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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1: Tension Comprssion and Shear"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.1, page no. 9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"Find compressive stress and strain in the post\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation\n",
"\n",
"d_1 = 4 # inner diameter (inch)\n",
"d_2 = 4.5 #outer diameter (inch)\n",
"P = 26000 # pressure in pound\n",
"L = 16 # Length of cylinder (inch)\n",
"my_del = 0.012 # shortening of post (inch)\n",
"\n",
"#calculation\n",
"A = (math.pi/4)*((d_2**2)-(d_1**2)) #Area (inch^2)\n",
"s = P/A # stress\n",
"\n",
"print \"compressive stress in the post is \", round(s), \"psi\"\n",
"\n",
"e = my_del/L # strain\n",
"\n",
"print \"compressive strain in the post is %e\" %e"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"compressive stress in the post is 7789.0 psi\n",
"compressive strain in the post is 7.500000e-04\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2, page no. 10"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"formula for maximum stress & calculating maximum stress\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"W = 1500 # weight (Newton)\n",
"d = 0.008 #diameter(meter) \n",
"g = 77000 # Weight density of steel\n",
"L = 40 # Length of bar (m)\n",
"\n",
"#calculation\n",
"\n",
"A = (math.pi/4)*(d**2) # Area\n",
"s_max = (1500/A) + (g*L) # maximum stress\n",
"\n",
"#result\n",
"print \"Therefore the maximum stress in the rod is \", round(s_max,1), \"Pa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Therefore the maximum stress in the rod is 32921551.8 Pa\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3. page no. 26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"calculating change in lenght of pipe, strain in pipe, increase in diameter & increase in wall thickness\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"d1 = 4.5 # diameter in inch\n",
"d2 = 6 # diameter in inch\n",
"A = (math.pi/4)*((d2**2)-(d1**2)) # Area\n",
"P = 140 # pressure in K\n",
"s = -P/A # stress (compression)\n",
"E = 30000 # young's modulus in Ksi\n",
"e = s/E # strain\n",
"\n",
"#calculation\n",
"\n",
"# Part (a)\n",
"my_del = e*4*12 # del = e*L \n",
"print \"Change in length of the pipe is\", round(my_del,3), \"inch\"\n",
"\n",
"# Part (b)\n",
"v = 0.30 # Poissio's ratio\n",
"e_ = -(v*e)\n",
"print \"Lateral strain in the pipe is %e\" %e_\n",
"\n",
"# Part (c)\n",
"del_d2 = e_*d2 \n",
"del_d1 = e_*d1\n",
"print \"Increase in the inner diameter is \", round(del_d1,6), \"inch\"\n",
"\n",
"# Part (d)\n",
"t = 0.75\n",
"del_t = e_*t\n",
"print \"Increase in the wall thicness is %f\" %del_t, \"inch\"\n",
"del_t1 = (del_d2-del_d1)/2 \n",
"print \"del_t1 = del_t\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in length of the pipe is -0.018 inch\n",
"Lateral strain in the pipe is 1.131768e-04\n",
"Increase in the inner diameter is 0.000509 inch\n",
"Increase in the wall thicness is 0.000085 inch\n",
"del_t1 = del_t\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4, page no. 35"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"calculate average shear stress and compressive stress\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"d = 0.02 # diameter in m\n",
"t = 0.008 # thickness in m\n",
"A = math.pi*d*t # shear area\n",
"P = 110000 # prassure in Newton\n",
"\n",
"#calculation\n",
"A1 = (math.pi/4)*(d**2) # Punch area\n",
"t_aver = P/A # Average shear stress \n",
"\n",
"\n",
"print \"Average shear stress in the plate is \", t_aver, \"Pa\"\n",
"s_c = P/A1 # compressive stress\n",
"print \"Average compressive stress in the plate is \", s_c, \"Pa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average shear stress in the plate is 218838046.751 Pa\n",
"Average compressive stress in the plate is 350140874.802 Pa\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Eample 1.5, page no. 36"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"calculate bearing stress, shear stress in pin,\n",
"bearing stress between pin and gussets,\n",
"shear stress in anchor bolts\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"\n",
"P = 12.0 # Pressure in K\n",
"t = 0.375 # thickness of wall in inch\n",
"theta = 40.0 # angle in degree\n",
"d_pin = 0.75 # diameter of pin in inch\n",
"t_G = 0.625 # thickness of gusset in inch\n",
"t_B = 0.375 #thickness of base plate in inch\n",
"d_b = 0.50 # diameter of bolt in inch\n",
"\n",
"#calculation\n",
"\n",
"#Part (a)\n",
"s_b1 = P/(2*t*d_pin) # bearing stress\n",
"print \"Bearing stress between strut and pin\", round(s_b1,1), \"ksi\"\n",
"\n",
"#Part (b)\n",
"t_pin = (4*P)/(2*math.pi*(d_pin**2)) # average shear stress in the \n",
"print \"Shear stress in pin is \", round(t_pin,1), \"ksi\"\n",
"\n",
"# Part (c)\n",
"s_b2 = P/(2*t_G*d_pin) # bearing stress between pin and gusset\n",
"print \"Bearing stress between pin and gussets is\", s_b2, \"ksi\"\n",
"\n",
"# Part (d)\n",
"s_b3 = (P*math.cos(math.radians(40))/(4*t_B*d_b)) # bearing stress between anchor bolt and base plate\n",
"print \"Bearing stress between anchor bolts & base plate\", round(s_b3,1), \"ksi\"\n",
"\n",
"# Part (e)\n",
"t_bolt = (4*math.cos(math.radians(40))*P)/(4*math.pi*(d_b**2)) # shear stress in anchor bolt\n",
"print \"Shear stress in anchor bolts is\", round(t_bolt,1), \"ksi\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Bearing stress between strut and pin 21.3 ksi\n",
"Shear stress in pin is 13.6 ksi\n",
"Bearing stress between pin and gussets is 12.8 ksi\n",
"Bearing stress between anchor bolts & base plate 12.3 ksi\n",
"Shear stress in anchor bolts is 11.7 ksi\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.7, page no. 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"determine stress at various parts\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation\n",
"b1 = 1.5 # width of recmath.tangular crosssection in inch\n",
"t = 0.5 # thickness of recmath.tangular crosssection in inch\n",
"b2 = 3.0 # width of enlarged recmath.tangular crosssection in inch\n",
"d = 1.0 # diameter in inch\n",
"\n",
"#calculation\n",
"\n",
"# Part (a)\n",
"s_1 = 16000 # maximum allowable tensile stress in Psi\n",
"P_1 = s_1*t*b1 \n",
"print \"The allowable load P1 is\", P_1, \"lb\"\n",
"\n",
"# Part (b)\n",
"s_2 = 11000 # maximum allowable tensile stress in Psi\n",
"P_2 = s_2*t*(b2-d) \n",
"print \"allowable load P2 at this section is\", P_2, \"lb\"\n",
"\n",
"#Part (c)\n",
"s_3 = 26000 # maximum allowable tensile stress in Psi\n",
"P_3 = s_3*t*d \n",
"print \"The allowable load based upon bearing between the hanger and the bolt is\", P_3, \"lb\"\n",
"\n",
"# Part (d)\n",
"s_4 = 6500 # maximum allowable tensile stress in Psi\n",
"P_4 = (math.pi/4)*(d**2)*2*s_4 \n",
"print \"the allowable load P4 based upon shear in the bolt is\", round(P_4), \"lb\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The allowable load P1 is 12000.0 lb\n",
"allowable load P2 at this section is 11000.0 lb\n",
"The allowable load based upon bearing between the hanger and the bolt is 13000.0 lb\n",
"the allowable load P4 based upon shear in the bolt is 10210.0 lb\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.8, page no. 46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"calculating the cross sectional area \n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"R_ah = (2700*0.8 + 2700*2.6)/2 # Horizontal component at A in N\n",
"R_ch = R_ah # Horizontal component at C in N\n",
"R_cv = (2700*2.2 + 2700*0.4)/3 # vertical component at C in N\n",
"R_av = 2700 + 2700 - R_cv # vertical component at A in N\n",
"R_a = math.sqrt((R_ah**2)+(R_av**2))\n",
"R_c = math.sqrt((R_ch**2)+(R_cv**2))\n",
"Fab = R_a # Tensile force in bar AB\n",
"Vc = R_c # Shear force acting on the pin at C\n",
"s_allow = 125000000 # allowable stress in tension \n",
"t_allow = 45000000 # allowable stress in shear\n",
"\n",
"#calculation\n",
"Aab = Fab / s_allow # required area of bar \n",
"Apin = Vc / (2*t_allow) # required area of pin\n",
"\n",
"\n",
"print \"Required area of bar is %f\" %Apin, \"m^2\"\n",
"d = math.sqrt((4*Apin)/math.pi) # diameter in meter\n",
"print \"Required diameter of pin is %f\" %d, \"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required area of bar is 0.000057 m^2\n",
"Required diameter of pin is 0.008537 m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
