{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 5 : Analysis and Design of Beams for Bending" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.1, Page number 322" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum normal stress in the beam = 60.000000 MPa\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "# Reactions\n", "Rb=40 # Reaction at B(kN)\n", "Rd=14 # Reaction at D(kN)\n", "\n", "# Calculations\n", "# Shear and Bending-Moment Diagrams\n", "V1=-20 # Force(kN) \n", "M1=0 # Moment(kN.m)\n", "V2=-20 # Force(kN) \n", "M2=-50 # Moment(kN.m) \n", "V3=26 # Force(kN) \n", "M3=-50 # Moment(kN.m)\n", "V4=26 # Force(kN)\n", "M4=28 # Moment(kN.m)\n", "V5=-14 # Force(kN) \n", "M5=28 # Moment(kN.m)\n", "V6=-14 # Force(kN)\n", "M6=0 # Moment(kN.m) \n", "# Maximum Normal Stress\n", "S=(1/6.0)*(0.080)*(pow(0.250,2)) # Section modulus of the beam(m**3)\n", "Mb=(50*pow(10,3)) # Moment(N.m)\n", "sM=(Mb/S)/((pow(10,6))) # Stress(Pa)\n", "\n", "# Result\n", "print ('Maximum normal stress in the beam = %lf MPa' %sM)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.2, Page number 323" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum normal stress to the left of point D = 16.000000 ksi\n", "Maximum normal stress to the right of point D = 14.095238 ksi\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "x=symbols('x')\n", "V=symbols('V')\n", "n=-1\n", "\n", "# Calculation\n", "#Equivalent Loading of Beam\n", "#Case(a) Shear and Bending-Moment Diagrams\n", "#From A to C\n", "V=n*(3*x) # Force(kips) \n", "M=n*3*(1/2)*x*x # Moment(kip.ft)\n", "#From C to D\n", "V=-24 # Force(kips) \n", "M=96-(24*x) # Moment(kip.ft)\n", "#From D to B\n", "V=-34 # Force(kips) \n", "M=226-(34*x) # Moment(kip.ft)\n", "#Case(b) Maximum Normal Stress to the Left and Right of Point D\n", "#To the left of D\n", "S=126 # Volume(in**3) \n", "M=2016 # Moment(kip.in)\n", "Sm=(M)/S # Stress(ksi)\n", "#To the right of D\n", "M=1776 # Moment(kip.in)\n", "Sm2=(M/S) # Stress(ksi)\n", "\n", "# Result\n", "print ('Maximum normal stress to the left of point D = %lf ksi' %Sm)\n", "print ('Maximum normal stress to the right of point D = %lf ksi' %Sm2)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 5.03, Page number 331" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum value of the bending moment :-\n", "0.5*l*w*x - 0.5*w*x**2\n" ] } ], "source": [ "import math\n", "from sympy import symbols,integrate\n", "\n", "#Variable declaration\n", "Ra=symbols('Ra') # Variable declaration\n", "Rb=symbols('Rb') # Variable declaration\n", "w=symbols('w') # Variable declaration\n", "l=symbols('l') # Variable declaration\n", "x=symbols('x') # Variable declaration\n", "Ma=0 # Bending moment \n", "n=-1\n", "\n", "#Calculation \n", "Ra=(1/2.0)*(w)*(l) # Reaction at A\n", "V=Ra+ n*integrate(w, (x, 0, x)) # Shear\n", "M=Ma + integrate(V, (x, 0, x)) # Bending moment\n", "\n", "# Result\n", "print ('Maximum value of the bending moment :-')\n", "print (M)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.3, Page number 333" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Bending moment at A = 0.000000 kip.ft\n", "Bending moment at B = 108.000000 kip.ft\n", "Bending moment at C = 92.000000 kip.ft\n", "Bending moment at D = 76.000000 kip.ft\n", "Bending moment at E = 60.000000 kip.ft\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "Mb=symbols('Mb') # Variable declaration\n", "Mc=symbols('Mc') # Variable declaration\n", "Md=symbols('Md') # Variable declaration\n", "Me=symbols('Me') # Variable declaration\n", "#Reactions\n", "D=((12)*(28) + (12)*(14) + (20)*(6))/24.0 # Force(kips)\n", "Ay=20+12-26+12 # Force(kips)\n", "Ax=0 # Force(kips)\n", "#Shear Diagram\n", "V=18-20 # Force(kips)\n", "#Bending-Moment Diagram\n", "Ma=0 # Moment(kip.ft)\n", "Mb=solve(Mb-108,Mb) # Moment(kip.ft)\n", "Mc=solve(Mc-Mb[0]+16,Mc) # Moment(kip.ft) \n", "Md=solve(Md-Mc[0]+16,Md) # Moment(kip.ft)\n", "Me=solve(Me-Md[0]+16,Me) # Moment(kip.ft)\n", "Vmax=18 # Force(kips)\n", "Mmax=108 # Moment(kip.ft)\n", "\n", "# Result\n", "print ('Bending moment at A = %lf kip.ft' %Ma)\n", "print ('Bending moment at B = %lf kip.ft' %Mb[0])\n", "print ('Bending moment at C = %lf kip.ft' %Mc[0])\n", "print ('Bending moment at D = %lf kip.ft' %Md[0])\n", "print ('Bending moment at E = %lf kip.ft' %Me[0])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.4, Page number 334" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Location of maximum normal stress = 4.000000 kip.ft\n", "Magnitude of maximum normal stress = 125.984252 kip.ft\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve\n", "\n", "#Variable declaration\n", "x=symbols('x') # Variable declaration\n", "Ra=80 # Reaction at A(kN)\n", "Rc=40 # Reaction at C(kN)\n", "Va=80 # Force at A(kN)\n", "Ma=0 # Moment at A \n", "S=1270\n", "\n", "#Calculation \n", "#Case(a)\n", "Vb=-120+80 # Force at B(kN)\n", "Vc=Vb # Force at C(kN) \n", "x=solve(0-80+(20*x),x) \n", "Md=Ma+160 # Bending moment at point D(kN.m)\n", "Mb=Md-40 # Bending moment at point B(kN.m) \n", "Mc=Mb-120 # Bending moment at point C(kN.m)\n", "sm=((Md*pow(10,3))/(S*pow(10,-6)*pow(10,6)*(1.0))) # Maximum normal stress(Pa)\n", "\n", "# Result\n", "print ('Location of maximum normal stress = %lf kip.ft' %x[0])\n", "print ('Magnitude of maximum normal stress = %lf kip.ft' %sm)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.7, Page number 342" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Minimum required depth of the beam = 14.546281 in\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "x=symbols('x') # Variable declaration \n", "Ra=80 \n", "Rc=40 \n", "sall=1.75 # Allowable stress(ksi)\n", "\n", "#Calculation \n", "#Reactions.\n", "B=((4.5*12)+(3.2*4))/8.0 # Reaction at B(kips)\n", "Ay=7.7-8.35 # Reaction at A(kips)\n", "A=-Ay\n", "#Shear Diagram\n", "Va=Ay # Force at A(kips) \n", "Vb=Va-((400*8)/1000.0) # Force at B(kips) \n", "#Determination of |M| max\n", "Ma=Mc=0 # Moment at C\n", "Mmax=18 # Maximum moment(kip.ft)\n", "\n", "#Minimum Allowable Section Modulus\n", "Smin=((Mmax*12)/sall) # Minimum allowable section modulus(in**3)\n", "\n", "\n", "#Minimum Required Depth of Beam\n", "h=math.sqrt((123.43*6)/3.5) # Minimum required depth of the beam\n", "\n", "# Result\n", "print ('Minimum required depth of the beam = %lf in' %h)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.8, Page number 343" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "We select the lightest shape available, namely W360*32.9\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "\n", "#Variable declaration\n", "x=symbols('x') # Variable declaration\n", "Ra=80 # Force on rod AC(lb)\n", "Rc=40 # Diameter at the upper junction of rod ABC(in)\n", "sall=160 # Normal stress for the grade of steel(MPa) \n", "V=0\n", "\n", "#Calculation \n", "#Reactions.\n", "D=((50*4)+(60*1.5))/5.0 # Reaction(kN)\n", "Ay=60+50-58 # Reaction(kN) \n", "\n", "#Shear Diagram\n", "Va=Ay=52 # Force at A(kN)\n", "Vb=52-60 # Force at B(kN)\n", "x=(52/20.0) # Distance(m)\n", "\n", "#Determination of |M| max\n", "Mmax=Me=67.6 # Maximum bending moment(kN.m)\n", "\n", "#Minimum Allowable Section Modulus\n", "Smin=(Mmax/sall)*(pow(10,-6))*(pow(10,9)) # Minimum Allowable Section Modulus(mm**3) \n", "\n", "#Selection of Wide-Flange Shape\n", "#W410*38.8 629\n", "#W360*32.9 475\n", "#W310*38.7 547\n", "#W250*44.8 531\n", "#W200*46.1 451\n", "\n", "#Result\n", "print('We select the lightest shape available, namely W360*32.9')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## Example 5.07, Page number 362" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The plate providing the most economical design is of triangular shape.\n", "Maximum depth h0 = 300.000000 in\n" ] } ], "source": [ "import math\n", "from sympy import symbols,integrate\n", "\n", "#Variable declaration\n", "x=symbols('x') # Variable declaration\n", "w=symbols('w') # Variable declaration \n", "b=symbols('b') # Variable declaration\n", "Sall=symbols('Sall') # Variable declaration \n", "h=symbols('h') # Variable declaration\n", "n=-1\n", "\n", "#Calculation \n", "# Bending Moment\n", "Vx=n*integrate(w,x) # Force \n", "Mx=n*integrate(w*x,x) # Moment\n", "\n", "\n", "#Case(a) Shape Of Plate\n", "h=(n*(6*(Mx))/(b*(Sall)))**(1/2.0) # Relation between h and x\n", "\n", "#Case(b) Shape Of Plate\n", "h0=(((3*135)/(0.040*1000.0*72.0))**(1/2.0))*(800) # Maximum depth h0(mm)\n", "\n", "#Result\n", "print('The plate providing the most economical design is of triangular shape.')\n", "print('Maximum depth h0 = %lf in' %h0)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.11, Page number 363" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The required number of pair of planks = 3.000000\n", "Length of plank 1 = 130.500000 in\n", "Length of plank 2 = 111.300000 in\n", "Length of plank 3 = 83.800000 in\n" ] } ], "source": [ "import math\n", "from sympy import symbols\n", "#Variable declaration\n", "x=symbols('x') # Variable declaration\n", "M=symbols('M') # Variable declaration\n", "b=symbols('b') # Variable declaration \n", "h=symbols('h') # Variable declaration\n", "sall=symbols('sall') # Variable declaration\n", "Mab=4.80*x # Moment AB(kip.in) \n", "Mbc=4.80*x-(4.80)*(x-48) # Moment BC(kip.in)\n", "\n", "\n", "#Calculation \n", "#Number of Pairs of Planks\n", "h=((6*M)/(b*sall))**(1/2.0) # Eq 1\n", "h=((6*230.4)/(4*2.40))**(1/2.0) # Height(in)\n", "No_Pl=3 # Since the original beam has a depth of 4.50 in., the planks must provide an additional depth of 7.50 in. Recalling that each pair of planks is 2.50 in. thick\n", "\n", "#Length of Planks\n", "h=symbols('h') # Variable declaration\n", "x=(((4)*(2.40))/(6.0*4.80))*(h**2) # from Eq 1\n", "x1=round(x.subs(h,4.50),2) # Distance(in)\n", "x2=round(x.subs(h,7.00),2) # Distance(in)\n", "x3=round(x.subs(h,9.50),2) # Distance(in)\n", "l1=144-2*x1 # Length(in)\n", "l2=round((144-2*x2),1) # Length(in)\n", "l3=round((144-2*x3),1) # Length(in)\n", "\n", "#Result\n", "print('The required number of pair of planks = %lf' %No_Pl)\n", "print('Length of plank 1 = %lf in' %l1)\n", "print('Length of plank 2 = %lf in' %l2)\n", "print('Length of plank 3 = %lf in' %l3)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## SAMPLE PROBLEM 5.12, Page number 364" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Length of plates = 3.530000 m\n", "Width of plates = 267.000000 mm\n" ] } ], "source": [ "import math\n", "from sympy import symbols,solve,Function\n", "\n", "#Variable declaration\n", "#Bending Moment\n", "x=symbols('x') # Variable declaration \n", "M=symbols('M') # Variable declaration\n", "I=symbols('I') # Variable declaration\n", "Ip=Function('Ip') # Variable declaration\n", "M=250*x # Bending moment \n", "S=(3.49*pow(10,-3)) # Section modulus(m**3)\n", "sall=160*(pow(10,3)) # Stress all(kN/(m**2)) \n", "Ib=1190*(pow(10,6)) # the moment of inertia Ib of the two plates with respect to the neutral axis\n", "d=678 # Depth(mm)\n", "t=16 # Thickness(mm)\n", "\n", "#Calculation \n", "#Case(a) Required Length of Plates\n", "M=round(S*sall,1) # Moment(kN.m)\n", "x=solve(M-250*x,x) # Distance(m)\n", "x=round(x[0],3) # Rounding\n", "l=round(8-(2*x),2) # Length of plates(m) \n", "\n", "\n", "#Case(b) Required Width of Plates\n", "M=250*4 # Maximum bending moment(kN.m)\n", "Ip=(((1/6.0)*b*(t**3)) + 2*b*t*(((1/2.0)*d)+((1/2)*t))**2)*(pow(10,3))# Moment of inertia\n", "I=Ib+Ip # Moment of inertia of beam and plates \n", "c=(1/2.0)*d + t # Distance from neutral axis to surface(mm)\n", "I=round((M*c)/(sall),3) # Moment of inertia(mm**4)\n", "I=I*(pow(10,9)) # Conversion\n", "b=round((I-Ib)/(3.854*(pow(10,6))),0) # Distance(mm)\n", "\n", "#Result\n", "print('Length of plates = %lf m' %l)\n", "print('Width of plates = %lf mm' %b)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.1" } }, "nbformat": 4, "nbformat_minor": 0 }