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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 5 General Case of Forces in a plane"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Example 5.2 Equations of equilibrium"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('The reaction at P is', 5656.85424949238, 'N')\n",
"('The reaction at Q is ', 4000.0, 'N')\n"
]
}
],
"source": [
"import math\n",
"\n",
"#Initialization of Variables\n",
"W=2000 #N\n",
"Lab=2 #m #length of the member from the vertical to the 1st load of 2000 N\n",
"Lac=5 #m #length of the member from the vertical to the 2nd load of 2000 N\n",
"Lpq=3.5 #m\n",
"\n",
"#Calculations\n",
"Rq=((W*Lab)+(W*Lac))/Lpq #N #take moment abt. pt P\n",
"Xp=Rq #N #sum Fx=0\n",
"Yp=2*W #N #sum Fy=0\n",
"Rp=math.sqrt(Xp**2+Yp**2) #N\n",
"\n",
"#Resuts\n",
"print('The reaction at P is' ,Rp ,'N')\n",
"print('The reaction at Q is ',Rq ,'N')"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Example 5.3 Equations of equilibrium"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('The reaction at A i.e Ra is ', matrix([[ 120.27406887]]), 'N')\n",
"('The reaction at B i.e Rb is ', matrix([[ 35.13703443]]), 'N')\n",
"('The required tension in the string is ', matrix([[ 40.57275258]]), 'N')\n"
]
}
],
"source": [
"import math,numpy\n",
"#Initilization of vaiables\n",
"W=25 #N # self weight of the ladder\n",
"M=75 #N # weight of the man standing o the ladder\n",
"theta=63.43 #degree # angle which the ladder makes with the horizontal\n",
"alpha=30 #degree # angle made by the string with the horizontal\n",
"Loa=2 #m # spacing between the wall and the ladder\n",
"Lob=4 #m #length from the horizontal to the top of the ladder touching the wall(vertical)\n",
"\n",
"#Calculations\n",
"#Using matrix to solve the simultaneous eqn's 3 & 4\n",
"A=numpy.matrix('2 -4; 1 -0.577')\n",
"B=numpy.matrix('100;100')\n",
"C=numpy.linalg.inv(A)*B\n",
"\n",
"#Results\n",
"print('The reaction at A i.e Ra is ',C[0] ,'N')\n",
"print('The reaction at B i.e Rb is ',C[1] ,'N')\n",
"\n",
"#Calculations\n",
"T=C[1]/math.cos(math.radians(alpha)) #N # from (eqn 1)\n",
"\n",
"#Results\n",
"print('The required tension in the string is ',T, 'N')"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Example 5.4 Equations of Equilibrium"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('The reaction at B i.e Rb is ', 25.0, 'N')\n",
"('The horizontal reaction at A i.e Xa is ', 21.650635094610966, 'N')\n",
"('The vertical reaction at A i.e Ya is ', 112.5, 'N')\n"
]
}
],
"source": [
"import math\n",
"#Initilization of variables\n",
"W=100 #N\n",
"theta=60 #degree angle made by the ladder with the horizontal\n",
"alpha=30 #degree angle made by the ladder with the vertical wall\n",
"Lob=4 #m length from the horizontal to the top of the ladder touching the wall(vertical)\n",
"Lcd=2 #m length from the horizontal to the centre of the ladder where the man stands\n",
"\n",
"#Calculations\n",
"Lab=Lob*(1/math.cos(math.radians(alpha))) #m length of the ladder\n",
"Lad=Lcd*math.tan(math.radians(alpha)) #m\n",
"Rb=(W*Lad)/Lab #N take moment at A\n",
"Xa=Rb*math.sin(math.radians(theta)) #N From eq'n 1\n",
"Ya=W+Rb*math.cos(math.radians(theta)) #N From eq'n 2\n",
"\n",
"#Results\n",
"print('The reaction at B i.e Rb is ',Rb, 'N')\n",
"print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n",
"print('The vertical reaction at A i.e Ya is ',Ya,'N')\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Example 5.5 Equations of Equilibrium"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('The horizontal reaction at A i.e Xa is ', 28.867513459481287, 'N')\n",
"('The vertical reaction at A i.e Ya is ', 100, 'N')\n",
"('The reaction at B i.e Rb is ', 28.867513459481287, 'N')\n"
]
}
],
"source": [
"import math\n",
"#Initilization of variables\n",
"W=100 #N self weight of the man\n",
"alpha=30 #degree angle made by the ladder with the wall\n",
"Lob=4 #m length from the horizontal to the top of the ladder touching the wall(vertical)\n",
"Lcd=2 #m\n",
"\n",
"#Calculations\n",
"# using the equiblirium equations\n",
"Ya=W #N From eq'n 2\n",
"Lad=Lcd*math.tan(math.radians(alpha)) #m Lad is the distance fom pt A to the point where the line from the cg intersects the horizontal\n",
"Rb=(W*Lad)/Lob #N Taking sum of moment abt A\n",
"Xa=Rb #N From eq'n 1\n",
"\n",
"#Results\n",
"print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n",
"print('The vertical reaction at A i.e Ya is ',Ya,'N' )\n",
"print('The reaction at B i.e Rb is ',Rb ,'N')\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Example 5.6 Equations of Equilibrium"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"('The horizontal reaction at A i.e Xa is ', 3.84, 'N')\n",
"('The vertical reaction at A i.e Ya is ', 7.12, 'N')\n",
"('Therefore the reaction at A i.e Ra is ', 8.089499366462674, 'N')\n",
"('The reaction at D i.e Rd is ', 4.8, 'N')\n"
]
}
],
"source": [
"import math\n",
"#Initilization of variables\n",
"d=0.09 #m diametre of the right circular cylinder\n",
"h=0.12 #m height of the cyinder\n",
"W=10 #N self weight of the bar\n",
"l=0.24 #m length of the bar\n",
"\n",
"#Calculations\n",
"theta=math.degrees(math.atan(h/d)) #angle which the bar makes with the horizontal\n",
"Lad=math.sqrt(d**2+h**2) #m Lad is the length of the bar from point A to point B\n",
"Rd=(W*h*(math.cos(theta*math.pi/180)))/Lad #N Taking moment at A\n",
"Xa=Rd*(math.sin(theta*math.pi/180)) #N sum Fx=0.... From eq'n 1\n",
"Ya=W-(Rd*(math.cos(theta*math.pi/180))) #N sum Fy=0..... From eq'n 2\n",
"Ra=math.sqrt(Xa**2+Ya**2) #resultant of Xa & Ya\n",
"\n",
"#Results\n",
"print('The horizontal reaction at A i.e Xa is ',Xa, 'N')\n",
"print('The vertical reaction at A i.e Ya is ',Ya, 'N')\n",
"print('Therefore the reaction at A i.e Ra is ',Ra,'N')\n",
"print('The reaction at D i.e Rd is ',Rd,'N')"
]
}
],
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"language": "python",
"name": "Python [Root]"
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"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
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|
