{
"metadata": {
"name": "",
"signature": "sha256:54c92a13ce853ac5fe5325a355c839b097f01e81340aa957e3602b8ea43de184"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter8-Axisymmetrically Loaded Members"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-pg239"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate pressure and for longitudinal stress \n",
"di=0.3 ##m\n",
"de=0.4 ##m\n",
"v=0.3\n",
"sigmathetamax=250*10**6 ##Pa\n",
"p0=0.\n",
"pi=0.\n",
"\n",
"##solution a:\n",
"a=0.15\n",
"b=0.2\n",
"r=a\n",
"##sigmathetamax=pi*((b**2+a**2)/(b**2-a**2))\n",
"pi=sigmathetamax*((b**2-a**2.)/(b**2+a**2.))\n",
"print'%s %.2f %s'%(\"in Pa is= \",pi,\"\")\n",
"\n",
"##solution b:\n",
"r=a\n",
"##sigmathetamax=-2*p0*(b**2/(b**2-a**2))\n",
"p0=-(-sigmathetamax)*((b**2.-a**2.)/(2.*b**2.))\n",
"print'%s %.2f %s'%(\"in Pa is= \",p0,\"\")\n",
"\n",
"##solution c:\n",
"u=((a**3*pi)/(b**2-a**2))*(0.7+1.3*(b**2./a**2.))\n",
"print'%s %.2f %s'%(\"in per E meter is= \",u,\"\")\n",
"sigmaz=(pi*a**2-p0*b**2)/(b**2-a**2)\n",
"print'%s %.2f %s'%(\"for longitudinal stress is\",sigmaz,\"\")\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"in Pa is= 70000000.00 \n",
"in Pa is= 54687500.00 \n",
"in per E meter is= 40650000.00 \n",
"for longitudinal stress is -35000000.00 \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg241"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate pressure \n",
"sigmayp=340. ##MPa\n",
"tauyp=sigmayp/2. ##MPa\n",
"print'%s %.2f %s'%(\"in MPa is=\",tauyp,\"\")\n",
"a=0.1 ##m\n",
"b=0.15 ##m\n",
"v=0.3 \n",
"##pi=4*p0\n",
"##sigmatheta=(pi*(a**2+b**2)-2*p0*b**2)/(b**2-a**2)\n",
"##sigmatheta=1.7*pi\n",
"\n",
"##sloution a: maxi principal stress theory\n",
"sigmatheta=1.7\n",
"pi=sigmayp/sigmatheta\n",
"print'%s %.2f %s'%(\"in MPa is= \",pi,\"\")\n",
"\n",
"##sloution b: maxi shearing stress theory\n",
"##(sigmatheta-sigmar)/2=1.35*pi\n",
"pi=tauyp/1.35\n",
"print'%s %.2f %s'%(\"in MPa is= \",pi,\"\")\n",
"\n",
"##solution c: energy of distortion theory\n",
"sigmar=-1\n",
"sigmayp1=math.sqrt(sigmatheta**2+sigmar**2-sigmatheta*sigmar)##*pi\n",
"print(sigmayp1)\n",
"pi=sigmayp/sigmayp1\n",
"print'%s %.2f %s'%(\"in MPa is=\",pi,\"\")\n",
"\n",
"##solution d: maxi principal strain theory\n",
"##(sigmatheta-v*sigmar)/E=sigmayp/E\n",
"pi=sigmayp/(sigmatheta-v*sigmar)\n",
"print'%s %.2f %s'%(\"in MPa is= \",pi,\"\")\n",
"\n",
"##solution e: octahedral shearing stress theory:\n",
"pi=(math.sqrt(2.)*sigmayp)/math.sqrt((sigmatheta-sigmar)**2.+sigmar**2.+(-sigmatheta)**2)\n",
"print'%s %.2f %s'%(\"in MPa is= \",pi,\"\")\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"in MPa is= 170.00 \n",
"in MPa is= 200.00 \n",
"in MPa is= 125.93 \n",
"2.36431808351\n",
"in MPa is= 143.80 \n",
"in MPa is= 170.00 \n",
"in MPa is= 143.80 \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg243"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate contact pressure and tangential stresses in the outer cylinder\n",
"a=0.15 ##m\n",
"b=0.2 ##m\n",
"c=0.25 ##m\n",
"E=200*10**9. ##Pa\n",
"delta=0.0001 ##m\n",
"140 ##MPa\n",
"\n",
"p=((E*delta)/8.)*(((b**2.-a**2.)*(c**2.-b**2.))/(2.*(b**2.)*(c**2.-a**2.)))\n",
"print'%s %.2f %s'%(\"the contact pressure in Pa is= \",p,\"\") ## textbook ans is wrong\n",
"\n",
"p=12.3*10**6\n",
"sigmatheta=p*((b**2+c**2.)/(c**2.-b**2.)) ## where r=0.2\n",
"print'%s %.2f %s'%(\"tangential stresses in the outer cylinder in Pa is= \",sigmatheta,\"\")\n",
"sigmatheta1=(2*p*b**2)/(c**2-b**2) ## where r=0.25\n",
"print'%s %.2f %s'%(\"tangential stresses in the outer cylinder in Pa is= \",sigmatheta1,\"\")\n",
"sigmatheta3=-(2*p*b**2)/(b**2-a**2) ## where r=0.15\n",
"print'%s %.2f %s'%(\"tangential stresses in the inner cylinder in Pa is= \",sigmatheta3,\"\")\n",
"sigmatheta4=-p*((b**2.+a**2.)/(b**2.-a**2.)) ## where r=0.2\n",
"print'%s %.2f %s'%(\"tangential stresses in the inner cylinder in Pa is= \",sigmatheta4,\"\")\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the contact pressure in Pa is= 307617.19 \n",
"tangential stresses in the outer cylinder in Pa is= 56033333.33 \n",
"tangential stresses in the outer cylinder in Pa is= 43733333.33 \n",
"tangential stresses in the inner cylinder in Pa is= -56228571.43 \n",
"tangential stresses in the inner cylinder in Pa is= -43928571.43 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4-pg246"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate radial displacement of disk and shaft\n",
"dn=0.1 ##m\n",
"do=0.5 ##m\n",
"t=0.08 ##m\n",
"w=6900*(2*math.pi/60.) ##rpm\n",
"row=7.8*10**3##Ns**2/m**4\n",
"E=200*10**9 ##Pa\n",
"v=0.3\n",
"b=0.05\n",
"c=0.25\n",
"\n",
"\n",
"##solution a:\n",
"##ud=((0.05*3.3*0.7)*(0.0025+0.0625-(1.3/3.3)*0.0025+(1.3/0.7)*0.0625)*row*w**2)/(8*E)\n",
"ud=((0.05*3.3*0.7)*(b**2+c**2-(1.3/3.3)*b**2+(1.3/0.7)*c**2))/(8)\n",
"print'%s %.4f %s'%(\"radial displacement of the disk in meter is= \",ud,\"\")\n",
"\n",
"##us=((0.05*0.7)*(3.3*0.0025-1.3*0.0025)*row*w**2)/(8*E)\n",
"us=((0.05*0.7)*(3.3*b**2-1.3*b**2))/(8)\n",
"print'%s %.6f %s'%(\"radial displacement of the shaft in meter is= \",us,\"\")\n",
"delta=(ud-us)*row*w**2./E\n",
"print(delta)\n",
"\n",
"##solution b:\n",
"##p=E*delta*(c**2-b**2)/(2*b*c**2)\n",
"p=E*delta*(c**2-b**2)/(2*b*c**2)\n",
"print'%s %.2f %s'%(\"in Pa is= \",p,\"\")\n",
"sigmathetamax=p*(c**2+b**2)/(c**2-b**2)\n",
"print'%s %.2f %s'%(\"in Pa is= \",sigmathetamax,\"\")\n",
"\n",
"##solution c:\n",
"sigmathetamax=3.3*(b**2.+c**2.-(1.9/3.3)*b**2.+c**2.)*row*w**2./8.\n",
"print'%s %.2f %s'%(\"in Pa is= \",sigmathetamax,\"\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"radial displacement of the disk in meter is= 0.0026 \n",
"radial displacement of the shaft in meter is= 0.000022 \n",
"5.24957318528e-05\n",
"in Pa is= 100791805.16 \n",
"in Pa is= 109191122.25 \n",
"in Pa is= 211764600.73 \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg250"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate sigma theta and sigma \n",
"ti=0.075 ##m\n",
"to=0.015##m\n",
"a=0.05##m\n",
"b=0.25##m\n",
"delta=0.05 ##mm\n",
"w=6900*(2*math.pi/60.) ##rpm\n",
"s=1.\n",
"row=7.8*10**3##Ns**2/m**4\n",
"E=200. ##GPa\n",
"\n",
"##solution a:\n",
"t1=ti*a**s\n",
"print'%s %.4f %s'%(\"t1 is=\",t1,\"\")\n",
"t1=to*b**2.\n",
"print'%s %.4f %s'%(\"t1 is=\",t1,\"\")\n",
"##(ti/to)=(t1*a**-s)/(t1*b**-s)=(b/a)**s\n",
"c=(b/a)**s\n",
"\n",
"(ti/to)==c\n",
"print'%s %.2f %s'%(\"ti/t0 is=\",c,\"\")\n",
"m1=-0.5+math.sqrt((0.5)**2+(1+0.3*1))\n",
"print'%s %.2f %s'%(\"m1 is=\",m1,\"\")\n",
"m2=-0.5-math.sqrt((0.5)**2.+(1.+0.3*1.))\n",
"print'%s %.2f %s'%(\"m2 is=\",m2,\"\")\n",
"\n",
"##sigmar=0=(c1/t1)*(0.05)**m1+(c2/t1)*(0.05)**(m2)-0.00176*row*w**2 ## r=0.05\n",
"##sigmar=0=(c1/t1)*(0.25)**m1+(c2/t1)*(0.25)**(m2)-0.0439*row*w**2 ## r=0.25\n",
"\n",
"c1=t1*0.12529*row*w**2.\n",
"print'%s %.2f %s'%(\"c1 is=\",c1,\"\")\n",
"c2=t1*-6.272*10**-5*row*w**2\n",
"print'%s %.2f %s'%(\"c2 is=\",c2,\"\")\n",
"\n",
"r=0.05\n",
"sigmar=(0.12529*r**0.745-6.272*10**-5*r**(-1.745)-0.70*r**2)##*row*w**2\n",
"print'%s %.5f %s'%(\"sigmar is= \",sigmar,\"\")\n",
"\n",
"sigmatheta=(0.09334*r**0.745+1.095*10**-4*r**(-1.745)-0.40*r**2)##*row*w**2\n",
"print'%s %.2f %s'%(\"sigmatheta is= \",sigmatheta,\"\")\n",
"\n",
"##solution b:\n",
"r=0.05\n",
"##ur=(r*sigmatheta)/E\n",
"ur=(r*sigmatheta)\n",
"print'%s %.7f %s'%(\"ur is= \",ur,\"\")\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"t1 is= 0.0037 \n",
"t1 is= 0.0009 \n",
"ti/t0 is= 5.00 \n",
"m1 is= 0.74 \n",
"m2 is= -1.74 \n",
"c1 is= 478341.11 \n",
"c2 is= -239.46 \n",
"sigmar is= 0.00001 \n",
"sigmatheta is= 0.03 \n",
"ur is= 0.0014711 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6-pg251"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate pressure\n",
"b=0.25 ##m\n",
"w=6900.*(2*math.pi/60.) ##rpm\n",
"t1=0.075 ##m\n",
"t2=0.015 ##m\n",
"row=7.8*10**3##Ns**2/m**4\n",
"c1=t1\n",
"\n",
"x=t2/t1\n",
"print(x)\n",
"\n",
"##(t2/t1)==(c1*exp(-(row*w**2/2*sigma)*b**2))/c1\n",
"##exp(-(row*w**2/2*sigma)*b**2)=x\n",
"##log(x)=-(row*w**2*b**2/2*sigma)\n",
"y=2.*math.log(x)\n",
"print(y)\n",
"sigma=-(row*w**2.*b**2.)/y\n",
"print'%s %.2f %s'%(\"in Pa is= \",sigma,\"\")\n",
"\n",
"##t=c1*exp(-row*(w**2/2*sigma)*r**2)\n",
"z=row*(w**2./(2.*sigma))\n",
"print(z)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"0.2\n",
"-3.21887582487\n",
"in Pa is= 79072562.70 \n",
"25.7510065989\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}