py))*180)/(pi*2)                        \n",
    "\n",
    "print\"theta=\",round(theta1,2),\"°\" \" clockwise\"\n",
    "\n",
    "#Plane of maximum shear makes 45° to it \n",
    "\n",
    "theta2=theta1+45\n",
    "print\"theta2=\",round(theta2,2),\"°\" \n",
    "\n",
    "#Normal stress on this plane is given by\n",
    "\n",
    "pn=((px+py)/2)+((px-py)/2)*cos(2*theta2*pi/180)+q*sin(2*theta2*pi/180)\n",
    "\n",
    "pt=qmax\n",
    "\n",
    "#Resultant stress\n",
    "p=sqrt(pow(pn,2)+pow(pt,2))\n",
    "\n",
    "print \"p=\",round(p,2),\"N/mm^2\"\n",
    "\n",
    "#Let ‘p’ make angle phi to tangential stress (maximum shear stress plane). \n",
    "\n",
    "phi=atan(pn/pt)*180/pi\n",
    "\n",
    "print \"phi=\",round(phi,1),\"°\"\n",
    "\n",
    "#there is mistake in book\n",
    "print\"mitake in book answer is wrong\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.9 page number 361"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " p1= 0.17 N/mm^2\n",
      " p2= -84.17 N/mm^2\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "\n",
    "#variable declaration\n",
    "\n",
    "w=float(100)                   #wide of rectangular beam,mm\n",
    "h=float(200)                   #height or rectangular beam dude,mm\n",
    "\n",
    "I=w*pow(h,3)/12\n",
    "\n",
    "#At point A, which is at 30 mm below top fibre \n",
    "y=100-30\n",
    "M=float(80*1000000)            #sagging moment,KN-m\n",
    "\n",
    "fx=M*y/I\n",
    "\n",
    "px=-fx\n",
    "F=float(100*1000 )            #shear force,N\n",
    "b=float(100)\n",
    "A=b*30\n",
    "y1=100-15\n",
    "\n",
    "q=(F*(A*y1))/(b*I)     #shearing stress,N/mm^2\n",
    "\n",
    "py=0\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "print \" p1=\",round(p1,2),\"N/mm^2\"\n",
    "print \" p2=\",round(p2,2),\"N/mm^2\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.10 page number 362\n",
    " "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(i) p1= 0.8333 N/mm^2\n",
      " p2= -0.8333 N/mm^2\n",
      "theta= 45.0 °  and  135.0 °\n",
      "(ii) p1= 0.0122 N/mm^2\n",
      " p2= -32.4196 N/mm^2\n",
      "theta= -1.0 °  and  89.0 °\n",
      "mistake in book\n",
      "(iii) p1= 0.0 N/mm^2\n",
      " p2= -64.8148 N/mm^2\n",
      "theta= -0.0 °  and  90.0 °\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt,atan\n",
    "\n",
    "P1=float(20)                        #vertical loading from A at distance of 1m,KN.\n",
    "P2=float(20)                        #vertical loading from A at distance of 2m,KN.\n",
    "P3=float(20)                        #vertical loading from A at distance of 3m,KN.\n",
    "Ra=(P1+P2+P3)/2              #Due to symmetry\n",
    "\n",
    "Rb=Ra                         \n",
    "#At section 1.5 m from A\n",
    "F=(Ra-P1)*1000\n",
    "M=float((Ra*1.5-P1*0.5)*1000000)\n",
    "b=float(100)\n",
    "h=float(180)\n",
    "\n",
    "I=float((b*pow(h,3))/12)\n",
    "\n",
    "# Bending stress \n",
    "#f=M*y/I\n",
    "y11=0\n",
    "f1=(-1)*M*y11/I\n",
    "y22=45\n",
    "f2=(-1)*M*y22/I\n",
    "y33=90\n",
    "f3=(-1)*M*y33/I\n",
    "#Shearing stress at a fibre ‘y’ above N–A is\n",
    "#q=(F/(b*I))*(A*y1)\n",
    "#at y=0,\n",
    "y1=45\n",
    "A1=b*90\n",
    "q1=(F/(b*I))*(A1*y1)\n",
    "#at y=45\n",
    "y2=float(90-45/2)\n",
    "A2=b*45\n",
    "q2=(F/(b*I))*(A2*y2)\n",
    "#at y=90\n",
    "q3=0\n",
    "\n",
    "#(a) At neutral axis (y = 0) : The element is under pure shear \n",
    "\n",
    "py=0\n",
    "\n",
    "p1=(f1+py)/2+sqrt(pow(((f1-py)/2),2)+pow(q1,2))\n",
    "\n",
    "p2=(f1+py)/2-sqrt(pow(((f1-py)/2),2)+pow(q1,2))\n",
    "print \"(i) p1=\",round(p1,4),\"N/mm^2\"\n",
    "print \" p2=\",round(p2,4),\"N/mm^2\"\n",
    "\n",
    "theta1=45\n",
    "theta2=theta1+90\n",
    "print\"theta=\",round(theta1),\"°\",\" and \",round(theta2),\"°\"\n",
    "\n",
    "#(b) At (y = 45)\n",
    "py=0 \n",
    "\n",
    "p1=(f2+py)/2+sqrt(pow(((f2-py)/2),2)+pow(q2,2))\n",
    "\n",
    "p2=(f2+py)/2-sqrt(pow(((f2-py)/2),2)+pow(q2,2))\n",
    "print \"(ii) p1=\",round(p1,4),\"N/mm^2\"\n",
    "print \" p2=\",round(p2,4),\"N/mm^2\"\n",
    "\n",
    "thetab1=(atan((2*q2)/(f2-py))*180)/(pi*2)\n",
    "thetab2=thetab1+90\n",
    "print\"theta=\",round(thetab1),\"°\",\" and \",round(thetab2),\"°\"\n",
    "#mistake in book\n",
    "print\"mistake in book\"\n",
    "\n",
    "#(c) At Y=90\n",
    "\n",
    "py=0\n",
    "\n",
    "p1=(f3+py)/2+sqrt(pow(((f3-py)/2),2)+pow(q3,2))\n",
    "\n",
    "p2=(f3+py)/2-sqrt(pow(((f3-py)/2),2)+pow(q3,2))\n",
    "print \"(iii) p1=\",round(p1,4),\"N/mm^2\"\n",
    "print \" p2=\",round(p2,4),\"N/mm^2\"\n",
    "\n",
    "thetac1=(atan((2*q3)/(f3-py))*180)/(pi*2)\n",
    "thetac2=thetac1+90\n",
    "print\"theta=\",round(thetac1),\"°\",\" and \",round(thetac2),\"°\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# example 11.11 page number 364\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " p1= 5.21 N/mm^2\n",
      " p2= -107.56 N/mm^2\n",
      "qmax= 56.38 N/mm^2\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "\n",
    "#variable declaration\n",
    "L=float(6)                     #m\n",
    "w=float(60)                    #uniformly distributed load,KN/m\n",
    "Rs=L*w/2                       #Reaction at support,KN\n",
    "\n",
    "#Moment at 1.5 m from support\n",
    "M =float( Rs*1.5-(w*pow(1.5,2)/2))\n",
    "#Shear force at 1.5 m from support \n",
    "F=Rs-1.5*w\n",
    "\n",
    "B=float(200)                   #width of I-beam,mm\n",
    "H=float(400)                   #height or I-beam,mm\n",
    "b=float(190)\n",
    "h=float(380)\n",
    "I= (B*pow(H,3)/12)-(b*pow(h,3)/12)\n",
    "\n",
    "#Bending stress at 100 mm above N–A\n",
    "y=100\n",
    "\n",
    "f=M*1000000*y/I\n",
    "\n",
    "#Thus the state of stress on an element at y = 100 mm, as px = f,py=0\n",
    "px=-f\n",
    "py=0\n",
    "A=200*10*195+10*90*145\n",
    "q=(F*1000*(A))/(10*I)     #shearing stress,N/mm^2\n",
    "\n",
    "p1=(px+py)/2+sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "p2=(px+py)/2-sqrt(pow(((px-py)/2),2)+pow(q,2))\n",
    "print \" p1=\",round(p1,2),\"N/mm^2\"\n",
    "print \" p2=\",round(p2,2),\"N/mm^2\"\n",
    "\n",
    "\n",
    "qmax=sqrt((pow((px-py)/2,2))+pow(q,2))\n",
    "\n",
    "print\"qmax=\",round(qmax,2),\"N/mm^2\"\n"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python [Root]",
   "language": "python",
   "name": "Python [Root]"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 2
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython2",
   "version": "2.7.12"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}