diff options
Diffstat (limited to 'Engineering_Mechanics_by_A_K_Tayal/10-Uniform_Flexible_Suspension_Cables.ipynb')
| -rw-r--r-- | Engineering_Mechanics_by_A_K_Tayal/10-Uniform_Flexible_Suspension_Cables.ipynb | 276 |
1 files changed, 276 insertions, 0 deletions
diff --git a/Engineering_Mechanics_by_A_K_Tayal/10-Uniform_Flexible_Suspension_Cables.ipynb b/Engineering_Mechanics_by_A_K_Tayal/10-Uniform_Flexible_Suspension_Cables.ipynb new file mode 100644 index 0000000..87f9f96 --- /dev/null +++ b/Engineering_Mechanics_by_A_K_Tayal/10-Uniform_Flexible_Suspension_Cables.ipynb @@ -0,0 +1,276 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: Uniform Flexible Suspension Cables" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1: Cable_subjected_to_concentrated_loads.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Initilization of variables\n", +"W1=400 // N // vertical load at pt C\n", +"W2=600 // N // vertical load at pt D\n", +"W3=400 // N // vertical load at pt E\n", +"l=2 // m // l= Lac=Lcd=Lde=Leb\n", +"h=2.25 // m // distance of the cable from top\n", +"L=2 // m // dist of A from top\n", +"// Calculations\n", +"// Solving eqn's 1&2 using MATRIX for Xb & Yb\n", +"A=[-L 4*l;-h 2*l]\n", +"B=[((W1*l)+(W2*2*l)+(W1*3*l));(W1*l)]\n", +"C=inv(A)*B\n", +"// Now consider the F.B.D of BE, Take moment at E\n", +"y_e=(C(2)*l)/C(1) // m / here y_e is the distance between E and the top\n", +"theta_1=atand(y_e/l) // degree // where theta_1 is the angle between BE and the horizontal\n", +"T_BE=C(1)/cosd(theta_1) // N (T_BE=T_max)\n", +"// Now consider the F.B.D of portion BEDC\n", +"// Take moment at C\n", +"y_c=((C(2)*6)-(W3*4)-(W2*2))/(C(1)) // m\n", +"theta_4=atand(((y_c)-(l))/(l)) // degree\n", +"T_CA=C(1)/cosd(theta_4) // N // Tension in CA\n", +"// Results\n", +"clc\n", +"printf('(i) The horizontal reaction at B (Xb) is %f N \n',C(1))\n", +"printf('(i) The vertical reaction at B (Yb) is %f N \n',C(2))\n", +"printf('(ii) The sag at point E (y_e) is %f m \n',y_e)\n", +"printf('(iii) The tension in portion CA (T_CA) is %f N \n',T_CA)\n", +"printf('(iv) The max tension in the cable (T_max) is %f N \n',T_BE)\n", +"printf('(iv) The max slope (theta_1) in the cable is %f degree \n',theta_1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2: Cables_subjected_to_concentrated_loads.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Initiization of variables\n", +"W1=100 // N // Pt load at C\n", +"W2=150 // N // Pt load at D\n", +"W3=200 // N // Pt load at E\n", +"l=1 // m // l=Lac=Lcd=Lde=Leb\n", +"h=2 // m // dist between Rb & top\n", +"Xa=200 // N\n", +"Xb=200 // N\n", +"// Calculations\n", +"// consider the F.B.D of entire cable\n", +"// Take moment at A\n", +"Yb=((W1*l)+(W2*2*l)+(W3*3*l)-(Xb*h))/(4*l) // N\n", +"Ya=W1+W2+W3-Yb // N // sum Fy=0\n", +"// Now consider the F.B.D of AC\n", +"// Take moment at C,\n", +"y_c=(Ya*l)/Xa // m\n", +"theta_1=atand(y_c/l) // degree\n", +"T_AC=Xa/cosd(theta_1) // N // T_AC*cosd(theta_1)=horizontal component of tension in the cable\n", +"// here, T_AC=T_max\n", +"T_max=T_AC // N\n", +"// Now consider the F.B.D of portion ACD\n", +"y_d=((Ya*2*l)-(W1*l))/(Xa) // m // taking moment at D\n", +"theta_2=atand(((y_d)-(y_c))/(l)) // degree\n", +"T_CD=Xa/(cosd(theta_2)) // N \n", +"// Results\n", +"clc\n", +"printf('(i) The component of support reaction at A (Ya) is %f N \n',Ya)\n", +"printf('(i) The component of support reaction at B (Yb) is %f N \n',Yb)\n", +"printf('(ii) The tension in portion AC (T_AC) of the cable is %f N \n',T_AC)\n", +"printf('(ii) The tension in portion CD (T_CD) of the cable is %f N \n',T_CD)\n", +"printf('(iii) The max tension in the cable is %f N \n',T_max)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3: Cables_uniformly_loaded_per_unit_horizontal_distance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Initilization of variables\n", +"w=75 // kg/m // mass per unit length of thw pipe\n", +"l=20 // m // dist between A & B\n", +"g=9.81 // m/s^2 // acc due to gravity\n", +"y=2 // m // position of C below B\n", +"// Calculations\n", +"// Let x_b be the distance of point C from B \n", +"// In eq'n x_b^2+32*x_b-320=0\n", +"a=1\n", +"b=32\n", +"c=-320\n", +"x_b=(-b+sqrt(b^2-(4*a*c)))/(2*a) // m // we get x_b by equating eqn's 1&2\n", +"// Now tension T_0\n", +"T_0=((w*g*x_b^2)/(2*y))*(10^-3) //kN // from eq'n 1\n", +"// Now the max tension occurs at point A,hence x is given as,\n", +"x=20-x_b // m\n", +"w_x=w*g*x*10^(-3) // kN \n", +"T_max=sqrt((T_0)^2+(w_x)^2) // kN // Maximum Tension\n", +"// Results\n", +"clc\n", +"printf('The lowest point C which is situated at a distance (x_b) from support B is %f m \n',x_b)\n", +"printf('The maximum tension (T_max) in the cable is %f kN \n',T_max)\n", +"printf('The minimum tension (T_0) in the cable is %f kN \n',T_0)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.4: Cables_uniformly_loaded_per_unit_horizontal_distance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Initilization of variables\n", +"m=0.5 // kg/m // mass of the cable per unit length\n", +"g=9.81 // m/s^2\n", +"x=30 // m // length AB\n", +"y=0.5 // m // dist between C & the horizontal\n", +"x_b=15 // m // dist of horizontal from C to B\n", +"// Calculations\n", +"w=m*g // N/m // weight of the cable per unit length\n", +"T_0=(w*x_b^2)/(2*y) // N // From eq'n 1\n", +"T_B=sqrt((T_0)^2+(w*x/2)^2) // N // Tension in the cable at point B\n", +"W=T_B // N // As pulley is frictionless the tension in the pulley on each side is same,so W=T_B\n", +"// Slope of the cable at B,\n", +"theta=acosd(T_0/T_B) // degree\n", +"// Now length of the cable between C & B is,\n", +"S_cb=x_b(1+((2/3)*(y/x_b)^2)) // m\n", +"// Now total length of the cable AB is,\n", +"S_ab=2*S_cb // m \n", +"// Results\n", +"clc\n", +"printf('(i) The magnitude of load W is %f N \n',W)\n", +"printf('(ii) The angle of the cable with the horizontal at B is %f degree \n',theta)\n", +"printf('(iii) The total length of the cable AB is %f m \n',S_ab)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5: Cables_uniformly_loaded_per_unit_horizontal_distance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Initilization of variables\n", +"x=30 // m // distance between two electric poles\n", +"Tmax=400 // N // Max Pull or tension\n", +"w=3 // N/m // weight per unit length of the cable\n", +"// Calculations\n", +"// The cable is assumed to be parabolic in shape, its eq'n is y=w*x^2/2*T_0.....(eq'n 1). Substuting the co-ordinates of point B (l/2,h), where h is the sag in the cable.This gives, T_0=(w*(l/2)^2)/(2*h)=wl^2/8*h\n", +"// Now the maximum pull or tension occurs at B,\n", +"T_B=Tmax // N \n", +"// Hence T_B=Tmax=sqrt(T_0^2+(w*l/2)^2). On simplyfyingthis eq'n we get, \n", +"h=sqrt(x^2/(16*(((Tmax*2)/(w*x))^2-(1)))) // m \n", +"// Results \n", +"clc\n", +"printf('The smallest value of the sag in the cable is %f m \n',h)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.6: Catenary_Cables.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Initilization of variables\n", +"l=200 // m // length of the cable\n", +"m=1000 // kg // mass of the cable\n", +"S=50 // m // sag in the cable\n", +"s=l/2 // m\n", +"g=9.81 // m/s^2\n", +"// Calculations\n", +"w=(m*g)/l // N/m // mass per unit length of the cable\n", +"// Substuting the values s=l/2 & y=c+S in eq'n 1 to get the value of c,\n", +"c=7500/100 // m \n", +"Tmax=sqrt((w*c)^2+(w*s)^2) // N // Maximum Tension\n", +"// To determine the span (2*x) let us use the eq'n of catenary, y=c*cosh(x/c), where y=c+50. On simplyfying we get y/c=cosh(x/c), here let y/c=A\n", +"y=c+50\n", +"A=y/c \n", +"x=c*(acosh(A)) // m \n", +"L=2*x // m // where L= span\n", +"// Results\n", +"clc\n", +"printf('The horizontal distance between the supports and the max Tension (L) is %f m \n',L)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |