{
"metadata": {
"name": ""
},
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8:Combined Loadings"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.1 Page No 408"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"di=4*12 #inch, diameter\n",
"ri=di/2.0 #Radius\n",
"t=0.5 #inch, thickness\n",
"sigma=20.0 #Ksi, stress\n",
"\n",
"#Calculation\n",
"#Cylindrical Pressure Vessel\n",
"p1=(t*sigma)/ri #sigma = pr/t\n",
"#Spherical Vessel\n",
"p2=(2*t*sigma)/(ri) #sigma = pr/2t\n",
"\n",
"#Display\n",
"print\"The maximum internal pressure the cylindrical pressure vessel can sustainis\",round(p1*1000,0),\"psi\"\n",
"print\"The maximum internal pressure a spherical pressure vessel can sustain is\",round(p2*1000,0),\"psi\"\n",
"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum internal pressure the cylindrical pressure vessel can sustainis 417.0 psi\n",
"The maximum internal pressure a spherical pressure vessel can sustain is 833.0 psi\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.2 Page No 414"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine stress at point B and C\n",
"\n",
"#Given\n",
"P = 15000.0 #N, force,\n",
"a = 40.0 #mm, length\n",
"b = 100.0 #mm, breadth\n",
"\n",
"#CAlculation\n",
"#Normal Force\n",
"A = a*b #Area\n",
"sigma = P/A\n",
"#Bending Moment\n",
"I = (a*b**3)/12.0 #I = (1/12)*bh**3\n",
"M = P*(b/2.0) \n",
"c = b/2.0\n",
"sigma_max =(M*c)/I\n",
"\n",
"#Superposition\n",
"x = ((sigma_max-sigma)*b)/((sigma_max+sigma)+(sigma_max-sigma))\n",
"sigma_b = (sigma_max-sigma)\n",
"sigma_c = (sigma_max + sigma)\n",
"\n",
"#Display\n",
"print\"The state of stress at B is(tensile)\",sigma_b,\"psi\"\n",
"print\"The state of stress at C is (compressive)\",sigma_c,\"psi\"\n",
"\n",
"\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The state of stress at B is(tensile) 7.5 psi\n",
"The state of stress at C is (compressive) 15.0 psi\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.3 Page No 415"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the stress\n",
"\n",
"#Given:\n",
"ri =24 #inch, radius\n",
"t = 0.5 #inch\n",
"ro = ri+t\n",
"sp_wt_water = 62.4 #lb/ft**3\n",
"sp_wt_steel = 490 #lb/ft**3\n",
"l_a = 3 #m depth of point A from the top\n",
"\n",
"#Internal Loadings:\n",
"import math\n",
"v = (math.pi*l_a)*((ro/12.0)**2 - (ri/12.0)**2)\n",
"W_st = sp_wt_steel*v\n",
"p = sp_wt_water*l_a #lb/ft**2,Pascal's Law\n",
"p_=p*0.0069 #psi\n",
"#Circumferential Stress:\n",
"sigma1 = (p_*ri)/t\n",
"#Longitudinal Stress:\n",
"A_st = (math.pi)*(ro**2 - ri**2)\n",
"sigma2 = W_st/A_st\n",
"\n",
"#Display:\n",
"print\"The state of stress at A (Circumferential)\",round(sigma1,0),\"KPa\"\n",
"print\"The state of stress at A (Longitudinal) \",round(sigma2,1),\"KPa\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The state of stress at A (Circumferential) 62.0 KPa\n",
"The state of stress at A (Longitudinal) 10.2 KPa\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.4 Page No 417"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the state of stress\n",
"\n",
"#Given\n",
"y_c = 125/1000.0 #m, length\n",
"x_c = 1.5 #m\n",
"y_b = 1.5 #m\n",
"x_b = 6.0 #m\n",
"udl = 50.0 #kN/m, force per unit length\n",
"l_udl = 2.5 #m\n",
"l = 250/1000.0 #m\n",
"width = 50/1000.0 #m \n",
"\n",
"\n",
"#Internal Loadings:\n",
"N = 16.45 #kN\n",
"V = 21.93 #kN\n",
"M = 32.89 #kNm\n",
"\n",
"#Stress Components:\n",
"#Normal Force:\n",
"A = l*width\n",
"sigma1 = N/(A*1000)\n",
"#Shear Force:\n",
"tou_c = 0\n",
"#Bending Moment:\n",
"c = y_c\n",
"I = (1/12.0)*(width*l**3)\n",
"sigma2 = (M*c)/(I*1000)\n",
"#Superposition:\n",
"sigmaC = sigma1+sigma2\n",
"\n",
"#Display:\n",
"print\"The stress due to normal force at C \",round(sigma1,2),\"MPa\"\n",
"print\"The stress due to shear force at C \",tou_c,\"MPa\"\n",
"print\"The stress due to bending moment at C \",round(sigma2,1),\"MPa\"\n",
"print\"The resultant stress at C \",round(sigmaC,1),\"MPa\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress due to normal force at C 1.32 MPa\n",
"The stress due to shear force at C 0 MPa\n",
"The stress due to bending moment at C 63.1 MPa\n",
"The resultant stress at C 64.5 MPa\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.5 Page No 418"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given:\n",
"r = 0.75*10 #mm, radius\n",
"f_x =40000 #N, force along x\n",
"f_y =800 #N force along y\n",
"l1 = 0.8 #mm\n",
"l2 = 0.4 #mm\n",
"\n",
"#Stress Components:\n",
"#Normal Force:\n",
"A1 =l1*l2\n",
"sigma1 = f_x/A1 #stress = P/A\n",
"\n",
"#Bending Moment:\n",
"M_y1 = 8000 #N\n",
"c1 = l2/2.0\n",
"I1 = (1/12.0)*(l1*l2**3)\n",
"sigma_A1 = (M_y1*c1)/I1 \n",
"M_y2 = 16000 #N\n",
"c2 = l2\n",
"I2 = (1/12.0)*(l2*l1**3)\n",
"sigma_A2 = (M_y2*c2)/I2 \n",
"\n",
"#Resultant:\n",
"res_normal= -sigma1-sigma_A1-sigma_A2\n",
"\n",
"#Display:\n",
"\n",
"print\"The stress due to normal force at A \",sigma1/1000,\"KPa\"\n",
"print\"The stress due to bending moment 8KN at A \",sigma_A1/1000,\"KPa\"\n",
"print\"The stress due to bending moment 16KN at A \",sigma_A2/1000,\"KPa\"\n",
"print\"The resultant normal stress component at A \",res_normal/1000,\"KPa\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress due to normal force at A 125.0 KPa\n",
"The stress due to bending moment 8KN at A 375.0 KPa\n",
"The stress due to bending moment 16KN at A 375.0 KPa\n",
"The resultant normal stress component at A -875.0 KPa\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.7 Page No 420"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate Stress at A\n",
"\n",
"#Given:\n",
"P = 500 #lb, load\n",
"r=0.75 #inch, radius\n",
"#Stress Components:\n",
"\n",
"#Normal Force:\n",
"import math\n",
"A = math.pi*r**2\n",
"sigma = P/A\n",
"\n",
"#Bendng Moments:\n",
"M_x =7000 #lb\n",
"cy = r\n",
"Ix = (1/4.0)*math.pi*(r**4) \n",
"sigma_max_1 = (M_x*cy)/Ix \n",
"\n",
"M_y = P*l_bc/2.0\n",
"cx = l_bc/2.0\n",
"Iy = (1/12.0)*(l_ab*l_bc**3) #I = (1/12)*(bh**3)\n",
"sigma_max_2 = (M_y*cx)/Iy #sigma = My/I\n",
"#Superposition\n",
"sigmaf=round(sigma/1000,3)+round(sigma_max_1/1000,1)\n",
"\n",
"#Display:\n",
"print\"The normal stress at corner A \",round(sigma/1000,3),\"ksi\"\n",
"print\"The normal stress at point A for Bending Moment \",round(sigma_max_1/1000,1),\"ksi\"\n",
"print\"The normal stress at point A for Superimposition \",round(sigmaf,1),\"ksi\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The normal stress at corner A 0.283 ksi\n",
"The normal stress at point A for Bending Moment 21.1 ksi\n",
"The normal stress at point A for Superimposition 21.4 ksi\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.8 Page No 421"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the state of stress at point A\n",
"\n",
"#Given\n",
"r=0.75 #radius,inch\n",
"V=800 #Forca, lb\n",
"\n",
"#Calculation\n",
"#shear force\n",
"import math\n",
"Q=(4*r/(3*math.pi))*(0.5*(math.pi*r**2))\n",
"Ix=(1/4.0)*math.pi*(r**4) \n",
"tau=V*Q/(Ix*2*r)\n",
"#Since point A is on neutral axis\n",
"sigmaA=0\n",
"T=11200 #lb inch, force \n",
"Iy=(1/2.0)*math.pi*(r**4) \n",
"sigma_a=T*r/Iy\n",
"#Superimposition\n",
"sigmayzA=tau+sigma_a\n",
"\n",
"\n",
"#Result\n",
"print\"The stress for shear stress distribution is\",round(tau/1000,3),\"ksi\"\n",
"print\"The stress for Bending moment is\",sigmaA,\"ksi\"\n",
"print\"The stress for torque\",round(sigma_a/1000,2),\"ksi\"\n",
"print\"The stress for Superimposition \",round(sigmayzA/1000,1),\"ksi\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress for shear stress distribution is 0.604 ksi\n",
"The stress for Bending moment is 0 ksi\n",
"The stress for torque 16.9 ksi\n",
"The stress for Superimposition 17.5 ksi\n"
]
}
],
"prompt_number": 67
}
],
"metadata": {}
}
]
}