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{
"metadata": {
"name": "",
"signature": "sha256:55c40faafb847932f0fdcda855b3af16f1a2e4ef45941baaf0d7ee692a22c20c"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Ch:30 Chain drive"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"exa 30-1 - Page 778"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from math import sqrt, pi\n",
"n1=17#\n",
"n2=51#\n",
"C=300#\n",
"p=9.52#\n",
"Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n",
"x=(Ln-((n2+n1)/(2)))**2#\n",
"y=8*(((n2-n1)/(2*pi))**2)#\n",
"z=Ln-((n1+n2)/2)#\n",
"C=(p/4)*(z+(sqrt(x-y)))\n",
"\n",
"\n",
" # printing data in scilab o/p window\n",
"print \"C is %0.2f mm \"%(C)#\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"C is 300.00 mm \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"exa 30-2 - Page 778"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import tan\n",
"G=4#\n",
"n1=17#\n",
"n2=n1*G#\n",
"N1=2300#\n",
"Kc=1.2# #from table 30-2\n",
"p=12.7# #fom table 30-1\n",
"D1=p*n1#\n",
"D2=p*n2#\n",
"phi=2*10.6#\n",
"x=tan(phi/2)# #phi/2 = 10.6deg, from table 30-3\n",
"Da1=(p/x)+(0.6*p)#\n",
"Da2=(p/x*4)+(0.6*p)#\n",
"Cmin=Kc*((Da1+Da2)/2)#\n",
"Ln1=(2*Cmin/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/Cmin))#\n",
"Ln1=80#\n",
"print \"Ln is %0.0f \"%(Ln1)#"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ln is 80 \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"exa 30-3 - Page 779"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"N1=1000#\n",
"N2=500#\n",
"P=2.03*10**3# #from table 30-8\n",
"K1=1.26#\n",
"Ks=1#\n",
"#let Pc be the power transmitting capacity of the chain\n",
"Pc=P*K1/Ks#\n",
"p=9.52#\n",
"n1=21#\n",
"n2=42#\n",
"V=n1*p*N1/(60*10**3)#\n",
"#Let the chain tension be T\n",
"T=Pc/V#\n",
"#Let the breaking load be BL\n",
"BL=10700#\n",
"FOS=BL/T#\n",
"C=50*p#\n",
"Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n",
"L=Ln*p#\n",
"Pc=Pc*10**-3#\n",
"print \" Pc is %0.2f KW \"%(Pc)#\n",
"print \"\\n V is %0.3f m/s \"%(V)#\n",
"print \"\\n T is %0.1f N \"%(T)#\n",
"print \"\\n FOS is %0.2f \"%(FOS)#\n",
"print \"\\n L is %0.2f mm \"%(L)#\n",
"\n",
"#The difference in the value of L and T is due to rounding-off the values."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Pc is 2.56 KW \n",
"\n",
" V is 3.332 m/s \n",
"\n",
" T is 767.6 N \n",
"\n",
" FOS is 13.94 \n",
"\n",
" L is 1254.01 mm \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"exa 30-5 - Page 780"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"G=2#\n",
"P=5000#\n",
"Ks=1.7#\n",
"Pd=P*Ks#\n",
"K2=1.7#\n",
"p=15.88#\n",
"n1=17#\n",
"n2=n1*G#\n",
"D1=n1*p#\n",
"D2=n2*p#\n",
"C=40*p#\n",
"Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#\n",
"L=Ln*p#\n",
"print \"L is %0.2f mm \"%(L)#\n",
"#The difference in the value of L is due to rounding-off the values."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"L is 1678.25 mm \n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
|
