{
"metadata": {
"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4:Torsion"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2 page number 183"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"dia = 10 #diameter of shaft(A-C)\n",
"c = dia/2 #mm - Radius\n",
"T = 30 #N/mm -Torque in the shaft \n",
"#Caliculations\n",
"\n",
"J = 3.14*(dia**4)/32 #mm4\n",
"shear_T = T*c*pow(10,3)/J # The torsion shear in the shaft AC\n",
"import numpy as np \n",
"print \"The maximum shear due to torsion is \",round(shear_T,2),\"Mpa\"\n",
"arr_T = np.zeros((3,3))\n",
"arr_T[0][1]=round(shear_T,1) #arranging the elements in array\n",
"arr_T[1][0]=round(shear_T,1)\n",
"print \"stress tensor matrix\",ceil(arr_T),\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum shear due to torsion is 152.87 Mpa\n",
"stress tensor matrix [[ 0. 153. 0.]\n",
" [ 153. 0. 0.]\n",
" [ 0. 0. 0.]]\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.3 page number 184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"dia_out = 20 #mm- outer diameter of shaft\n",
"dia_in = 16 #mm- inner diameter of shaft \n",
"c_out = dia_out/2 #mm - outer Radius of shaft \n",
"c_in = dia_in/2 #mm - inner radius of shaft \n",
"T = 40 #N/mm -Torque in the shaft \n",
"#caliculations\n",
"\n",
"J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4\n",
"shear_T_max = T*c_out*pow(10,3)/J # The maximum torsion shear in the shaft\n",
"shear_T_min = T*c_in*pow(10,3)/J # The maximum torsion shear in the shaft\n",
"print \"The maximum shear due to torsion is \",round(shear_T_max,2),\"Mpa\"\n",
"print \"The minimum shear due to torsion is \",round(shear_T_min,2),\"Mpa\"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum shear due to torsion is 43.15 Mpa\n",
"The minimum shear due to torsion is 34.52 Mpa\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4 page number 187"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"hp = 10 # horse power of motor \n",
"f = 30 # given \n",
"shear_T = 55 #Mpa - The maximum shearing in the shaft \n",
"#caliculations\n",
"\n",
"T = 119*hp/f # N.m The torsion in the shaft \n",
"#j/c=T/shear_T=K\n",
"k = T*pow(10,3)/shear_T #mm3\n",
"#c3=2K/3.14\n",
"c = pow((2*k/3),0.33) #mm - The radius of the shaft \n",
"diamter = 2*c #mm - The diameter of the shaft\n",
"print \"The Diameter of the shaft used is\",round(diamter,2),\"mm\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Diameter of the shaft used is 15.26 mm\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.5 page number 188"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"hp = 200 #Horse power\n",
"stress_sh = 10000 #psi- shear stress\n",
"rpm_1 = 20.0 # The rpm at which this shaft1 operates \n",
"rpm_2 = 20000.0 # The rpm at which this shaft2 operates\n",
"T_1= hp*63000.0/rpm_1 #in-lb Torsion due to rpm1\n",
"T_2= hp*63000/rpm_2 #in-lb Torsion due to rpm1\n",
"#caliculations \n",
"\n",
"#j/c=T/shear_T=K\n",
"k_1= T_1/stress_sh #mm3\n",
"#c3=2K/3.14\n",
"c_1= pow((2*k_1/3),0.33) #mm - The radius of the shaft \n",
"diamter_1 = 2*c_1 #mm - The diameter of the shaft\n",
"print \"The Diameter of the shaft1 is\",round(diamter_1,2),\"mm\"\n",
"\n",
"#j/c=T/shear_T=K\n",
"k_2= T_2/stress_sh #mm3\n",
"#c3=2K/3.14\n",
"c_2= pow((2*k_2/3),0.33) #mm - The radius of the shaft \n",
"diamter_2 = 2*c_2 #mm - The diameter of the shaft\n",
"print \"The Diameter of the shaft2 is\",diamter_2,\"mm\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Diameter of the shaft1 is 6.87 mm\n",
"The Diameter of the shaft2 is 0.702590481015 mm\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.7 page number 193"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"T_ab = 0 #N.m - torsion in AB \n",
"T_bc = 150 #N.m - torsion in BC\n",
"T_cd = 150 #N.m - torsion in CD\n",
"T_de = 1150 #N.m - torsion in DE\n",
"l_ab = 250 #mm - length of AB\n",
"l_bc = 200 #mm - length of BC\n",
"l_cd = 300 #mm - length of cd \n",
"l_de = 500.0 #mm - length of de\n",
"d_1 = 25 #mm - outer diameter \n",
"d_2 = 50 #mm - inner diameter\n",
"G = 80 #Gpa -shear modulus\n",
"#Caliculations \n",
"\n",
"J_ab = 3.14*(d_1**4)/32 #mm4\n",
"J_bc = 3.14*(d_1**4)/32 #mm4\n",
"J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4\n",
"J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4\n",
"rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) # adding the maximum radians roteted in each module\n",
"print \"The maximum angle rotated is \",rad,\"radians \" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum angle rotated is 0.0232628450106 radians \n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.9 Pagenumber 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given \n",
"#its a statistally indeterminant \n",
"#we will take of one of the support \n",
"#Given \n",
"T_ab = 0 #N.m - torsion in AB \n",
"T_bc = 150 #N.m - torsion in BC\n",
"T_cd = 150 #N.m - torsion in CD\n",
"T_de = 1150 #N.m - torsion in DE\n",
"l_ab = 250 #mm - length of AB\n",
"l_bc = 200 #mm - length of BC\n",
"l_cd = 300 #mm - length of cd \n",
"l_de = 500.0#mm - length of de\n",
"d_1 = 25 #mm - outer diameter \n",
"d_2 = 50 #mm - inner diameter\n",
"#Caliculations \n",
"\n",
"J_ab = 3.14*(d_1**4)/32 #mm4\n",
"J_bc = 3.14*(d_1**4)/32 #mm4\n",
"J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4\n",
"J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4\n",
"G = 80 #Gpa -shear modulus\n",
"rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) \n",
"#now lets consider T_A then the torsion is only T_A\n",
"# T_A*(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) +rad = 0\n",
"# since there will be no displacement \n",
"T_A =-rad/(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) #Torsion at A\n",
"T_B = 1150 - T_A #n-m F_X = 0 torsion at B\n",
"print \"The Torsion at rigid end A is\",round(T_A,2),\"N-m\"\n",
"print \"The Torsion at rigid end B is\",round(T_B,2),\"N-m\"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Torsion at rigid end A is -141.72 N-m\n",
"The Torsion at rigid end B is 1291.72 N-m\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.12 Pagenumber 202"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"dai_bc = 240 #mm- daimeter of '8'bolt circle \n",
"dia = dai_bc/8 #Diameter of each bolt\n",
"A = 0.25*(dia**2)*3.14 # Area of a bolt\n",
"S_allow = 40 #Mpa - The maximum allowable allowable shear stress \n",
"P_max = (S_allow)*A #N - The maximum allowable force \n",
"D = 120.0 #mm - the distance from central axis \n",
"T_allow =P_max*D*8 #N-m The allowable torsion on the 8 bolt combination \n",
"print \"The allowable torsion on the 8 bolt combination\",T_allow ,\"N-m\"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The allowable torsion on the 8 bolt combination 27129600.0 N-m\n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.15 page number 211"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#AISC MANUALS\n",
"#approximated by three narrow tubes \n",
"#J = Bbt^3\n",
"B = 0.33 # constant mentiones in AISC\n",
"#three rods \n",
"\n",
"#rod_1\n",
"t_1 = 0.605 #inch - Thickness \n",
"b = 12.0 #inches - width \n",
"J_1 = B*b*(t_1**3) #in4 - Torsion constant \n",
"\n",
"#rod_2\n",
"t_2 = 0.605 #inch - Thickness \n",
"b = 12 #inches - width \n",
"J_2 = B*b*(t_2**3) #in4 - Torsion constant \n",
"\n",
"#rod_3\n",
"t_3 = 0.390 #inch - Thickness \n",
"b = 10.91 #inches - width \n",
"J_3 = B*b*(t_3**3) #in4 - Torsion constant \n",
"\n",
"#Equivalent\n",
"J_eq = J_1+J_2+J_3 #in4 - Torsion constant \n",
"print \"the Equivalent Torsion constant is \",round(J_eq,2), \"in4\"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the Equivalent Torsion constant is 1.97 in4\n"
]
}
],
"prompt_number": 57
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.16 page number 214"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"dia_out = 10 #mm- outer diameter of shaft\n",
"dia_in = 8 #mm- inner diameter of shaft \n",
"c_out = dia_out/2 #mm - outer Radius of shaft \n",
"c_in = dia_in/2 #mm - inner radius of shaft \n",
"T = 40 #N/mm -Torque in the shaft \n",
"#caliculations\n",
"\n",
"J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4\n",
"shear_T_max = T*c_out*pow(10,3)/J # The maximum torsion shear in the shaft\n",
"shear_T_min = T*c_in*pow(10,3)/J # The maximum torsion shear in the shaft\n",
"print \"The maximum shear due to torsion is \",round(shear_T_max,2),\"Mpa\"\n",
"print \"The minimum shear due to torsion is \",round(shear_T_min,2),\"Mpa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum shear due to torsion is 345.23 Mpa\n",
"The minimum shear due to torsion is 276.18 Mpa\n"
]
}
],
"prompt_number": 58
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}