{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Examples 5 -- Body Equilibrium\n", "\n", "Contents:\n", "- [A. Spool and sprocket](#A)\n", "- [B. Plane wing](#B)\n", "- [C. Antenna](#C)\n", "- [D. Device with pulleys](#D) \n", "- [E. Clipboard clamp](#E) \n", "- [F. Bar with self-aligning bearing (3D)](#F) \n", "- [G. Bar with cable (3D)](#G) \n", "- [H. Plane wing (3D)](#H) " ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "from numpy import *\n", "dtr = pi/180 # degree-to-radian conversion" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## A. Spool and sprocket\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": true }, "outputs": [], "source": [ "R_spool = 8.0 # in\n", "R_sproc = 5.0 # in\n", "CDx,CDy = 3.0,6.0 # in\n", "BDx = CDx\n", "BDy = 4.0 # in\n", "l_BA = 13. # in\n", "Pdirx,Pdiry = -4.,-3.\n", "T_angle = 15*dtr # radians below -x\n", "# part a\n", "P_max = 50.0 # lb\n", "# part b\n", "T_b = 80.0 # lb" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "For part a, we balance two moments whose forces are already perpendicular to the radial direction:\n", "\n", "$$T_{max} R_{spool} = P_{max} L_{BA}$$\n", "\n", "For part b, moment sum in $z$-direction gives\n", "\n", "$$ T R_{spool} = |( R_{sproc} \\hat{u}_{BD}) \\times (F_{CD} \\hat{u}_{CD})| $$\n", "\n", "so\n", "\n", "$$ F_{CD} = \\frac{ T R_{spool}}{ R_{sproc} \\left |\\hat{u}_{BD} \\times \\hat{u}_{CD} \\right |} $$" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "part a\n", " T = 81.25 lb\n", " Bx= 118.481473386 lb\n", " By= 51.0290474146 lb\n", "part b\n", " FCD= -143.10835056 lb\n", " Bx= 13.2740661031 lb\n", " By= 148.705523608 lb\n" ] } ], "source": [ "# part a\n", "T_max = P_max*l_BA/R_spool\n", "P = P_max * array((Pdirx,Pdiry)) / sqrt(Pdirx**2+Pdiry**2)\n", "T = T_max * array((-cos(T_angle),-sin(T_angle)))\n", "Bx = -P[0] - T[0]\n", "By = -P[1] - T[1]\n", "print ('part a')\n", "print (' T =',T_max,'lb')\n", "print (' Bx=',Bx,'lb')\n", "print (' By=',By,'lb')\n", "# part b\n", "T = T_b * array((-cos(T_angle),-sin(T_angle)))\n", "uCD = array((CDx,-CDy)) / sqrt(CDx**2+CDy**2)\n", "uBD = array((BDx, BDy)) / sqrt(BDx**2+BDy**2)\n", "FCD_mag = T_b * R_spool / ( R_sproc*abs(cross(uBD,uCD)))\n", "FCD = FCD_mag * uCD\n", "Bx = -FCD[0] - T[0]\n", "By = -FCD[1] - T[1]\n", "print ('part b')\n", "print (' FCD=',-FCD_mag,'lb')\n", "print (' Bx=',Bx,'lb')\n", "print (' By=',By,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## B. Plane wing\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": true }, "outputs": [], "source": [ "M = 1150. # kg\n", "M_wing = 240. # kg\n", "AEy = 1.4 # m\n", "ABx = 3.0 # m\n", "BCx = 0.8 # m\n", "CDx = 2.5 # m\n", "g = 9.81 # N/kg" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "For part b, consider the free-body diagram and sum moments about point A:\n", "\n", "$$ - \\left | \\vec{r}_{AB} \\right | M_{wing}g - \\left |\\vec{r}_{AB} \\times \\left (T_{BE} \\hat{u}_{BE} \\right ) \\right | + \\left | \\vec{r}_{AC} \\right | P = 0 $$\n", "\n", "So\n", "\n", "$$ T_{BE} = \\frac{\\left | \\vec{r}_{AC} \\right | P- \\left | \\vec{r}_{AB} \\right | M_{wing}g}{\\left |\\vec{r}_{AB} \\hat{\\imath} \\times \\hat{u}_{BE} \\right |} $$\n", "\n", "$$T_{BE} = \\frac{\\left | \\vec{r}_{AC} \\right | P- \\left | \\vec{r}_{AB} \\right | M_{wing}g}{\\left |\\vec{r}_{AB} \\right | \\frac{\\left |\\vec{r}_{AE}\\right |}{\\left |\\vec{r}_{BE} \\right |} } $$\n", "\n", "$$T_{BE} = \\left (\\frac{\\left |\\vec{r}_{AC}\\right |}{\\left |\\vec{r}_{AB} \\right |} P- M_{wing}g \\right ) \\frac{\\left |\\vec{r}_{BE}\\right |}{\\left |\\vec{r}_{AE} \\right |}$$" ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "part a\n", " P= 5640.75 N\n", "part b\n", " TBE= 11328.2446259 N\n", " Ax= 10265.4642857 N\n", " Ay= 1504.2 N\n" ] } ], "source": [ "# part a\n", "P = 0.5 * M*g\n", "print ('part a')\n", "print (' P=',P,'N')\n", "# part b\n", "TBE_tension = ( (ABx+BCx)/ABx * P - M_wing*g ) * sqrt(ABx**2+AEy**2)/AEy\n", "uBE = array((-ABx,-AEy))/sqrt(ABx**2+AEy**2)\n", "TBE = TBE_tension * uBE\n", "FAx = -TBE[0]\n", "FAy = -TBE[1] + M_wing*g - P\n", "print ('part b')\n", "print (' TBE=',TBE_tension,'N')\n", "print (' Ax=',FAx,'N')\n", "print (' Ay=',FAy,'N')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## C. Antenna\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [], "source": [ "Wa, Wg = 327., 123. # lb\n", "F_load = 92. # lb\n", "Y = 29.*12 # in\n", "X1,X2 = 5.,6. # in\n", "R = 20. # in (gear radius)\n", "Q = 16*dtr # radians" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "\n", "\n", "\n", "Define positive $G$ to be in lower-right direction (because of arrow in answer box).\n", "\n", "$$\\sum F_x = A_x + G \\cos \\theta - w = 0$$\n", "\n", "$$\\sum F_y = -W_a - W_g + A_y - G \\sin \\theta = 0$$\n", "\n", "$$\\sum M_A = R G + X_1 W_g - X_2 W_a + Yw = 0$$\n", "\n", "Solving moment equation for $G$, then force sums for $A_x$ and $A_y$:" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "answer:\n", " G = -1533.45 lb\n", " Ay= 27.3238967224 lb\n", " Ax= 1566.04674764 lb\n" ] } ], "source": [ "w = F_load\n", "G = -(X1*Wg - X2*Wa + Y*w)/R\n", "Ax = w - G*cos(Q) \n", "Ay = Wa + Wg + G*sin(Q)\n", "print ('answer:')\n", "print (' G =',G,'lb')\n", "print (' Ay=',Ay,'lb')\n", "print (' Ax=',Ax,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "checked; works." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Note the direction for $G$ in the answer box defines the positive direction for that answer." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## D. Device with pulleys\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": true }, "outputs": [], "source": [ "R = 4. # in (pulley radii)\n", "W = 40. # lb\n", "X1,X2,X3 = 12.,26.,12. # in (left to right)\n", "Y1,Y2,Y3,Y4 = 17.,5.,10.,5. # in (bottom to top)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "\n", "\n", "\n", "First note that $T=W/2$. Also, we may shift forces to pulley bearings as you can verify with an FBD of the pulley at *C*.\n", "\n", "Sum of moments about $A$:\n", "\n", "$$\\sum M_A = B_y X_1 -B_X Y_2 +T(Y_2+Y_3)-T(X_1+X_2)-T(X_1+X_2+X_3)=0$$\n", "\n", "where $B_x = \\frac{X_1}{Z}B$ and $B_y = \\frac{Y_1+Y_2}{Z}B$\n", "where $Z = \\sqrt{X_1^2+(Y_1+Y_2)^2}$.\n", "\n", "Solve for $B$:\n", "\n", "$$B = T(2X_1+2X_2+X_3-Y_2-Y_3)\\frac{Z}{X_1Y_1} $$" ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "answer:\n", " Ax= -65.8823529412 lb\n", " Ay= -117.450980392 lb\n" ] } ], "source": [ "T=W/2\n", "Z = sqrt(X1**2+(Y1+Y2)**2)\n", "B = T*( 2*X1 + 2*X2 + X3 - Y2 - Y3) * Z/(X1*Y1)\n", "# force sum:\n", "Ax = T - B*X1/Z\n", "Ay = 2*T - B*(Y1+Y2)/Z\n", "print ('answer:')\n", "print (' Ax=',Ax,'lb')\n", "print (' Ay=',Ay,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## E. Clipboard clamp\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": true }, "outputs": [], "source": [ "kt = 3.0 # in.lb/rad\n", "X1,X2 = 2.1,0.8 # in (left to right)\n", "Ay = 4.0 # lb\n", "# vertical distances not required" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Moment sum about C when $P=0$:\n", "\n", "$$\\sum M_C = k_t \\theta - X_1 Ay = 0$$\n", "\n", "Moment sum about C when $A_y=0$:\n", "\n", "$$\\sum M_C = k_t \\theta - X_2 P = 0$$" ] }, { "cell_type": "code", "execution_count": 13, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "answer:\n", " P= 10.5 lb\n", " Q= 0.445633840657307 turns\n" ] } ], "source": [ "Q = X1*Ay/kt\n", "Q_turns = Q/(2*pi)\n", "P = kt*Q/X2\n", "print ('answer:')\n", "print (' P=',P,'lb')\n", "print (' Q=',Q_turns,'turns')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "\n", "\n", "## F. Bar with self-aligning bearing (3D)\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": true }, "outputs": [], "source": [ "X = 4.0 # ft\n", "Y1,Y2 = 4.,4. # ft\n", "ZE,ZF = 2.,4. # ft\n", "F = 102. # lb" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Force sum:\n", "\n", "$\\sum \\vec{F} = \\vec{A} + T_{BF} (-\\tfrac{1}{\\sqrt{2}}\\hat{\\imath}+\\tfrac{1}{\\sqrt{2}}\\hat{k}) -F\\hat{k} +(E_x\\hat{\\imath}+E_z\\hat{k})=0$\n", "\n", "x-direction: $A_x - T_{BF} \\tfrac{1}{\\sqrt{2}} + E_x=0$\n", "\n", "y-direction: $A_y =0$\n", "\n", "z-direction: $A_z + T_{BF} \\tfrac{1}{\\sqrt{2}} -F +E_z=0$\n", "\n", "Moment sum about *A*:\n", "\n", "$\\sum \\vec{M}_A = XT_{BF}\\tfrac{1}{\\sqrt{2}}(-\\hat{\\jmath}) + (X\\hat{\\imath}+Y_1\\hat{\\jmath})\\times(-F\\hat{k})+(X\\hat{\\imath}+(Y_1+Y_2)\\hat{\\jmath}+Z_E\\hat{k})\\times(E_x\\hat{\\imath}+E_z\\hat{k})=0$\n", "\n", "$\\sum \\vec{M}_A = -\\frac{XT_{BF}}{\\sqrt{2}}\\hat{\\jmath} + XF\\hat{\\jmath} - Y_1F\\hat{\\imath}+(Y_1+Y_2)E_z\\hat{\\imath}+ (Z_EE_x-XE_z) \\hat{\\jmath} - (Y_1+Y_2)E_x\\hat{k}=0\n", "$\n", "\n", "x-direction: $-Y_1F+(Y_1+Y_2)E_z =0$\n", "\n", "y-direction: $-\\frac{XT_{BF}}{\\sqrt{2}} + XF +Z_EE_x-XE_z =0$\n", "\n", "z-direction: $-(Y_1+Y_2)E_x=0$\n", "\n", "So $A_y=0$ and $E_x=0$ and $E_z=\\frac{Y_1F}{Y_1+Y_2}$. \n", "\n", "Then $T_{BF}=-(XE_z-XF)\\sqrt{2}/X$. \n", "\n", "And $A_x = T_{BF} \\tfrac{1}{\\sqrt{2}}$. \n", "\n", "And $A_z = -T_{BF} \\tfrac{1}{\\sqrt{2}} +F -E_z$." ] }, { "cell_type": "code", "execution_count": 17, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "answer:\n", " Ax= 51.0 lb\n", " Ay= 0.0 lb\n", " Az= 7.1054273576e-15 lb\n", " TBF= 72.124891681 lb\n", " Ex= 0.0 lb\n", " Ez= 51.0 lb\n" ] } ], "source": [ "Ay=Ex=0.\n", "Ez = Y1*F/(Y1+Y2)\n", "TBF = -(X*Ez-X*F)*sqrt(2)/X\n", "Ax = TBF/sqrt(2)\n", "Az = -TBF/sqrt(2)+F-Ez\n", "print ('answer:')\n", "print (' Ax=',Ax,'lb')\n", "print (' Ay=',Ay,'lb')\n", "print (' Az=',Az,'lb')\n", "print (' TBF=',TBF,'lb')\n", "print (' Ex=',Ex,'lb')\n", "print (' Ez=',Ez,'lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## G. Bar with cable (3D)\n", "\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [], "source": [ "F = 27. # lb\n", "X,Y,Z = 8.,8.,4. # in." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Force sum:\n", "\n", "$\\sum \\vec{F} = A_x \\hat{\\imath}+A_z\\hat{k}-F\\hat{\\jmath}+T_{CD}\\hat{u}_{CD}=0$\n", "\n", "x-direction: $A_x -\\tfrac{X}{\\alpha}T_{CD}=0$\n", "\n", "y-direction: $-F +\\tfrac{Y}{\\alpha}T_{CD}=0$\n", "\n", "z-direction: $A_z -\\tfrac{Z}{\\alpha}T_{CD}=0$\n", "\n", "Let $\\alpha = \\sqrt{X^2+Y^2+Z^2}$.\n", "\n", "Moment sum about *A*:\n", "\n", "$\\sum \\vec{M}_A = M_{Ax}\\hat{\\imath} +M_{Az}\\hat{k} -XF\\hat{k}+T_{CD}\\vec{r}_{AC}\\times\\hat{u}_{CD}=0$\n", "\n", "$\\sum \\vec{M}_A = M_{Ax}\\hat{\\imath} +M_{Az}\\hat{k} -XF\\hat{k}+\\frac{T_{CD}}{\\alpha}\\left ( -ZY\\hat{\\imath}+XY\\hat{k} \\right ) =0$\n", "\n", "x-direction: $M_{Ax} -ZY\\frac{T_{CD}}{\\alpha} =0$\n", "\n", "y-direction: (no equation)\n", "\n", "z-direction: $M_{Az} -XF+\\frac{T_{CD}}{\\alpha} XY =0$\n", "\n", "The force sum in the *y* direction yields $T_{CD}$, then the rest of the equations are just in terms of $T_{CD}$." ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "answer:\n", " TCD= 40.5 lb\n", " Ax= 27.0 lb\n", " Az= 13.5 lb\n", " MAx= 108.0 in.lb\n", " MAz= 0.0 in.lb\n" ] } ], "source": [ "alpha = sqrt(X**2+Y**2+Z**2)\n", "TCD = alpha*F/Y\n", "Ax = TCD*X/alpha\n", "Az = TCD*Z/alpha\n", "MAx = Z*Y*TCD/alpha\n", "MAz = X*F - X*Y*TCD/alpha\n", "print ('answer:')\n", "print (' TCD=',TCD,'lb')\n", "print (' Ax=',Ax,'lb')\n", "print (' Az=',Az,'lb')\n", "print (' MAx=',MAx,'in.lb')\n", "print (' MAz=',MAz,'in.lb')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "## H. Plane wing (3D)\n", "\n", "" ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "collapsed": true }, "outputs": [], "source": [ "M = 800. # kg\n", "Mwing = 170. # kg\n", "XA,XF = 0.3,0.1 # m\n", "YB,YBC = 2.2,0.7 # m\n", "ZE = 1.3 # m\n", "f = 0.1 # fraction of P that is Q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "\n", "\n", "Let $\\alpha=\\sqrt{Y_B^2+Z_E^2}$.\n", "\n", "Force sum:\n", "\n", "x-direction: $A_x-Q=0$\n", "\n", "y-direction: $A_y+F_y-T_{BE}\\frac{Y_B}{\\alpha}=0$\n", "\n", "z-direction: $A_z+F_z-T_{BE}\\frac{Z_E}{\\alpha}+P-M_wg=0 $\n", "\n", "Moment sum about *F*:\n", "\n", "$\\sum \\vec{M}_F = -(X_A+X_F)A_z\\hat{\\jmath}+(X_A+X_F)A_y\\hat{k}+(X_F\\hat{\\imath}+Y_B\\hat{\\jmath}) \\times \\left [-T_{BE}\\frac{Y_B}{\\alpha}\\hat{\\jmath}+\\left (-T_{BE}\\frac{Z_E}{\\alpha}-M_wg \\right )\\hat{k} \\right ]+Y_CQ\\hat{k}+Y_CP\\hat{\\imath}=0$\n", "\n", "$\\sum \\vec{M}_F = -(X_A+X_F)A_z\\hat{\\jmath}+(X_A+X_F)A_y\\hat{k}+\\left [Y_B\\left (-\\frac{T_{BE}Z_E}{\\alpha}-M_wg \\right )\\hat{\\imath}-X_F\\left (-\\frac{T_{BE}Z_E}{\\alpha}-M_wg \\right )\\hat{\\jmath}-\\frac{T_{BE}X_FY_B}{\\alpha}\\hat{k} \\right ]+Y_CQ\\hat{k}+Y_CP\\hat{\\imath}=0$\n", "\n", "x-direction: $Y_B\\left (-\\frac{T_{BE}Z_E}{\\alpha} -M_wg\\right )+Y_CP=0$\n", "\n", "y-direction: $-(X_A+X_F)A_z-X_F\\left (-\\frac{T_{BE}Z_E}{\\alpha}-M_wg\\right )=0$\n", "\n", "z-direction: $(X_A+X_F)A_y-\\frac{T_{BE}}{\\alpha}X_FY_B+Y_CQ=0$" ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "answer:\n", " P= 3924.0 N\n", " TBE= 6889.41126646 N\n", " Fy= 7293.35769231 N\n", " Fz= -44.5909090909 N\n", " Ax= 392.40000000000003 N\n", " Ay= -1362.08076923 N\n", " Az= 1293.13636364 N\n" ] } ], "source": [ "# part a\n", "g = 9.81 # N/kg\n", "P = 0.5*M*g\n", "# part b\n", "Q = f*P\n", "Ax = Q\n", "alpha = sqrt(YB**2+ZE**2)\n", "YC = YB+YBC\n", "TBE = -(-YC*P/YB+Mwing*g)*alpha/ZE\n", "tmp = XF*(-TBE*ZE/alpha - Mwing*g)\n", "Az = -tmp/(XA+XF)\n", "Ay = -(-TBE*XF*YB/alpha+YC*Q)/(XA+XF)\n", "Fy = TBE*YB/alpha - Ay\n", "Fz = Mwing*g - P - Az + TBE*ZE/alpha\n", "print ('answer:')\n", "print (' P=',P,'N')\n", "print (' TBE=',TBE,'N')\n", "print (' Fy=',Fy,'N')\n", "print (' Fz=',Fz,'N')\n", "print (' Ax=',Ax,'N')\n", "print (' Ay=',Ay,'N')\n", "print (' Az=',Az,'N')" ] } ], "metadata": { "anaconda-cloud": {}, "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": "3.7.1" } }, "nbformat": 4, "nbformat_minor": 1 }