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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 18: Torsion of Beams"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 18.1 Pg.No.527"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#variable declaration\n",
"r=100 #radius of cylinder (mm)\n",
"l=2*10**3 #length of cylinder (mm)\n",
"torque=30*10**6 #torque at mid point (N.mm)\n",
"Tmax=15*10**6 \n",
"Tau_max=200 #max shear stress (N/mm^2)\n",
"G=25000 #shear modulus (N/mm^2)\n",
"theta =2 # rotation at mid point (degree)\n",
"\n",
"#ref equation 18.1 T=2Aq\n",
"tmin1=Tmax/(2*math.pi*r**2*Tau_max)\n",
"print \"\\nminimum thickness of beam due to stress limited to 200N/mm^2 = %3.1f mm\"%(tmin1)\n",
"\n",
"z=1*10**3 # twist at mid point (mm)\n",
"#equation 18.4 after integration\n",
"tmin2=Tmax*200*math.pi*z/(4*(math.pi*r**2)**2*G*theta*math.pi/180)\n",
"print \"\\nminimum thickness due to constraint on maximum angle twist = %3.1f mm\"%(tmin2)\n",
"t=max(tmin1,tmin2)\n",
"print \"\\nthickness which satisfies both conditions = %3.1f mm\"%(t)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"minimum thickness of beam due to stress limited to 200N/mm^2 = 1.2 mm\n",
"\n",
"minimum thickness due to constraint on maximum angle twist = 2.7 mm\n",
"\n",
"thickness which satisfies both conditions = 2.7 mm\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 18.3 Pg.No 540"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#reference Fig 18.12\n",
"from __future__ import division\n",
"import math\n",
"from sympy import symbols\n",
"from sympy import integrate\n",
"\n",
"S1,S2,S3=symbols('S1,S2,S3')\n",
"\n",
"#variable declaration\n",
"s1=50 #length of channel (mm)\n",
"s2=25 #width of channel (mm)\n",
"t1=2.5 #thickness of web(mm)\n",
"t2=1.5 #thickness of flange(mm)\n",
"T=10*10**3 #torque applied (N.mm)\n",
"zeta_s=8.04 \n",
"G=25000 #shear modulus (N/mm^2)\n",
"\n",
"#J=Sum(st^3/3)\n",
"J=1/3*(s1*t1**3+2*s2*t2**3)\n",
"tau_max=t1*T/J\n",
"print \"maximum shear stress = %3.1f N/mm^2\"%(tau_max)\n",
"\n",
"#in region O2\n",
"s1=25\n",
"ARf=1/2*zeta_s*s1\n",
"ws=-2*ARf*T/(G*J)\n",
"print \"warping in O2 region is linear = %3.2f mm\"%(ws)\n",
"\n",
"#in the wall 21\n",
"AR=1/2*zeta_s*s2-0.5*25*s2\n",
"w21=-0.03*(zeta_s-s2)\n",
"s1=50\n",
"tds=2*s2*t2+s1*t1\n",
"\n",
"A12=25/2*S1\n",
"A23=312.5-4.02*S2\n",
"A34=111.5+25/2*S3\n",
"AR1=1/2*(integrate(A12,(S1,0,25))+integrate(5*(A23),(S2,0,50))+integrate(3*A34,(S3,0,25)))/200\n",
"\n",
"#equation 18.20\n",
"ws=-2*ARf*T/G/J\n",
"print \"warping distribution in flange 34 = %1.2f \\n\"%(ws)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum shear stress = 78.9 N/mm^2\n",
"warping in O2 region is linear = -0.25 mm\n",
"warping distribution in flange 34 = -0.25 \n",
"\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
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}
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