{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 3 - Design Against Static Load"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.1 Pg 62"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " dimension of cross section of link, t=19 mm. Adopt t=21 mm. \n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from sympy import symbols,solve\n",
    "# Given Data\n",
    "P=30## kN\n",
    "Sut=350## MPa\n",
    "n=2.5## factor of safety\n",
    "\n",
    "sigma_w=Sut/n## MPa (Working stress for the link)\n",
    "\n",
    "t=symbols('t')## thickness of link\n",
    "A=2.5*t**2## mm.sq. \n",
    "I=t*(2.5*t)**3/12## mm**4 (Moment of Inertia about N-A)\n",
    "sigma_d=P/A## N/mm.sq.\n",
    "e=10+1.25*t##mm\n",
    "M=P*10**3*e## N.mm\n",
    "sigma_t=M*1.25*t/I## N/mm.sq.\n",
    "#maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w  ...eqn(1)\n",
    "expr=sigma_d+sigma_t-sigma_w ## expression of polynomial from above eqn.\n",
    "t=solve(expr)## solving the equation (as denominator will me be multiplied by zero on R.H.S)\n",
    "t=t[0]## mm # discarding -ve roots\n",
    "print ' dimension of cross section of link, t=%.f mm. Adopt t=21 mm. '%(t)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.2 Pg 63"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " dimension of cross section of link, t=27 mm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from sympy import symbols,solve\n",
    "from math import sin,cos,pi\n",
    "# Given Data\n",
    "P=6## kN\n",
    "alfa=30## degree\n",
    "Sut=250## MPa\n",
    "n=2.5## factor of safety\n",
    "\n",
    "sigma_w=Sut/n## MPa (Working stress for the link)\n",
    "PH=P*10**3*cos(pi/180*alfa)## kN\n",
    "PV=P*10**3*sin(pi/180*alfa)## kN\n",
    "\n",
    "t=symbols('t')## thickness of link\n",
    "A=2*t*t## mm.sq. \n",
    "sigma_d=PH/A## N/mm.sq.\n",
    "M=PH*100+PV*250## N.mm\n",
    "I=t*(2*t)**3/12## mm**4 (Moment of Inertia)\n",
    "sigma_t=M*t/I## N/mm.sq.\n",
    "#maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w  ...eqn(1)\n",
    "expr=sigma_d+sigma_t-sigma_w ## expression of polynomial from above eqn.\n",
    "t=solve(expr,'t')## solving the equation (as denominator will me be multiplied by zero on R.H.S)\n",
    "t=t[0]## mm # discarding -ve roots\n",
    "print ' dimension of cross section of link, t=%.f mm.'%(t)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.3 Pg 64"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " dimension of cross section of link, t=22.36 mm. Use 23 mm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from sympy import symbols,solve\n",
    "# Given Data\n",
    "P=20## kN\n",
    "Sut=300## MPa\n",
    "n=3## factor of safety\n",
    "\n",
    "sigma_w=Sut/n## MPa (Working stress for the link)\n",
    "\n",
    "t=symbols('t')## thickness of link\n",
    "A=4*t*t## mm.sq. \n",
    "sigma_d=P*10**3/A## N/mm.sq.\n",
    "e=6*t##mm\n",
    "M=P*10**3*e## N.mm\n",
    "z=t*(4*t)**2/6## mm**3 (section modulus at x1-x2)\n",
    "sigma_b=M/z## N/mm.sq.\n",
    "#maximum tensile stress at x1 = sigma_d+sigma_b=sigma_w  ...eqn(1)\n",
    "expr=sigma_d+sigma_b-sigma_w ## expression of polynomial from above eqn.\n",
    "t=solve(expr,'t')## solving the equation (as denominator will me be multiplied by zero on R.H.S)\n",
    "t=t[1]## mm # discarding -ve roots\n",
    "print ' dimension of cross section of link, t=%.2f mm. Use 23 mm.'%(t)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.4 Pg 65"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " Equating resultant tensile stress gives, a = 21.85 mm\n",
      " \n",
      " Equating resultant compressive stress gives, a = 4.77 mm\n",
      " \n",
      " dimension of cross section of link, a=21.85 mm. adopt a=22 mm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from sympy import symbols,solve\n",
    "from math import ceil\n",
    "# Given Data\n",
    "P=15## kN\n",
    "sigma_t=20## MPa\n",
    "sigma_c=60## MPa\n",
    "n=3## factor of safety\n",
    "\n",
    "a=symbols('a')## from the diagram.\n",
    "# Area of cross section\n",
    "A1=2*a*a## mm.sq.\n",
    "A2=2*a*a/2## mm.sq.\n",
    "A=A1+A2## mm.sq. \n",
    "\n",
    "# Location of neutral axis\n",
    "#3*a**2*y_bar=2*a**2*a/2+a**2*(a+a/2)\n",
    "y_bar=(2*a**2*a/2+a**2*(a+a/2))/(3*a**2)## mm\n",
    "\n",
    "# Moment of Inertia about neutral axis N-A\n",
    "I=2*a*a**3/12+2*a**2*(y_bar-0.5*a)**2+2*((a/2)*(a**3/12)+(a**2/2)*(1.5*a-y_bar)**2)## mm**4\n",
    "yt=y_bar##mm\n",
    "yc=2*a-y_bar## mm\n",
    "e=y_bar-0.5*a##mm\n",
    "M=P*10**3*e## N.mm\n",
    "sigma_d=P*10**3/A## N/mm.sq.\n",
    "sigma_t1=M*yt/I##N/mm.sq.\n",
    "sigma_c1=M*yc/I##N/mm.sq.\n",
    "sigma_r_t=sigma_d+sigma_t1## N/mm.sq. (sigma_r_t=resultant tensile stress at AB=sigma_d+sigma_t)\n",
    "sigma_r_c=sigma_c1-sigma_d## N/mm.sq. (sigma_r_t=resultant tensile stress at AB=sigma_d+sigma_t)\n",
    "\n",
    "#equating resulting tensile stress with given value sigma_t-sigma_r_t=0...eqn(1)\n",
    "expr1=sigma_t-sigma_r_t## expression of polynomial from above eqn.\n",
    "a1=solve(expr1,'a')## solving the equation (as denominator will me be multiplied by zero on R.H.S)\n",
    "a1=a1[1]## mm # discasrding -ve roots\n",
    "print ' Equating resultant tensile stress gives, a = %.2f mm'%(a1)\n",
    "\n",
    "#equating resulting compressive stress with given value sigma_c-sigma_c_t=0...eqn(1)\n",
    "expr2=sigma_c-sigma_r_c## expression of polynomial from above eqn.\n",
    "a2=solve(expr2,'a')## solving the equation (as denominator will me be multiplied by zero on R.H.S)\n",
    "a2=a2[1]## mm # discarding -ve roots\n",
    "print ' \\n Equating resultant compressive stress gives, a = %.2f mm'%(a2)\n",
    "a=ceil(a1)##mm\n",
    "print ' \\n dimension of cross section of link, a=%.2f mm. adopt a=%.f mm.'%(a1,a)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.5 Pg 67"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " (i) Maximum shear stress theory\n",
      " diameter of shaft, d=99.2 mm or 100 mm\n",
      " \n",
      " (ii) Maximum strain energy theory\n",
      " diameter of shaft, d=94.0 mm\n",
      " \n",
      " Adopt d=100mm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import pi,sqrt,ceil\n",
    "# Given Data\n",
    "Syt=760## MPa\n",
    "M=15## kN.m\n",
    "T=25##kN.m\n",
    "n=2.5## factor of safety\n",
    "E=200## GPa\n",
    "v=0.25## Poisson's ratio\n",
    "\n",
    "sigma_d=Syt/n## MPa\n",
    "# let d is diameter of the shaft\n",
    "sigma_b_into_d_cube=32*M*10**6/pi## N/mm.sq. (where sigma_b_into_d_cube = sigma_d*d**3)\n",
    "tau_into_d_cube=16*T*10**6/pi#d**3## N/mm.sq. (where tau_into_d_cube = tau*d**3)\n",
    "sigma1_into_d_cube=sigma_b_into_d_cube/2+1/2*sqrt(sigma_b_into_d_cube**2+4*tau_into_d_cube**2) # # (where sigma1_into_d_cube=sigma1*d**3)\n",
    "sigma2_into_d_cube=sigma_b_into_d_cube/2-1/2*sqrt(sigma_b_into_d_cube**2+4*tau_into_d_cube**2)# # (where sigma2_into_d_cube=sigma2*d**3)\n",
    "print ' \\n (i) Maximum shear stress theory'\n",
    "tau_max_into_d_cube=(sigma1_into_d_cube-sigma2_into_d_cube)/2# #(where tau_max_into_d_cube = tau_max*d**3)\n",
    "d=(tau_max_into_d_cube/(sigma_d/2))**(1/3)##mm\n",
    "print ' diameter of shaft, d=%.1f mm or %.f mm'%(d,ceil(d))\n",
    "\n",
    "print ' \\n (ii) Maximum strain energy theory'\n",
    "#sigma1**2+sigma2**2-2*v*sigma1*sigma2=sigma_d**2\n",
    "d=((sigma1_into_d_cube**2+sigma2_into_d_cube**2-2*v*sigma1_into_d_cube*sigma2_into_d_cube)/sigma_d**2)**(1/6)\n",
    "print ' diameter of shaft, d=%.1f mm'%(d)\n",
    "print ' \\n Adopt d=100mm'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.6 Pg 69"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " Using equivalent torque equation,\n",
      " shaft diameter d = 105 mm\n",
      " \n",
      " Using equivalent bending moment equation,\n",
      " shaft diameter d = 97.68 mm or 98 mm\n",
      " \n",
      " Adopt d=105 mm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import sqrt,pi\n",
    "# Given Data\n",
    "N=200## rpm\n",
    "P=200## kW\n",
    "tau_d=42## Mpa\n",
    "W=900## N\n",
    "L=3## m\n",
    "sigma_t=56## MPa\n",
    "sigma_c=56## MPa\n",
    "\n",
    "T=P*60*10**3/(2*pi*N)## N.m\n",
    "M=W*L/4## N.m\n",
    "Te=sqrt(M**2+T**2)## N.m\n",
    "#Te=(pi/16)*d**3*tau_d\n",
    "d=(Te/((pi/16)*tau_d)*1000)**(1/3)## mm\n",
    "print ' \\n Using equivalent torque equation,\\n shaft diameter d = %.f mm'%(d)\n",
    "\n",
    "Me=(1/2)*(M+sqrt(M**2+T**2))## N.m\n",
    "#Me=(pi/32)*d**3*sigma_d\n",
    "d=(Me/((pi/32)*sigma_c)*10**3)**(1/3)##mm\n",
    "print ' \\n Using equivalent bending moment equation,\\n shaft diameter d = %.2f mm or %.f mm'%(d, ceil(d))\n",
    "print ' \\n Adopt d=105 mm.'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.8 Pg 70"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " Using equivalent torque equation,\n",
      " shaft diameter d = 23 mm\n",
      " \n",
      " Using equivalent bending moment equation,\n",
      " shaft diameter d = 21.40 mm or 22 mm\n",
      " \n",
      " Adopt d=23 mm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import sqrt,pi\n",
    "# Given Data\n",
    "M=15## N.m\n",
    "P=5## kW\n",
    "N=500## rpm\n",
    "tau_d=40## Mpa\n",
    "sigma_d=58## MPa\n",
    "\n",
    "T=P*60*10**3/(2*pi*N)## N.m\n",
    "Te=sqrt(M**2+T**2)## N.m\n",
    "#Te=(pi/16)*d**3*tau_d\n",
    "d=(Te/((pi/16)*tau_d)*1000)**(1/3)## mm\n",
    "print ' \\n Using equivalent torque equation,\\n shaft diameter d = %.f mm'%(d)\n",
    "\n",
    "Me=(1/2)*(M+sqrt(M**2+T**2))## N.m\n",
    "#Me=(pi/32)*d**3*sigma_d\n",
    "d=(Me/((pi/32)*sigma_d)*10**3)**(1/3)##mm\n",
    "print ' \\n Using equivalent bending moment equation,\\n shaft diameter d = %.2f mm or %.f mm'%(d, ceil(d))\n",
    "print ' \\n Adopt d=23 mm.'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.10 Pg 71"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " (i) Maximum Principal Stress Theory-\n",
      " \n",
      " Maximum value of torque, T = 235851 N.cm.\n",
      " \n",
      " (ii) Maximum Shear Stress Theory\n",
      " \n",
      " Maximum value of torque, T = 124765 N.cm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import sqrt,pi\n",
    "from sympy import symbols,solve\n",
    "# Given Data\n",
    "d=4## cm\n",
    "M=15000## N.cm\n",
    "Syt=20000## N/cm.sq.\n",
    "\n",
    "print ' \\n (i) Maximum Principal Stress Theory-'\n",
    "z=pi*d**3/32## cm.cube.\n",
    "sigma_b=M/z## N/cm.sq.\n",
    "T=symbols('T')\n",
    "tau=16*T/(pi*d**3)## N/cm.sq.\n",
    "#sigma1=(1/2)*(sigma_b+sqrt(sigma_b**2+4*tau**2)) # Maximum principal stress\n",
    "#sigma1=(sigma_b/2+sqrt(sigma_b**2/4+tau**2)) # on solving\n",
    "#tau=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)\n",
    "sigma1=Syt## N/cm.sq.\n",
    "T=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)*(pi*d**3)/16## N.cm.\n",
    "print ' \\n Maximum value of torque, T = %.f N.cm.'%(T)\n",
    "\n",
    "print ' \\n (ii) Maximum Shear Stress Theory'\n",
    "tau_d=0.5*Syt## N.cm.\n",
    "#Te=sqrt(M**2+T**2)=(pi/16)*d**3*tau_d\n",
    "T=sqrt(((pi/16)*d**3*tau_d)**2-M**2)## N.cm.\n",
    "print ' \\n Maximum value of torque, T = %.f N.cm.'%(T)\n",
    "# Answer in the textbook is not accurate."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.11 Pg 72"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " shaft diameter(using equivalent torque)-\n",
      " d=55 mm.\n",
      " \n",
      " shaft diameter(using equivalent bending moment)-\n",
      " d=57 mm.\n",
      " \n",
      " adopt d=57 mm.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import sqrt,pi\n",
    "# Given Data\n",
    "N=200## rpm\n",
    "P=25## kW\n",
    "tau_d=42## MPa\n",
    "W=900## N\n",
    "L=3## m\n",
    "Syt=56## MPa\n",
    "Syc=56## MPa\n",
    "sigma_d=56## MPa\n",
    "\n",
    "T=P*60*10**3/(2*pi*N)## N.m\n",
    "M=W*L/4## N.m\n",
    "Te=sqrt(M**2+T**2)## N.m\n",
    "# Te=(pi/16)*d**3*tau_d\n",
    "d=(Te*10**3/((pi/16)*tau_d))**(1/3)## mm\n",
    "print ' \\n shaft diameter(using equivalent torque)-\\n d=%.f mm.'%(d)\n",
    "\n",
    "Me=(1/2)*(M+sqrt(M**2+T**2))##N.m\n",
    "# Me=(pi/32)*d**3*sigma_d\n",
    "d=(Me*10**3/((pi/32)*sigma_d))**(1/3)## mm\n",
    "print ' \\n shaft diameter(using equivalent bending moment)-\\n d=%.f mm.'%(d)\n",
    "print ' \\n adopt d=57 mm.'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.12 Pg 72"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " value of t = 36.9 mm\n",
      " \n",
      " Area of cross-section of Hanger, A = 2716 mm.sq.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import sqrt,pi,cos,sin\n",
    "from sympy import symbols,solve\n",
    "# Given Data\n",
    "sigma_w=60## MPa\n",
    "F=10## kN\n",
    "alfa=30## degree\n",
    "\n",
    "FH=F*sin(pi/180*alfa)## kN\n",
    "FV=F*cos(pi/180*alfa)## kN\n",
    "t=symbols('t')## mm\n",
    "A=t*t## mm.sq.\n",
    "sigma_d=FV*10**3/A\n",
    "M=FV*10**3*120+FH*10**3*150## N.mm\n",
    "I=t*(2*t)**3/12## mm**4\n",
    "sigma_t=M*t/I## N/mm.sq.\n",
    "# Tensile stress at A=sigma_d+sigma_t=sigma_w ...eqn(1)\n",
    "expr = sigma_d+sigma_t-sigma_w## polynomial from above eqn.\n",
    "t=solve(expr,'t')## roots of the polynomial\n",
    "t=t[0]## mm # discarding -ve roots\n",
    "print ' \\n value of t = %.1f mm'%(t)\n",
    "A=2*t**2## mm.sq.\n",
    "print ' \\n Area of cross-section of Hanger, A = %.f mm.sq.'%(A)\n",
    "# Note-Answer in the textbook is slighly wrong and cross section not calculated."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## exa 3.13 Pg 74"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      " shaft diameter is : 63 mm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import tan,pi,sqrt\n",
    "# Given Data\n",
    "P=15## kW\n",
    "n1=200## rpm\n",
    "l=600## mm\n",
    "z2=18## no. of teeth\n",
    "m2=5## mm\n",
    "alfa2=14.5## degree\n",
    "l2=120## mm\n",
    "z1=72## no. of teeth\n",
    "m1=5## mm\n",
    "alfa1=14.5## degree\n",
    "l1=150## mm\n",
    "sigma_d=80## MPa\n",
    "\n",
    "d1=m1*z1## mm\n",
    "v1=pi*d1*n1/(60*10**3)## m/s\n",
    "Ft1=10**3*P/v1## N (outwards)\n",
    "Fr1=Ft1*tan(pi/180*alfa1)## N (Downwards)\n",
    "d2=m2*z2## mm\n",
    "v2=pi*d2*n1/(60*10**3)## m/s\n",
    "Ft2=10**3*P/v2## N (outwards)\n",
    "Fr2=Ft2*tan(pi/180*alfa2)## N (Upwards)\n",
    "\n",
    "# RAV*600=Fr1*450+Fr2*120 (Taking moments about bearing B)\n",
    "RAV=(Fr1*450+Fr2*120)/600## N (Downwards)\n",
    "RBV=(Fr1-Fr2-RAV)## N (upwards)\n",
    "MCV=RAV*l1## N.mm\n",
    "MBV=Fr2*l2## N.mm\n",
    "\n",
    "# RAH*600=-Ft1*450+Ft2*120 (Taking moments about bearing B)\n",
    "RAH=(-Ft1*450+Ft2*120)/600## N (Outwards)\n",
    "RBH=Ft1+Ft2+RAH## N (inwards)\n",
    "MCH=RAH*l1## N.mm\n",
    "MBH=Ft2*l2## N.mm\n",
    "\n",
    "# Resultant Bending Moments\n",
    "MC=sqrt(MCV**2+MCH**2)## N.mm\n",
    "MB=sqrt(MBV**2+MBH**2)## N.mm\n",
    "Mmax=max(MC,MB)## N.mm\n",
    "T=10**3*P/(2*pi*n1)## N.m\n",
    "Me=(1/2)*(Mmax+sqrt(Mmax**2+T**2))## N.mm\n",
    "# Me=(pi/32)*d**3*sigma_d\n",
    "d=(Me/((pi/32)*sigma_d))**(1/3)\n",
    "print ' \\n shaft diameter is : %.f mm'%(d)"
   ]
  }
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