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diff --git a/Test/README.txt b/Test/README.txt
deleted file mode 100755
index b9a36bc6..00000000
--- a/Test/README.txt
+++ /dev/null
@@ -1,10 +0,0 @@
-Contributed By: Hardik Ghaghada
-Course: mca
-College/Institute/Organization: FOSSEE - Indian Institute of Technology
-Department/Designation: Aerospace
-Book Title: Test
-Author: Subramanium & Brij Lal
-Publisher: McGraw Hill Education (India) Private Limited, New Delhi
-Year of publication: 2000
-Isbn: 2548872474
-Edition: 2nd \ No newline at end of file
diff --git a/Test/chapter1.ipynb b/Test/chapter1.ipynb
deleted file mode 100755
index cf45a409..00000000
--- a/Test/chapter1.ipynb
+++ /dev/null
@@ -1,423 +0,0 @@
-{
- "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": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Test/chapter1_1.ipynb b/Test/chapter1_1.ipynb
deleted file mode 100755
index cf45a409..00000000
--- a/Test/chapter1_1.ipynb
+++ /dev/null
@@ -1,423 +0,0 @@
-{
- "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": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Test/chapter1_2.ipynb b/Test/chapter1_2.ipynb
deleted file mode 100755
index cf45a409..00000000
--- a/Test/chapter1_2.ipynb
+++ /dev/null
@@ -1,423 +0,0 @@
-{
- "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": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Test/chapter2.ipynb b/Test/chapter2.ipynb
deleted file mode 100755
index c4e1ad0f..00000000
--- a/Test/chapter2.ipynb
+++ /dev/null
@@ -1,501 +0,0 @@
-{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Axially Loaded Members"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.1, page no. 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculating the number of revolutions for the nut\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "\n",
- "W = 2.0 #lb\n",
- "b = 10.5 #inch\n",
- "c = 6.4 #inch\n",
- "k = 4.2 #inch\n",
- "p = 1.0/16.0 #inch\n",
- "\n",
- "#calculation\n",
- "\n",
- "n = (W*b)/(c*k*p) #inch\n",
- "\n",
- "#result\n",
- "\n",
- "print \" No. of revolution required = \", n, \"revolutions\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " No. of revolution required = 12.5 revolutions\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2, page no. 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "finding maximum allowable load\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "import numpy\n",
- "\n",
- "#initialisation\n",
- "\n",
- "Fce_ = 2.0 #dummy variable\n",
- "Fbd_ = 3.0 #dummy variable \n",
- "Lbd = 480.0 #mm\n",
- "Lce = 600.0 #mm\n",
- "E = 205e6 #205Gpa\n",
- "Abd = 1020.0 #mm\n",
- "Ace = 520.0 #mm\n",
- "\n",
- "#calculation\n",
- "Dbd_ = (Fbd_*Lbd)/(E*Abd) #dummy variable\n",
- "Dce_ = (Fce_*Lce)/(E*Ace) #dummy variable\n",
- "Da = 1 #limiting value\n",
- "P = ((((450+225)/225)*(Dbd_ + Dce_) - Dce_ )**(-1)) * Da \n",
- "Fce = 2*P # Real value in newton\n",
- "Fbd = 3*P #real value in newton\n",
- "Dbd = (Fbd*Lbd)/(E*Abd) #print lacement in mm\n",
- "Dce = (Fce*Lce)/(E*Ace) # print lacement in mm\n",
- "a = numpy.degrees(numpy.arctan(((Da+Dce)/675))) #alpha in degree\n",
- "\n",
- "#result\n",
- "print \"alpha = \", round(a,2), \"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "alpha = 0.11 degree\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.3, page no. 80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculation if vertical displacement\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "P1 = 2100.0 #lb\n",
- "P2 = 5600.0 #lb\n",
- "b = 25.0 #inch\n",
- "a = 28.0 #inch\n",
- "A1 = 0.25 #inch^2\n",
- "A2 = 0.15 #inch^2\n",
- "L1 = 20.0 #inch\n",
- "L2 = 34.8 #inch\n",
- "E = 29e6 #29Gpa\n",
- "\n",
- "#Calculations\n",
- "P3 = (P2*b)/a \n",
- "Ra = P3-P1\n",
- "N1 = -Ra \n",
- "N2 = P1 \n",
- "D = ((N1*L1)/(E*A1)) + ((N2*L2)/(E*A2)) #print lacement\n",
- "\n",
- "#Result\n",
- "print \"Downward print lacement is = \", D, \"inch\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Downward print lacement is = 0.0088 inch\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.6, page no. 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "obtaing formula and calculating allowable load\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#Numerical calculation of allowable load\n",
- "\n",
- "d1 = 4.0 #mm\n",
- "d2 = 3.0 #mm\n",
- "A1 = (math.pi*(d1**2))/4 #area\n",
- "A2 = (math.pi*(d2**2))/4 #area\n",
- "L1 = 0.4 #meter\n",
- "L2 = 0.3 #meter\n",
- "E1 = 72e9 #Gpa\n",
- "E2 = 45e9 #Gpa\n",
- "f1 = L1/(E1*A1) * 1e6 # To cpmpensate for the mm**2\n",
- "f2 = L2/(E2*A2) * 1e6 \n",
- "s1 = 200e6 #stress\n",
- "s2 = 175e6 #stress\n",
- "\n",
- "#Calculations\n",
- "P1 = ( (s1*A1*(4*f1 + f2))/(3*f2) ) * 1e-6 # To cpmpensate for the mm**2\n",
- "P2 = ( (s2*A2*(4*f1 + f2))/(6*f1) ) * 1e-6 \n",
- "\n",
- "#Result\n",
- "print \"Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = \", P2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = 1264.49104307\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.10, page no. 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculate stress acting on inclined section &\n",
- "the complete state of stress\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "P = 90000.0 #newton\n",
- "A = 1200e-6 # meter^2\n",
- "s_x = -P/A #stress\n",
- "t_1 = 25.0 #for the stresses on ab and cd plane\n",
- "\n",
- "#Calculations\n",
- "s_1 = s_x*(math.cos(math.radians(t_1))**2)\n",
- "T_1 = -s_x*math.cos(math.radians(t_1))*math.sin(math.radians(t_1))\n",
- "t_2 = -65.0 #for the stresses on ad and bc plane\n",
- "s_2 = s_x*(math.cos(math.radians(t_2))**2)\n",
- "T_2 = -s_x*math.cos(math.radians(t_2))*math.sin(math.radians(t_2))\n",
- "\n",
- "#Result\n",
- "print \"The normal and shear stresses on the plane ab and cd are\", round((T_1/1E+6),2), round((s_1/1E+6),2), \"MPa respecively\" \n",
- "print \"respecively The normal and shear stresses on the plane ad and bc are\", round((T_2/1E+6),2), round((s_2/1E+6),2), \"MPa respecively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal and shear stresses on the plane ab and cd are 28.73 -61.6 MPa respecively\n",
- "respecively The normal and shear stresses on the plane ad and bc are -28.73 -13.4 MPa respecively\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.11, page no. 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Calculate the vertical displacement of the joint\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "# Value of s_x based on allowable stresses on glued joint\n",
- "\n",
- "#initialisation\n",
- "s_t = -750.0 #psi\n",
- "t = -50.0 #degree\n",
- "T_t = -500.0 #psi\n",
- "\n",
- "sg_x_1 = s_t/(math.cos(math.radians(t))**2)\n",
- "sg_x_2 = -T_t/(math.cos(math.radians(t))*math.sin(math.radians(t)))\n",
- "\n",
- "# Value of s_x based on allowable stresses on plastic\n",
- "\n",
- "sp_x_1 = -1100.0 #psi\n",
- "T_t_p = 600.0 #psi\n",
- "t_p = 45.0 #degree\n",
- "sp_x_2 = -T_t_p/(math.cos(math.radians(t_p))*math.sin(math.radians(t_p)))\n",
- "\n",
- "# Minimum width of bar\n",
- "\n",
- "P = 8000.0 #lb\n",
- "A = P/sg_x_2\n",
- "b_min = math.sqrt(abs(A)) #inch\n",
- "print \"The minimum width of the bar is\", round(b_min,2), \"inch\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum width of the bar is 2.81 inch\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.15, page no. 126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Comparison of energy-absorbing capacity with different type of bolts\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "#Bolt with reduced shank diameter\n",
- "\n",
- "#initialisation\n",
- "g = 1.50 # inch\n",
- "d = 0.5 #inch\n",
- "t = 0.25 #inch\n",
- "d_r = 0.406 #inch\n",
- "L = 13.5 #inch\n",
- "\n",
- "#calculation\n",
- "ratio = ((g*(d**2))/(((g-t)*(d_r**2))+(t*(d**2)))) #U2/U1\n",
- "\n",
- "print \"The energy absorbing capacity of the bolts with reduced shank diameter\", round(ratio,2)\n",
- "ratio_1 = ( (((L-t)*(d_r**2))+(t*(d**2))) / ((2*(g-t)*(d_r**2))+2*(t*(d**2))) ) #U3/2U1\n",
- "print \"The energy absorbing capacity of the long bolts\", round(ratio_1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The energy absorbing capacity of the bolts with reduced shank diameter 1.4\n",
- "The energy absorbing capacity of the long bolts 4.18\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- " Example 2.16, page no. 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Determine the maximum elongation and tensile stress\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "# Maximum elongation\n",
- "M = 20 #kg\n",
- "g = 9.81 #m/s^2\n",
- "L = 2 #meter\n",
- "E = 210e9 #210Gpa\n",
- "h = 0.15 #meter\n",
- "diameter = 0.015 #milimeter\n",
- "\n",
- "#Calculations & Result\n",
- "A = (math.pi/4)*(diameter**2) #area\n",
- "D_st = ((M*g*L)/(E*A)) \n",
- "D_max = D_st*(1+(1+(2*h/D_st))**0.5) \n",
- "D_max_1 = math.sqrt(2*h*D_st) # another approach to find D_max\n",
- "i = D_max / D_st # Impact factor\n",
- "print \"Maximum elongation is\",round((D_max/1E-3),2), \"mm\" # Maximum tensile stress\n",
- "s_max = (E*D_max)/L #Maximum tensile stress\n",
- "s_st = (M*g)/A #static stress\n",
- "i_1 = s_max / s_st #Impact factor \n",
- "print \"Maximum tensile stress is \", round((s_max/1E+6),2), \"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum elongation is 1.79 mm\n",
- "Maximum tensile stress is 188.13 MPa\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.18, page no. 148"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "determine displacement at the lower end of bar in various conditions\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "\n",
- "#initialisation\n",
- "P1 = 108000.0 #Newton\n",
- "P2 = 27000.0 #Newton\n",
- "L = 2.2 #meter\n",
- "A = 480.0 #mm^2\n",
- "\n",
- "\n",
- "#calculations\n",
- "\n",
- "# Displacement due to load P1 acting alone\n",
- "s = (P1/A) #stress in MPa\n",
- "e = (s/70000) + (1/628.2)*((s/260)**10) #strain\n",
- "D_b = e*L*1e3 #elongation in mm\n",
- "print \"elongation when only P1 load acting is = \", round(D_b,2), \" mm\"\n",
- "\n",
- "# Displacement due to load P2 acting alone\n",
- "s_1 = (P2/A) #stress in MPa\n",
- "e_1 = (s_1/70000) + (1/628.2)*((s_1/260)**10) #strain\n",
- "D_b_1 = e_1*(L/2)*1e3 #elongation in mm (no elongation in lower half)\n",
- "print \"elongation when only P2 load acting is = \", round(D_b_1,2), \" mm\"\n",
- "\n",
- "# Displacement due to both load acting simonmath.taneously\n",
- "#upper half\n",
- "s_2 = (P1/A) #stress in MPa\n",
- "e_2 = (s_2/70000) + (1/628.2)*((s_2/260)**10) #strain\n",
- "\n",
- "#lower half\n",
- "s_3 = (P1+P2)/A #stress in MPa\n",
- "e_3 = (s_3/70000) + (1/628.2)*((s_3/260)**10) #strain\n",
- "D_b_2 = ((e_2*L)/2 + (e_3*L)/2) * 1e3 # elongation in mm\n",
- "print \"elongation when P1 and P2 both loads are acting is = \", round(D_b_2,2), \" mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "elongation when only P1 load acting is = 7.9 mm\n",
- "elongation when only P2 load acting is = 0.88 mm\n",
- "elongation when P1 and P2 both loads are acting is = 12.21 mm\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Test/chapter2_1.ipynb b/Test/chapter2_1.ipynb
deleted file mode 100755
index c4e1ad0f..00000000
--- a/Test/chapter2_1.ipynb
+++ /dev/null
@@ -1,501 +0,0 @@
-{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Axially Loaded Members"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.1, page no. 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculating the number of revolutions for the nut\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "\n",
- "W = 2.0 #lb\n",
- "b = 10.5 #inch\n",
- "c = 6.4 #inch\n",
- "k = 4.2 #inch\n",
- "p = 1.0/16.0 #inch\n",
- "\n",
- "#calculation\n",
- "\n",
- "n = (W*b)/(c*k*p) #inch\n",
- "\n",
- "#result\n",
- "\n",
- "print \" No. of revolution required = \", n, \"revolutions\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " No. of revolution required = 12.5 revolutions\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2, page no. 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "finding maximum allowable load\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "import numpy\n",
- "\n",
- "#initialisation\n",
- "\n",
- "Fce_ = 2.0 #dummy variable\n",
- "Fbd_ = 3.0 #dummy variable \n",
- "Lbd = 480.0 #mm\n",
- "Lce = 600.0 #mm\n",
- "E = 205e6 #205Gpa\n",
- "Abd = 1020.0 #mm\n",
- "Ace = 520.0 #mm\n",
- "\n",
- "#calculation\n",
- "Dbd_ = (Fbd_*Lbd)/(E*Abd) #dummy variable\n",
- "Dce_ = (Fce_*Lce)/(E*Ace) #dummy variable\n",
- "Da = 1 #limiting value\n",
- "P = ((((450+225)/225)*(Dbd_ + Dce_) - Dce_ )**(-1)) * Da \n",
- "Fce = 2*P # Real value in newton\n",
- "Fbd = 3*P #real value in newton\n",
- "Dbd = (Fbd*Lbd)/(E*Abd) #print lacement in mm\n",
- "Dce = (Fce*Lce)/(E*Ace) # print lacement in mm\n",
- "a = numpy.degrees(numpy.arctan(((Da+Dce)/675))) #alpha in degree\n",
- "\n",
- "#result\n",
- "print \"alpha = \", round(a,2), \"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "alpha = 0.11 degree\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.3, page no. 80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculation if vertical displacement\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "P1 = 2100.0 #lb\n",
- "P2 = 5600.0 #lb\n",
- "b = 25.0 #inch\n",
- "a = 28.0 #inch\n",
- "A1 = 0.25 #inch^2\n",
- "A2 = 0.15 #inch^2\n",
- "L1 = 20.0 #inch\n",
- "L2 = 34.8 #inch\n",
- "E = 29e6 #29Gpa\n",
- "\n",
- "#Calculations\n",
- "P3 = (P2*b)/a \n",
- "Ra = P3-P1\n",
- "N1 = -Ra \n",
- "N2 = P1 \n",
- "D = ((N1*L1)/(E*A1)) + ((N2*L2)/(E*A2)) #print lacement\n",
- "\n",
- "#Result\n",
- "print \"Downward print lacement is = \", D, \"inch\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Downward print lacement is = 0.0088 inch\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.6, page no. 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "obtaing formula and calculating allowable load\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#Numerical calculation of allowable load\n",
- "\n",
- "d1 = 4.0 #mm\n",
- "d2 = 3.0 #mm\n",
- "A1 = (math.pi*(d1**2))/4 #area\n",
- "A2 = (math.pi*(d2**2))/4 #area\n",
- "L1 = 0.4 #meter\n",
- "L2 = 0.3 #meter\n",
- "E1 = 72e9 #Gpa\n",
- "E2 = 45e9 #Gpa\n",
- "f1 = L1/(E1*A1) * 1e6 # To cpmpensate for the mm**2\n",
- "f2 = L2/(E2*A2) * 1e6 \n",
- "s1 = 200e6 #stress\n",
- "s2 = 175e6 #stress\n",
- "\n",
- "#Calculations\n",
- "P1 = ( (s1*A1*(4*f1 + f2))/(3*f2) ) * 1e-6 # To cpmpensate for the mm**2\n",
- "P2 = ( (s2*A2*(4*f1 + f2))/(6*f1) ) * 1e-6 \n",
- "\n",
- "#Result\n",
- "print \"Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = \", P2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = 1264.49104307\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.10, page no. 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculate stress acting on inclined section &\n",
- "the complete state of stress\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "P = 90000.0 #newton\n",
- "A = 1200e-6 # meter^2\n",
- "s_x = -P/A #stress\n",
- "t_1 = 25.0 #for the stresses on ab and cd plane\n",
- "\n",
- "#Calculations\n",
- "s_1 = s_x*(math.cos(math.radians(t_1))**2)\n",
- "T_1 = -s_x*math.cos(math.radians(t_1))*math.sin(math.radians(t_1))\n",
- "t_2 = -65.0 #for the stresses on ad and bc plane\n",
- "s_2 = s_x*(math.cos(math.radians(t_2))**2)\n",
- "T_2 = -s_x*math.cos(math.radians(t_2))*math.sin(math.radians(t_2))\n",
- "\n",
- "#Result\n",
- "print \"The normal and shear stresses on the plane ab and cd are\", round((T_1/1E+6),2), round((s_1/1E+6),2), \"MPa respecively\" \n",
- "print \"respecively The normal and shear stresses on the plane ad and bc are\", round((T_2/1E+6),2), round((s_2/1E+6),2), \"MPa respecively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal and shear stresses on the plane ab and cd are 28.73 -61.6 MPa respecively\n",
- "respecively The normal and shear stresses on the plane ad and bc are -28.73 -13.4 MPa respecively\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.11, page no. 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Calculate the vertical displacement of the joint\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "# Value of s_x based on allowable stresses on glued joint\n",
- "\n",
- "#initialisation\n",
- "s_t = -750.0 #psi\n",
- "t = -50.0 #degree\n",
- "T_t = -500.0 #psi\n",
- "\n",
- "sg_x_1 = s_t/(math.cos(math.radians(t))**2)\n",
- "sg_x_2 = -T_t/(math.cos(math.radians(t))*math.sin(math.radians(t)))\n",
- "\n",
- "# Value of s_x based on allowable stresses on plastic\n",
- "\n",
- "sp_x_1 = -1100.0 #psi\n",
- "T_t_p = 600.0 #psi\n",
- "t_p = 45.0 #degree\n",
- "sp_x_2 = -T_t_p/(math.cos(math.radians(t_p))*math.sin(math.radians(t_p)))\n",
- "\n",
- "# Minimum width of bar\n",
- "\n",
- "P = 8000.0 #lb\n",
- "A = P/sg_x_2\n",
- "b_min = math.sqrt(abs(A)) #inch\n",
- "print \"The minimum width of the bar is\", round(b_min,2), \"inch\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum width of the bar is 2.81 inch\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.15, page no. 126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Comparison of energy-absorbing capacity with different type of bolts\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "#Bolt with reduced shank diameter\n",
- "\n",
- "#initialisation\n",
- "g = 1.50 # inch\n",
- "d = 0.5 #inch\n",
- "t = 0.25 #inch\n",
- "d_r = 0.406 #inch\n",
- "L = 13.5 #inch\n",
- "\n",
- "#calculation\n",
- "ratio = ((g*(d**2))/(((g-t)*(d_r**2))+(t*(d**2)))) #U2/U1\n",
- "\n",
- "print \"The energy absorbing capacity of the bolts with reduced shank diameter\", round(ratio,2)\n",
- "ratio_1 = ( (((L-t)*(d_r**2))+(t*(d**2))) / ((2*(g-t)*(d_r**2))+2*(t*(d**2))) ) #U3/2U1\n",
- "print \"The energy absorbing capacity of the long bolts\", round(ratio_1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The energy absorbing capacity of the bolts with reduced shank diameter 1.4\n",
- "The energy absorbing capacity of the long bolts 4.18\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- " Example 2.16, page no. 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Determine the maximum elongation and tensile stress\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "# Maximum elongation\n",
- "M = 20 #kg\n",
- "g = 9.81 #m/s^2\n",
- "L = 2 #meter\n",
- "E = 210e9 #210Gpa\n",
- "h = 0.15 #meter\n",
- "diameter = 0.015 #milimeter\n",
- "\n",
- "#Calculations & Result\n",
- "A = (math.pi/4)*(diameter**2) #area\n",
- "D_st = ((M*g*L)/(E*A)) \n",
- "D_max = D_st*(1+(1+(2*h/D_st))**0.5) \n",
- "D_max_1 = math.sqrt(2*h*D_st) # another approach to find D_max\n",
- "i = D_max / D_st # Impact factor\n",
- "print \"Maximum elongation is\",round((D_max/1E-3),2), \"mm\" # Maximum tensile stress\n",
- "s_max = (E*D_max)/L #Maximum tensile stress\n",
- "s_st = (M*g)/A #static stress\n",
- "i_1 = s_max / s_st #Impact factor \n",
- "print \"Maximum tensile stress is \", round((s_max/1E+6),2), \"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum elongation is 1.79 mm\n",
- "Maximum tensile stress is 188.13 MPa\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.18, page no. 148"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "determine displacement at the lower end of bar in various conditions\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "\n",
- "#initialisation\n",
- "P1 = 108000.0 #Newton\n",
- "P2 = 27000.0 #Newton\n",
- "L = 2.2 #meter\n",
- "A = 480.0 #mm^2\n",
- "\n",
- "\n",
- "#calculations\n",
- "\n",
- "# Displacement due to load P1 acting alone\n",
- "s = (P1/A) #stress in MPa\n",
- "e = (s/70000) + (1/628.2)*((s/260)**10) #strain\n",
- "D_b = e*L*1e3 #elongation in mm\n",
- "print \"elongation when only P1 load acting is = \", round(D_b,2), \" mm\"\n",
- "\n",
- "# Displacement due to load P2 acting alone\n",
- "s_1 = (P2/A) #stress in MPa\n",
- "e_1 = (s_1/70000) + (1/628.2)*((s_1/260)**10) #strain\n",
- "D_b_1 = e_1*(L/2)*1e3 #elongation in mm (no elongation in lower half)\n",
- "print \"elongation when only P2 load acting is = \", round(D_b_1,2), \" mm\"\n",
- "\n",
- "# Displacement due to both load acting simonmath.taneously\n",
- "#upper half\n",
- "s_2 = (P1/A) #stress in MPa\n",
- "e_2 = (s_2/70000) + (1/628.2)*((s_2/260)**10) #strain\n",
- "\n",
- "#lower half\n",
- "s_3 = (P1+P2)/A #stress in MPa\n",
- "e_3 = (s_3/70000) + (1/628.2)*((s_3/260)**10) #strain\n",
- "D_b_2 = ((e_2*L)/2 + (e_3*L)/2) * 1e3 # elongation in mm\n",
- "print \"elongation when P1 and P2 both loads are acting is = \", round(D_b_2,2), \" mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "elongation when only P1 load acting is = 7.9 mm\n",
- "elongation when only P2 load acting is = 0.88 mm\n",
- "elongation when P1 and P2 both loads are acting is = 12.21 mm\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
diff --git a/Test/chapter2_2.ipynb b/Test/chapter2_2.ipynb
deleted file mode 100755
index c4e1ad0f..00000000
--- a/Test/chapter2_2.ipynb
+++ /dev/null
@@ -1,501 +0,0 @@
-{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2: Axially Loaded Members"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.1, page no. 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculating the number of revolutions for the nut\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "\n",
- "W = 2.0 #lb\n",
- "b = 10.5 #inch\n",
- "c = 6.4 #inch\n",
- "k = 4.2 #inch\n",
- "p = 1.0/16.0 #inch\n",
- "\n",
- "#calculation\n",
- "\n",
- "n = (W*b)/(c*k*p) #inch\n",
- "\n",
- "#result\n",
- "\n",
- "print \" No. of revolution required = \", n, \"revolutions\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " No. of revolution required = 12.5 revolutions\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2, page no. 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "finding maximum allowable load\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "import numpy\n",
- "\n",
- "#initialisation\n",
- "\n",
- "Fce_ = 2.0 #dummy variable\n",
- "Fbd_ = 3.0 #dummy variable \n",
- "Lbd = 480.0 #mm\n",
- "Lce = 600.0 #mm\n",
- "E = 205e6 #205Gpa\n",
- "Abd = 1020.0 #mm\n",
- "Ace = 520.0 #mm\n",
- "\n",
- "#calculation\n",
- "Dbd_ = (Fbd_*Lbd)/(E*Abd) #dummy variable\n",
- "Dce_ = (Fce_*Lce)/(E*Ace) #dummy variable\n",
- "Da = 1 #limiting value\n",
- "P = ((((450+225)/225)*(Dbd_ + Dce_) - Dce_ )**(-1)) * Da \n",
- "Fce = 2*P # Real value in newton\n",
- "Fbd = 3*P #real value in newton\n",
- "Dbd = (Fbd*Lbd)/(E*Abd) #print lacement in mm\n",
- "Dce = (Fce*Lce)/(E*Ace) # print lacement in mm\n",
- "a = numpy.degrees(numpy.arctan(((Da+Dce)/675))) #alpha in degree\n",
- "\n",
- "#result\n",
- "print \"alpha = \", round(a,2), \"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "alpha = 0.11 degree\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.3, page no. 80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculation if vertical displacement\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "P1 = 2100.0 #lb\n",
- "P2 = 5600.0 #lb\n",
- "b = 25.0 #inch\n",
- "a = 28.0 #inch\n",
- "A1 = 0.25 #inch^2\n",
- "A2 = 0.15 #inch^2\n",
- "L1 = 20.0 #inch\n",
- "L2 = 34.8 #inch\n",
- "E = 29e6 #29Gpa\n",
- "\n",
- "#Calculations\n",
- "P3 = (P2*b)/a \n",
- "Ra = P3-P1\n",
- "N1 = -Ra \n",
- "N2 = P1 \n",
- "D = ((N1*L1)/(E*A1)) + ((N2*L2)/(E*A2)) #print lacement\n",
- "\n",
- "#Result\n",
- "print \"Downward print lacement is = \", D, \"inch\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Downward print lacement is = 0.0088 inch\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.6, page no. 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "obtaing formula and calculating allowable load\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#Numerical calculation of allowable load\n",
- "\n",
- "d1 = 4.0 #mm\n",
- "d2 = 3.0 #mm\n",
- "A1 = (math.pi*(d1**2))/4 #area\n",
- "A2 = (math.pi*(d2**2))/4 #area\n",
- "L1 = 0.4 #meter\n",
- "L2 = 0.3 #meter\n",
- "E1 = 72e9 #Gpa\n",
- "E2 = 45e9 #Gpa\n",
- "f1 = L1/(E1*A1) * 1e6 # To cpmpensate for the mm**2\n",
- "f2 = L2/(E2*A2) * 1e6 \n",
- "s1 = 200e6 #stress\n",
- "s2 = 175e6 #stress\n",
- "\n",
- "#Calculations\n",
- "P1 = ( (s1*A1*(4*f1 + f2))/(3*f2) ) * 1e-6 # To cpmpensate for the mm**2\n",
- "P2 = ( (s2*A2*(4*f1 + f2))/(6*f1) ) * 1e-6 \n",
- "\n",
- "#Result\n",
- "print \"Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = \", P2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Newton Minimum allowable stress aomong the two P1 and P2 is smaller one, therefore MAS = 1264.49104307\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.10, page no. 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "calculate stress acting on inclined section &\n",
- "the complete state of stress\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "P = 90000.0 #newton\n",
- "A = 1200e-6 # meter^2\n",
- "s_x = -P/A #stress\n",
- "t_1 = 25.0 #for the stresses on ab and cd plane\n",
- "\n",
- "#Calculations\n",
- "s_1 = s_x*(math.cos(math.radians(t_1))**2)\n",
- "T_1 = -s_x*math.cos(math.radians(t_1))*math.sin(math.radians(t_1))\n",
- "t_2 = -65.0 #for the stresses on ad and bc plane\n",
- "s_2 = s_x*(math.cos(math.radians(t_2))**2)\n",
- "T_2 = -s_x*math.cos(math.radians(t_2))*math.sin(math.radians(t_2))\n",
- "\n",
- "#Result\n",
- "print \"The normal and shear stresses on the plane ab and cd are\", round((T_1/1E+6),2), round((s_1/1E+6),2), \"MPa respecively\" \n",
- "print \"respecively The normal and shear stresses on the plane ad and bc are\", round((T_2/1E+6),2), round((s_2/1E+6),2), \"MPa respecively\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The normal and shear stresses on the plane ab and cd are 28.73 -61.6 MPa respecively\n",
- "respecively The normal and shear stresses on the plane ad and bc are -28.73 -13.4 MPa respecively\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.11, page no. 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Calculate the vertical displacement of the joint\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "# Value of s_x based on allowable stresses on glued joint\n",
- "\n",
- "#initialisation\n",
- "s_t = -750.0 #psi\n",
- "t = -50.0 #degree\n",
- "T_t = -500.0 #psi\n",
- "\n",
- "sg_x_1 = s_t/(math.cos(math.radians(t))**2)\n",
- "sg_x_2 = -T_t/(math.cos(math.radians(t))*math.sin(math.radians(t)))\n",
- "\n",
- "# Value of s_x based on allowable stresses on plastic\n",
- "\n",
- "sp_x_1 = -1100.0 #psi\n",
- "T_t_p = 600.0 #psi\n",
- "t_p = 45.0 #degree\n",
- "sp_x_2 = -T_t_p/(math.cos(math.radians(t_p))*math.sin(math.radians(t_p)))\n",
- "\n",
- "# Minimum width of bar\n",
- "\n",
- "P = 8000.0 #lb\n",
- "A = P/sg_x_2\n",
- "b_min = math.sqrt(abs(A)) #inch\n",
- "print \"The minimum width of the bar is\", round(b_min,2), \"inch\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum width of the bar is 2.81 inch\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.15, page no. 126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Comparison of energy-absorbing capacity with different type of bolts\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "#Bolt with reduced shank diameter\n",
- "\n",
- "#initialisation\n",
- "g = 1.50 # inch\n",
- "d = 0.5 #inch\n",
- "t = 0.25 #inch\n",
- "d_r = 0.406 #inch\n",
- "L = 13.5 #inch\n",
- "\n",
- "#calculation\n",
- "ratio = ((g*(d**2))/(((g-t)*(d_r**2))+(t*(d**2)))) #U2/U1\n",
- "\n",
- "print \"The energy absorbing capacity of the bolts with reduced shank diameter\", round(ratio,2)\n",
- "ratio_1 = ( (((L-t)*(d_r**2))+(t*(d**2))) / ((2*(g-t)*(d_r**2))+2*(t*(d**2))) ) #U3/2U1\n",
- "print \"The energy absorbing capacity of the long bolts\", round(ratio_1,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The energy absorbing capacity of the bolts with reduced shank diameter 1.4\n",
- "The energy absorbing capacity of the long bolts 4.18\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- " Example 2.16, page no. 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "Determine the maximum elongation and tensile stress\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "#initialisation\n",
- "# Maximum elongation\n",
- "M = 20 #kg\n",
- "g = 9.81 #m/s^2\n",
- "L = 2 #meter\n",
- "E = 210e9 #210Gpa\n",
- "h = 0.15 #meter\n",
- "diameter = 0.015 #milimeter\n",
- "\n",
- "#Calculations & Result\n",
- "A = (math.pi/4)*(diameter**2) #area\n",
- "D_st = ((M*g*L)/(E*A)) \n",
- "D_max = D_st*(1+(1+(2*h/D_st))**0.5) \n",
- "D_max_1 = math.sqrt(2*h*D_st) # another approach to find D_max\n",
- "i = D_max / D_st # Impact factor\n",
- "print \"Maximum elongation is\",round((D_max/1E-3),2), \"mm\" # Maximum tensile stress\n",
- "s_max = (E*D_max)/L #Maximum tensile stress\n",
- "s_st = (M*g)/A #static stress\n",
- "i_1 = s_max / s_st #Impact factor \n",
- "print \"Maximum tensile stress is \", round((s_max/1E+6),2), \"MPa\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum elongation is 1.79 mm\n",
- "Maximum tensile stress is 188.13 MPa\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.18, page no. 148"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\"\"\"\n",
- "determine displacement at the lower end of bar in various conditions\n",
- "\"\"\"\n",
- "\n",
- "import math \n",
- "\n",
- "\n",
- "#initialisation\n",
- "P1 = 108000.0 #Newton\n",
- "P2 = 27000.0 #Newton\n",
- "L = 2.2 #meter\n",
- "A = 480.0 #mm^2\n",
- "\n",
- "\n",
- "#calculations\n",
- "\n",
- "# Displacement due to load P1 acting alone\n",
- "s = (P1/A) #stress in MPa\n",
- "e = (s/70000) + (1/628.2)*((s/260)**10) #strain\n",
- "D_b = e*L*1e3 #elongation in mm\n",
- "print \"elongation when only P1 load acting is = \", round(D_b,2), \" mm\"\n",
- "\n",
- "# Displacement due to load P2 acting alone\n",
- "s_1 = (P2/A) #stress in MPa\n",
- "e_1 = (s_1/70000) + (1/628.2)*((s_1/260)**10) #strain\n",
- "D_b_1 = e_1*(L/2)*1e3 #elongation in mm (no elongation in lower half)\n",
- "print \"elongation when only P2 load acting is = \", round(D_b_1,2), \" mm\"\n",
- "\n",
- "# Displacement due to both load acting simonmath.taneously\n",
- "#upper half\n",
- "s_2 = (P1/A) #stress in MPa\n",
- "e_2 = (s_2/70000) + (1/628.2)*((s_2/260)**10) #strain\n",
- "\n",
- "#lower half\n",
- "s_3 = (P1+P2)/A #stress in MPa\n",
- "e_3 = (s_3/70000) + (1/628.2)*((s_3/260)**10) #strain\n",
- "D_b_2 = ((e_2*L)/2 + (e_3*L)/2) * 1e3 # elongation in mm\n",
- "print \"elongation when P1 and P2 both loads are acting is = \", round(D_b_2,2), \" mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "elongation when only P1 load acting is = 7.9 mm\n",
- "elongation when only P2 load acting is = 0.88 mm\n",
- "elongation when P1 and P2 both loads are acting is = 12.21 mm\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
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generated by cgit (git 2.34.1) at 2025-04-02 11:28:52 +0000