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KiranUofG authored and KiranUofG committed Aug 5, 2016
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332 changes: 332 additions & 0 deletions Hinge_Moment.ipynb
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{
"cells": [
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"from sympy import *"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"var('A0, A1, c, A,n')\n",
"x = symbols('x')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The following equation must be evaluated within LDVM:"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$M_{\\beta} = \\int_A^c\\Delta p (x_b - x)dx=\\rho\\int_A^c\\Big[\\sqrt{\\Big(1+\\Big(\\dfrac{\\partial \\eta}{\\partial x}\\Big)^2\\Big)}(U\\cos\\alpha + \\dot{h}\\sin\\alpha - \\dot{\\alpha}\\eta + \\dfrac{\\partial \\phi_{LEV}}{\\partial x}+\\dfrac{\\partial \\phi_{TEV}}{\\partial x}\\Big)\\gamma(x) + \\dfrac{\\partial}{\\partial t}\\int_0^x\\gamma(x_0)dx_0\\Big](x_b-x)dx$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The first terms are already set up within LDVM and can be integrated easily by switching the lower index from 1 to the index corresponding to the hinge moment. The time derivative term which was analytically easy to derive for the pitching moment is no longer easy due to the lower bound no longer being zero. The infinite summation doesn't evaluate to the simple expressions given in Ramesh et al. and Katz and Plotkin."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$\\dfrac{\\partial}{\\partial t} \\int_A^c \\int_0^x \\gamma(x_0,t)dx_0 dx\n",
"=\\dfrac{\\partial}{\\partial t} \\int_A^c \\int_0^\\theta 2U \\Bigg[A_0(t) \\dfrac{1 + \\cos\\theta_0}{\\sin\\theta_0} + \\sum_{n=1}^\\infty A_n(t) \\sin(n\\theta_0)\\Bigg] \\dfrac{c}{2} \\sin\\theta_0 d\\theta_0 dx $$\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$x_b\\dfrac{\\partial}{\\partial t} \\int_A^c \\int_0^\\theta cU \\Bigg[A_0(t) \\dfrac{1 + \\cos\\theta_0}{\\sin\\theta_0} + A_1(t) \\sin\\theta_0 +\\sum_{n=2}^\\infty A_n(t) \\sin(n\\theta_0)\\Bigg]\\sin\\theta_0 d\\theta_0 dx$$"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"A0*(x + sin(x))"
]
},
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(A0*(1 + cos(x)),x)"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"A1*(x/2 - sin(x)*cos(x)/2)"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(A1*sin(x)**2,x)"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"Piecewise((-x*sin(x)**2/2 - x*cos(x)**2/2 + sin(x)*cos(x)/2, Eq(n, -1)), (x*sin(x)**2/2 + x*cos(x)**2/2 - sin(x)*cos(x)/2, Eq(n, 1)), (-n*sin(x)*cos(n*x)/(n**2 - 1) + sin(n*x)*cos(x)/(n**2 - 1), True))"
]
},
"execution_count": 20,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(sin(n*x)*sin(x),x)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$x_b\\dfrac{\\partial}{\\partial t}cU \\int_A^c\\Bigg\\{A_0(t)(\\theta+\\sin\\theta) + A_1\\Big(\\dfrac{\\theta}{2} - \\dfrac{\\sin\\theta\\cos\\theta}{2}\\Big)+ \\sum_{n=2}^\\infty A_n(t) \\Big[\\dfrac{\\sin(n\\theta)\\cos(\\theta)-n\\sin(\\theta)\\cos(n\\theta)}{n^2 - 1}\\Big]\\Bigg\\}dx$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$x_b\\dfrac{\\partial}{\\partial t}\\dfrac{c^2U}{2} \\int_A^c\\Bigg\\{A_0(t)(\\theta+\\sin\\theta) + A_1\\Big(\\dfrac{\\theta}{2} - \\dfrac{\\sin\\theta\\cos\\theta}{2}\\Big)+ \\sum_{n=2}^\\infty A_n(t) \\Big[\\dfrac{\\sin(n\\theta)\\cos(\\theta)-n\\sin(\\theta)\\cos(n\\theta)}{n^2 - 1}\\Big]\\Bigg\\}\\sin\\theta d\\theta$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The definite integral is evaluated using symbolic math in Python and can be seen below. The first two terms generate relatively simple expressions while the latter has conditions placed upon it. Neglecting the first output in piecewise as it relates to n = -2, the remainder of the output dictates what the series will generate for n >= 2. "
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"-A0*(A*sin(A)**2/2 + A*cos(A)**2/2 - A*cos(A) - sin(A)*cos(A)/2 + sin(A)) + 3*pi*A0/2"
]
},
"execution_count": 33,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(A0*(x+sin(x))*sin(x),(x,A,pi))"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"-A1*(-A*cos(A)/2 - sin(A)**3/6 + sin(A)/2) + pi*A1/2"
]
},
"execution_count": 34,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(A1*(x/2 - sin(x)*cos(x)/2)*sin(x),(x,A,pi))"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false,
"scrolled": true
},
"outputs": [
{
"data": {
"text/plain": [
"Piecewise((-A*sin(A)**2*cos(2*A)/4 + A*sin(A)*sin(2*A)*cos(A)/2 + A*cos(A)**2*cos(2*A)/4 - 5*sin(A)**2*sin(2*A)/24 - sin(2*A)*cos(A)**2/8 - pi/4, Eq(n, -2)), (A*sin(A)**2*cos(2*A)/4 - A*sin(A)*sin(2*A)*cos(A)/2 - A*cos(A)**2*cos(2*A)/4 + 5*sin(A)**2*sin(2*A)/24 + sin(2*A)*cos(A)**2/8 + pi/4, Eq(n, 2)), (n**2*sin(A)**2*sin(A*n)/(n**4 - 5*n**2 + 4) + 3*n*sin(A)*cos(A)*cos(A*n)/(n**4 - 5*n**2 + 4) - sin(A)**2*sin(A*n)/(n**4 - 5*n**2 + 4) + 3*sin(pi*n)/(n**4 - 5*n**2 + 4) - 3*sin(A*n)*cos(A)**2/(n**4 - 5*n**2 + 4), True))"
]
},
"execution_count": 35,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate((-n*sin(x)*cos(n*x)/(n**2 - 1) + sin(n*x)*cos(x)/(n**2 - 1))*sin(x),(x,A,pi))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The last integration that needs to be determined is the following:"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$\\dfrac{\\partial}{\\partial t}\\dfrac{c^2U}{2} \\int_A^c x\\Bigg\\{A_0(t)(\\theta+\\sin\\theta) + A_1\\Big(\\dfrac{\\theta}{2} - \\dfrac{\\sin\\theta\\cos\\theta}{2}\\Big)+ \\sum_{n=2}^\\infty A_n(t) \\Big[\\dfrac{\\sin(n\\theta)\\cos(\\theta)-n\\sin(\\theta)\\cos(n\\theta)}{n^2 - 1}\\Big]\\Bigg\\}\\sin\\theta d\\theta$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$$\\dfrac{\\partial}{\\partial t} \\dfrac{c^3U}{4}\\int_A^c\\Bigg\\{A_0(t)(\\theta+\\sin\\theta) + A_1\\Big(\\dfrac{\\theta}{2} - \\dfrac{\\sin\\theta\\cos\\theta}{2}\\Big)+ \\sum_{n=2}^\\infty A_n(t) \\Big[\\dfrac{\\sin(n\\theta)\\cos(\\theta)-n\\sin(\\theta)\\cos(n\\theta)}{n^2 - 1}\\Big]\\Bigg\\}\\sin\\theta(1-\\cos\\theta) d\\theta$$"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"-A0*(A*sin(A)**2/4 + 3*A*cos(A)**2/4 - A*cos(A) - sin(A)**3/3 - 3*sin(A)*cos(A)/4 + sin(A)) + 7*pi*A0/4"
]
},
"execution_count": 36,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(A0*(x+sin(x))*sin(x)*(1-cos(x)),(x,A,pi))"
]
},
{
"cell_type": "code",
"execution_count": 37,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"-A1*(A*sin(A)**4/16 + A*sin(A)**2*cos(A)**2/8 - A*sin(A)**2/8 + A*cos(A)**4/16 + A*cos(A)**2/8 - A*cos(A)/2 + sin(A)**3*cos(A)/16 - sin(A)**3/6 - sin(A)*cos(A)**3/16 - sin(A)*cos(A)/8 + sin(A)/2) + 11*pi*A1/16"
]
},
"execution_count": 37,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate(A1*(x/2 - sin(x)*cos(x)/2)*sin(x)*(1-cos(x)),(x,A,pi))"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"Piecewise((A*sin(A)**3*sin(3*A)/16 + 3*A*sin(A)**2*cos(A)*cos(3*A)/16 - 3*A*sin(A)*sin(3*A)*cos(A)**2/16 - A*cos(A)**3*cos(3*A)/16 + sin(A)**2*sin(3*A)*cos(A)/16 - sin(A)**2*sin(3*A)/5 - 9*sin(A)*cos(A)*cos(3*A)/40 + sin(3*A)*cos(A)**3/48 + 3*sin(3*A)*cos(A)**2/40 + pi/16, Eq(n, -3)), (-A*sin(A)**2*cos(2*A)/4 + A*sin(A)*sin(2*A)*cos(A)/2 + A*cos(A)**2*cos(2*A)/4 + 2*sin(A)**3*cos(2*A)/15 - 4*sin(A)**2*sin(2*A)*cos(A)/15 - 5*sin(A)**2*sin(2*A)/24 - 2*sin(A)*cos(A)**2*cos(2*A)/5 + sin(2*A)*cos(A)**3/5 - sin(2*A)*cos(A)**2/8 - pi/4, Eq(n, -2)), (A*sin(A)**2*cos(2*A)/4 - A*sin(A)*sin(2*A)*cos(A)/2 - A*cos(A)**2*cos(2*A)/4 - 2*sin(A)**3*cos(2*A)/15 + 4*sin(A)**2*sin(2*A)*cos(A)/15 + 5*sin(A)**2*sin(2*A)/24 + 2*sin(A)*cos(A)**2*cos(2*A)/5 - sin(2*A)*cos(A)**3/5 + sin(2*A)*cos(A)**2/8 + pi/4, Eq(n, 2)), (-A*sin(A)**3*sin(3*A)/16 - 3*A*sin(A)**2*cos(A)*cos(3*A)/16 + 3*A*sin(A)*sin(3*A)*cos(A)**2/16 + A*cos(A)**3*cos(3*A)/16 - sin(A)**2*sin(3*A)*cos(A)/16 + sin(A)**2*sin(3*A)/5 + 9*sin(A)*cos(A)*cos(3*A)/40 - sin(3*A)*cos(A)**3/48 - 3*sin(3*A)*cos(A)**2/40 - pi/16, Eq(n, 3)), (-n**4*sin(A)**2*sin(A*n)*cos(A)/(n**6 - 14*n**4 + 49*n**2 - 36) + n**4*sin(A)**2*sin(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + n**3*sin(A)**3*cos(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) - 3*n**3*sin(A)*cos(A)**2*cos(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + 3*n**3*sin(A)*cos(A)*cos(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + 4*n**2*sin(A)**2*sin(A*n)*cos(A)/(n**6 - 14*n**4 + 49*n**2 - 36) - 10*n**2*sin(A)**2*sin(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + 6*n**2*sin(pi*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + 3*n**2*sin(A*n)*cos(A)**3/(n**6 - 14*n**4 + 49*n**2 - 36) - 3*n**2*sin(A*n)*cos(A)**2/(n**6 - 14*n**4 + 49*n**2 - 36) - 4*n*sin(A)**3*cos(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + 12*n*sin(A)*cos(A)**2*cos(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) - 27*n*sin(A)*cos(A)*cos(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) + 9*sin(A)**2*sin(A*n)/(n**6 - 14*n**4 + 49*n**2 - 36) - 39*sin(pi*n)/(n**6 - 14*n**4 + 49*n**2 - 36) - 12*sin(A*n)*cos(A)**3/(n**6 - 14*n**4 + 49*n**2 - 36) + 27*sin(A*n)*cos(A)**2/(n**6 - 14*n**4 + 49*n**2 - 36), True))"
]
},
"execution_count": 38,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"integrate((-n*sin(x)*cos(n*x)/(n**2 - 1) + sin(n*x)*cos(x)/(n**2 - 1))*sin(x)*(1-cos(x)),(x,A,pi))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Again, the expressions generated are very large and could be computationally intensive. The first two piecewise generated solutions are neglected as they deal with n = -2,-3."
]
}
],
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"display_name": "Python [Root]",
"language": "python",
"name": "Python [Root]"
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"language_info": {
"codemirror_mode": {
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"file_extension": ".py",
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"name": "python",
"nbconvert_exporter": "python",
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