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Diffstat (limited to 'Mechanics_of_Structures')
| -rwxr-xr-x | Mechanics_of_Structures/Chapter1.ipynb | 745 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter10.ipynb | 661 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter11.ipynb | 527 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter12.ipynb | 850 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter13.ipynb | 366 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter2.ipynb | 309 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter3.ipynb | 283 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter5.ipynb | 1225 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter6.ipynb | 491 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter8.ipynb | 489 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/Chapter9.ipynb | 523 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/README.txt | 10 | ||||
| -rwxr-xr-x | Mechanics_of_Structures/screenshots/Chapter12.png | bin | 0 -> 159535 bytes | |||
| -rwxr-xr-x | Mechanics_of_Structures/screenshots/Chapter13.png | bin | 0 -> 135903 bytes | |||
| -rwxr-xr-x | Mechanics_of_Structures/screenshots/chapter3.png | bin | 0 -> 158772 bytes |
15 files changed, 6479 insertions, 0 deletions
diff --git a/Mechanics_of_Structures/Chapter1.ipynb b/Mechanics_of_Structures/Chapter1.ipynb new file mode 100755 index 00000000..339836e5 --- /dev/null +++ b/Mechanics_of_Structures/Chapter1.ipynb @@ -0,0 +1,745 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9ca9b2e0d1cc4391d74881e14e5e9fe9e8aa0cd9b6ad3418d2abbf628f065e94"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Simple stresses and strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg 5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ "#calculate the elongnation of the bar \n",
+ "P = 5.5 ;##Axial pull in tons\n",
+ "E = 13000 ;##modulus of elasticity tons/in^2\n",
+ "l = 120 ;##length in inches\n",
+ "A = math.pi/4. ;##Area of resisting section in^2\n",
+ "p = P/A ;##Intensity of stress in tons/in^2\n",
+ "e = p/E ;##strain\n",
+ "delta_l = l*e;##elongation of the bar in inches\n",
+ "print'%s %.4f %s'%('The elongation of the bar is',delta_l,'inch');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The elongation of the bar is 0.0646 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the minimum diameter d of each stay bolt\n",
+ "s_p = 200.;##steam pressure in lb/in^2\n",
+ "l = 4.;##length in inches\n",
+ "b = 4.;##breadth in inches\n",
+ "p = 14000.;##permissible streaa in lb/in^2\n",
+ "P = s_p*l*b;##Pull on each bolt in lb-wt\n",
+ "A = P/p ;##necessary area of bolt-section\n",
+ "d = math.sqrt(4*A/math.pi) ;##minimum diameter in inches\n",
+ "print'%s %.2f %s'%('The minimum diameter d of each stay bolt =',d,'inch');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum diameter d of each stay bolt = 0.54 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the safe load in tons\n",
+ "D = 8.;##external diameter in inches \n",
+ "d = 6.;##internal diameter in inches\n",
+ "sigma = 36.;##ultimate stress in tons/in^2\n",
+ "n = 6.;##safety factor\n",
+ "A = 0.25*math.pi*(D**2 - d**2);##Area of section in in^2\n",
+ "P = sigma*A; ##crushing load for the column in tons \n",
+ "P_safe = P/n ;##safe load in tons\n",
+ "print'%s %.2f %s'%('Safe load =',P_safe,'tons'); \n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe load = 131.95 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate compressive stress of the punch\n",
+ "sigma = 20.;##ultimate sheat stress in tons/in^2\n",
+ "d = 1./2.;##diameter of the hole in inches\n",
+ "t = 3./8.;##thickness of the plate in inches\n",
+ "A = 0.25*math.pi*d**2;##area of the cross-section of the punch in^2\n",
+ "P = math.pi*d*t*sigma;##necessary force in tons\n",
+ "sigma_comp = P/A;##compressive stress on the punch\n",
+ "print'%s %.1f %s'%('The compressive stress of the punch =',sigma_comp,'tons/in^2');\n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compressive stress of the punch = 60.0 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate change in volume \n",
+ "b = 8.;##width in inches\n",
+ "t = 3./8.;##thickness in inches\n",
+ "l = 20.;##length in feets\n",
+ "P = 22.;##pull in tons\n",
+ "E = 13500.;##modulus of elasticity in tons/in^2\n",
+ "sigma = 0.3;##poisson/s ratio\n",
+ "A = b*t;##in in^2\n",
+ "V = l*A*12;##in cub.inch\n",
+ "p = P/A;##in tons/in^2\n",
+ "e = p/E;\n",
+ "delta_l = e*l*12;##stretch of the bar in inches\n",
+ "Lateral_strain = e*sigma ;##lateral strain\n",
+ "del_b = b*Lateral_strain;##in inches\n",
+ "del_t = t*Lateral_strain;##in inches\n",
+ "k = e*(1-2*sigma);##(del_V)/(V)\n",
+ "del_V = k*V;##change in volume in cub.inch\n",
+ "print'%s %.3f %s'%('The change in volume is',del_V,'cub.inch');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in volume is 0.156 cub.inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate contraction in diameter and change in volume and work done in streching the bar \n",
+ "d = 7./8.;##diameter of the bar in inches\n",
+ "l = 10.;##length in feets\n",
+ "P = 6.;##axial pull in tons\n",
+ "E = 13000.;##modulus of elsticity in tons/in^2\n",
+ "m = 4.;\n",
+ "A = 0.25*math.pi*d**2;##in in^2\n",
+ "V = 0.25*math.pi*d**2*l*12;##volume in cub.inches\n",
+ "p = P/A;##in tons/in^2\n",
+ "e = p/E;\n",
+ "del_l = e*l*12;##stretchof the bar in inches\n",
+ "Lateral_strain = e/m ;##lateral strain\n",
+ "del_d = Lateral_strain*d;##Contraction in diameter in inches\n",
+ "print'%s %.4f %s'%('The Contraction in diameter is',del_d,'inches');\n",
+ "k = e*(1-2/m);##(del_V)/(V)\n",
+ "del_V = k*V;##change in volume in cub.inch\n",
+ "print'%s %.4f %s'%('The change in volume is',del_V,'cub. inch');\n",
+ "W = 0.5*P*del_l;##work done in stretching the bar in in-ton\n",
+ "print'%s %.4f %s'%('The work done in stretching the bar is',W,'in-ton');\n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Contraction in diameter is 0.0002 inches\n",
+ "The change in volume is 0.0277 cub. inch\n",
+ "The work done in stretching the bar is 0.2763 in-ton\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate total change in length of the bar and energy stored in bar \n",
+ "L = 24.;##length of the bar in ft\n",
+ "d1 = 9./8.;##diameter of the bar in inches\n",
+ "l1 = 6.;##in ft\n",
+ "d2 = 1.;##in inches\n",
+ "l2 = 12.;##in ft\n",
+ "d3 = 5./4.;##in inches\n",
+ "l3 = L-l1-l2;##in ft\n",
+ "P = 10000.;##axial compression in lb-wt\n",
+ "E = 28.*10**6;##modulus of elasticity in lb/in^2\n",
+ "A1 = 0.25*math.pi*d1**2;##in in^2\n",
+ "A2 = 0.25*math.pi*d2**2;##in in^2\n",
+ "A3 = 0.25*math.pi*d3**2;##in in^2\n",
+ "p1 = P/A1 ;##in lb/in^2\n",
+ "e1 = p1/E;\n",
+ "p2 = P/A2 ;##in lb/in^2\n",
+ "e2 = p2/E;\n",
+ "p3 = P/A3 ;##in lb/in^2\n",
+ "e3 = p3/E;\n",
+ "del_l1 = e1*l1*12;##in inches\n",
+ "del_l2 = e2*l2*12;##in inches\n",
+ "del_l3 = e3*l3*12;##in inches\n",
+ "del_l = del_l1+del_l2+del_l3;##total change in length in ft\n",
+ "W = 0.5*P*del_l/12;##energy stored in the bar in ft-lbs\n",
+ "print'%s %.3f %s'%('Total change in length of the bar is',del_l,'inches');\n",
+ "print'%s %.f %s'%('The energy stored in the bar is',W,'ft-lbs');\n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total change in length of the bar is 0.112 inches\n",
+ "The energy stored in the bar is 47 ft-lbs\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate change in lenght of the rod \n",
+ "P = 1200.;##axial pull in lb-wt\n",
+ "d1 = 1.;##diameter of one end in inches\n",
+ "d2 = 0.5;##diameter of other end in inches\n",
+ "l = 10.;##length of the rod in inches\n",
+ "E = 14.*10**6;##modulus of elsticity in lb/in^2\n",
+ "del_l = 4*P*l/(math.pi*E*d1*d2);##change in length in inches \n",
+ "print'%s %.4f %s'%('The change in length of the rod is',del_l,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in length of the rod is 0.0022 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the strain and extension of the bar and the work done in streching \n",
+ "d = 1.;##diameter of the steel bar in inches\n",
+ "l = 12.;##length of the steel bar in inches\n",
+ "d1 = 3./2.;##external diameter in inches\n",
+ "d2 = 1.;##internal diameter in inches\n",
+ "P = 5.;##axial pull in tons\n",
+ "E_s = 30.*10**6;##modulus of elasticity of steel in lb/in^2\n",
+ "E_b = 14.*10**6;##modulus of elasticity of brass in lb/in^2\n",
+ "A_s = 0.25*math.pi*d**2;##area of the steel section in in^2\n",
+ "A_b = 0.25*math.pi*(d1**2-d2**2);##area of the brass section in in^2\n",
+ "P_b = (P/((E_s/E_b)*A_s+A_b))*A_b;##load resisted by the brass tube in tons\n",
+ "P_s = P-P_b;##bal;ance load resisted by the steel tube\n",
+ "e = (P_b/A_b)*2240./E_b ;##strain\n",
+ "print'%s %.4f %s'%('The strain e =',e,'');\n",
+ "del_l = e*l ;##extension of the bar in inches\n",
+ "print'%s %.4f %s'%('The extension of the bar =',del_l,'inches');\n",
+ "W = 0.5*P*del_l;##work done in stretching in inch-ton\n",
+ "print'%s %.3f %s'%('The work done in stretching is',W,'inch-ton');"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The strain e = 0.0003 \n",
+ "The extension of the bar = 0.0036 inches\n",
+ "The work done in stretching is 0.009 inch-ton\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate safe central load and the reinforcing bar\n",
+ "a = 12.;##length of each side in inches\n",
+ "d = 9./8.;##diameter of each reinforced bar in inches\n",
+ "r = 3.;##distance of centre from the edges in inches\n",
+ "p_c = 600.;##in lb/in^2\n",
+ "n = 18.;##modular ration E_s/E_c\n",
+ "A_s = 4.*0.25*math.pi*d**2;##in in^2\n",
+ "A_c = a**2 - A_s;##in in^2\n",
+ "p_s = n*p_c;##in lb/in^2\n",
+ "P = p_s*A_s+p_c*A_c;##safe central load in lb-wt\n",
+ "print'%s %.f %s'%('Safe central load =',P,'lb-wt');\n",
+ "print'%s %.d %s'%('Of this, the reinforcing bars carry',p_s*A_s,'lb-wt',);\n",
+ "\n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe central load = 126956 lb-wt\n",
+ "Of this, the reinforcing bars carry 42941 lb-wt\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate P_sand p_b and x\n",
+ "l = 8.;##length in feet\n",
+ "d = 0.5;##diameter in inches\n",
+ "r = 30.;##distance between two rods in inches\n",
+ "P = 2000.;##load in lb-wt\n",
+ "E_s = 30.*10**6;##modulus of elsticity of steel rod\n",
+ "E_b = 16.*10**6;##modulus of elsticity of brass rod\n",
+ "A_s = 0.25*math.pi*d**2;##section area in in**2\n",
+ "p_b = P/(A_s*(1+(E_s/E_b)));\n",
+ "p_s = (P/A_s) - p_b ;\n",
+ "P_b = A_s*p_b;\n",
+ "P_s = A_s*p_s;\n",
+ "print'%s %.1f %s'%('P_s =',P_s,'lb/in**2')\n",
+ "print'%s %.1f %s'%('and P_b =',P_b,'lb/in**2');\n",
+ "x = r*P_b/P ;##\n",
+ "print'%s %.2f %s'%('x =',x,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "P_s = 1304.3 lb/in**2\n",
+ "and P_b = 695.7 lb/in**2\n",
+ "x = 10.43 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#calculate tensile strain imposed by end grips and p and P\n",
+ "alpha = 0.0000062 ;##co-efficient of the expansion in \"per F\"\n",
+ "t = 100.;##in F\n",
+ "d = 3/4.;##in inches\n",
+ "D = 0.02;##in inches\n",
+ "l = 15.;##in ft\n",
+ "E = 13000.;##in tons/in**2\n",
+ "e = alpha*t - (D/(l*12));\n",
+ "p = E*e;##in tons/in**2\n",
+ "A = 0.25*math.pi*d**2;##in in**2\n",
+ "P = p*A ;##in tons\n",
+ "print'%s %.4f %s'%('Tensile strain imposed by end-grips,e =',e,'');\n",
+ "print'%s %.2f %s'%('p =',p,'tons/in**2')\n",
+ "print'%s %.4f %s'%('P =',P,'tons');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Tensile strain imposed by end-grips,e = 0.00051 \n",
+ "p = 6.62 tons/in**2\n",
+ "P = 2.9227 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the stresses induced in each metal and p_s\n",
+ "d = 1.;##diameter of steel bar in inches\n",
+ "d1 = 3./2.;##external diameter of brass tube in inches\n",
+ "d2 = 1.;##internal diameter of brass tube in inches\n",
+ "t = 100.;##in F\n",
+ "alpha_s = 0.0000062;##alpha of steel in \"per F\"\n",
+ "alpha_b = 0.000010;##alpha of brass in \"per F\"\n",
+ "E_s = 30.*10**6;##in lb/in^2\n",
+ "E_b = 14.*10**6;##in lb/in^2\n",
+ "A_s = 0.25*math.pi*d**2;##section area of steel bar in in^2\n",
+ "A_b = 0.25*math.pi*(d1**2-d2**2);##section area of brass tube in in^2\n",
+ "p_b = t*(alpha_b-alpha_s)*E_s/((A_b/A_s)+(E_s/E_b));\n",
+ "p_s = (A_b/A_s)*p_b;\n",
+ "print'%s %.d %s'%('The stresses induced in each metal are, p_b =',p_b,'lb/in^2')\n",
+ "print'%s %.d %s'%('p_s =',p_s,'lb/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stresses induced in each metal are, p_b = 3360 lb/in^2\n",
+ "p_s = 4200 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the least temperature the tube must be heated\n",
+ "D = 4.;##diameter of the wheel in ft\n",
+ "p = 6. ;##hoop stress in tons/in^2\n",
+ "alpha = 0.0000062;##in \"per F\" \n",
+ "E = 13000.;##in tons/in^2\n",
+ "d = (1./(1.+(p/E)))*D*12.;##internal diameter in inches\n",
+ "t = (D*12.-d)/(d*alpha);\n",
+ "print'%s %.1f %s'%('The least temperature the tube must be heated is, t =',t,'F');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The least temperature the tube must be heated is, t = 74.4 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Resultant stress intensity and normal stress intensity and tangential stress intensity and The maximum possible shear on any plane and these planes are inclined at',angle,'degrees to the normal section.\n",
+ "p = 8.;##normal stress intensity in tons/in^2\n",
+ "theta = 35.*math.pi/180.;##inclination of the section in degrees\n",
+ "P = p*math.cos(theta);##resultant stress intensity in tons/in^2\n",
+ "p_n = P*math.cos(theta);##normal stress intensity in tons/in^2\n",
+ "p_t = P*math.sin(theta);##tangential stress intensity in tons/in^2\n",
+ "p_max = 0.5*p;##maximum possible shear in tons/in^2\n",
+ "angle = 45.;##inclination of these planes in degrees\n",
+ "print'%s %.2f %s'%('Resultant stress intensity =',P,'tons/in^2');\n",
+ "print'%s %.2f %s'%('normal stress intensity =',p_n,'tons/in^2');\n",
+ "print'%s %.2f %s'%('tangential stress intensity =',p_t,'tons/in^2');\n",
+ "print'%s %.d %s'%('The maximum possible shear on any plane is',p_max,'tons/in^2');\n",
+ "print'%s %.d %s'%('and these planes are inclined at',angle,'degrees to the normal section.');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant stress intensity = 6.55 tons/in^2\n",
+ "normal stress intensity = 5.37 tons/in^2\n",
+ "tangential stress intensity = 3.76 tons/in^2\n",
+ "The maximum possible shear on any plane is 4 tons/in^2\n",
+ "and these planes are inclined at 45 degrees to the normal section.\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate poisson ratio and E,N,K,\n",
+ "d = 9./8.;##diameter of the steel bar in inches\n",
+ "P = 6.;##tensile load in tons\n",
+ "del_l = 0.0036 ;##extension of length inches\n",
+ "l = 8.;##gauge length in inches\n",
+ "del_d = 0.00015;##change in diameter in inches\n",
+ "A = 0.25*math.pi*d**2;##section area in in^2\n",
+ "p = P/A;##stress in tons/in^2\n",
+ "e = del_l/l;##strain\n",
+ "E = p/e;##modulus of elasticity in tons/in^2\n",
+ "LS = del_d/d;##lateral strain \n",
+ "PR = LS/e;##poisson's ratio \n",
+ "N = E/(2.*(1.+PR));##rigidity modulus in tons/in^2\n",
+ "K = E/(3.*(1.-2.*PR));##bulk modulus in tons/in^2\n",
+ "print'%s %.4f %s'%('Poisson ratio 1/m =',PR,'');\n",
+ "print'%s %.d %s'%('E =',E,'tons/in^2');\n",
+ "print'%s %.d %s'%('N =',N,'tons/in^2');\n",
+ "print'%s %.d %s'%('K =',K,'tons/in^2');\n",
+ "\n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson ratio 1/m = 0.2963 \n",
+ "E = 13413 tons/in^2\n",
+ "N = 5173 tons/in^2\n",
+ "K = 10974 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Poisson ratio and E\n",
+ "N = 2640.;##rigidity modulus in tons/in^2\n",
+ "d = 3./8.;##diameter of the rod in inches\n",
+ "P = 1./2.;##axial pull in tons\n",
+ "del_d = 0.000078;##change in diameter in inches\n",
+ "A = 0.25*math.pi*d**2;##section area in in^2\n",
+ "p = P/A ;##stress tons/in^2\n",
+ "LS = del_d/d;##lateral strain\n",
+ "m = p/(LS*2.*N) - 1.;\n",
+ "E = 2.*N*(1. + 1./m);##modulus of elasticity in ton/in^2\n",
+ "PR = 1./m;##poisson's ratio \n",
+ "print'%s %.3f %s'%('Poisson ratio 1/m =',PR,'');\n",
+ "print'%s %.d %s'%('E =',E,'ton/in^2');\n",
+ "\n",
+ "##there is an error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson ratio 1/m = 0.320 \n",
+ "E = 6971 ton/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter10.ipynb b/Mechanics_of_Structures/Chapter10.ipynb new file mode 100755 index 00000000..5ba91400 --- /dev/null +++ b/Mechanics_of_Structures/Chapter10.ipynb @@ -0,0 +1,661 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:65543f2d4bb6fb1415ff8fc05c077edf5e538efef4c593534466949561fc7f05"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Radial pressure-cylindrical and spherical shells"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The necessary thickness of metal for seamless pipe \n",
+ "d = 2.;## feet\n",
+ "p = 250.;## lb/in^2\n",
+ "f = 12000.;## lb/in^2\n",
+ "t_limit = p*d*12/(2*f) ;## inches\n",
+ "print'%s %.2f %s'%('The necessary thickness of metal for seamless pipe is',t_limit,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The necessary thickness of metal for seamless pipe is 0.25 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Maximum intensity of shear stress induced \n",
+ "l = 8.;##feet\n",
+ "d = 3.;## feet\n",
+ "t = 1/2.;## inches\n",
+ "p = 200.;## lb/in**2\n",
+ "E = 30.*10**6.;## lb/in**2\n",
+ "PR = 0.3;## poisson's ratio\n",
+ "f1 = p*d*12./(2.*t);## lb/in**2\n",
+ "f2 = p*d*12./(4.*t);## lb/in**2\n",
+ "f_s = 0.5*(f1-f2);## lb/in**2\n",
+ "e1 = (f1/E)-(PR*f2/E);## lb/in**2\n",
+ "e2 = (f2/E)-(PR*f1/E);## lb/in**2\n",
+ "del_d = e1*d*12.;## inches\n",
+ "del_l = e2*l*12.;## inches\n",
+ "del_V = (e2+2.*e1)*0.25*math.pi*(12*d)**2. * l*12.;## cub. inches\n",
+ "print'%s %.d %s'%('Maximum intensity of shear stress induced =',f_s,'lb/in**2')\n",
+ "\n",
+ "print'%s %.6f %s'%('del_d =',del_d,'inches')\n",
+ "print'%s %.6f %s'%('del_l = ',del_l,'inches')\n",
+ "print'%s %.1f %s'%('del_V =',del_V,'cub. inches')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum intensity of shear stress induced = 1800 lb/in**2\n",
+ "del_d = 0.007344 inches\n",
+ "del_l = 0.004608 inches\n",
+ "del_V = 44.6 cub. inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Thickness of metal required\n",
+ "d = 30.;## inches\n",
+ "H = 300.;## feet\n",
+ "w = 62.5;\n",
+ "f = 2800.;\n",
+ "##intensity of water pressur\n",
+ "p = w*H/144.;## lb/in**2\n",
+ "t_limit = p*d/(2*f);## inches\n",
+ "print'%s %.4f %s'%('Thickness of metal required is',t_limit, 'inches')\n",
+ "\n",
+ "\n",
+ "##the answer is correct only, but it is approximated in the text book.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of metal required is 0.6975 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calcualte The permissible steam pressure and circumferential stress and longitudinal sretss \n",
+ "d = 78.;## inches\n",
+ "t = 3/4.;## inches\n",
+ "n1 = 70/100.;## efficiency of the longitudinal riveted joint\n",
+ "f = 6.;## tons/in^2\n",
+ "n2 = 60/100.;## efficiency of the circumferential riveted joint\n",
+ "p = f*2240./(d/(2.*t*n1));##lb/in^2\n",
+ "p = round(p-1);\n",
+ "f1 = p*d/(2.*t);## lb/in^2\n",
+ "f2 = p*d/(4.*t*n2);## lb/in^2\n",
+ "print'%s %.d %s'%('The permissible steam pressure, p = ',p,'lb/in^2');\n",
+ "print'%s %.d %s'%('The circumferential stress, f1 =',f1,'lb/in^2')\n",
+ "print'%s %.2f %s'%('The circumferential stress, f1= ',f1/2240,'tons/in^2');\n",
+ "print'%s %.d %s'%('The longitudinal stress, f2 =',f2,'lb/in^2') \n",
+ "print'%s %.2f %s'%('The longitudinal sretss, f2 =',f2/2240,'lb/in^2')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The permissible steam pressure, p = 180 lb/in^2\n",
+ "The circumferential stress, f1 = 9360 lb/in^2\n",
+ "The circumferential stress, f1= 4.18 tons/in^2\n",
+ "The longitudinal stress, f2 = 7800 lb/in^2\n",
+ "The longitudinal stress, f2 = 3.48 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The thickness of the plate required\n",
+ "d = 4.;## feet\n",
+ "p = 200.;## lb/in^2\n",
+ "f = 15000.;## lb/in^2\n",
+ "n = 0.7;## efficiency\n",
+ "t_limit = p*d*12/(4*f*n);## inches\n",
+ "print'%s %.2f %s'%('The thickness of the plate required =',t_limit,'inches');\n",
+ "\n",
+ "##the answer is correct only, but it is approximated in the text book.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of the plate required = 0.23 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The pressure exerted by fluid on the shell\n",
+ "d = 3.;## feet\n",
+ "t = 1/4.;## inches\n",
+ "del_V = 9.;## cub. inches\n",
+ "E = 30*10**6;## lb/in^2\n",
+ "PR = 0.3;## poisson's ratio\n",
+ "V = (math.pi/6.)*(12.*d)**3;## in^3\n",
+ "k = del_V/V;\n",
+ "f = k*E/(3.*(1.-PR));## lb/in^2\n",
+ "p = 4.*f*t/(12.*d);## lb/in^2\n",
+ "print'%s %.d %s'%('The pressure exerted by fluid on the shell, p =',p,'lb/in^2');\n",
+ "\n",
+ "##there is a minute calculation error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure exerted by fluid on the shell, p = 146 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calcualte the Pipe and Steel-wire of intiually due to p and finally due to p\n",
+ "d = 12.;## feet\n",
+ "t = 1/2.;## inches\n",
+ "d1 = 1/4.;## inches\n",
+ "p = 500.;## lb/in^2\n",
+ "E_c = 6000.;## tons/in^2\n",
+ "PR = 0.3;## Poisson's ratio\n",
+ "E_s = 13000.;## tons/in^2\n",
+ "f_t = 8000.;## lb/in^2\n",
+ "l = 8.;\n",
+ "P_c = l*(math.pi/64.)*f_t;## lb-wt\n",
+ "f_c = P_c/(2.*t);## lb/in^2\n",
+ "##bursting force per inch unit length\n",
+ "f_b = p*d;## lb-wt\n",
+ "f_p = (f_b + (l*0.049*PR*p*d/(4.*t))*(E_s/E_c))/(1 + (l*0.049)*E_s/E_c);## lb/in^2 \n",
+ "f_w = (f_p - PR*p*d/(4.*t))*E_s/E_c;## lb/in^2\n",
+ "\n",
+ "print(' Pipe Steel-wire');\n",
+ "print'%s %.d %s'%(' Initually,',f_c,'lb/in^2.,compr.')\n",
+ "print'%s %.d %s'%('Intiually',f_t,'lb/in^2., tensile')\n",
+ "print'%s %.d %s'%( 'Due to p, ',f_p,'lb/in^2.,tensile.')\n",
+ "print'%s %.d %s'%('Due to p',f_t,'lb/in^2., tensile')\n",
+ "print'%s %.d %s'%(' Finally, ',f_p-f_c,'lb/in^2.,tensile.')\n",
+ "print'%s %.d %s'%('Finally',f_w+f_t,'lb/in^2., tensile')\n",
+ "\n",
+ "##there is a calculation error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Pipe Steel-wire\n",
+ " Initually, 3141 lb/in^2.,compr.\n",
+ "Intiually 8000 lb/in^2., tensile\n",
+ "Due to p, 3657 lb/in^2.,tensile.\n",
+ "Due to p 8000 lb/in^2., tensile\n",
+ " Finally, 516 lb/in^2.,tensile.\n",
+ "Finally 13975 lb/in^2., tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'The maximum intensities of circumferential stresses are\n",
+ "d = 12.;## inches\n",
+ "t = 3.;## inches\n",
+ "p_x1 = 900.;## lb/in^2\n",
+ "x1 = 0.5*d;## inches\n",
+ "p_x2 = 0.;\n",
+ "x2 = 0.5*d+t;## inches\n",
+ "##from Lame's formulae\n",
+ "b = (p_x1-p_x2)/((1/x1**2)-(1/x2**2));\n",
+ "a = (b/x1**2)- p_x1;\n",
+ "f_x1 = (b/x1**2)+a;## lb/in^2\n",
+ "f_x2 = (b/x2**2)+a;## lb/in^2\n",
+ "print'%s %.d %s'%('The maximum intensities of circumferential stresses are: f_6 =',f_x1,'lb/in^2.,tensile') \n",
+ "print'%s %.d %s'%('The minimum intensities of circumferential stresses are: f_6f_9 =',f_x2,'lb/in^2., tensile')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum intensities of circumferential stresses are: f_6 = 2340 lb/in^2.,tensile\n",
+ "The minimum intensities of circumferential stresses are: f_6f_9 = 1440 lb/in^2., tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# calculate 'The maximum intensities of circumferential stresses are and the outer diameter\n",
+ "d = 5.;## inches \n",
+ "p = 3.;## tons/in^2\n",
+ "f = 8.;## tons/in^2\n",
+ "x = 0.5*d;## inches\n",
+ "b = (p+f)/(2/x**2);\n",
+ "a = f-(b/x**2); \n",
+ "r = math.sqrt(b/a);##outer radius\n",
+ "t = r-0.5*d;##thickness\n",
+ "D = 2.*t+d;##outer diameter\n",
+ "print'%s %.3f %s'%('The thickness of metal necessary, t =',t,'inches');\n",
+ "print'%s %.1f %s'%('the outer diameter will be, D =',D,'inches');\n",
+ "\n",
+ "##the answer is correct, but it is approximated in the text book.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of metal necessary, t = 1.208 inches\n",
+ "the outer diameter will be, D = 7.4 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg399"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# calculate The outer radius in all four cases\n",
+ "d = 9.;## inches\n",
+ "p = 5000/2240.;## lb/in**2\n",
+ "f = 8.;## tons/in**2\n",
+ "PR = 0.3;## Poisson's ratio\n",
+ "##(i) Maximum principal stress hypothesis:\n",
+ "k_limit1 = math.sqrt((f + p)/(f - p));##k_limit = r1/r2\n",
+ "r_limit1 = k_limit1*0.5*d;##inches\n",
+ "print'%s %.3f %s'%('The outer radius in case(i), r2 =',r_limit1,'inches');\n",
+ "##(ii) Maximum principal strain:\n",
+ "k_limit2 = math.sqrt(((f/p - PR)+1)/(f/p - PR -1));\n",
+ "r_limit2 = k_limit2*0.5*d;## inches\n",
+ "print'%s %.3f %s'%(' The outer radius in case(ii), r2 =',r_limit2,'inches');\n",
+ "##(iii) Maximum shear stress:\n",
+ "k_limit3 = math.sqrt(f/(2*p) /((f/(2*p)) - 1));\n",
+ "r_limit3 = k_limit3*0.5*d;## inches\n",
+ "print'%s %.3f %s'%('The outer radius in case(iii), r2 =',r_limit3,' inches');\n",
+ "##(iv) Maximum strain energy\n",
+ "K1 = (f**2 /p**2)/(2*((f**2 /(2*p**2)) - (1+PR)));\n",
+ "K2 = K1**2;\n",
+ "K3 = ((f**2 /(2*p**2)) - (1-PR))/((f**2 /(2*p**2)) - (1+PR));\n",
+ "k_limit4 =math. sqrt(K1+math.sqrt(K2-K3));\n",
+ "r_limit4 = k_limit4*0.5*d;## inches\n",
+ "print'%s %.3f %s'%(' The outer radius in case(iv), r2 =',r_limit4,' inches');\n",
+ "\n",
+ "##there are calculation errors in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The outer radius in case(i), r2 = 5.994 inches\n",
+ " The outer radius in case(ii), r2 = 6.163 inches\n",
+ "The outer radius in case(iii), r2 = 6.769 inches\n",
+ " The outer radius in case(iv), r2 = 6.249 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# calculate hoop stress at x = 3,6 inches,x = 5 inches initially on inner tube and outer tube\n",
+ "r1 = 0.5*6.;## inches\n",
+ "r2 = 0.5*12.;## inches\n",
+ "r3 = 0.5*10.;## inches\n",
+ "p = 1500.;## lb/in**2\n",
+ "p_f = 12000.;## lb/in**2\n",
+ "##Initially, for the inner tube\n",
+ "b = -p/((1./r1**2.) - (1./r3**2.));\n",
+ "a = b/r1**2.;\n",
+ "f_3 = (b/r1**2.) +a;## lb/in**2\n",
+ "f_5 = (b/r3**2.) +a;## lb/in**2\n",
+ "##for the outer tube\n",
+ "b1 = p/((1/r3**2.)-(1/r2**2.));\n",
+ "a1 = b1/r2**2.;\n",
+ "f1_5 = (b1/r3**2.)+a1;## lb/in**2\n",
+ "f1_6 = (b1/r2**2.)+a1;## lb/in**2\n",
+ "##When the fluid pressure of 12000 lb/in**2, is admitted into the compound tube\n",
+ "B = p_f/((1./r1**2.)-(1/r2**2.));\n",
+ "A = B/(r2**2.);\n",
+ "f_3_ = (B/r1**2.)+A;## lb/in**2\n",
+ "f_5_ = (B/r3**2.)+A;## lb/in**2\n",
+ "f_6_ = (B/r2**2.)+A;## lb/in**2\n",
+ "\n",
+ "print('The hoop stresse are');\n",
+ "print'%s %.1f %s %.1f %s '%(' at x = 3 inches,x = 5 inches initially on inner tube are ',-f_3,' lb/in^2.., compressive, ',-f_5,' lb/in^2..,compressive respectively');\n",
+ "print'%s %.1f %s %.1f %s'%(' at x = 5 inches,x = 6 inches initially on outer tube are ',f1_5,' lb/in^2.., tensile, ',f1_6,' lb/in^2..,tensile respectively');\n",
+ "print'%s %.d %s %.d %s %.d %s'%(' at x = 3 inches,x = 5 inches and x = 6 inches due to fluid pressure are ',f_3_,' lb/in^2.., tensile, ',f_5_,' lb/in^2..,tensile, and',f_6_,' lb/in^2..,tensile respectively')\n",
+ "print'%s %.1f %s %.1f %s'%(' at x = 3 inches,x = 5 inches finally on inner tube are ',f_3_+f_3,' lb/in^2.., tensile, ',f_5_+f_5,' lb/in^2..,tensile respectively');\n",
+ "print'%s %.d %s %.d %s'%(' at x = 5 inches,x = 6 inches finally on outer tube are',f1_5+f_5_,' lb/in^2.., tensile, ',f1_6+f_6_,' lb/in^2.., tensile respectively');"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hoop stresse are\n",
+ " at x = 3 inches,x = 5 inches initially on inner tube are 4687.5 lb/in^2.., compressive, 3187.5 lb/in^2..,compressive respectively \n",
+ " at x = 5 inches,x = 6 inches initially on outer tube are 8318.2 lb/in^2.., tensile, 6818.2 lb/in^2..,tensile respectively\n",
+ " at x = 3 inches,x = 5 inches and x = 6 inches due to fluid pressure are 20000 lb/in^2.., tensile, 9760 lb/in^2..,tensile, and 8000 lb/in^2..,tensile respectively\n",
+ " at x = 3 inches,x = 5 inches finally on inner tube are 15312.5 lb/in^2.., tensile, 6572.5 lb/in^2..,tensile respectively\n",
+ " at x = 5 inches,x = 6 inches finally on outer tube are 18078 lb/in^2.., tensile, 14818 lb/in^2.., tensile respectively\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The minimum temperature to which outer tube should be heated before it can be slipped\n",
+ "p = 1500.;## lb/in^2\n",
+ "E = 30*10**6;## lb/in^2\n",
+ "f1_5 = 8318.;## lb/in^2\n",
+ "f2_5 = 3187.5;## lb/in^2\n",
+ "alpha = 0.0000062;## per F\n",
+ "r3 = 6;## inches\n",
+ "del_r3 = r3*(f1_5+f2_5)/E;## inches\n",
+ "t = ((f1_5+f2_5)/E)/(alpha);## inches\n",
+ "print'%s %.2f %s'%('The minimum temperature to which outer tube should be heated before it can be slipped on, t =',t,'F');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum temperature to which outer tube should be heated before it can be slipped on, t = 61.86 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The hoop stresses are as under for inner tube at x=1/5 and 3 and 4.5 inches \n",
+ "r1 = 0.5*9.;## inches\n",
+ "r2 = 0.5*3.;## inches\n",
+ "r3 = 0.5*6.;## inches\n",
+ "del_r3 = 0.5*0.003;## inches\n",
+ "E = 13000.;## tons/in^2\n",
+ "k1 = r1/r3;\n",
+ "k2 = r2/r3;\n",
+ "a1 = (del_r3/r3)*E/((k1**2 +1)- (k2**2 +1)*(k1**2 -1)/(k2**2 -1));\n",
+ "a = a1*(k1**2 -1)/(k2**2 -1);\n",
+ "b1 = a1*r1**2;\n",
+ "b = a*r2**2;\n",
+ "p_ = (b/r3**2) -a;## tons/in^2\n",
+ "## for the inner tube\n",
+ "f_x1 = (b/r2**2) +a;## tons/in^2\n",
+ "f_x2 = (b/r3**2) +a;## tons/in^2\n",
+ "## for the outer tube\n",
+ "f_x3 = (b1/r3**2) +a1;## tons/in^2\n",
+ "f_x4 = (b1/r1**2) +a1;## tons/in^2\n",
+ "print ('The hoop stresses are as under:');\n",
+ "print'%s %.2f %s'%('For the inner tube, at x = 1/5 inches, f =',-f_x1,'tons/in^2., compressive')\n",
+ "print'%s %.2f %s'%('at x = 3 inches, f =',-f_x2,'tons/in^2.,compressive')\n",
+ "print'%s %.2f %s'%(' For the outer tube, at x = 3 inches, f =',f_x3,'tons/in^2., tensile')\n",
+ "\n",
+ "print'%s %.2f %s'%('at x = 4.5 inches, f =',f_x4,' tons/in^2.,tensile');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hoop stresses are as under:\n",
+ "For the inner tube, at x = 1/5 inches, f = 4.06 tons/in^2., compressive\n",
+ "at x = 3 inches, f = 2.54 tons/in^2.,compressive\n",
+ " For the outer tube, at x = 3 inches, f = 3.96 tons/in^2., tensile\n",
+ "at x = 4.5 inches, f = 2.44 tons/in^2.,tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the thickness of the shell required\n",
+ "r1 = 0.5*5.;## inches\n",
+ "p = 5000.;## lb/in^2\n",
+ "f = 5.;## tons/in^2\n",
+ "b = (f + p/2240.)/((1/r1**3) + (2/r1**3));\n",
+ "a = f - (b/r1**3);\n",
+ "##external diameter\n",
+ "r = (2*b/a)**(1/3);## inches\n",
+ "t = r - r1;## inches\n",
+ "print'%s %.3f %s'%('The thickness of the shell required, t =',t,'inches');\n",
+ "\n",
+ "##the answer is approximated in the text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of the shell required, t = -1.500 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter11.ipynb b/Mechanics_of_Structures/Chapter11.ipynb new file mode 100755 index 00000000..be126298 --- /dev/null +++ b/Mechanics_of_Structures/Chapter11.ipynb @@ -0,0 +1,527 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:efd3a63c379116e766993d1d8b2baacab12f4039280250b30fc0874f009306a1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Chapter11-Reveted joints"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg424"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The efficiency of the joint\n",
+ "t = 5/8.;## inch\n",
+ "d = 1.;## inch\n",
+ "p = 4.;## inches\n",
+ "f_t = 28.; ##tons/in^2\n",
+ "f_s = 20.; ##tons/in^2\n",
+ "f_b = 40.; ##tons/in^2\n",
+ "P_t = (p-d)*t*f_t;## tons\n",
+ "P_s = 2*2*0.25*math.pi*d**2 *f_s;## tons\n",
+ "P_b = 2*d*t*f_b;##tons\n",
+ "P = p*t*f_t;## tons\n",
+ "n = min(P_t,P_s,P_b)/P ;## efficiency\n",
+ "print'%s %.1f %s'%('The efficiency of the joint =',n,'f')\n",
+ "print'%s %.1f %s'%('or',n*100,'percentage')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiency of the joint = 0.7 f\n",
+ "or 71.4 percentage\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg425"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The efficiency of first joint with required conditions\n",
+ "t = 1/2.;## inches\n",
+ "d1= 7/8.;## inches\n",
+ "p1 = 5/2.;## inches\n",
+ "d2= 9/8.;## inches\n",
+ "p2 = 7/2.;## inches\n",
+ "f_t = 8.;## tons/in^2\n",
+ "f_s = 6.;## tons/in^2\n",
+ "f_b = 10.;## tons/in^2\n",
+ "\n",
+ "P_t1 = (p1-d1)*t*f_t;## tons\n",
+ "P_s1 = 0.25*math.pi*d1**2 *f_s;## tons\n",
+ "P_b1 = d1*t*f_b;##tons\n",
+ "P1 = p1*t*f_t;## tons\n",
+ "n1 = min(P_t1,P_s1,P_b1)/P1 ;## efficiency\n",
+ "print'%s %.3f %s'%('The efficiency of first joint = ',n1,'')\n",
+ "print'%s %.1f %s'%('or =',n1*100,'percentage')\n",
+ "\n",
+ "P_t2 = (p2-d2)*t*f_t;## tons\n",
+ "P_s2 = 0.25*math.pi*d2**2 *f_s;## tons\n",
+ "P_b2 = d2*t*f_b;##tons\n",
+ "P2 = p2*t*f_t;## tons\n",
+ "n2 = min(P_t2,P_s2,P_b2)/P2 ;## efficiency\n",
+ "print'%s %.3f %s'%('The efficiency of second joint = ',n2,'f') \n",
+ "print'%s %.1f %s'%('or= ',n2*100,'percentage')\n",
+ "\n",
+ "if n2 > n1: \n",
+ " print(' The second joint, with its higher efficiency, is stronger');\n",
+ "else:\n",
+ " print(' The first joint, with its higher efficiency, is stronger');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiency of first joint = 0.361 \n",
+ "or = 36.1 percentage\n",
+ "The efficiency of second joint = 0.402 f\n",
+ "or= 40.2 percentage\n",
+ " The second joint, with its higher efficiency, is stronger\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'The efficiency of the joint\n",
+ "t = 3/8.;## inches\n",
+ "p2 = 7/2.;## inches\n",
+ "f_t = 11/2.;## tons/in^2\n",
+ "f_s = 5.;## tons/in^2\n",
+ "f_b = 12.;## tons/in^2\n",
+ "d = 1.2*math.sqrt(t);## inches\n",
+ "##d = 0.735, say 0.75 inches\n",
+ "d = 0.75;## inches\n",
+ "P_s = 0.25*math.pi*d**2 *f_s;## tons\n",
+ "P_b = d*t*f_b;##tons\n",
+ "P_t_limit = P_s;##tons\n",
+ "p_limit = P_s/(t*f_t) + d;## inches\n",
+ "##p_limit = 1.763, take p = 1.75\n",
+ "p = 1.75;## inches\n",
+ "n = (p-d)/p;## efficiency\n",
+ "print'%s %.1f %s'%('The efficiency of the joint = ',n*100,' percentage')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiency of the joint = 57.1 percentage\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg428"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Pitch and efficency\n",
+ "d = 7/8.;## inches\n",
+ "t = 1/2.;## inches \n",
+ "f_t = 6.;## tons/in^2\n",
+ "f_s = 5.;## tons/in^2\n",
+ "f_b = 10.;## tons/in^2\n",
+ "p_s = 2*0.25*math.pi*d**2*f_s;## tons\n",
+ "P_b = d*t*f_b;## tons\n",
+ "p_t_limit = 2*P_b/3 + d;## inches\n",
+ "n = (p_t_limit-d)/p_t_limit;## efficiency\n",
+ "print'%s %.3f %s'%('Pitch, p =',p_t_limit,' inches')\n",
+ "\n",
+ "print'%s %.1f %s'%('Efficiency = ',n,' ')\n",
+ "print'%s %.d %s'%('or',n*100,' percentage');\n",
+ "\n",
+ "##the answer is approximated in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch, p = 3.792 inches\n",
+ "Efficiency = 0.8 \n",
+ "or 76 percentage\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg430"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Pitch and The efficiency of the joint \n",
+ "d = 6.;## feet\n",
+ "p = 180.;## lb/in^2\n",
+ "f = 6.;## tons\n",
+ "n = 70./100.;## efficiency\n",
+ "d1 = 1.;## inches\n",
+ "f_s = 5.;## tons/in^2\n",
+ "f_b = 10.;## tons/in^2\n",
+ "t = p*d/(2.*f*n);## inches\n",
+ "## t = 0.6889 inches, say 0.75 inches\n",
+ "t = 0.75;##inches\n",
+ "P_s = 2*0.25*math.pi*d1**2*f_s;## tons\n",
+ "P_b = d1*t*f_b;## tons\n",
+ "p_limit = 2*P_b/(t*f) + d1;## inches\n",
+ "##p_limit = 4.33 inches, make it 4 inches\n",
+ "p = round(p_limit);## inches\n",
+ "n1 = (p-d1)/p;## efficiency\n",
+ "print'%s %.2f %s'%('Pitch = ',p_limit,'inches')\n",
+ "print'%s %.d %s'%('make it',p,' inches.')\n",
+ "print'%s %.d %s'%(' The efficiency of the joint will be',n1*100,'percentage')\n",
+ "print'%s %.d %s'%('percentage aganist the assumed value of ',n*100,' percentage.')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch = 4.33 inches\n",
+ "make it 4 inches.\n",
+ " The efficiency of the joint will be 75 percentage\n",
+ "percentage aganist the assumed value of 70 percentage.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg433"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The number of rivets required and The efficiency of the joint and The actual stresses induce in the rivet and The tensile stress at section 11 ,22 33 44\n",
+ "t = 1./2.;## inches\n",
+ "a = 1./2.;## inches\n",
+ "P = 42.;## tons\n",
+ "d = 3/4.;## inches\n",
+ "f_t = 7.5;## tons/in^2\n",
+ "f_s = 6.;## tons/in^2\n",
+ "f_b = 12.;## tons/in^2\n",
+ "P_s = 2*0.25*math.pi*d**2 *f_s;## tons\n",
+ "P_b = d*t*f_b;## tons\n",
+ "n = P/min(P_s,P_b);\n",
+ "n = round(n+1);\n",
+ "b1 = P/(t*f_t) + d;## inches\n",
+ "b = round(b1);\n",
+ "e = (b-d)/b;## efficiency\n",
+ "f_s = (P/n)/(2*0.25*math.pi*d**2) ;## tons/in^2\n",
+ "f_b = (P/n)/(d*t);## tons/in^2\n",
+ "f1 = P/(a*(b-d));## tons/in^2\n",
+ "f2 = (P-(P/n))/((b-2*d)*t);## tons/in^2\n",
+ "f3 = (P-(3.*P/n))/((b-3.*d)*t);## tons/in^2\n",
+ "f4 = (P-(6.*P/n))/((b-4.*d)*t);## tons/in^2\n",
+ "print'%s %.d %s'%('The number of rivets required, n = ',n,'');\n",
+ "print'%s %.2f %s %.d %s'%(' The width of the flat required, b = ',b1,'inches'and' ',b,' inches');\n",
+ "print'%s %.2f %s '%(' The efficiency of the joint = ',e*100,' percentage');\n",
+ "print'%s %.2f %s %.2f %s'%(' The actual stresses induce in the rivet are, f_s = ',f_s,' tons/in^2' and ' f_b = ',f_b ,'tons/in^2');\n",
+ "print'%s %.3f %s'%(' The tensile stress at section 11, f1 = ',f1,'rons/in^2');\n",
+ "print'%s %.3f %s'%(' The tensile stress at section 22, f2 = ',f2,' rons/in^2');\n",
+ "print'%s %.3f %s'%(' The tensile stress at section 33, f3 = ',f3,' rons/in^2');\n",
+ "print'%s %.3f %s'%(' The tensile stress at section 44, f4 = ',f4,' rons/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of rivets required, n = 10 \n",
+ " The width of the flat required, b = 11.95 12 inches\n",
+ " The efficiency of the joint = 93.75 percentage \n",
+ " The actual stresses induce in the rivet are, f_s = 4.75 f_b = 11.20 tons/in^2\n",
+ " The tensile stress at section 11, f1 = 7.467 rons/in^2\n",
+ " The tensile stress at section 22, f2 = 7.200 rons/in^2\n",
+ " The tensile stress at section 33, f3 = 6.031 rons/in^2\n",
+ " The tensile stress at section 44, f4 = 3.733 rons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate The number of rivets required and The efficiency of the joint and The pull section at different points and The maximum possible pull which the flat will safely transmit\n",
+ "import math\n",
+ "\n",
+ "b = 9.;## inches\n",
+ "t = 3./4.;## inches\n",
+ "f_t = 8.;## tons/in**2\n",
+ "f_s = 5.;## tons/in**2\n",
+ "f_b = 10.;## tons/in**2\n",
+ "d = 7/8.;## inches\n",
+ "P = (b-d)*t*f_t;## tons\n",
+ "P_s = 2.*0.25*math.pi*d**2 *f_s;## tons\n",
+ "P_b = d*t*f_b;## tons\n",
+ "n = P/min(P_s,P_b);\n",
+ "e = (b-d)/b;## efficiency\n",
+ "P1 = f_t*(b-d)*t;## tons\n",
+ "P2 = f_t*(b-2*d)*t+P_s;## tons\n",
+ "P3 = f_t*(b-3*d)*t+3*P_s;## tons\n",
+ "P4 = f_t*(b-3*d)*t+6*P_s;## tons\n",
+ "print'%s %.d %s'%('The number of rivets required, n = ',round(n+1),'')\n",
+ "print'%s %.1f %s'%(' The efficiency of the joint =',e*100,' percentage');\n",
+ "print'%s %.2f %s'%(' The pull at section 11, P1 = ',P1,' rons/in**2');\n",
+ "print'%s %.1f %s'%( 'The pull at section 22, P2 = ',P2,'rons/in**2');\n",
+ "print'%s %.2f %s'%(' The pull at section 33, P3 = ',P3,' rons/in**2');\n",
+ "print'%s %.2f %s'%(' The pull at section 44, P4 = ',P4,' rons/in**2');\n",
+ "if P1 == min(P1,P2,P3,P4) :\n",
+ " print'%s %.2f %s'%(' The maximum possible pull which the flat will safely transmit is P1 = ',P1,' tons/in^2 at section 11');\n",
+ "elif P2 == min(P1,P2,P3,P4):\n",
+ " print'%s %.1f %s'%(' The maximum possible pull which the flat will safely transmit is P2 = ',P2,' tons/in^2 at section 22'); \n",
+ "elif P3 == min(P1,P2,P3,P4): \n",
+ " print'%s %.2f %s'%('The maximum possible pull which the flat will safely transmit is P3 = ',P3,' tons/in^2 at section 33');\n",
+ "else:\n",
+ " print'%s %.2f %s'%(' The maximum possible pull which the flat will safely transmit is P4 = ',P4,' tons/in^2 at section 44'); \n",
+ "\n",
+ "\n",
+ "##there is a minute error in the answer given textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of rivets required, n = 9 \n",
+ " The efficiency of the joint = 90.3 percentage\n",
+ " The pull at section 11, P1 = 48.75 rons/in**2\n",
+ "The pull at section 22, P2 = 49.5 rons/in**2\n",
+ " The pull at section 33, P3 = 56.29 rons/in**2\n",
+ " The pull at section 44, P4 = 74.33 rons/in**2\n",
+ " The maximum possible pull which the flat will safely transmit is P1 = 48.75 tons/in^2 at section 11\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg437"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The number of rivets required \n",
+ "P = 150.; ##tons\n",
+ "t = 3./4.;## inches\n",
+ "d = 1.;## inches\n",
+ "f_s = 6.;## tons/in**2\n",
+ "f_b = 12.;## tons/in**2\n",
+ "P_s = 0.25*math.pi*d**2 *f_s;## tons\n",
+ "P_b = t*d*f_b;## tons\n",
+ "n = P/min(P_s,P_b);## no. of rivets required\n",
+ "print'%s %.2f %s %.d %s'%('The number of rivets required, n = ',n,', say ',round(n),'');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of rivets required, n = 31.83 , say 32 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg440\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The minimum pitch required\n",
+ "l = 50.;## feet\n",
+ "b = 4.;## feet\n",
+ "P = 3.;## tons per foot run\n",
+ "t = 1/2.;## inches\n",
+ "b1 = 4.;## inches\n",
+ "d1 = 4.;## inches\n",
+ "h1 = 1/2.;## inches\n",
+ "d = 7/8.;## inches\n",
+ "f_s = 6.;## tons/in**2\n",
+ "f_b = 12.;## tons/in**2\n",
+ "P_s = 2.*0.25*math.pi*d**2 *f_s;## tons\n",
+ "P_b = t*d*f_b;## tons\n",
+ "R = P_b;## tons\n",
+ "F = l*P*2./d1;## tons\n",
+ "p_min = R*(l-0.5*b1)/F ;## inches\n",
+ "print'%s %.2f %s %.d %s'%('The minimum pitch required is p = ',p_min,' inches, say ',p_min,' inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum pitch required is p = 3.36 inches, say 3 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg442"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum shear intensity induced at any rivet\n",
+ "P = 2.4;## tons\n",
+ "e = 18.;## inches\n",
+ "n = 8.;##no. of rivets\n",
+ "d = 7/8.;## inches\n",
+ "h = 4.;## inches\n",
+ "M = P*e;## ton-inches\n",
+ "d1 = 2.;## \n",
+ "d2 = 6.;##\n",
+ "square_r_sum = h*((0.5*h)**2. + d2**2.) + h*((0.5*h)**2 + d1**2.);##\n",
+ "r = math.sqrt(40.);\n",
+ "F = M*r/square_r_sum;## tons\n",
+ "theta = math.atan(d2/d1)##radians\n",
+ "theta1 = theta*180./math.pi## degrees\n",
+ "V = (P/n) + F*math.cos(theta);## tons\n",
+ "H = F*math.sin(theta);## tons\n",
+ "R = math.sqrt(V**2 + H**2);## tons\n",
+ "f_s = R/(0.25*math.pi*d**2);## tons/in**2\n",
+ "print('The maximum shear intensity induced at any rivet is' )\n",
+ "print'%s %.2f %s'%('f_s = ',f_s,' tons/in**2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum shear intensity induced at any rivet is\n",
+ "f_s = 2.57 tons/in**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter12.ipynb b/Mechanics_of_Structures/Chapter12.ipynb new file mode 100755 index 00000000..794c3e4e --- /dev/null +++ b/Mechanics_of_Structures/Chapter12.ipynb @@ -0,0 +1,850 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f6a5fb5a9bb50976164cfcdf430ee1b396b67dbf2ffbf7763fc02510b90290d3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12-Shafts and springs in torsion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg450"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum intensity of shear stress induced and The angle of twist in degree\n",
+ "d = 3.;##inches\n",
+ "HP = 120.;##horse power\n",
+ "RPM = 180.;\n",
+ "l = 25.;##feet\n",
+ "N = 12.*10**6.;## lb/in**2\n",
+ "T = 33000.*HP/(2.*math.pi*RPM);## lb-feet\n",
+ "f_s = 16*T*12/(math.pi*d**3.);## lb/in**2\n",
+ "theta = f_s*l*12./(0.5*d*N);## radian\n",
+ "print'%s %.d %s'%('The maximum intensity of shear stress induced is f_s =',f_s,' lb/in**2');\n",
+ "print'%s %.2f %s'%('The angle of twist in degrees is theta = ',theta*180/math.pi,'');\n",
+ "\n",
+ "##there is a minute error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum intensity of shear stress induced is f_s = 7925 lb/in**2\n",
+ "The angle of twist in degrees is theta = 7.57 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg451"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate H.P transmitted is\n",
+ "D = 2.;## inches\n",
+ "N = 150.;## RPM\n",
+ "f_s = 9000.;## lb/in^2\n",
+ "M_r = f_s*(math.pi/16.)*D**3.;## lb-inches\n",
+ "HP = M_r*2.*math.pi*N/(12.*33000.);##\n",
+ "print'%s %.2f %s'%('H.P transmitted is',HP,'');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H.P transmitted is 33.65 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg451"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Suitable diameter \n",
+ "HP = 80.;\n",
+ "N = 200.;## RPM\n",
+ "m = 30/100.;\n",
+ "f = 12000.;## lb/in^2\n",
+ "T = HP*33000./(2.*math.pi*N);## lb-feet\n",
+ "T_max = (1+m)*T;## lb-feet\n",
+ "D = (T_max*12.*16./(math.pi*f))**(1/3.);## inches\n",
+ "print'%s %.3f %s'%('Suitable diameter is D =',D,'inches');\n",
+ "\n",
+ "##the answer is approximated in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Suitable diameter is D = 2.405 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg451"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#check whether it is satisfactory or not for given st A shaft',round(D),'inches external diameter and',round(d),' inches internal diameter\n",
+ "HP = 750.;\n",
+ "N = 90.;## RPM\n",
+ "m = 40/100.;\n",
+ "f = 12000.;## lb/in^2\n",
+ "t = 1.;## inch\n",
+ "T = HP*33000./(2.*math.pi*N);## lb-inches\n",
+ "T_max = (1.+m)*T;## lb-inches\n",
+ "##On solving (4*t)D^3 - (6*t^2)D^2 +(4*t^3 -(16*M/f*%pi))D -t^4 = 0, we get D\n",
+ "D = 7.6;##inches\n",
+ "d = D - 2.;##inches\n",
+ "print'%s %.d %s %d %s '%('A shaft',round(D),'inches external diameter and',round(d),' inches internal diameter will be satisfactory.');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A shaft 8 inches external diameter and 6 inches internal diameter will be satisfactory. \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg452"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate'D = ',round(D1),' inches will be suitable for the shaft\n",
+ "RPM = 180.;## RPM\n",
+ "HP = 130.;\n",
+ "f = 9000.;## lb/in^2\n",
+ "alpha = 1.;##degree\n",
+ "l = 10.;## feet\n",
+ "N = 6000.;## tons/in^2\n",
+ "T = 33000.*HP/(2*math.pi*RPM);## lb-feet\n",
+ "D1 = (16*T*12/(f*math.pi))**(1/3);## inches\n",
+ "D2 = (T*12*l*12*32*alpha*180/(math.pi*N*math.pi*2240.))**(1/4.);## inches\n",
+ "if D1 > D2 :\n",
+ " print'%s %.d %s'%('D = ',round(D1),' inches will be suitable for the shaft');\n",
+ "else:\n",
+ " print'%s %.d %s'%('D =',round(D2),'inches will be suitable for the shaft');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "D = 4 inches will be suitable for the shaft\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg453"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'The saving in weight per foot run and percentage saving\n",
+ "HP = 3000.;\n",
+ "RPM = 60.;\n",
+ "f = 12000.;##lb/in^2\n",
+ "rho = 480.;##lb. per sq. foot\n",
+ "k = 3./4.;## k = d/D \n",
+ "T = HP*33000.*12./(2.*math.pi*RPM);## lb-inches\n",
+ "D1 = (T*16/(f*math.pi))**(1/3.);##inches\n",
+ "D2 = (T/((1.+k**2.)*(1.-k**2.)*math.pi*f/16.))**(1/3.);##inches\n",
+ "d = k*D2;## inches\n",
+ "w1 = 0.25*math.pi*D1**2 *rho/144. ;## lb-wt\n",
+ "w2 = 0.25*math.pi*(D2+d)*(D2-d)*rho/144. ;## lb-wt\n",
+ "w = w1-w2;## lb-wt\n",
+ "n = (w/w1)*100.;\n",
+ "print'%s %.d %s'%('The saving in weight per foot run is w =',w,' lb-wt');\n",
+ "print'%s %.2f %s'%('Percentage saving is ',n,'');\n",
+ "\n",
+ "##there is a minute error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The saving in weight per foot run is w = 138 lb-wt\n",
+ "Percentage saving is 43.62 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg454"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate T1 and T2 \n",
+ "l1 = 3.;## feet\n",
+ "d1 = 1.;## feet\n",
+ "l2 = 9.;## feet\n",
+ "M = 200.;## lb-wt\n",
+ "l = 9.;## inches\n",
+ "N = 12.*10**6;## lb/in^2\n",
+ "k = l2/l1;\n",
+ "T1 = M/(1.+k);## lb-feet\n",
+ "T2 = k*T1;## lb-feet\n",
+ "f_s = T2*12./(math.pi/16.);## lb/in^2\n",
+ "theta = f_s*l/(0.5*d1*N);## radians\n",
+ "print'%s %.d %s %.d %s '%('T1 =',T1,'lb-feet, T2 = ',T2,' lb-feet');\n",
+ "print'%s %.d %s'%('f_s = ',f_s,' lb/in^2');\n",
+ "print'%s %.4f %s %.3f %s'%(' theta = ',theta,' radian , theta = ',theta*180/math.pi,' degrees');\n",
+ "##there is a minute error in the answer given in twxtbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "T1 = 50 lb-feet, T2 = 150 lb-feet \n",
+ "f_s = 9167 lb/in^2\n",
+ " theta = 0.0138 radian , theta = 0.788 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg456"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate forces \n",
+ "D = 5.;## inches\n",
+ "HP = 120.;\n",
+ "RPM = 150.;\n",
+ "b = 5.;## inches\n",
+ "h = 1.;##inch\n",
+ "n = 6.;## no. of bolts\n",
+ "d = 3/4.;## inches\n",
+ "T = HP*33000.*12./(2.*math.pi*RPM);## lb-inches\n",
+ "f_s = T*16./(math.pi*27.);\n",
+ "f_k = T/(b*h*2.*d);\n",
+ "f_b = T/(n*0.25*math.pi*d**2 * b);## lb-inches\n",
+ "print'%s %.d %s %d %s %d %s'%('f_s =',f_s,'lb/in^2, f_k =',f_k,'lb/in^2, f_b =',f_b,' lb/in^2');\n",
+ "\n",
+ "##there are errors given in the answers given in the textbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "f_s = 9510 lb/in^2, f_k = 6722 lb/in^2, f_b = 3804 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg460"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'Maximum strain energys\n",
+ "d = 4.;##inches\n",
+ "T = 30.;## ton-inches\n",
+ "M = 20.;##ton-inches\n",
+ "m = 1/0.3;\n",
+ "f_s = 16.*T/(math.pi*d**3);## tons/in^2\n",
+ "f_b = 32*M/(math.pi*d**3);## tons/in^2\n",
+ "theta = 0.5*math.atan(T/M);## radians\n",
+ "theta1 = theta*180./math.pi;\n",
+ "theta2 = theta1+90.;\n",
+ "f1 = 0.5*f_b + math.sqrt(f_s**2 + 0.25*f_b**2);## tons/in^2\n",
+ "f2 = 0.5*f_b - math.sqrt(f_s**2 + 0.25*f_b**2);## tons/in^2\n",
+ "Ee = f1 - (f2/m);## tons/in^2\n",
+ "f = math.sqrt(f1**2 + f2**2 - 2*f1*f2/m);## tons/in^2\n",
+ "print'%s %.3f %s'%('Maximum strain is Ee = ',Ee,' tons/in^2');\n",
+ "print'%s %.3f %s'%(' Maximum strain energy is f = ',f,' tons/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum strain is Ee = 4.844 tons/in^2\n",
+ " Maximum strain energy is f = 4.995 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg461"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate diameter of the shaft in all different cases\n",
+ "HP = 80.;\n",
+ "RPM = 120.;\n",
+ "b = 10.;## feet\n",
+ "h = 3.;## feet\n",
+ "F = 8000.;## lb-wt\n",
+ "m = 4.;\n",
+ "T = HP*33000.*12./(2.*math.pi*RPM*2240.);## ton-inches\n",
+ "M = F*h*(b-h)*12./(b*2240.);## ton-inches\n",
+ "##(i) The major principal stress f1 is given by\n",
+ "f1 = 6;## tons/in^2\n",
+ "d1 = ((M+math.sqrt(M**2 + T**2))*16/(math.pi*f1))**(1/3.);## inches\n",
+ "\n",
+ "##(ii) If f_s_dash is the maximum intensity of shear stress\n",
+ "f_s_dash = 3;## tons/in^2\n",
+ "d2 = (math.sqrt(M**2 + T**2) * 16./(math.pi*f_s_dash))**(1/3.);## inches\n",
+ "\n",
+ "##(iii) If e is the major principal strain\n",
+ "Ee = 6.;## tons/in^2\n",
+ "d3 = (((1.-(1./m))*M + (1.+(1./m))*math.sqrt(M**2 + T**2))*16./(math.pi*Ee))**(1/3.);## inches\n",
+ "\n",
+ "##(iv) If f is the direct stress which, acting alone will produce the same maximum strain energy \n",
+ "f = 6.;## tons/in^2\n",
+ "d4 = ((math.sqrt(4*M**2 + 2.*(m+1.)*(T**2)/m))*16./(math.pi*f))**(1/3.);## inches\n",
+ "print'%s %.3f %s %.3f %s %.3f %s %.3f %s'%('The diameter of the shaft in different cases will be, (i) d = ',d1,' inches,(ii) d = ',d2,' inches , (iii) d = ',d3,' inches ,(iv) d = ',d4,'inches');\n",
+ "##there are round-off errors in the answers given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diameter of the shaft in different cases will be, (i) d = 5.365 inches,(ii) d = 5.384 inches , (iii) d = 5.370 inches ,(iv) d = 5.370 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum principal stress and The maximum shear intensity and maximum strain\n",
+ "D = 12.;## inches\n",
+ "d = 6.;## inches\n",
+ "HP = 2400.;\n",
+ "RPM = 80.;\n",
+ "M = 40.;## ton-feet\n",
+ "P = 25.;## tons\n",
+ "PR = 0.3;##poisson's ratio\n",
+ "A = 0.25*math.pi*(D**2 - d**2);## in**2\n",
+ "Z = (math.pi/32)*(D**4 - d**4)/D;## in**3\n",
+ "J = (math.pi/16)*(D**4 - d**4)/D;## in**3\n",
+ "p_0 = P/A ;## ton/in**2\n",
+ "p_b = M*12/Z ;## tons/in**2\n",
+ "f_b = p_0 + p_b;##tons/in**2\n",
+ "f_s = HP*33000*12/(2*math.pi*RPM*2240*J);## tons/in**2\n",
+ "theta = 0.5*math.atan(2*f_s/f_b);## radians\n",
+ "theta1 = theta*180/math.pi;## degrees\n",
+ "theta2 = theta1+90;##degrees\n",
+ "f_1 = 0.5*f_b + math.sqrt(f_s**2 + 0.25*f_b**2);##tons/in**2\n",
+ "f_2 = 0.5*f_b - math.sqrt(f_s**2 + 0.25*f_b**2);##tons/in**2\n",
+ "f = math.sqrt(0.25*f_b**2 + f_s**2);## tons/in**2\n",
+ "Ee = f_1 - PR*f_2;## tons/in**2\n",
+ "print'%s %.2f %s %.2f %s'%('The maximum principal stresse are f_1 = ',f_1,' tons/in**2.,compressive , f_2 = ',-f_2,' tons/in**2., tensile');\n",
+ "print'%s %.1f %s %.1f %s'%('theta1 = ',theta1,' degrees,theta2 =',theta2,'degrees');\n",
+ "print'%s %.1f %s '%(' The maximum shear intensity = ',f,'tons/in**2');\n",
+ "print'%s %.3f %s'%(' Maximum strain is, Ee =',Ee,'tons/in**2');\n",
+ "\n",
+ "##there are minute errors in the answers given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum principal stresse are f_1 = 4.78 tons/in**2.,compressive , f_2 = 1.47 tons/in**2., tensile\n",
+ "theta1 = 29.0 degrees,theta2 = 119.0 degrees\n",
+ " The maximum shear intensity = 3.1 tons/in**2 \n",
+ " Maximum strain is, Ee = 5.226 tons/in**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg466"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'The size of the shaft and angle of the twist in the shaft for a length \n",
+ "RPM = 180.;\n",
+ "P = 10.;## tons\n",
+ "v = 25.;## feet per minute\n",
+ "n = 64/100.;## efficiency of the crane\n",
+ "f = 5500.;## lb/in^2\n",
+ "l = 10.;## feet\n",
+ "N = 12*10**6.;## lb/in^2\n",
+ "W = P*v*2240./n;## ft-lbs\n",
+ "T = W*12./(2.*math.pi*RPM);## lb-inches\n",
+ "s = (T/(0.208*f))**(1/3.);## inches\n",
+ "theta = 7.11*T*l*12.*180./(math.pi*N*s**4.);## degrees\n",
+ "print'%s %.3f %s'%('The size of the shaft is s = ',s,' inches');\n",
+ "print'%s %.1f %s %.2f %s'%('The angle of the twist in the shaft for a length of ',l,' feet, theta =',theta,' degrees');\n",
+ "##there is a round-off error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The size of the shaft is s = 2.010 inches\n",
+ "The angle of the twist in the shaft for a length of 10.0 feet, theta = 2.32 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Shear stress induced and Shear stress induced and Energy stored\n",
+ "d = 3/8.;## inches\n",
+ "n = 12.;##no. of complete turns\n",
+ "D = 4.;## inches\n",
+ "W = 50.;## lb-wt\n",
+ "N = 12*10**6;## lb/in^2\n",
+ "T = W*0.5*D;## lb-inches\n",
+ "f_s = T*16./(math.pi*d*83);##lb/in^2\n",
+ "delta = 64.*W*(D**3 /8.)*n/(N*d**4.);## inches\n",
+ "E = 0.5*W*delta;## inch-lbs\n",
+ "print'%s %.d %s'%('Shear stress induced is f_s =',f_s,' lb/in^2');\n",
+ "print'%s %.3f %s'%(' Deflection under the pull is delta = ',delta,' inches');\n",
+ "print'%s %.3f %s'%(' Energy stored = ',E,' lb-inches');\n",
+ "\n",
+ "##there is a minute error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress induced is f_s = 16 lb/in^2\n",
+ " Deflection under the pull is delta = 1.295 inches\n",
+ " Energy stored = 32.363 lb-inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The number of springs required\n",
+ "W = 2.;## tons\n",
+ "v = 4.;## miles per hour\n",
+ "n = 18.;## no. of coils\n",
+ "delta = 9.;## inches\n",
+ "N = 6000.;## tons/in^2\n",
+ "d = 1.;## inch\n",
+ "D = 8.;## inches\n",
+ "KE = 12.*(W*(v*44./30.)**2.)/(2.*32.);## inch-tons\n",
+ "P = (delta*N*d**4)/(64.*n*(0.5*D)**3);## tons\n",
+ "E = 0.5*P*delta;## inch-tons\n",
+ "m = KE/E ;## no. of springs required\n",
+ "print'%s %.d %s'%('The number of springs required m =',round(m),'');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of springs required m = 4 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-470"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate height of drop\n",
+ "W = 5.;## cwt\n",
+ "n = 18.;## no. of coils\n",
+ "delta = 9.;## inches\n",
+ "d = 1.;## inch\n",
+ "D = 8.;## inches\n",
+ "N = 6000.;## tons/in^2\n",
+ "P = (delta*N*d**4.)/(64.*n*(0.5*D)**3);## tons\n",
+ "h = (0.5*P*delta*20./W)-delta;## inches\n",
+ "print'%s %.3f %s'%('The height of drop h = ',h,' inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The height of drop h = 4.184 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg471"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Maximum possible axial load and Deflection, delta\n",
+ "s = 1/4.;## inch\n",
+ "n = 12.;## no. of coils\n",
+ "D = 3.;## inches\n",
+ "f_s = 45000.;## lb/in^2\n",
+ "N = 12*10**6;## lb/in^2\n",
+ "T = 0.208*f_s*s**3;## lb-inches\n",
+ "W = T/(0.5*D);## lb-wt\n",
+ "theta = 7.11*T*math.pi*D*12./(N*s**4);##rdaians\n",
+ "delta = 0.5*D*theta;## inches\n",
+ "print'%s %.1f %s'%('Maximum possible axial load is W = ',W,' lb-wt');\n",
+ "print'%s %.3f %s'%(' Deflection, delta = ',delta,' inches');\n",
+ "\n",
+ "##there is a minute error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum possible axial load is W = 97.5 lb-wt\n",
+ " Deflection, delta = 3.763 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg472"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The bending stresses\n",
+ "d = 3/8.;## inches\n",
+ "n = 12.;##no. of complete turns\n",
+ "D = 4.;## inches\n",
+ "W = 50.;## lb-wt\n",
+ "N = 12.*10**6;## lb/in^2\n",
+ "E = 30.*10**6;## lb/in^2\n",
+ "M = 75.;## lb-inches\n",
+ "I = (math.pi/64.)*d**4;## in^4\n",
+ "Z = 2.*I/d;## in^3\n",
+ "f = M/Z ;## lb/in^2\n",
+ "phi = M*math.pi*D*12./(E*I);## radians\n",
+ "n_ = (phi/(2*math.pi)) + n;## increase in no. of turns\n",
+ "print'%s %.d %s'%('The bending stress is f = ',f,' lb/in^2');\n",
+ "print'%s %.3f %s'%(' n_new = ',n_,' turns');\n",
+ "\n",
+ "##there are minute errors in the answers given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bending stress is f = 14486 lb/in^2\n",
+ " n_new = 12.062 turns\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg476"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum intensity of shear stress and 'Deflection\n",
+ "d = 3/8.;## inches\n",
+ "n = 12.;##no. of complete turns\n",
+ "D = 4.;## inches\n",
+ "W = 50.;## lb-wt\n",
+ "N = 12.*10**6.;## lb/in^2\n",
+ "alpha = 15*math.pi/180.;## degrees\n",
+ "E = 30.*10**6.;## lb/in^2\n",
+ "T = W*0.5*D*math.cos(alpha);## lb-inches\n",
+ "M = W*0.5*D*math.sin(alpha);## lb-inches\n",
+ "J = math.pi*d**4 /32.;## in^4\n",
+ "I = math.pi*d**4 /64.;## in^4\n",
+ "delta = 64.*W*((D/2.)**3.)*n/math.cos(alpha)*((math.cos(alpha)**2.)/N + (2.*math.sin(alpha)**2.)/E)/d**4. ;## inches\n",
+ "f = 32.*W*0.5*D*math.sin(alpha)/(math.pi*d**3) ;## lb/in^2\n",
+ "f_s = T*16./(math.pi*d**3);## lb/in^2\n",
+ "f_1 = 0.5*f + math.sqrt(f_s**2. + 0.25*f**2.);## lb/in^2\n",
+ "f_2 = 0.5*f - math.sqrt(f_s**2. + 0.25*f**2.);## lb/in^2\n",
+ "f_s_dash = math.sqrt(f_s**2. + 0.25*f**2.);## lb/in^2\n",
+ "print'%s %.3f %s'%('Deflection, delta = ',delta,' inches');\n",
+ "print'%s %.d %s %.d %s'%(' f = ',f,' lb/in^2, f_s = ',f_s,' lb/in^2');\n",
+ "print'%s %.d %s'%('The maximum intensity of shear stress =',f_s_dash,' lb/in^2');\n",
+ "\n",
+ "##there are calculation errors in the answers given in textbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection, delta = 1.322 inches\n",
+ " f = 4999 lb/in^2, f_s = 9328 lb/in^2\n",
+ "The maximum intensity of shear stress = 9657 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg477"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'Angle of rotation and 'Angle of rotation\n",
+ "d = 3/8.;## inches\n",
+ "n = 12.;##no. of complete turns\n",
+ "D = 4.;## inches\n",
+ "M = 75.;## lb-inches\n",
+ "N = 12.*10**6.;## lb/in**2\n",
+ "alpha = 15*math.pi/180.;## degrees\n",
+ "E = 30.*10.**6.;## lb/in**2\n",
+ "phi_dash = (64./d**4.)*M*0.5*D*n/math.cos(alpha)*((2.*(math.cos(alpha))**2.)/E + ((math.sin(alpha))**2)/N);## radians\n",
+ "DELTA = 64.*M*((0.5*D)**2.)*n*math.sin(alpha)*((1./N) - (2./E))/d**4.;## inches\n",
+ "print'%s %.2f %s %.3f %s'%('Angle of rotation phi_dash = ',phi_dash,' radians or ',phi_dash*180/math.pi,' degrees');\n",
+ "print'%s %.4f %s '%(' The axial deflection = ',DELTA,' inches');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of rotation phi_dash = 0.41 radians or 23.422 degrees\n",
+ " The axial deflection = 0.0503 inches \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter13.ipynb b/Mechanics_of_Structures/Chapter13.ipynb new file mode 100755 index 00000000..fb9b7fe7 --- /dev/null +++ b/Mechanics_of_Structures/Chapter13.ipynb @@ -0,0 +1,366 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b60753ad5256b26ed900efc57b1a6de06dc4217fa7384fe1bb9afe99a185fc77"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter13-Elements of reinforced concrete"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-486"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The stress induced in the concrete and steel and \n",
+ "b = 10.;##inches\n",
+ "d = 1.;##inches\n",
+ "h = 20.;##inches\n",
+ "r = 2.;##inches\n",
+ "M_r = 500000.;##lb-inches\n",
+ "m = 15.;\n",
+ "A_r = 4.*0.25*math.pi*d**2;##in^2\n",
+ "h_eff = h-r;##inches\n",
+ "K = m*A_r/(b*h_eff);##inches\n",
+ "n1 = math.sqrt((K)**2+(2*K))-K;\n",
+ "n = n1*h_eff;##inches\n",
+ "a = h_eff-(n/3);##inches\n",
+ "c = 2.*M_r/(b*n*a);##lb/in^2\n",
+ "t = (h_eff-n)*m*c/n;##lb/in^2\n",
+ "print'%s %.d %s'%('The stress induced in the concrete and steel, t =',t,'lb/in^2');\n",
+ "\n",
+ "##there is a minute error in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress induced in the concrete and steel, t = 10643 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The distance of the N.A from the top edge and The safe moment of inertia and The safe moment of inertia\n",
+ "b = 8.;##inches\n",
+ "d = 7/8.;##inches\n",
+ "h = 18.;##inches\n",
+ "r = 2.;##inches\n",
+ "c = 750.;##lb/in^2\n",
+ "t_limit = 18000.;##lb/in^2\n",
+ "m = 8.;\n",
+ "h_eff = 16.;##inches\n",
+ "m = 18;\n",
+ "A_t = 3*0.25*math.pi*d**2;##in^2\n",
+ "K = m*A_t/(b*h_eff);##inches\n",
+ "n1 = math.sqrt((K)**2.+(2.*K))-K;\n",
+ "n = n1*h_eff;##inches\n",
+ "a = h_eff - (n/3.);##inches\n",
+ "t = m*c*(h_eff-n)/n;##lb/in^2\n",
+ "\n",
+ "if t<t_limit:\n",
+ " t = t;\n",
+ "else:\n",
+ " t = t_limit;\n",
+ "\n",
+ "M_r = t*A_t*a;## lb/inches\n",
+ "W = M_r*8./(12.*h_eff);##lb-wt\n",
+ "print'%s %.3f %s'%('The distance of the N.A from the top edge, n =',n,'inches.');\n",
+ "print'%s %.d %s'%('The safe moment of inertia is, t =',t,'lb/in^2.');\n",
+ "print'%s %.d %s %d %s'% ('Unifromly distributed load over the beam, W =',W,'lb-wt. or',W/16.011,'lb. per foot run');\n",
+ "\n",
+ "##there are calculation errors given in the answer in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance of the N.A from the top edge, n = 8.039 inches.\n",
+ "The safe moment of inertia is, t = 13368 lb/in^2.\n",
+ "Unifromly distributed load over the beam, W = 13385 lb-wt. or 836 lb. per foot run\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Area of steel reinforcement required and Corresponding stress in steel \n",
+ "import numpy\n",
+ "from numpy import roots\n",
+ "def quadratic(d,M_c,c_limit,b):\n",
+ " p = ([1 , -(d*3), + M_c*3/(0.5*c_limit*b)]);\n",
+ " Z = numpy.roots(p);\n",
+ " return Z;\n",
+ "\n",
+ "b = 12.;##inches\n",
+ "h = 22.;##inches\n",
+ "r = 2.;##inches\n",
+ "W = 1500.;##lb per foot run\n",
+ "d = h-r;##feet\n",
+ "l = 20.;##inches\n",
+ "c_limit = 700.;## lb/in^2\n",
+ "m = 15.;\n",
+ "M_c = W*20.*l*b/8.;## lb-inches\n",
+ "Z = quadratic(d,M_c,c_limit,b);\n",
+ "n = round(Z[1]);\n",
+ "t = m*c_limit*(d-n)/n;## lb/in62\n",
+ "A_t = 0.5*c_limit*b*n/t;## in^2\n",
+ "print'%s %.d %s'%('Area of steel reinforcement required is, A_t =',A_t,'in^2');\n",
+ "print'%s %.d %s'%(' Corresponding stress in steel is, t =',t,'lb/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area of steel reinforcement required is, A_t = 13 in^2\n",
+ " Corresponding stress in steel is, t = 4500 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg492"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Effective deapth\n",
+ "m = 15.;\n",
+ "t = 18000.;## lb/in^2\n",
+ "c = 700.;## lb/in^2\n",
+ "b = 12.;## inches\n",
+ "M = 900000.;##bending moment lb/inches\n",
+ "k1 = 1./((t/(m*c))+1.);##k = n/d\n",
+ "k2 = 1.-k1/3.;##k2 = a/d\n",
+ "p = 0.5*c*k1/(t);\n",
+ "d = math.sqrt(M/(0.5*c*b*k1*k2));##inches\n",
+ "A_t = p*b*d;## sq.inches\n",
+ "A_t_previous = 0.25*math.pi*(7/8.)**2;##section area with diameter 7/8 inches\n",
+ "n = A_t/A_t_previous;\n",
+ "print'%s %.2f %s'%('Effective deapth is d =',d,'inches');\n",
+ "print'%s %.3f %s'%('A_t =',A_t,'sq.inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective deapth is d = 25.75 inches\n",
+ "A_t = 2.214 sq.inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg492"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the effective death\n",
+ "l = 20.;##feet\n",
+ "W = 500.;## lb per foot run\n",
+ "c = 750.;## lb/in^2\n",
+ "t = 18000.;## lb/in^2\n",
+ "m = 15.;\n",
+ "BM_max = W*l*l*12./8. ;## lb-inches\n",
+ "##by making the effective deapth d twice the width b\n",
+ "d = (BM_max/(126.*0.5))**(1/3.);##inches\n",
+ "b = 0.5*d;##inches\n",
+ "##necessary reinforcement is 0.8% of concrete section\n",
+ "A_t = 0.008*b*d;## in^2\n",
+ "print'%s %.2f %s %.2f %s'%('d =',d,'inches ,b =',b,'inches');\n",
+ "print'%s %.3f %s'%('A_t =',A_t,'in^2');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d = 16.82 inches ,b = 8.41 inches\n",
+ "A_t = 1.132 in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate d1 and d2 and r1\n",
+ "W = 180.;## lb per sq.foot\n",
+ "l = 10.;## feet\n",
+ "b = 12.;##inches\n",
+ "c = 750.;## lb/in^2\n",
+ "m = 15.;\n",
+ "M = W*l*l*12./8.;##lb-inches\n",
+ "d_new = math.sqrt(M/(126.*b));##inches\n",
+ "A_t = 0.8*b*d_new/100.;##in^2\n",
+ "##using 3/8 inch rods \n",
+ "d1 = 3/8.;##inches\n",
+ "A1 = 0.25*math.pi*(d1)**2;##in^2 \n",
+ "r1 = A1*b/A_t;##inch\n",
+ "##using 1/2 inch rods \n",
+ "d2 = 1/2.;##inches\n",
+ "A2 = 0.25*math.pi*(d2)**2;##in^2 \n",
+ "r2 = A2*b/A_t;##inches\n",
+ "print'%s %.3f %s'%('d =',d_new,'inches');\n",
+ "print'%s %.3f %s'%('A_t =',A_t,'in^2');\n",
+ "print'%s %.3f %s %.2f %s'%('Using',d1,'inch rods, spacing centre to centre will be',r1,'inches');\n",
+ "print'%s %.2f %s %.1f %s'%('Using ',d2,' inch rods, spacing centre to centre will be ',r2,' inches');\n",
+ "##there are round-off errors in the answer given in textbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d = 4.226 inches\n",
+ "A_t = 0.406 in^2\n",
+ "Using 0.375 inch rods, spacing centre to centre will be 3.27 inches\n",
+ "Using 0.50 inch rods, spacing centre to centre will be 5.8 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate foot width of slab and spacing centre to centre \n",
+ "l = 12.;##feet\n",
+ "w = 150.;## lb per sq.foot\n",
+ "##Live load\n",
+ "LL = w*l;##lb-wt\n",
+ "##Dead Load assuming the slab thickness to be 6 inches\n",
+ "t = 6.;##inches\n",
+ "DL = t*l*12.;##lb-wt\n",
+ "##total load\n",
+ "W = LL+DL;##lb-wt\n",
+ "M = W*l*12./10.;##lb-inches\n",
+ "d = math.sqrt(M/(12.*126.));\n",
+ "print'%s %.4f %s'%('d =',d,'inches');\n",
+ "##With about an inch to cover the slab will be 6 inch thick\n",
+ "A_t = 0.8*l*d/100.;## in^2\n",
+ "##using 1/2 inch rods \n",
+ "d1 = 1/2.;##inches\n",
+ "A1 = 0.25*math.pi*(d1)**2;##in^2 \n",
+ "r1 = A1*l/A_t;##inches\n",
+ "print'%s %.4f %s'%(' Per foot width of slab, A_t =',A_t,'in^2');\n",
+ "print'%s %.2f %s %.3f %s'%('Using',d1,'inch rods, spacing centre to centre will be ',r1,' inches');\n",
+ "##there are minute calculation errors in the answer given in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d = 5.0370 inches\n",
+ " Per foot width of slab, A_t = 0.4836 in^2\n",
+ "Using 0.50 inch rods, spacing centre to centre will be 4.873 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter2.ipynb b/Mechanics_of_Structures/Chapter2.ipynb new file mode 100755 index 00000000..42876b36 --- /dev/null +++ b/Mechanics_of_Structures/Chapter2.ipynb @@ -0,0 +1,309 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1983ff28cbb6fa2f27736543d13290cc4b84b616fb5e9a5a775ca4d190cec891"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Principal planes and principal stress"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Normal stress intensity and Tangential stress intensity and Resultant stress intensity and angle\n",
+ "p_1 = 5.;##principal stress in tons/in^2\n",
+ "p_2 = 5./2.;##principal stress in tons/in^2\n",
+ "theta = 50*math.pi/180;##angle in degrees\n",
+ "p_n = p_1*math.cos(theta)**2+p_2*math.sin(theta)**2;##normal stress intensity\n",
+ "p_t = (p_1-p_2)*math.sin(theta)*math.cos(theta);##tangential stress intensity\n",
+ "p = math.sqrt((p_1*math.cos(theta))**2+(p_2*math.sin(theta))**2);##resultant intensity of stress\n",
+ "alpha = math.atan((p_2*math.sin(theta))/(p_1*math.cos(theta)));##in radians\n",
+ "alpha = alpha*180/math.pi;##in degrees\n",
+ "print'%s %.2f %s'%('Normal stress intensity p_n = ',p_n,' tons/in^2');\n",
+ "print'%s %.2f %s'%('h Tangential stress intensity p_t = ',p_t,' tons/in^2');\n",
+ "print'%s %.2f %s'%(' Resultant stress intensity p = ',p,'tons/in^2');\n",
+ "print'%s %.2f %s'%(' angle alpha p_n = ',alpha,' degrees');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal stress intensity p_n = 3.53 tons/in^2\n",
+ "h Tangential stress intensity p_t = 1.23 tons/in^2\n",
+ " Resultant stress intensity p = 3.74 tons/in^2\n",
+ " angle alpha p_n = 30.79 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The inclination of principal planes to the axis of the bolt and p_2 and Maximum shear stress is p_max and stress which acting alone will produce the same maximum strain \n",
+ "d = 3./4. ;##inches\n",
+ "P = 2.;##tons\n",
+ "Q = 0.5;##tons\n",
+ "m = 4.;\n",
+ "A = 0.25*math.pi*d**2;##in^2\n",
+ "p = P/A ;##tons/in^2\n",
+ "q = Q/A;##tons/in^2\n",
+ "theta = 0.5*math.atan(2.*q/p);##radians\n",
+ "theta1 = theta*180/math.pi;##degrees\n",
+ "theta2 = theta1+90;##degrees\n",
+ "print'%s %.2f %s'%('The inclination of principal planes to the axis of the bolt will be',theta1,'degres') \n",
+ "print'%s %.2f %s'%('The inclination of principal planes to the axis of the bolt will be',180-theta2,'degrees') \n",
+ "print'%s %.2f %s'%('The inclination of maximum shear planes to the axis of the bolt will be',theta1+45,'degress')\n",
+ "print'%s %.2f %s'%('The inclination of principal planes to the axis of the bolt will be',180-theta2-45,'degrees')\n",
+ "\n",
+ "p_1 = 0.5*p+math.sqrt(0.25*p**2+q**2);##tons/in^2\n",
+ "p_2 = 0.5*p-math.sqrt(0.25*p**2+q**2);##tons/in^2\n",
+ "p_max = 0.5*(p_1-p_2);##tons/in^2\n",
+ "p_s = p_1-(p_2/m);##tons/in^2 \n",
+ "print'%s %.2f %s'%('The principal stresse are given by p_1 =',p_1,'tons/in^2.,tensile')\n",
+ "print'%s %.2f %s'%('p_2 =',p_2,'tons/in^2.,compressive')\n",
+ "print'%s %.2f %s'%('p_2 =',p_2,'tons/in^2 .,compressive');\n",
+ "print'%s %.2f %s'%('Maximum shear stress is p_max =',p_max,'tons/in^2');\n",
+ "print'%s %.2f %s'%('The stress which acting alone will produce the same maximum strain is given by,',p_s,'tons/in^2');\n",
+ "\n",
+ "##there is an error in the answer given in text book"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The inclination of principal planes to the axis of the bolt will be 13.28 degres\n",
+ "The inclination of principal planes to the axis of the bolt will be 76.72 degrees\n",
+ "The inclination of maximum shear planes to the axis of the bolt will be 58.28 degress\n",
+ "The inclination of principal planes to the axis of the bolt will be 31.72 degrees\n",
+ "The principal stresse are given by p_1 = 4.79 tons/in^2.,tensile\n",
+ "p_2 = -0.27 tons/in^2.,compressive\n",
+ "p_2 = -0.27 tons/in^2 .,compressive\n",
+ "Maximum shear stress is p_max = 2.53 tons/in^2\n",
+ "The stress which acting alone will produce the same maximum strain is given by, 4.86 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The principal stresses and The maximum shear stress and the planes offering it being inclined and the normal stress intensity \n",
+ "q = 2.;##tons/in^2\n",
+ "p = 5.;##tons/in^2\n",
+ "p_dash = 2.;##tons/in^2\n",
+ "theta = 0.5*math.atan(2*q/(p-p_dash));##radians\n",
+ "theta1 = theta*180/math.pi;##degrees\n",
+ "theta2 = theta1+90;##degrees\n",
+ "p_1 = 0.5*(p+p_dash)+math.sqrt(q**2 + 0.25*(p-p_dash)**2);##tons/in^2\n",
+ "p_2 = 0.5*(p+p_dash)-math.sqrt(q**2 + 0.25*(p-p_dash)**2);##tons/in^2\n",
+ "q_max = 0.5*(p_1-p_2);##tons/in^2\n",
+ "print'%s %.2f %s'%('The principal stresses are p_1 =',p_1,'tons/in^2 .,tensile')\n",
+ "print'%s %.2f %s'%('The principal stresses arep_2 =',p_2,'tons/in^2., tensile');\n",
+ "print'%s %.1f %s'%('The maximum shear stress is',q_max,'tons/in^2.,') \n",
+ "print'%s %.2f %s'%('the planes offering it being inclined at',theta1+45,'degrees') \n",
+ "print'%s %.2f %s'%('the planes offering it being inclined at',theta2+45,'degrees')\n",
+ "print'%s %.2f %s'%('to the plane having the normal stress intensity of',p,'tons/in^2.')\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The principal stresses are p_1 = 6.00 tons/in^2 .,tensile\n",
+ "The principal stresses arep_2 = 1.00 tons/in^2., tensile\n",
+ "The maximum shear stress is 2.5 tons/in^2.,\n",
+ "the planes offering it being inclined at 71.57 degrees\n",
+ "the planes offering it being inclined at 161.57 degrees\n",
+ "to the plane having the normal stress intensity of 5.00 tons/in^2.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate theta1 and theta2 and P_1 and P_2 and maximum shear intensity\n",
+ "p_res = 6.;##tons/in^2\n",
+ "p_dash = 4.;##tons/in^2\n",
+ "theta = 30.*math.pi/180.;##degrees\n",
+ "p_n = 4.;##tons/in^2\n",
+ "p = p_res*math.cos(theta);##tons/in^2\n",
+ "q = p_res*math.sin(theta);##tons/in^2\n",
+ "L = 2*q/(p-p_dash);\n",
+ "theta = 0.5*math.atan(2*q/(p-p_dash));\n",
+ "theta1 = theta*180/math.pi;##degrees\n",
+ "theta2 = theta1+90;##degrees\n",
+ "p_1 = 0.5*(p+p_dash)+math.sqrt(q**2 + 0.25*(p-p_dash)**2);##tons/in^2\n",
+ "p_2 = 0.5*(p+p_dash)-math.sqrt(q**2 + 0.25*(p-p_dash)**2);##tons/in^2\n",
+ "p_max = 0.5*(p_1-p_2);##tons/in^2\n",
+ "print'%s %.2f %s'%('Theta1 =',theta1,'degrees') \n",
+ "print'%s %.2f %s'%('Theta2 =',theta2,'degrees')\n",
+ "print'%s %.2f %s'%('p_1 =',p_1,'tons/in^2.,tensile')\n",
+ "print'%s %.2f %s'%('p_2 =',p_2,'tons/in^2.,tensile')\n",
+ "print'%s %.2f %s'%('The maximum shear intensity will be',p_max,'tons/in^2 across the planes of maximum shear.');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Theta1 = 39.36 degrees\n",
+ "Theta2 = 129.36 degrees\n",
+ "p_1 = 7.66 tons/in^2.,tensile\n",
+ "p_2 = 1.54 tons/in^2.,tensile\n",
+ "The maximum shear intensity will be 3.06 tons/in^2 across the planes of maximum shear.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate tensiles and compressive \n",
+ "p_1 = 7.;##tons/in^2\n",
+ "p_2 = 4.;##tons/in^2\n",
+ "p_3 = 3.;##tons/in^2\n",
+ "m = 4.;\n",
+ "E = 13000.;##tons/in^2\n",
+ "e_1 = (p_1/E)+(p_2/(m*E))-(p_3/(m*E));\n",
+ "e_2 = (p_2/E)+(p_1/(m*E))+(p_3/(m*E));\n",
+ "e_3 = (p_3/E)-(p_1/(m*E))+(p_2/(m*E));\n",
+ "print'%s %.4f %s'%('e_1 =',e_1,'tensile')\n",
+ "print'%s %.4f %s'%('e_2 =',e_2,'compressive')\n",
+ "print'%s %.4f %s'%('e_3 =',e_3,'tensile')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e_1 = 0.0006 tensile\n",
+ "e_2 = 0.0005 compressive\n",
+ "e_3 = 0.0002 tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the contraction in the length \n",
+ "a = 2.;##inches\n",
+ "l = 6.;##inches\n",
+ "E = 13000.;##tons/In^2\n",
+ "m = 1./0.3;\n",
+ "P = 20.;##tons\n",
+ "p_1 = P/a**2;##tons/in^2\n",
+ "p_2 = p_1/(2.*(m-1));##tons/in^2\n",
+ "e_1 = (5.-0.6*p_2)/E;##tons/in^2\n",
+ "del_l = e_1*l;##inches\n",
+ "print'%s %.4f %s'%('The contraction in the length del_l =',del_l,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The contraction in the length del_l = 0.0020 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter3.ipynb b/Mechanics_of_Structures/Chapter3.ipynb new file mode 100755 index 00000000..59d8003c --- /dev/null +++ b/Mechanics_of_Structures/Chapter3.ipynb @@ -0,0 +1,283 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fda2efbde9309c997fd1c5768ac3e3ef47209d8a8fc8385592f593bd1834cc52"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Impact or shock loading:strain energy"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Ex1-pg63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum instantaneous stress produced and corresponding strain work done on the rod and stored\n",
+ "d = 3./2.;##inches\n",
+ "l = 10.;##feet\n",
+ "P = 8.;##tons\n",
+ "E = 13500.;##tons/in^2\n",
+ "A = 0.25*math.pi*d**2;##in^2\n",
+ "p = 2.*P/A;## tons/in^2\n",
+ "e = p/E;\n",
+ "del_l = e*l*12.;##inches\n",
+ "W = P*del_l;##inch-ton\n",
+ "print'%s %.2f %s'%('The maximum instantaneous stress produced is, p =',p,'tons/in^2');\n",
+ "print'%s %.4f %s'%('The corresponding strain is e =',e,'') \n",
+ "print'%s %.2f %s'%('del_l =',del_l,'inches')\n",
+ "print'%s %.2f %s'%('The work done on the rod and stored by it is',W,'inch-ton.');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum instantaneous stress produced is, p = 9.05 tons/in^2\n",
+ "The corresponding strain is e = 0.0007 \n",
+ "del_l = 0.08 inches\n",
+ "The work done on the rod and stored by it is 0.64 inch-ton.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Maximum instantaneous and extension stress produced\n",
+ "l = 6.;##feet\n",
+ "d = 1.;##inches\n",
+ "h = 4.;##inches\n",
+ "E = 30*10**6;##lb/in^2\n",
+ "A = 0.25*math.pi*d**2;##in^2\n",
+ "P = 50.;##lb\n",
+ "p = (P/A)+math.sqrt((P**2/A**2)+(2*E*P*h)/(A*l*12.));\n",
+ "e = p/E;\n",
+ "del_l = e*l*12.##inches\n",
+ "print'%s %.d %s'%('Maximum instantaneous stress produced is p =',p,'lb/in^2');\n",
+ "print'%s %.3f %s'%('Maximum instantaneous extension produced is del_l =',del_l,'lb/in^2');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum instantaneous stress produced is p = 14631 lb/in^2\n",
+ "Maximum instantaneous extension produced is del_l = 0.035 lb/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Instantaneous elongation and Height of the drop\n",
+ "l = 10.;##feet\n",
+ "d = 5./4.;##inches\n",
+ "p = 8.;##tons/in^2\n",
+ "E = 13000.;##tons/in^2\n",
+ "A = 0.25*math.pi*d**2;##in^2\n",
+ "e = p/E;\n",
+ "del_l = e*l*12.;##inches\n",
+ "W = 0.5*p**2*A*l*12./E;## inch-ton\n",
+ "h = W*10.-del_l;##inches\n",
+ "print'%s %.3f %s'%('Instantaneous elongation is del_l =',del_l,'inches')\n",
+ "print'%s %.2f %s'%('Height of the drop is h =',h,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous elongation is del_l = 0.074 inches\n",
+ "Height of the drop is h = 3.55 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Maximum instantaneous extension and stress\n",
+ "w = 4.;##tons\n",
+ "v = 2.;##miles per hour\n",
+ "l = 150.;##feet\n",
+ "d = 3./2.;##inches\n",
+ "E = 13000.;## tons/in^2\n",
+ "g = 32.;## ft/sec^2\n",
+ "A = 0.25*math.pi*d**2;## in^2\n",
+ "KE = w*(v*17.6)**2/(2*g*12);## inch-tons\n",
+ "p = math.sqrt(KE*(2.*E)/(A*l*12.));## tons/in^2\n",
+ "del_l = p*l*12./E;## inches\n",
+ "print'%s %.2f %s'%('Maximum instantaneous stress produced is p =',p,'tons/in^2')\n",
+ "print'%s %.f %s'%('Maximum instantaneous extension produced is del_l =',del_l,'inch')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum instantaneous stress produced is p = 7.26 tons/in^2\n",
+ "Maximum instantaneous extension produced is del_l = 1 inch\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Maximum instantaneous stress and extension \n",
+ "d = 2.;##inches\n",
+ "l = 8.;##feet\n",
+ "U = 50.;## ft-lbs\n",
+ "E = 28.*10**6;## lb/in^2\n",
+ "V = 0.25*math.pi*d**2*l*12.;## in^3\n",
+ "p = math.sqrt(2*U*12.*E/V);## lb/in^2\n",
+ "e = p/E;\n",
+ "del_l = e*l*12.;## inches\n",
+ "print'%s %.1f %s'%('Maximum instantaneous stress produced is p =',p,'lb/in^2');\n",
+ "print'%s %.4f %s'%('Maximum instantaneous extension produced is del_l =',del_l,'inches');\n",
+ "##there is an error in the answer given in text book\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum instantaneous stress produced is p = 10555.0 lb/in^2\n",
+ "Maximum instantaneous extension produced is del_l = 0.0362 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# calculate Maximum instantaneous stress and Ratio of energies stored if both bars are allowed to reach the proof stress and Ratio of energies stored at the same stress per unit volume\n",
+ "l = 12.;##feet\n",
+ "d1_A = 1.;##inch\n",
+ "d2_A = 2.;##inches\n",
+ "l1_A = 4.;##inches\n",
+ "l2_A = 8.;##inches\n",
+ "d1_B = 1.;##inch\n",
+ "d2_B = 2.;##inches\n",
+ "l1_B = 8.;##inches\n",
+ "l2_B = 4.;##inches\n",
+ "p_A = 15./2.;## tons/in^2\n",
+ "p_B = math.sqrt((2./3.)*p_A**2);## tons/in^2\n",
+ "r1 = (9*math.pi/8.)/(3*math.pi/4.);##ratio of energies if both bars are allowed to reach the proof stress\n",
+ "V_A = 0.25*math.pi*d1_A**2*l1_A+ 0.25*math.pi*d2_A**2*l2_A;## in^3\n",
+ "V_B = 0.25*math.pi*d1_B**2*l1_B+ 0.25*math.pi*d2_B**2*l2_B;## in^3\n",
+ "r2 = ((3./16.)*p_B**2)/((1./12.)*p_B**2);##ratio of enrgies\n",
+ "print'%s %.2f %s'%('Maximum instantaneous stress produced is p_B =',p_B,'tons/in^2')\n",
+ "print'%s %.2f %s'%('Ratio of energies stored if both bars are allowed to reach the proof stress is r1 =',r1,'');\n",
+ "print'%s %.2f %s'%('Ratio of energies stored at the same stress per unit volume, is r2 =',r2,'');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum instantaneous stress produced is p_B = 6.12 tons/in^2\n",
+ "Ratio of energies stored if both bars are allowed to reach the proof stress is r1 = 1.50 \n",
+ "Ratio of energies stored at the same stress per unit volume, is r2 = 2.25 \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter5.ipynb b/Mechanics_of_Structures/Chapter5.ipynb new file mode 100755 index 00000000..8282741d --- /dev/null +++ b/Mechanics_of_Structures/Chapter5.ipynb @@ -0,0 +1,1225 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:369e99bb86bfee04526d8c71fa878425d0c2630ec6c7783b1e559db8ef69fb9f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Beams and bending"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum intensity of stress\n",
+ "b = 6.;## inches \n",
+ "t = 1./2.;## inch\n",
+ "R = 40.;## feet\n",
+ "E = 13000.;## tons/in^2\n",
+ "y = t/2.;## inch\n",
+ "f = (E/(R*12.))*(y);## tons/in^2\n",
+ "print'%s %.2f %s'%('The maximum intensity of stress induced is f =',f,'tons/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum intensity of stress induced is f = 6.77 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate moment of resistance and The radius to which it should be bent \n",
+ "d = 14.;## inches\n",
+ "I = 442.57;##inch units\n",
+ "f = 8.;## tons/in^2\n",
+ "E = 13000.;## tons/in^2\n",
+ "R = E*d/(2.*f);## inches\n",
+ "M_r = f*(I/(d/2.));## ton-inches\n",
+ "print'%s %.d %s'%('The radius to which it should be bent is R =',R,'inches')\n",
+ "print'%s %.1f %s'%('The radius to which it should be bent is R =',R/12,'feet');\n",
+ "print'%s %.1f %s'%('The moment of resistance is M_r =',M_r,'ton-inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius to which it should be bent is R = 11375 inches\n",
+ "The radius to which it should be bent is R = 947.9 feet\n",
+ "The moment of resistance is M_r = 505.8 ton-inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg141\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the load is uniformly spread over its span and the load is concentrated at the centre \n",
+ "d = 16.;## inches\n",
+ "I = 618.;## inch units\n",
+ "l = 24.;## feet\n",
+ "f = 15./2.;## tons/in^2\n",
+ "Z = I/(d/2.);## inch-units\n",
+ "M_r = f*Z;## ton-inches\n",
+ "##If the load is uniformly spread over its span,BM = W*l/8\n",
+ "W1 = 8.*M_r/(12.*l);##tons\n",
+ "##If the load is concentrated at the centre,BM = W*l/4\n",
+ "W2 = 4.*M_r/(12.*l);##tons\n",
+ "print'%s %.1f %s'%('If the load is uniformly spread over its span, then W is given by W1=',W1,'tons')\n",
+ "print'%s %.3f %s'%('If the load is uniformly spread over its span, then W is given by W1 =',W1/l,'ton per foot run');\n",
+ "print'%s %.2f %s'%('If the load is concentrated at the centre, then W is given by W =',W2,'tons')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "If the load is uniformly spread over its span, then W is given by W1= 16.1 tons\n",
+ "If the load is uniformly spread over its span, then W is given by W1 = 0.671 ton per foot run\n",
+ "If the load is concentrated at the centre, then W is given by W = 8.05 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculalte The maximum permissible span for this beam\n",
+ "d = 20.;## inches\n",
+ "I = 1673.;## inch units\n",
+ "W = 3./4.;## ton per foot run\n",
+ "f = 8.;## tons/in^2\n",
+ "Z = I/(d/2.);## inch-units\n",
+ "M_r = f*Z;## ton-inches\n",
+ "l = math.sqrt(M_r*32/(3*12));##feet\n",
+ "print'%s %.2f %s'%('The maximum permissible span for this beam is l =',l,'feet')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum permissible span for this beam is l = 34.49 feet\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate X\n",
+ "w = 160.;## lb. per sq. foot\n",
+ "b = 3.;## inches\n",
+ "d = 9.;## inches\n",
+ "l = 15.;## feet\n",
+ "f = 1200.;## lb. per sq. inch\n",
+ "Z = (1./6.)*b*d**2;## in^3\n",
+ "M_r = f*Z;## lb-inches\n",
+ "x = M_r/(w*l**2*12./8.);## feet\n",
+ "print'%s %.1f %s'%('x =',x,'feet');\n",
+ "\n",
+ "##The answer is correct only, but it is approximated in the text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x = 0.9 feet\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate area of cross section of beam\n",
+ "l = 20.;## feet\n",
+ "b = 9.;## inches\n",
+ "h = 10.;## feet\n",
+ "w = 120.;## lb. per cub. foot\n",
+ "f = 1100.;## lb/in^2\n",
+ "W = w*(3./4.)*l*h;## lb-wt\n",
+ "BM_max = W*l*12./8.;## lb-inches\n",
+ "##assumnig d = 2b\n",
+ "b = (6.*BM_max/(f*4.))**(1/3.);## inches\n",
+ "d = 2.*b;## inches\n",
+ "print'%s %.3f %s'%('b =',b,'inches')\n",
+ "print'%s %.2f %s'%('d =',d,'inches',)\n",
+ "print'%s %.d %s'%('A section',b,'') \n",
+ "print'%s %.d %s'%('X',d,'will therfore do.')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "b = 9.030 inches\n",
+ "d = 18.06 inches\n",
+ "A section 9 \n",
+ "X 18 will therfore do.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate beam of area\n",
+ "B = 5.;## inches\n",
+ "D = 12.;## inches\n",
+ "t1 = 0.55;## inches\n",
+ "t2 = 0.35;## inches\n",
+ "f = 15./2.;## tons/in^2\n",
+ "l = 16.;## feet\n",
+ "b = B-t2;## inches\n",
+ "d = D-2.*t1;## inches\n",
+ "I_xx = (B*D**3 - b*d**3)/12.;## in^4\n",
+ "Z = I_xx/6.;## in^3\n",
+ "M_r = f*Z;## ton-inches\n",
+ "W = M_r/(l*12./8.);## tons\n",
+ "w = W/l;## ton per foot run\n",
+ "print'%s %.2f %s'%('W =',W,'tons')\n",
+ "print'%s %.2f %s'%('w =',w,'ton per foot run')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "W = 11.36 tons\n",
+ "w = 0.71 ton per foot run\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum stress\n",
+ "D = 19.5;## inches\n",
+ "d = 18.;## inches\n",
+ "l = 30.;## feet\n",
+ "t1 = 3./4.;## inch\n",
+ "rho1 = 450.;## lb. per cub. foot\n",
+ "rho2 = 62.5;## lb. per cub. foot\n",
+ "A = 0.25*math.pi*(D**2 - d**2);## sq. in\n",
+ "DW = rho1*l*A/144.;## lb-wt\n",
+ "WW = rho2*0.25*math.pi*(D-d)**2*l;## lb-wt\n",
+ "W = DW+WW;## lb-wt\n",
+ "BM_max = W*l*0.0004467202*12./8.;## ton-inches\n",
+ "I_xx = (math.pi/64)*(D**4 - d**4);## in^4\n",
+ "Z_xx = I_xx/(0.5*d+t1);## ton/in^2\n",
+ "f = BM_max/Z_xx;## ton/in^2\n",
+ "print'%s %.3f %s'%('The maximum stress f =',f,'ton/in^2');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum stress f = 0.751 ton/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum permissible span for this beam\n",
+ "b = 6.;## inches\n",
+ "d = 12.;## inches\n",
+ "t1 = 7./8;## inch\n",
+ "t2 = 1./2.;## inch\n",
+ "I_xx = (1./12.)*(b*d**3 - (b-t2)*(d-2*t1)**3);## in^4\n",
+ "Z1 = I_xx/b;## in ^3\n",
+ "A = 2*b*t1 + 0.5*(d-2*t1);## in^2\n",
+ "b = math.sqrt(A/2);## inches\n",
+ "d = 2.*b;## inches\n",
+ "Z2 = (1./6.)*b*d**2;## in^3\n",
+ "k = Z1/Z2 ;\n",
+ "print'%s %.2f %s'%('The ratio of strengths Z1/Z2 =',k,'');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of strengths Z1/Z2 = 4.24 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the strength of the solid circular section is taken as unity\n",
+ "A = 15.625;## in**2\n",
+ "Z1 = 61.75;## in**3\n",
+ "Z2 = 14.63;## in**3\n",
+ "d = math.sqrt(4.*A/math.pi);## inches\n",
+ "Z3 = (math.pi/32.)*d**3.;## in**3\n",
+ "R1 = Z1/Z3;\n",
+ "R2 = Z2/Z3;\n",
+ "print('If the strength of the solid circular section is taken as unity') \n",
+ "print'%s %.2f %s %.2f %s'%('that of the rectangular section is',R1,' and of the I-section it is ',R2,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "If the strength of the solid circular section is taken as unity\n",
+ "that of the rectangular section is 7.09 and of the I-section it is 1.68 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Position of the c.g of the sections\n",
+ "D = 8.;##h inches\n",
+ "B = 3.;## inches\n",
+ "t1 = 1/2.;## inch\n",
+ "t2 = 3/8.;## inch\n",
+ "b = B-t2;## inches\n",
+ "d = D-2.*t1;## inches\n",
+ "a1 = t1*B;## in**2\n",
+ "x1 = 0.5*B;## inches\n",
+ "a2 = t2*(D-2.*t1);## in**2\n",
+ "x2 = 0.5*t2;## inches\n",
+ "a3 = B*t1;## in**2\n",
+ "x3 = 0.5*B;## inches\n",
+ "a = a1+a2+a3;## in**2\n",
+ "P = (a1*x1+a2*x2+a3*x3)/(a1+a2+a3);## inches\n",
+ "I_xx = (1/12.)*(B*D**3. - b*d**3.);## in**4\n",
+ "I_AB = (1/3.)*t1*B**3. + (1/3.)*d*t2**3. + (1/3.)*t1*B**3.;## in**4\n",
+ "I_yy = I_AB - a*P**2.;## in**4\n",
+ "print'%s %.3f %s'%('Position of the c.g of the section P = ',P,' inches');\n",
+ "print'%s %.4f %s'%('I_xx = ',I_xx,'in**4')\n",
+ "print'%s %.3f %s' %('I_yy =',I_yy,'in**4');\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the c.g of the section P = 0.887 inches\n",
+ "I_xx = 52.9688 in**4\n",
+ "I_yy = 4.692 in**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The position of the c.g and Maximum stresses induced\n",
+ "b = 6.;## inches\n",
+ "d = 4.;## inches\n",
+ "t = 5/8.;## inch\n",
+ "a1 = d*t;## in^2\n",
+ "y1 = d/2.;## inches\n",
+ "a2 = (b-t)*t;## in^2\n",
+ "y2 = t/2.;## inch\n",
+ "a = a1+a2;## in^2\n",
+ "J = (a1*y1+a2*y2)/(a1+a2);## inches\n",
+ "I_AB = (1/3.)*t*d**3 + (1/3.)*(b-t)*t**3;## in^4\n",
+ "I_xx = I_AB - a*J**2.;## in^4\n",
+ "I_yy = (1/12.)*t*b**3. + (1/12.)*(d-t)*t**3.;## in^4\n",
+ "print'%s %.1f %s'%('The position of the c.g is J = ',J,' inches');\n",
+ "print'%s %.2f %s %.2f %s'%(' I_xx =',I_xx,' in^4,'' I_yy = ',I_yy,'in^4');\n",
+ "\n",
+ "\n",
+ "##14(a)\n",
+ "H = 18.;## feet\n",
+ "l = 10.;## feet\n",
+ "w = 3/2.;## cwt/ per .sq. foot\n",
+ "y_c = 2.97;## inches\n",
+ "y_t = 1.03;## inches\n",
+ "W = (3/40.)*(w*l);## tons\n",
+ "BM_max = W*l*12./8.;## ton-inches\n",
+ "M_r = BM_max;## ton-inches\n",
+ "f_c = M_r*y_c/I_xx ;## tons/in^2\n",
+ "f_t = M_r*y_t/I_xx ;## tons/in^2\n",
+ "print'%s %.2f %s %.2f %s'%(' Maximum stresses induced are f_c = ',f_c,' tons/in^2,''f_t = ',f_t,' tons/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of the c.g is J = 1.0 inches\n",
+ " I_xx = 7.52 in^4, I_yy = 11.32 in^4\n",
+ " Maximum stresses induced are f_c = 6.66 tons/in^2,f_t = 2.31 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The position of c.g is x_bar and Total uniformly distribute load over the span\n",
+ "b = 5.;## inches\n",
+ "d = 4.;## inches\n",
+ "t = 1/2.;## inches\n",
+ "a1 = b*t;## in**2\n",
+ "x1 = t/2.;## inches\n",
+ "y1 = b/2.;## inches\n",
+ "a2 = (d-t)*t;## in**2\n",
+ "y2 = t/2.;## inch\n",
+ "x2 = t + 0.5*(d-t);## inches\n",
+ "x_bar = (a1*x1+a2*x2)/(a1+a2);## inches\n",
+ "y_bar = (a1*y1+a2*y2)/(a1+a2);## inches\n",
+ "I_AB = (1/3.)*t*b**3. + (1/3.)*(d-t)*t**3.;## in**4\n",
+ "I_xx = I_AB - (a1+a2)*y_bar**2.;## in**4\n",
+ "I_yy = (1/3.)*t*d**3. + (1/3.)*(b-t)*t**3. - (a1+a2)*x_bar**2.;## in**4\n",
+ "print'%s %.3f %s %.2f %s'%('The position of c.g is x_bar = ',x_bar,'inches'and'y_bar = ',y_bar,'inches');\n",
+ "print'%s %.3f %s %.2f %s'%('I_xx = ',I_xx,'in**4'and 'I_yy = ',I_yy,' in**4');\n",
+ "\n",
+ "\n",
+ "\n",
+ "##Example 15(a)\n",
+ "l = 12.;## feet\n",
+ "y_c = y_bar;##inches\n",
+ "y_t = b - y_c;##inches\n",
+ "f_t_max = 7.;## tons/in**2\n",
+ "f_c = y_c*f_t_max/y_t;## tons/in**2\n",
+ "M_r = f_t_max*I_xx/y_t;## ton-inches\n",
+ "W = M_r/(l*12./8.);## tons\n",
+ "print'%s %.2f %s'%(' Total uniformly distribute load over the span is W = ',W,' tons');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of c.g is x_bar = 1.074 y_bar = 1.57 inches\n",
+ "I_xx = 10.456 I_yy = 5.96 in**4\n",
+ " Total uniformly distribute load over the span is W = 1.19 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The c.g of the section is y_bar and Load required\n",
+ "b1 = 6.;##inches\n",
+ "d1 = 1.;##inch\n",
+ "b2 = 9.;##inches\n",
+ "d2 = 1.;##inch\n",
+ "b3 = 10.;##inches\n",
+ "d3 = 2.;##inch\n",
+ "a1 = b3*d3;## in**2\n",
+ "y1 = d3/2.;## inches\n",
+ "a2 = b2*d2;## in**2\n",
+ "y2 = d3 + b2/2.;## inches\n",
+ "a3 = b1*d1;## in**2\n",
+ "y3 = b2+d3+d1/2.;## inches\n",
+ "y_bar = (a1*y1+a2*y2+a3*y3)/(a1+a2+a3);##inches\n",
+ "I_AB = (1/3.)*b3*d3**3 + (1/12.)*d2*b2**3. +b2*(d3+b2/2.)**2 + (1/12.)*b1*d1**3 + b1*(b2+d3+d1/2.)**2.;## in**4\n",
+ "I_xx = I_AB - (a1+a2+a3)*y_bar**2;## in**4\n",
+ "I_yy = (1/12.)*(d3*b3**3 + b2*d2**3 +d1*b1**3.);## in**4\n",
+ "print'%s %.3f %s'%('The c.g of the section is y_bar = ',y_bar,' inches');\n",
+ "print'%s %.2f %s %.2f %s'%(' I_xx = ',I_xx,' in**4' 'I_yy = ',I_yy,' in**4');\n",
+ "\n",
+ "##Example 16(a)\n",
+ "l = 20.;## feet\n",
+ "y_t = y_bar;## inches\n",
+ "y_c = d1+b2+d3-y_t;## inches\n",
+ "f_t = 1.5;## tons/in**2\n",
+ "f_c = y_c*f_t/y_t;## tons/in**2\n",
+ "M_r = f_c*I_xx/y_c;## ton-inches\n",
+ "W = M_r*8./(l*12.);## tons\n",
+ "w = W/l;## ton per foot run\n",
+ "print'%s %.2f %s'%('Load required is w = ',w,' ton per foot run');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The c.g of the section is y_bar = 4.214 inches\n",
+ " I_xx = 640.06 in**4I_yy = 185.42 in**4\n",
+ "Load required is w = 0.38 ton per foot run\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate stresses\n",
+ "b = 12.;## inches\n",
+ "d = 6.;## inches\n",
+ "h = 14.;## inches\n",
+ "t = 1/2.;## inch\n",
+ "A = 12.94;## in^2\n",
+ "##section moment of inertia\n",
+ "I_xx_s = 315.3;## in^4\n",
+ "I_yy_s = 22.27;## in^4\n",
+ "\n",
+ "I_xx = 2.*I_xx_s + 2.*((1./12.)*h*(2.*t)**3 + h*2.*t*(d+t)**2.);## in^4\n",
+ "I_yy = 2.*(I_yy_s + A*(d/2.)**2.) + 2.*((1/12.)*2*t*h**3.);## in^4\n",
+ "print'%s %.2f %s %.2f %s'%('I_xx = ',I_xx,' in^4 I_yy = ',I_yy,' in^4');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_xx = 1815.93 in^4 I_yy = 734.79 in^4\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum loads\n",
+ "b = 15.;## inches\n",
+ "d = 15/2.;## inches\n",
+ "h = 16.;## inches\n",
+ "t = 1/2.;## inch\n",
+ "P = 0.935;## inches\n",
+ "A = 12.33;## in**2\n",
+ "##section moment of inertia\n",
+ "I_xx_s = 377.;## in**4\n",
+ "I_yy_s = 14.55;## in**4\n",
+ "\n",
+ "I_xx = 2.*I_xx_s + 2.*((1/12.)*h*(2.*t)**3. + h*2.*t*(d+t)**2.);## in**4\n",
+ "I_yy = 2.*(I_yy_s + A*(d/2. + P)**2.) + 2.*((1/12.)*2.*t*h**3.);## in**4\n",
+ "print'%s %.2f %s %.2f %s'%('I_xx = ',I_xx,'in**4 I_yy = ',I_yy,'in**4')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_xx = 2804.67 in**4 I_yy = 1253.03 in**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum stresses\n",
+ "b1 = 16.;## inches\n",
+ "d1 = 6.;## inches\n",
+ "b2 = 9.;## inches\n",
+ "d2 = 7.;## inches\n",
+ "A = 14.71;## in**2\n",
+ "I_xx1 = 618.09;## in**4\n",
+ "I_yy1 = 22.47;## in**4\n",
+ "I_xx2 = 208.13;## in**4\n",
+ "I_yy2 = 40.17;## in**4\n",
+ "I_xx = I_xx1 + 2.*I_yy2;## in**4\n",
+ "I_yy = I_yy1 + 2.*(I_xx2 + A*(b2/2. + 0.5*0.4)**2.);## in**4\n",
+ "k_xx = math.sqrt(I_xx/(3.*A));## inches\n",
+ "k_yy = math.sqrt(I_yy/(3.*A));## inches\n",
+ "print'%s %.2f %s %.2f %s %.2f %s %.2f %s'%('I_xx =',I_xx,' in**4 I_yy = ',I_yy,' in**4' 'k_xx =',k_xx,' inches'' k_yy = ',k_yy,' inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_xx = 698.43 in**4 I_yy = 1088.62 in**4k_xx = 3.98 inches k_yy = 4.97 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum load stresses\n",
+ "b1 = 7/2.;## inches\n",
+ "d1 = 7/2.;## inches\n",
+ "t1 = 3/8.;## inches\n",
+ "l = 18.;## inches\n",
+ "I_xx1 = 2.80;## in^4\n",
+ "I_yy1 = 2.80;## in^4\n",
+ "J = 1.;## inch\n",
+ "A = 2.49;## in^2\n",
+ "I_xx = 4.*(I_xx1 + A*(l/2. - J)**2);## in^4\n",
+ "k_xx = math.sqrt(I_xx/(4.*A));## inches\n",
+ "print'%s %.2f %s %.1f %s'%('I_xx = ',I_xx,' in^4'' k_xx = ',k_xx,' inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_xx = 648.64 in^4 k_xx = 8.1 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex21-pg170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum stresses\n",
+ "b1 = 12.;## inches\n",
+ "d1 = 4.;## inches\n",
+ "A = 9.21;## in^2\n",
+ "I_xx1 = 200.1;## in^4\n",
+ "P = 1.055;## inches\n",
+ "I_yy1 = 12.12;## in^4\n",
+ "I_xx = 2.*I_xx1;\n",
+ "##for equal strength I_xx = I_yy\n",
+ "x = 2.*(math.sqrt(((I_xx/2.)-I_yy1)/A) - P);## in^4\n",
+ "print'%s %.2f %s'%('x = ',x,' inches');\n",
+ "\n",
+ "##answer is corrct only, but it is approximated in the text book.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x = 6.93 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex22-pg184"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The moment of resistance offered by the flanges and The moment of resistance offered by the flanges\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "d =10.;## inches\n",
+ "b = 8.;## inches\n",
+ "t1 = 1.;## inch\n",
+ "t2 = 0.6;## inch\n",
+ "I = (1./12.)*(b*d**3. - (b-t2)*b**3.);## in^4\n",
+ "##(i) Resistance to M\n",
+ "def fun(y):\n",
+ "\tx=(t2/I)*y**2.\n",
+ "\treturn x\n",
+ "\n",
+ "R1 = scipy.integrate.quad(fun,-4.,4.);\n",
+ "R1=R1[0]\n",
+ "##(ii) Resistance to F\n",
+ "def fun(y):\n",
+ "\tz=(4./I)*(25.-y**2.)\n",
+ "\treturn z\n",
+ "\n",
+ "R2 = scipy.integrate.quad(fun,4.,5.);\n",
+ "R2=R2[0]\n",
+ "print'%s %.2f %s %.2f %s %.1f %s '%('The moment of resistance offered by the flanges is',1-R1,'M' 'The flanges take up ',(1-R1)*100,'percentage of the B.M.,''the web resisting only ',R1*100,'percentage of the B.M');\n",
+ "print'%s %.3f %s %.3f %s %.3f %s '%(' The shear borne by the web is',(1-2*R2),'F' and ' The web thus takes up ',(1-2*R2)*100,' ' and 'percentage of the shear force.'and 'the flanges resisting only',2*R2*100,' percentage of the shear force');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The moment of resistance offered by the flanges is 0.93 MThe flanges take up 92.71 percentage of the B.M.,the web resisting only 7.3 percentage of the B.M \n",
+ " The shear borne by the web is 0.894 The web thus takes up 89.362 the flanges resisting only 10.638 percentage of the shear force \n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex23-pg186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum intensity of shear stress\n",
+ "b = 14.;## inches\n",
+ "d = 6.;## inches\n",
+ "t1 = 0.7;## inch\n",
+ "t2 = 0.4;## inch\n",
+ "F = 20.;## tons\n",
+ "I = (1./12.)*(d*b**3. - (d-t2)*(b-2.*t1)**3.);## in^4\n",
+ "q = F*t1*0.5*(0.5*b + (0.5*b-t1))/I;## ton/in^2\n",
+ "q_max = (F/(I*t2))*(d*t1*0.5*(0.5*b + (0.5*b-t1)) + t2*(0.5*b-t1)*(0.5*b-t1)*0.5);## tons/in^2\n",
+ "print'%s %.2f %s'%('The maximum intensity of shear stress is q_max = ',q_max,' tons/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum intensity of shear stress is q_max = 4.09 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex24-pg186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum intensity of shear stress\n",
+ "b = 4.;## inches\n",
+ "d = 13/2.;## inches\n",
+ "t = 1/2.;## inch\n",
+ "a = 4.;## inches\n",
+ "F = 10.;## tons\n",
+ "a1 = b*t;## in^2\n",
+ "y1 = t/2.;## inch\n",
+ "a2 = (d-t)*t;## in^2\n",
+ "y2 = t+0.5*(d-t);## inches\n",
+ "y_bar = (a1*y1+a2*y2)/(a1+a2);## inches\n",
+ "I_AB = (1/3.)*b*t**3. + (1./12.)*t*(d-t)**3 + (b-2*t)*(b-t)**2.;## in^4\n",
+ "I_xx = I_AB - (a1+a2)*y_bar**2.;## in^4\n",
+ "q = (F/(b*I_xx))*b*t*(y_bar-0.5*t);## ton/in^2\n",
+ "q_max = (F/(t*I_xx))*(b*t*(y_bar-0.5*t) + 0.5*t*(y_bar-t)*(y_bar-t));## tons/in^2\n",
+ "print'%s %.2f %s'%('The maximum intensity of shear stress at the N.A is q_max = ',q_max,' tons/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum intensity of shear stress at the N.A is q_max = 4.26 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex25-pg190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate tensile and compressive various inches at 2,4, 6 and at N.A points\n",
+ "import math\n",
+ "def func(p,q):\n",
+ " p1 = 0.5*p + math.sqrt(q**2. + 0.25*p**2);\n",
+ " p2 = 0.5*p - math.sqrt(q**2. + 0.25*p**2);\n",
+ " theta = 0.5*math.atan(2.*q/p) * 180/math.pi;\n",
+ " return p1,p2,theta\n",
+ "b = 5.;## inches\n",
+ "d = 12.;## inches\n",
+ "F = 4800. ;## lb-wt\n",
+ "M = 192000.;## lb-inches\n",
+ "I = (1/12.)*b*d**3.;## in**4\n",
+ "\n",
+ "##At 6 inches above the N.A\n",
+ "p6 = M*6./I ; ## lb/in**2\n",
+ "q6 = 0;\n",
+ "[p1_6,p2_6,theta6] = func(p6,q6);\n",
+ "\n",
+ "##At 4 inches above the N.A\n",
+ "p4 = M*4/I;## lb/in**2\n",
+ "q4 = (F/(I*b))*b*(0.5*d-4)*b;\n",
+ "[p1_4,p2_4,theta4] = func(p4,q4);\n",
+ "\n",
+ "##At 2 inches above the N.A\n",
+ "p2 = M*2./I;## lb/in**2\n",
+ "q2 = (F/(I*b))*b*(0.5*d-2)*4.;\n",
+ "[p1_2,p2_2,theta2] = func(p2,q2);\n",
+ "\n",
+ "##At the N.A\n",
+ "p = 0.;##\n",
+ "q = F*b*0.5**3*d**2./(I*b);## lb/in**2\n",
+ "p1 = q;## lb/in**2\n",
+ "p2 = -q;## lb/in**2\n",
+ "\n",
+ "print'%s %.2f %s %.2f %s '%('At 6 inches above the N.A, p1 =',p1_6,' lb/in**2.'and 'compressive, and p2 = ',p2_6,'');\n",
+ "print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%(' At 4 inches above the N.A, p1 =',p1_4,' lb/in**2.'and ' compressive, and p2 = ',-p2_4,' lb/in**2 'and'tensile theta1 = ',theta4,' degrees' 'theta2 = ',theta4+90,' degrees');\n",
+ "print'%s %.2f %s %.2f %s %.2f %s %.2f %s'%(' At 2 inches above the N.A, p1 = ',p1_2,' lb/in**2.'and' compressive' and 'p2 = ',-p2_2,' lb/in**2.,'and 'tensile theta1 = ',theta2,' degrees 'and ' theta2 = ',theta2+90,' degrees');\n",
+ "print'%s %.2f %s %.2f %s'%(' At the N.A, p1 = ',p1,' lb/in**2.,'and ' compressive, and p2 = ',-p2,'.,tensile ');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At 6 inches above the N.A, p1 = 1600.00 compressive, and p2 = 0.00 \n",
+ " At 4 inches above the N.A, p1 = 1070.82 compressive, and p2 = 4.15 tensile theta1 = 3.56 degreestheta2 = 93.56 degrees \n",
+ " At 2 inches above the N.A, p1 = 553.88 p2 = 20.54 tensile theta1 = 10.90 theta2 = 100.90 degrees\n",
+ " At the N.A, p1 = 120.00 compressive, and p2 = 120.00 .,tensile \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex26-pg194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate At the top, principal stresses and The principal stresse across the diagonal and compressive on one plane\n",
+ "b = 10.;## inches\n",
+ "d = 8.;## inches\n",
+ "t1 = 1.;## inch\n",
+ "t2 = 0.6;## inch\n",
+ "M = 500.;## ton-inches\n",
+ "F = 25.;## tons\n",
+ "I = (1./12.)*(d*b**3. - (d-t2)*d**3.);## in**4\n",
+ "\n",
+ "##At the top\n",
+ "p = M*b/(2.*I);## tons/in**2\n",
+ "q = 0.;\n",
+ "p1 = p;## tons/in**2\n",
+ "p2 = 0.;\n",
+ "print'%s %.1f %s %.1f %s'%('At the top, principal stresses are p1 =',p1,'tons/in**2'' p2 = ',p2,' tons/in**2')\n",
+ "\n",
+ "##In the web, 4 inches from the N.A\n",
+ "p = M*d/(2.*I);## tons/in**2\n",
+ "q = F*d*t1*0.5*(d+t1)/(I*t2);## tons/in**2\n",
+ "theta = 0.5*math.atan(2*q/p);\n",
+ "theta1 = theta*180/math.pi;\n",
+ "theta2 = theta1+90;\n",
+ "p1 = 0.5*p + math.sqrt(q**2 + 0.25*p**2);## tons/in**2\n",
+ "p2 = 0.5*p - math.sqrt(q**2 + 0.25*p**2);## tons/in**2\n",
+ "print'%s %.3f %s %.1f %s %.1f %s %.1f %s '%(' In the web, 4 inches from the N.A.:'' The principal stresse are p1 = ',p1,' tons/in**2.,compressive\\n p2 = ',-p2,' tons/in**2.,tensile\\n theta1 =',theta1,' degrees\\n theta2 = ',theta2,' degrees')\n",
+ "\n",
+ "\n",
+ "##At the N.A\n",
+ "p = 0.;\n",
+ "q = (F/(I*t2))*(d*t1*0.5*(d+t1) + t2*0.5*d*2*t1);\n",
+ "p1 = q;## tons/in**2\n",
+ "p2 = -q;##tons/in**2\n",
+ "print'%s %.1f %s %.1f %s'%(' The principal stresse across the diagonal are ',q,' tons/in**2., compressive on one plane and',q,' tons/in**2., tensile on the other.');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At the top, principal stresses are p1 = 7.1 tons/in**2 p2 = 0.0 tons/in**2\n",
+ " In the web, 4 inches from the N.A.: The principal stresse are p1 = 7.987 tons/in**2.,compressive\n",
+ " p2 = 2.3 tons/in**2.,tensile\n",
+ " theta1 = 28.2 degrees\n",
+ " theta2 = 118.2 degrees \n",
+ " The principal stresse across the diagonal are 4.8 tons/in**2., compressive on one plane and 4.8 tons/in**2., tensile on the other.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex27-pg195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate maximum intensity of shear stress \n",
+ "W = 10.;## tons\n",
+ "l = 16.;## feet\n",
+ "f = 15/2.;## tons/in^2\n",
+ "##section modulus required\n",
+ "SM = W*l*12./(8.*f);## in^3\n",
+ "##for this section modulus \n",
+ "l1 = 12.;## inches\n",
+ "b1 = 5.;## inches\n",
+ "t1 = 0.55;## inches\n",
+ "t2 = 0.35;## inches\n",
+ "I_xx = 220.;## in^4\n",
+ "F_max = 5.;## tons\n",
+ "q_max = (F_max/(I_xx*t2))*(F_max*t1*(0.5*l1-0.5*t1) + t2*0.5*(0.5*l1-t1)**2);## tons/in^2\n",
+ "print'%s %.2f %s'%('The maximum intensity of shear stress is q_max = ',q_max,' tons/in^2');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum intensity of shear stress is q_max = 1.36 tons/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex28-pg198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Skin stresse in steel plate and The total moment of resistance\n",
+ "b = 9/2.;## inches\n",
+ "D = 12.;## inches\n",
+ "d = 10.;## inches\n",
+ "t = 1/2.;## inches\n",
+ "f_w = 1000.;## lb/in**2\n",
+ "m = 18.;##m = E_s/E_w\n",
+ "f_t = m*d*f_w/D ;## lb/in**2\n",
+ "M_w = f_w*(1./6.)*2.*b*D**2.;## lb-inches\n",
+ "M_s = f_t*(1/6.)*t*d**2;## lb-inches\n",
+ "M = M_w + M_s;## lb-inches\n",
+ "print'%s %.1f %s %.1f %s'%('Skin stresse in steel plate is, M_s = ',M_s,' lb-inches'' The total moment of resistance is M =',M,' lb-inches')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Skin stresse in steel plate is, M_s = 125000.0 lb-inches The total moment of resistance is M = 341000.0 lb-inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter6.ipynb b/Mechanics_of_Structures/Chapter6.ipynb new file mode 100755 index 00000000..70a08d4a --- /dev/null +++ b/Mechanics_of_Structures/Chapter6.ipynb @@ -0,0 +1,491 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c177c52862f24e7aa2f65bd196b1696db90de567ae2d80d6061ed266557f0af1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Beams and bending part 3"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The deflection at the free\n",
+ "l = 5.;## feet\n",
+ "W = 150.;## lb\n",
+ "w = 120.;## lb. per foot run\n",
+ "l1 = 3.;## feet\n",
+ "b = 3.;## inches\n",
+ "d = 6.;## inches\n",
+ "E = 1.5*10**6;## lb/in^2\n",
+ "I = (1./12.)*b*d**3;## in^4\n",
+ "y_B1 = (W*l**3)/(3.*E*I);## feet\n",
+ "y_B2 = (w*l1*l1**3)/(8.*E*I) + (l-l1)*(w*l1*l1**2)/(6.*E*I);## feet\n",
+ "y_B = (12**3)*(y_B1+y_B2);## inches\n",
+ "print'%s %.4f %s'%('The deflection at the free end =',y_B,'inches')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The deflection at the free end = 0.1823 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Ex2-pg218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'Uniform distributed load\n",
+ "b = 4.;## inches\n",
+ "d = 9.;## inches\n",
+ "l = 12.;## feet\n",
+ "y_c = 1/4.;## inches\n",
+ "E = 1.5*10**6;## lb/in^2\n",
+ "I = (1/12.)*b*d**3;## in^4\n",
+ "W = y_c*384.*E*I/(5.*12**3.*l**3);## inches\n",
+ "print'%s %.d %s'%('Uniform distributed load, the beam should carry is, W =',W,'lb-wt');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniform distributed load, the beam should carry is, W = 2343 lb-wt\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg224"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'Uniform distributed load the beam should carry\n",
+ "b = 4.;## inches\n",
+ "d = 9.;## inches\n",
+ "l = 12.;## feet\n",
+ "y_c = 1/4.;## inches\n",
+ "E = 1.5*10**6;## lb/in^2\n",
+ "I = (1/12.)*b*d**3;## in^4\n",
+ "W = y_c*384.*E*I/(5.*12**3.*l**3);## inches\n",
+ "print'%s %.d %s'%('Uniform distributed load, the beam should carry is, W =',W,'lb-wt');\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniform distributed load, the beam should carry is, W = 2343 lb-wt\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculateThe deflection for a uniformly distributed load\n",
+ "d = 6.;## feet\n",
+ "l = 60.;## feet\n",
+ "f = 15./2.;## tons/in^2\n",
+ "E = 13000.;## tons/in^2\n",
+ "k1 = 2.*f/(12.*d);## k1 = M_r/I \n",
+ "k2 = k1/(l*12./8.);##k2 = W/I\n",
+ "y_c = (5./384.)*k2*l**3 *12**3 /E;## inches\n",
+ "##If the giredr is of constant deapth and uniform strength, it bends to an arc of a circle of radius R \n",
+ "R = E*d*12/(2*f);## inches\n",
+ "delta = (l*12)**2 /(8*R);## inches\n",
+ "print'%s %.2f %s'%('The deflection for a uniformly distributed load on it is,delta =',delta,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The deflection for a uniformly distributed load on it is,delta = 1.04 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The ratio of deapth to span\n",
+ "f = 8.;##tons/in^2\n",
+ "E = 12800.;## tons/in^2\n",
+ "k1 = 1./480.;##central deflection = k = delta/l\n",
+ "k2 = (5./24.)*(f/E)/k1 ;##k2 = d/l = deapth to span ratio\n",
+ "print'%s %.3f %s'%('The ratio of deapth to span, d/l =',k2,'');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of deapth to span, d/l = 0.062 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate 'A section with d and If the deapth of section is limited\n",
+ "w = 550.;## lb. per foot run\n",
+ "f = 1000.;## lb/in^2\n",
+ "l = 20.;## feet\n",
+ "d_limit = 15.;## inches\n",
+ "E = 1.5*10**6;## lb/in^2\n",
+ "##central ddeflection\n",
+ "delta = (1./2.);## inches\n",
+ "d = (5./24.)*(f/E)*20.*12./(1./(2.*20.*12.));## inches\n",
+ "M = w*l*l*12./8.;## lb-inches\n",
+ "b = M/(f*(1./6.)*d**2);## inches\n",
+ "print'%s %.d %s'%('A section with d =',round(d),'inches')\n",
+ "print'%s %.d %s'%('b =',round(b),'inches will do.')\n",
+ "f1 = (1./(2.*20.*12.))*(d_limit/(l*12.))*E/(5./24.);## lb/in^2\n",
+ "b = M/(f1*(1./6.)*d_limit**2);## inches\n",
+ "print'%s %.d %s'%('If the deapth of section is limited to',d_limit,'inches,then')\n",
+ "print'%s %.1f %s'%('f =',f1,'lb/in^2')\n",
+ "print'%s %.1f %s'%('b =',b,'inches');\n",
+ "\n",
+ "##tha answer is correct only, but it is approximated in the text book.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A section with d = 16 inches\n",
+ "b = 8 inches will do.\n",
+ "If the deapth of section is limited to 15 inches,then\n",
+ "f = 937.5 lb/in^2\n",
+ "b = 9.4 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The position of maximum deflection occurs and The maximum deflection and The deflection at the centre\n",
+ "l = 20.;## feet\n",
+ "b = 4.;## feet\n",
+ "W = 5.;## tons\n",
+ "d = 12.;## inches\n",
+ "h = 5.;## inches\n",
+ "I_xx = 220.;## in^4\n",
+ "E = 13000.;## tons/in^2\n",
+ "a = l-b;## feet\n",
+ "##for maximum deflection\n",
+ "x = math.sqrt((a**2 + 2.*a*b)/3.);## feet\n",
+ "y_max = x*12**3 *((a**2 + 2.*a*b) - x**2)/(6.*E*I_xx);## inches\n",
+ "##for deflection at the centre\n",
+ "x1 = 0.5*l;## inches\n",
+ "y_x1 = x1*12**3 *((a*82. + 2.*a*b) - x1*82.)/(6.*E*I_xx);## inches\n",
+ "print'%s %.2f %s'%('The position of maximum deflection occurs at x =',x,'feet')\n",
+ "print'%s %.4f %s'%('The maximum deflection is, y_max =',y_max,'inches');\n",
+ "print'%s %.3f %s'%('The deflection at the centre,',y_x1,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of maximum deflection occurs at x = 11.31 feet\n",
+ "The maximum deflection is, y_max = 0.2917 inches\n",
+ "The deflection at the centre, 0.624 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The position of the maximum deflection and the maximum deflection\n",
+ "d = 12.;## inches\n",
+ "h = 5.;## inches\n",
+ "l = 20.;## feet\n",
+ "E = 13000.; ##tons/in**2\n",
+ "I_xx = 220.;## in**4\n",
+ "W = 4.;## tons\n",
+ "W1 = 3.;## tons\n",
+ "a = 15.;## feet\n",
+ "b = l-a;## feet\n",
+ "a1 = 16.;## feet \n",
+ "b1 = l-a1;## feet\n",
+ "K1 = (-2*W1*b1*l)/(W1*b1-W*b);\n",
+ "K2 = (W*b*a**2 + 2*a*W*b**2 + 2*W1*b1*l**2 - W1*b1*a1**2 -2*W1*a1*b1**2 +W1*b1*l**2)/(3*(W1*b1 - W*b));\n",
+ "x = -0.5*K1 + math.sqrt(-K2 + 0.25*K1**2);## feet\n",
+ "x1 = l-x;## feet\n",
+ "y_max = W*b*x*1728.*(a**2 +2*a*b -x**2)/(6.*E*I_xx*l) + W1*b1*x1*1728.*(a1**2 +2.*a1*b1 -x1**2)/(6.*E*I_xx*l);## inches\n",
+ "print'%s %.2f %s'%('The position of the maximum deflection is, x =',x,'feet.');\n",
+ "print'%s %.4f %s'%('And the maximum deflection is, y_max =',y_max,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of the maximum deflection is, x = 10.26 feet.\n",
+ "And the maximum deflection is, y_max = 0.4488 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The position of the maximum deflection and the maximum deflection\n",
+ "b = 18.;## inches\n",
+ "d = 7.;## inches\n",
+ "w1 = 1.;## ton per foot run\n",
+ "w2 = 3.;## ton per foot run\n",
+ "I_xx = 1149.;## in**4\n",
+ "E = 13000.;## tons/in**2\n",
+ "R_A = 0.5*b + (b/3.);## tons\n",
+ "R_B = 0.5*b + (2.*b/3.);## tons\n",
+ "##integrating M = E*I*y'', to get E*I*y' and making y' = 0;, we get maximu deflection\n",
+ "x = 9.18;## by trial and error method\n",
+ "y_derivative = -R_A*0.5*x**3 + x**4 /6. +0.5*(2./3.)*(1./b)*(1./4.)*x**5 + 469.8;\n",
+ "y = -R_A*0.5*x**3 /3. + x**4 /24. +0.5*(2./3.)*(1./b)*(1./(4.*5.))*x**5 + 469.8*x;\n",
+ "y_max = y;## inches\n",
+ "print'%s %.2f %s'%('The position of maximum deflection from the end A, x =',x,'inches')\n",
+ "print'%s %.4f %s'%('Maximum deflection, y_max =',y_max*12**3 /(E*I_xx),'inches')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of maximum deflection from the end A, x = 9.18 inches\n",
+ "Maximum deflection, y_max = 0.3164 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The position of the maximum deflection and the maximum deflection and stress induced\n",
+ "b = 18.;## inches\n",
+ "d = 6.;## inches\n",
+ "l = 16.;## feet\n",
+ "W = 2.;## tons\n",
+ "h = 1./2.;## inches\n",
+ "I_xx = 841.76;## in**4\n",
+ "E = 13000.;## tons/in**2\n",
+ "P = W + math.sqrt(2.*W*h*48.*E*I_xx/(l*12.)**3 + 2.*W);## tons\n",
+ "M_max = P*l*12./4.;## ton-inches\n",
+ "Z = 2.*I_xx/b ;## in**3\n",
+ "f = M_max/Z ;## tons/in**2\n",
+ "delta = P*(l*12.)**3 /(48.*E*I_xx);## inches\n",
+ "print'%s %.3f %s'%('The maximum instantaneous deflection delta =',delta,'inches') \n",
+ "print'%s %.3f %s'%('stress induced, f =',f,'tons/in**2');\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum instantaneous deflection delta = 0.193 inches\n",
+ "stress induced, f = 7.362 tons/in**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Number of plates required and Number of plates required and The initial radius to which the plates must be bent\n",
+ "l = 3.;## feet\n",
+ "b = 3.;## inches\n",
+ "t = 3./8.;## inches\n",
+ "W = 1500.;## lb.\n",
+ "f = 12.;## tons/in**2\n",
+ "E = 30.*10**6;## tons/in**2\n",
+ "M_max = W*l*12./4. ;## lb-inches\n",
+ "M_r = f*(1./6.)*b*t**2 *2240.;## lb-inches\n",
+ "n = M_max/M_r ;## no. of plates\n",
+ "n = round(n+1);\n",
+ "f = M_max/(n*(1./6.)*b*t**2);## lb/in**2\n",
+ "R = E/(2.*f/t) ;## inches\n",
+ "delta = (l*12.)**2 /(8.*R);## inches\n",
+ "print'%s %.d %s'%('Number of plates required, n =',n,'');\n",
+ "print'%s %.4f %s'%(' The central deflection, delta =',delta,'inch.');\n",
+ "print'%s %.3f %s'%('The initial radius to which the plates must be bent, R =',R,'inches');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of plates required, n = 8 \n",
+ " The central deflection, delta = 0.6912 inch.\n",
+ "The initial radius to which the plates must be bent, R = 234.375 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter8.ipynb b/Mechanics_of_Structures/Chapter8.ipynb new file mode 100755 index 00000000..ea8f2292 --- /dev/null +++ b/Mechanics_of_Structures/Chapter8.ipynb @@ -0,0 +1,489 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:48abd6a26c994c0391cd48807ab25037b2bca2d3bc885b33747ef231a0cbe4f3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Direct and bending stresses"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "d = 6.;##inches\n",
+ "b = 3/4.;##inch\n",
+ "P = 18.;##tons\n",
+ "e = 1/8.;##inch\n",
+ "A = b*d;##sq.in\n",
+ "M = P*e;##ton-in\n",
+ "Z = (1./6.)*b*d**2;##in^3\n",
+ "p_0 = P/A;## tons/in^2\n",
+ "p_b = M/Z;## ton/in^2\n",
+ "p_max = p_0+p_b;## tons/in^2\n",
+ "p_min = p_0-p_b;## tons/in^2\n",
+ "print'%s %.1f %s'%('p_max =',p_max,'tons/in^2 tensile')\n",
+ "print'%s %.1f %s'%('p_min = ',p_min,'tons/in^2.,tensile')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 4.5 tons/in^2 tensile\n",
+ "p_min = 3.5 tons/in^2.,tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "d1 = 12.;##inches\n",
+ "t = 1.;##inch\n",
+ "d2 = d1-2.*t;##inches\n",
+ "P = 5.;##tons\n",
+ "e = 12.;##inch\n",
+ "A = 0.25*math.pi*(d1**2-d2**2);##sq.in\n",
+ "M = P*e;##ton-in\n",
+ "Z = math.pi*(d1**4-d2**4)/(32*d1);##in^3\n",
+ "p_0 = P/A;## tons/in^2\n",
+ "p_b = M/Z;## ton/in^2\n",
+ "p_max = p_0+p_b;## tons/in^2\n",
+ "p_min = p_0-p_b;## tons/in^2\n",
+ "print'%s %.4f %s'%('p_max =',p_max,'ton/in^2.,cmopressive')\n",
+ "print'%s %.4f %s'%('p_min =',p_min,'ton/in^2., tensile')\n",
+ "print'%s %.4f %s'%('p_min = ',-p_min,'ton/in^2., tensile')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 0.8278 ton/in^2.,cmopressive\n",
+ "p_min = -0.5384 ton/in^2., tensile\n",
+ "p_min = 0.5384 ton/in^2., tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "l = 6.;##inches\n",
+ "b = 4.;##inches\n",
+ "d = 1/2.;##inch\n",
+ "P = 10.;##tons\n",
+ "r = 1.5;##inches\n",
+ "A = 4.771;## in^2\n",
+ "J = 0.968;## inches\n",
+ "I_xx = 6.07;## in^4\n",
+ "I_yy = 8.64;## in^4\n",
+ "e = r-J;## inches\n",
+ "M = P*e;## ton-inches\n",
+ "y_t = J;##inches \n",
+ "y_c = b-y_t;##inches\n",
+ "##compressive\n",
+ "p_c = M*y_c/I_xx;## tons/in^2\n",
+ "##tensile\n",
+ "p_t = M*y_t/I_xx;## tons/in^2\n",
+ "##compressive\n",
+ "p_0 = P/A ;## tons/in^2\n",
+ "p_max = p_0+p_c;## tons/in^2\n",
+ "p_min = p_0-p_t;## tons/in^2\n",
+ "print'%s %.3f %s'%('p_max =',p_max,'tons/in^2.,compressive')\n",
+ "print'%s %.3f %s'%('p_min =',p_min,'tons/in^2.,compressive')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 4.753 tons/in^2.,compressive\n",
+ "p_min = 1.248 tons/in^2.,compressive\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "b = 5.;##inches\n",
+ "t = 1/2.;##inch\n",
+ "P = 12.;##tons\n",
+ "d = 1/2.;##inch\n",
+ "r = 3/2.;##inch\n",
+ "A = (b-d)*t;## in^2\n",
+ "p_0 = P/A;## tons/in^2\n",
+ "a1 = b*d;## in^2\n",
+ "x1 = 0.;\n",
+ "a2 = d*t;## in^2\n",
+ "x2 = -3/2.;\n",
+ "e = (a1*x1-a2*x2)/(a1-a2);##inches\n",
+ "M = P*e;## ton-inches\n",
+ "y_c = a1-e;## inches\n",
+ "y_t = a1+e;## inches\n",
+ "I_yy = (t*b**3)/12. - ((t*d**3)/12. + 0.5*d*r**2);## in^4\n",
+ "I_GG = I_yy - (a1-a2)*(e**2);## in^4\n",
+ "p_c = M*y_c/I_GG;## tons/in^2\n",
+ "p_t = M*y_t/I_GG;## tons/in^2\n",
+ "p_max = p_0 + p_t;## tons/in^2\n",
+ "p_min = p_0 - p_c;## tons/in^2\n",
+ "print'%s %.1f %s'%('p_max = ',p_max,' tons/in^2., tensile')\n",
+ "print'%s %.2f %s'%('p_min =',p_min,'tons/in^2., tensile')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 6.5 tons/in^2., tensile\n",
+ "p_min = 4.31 tons/in^2., tensile\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "h = 20.;## feet\n",
+ "b = 12.;## feet\n",
+ "d = 4.;## feet\n",
+ "p = 30.;## lb. per sq.foot\n",
+ "rho = 140.;## lb. per cubic foot\n",
+ "p_0 = rho*h;## lb-ft^2\n",
+ "P = p*b*h;## lb-wt\n",
+ "M = P*h/2;##lb-ft\n",
+ "Z = b*d**2/6;## ft^3\n",
+ "p_b = M/Z;## lb/ft^2\n",
+ "p_max = p_0 + p_b;## lb/ft^2\n",
+ "p_min = p_0 - p_b;## lb/ft^2\n",
+ "print'%s %.d %s'%('p_max =',p_max,'lb/ft^2., compressive')\n",
+ "print'%s %.d %s'%('p_min =',p_min,'lb/ft^2., compressive')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 5050 lb/ft^2., compressive\n",
+ "p_min = 550 lb/ft^2., compressive\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The necessary thickness\n",
+ "h = 80.;## feet\n",
+ "p = 28.;## lb. per sq.foot\n",
+ "rho = 126.;## lb. per cubic foot\n",
+ "p_0 = rho*h/2240.;## tons-ft^2\n",
+ "p_max = 7.;## tons/ft^2\n",
+ "d = 4.;## feet\n",
+ "p_b = p_max - p_0;## tons/ft^2\n",
+ "D = math.sqrt(3*p*h**2 /(2*p_b*2240.) +math.sqrt(d**4 + (3*p*h**2 /(2.*p_b*2240.))**2));## feet\n",
+ "t = 0.5*(D-d);## feet\n",
+ "print'%s %.d %s'%('The necessary thickness is, t =',round(t),'feet');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The necessary thickness is, t = 3 feet\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "h = 60.;##feet\n",
+ "rho = 130.;## lb. per cubic foot\n",
+ "D = 12.;##feet\n",
+ "d = 5.;##feet\n",
+ "P_h = 24.;## lb. per sq. foot\n",
+ "p_0 = rho*h;## lb-ft^2\n",
+ "P = P_h*D*h;## lb-wt\n",
+ "M = P*h/2;## lb-feet\n",
+ "Z = math.pi*(D**4 - d**4)/(32*D);##lb-ft^3\n",
+ "p_b = M/Z;## lb/ft^2\n",
+ "p_max = p_0 + p_b;## lb/ft^2\n",
+ "p_min = p_0 - p_b;## lb/ft^2\n",
+ "print'%s %.d %s'%('p_max =',p_max,'lb/ft^2., compressive')\n",
+ "print'%s %.d %s'%('p_min =',p_min,'lb/ft^2., compressive')\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 10950 lb/ft^2., compressive\n",
+ "p_min = 4649 lb/ft^2., compressive\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import scipy\n",
+ "#calculate Bending moment at the foot of the chimney \n",
+ "from scipy import integrate\n",
+ "h = 120.;## feet\n",
+ "d = 5.;## feet\n",
+ "h1 = 49.;## feet\n",
+ "p = 42.;## lb. per square foot\n",
+ "c = 0.6;\n",
+ "k = p/math.sqrt(h1);\n",
+ "##p = k*math.sqrt(x)\n",
+ "\n",
+ "def fun(x):\n",
+ "\ty=18.*x**(3./2.)\n",
+ "\treturn y\n",
+ "\n",
+ "M = scipy.integrate.quad(fun,0.,120.);\n",
+ "M=M[0]\n",
+ "print'%s %.1f %s'%('Bending moment at the foot of the chimney is, M =',M,'lb-ft');\n",
+ "##there is an error in the answer given in text book\n",
+ "print(M)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending moment at the foot of the chimney is, M = 1135757.5 lb-ft\n",
+ "1135757.49523\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Bending moment at the foot of the chimney \n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "h = 100.;## feet\n",
+ "d = 4.;## feet\n",
+ "p = 50.;## lb. per square foot\n",
+ "c = 2/3.;\n",
+ "def fun(x):\n",
+ "\ty=(100*x/3.)*(10-(6*x/100.))\n",
+ "\treturn y;\n",
+ "\n",
+ "M=scipy.integrate.quad(fun,0,100)\t\n",
+ "M = M[0]\n",
+ "print'%s %.d %s'%('Bending moment at the foot of the chimney is, M =',M,'lb-feet');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending moment at the foot of the chimney is, M = 1000000 lb-feet\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate p_max and p_min\n",
+ "h = 20.;## feet\n",
+ "b = 4.;## feet\n",
+ "d = 12.;## feet\n",
+ "h1 = 18.;## feet\n",
+ "##density of masonary\n",
+ "rho_m = 140.;## lb-ft^3\n",
+ "##density of water\n",
+ "rho_w = 62.5;## lb-ft^3\n",
+ "W = rho_m*0.5*(b+d)*h;## lb-wt\n",
+ "##to locate its line of action divide the dam section into a rectangle and a triangle \n",
+ "x1 = b/2.;##feet\n",
+ "a1 = b*h;## sq. feet\n",
+ "a2 = 0.5*(d-b)*h;## sq. feet\n",
+ "x2 = b+((d-b)/3);## feet\n",
+ "x_bar = (a1*x1+a2*x2)/(a1+a2);## feet\n",
+ "P = rho_w*h1**2/2;## lb-wt\n",
+ "z = x_bar + (h1/3)*(P/W);## feet\n",
+ "e = z - d/2.;## feet\n",
+ "p_0 = W/d;## lb/ft^2\n",
+ "M = W*e;## lb-feet\n",
+ "Z = 1*d**2*1/6.;## ft^3\n",
+ "p_b = M/Z;## lb-ft^2\n",
+ "p_max = p_0 + p_b;## lb-ft^2\n",
+ "p_min = p_0 - p_b;## lb-ft^2\n",
+ "print'%s %.1f %s'%('p_max =',p_max,'lb/ft^2., compressive at B.')\n",
+ "print'%s %.1f %s'%('p_min =',p_min,'lb/ft^2., compressive at A.')\n",
+ "\n",
+ "\n",
+ "##there is an error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p_max = 2842.4 lb/ft^2., compressive at B.\n",
+ "p_min = 891.0 lb/ft^2., compressive at A.\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/Chapter9.ipynb b/Mechanics_of_Structures/Chapter9.ipynb new file mode 100755 index 00000000..27f2d79d --- /dev/null +++ b/Mechanics_of_Structures/Chapter9.ipynb @@ -0,0 +1,523 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:de66edc6628557b633dd9783c4dabb08e00d657b0d4a1fadad535f3bd5215433"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-Columns and struts of uniform section"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate m1 and m2,m3,m4,m5,m6,m7,m8,m9\n",
+ "## n =l/k\n",
+ "n1 = 40.;\n",
+ "n2 = 60.;\n",
+ "n3 = 80.;\n",
+ "n4 = 100.;\n",
+ "n5 = 120.;\n",
+ "n6 = 140.;\n",
+ "n7 = 160.;\n",
+ "n8 = 180.;\n",
+ "n9 = 200.;\n",
+ "E = 13000.;## tons/in**2\n",
+ "##m = P/A\n",
+ "m1 = 4*math.pi**2 *E/n1**2;## tons per sq. inch\n",
+ "m2 = 4*math.pi**2 *E/n2**2;## tons per sq. inch\n",
+ "m3 = 4*math.pi**2 *E/n3**2;## tons per sq. inch\n",
+ "m4 = 4*math.pi**2 *E/n4**2;## tons per sq. inch\n",
+ "m5 = 4*math.pi**2 *E/n5**2;## tons per sq. inch\n",
+ "m6 = 4*math.pi**2 *E/n6**2;## tons per sq. inch\n",
+ "m7 = 4*math.pi**2 *E/n7**2;## tons per sq. inch\n",
+ "m8 = 4*math.pi**2 *E/n8**2;## tons per sq. inch\n",
+ "m9 = 4*math.pi**2 *E/n9**2;## tons per sq. inch\n",
+ "print'%s %.d %s %.d %s %.d %s %.d %s %.d %s %.d %s %.d %s %.d %s %.d %s'%('l/k : ',n1,' ',n2,'',n3,' ',n4,' ',n5,'',n6,'',n7,'',n8,'',n9,'' )\n",
+ "print'%s %.1f %s %.1f %s %.1f %s %.1f %s %.1f %s %.1f %s %.1f %s %.1f %s %.1f %s '%('P/A:',m1,'',m2,'',m3,'',m4,'',m5,'',m6,'',m7,'',m8,'',m9,'')\n",
+ "\n",
+ "##there is a minute error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "l/k : 40 60 80 100 120 140 160 180 200 \n",
+ "P/A: 320.8 142.6 80.2 51.3 35.6 26.2 20.0 15.8 12.8 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The collapsing load\n",
+ "d = 1.;## inches\n",
+ "t = 1/8.;## inches\n",
+ "l = 10.;## feet\n",
+ "E = 13500.;## tons/in**2\n",
+ "D = d+2.*t;## inches\n",
+ "I = (math.pi/64.)*(D**4. - d**4.);## in**4\n",
+ "P = 20.25*E*I/(12.*l)**2. ;## tons\n",
+ "print'%s %.2f %s'%('The collapsing load, P =',P,'tons')\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The collapsing load, P = 1.34 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The safe axial load\n",
+ "b = 10.;## inches\n",
+ "d = 6.;##inches\n",
+ "l = 15.;## feet\n",
+ "A = 11.77;## in^2\n",
+ "I_xx = 204.80;## in^4\n",
+ "I_yy = 21.76;## in^4\n",
+ "f_c = 21.;## tons/in^2\n",
+ "a = 1/7500.;\n",
+ "n = 3.;##factor of safety\n",
+ "k = math.sqrt(I_yy/A);## radius of gyration\n",
+ "P = f_c*A/(1.+(a/2.)*(l*12./k)**2);## tons\n",
+ "P_s = P/n;## safe load\n",
+ "print'%s %.3f %s'%('The safe axial load =',P_s,'tons');\n",
+ "\n",
+ "##there is a minute calculation error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe axial load = 37.997 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The internal diameter and The thickness of the metal\n",
+ "l = 16.;## feet\n",
+ "F = 30.;## tons\n",
+ "n = 8.;## factor of safety\n",
+ "k = 0.8;##k = d/D\n",
+ "f_c = 36.;## tons/in^2\n",
+ "a = 1/1600.;\n",
+ "r = 0.25*math.pi*(1-k**2);##r = A/D^2\n",
+ "P = n*F;## tons\n",
+ "D1 = math.sqrt(P/(f_c*r*2) +math.sqrt((P/(f_c*r))*((a/4)*(l*12.)**2.)/((1+k**2.)/16.) + (P/(f_c*r*2.))**2.));## inches\n",
+ "D = round(D1);## inches\n",
+ "d = k*D;## inches\n",
+ "t = (D-d)/2.;## inches\n",
+ "print'%s %.1f %s'%('The internal diameter d =',d,'inches');\n",
+ "print'%s %.2f %s'%('The thickness of the metal will be',t,'inches');\n",
+ "## the answer is correct only, but it is approximated in the text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal diameter d = 5.6 inches\n",
+ "The thickness of the metal will be 0.70 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg356"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The safe axial load \n",
+ "l = 5.;## feet\n",
+ "b = 5./2.;## inches\n",
+ "d = 5/2.;## inches\n",
+ "h = 1/4.;## inches\n",
+ "n = 3.;## factor of safety\n",
+ "A = 1.19;## in^2\n",
+ "k = 0.49;## minimum radius of gyration\n",
+ "f_c = 21.;## lb/in^2\n",
+ "a = 1/7500.;\n",
+ "P = f_c*A/(1+(a/2)*((l*12)**2)/k**2);## tons\n",
+ "P_safe = P/n;## tons\n",
+ "print'%s %.2f %s'%('The safe axial load =',P_safe,'tons');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe axial load = 4.17 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg356"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The safe axial load\n",
+ "b1 = 10.;## inches\n",
+ "d1 = 7/2.;## inches\n",
+ "r = 9/2.;## inches\n",
+ "b2 = 12.;## inches\n",
+ "d2 = 1/2.;## inches\n",
+ "l = 20.;## feet\n",
+ "n = 4.;## factor of safety\n",
+ "A_s = 7.19;## in^2\n",
+ "I_xx1 = 109.42;## in^4\n",
+ "I_yy1 = 7.42;## in^4\n",
+ "d = 0.97;## inches\n",
+ "f_c = 21.;## lb/in^2\n",
+ "a = 1/7500.;\n",
+ "A = 2*A_s + 4*b2*d2;## in^2\n",
+ "I_xx = 2.*I_xx1 + 2.*((1/12.)*b2*(2*d2)**3. + b2*(r+2.*d2)**2.);## in^4\n",
+ "I_yy = 2.*(1/12.)*(2*d2)*b2**3. + 2.*(I_yy1 + A_s*(0.5*r+d)**2.);## in^4\n",
+ "k = math.sqrt(min(I_xx,I_yy)/A);## minimum radius of gyration\n",
+ "P = f_c*A/(1.+ a*((l*12.)**2./k**2));## tons\n",
+ "P_safe = P/n;## tons\n",
+ "print'%s %.1f %s'%('The safe axial load =',round(P_safe),'tons');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe axial load = 122.0 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg357"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate the safe axial load\n",
+ "m = 4.;## no. of angles\n",
+ "b = 7/2.;## inches\n",
+ "d = 7/2.;## inches\n",
+ "h = 3/8.;## inches\n",
+ "s = 18.;## inches\n",
+ "l = 30.;## feet\n",
+ "n = 3.;## factor of safety\n",
+ "A = 2.49;## in^2\n",
+ "J = 1.;## inches\n",
+ "I_xxs = 2.80;## in^4\n",
+ "I_yys = I_xxs;## in^4\n",
+ "##from the chapter V. \n",
+ "I = 648.64;## in^4\n",
+ "k = math.sqrt(65.2);## in^2\n",
+ "f_c = 21.;## lb/in^2\n",
+ "a = 1/7500.;\n",
+ "P = m*f_c*A/(1.+a*((l*12)**2)/k**2);## tons\n",
+ "P_safe = P/n;## tons\n",
+ "print'%s %.1f %s'%('The safe axial load =',P_safe,'tons');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe axial load = 55.1 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg365"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate Stress intensities and Maximum possible eccentricity\n",
+ "D = 7.;## inches\n",
+ "t = 3/4.;## inches\n",
+ "l = 16.;## feet\n",
+ "P = 12.;## tons\n",
+ "e = 3/4.;## inches\n",
+ "E = 6000.;## tons/in^2\n",
+ "d = D-2.*t;## inches\n",
+ "A = 0.25*math.pi*(D**2. - d**2.);## in^2\n",
+ "I = (math.pi/64.)*(D**4. - d**4.);## in^4\n",
+ "p_0 = P/A;## tons/in^2\n",
+ "Z = 2.*I/D;## in^3\n",
+ "M = P*e/math.cos(0.25*l*12.*math.sqrt(P/(E*I)));## ton-inches\n",
+ "p_b = M/Z;## tons/in^2\n",
+ "p_max = p_0+p_b;## tons/in^2\n",
+ "p_min = p_0-p_b;## tons/in^2\n",
+ "##if tension is just on the point being induced in the section, p_b = p_0\n",
+ "e = p_0*t*Z/M;## inches\n",
+ "print'%s %.3f %s'%('Stress intensities, p_max =',p_max,'tons/in^2.,compressive')\n",
+ "print'%s %.3f %s'%('p_min =',p_min,'tons/in^2., compressive');\n",
+ "print'%s %.2f %s'%('Maximum possible eccentricity, e =',e,'inches')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress intensities, p_max = 1.261 tons/in^2.,compressive\n",
+ "p_min = 0.369 tons/in^2., compressive\n",
+ "Maximum possible eccentricity, e = 1.37 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg366"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The maximum possible eccentricity\n",
+ "P = 80;## tons\n",
+ "p_max = 5;## tons/in^2\n",
+ "E = 13000;## tons/in^2\n",
+ "A = 38.38;## in^2\n",
+ "I_yy = 451.94;## in^4\n",
+ "y_c = 6;## inches\n",
+ "l = 20;## inches\n",
+ "k = math.sqrt(I_yy/A);## inches\n",
+ "Z_yy = I_yy/y_c;## in^3\n",
+ "p_0 = P/A;## tons/in^2\n",
+ "p_b = p_max-p_0;## tons/in^2\n",
+ "M_max = p_b*Z_yy;## ton-inches\n",
+ "e = M_max/(P/math.cos(0.5*l*12*math.sqrt(P/(E*I_yy))));##inches\n",
+ "print'%s %.2f %s'%('The maximum possible eccentricity, e =',e,'inches')\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum possible eccentricity, e = 2.48 inches\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg368"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The safe load\n",
+ "e = 7/4.;## inches\n",
+ "E = 13000.;## tons/in^2\n",
+ "p = 5.;## tons/in^2\n",
+ "y_c = 6.;## inches\n",
+ "l = 20.;## feet\n",
+ "A = 38.38;## in^2\n",
+ "k = math.sqrt(11.78);## inches\n",
+ "I = 11.78;## in^4\n",
+ "p_e = (math.pi)**2 *E*k**2 /(l*12)**2;## tons/in^2\n",
+ "##from Perry's formula\n",
+ "p_0 = 0.5*((p_e*1.2*e*y_c/k**2)+p_e+p)-math.sqrt((0.5*((p_e*1.2*e*y_c/k**2)+p_e+p))**2 - p_e*p);## tons/in^2\n",
+ "P = p_0*A;## tons\n",
+ "print'%s %.2f %s'%('The safe load, P =',P,'tons');\n",
+ "\n",
+ "##there is a minute calculation error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe load, P = 88.33 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#calculate The safe load\n",
+ "b1 = 10.;## inches\n",
+ "d1 = 6.;## inches\n",
+ "b2 = 12.;## inches\n",
+ "d2 = 1/2.;## inches\n",
+ "l = 16.;## feet\n",
+ "A_s = 11.77;## in^2\n",
+ "I_xxs = 204.80;## in^4\n",
+ "I_yys = 21.76;## in^4\n",
+ "A = A_s + 2*b2*d2;## in^2\n",
+ "I_yy = I_yys + 2.*(1./12.)*d2*b2**3.;## in^4\n",
+ "k = math.sqrt(I_yy/A);## inches\n",
+ "##from the Perry-Robertson formula\n",
+ "n = 0.003*l*12/k;\n",
+ "p_e = 13000*math.pi**2/((l*12)/k)**2 ;## tons/in^2\n",
+ "f = 18.;## tons/in^2\n",
+ "x = 0.5*(f+p_e*(1.+n));\n",
+ "p_0 = x - math.sqrt(x**2 - f*p_e);## tons/in^2\n",
+ "P = p_0*A;## tons\n",
+ "P_safe = P/2.36;## tons\n",
+ "print'%s %.1f %s'%('The safe load, P =',P_safe,'tons');\n",
+ "\n",
+ "##there is a minute calculation error in the answer given in text book\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The safe load, P = 125.3 tons\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Mechanics_of_Structures/README.txt b/Mechanics_of_Structures/README.txt new file mode 100755 index 00000000..447566b1 --- /dev/null +++ b/Mechanics_of_Structures/README.txt @@ -0,0 +1,10 @@ +Contributed By: bikash behera +Course: btech +College/Institute/Organization: iitbombay +Department/Designation: aerospace engnieering +Book Title: Mechanics of Structures +Author: S. B. Junnarkar +Publisher: N. Hernandez +Year of publication: 1957 +Isbn: 8185594678 +Edition: 2
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