{
"metadata": {
"name": "chapter no.6.ipynb"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter No.6:Torsion"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.1,Page No.225"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"L=10000 #mm #Length of solid shaft\n",
"d=100 #mm #Diameter of shaft\n",
"n=150 #rpm\n",
"P=112.5*10**6 #N-mm/sec #Power Transmitted\n",
"G=82*10**3 #N/mm**2 #modulus of Rigidity\n",
"\n",
"#Calculations\n",
"\n",
"J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n",
"T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
"\n",
"r=50 #mm #Radius\n",
"\n",
"q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",
"Theta=T*L*(G*J)**-1 #angle of twist\n",
"\n",
"#Result\n",
"print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",
"print\"Angle of Twist\",round(Theta,3),\"radian\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Max shear stress intensity 36.48 N/mm**2\n",
"Angle of Twist 0.089 radian\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.2,Page No.226"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"P=440*10**6 #N-m/sec #Power transmitted\n",
"n=280 #rpm\n",
"theta=pi*180**-1 #radian #angle of twist\n",
"L=1000 #mm #Length of solid shaft\n",
"q_s=40 #N/mm**2 #Max torsional shear stress\n",
"G=84*10**3 #N/mm**2 #Modulus of rigidity\n",
"\n",
"#Calculations\n",
"\n",
"#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n",
"T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",
"\n",
"#From Consideration of shear stress\n",
"d1=(T*16*(pi*40)**-1)**0.333333 \n",
"\n",
"#From Consideration of angle of twist\n",
"d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",
"\n",
"#result\n",
"print\"Diameter of solid shaft is\",round(d1,2),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of solid shaft is 124.09 mm\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.3,Page No.227"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
"q_s=80 #N/mm**2 #Max sheare stress\n",
"P=736*10**6 #N-mm/sec #Power transmitted\n",
"n=200\n",
"\n",
"#Calculations\n",
"\n",
"T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
"\n",
"#Now From consideration of angle of twist\n",
"theta=pi*180**-1\n",
"#L=15*d\n",
"\n",
"d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",
"\n",
"#Now corresponding stress at the surface is\n",
"q_s2=T*32*d*(pi*2*d**4)**-1\n",
"\n",
"#Result\n",
"print\"Max diameter required is\",round(d,2),\"mm\"\n",
"print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Max diameter required is 156.66 mm\n",
"Corresponding shear stress is 46.55 N/mm**2\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.4,Page No.228"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"d=25 #mm #Diameter of steel bar\n",
"p=50*10**3 #N #Pull\n",
"dell_1=0.095 #mm #Extension of bar\n",
"l=200 #mm #Guage Length\n",
"T=200*10**3 #N-mm #Torsional moment\n",
"theta=0.9*pi*180**-1 #angle of twist\n",
"L=250 #mm Length of steel bar\n",
"\n",
"#Calculations\n",
"\n",
"A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",
"E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",
"\n",
"J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",
"\n",
"G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",
"\n",
"#Now from the relation of Elastic constants\n",
"mu=E*(2*G)**-1-1\n",
"\n",
"#result\n",
"print\"The Poissoin's ratio is\",round(mu,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Poissoin's ratio is 0.292\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.5,Page No.229"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"L=6000 #mm #Length of circular shaft\n",
"d1=100 #mm #Outer Diameter\n",
"d2=75 #mm #Inner Diameter\n",
"R=100*2**-1 #Radius of shaft\n",
"T=10*10**6 #N-mm #Torsional moment\n",
"G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
"\n",
"#Calculations\n",
"\n",
"J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
"\n",
"#Max Shear stress produced\n",
"q_s=T*R*J**-1 #N/mm**2\n",
"\n",
"#Angle of twist\n",
"theta=T*L*(G*J)**-1 #Radian\n",
"\n",
"#Result\n",
"print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",
"print\"Angle of Twist is\",round(theta,2),\"Radian\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"MAx shear stress produced is 74.5 N/mm**2\n",
"Angle of Twist is 0.11 Radian\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.6,Page No.229"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"d1=200 #mm #External Diameter of shaft\n",
"t=25 #mm #Thickness of shaft\n",
"n=200 #rpm\n",
"theta=0.5*pi*180**-1 #Radian #angle of twist\n",
"L=2000 #mm #Length of shaft\n",
"G=84*10**3 #N/mm**2\n",
"d2=d1-2*t #mm #Internal Diameter of shaft\n",
"\n",
"#Calculations\n",
"\n",
"J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
"\n",
"#Torsional moment\n",
"T=G*J*theta*L**-1 #N/mm**2 \n",
"\n",
"#Power Transmitted\n",
"P=2*pi*n*T*60**-1*10**-6 #N-mm\n",
"\n",
"#Max shear stress transmitted\n",
"q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",
"\n",
"#Result\n",
"print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",
"print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Power Transmitted is 824.28 N-mm\n",
"Max Shear stress produced is 36.65 N/mm**2\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.7,Page No.230"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"P=3750*10**6 #N-mm/sec\n",
"n=240 #Rpm\n",
"q_s=160 #N/mm**2 #Max shear stress\n",
"\n",
"#Calculations\n",
"\n",
"#d2=0.8*d2 #mm #Internal Diameter of shaft\n",
"\n",
"#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",
"#After substituting value in above Equation we get\n",
"#J=0.05796*d1**4\n",
"\n",
"T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
"\n",
"#Now from Torsion Formula\n",
"#T*J**-1=q_s*R**-1 ......................................(1)\n",
"\n",
"#But R=d1*2**-1 \n",
"\n",
"#Now substituting value of R and J in Equation (1) we get\n",
"d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",
"\n",
"d2=d1*0.8\n",
"\n",
"#Result\n",
"print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",
"print\" :d2\",round(d2,3),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The size of the Shaft is:d1 200.362 mm\n",
" :d2 160.289 mm\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.8,Page No.231"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"P=245*10**6 #N-mm/sec #Power transmitted\n",
"n=240 #rpm\n",
"q_s=40 #N/mm**2 #Shear stress\n",
"theta=pi*180**-1 #radian #Angle of twist\n",
"L=1000 #mm #Length of shaft\n",
"G=80*10**3 #N/mm**2\n",
"\n",
"#Tmax=1.5*T\n",
"\n",
"#Calculations\n",
"\n",
"T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",
"Tmax=1.5*T\n",
"\n",
"#Now For Solid shaft\n",
"#J=pi*32*d**4\n",
"\n",
"#Now from the consideration of shear stress we get\n",
"#T*J**-1=q_s*(d*2**-1)**-1\n",
"#After substituting value in above Equation we get\n",
"#T=pi*16**-1*d**3*q_s\n",
"\n",
"#Designing For max Torque\n",
"d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",
"\n",
"#For max Angle of Twist\n",
"#Tmax*J**-1=G*theta*L**-1 \n",
"#After substituting value in above Equation we get\n",
"d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",
"\n",
"#For Hollow Shaft\n",
"\n",
"#d1_2=Outer Diameter\n",
"#d2_2=Inner Diameter\n",
"\n",
"#d2_2=0.5*d1_2\n",
"\n",
"# Polar modulus\n",
"#J=pi*32**-1*(d1_2**4-d2_2**4)\n",
"#After substituting values we get\n",
"#J=0.092038*d1_2**4\n",
"\n",
"#Now from the consideration of stress\n",
"#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",
"#After substituting values and further simplifying we get\n",
"d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",
"\n",
"#Now from the consideration of angle of twist\n",
"#Tmax*J**-1=G*theta*L**-1\n",
"#After substituting values and further simplifying we get\n",
"d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",
"\n",
"d2_2=0.5*d1_2\n",
"\n",
"#result\n",
"print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",
"print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",
"print\" : :d2_2\",round(d2_2,3),\"mm\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of shaft is:For solid shaft:d 123.01 mm\n",
" :For Hollow shaft:d1_2 125.69 mm\n",
" : :d2_2 62.845 mm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.11,Page No.235"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"P=250*10**6 #N-mm/sec #Power transmitted\n",
"n=100 #rpm\n",
"q_s=75 #N/mm**2 #Shear stress\n",
"\n",
"#Calculations\n",
"\n",
"#From Equation of Power we have\n",
"T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
"\n",
"#Now from torsional moment equation we have\n",
"#T=j*q_s*(d/2**-1)**-1\n",
"#After substituting values in above equation and further simplifying we get\n",
"#T=pi*16**-1**d**3*q_s\n",
"d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",
"\n",
"#PArt-2\n",
"\n",
"#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",
"#d2=0.6*d1\n",
"\n",
"#Again from torsional moment equation we have\n",
"#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",
"d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",
"d2=0.6*d1\n",
"\n",
"#Cross sectional area of solid shaft\n",
"A1=pi*4**-1*d**2 #mm**2\n",
"\n",
"#cross sectional area of hollow shaft\n",
"A2=pi*4**-1*(d1**2-d2**2)\n",
"\n",
"#Now percentage saving in weight\n",
"#Let W be the percentage saving in weight\n",
"W=(A1-A2)*100*A1**-1\n",
"\n",
"#Result\n",
"print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",
"print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",
"print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n",
"print\" : :d2\",round(d2,3),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage saving in Weight is 29.735 %\n",
"Size of shaft is:solid shaft:d 117.418 mm\n",
" :Hollow shaft:d1 123.031 mm\n",
" : :d2 73.818 mm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.12,Page No.237"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"d=100 #mm #Diameter of solid shaft\n",
"d1=100 #mm #Outer Diameter of hollow shaft\n",
"d2=50 #mm #Inner Diameter of hollow shaft\n",
"\n",
"#Calculations\n",
"\n",
"#Torsional moment of solid shaft\n",
"#T_s=J*q_s*(d*2**-1)**-1 \n",
"#After substituting values in above equation and further simplifying we get\n",
"#T_s=pi*16*d**3*q_s ...............(1)\n",
"\n",
"#torsional moment for hollow shaft is\n",
"#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",
"#After substituting values in above equation and further simplifying we get\n",
"#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n",
"\n",
"#Dividing Equation 2 by 1 we get\n",
"#Let the ratio of T_h*T_s**-1 Be X\n",
"X=1-0.5**4\n",
"\n",
"#Loss in strength \n",
"#Let s be the loss in strength\n",
"#s=T_s*T_h*100*T_s**-1\n",
"#After substituting values in above equation and further simplifying we get\n",
"s=(1-0.9375)*100\n",
"\n",
"#Weight Ratio \n",
"#Let w be the Weight ratio\n",
"#w=W_h*W_s**-1\n",
"\n",
"A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",
"A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",
"\n",
"w=A_h*A_s**-1 \n",
"\n",
"#Result\n",
"print\"Loss in strength is\",round(s,2)\n",
"print\"Weight ratio is\",round(w,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Loss in strength is 6.25\n",
"Weight ratio is 0.75\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.13,Page No.239"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"T=8 #KN-m #Torque \n",
"d=100 #mm #Diameter of portion AB\n",
"d1=100 #mm #External Diameter of Portion BC\n",
"d2=75 #mm #Internal Diameter of Portion BC\n",
"G=80 #KN/mm**2 #Modulus of Rigidity\n",
"L1=1500 #mm #Radial Distance of Portion AB\n",
"L2=2500 #mm #Radial Distance ofPortion BC\n",
"\n",
"#Calculations\n",
"\n",
"R=d*2**-1 #mm #Radius of shaft\n",
"\n",
"#For Portion AB,Polar Modulus\n",
"J1=pi*32**-1*d**4 #mm**4 \n",
"\n",
"#For Portion BC,Polar modulus \n",
"J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
"\n",
"#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",
"q_max=T*J2**-1*R*10**6 #N/mm**2 \n",
"\n",
"#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",
"theta1=T*L1*(G*J1)**-1 #Radians\n",
"theta2=T*L2*(G*J2)**-1 #Radians\n",
"\n",
"#Total Rotational at end C\n",
"theta=(theta1+theta2)*10**3 #Radians\n",
"\n",
"#Result\n",
"print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",
"print\"Angle of Twist is\",round(theta,3),\"radians\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Max stress induced is 59.6 N/mm**2\n",
"Angle of Twist is 0.053 radians\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.14,Page No.240"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"q_b=80 #N/mm**2 #Shear stress in Brass\n",
"q_s=100 #N/mm**2 #Shear stress in Steel\n",
"G_b=40*10**3 #N/mm**2 \n",
"G_s=80*10**3 \n",
"L_b=1000 #mm #Length of brass shaft\n",
"L_s=1200 #mm #Length of steel shaft\n",
"d1=80 #mm #Diameter of brass shaft\n",
"d2=60 #mm #Diameter of steel shaft\n",
"\n",
"#Calculations\n",
"\n",
"#Polar modulus of brass rod\n",
"J_b=pi*32**-1*d1**4 #mm**4 \n",
"\n",
"#Polar modulus of steel rod\n",
"J_s=pi*32**-1*d2**4 #mm**4\n",
"\n",
"#Considering bras Rod:AB\n",
"T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",
"\n",
"#Considering Steel Rod:BC\n",
"T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",
"\n",
"#Max Torque that can be applied\n",
"T2\n",
"\n",
"#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",
"theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",
"theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",
"\n",
"theta=theta_b+theta_s #Radians #Rotation of free end\n",
"\n",
"#Result\n",
"print\"Total of free end is\",round(theta,3),\"Radians\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total of free end is 0.076 Radians\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.15,Page No.241"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
"d1=100 #mm #Outer diameter of hollow shft\n",
"d2=80 #mm #Inner diameter of hollow shaft\n",
"d=80 #mm #diameter of Solid shaft\n",
"d3=60 #mm #diameter of Solid shaft having L=0.5m\n",
"L1=300 #mm #Length of Hollow shaft\n",
"L2=400 #mm #Length of solid shaft\n",
"L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",
"T1=2*10**6 #N-mm #Torsion in Shaft AB\n",
"T2=1*10**6 #N-mm #Torsion in shaft BC\n",
"T3=1*10**6 #N-mm #Torsion in shaft CD\n",
"\n",
"#Calculations\n",
"\n",
"#Now Polar modulus of section AB\n",
"J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
"\n",
"#Polar modulus of section BC\n",
"J2=pi*32**-1*d**4 #mm**4\n",
"\n",
"#Polar modulus of section CD\n",
"J3=pi*32**-1*d3**4 #mm**4\n",
"\n",
"#Now angle of twist of AB\n",
"theta1=T1*L1*(G*J1)**-1 #radians\n",
"\n",
"#Angle of twist of BC\n",
"theta2=T2*L2*(G*J2)**-1 #radians\n",
"\n",
"#Angle of twist of CD\n",
"theta3=T3*L3*(G*J3)**-1 #radians\n",
"\n",
"#Angle of twist\n",
"theta=theta1-theta2+theta3 #Radians\n",
"\n",
"#Shear stress in AB From Torsion Equation\n",
"q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",
"\n",
"#Shear stress in BC\n",
"q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",
"\n",
"#Shear stress in CD\n",
"q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",
"\n",
"#As max shear stress occurs in portion CD,so consider CD\n",
"\n",
"#Result\n",
"print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",
"print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Angle of twist at free end is 0.00496 Radian\n",
"Max Shear stress 23.58 N/mm**2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.16,Page No.242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"L=1000 #mm #Length of bar\n",
"L1=600 #mm #Length of Bar AB\n",
"L2=400 #mm #Length of Bar BC\n",
"d1=60 #mm #Outer Diameter of bar BC\n",
"d2=30 #mm #Inner Diameter of bar BC\n",
"d=60 #mm #Diameter of bar AB\n",
"T=2*10**6 #N-mm #Total Torque\n",
"\n",
"#Calculations\n",
"\n",
"#Polar Modulus of Portion AB\n",
"J1=pi*32**-1*d**4 #mm*4\n",
"\n",
"#Polar Modulus of Portion BC\n",
"J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
"\n",
"#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",
"#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",
"\n",
"#theta1=T1*L1*(G*J1)**-1 #radians\n",
"#After substituting values and further simplifying we get \n",
"#theta1=32*600*T1*(pi*60**4*G)**-1\n",
"\n",
"#theta2=T2*L*(J2*G)**-1 #Radians\n",
"#After substituting values and further simplifying we get \n",
"#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",
"\n",
"#Now For consistency of Deformation,theta1=theta2\n",
"#After substituting values and further simplifying we get \n",
"#T1=0.7111*T2 ..................................................(1)\n",
"\n",
"#But T1+T2=T=2*10**6 ...........................................(2)\n",
"#Substituting value of T1 in above equation\n",
"\n",
"T2=T*(0.7111+1)**-1\n",
"T1=0.71111*T2\n",
"\n",
"#Max stress in Portion AB\n",
"q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",
"\n",
"#Max stress in Portion BC\n",
"q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",
"\n",
"#Result\n",
"print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",
"print\" :BC\",round(q_s2,2),\"N/mm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Stresses Developed in Portion:AB 19.6 N/mm**2\n",
" :BC 29.4 N/mm**2\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.17,Page No.243"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"d1=80 #mm #External Diameter of Brass tube\n",
"d2=50 #mm #Internal Diameter of Brass tube\n",
"d=50 #mm #Diameter of steel Tube\n",
"G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",
"G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",
"T=6*10**6 #N-mm #Torque\n",
"L=2000 #mm #Length of Tube\n",
"\n",
"#Calculations\n",
"\n",
"#Polar Modulus of brass tube\n",
"J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
"\n",
"#Polar modulus of steel Tube\n",
"J2=pi*32**-1*d**4 #mm**4\n",
"\n",
"#Let T_s & T_b be the torque resisted by steel and brass respectively\n",
"#Then, T_b+T_s=T ............................................(1)\n",
"\n",
"#Since the angle of twist will be the same\n",
"#Theta1=Theta2\n",
"#After substituting values and further simplifying we get \n",
"#Ts=0.360*Tb ...........................................(2)\n",
"\n",
"#After substituting value of Ts in eqn 1 and further simplifying we get \n",
"T_b=T*(0.36+1)**-1 #N-mm\n",
"T_s=0.360*T_b\n",
"\n",
"#Let q_s and q_b be the max stress in steel and brass respectively\n",
"q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",
"q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",
"\n",
"#Since angle of twist in brass=angle of twist in steel\n",
"theta_s=T_s*L*(J2*G_s)**-1\n",
"\n",
"#Result\n",
"print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",
"print\" :Steel\",round(q_s,2),\"N/mm**2\"\n",
"print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",
" :Steel 64.71 N/mm**2\n",
"Angle of Twist in 2m Length 0.065 Radians\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.18,Page No.245"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"d1=60 #mm #External Diameter of aluminium Tube\n",
"d2=40 #mm #Internal Diameter of aluminium Tube\n",
"d=40 #mm #Diameter of steel tube\n",
"q_a=60 #N/mm**2 #Permissible stress in aluminium\n",
"q_s=100 #N/mm**2 #Permissible stress in steel tube\n",
"G_a=27*10**3 #N/mm**2 \n",
"G_s=80*10**3 #N/mm**2 \n",
"\n",
"#Calculations\n",
"\n",
"#Polar modulus of aluminium Tube\n",
"J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",
"\n",
"#Polar Modulus of steel Tube\n",
"J_s=pi*32**-1*d**4 #mm**4\n",
"\n",
"#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",
"#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",
"#After substituting values in above Equation and Further simplifyin we get\n",
"#T_s=0.7293*T_a .....................(1)\n",
"\n",
"#If steel Governs the resisting capacity\n",
"T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",
"T_a1=T_s1*0.7293**-1 #N-mm\n",
"T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",
"\n",
"#If aluminium Governs the resisting capacity \n",
"T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",
"T_s2=T_a2*0.7293 #N-mm\n",
"T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",
"\n",
"#Result\n",
"print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Steel Governs the torque carrying capacity 2.98 KN-m\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.19,Page No.247"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"P=225*10**6 #N-mm/sec #Power Trasmitted\n",
"q_b=80 #N/mm**2 #Shear stress\n",
"n=200 #Rpm\n",
"q_k=100 #N/mm**2 #PErmissible stress in Keys\n",
"D=300 #mm #Diameter of bolt circle\n",
"L=150 #mm #Length of shear key\n",
"d=16 #mm #Diameterr of bolt\n",
"\n",
"#Calculations\n",
"T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",
"\n",
"#Now From Torsion Formula\n",
"#T*J**-1=q_s*R**-1\n",
"#After substituting values we get\n",
"#T=pi*16*d**3*n\n",
"#After further simplifying we get\n",
"d1=(T*16*(pi*q_s)**-1)**0.33333\n",
"\n",
"#Let b be the width of shear Key\n",
"#T=q_k*L*b*R\n",
"#After simplifying further we get\n",
"b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",
"\n",
"#Let n2 be the no. of bolts required at bolt circle of radius\n",
"R_b=D*2**-1 #mm \n",
"\n",
"n2=T*4*(q_b*pi*d**2*R_b)**-1\n",
"\n",
"#result\n",
"print\"Minimum no. of Bolts Required are\",round(n2,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum no. of Bolts Required are 4.45\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.20,Page No.250"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"T=2*10**6 #N-mm #Torque transmitted\n",
"G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
"d1=40 #mm \n",
"d2=80 #mm\n",
"r1=20 #mm\n",
"r2=40 #mm\n",
"L=2000 #mm #Length of shaft\n",
"\n",
"#Calculations\n",
"\n",
"#Angle of twist \n",
"theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",
"\n",
"#If the shaft is treated as shaft of average Diameter\n",
"d_avg=(d1+d2)*2**-1 #mm\n",
"\n",
"theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",
"\n",
"#Percentage Error\n",
"#Let Percentage Error be E\n",
"X=theta-theta1\n",
"E=(X*theta**-1)*100 \n",
"\n",
"#Result\n",
"print\"Percentage Error is\",round(E,2),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage Error is 32.28 %\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.21,Page No.252"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"G=80*10**3 #N/mm**2 \n",
"P=1*10**9 #N-mm/sec #Power\n",
"n=300 \n",
"d1=150 #mm #Outer Diameter\n",
"d2=120 #mm #Inner Diameter\n",
"L=2000 #mm #Length of circular shaft\n",
"\n",
"#Calculations\n",
"\n",
"T=P*60*(2*pi*n)**-1 #N-mm\n",
"\n",
"#Polar Modulus \n",
"J=pi*32**-1*(d1**4-d2**4) #mm**4\n",
"\n",
"q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",
"\n",
"\n",
"#Strain ENergy\n",
"U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",
"\n",
"#Result\n",
"print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
"print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Max shear stress is 81.36 N/mm**2\n",
"Strain Energy stored in the shaft is 263181.37 N-mm\n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.22,Page No.254"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"d=12 #mm #Diameter of helical spring\n",
"D=150 #mm #Mean Diameter\n",
"R=D*2**-1 #mm #Radius of helical spring\n",
"n=10 #no.of turns\n",
"G=80*10**3 #N/mm**2 \n",
"W=450 #N #Load\n",
"\n",
"#Calculations\n",
"\n",
"#Max shear stress \n",
"q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",
"\n",
"#Strain Energy stored\n",
"U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",
"\n",
"#Deflection Produced\n",
"dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",
"\n",
"#Stiffness Spring\n",
"k=W*dell**-1 #N/mm\n",
"\n",
"#Result\n",
"print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
"print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",
"print\"Deflection Produced is\",round(dell,2),\"mm\"\n",
"print\"Stiffness spring is\",round(k,2),\"N/mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Max shear stress is 99.47 N/mm**2\n",
"Strain Energy stored is 16479.49 N-mm\n",
"Deflection Produced is 73.24 mm\n",
"Stiffness spring is 6.14 N/mm\n"
]
}
],
"prompt_number": 53
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.23,Page No.255"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"K=5 #N/mm #Stiffness\n",
"L=100 #mm #Solid Length\n",
"q_s=60 #N/mm**2 #Max shear stress\n",
"W=200 #N #Max Load\n",
"G=80*10**3 #N/mm**2\n",
"\n",
"#Calculations\n",
"\n",
"#K=W*dell**-1\n",
"#After substituting values and further simplifying we get\n",
"#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n",
"#n=L*d**-1 ........(2)\n",
"\n",
"#From Shearing stress\n",
"#q_s=16*W*R*(pi*d**3)**-1 \n",
"#After substituting values and further simplifying we get\n",
"#d**4=0.004*R**3*n .................(4)\n",
"\n",
"#From Equation 1,2,3\n",
"#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",
"#after further simplifying we get\n",
"d=5168.101**0.25\n",
"n=100*d**-1\n",
"R=(d**4*(0.004*n)**-1)**0.3333\n",
"\n",
"#Result\n",
"print\"Diameter of Wire is\",round(d,2),\"mm\"\n",
"print\"No.of turns is\",round(n,2)\n",
"print\"Mean Radius of spring is\",round(R,2),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of Wire is 8.48 mm\n",
"No.fo turns is 11.79\n",
"Mean Radius of spring is 47.83 mm\n"
]
}
],
"prompt_number": 54
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.24,Page No.255"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"m=5*10**5 #Wagon Weighing\n",
"v=18*1000*36000**-1 \n",
"d=300 #mm #Diameter of Beffer springs\n",
"n=18 #no.of turns\n",
"G=80*10**3 #N/mm**2\n",
"dell=225\n",
"R=100 #mm #Mean Radius\n",
"\n",
"#Calculations\n",
"\n",
"#Energy of Wagon\n",
"E=m*v**2*(9.81*2)**-1 #N-mm\n",
"\n",
"#Load applied\n",
"W=dell*G*d**4*(64*R**3*n)**-1 #N \n",
"\n",
"#Energy each spring can absorb is\n",
"E2=W*dell*2**-1 #N-mm\n",
"\n",
"#No.of springs required to absorb energy of Wagon\n",
"n2=E*E2**-1 *10**7\n",
"\n",
"#Result\n",
"print\"No.of springs Required for Buffer is\",round(n2,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"No.of springs Required for Buffer is 4.47\n"
]
}
],
"prompt_number": 66
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.25,Page No.259"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"b=180 #mm #width of flange\n",
"d=10 #mm #Depth of flange\n",
"t=10 #mm #Thickness of flange\n",
"D=400 #mm #Overall Depth \n",
"\n",
"#Calculations\n",
"\n",
"I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",
"I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",
"\n",
"#If warping is neglected\n",
"J=I_xx+I_yy #mm**4\n",
"\n",
"#Since b/d>1.6,we get\n",
"J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",
"\n",
"#Over Estimation of torsional Rigidity would have been \n",
"T=J*J2**-1\n",
"\n",
"#Result\n",
"print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"
]
}
],
"prompt_number": 68
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.6.26,Page No.261"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Initilization of Variables\n",
"\n",
"d1=100 #mm #Outer Diameter\n",
"d2=95 #mm #Inner Diameter\n",
"T=2*10**6 #N-mm #Torque\n",
"\n",
"#Calculations\n",
"\n",
"J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",
"\n",
"#Shear stress\n",
"q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",
"\n",
"#Now theta*L**-1=T*(G*J)**-1\n",
"#After substituting values and further simplifying we get\n",
"#Let theta*L**-1=X\n",
"X=T*J**-1\n",
"\n",
"#Now Treating it as very thin walled tube\n",
"d=(d1+d2)*2**-1 #mm\n",
"\n",
"r=d*2**-1 \n",
"t=(d1-d2)*2**-1\n",
"q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",
"\n",
"X2=T*(2*pi*r**3*t)**-1 \n",
"\n",
"#Result\n",
"print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",
"print\" :Angle of Twist per unit Length\",round(X,3)\n",
"print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",
"print\" :Angle of twist per Unit Length\",round(X2,3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",
" :Angle of Twist per unit Length 1.098\n",
"When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n",
" :Angle of twist per Unit Length 1.099\n"
]
}
],
"prompt_number": 72
}
],
"metadata": {}
}
]
}