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diff --git a/Mechanics_of_Materials_by_R_C_Hibbeler/1-Stress.ipynb b/Mechanics_of_Materials_by_R_C_Hibbeler/1-Stress.ipynb new file mode 100644 index 0000000..0da90e7 --- /dev/null +++ b/Mechanics_of_Materials_by_R_C_Hibbeler/1-Stress.ipynb @@ -0,0 +1,1022 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Stress" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.10: S10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.10 : ')\n", +"\n", +"//Given:\n", +" af = 800; //N Axial force along centroidal axis\n", +"t = 0.040; //m thickness of square cross section\n", +"ang_b = 30 *(%pi/180) ;\n", +"ang_b_comp = 60 *(%pi/180);\n", +"a = t^2; //m^2 Area of cross section\n", +"a_new = ((t*1000)^2)/(sin(ang_b_comp)); // mm^2 Area of section at b-b\n", +"\n", +"//Part(a)\n", +"\n", +"//Internal Loading: The bar is sectioned, Fig 1-24b, and the internal resultant loading consists of only axial force.\n", +"\n", +"// Average Stress: \n", +"avg_stress = af/(a* 1000);\n", +"\n", +"//Shear Force at the section is zero.\n", +"//The average normal stress distribution over the cross section is shown in Fig 1-24c.\n", +"\n", +"\n", +"//Part(b)\n", +"\n", +"\n", +"//solve the two equations for two unknowns:\n", +"\n", +"N = af * cos(ang_b); \n", +"V = af * sin(ang_b);\n", +"avg_normal_stress = (N*1000)/ a_new; // kPa\n", +"avg_shear_stress = (V*1000)/a_new; //kPa\n", +"\n", +"//Display\n", +"\n", +"printf('\n\nThe average stress for section a-a = %.2f kPa',avg_stress);\n", +"printf('\nThe Normal Force for section b-b = %.2f N',N);\n", +"printf('\nThe Shear Force for section b-b = %.2f N',V);\n", +"printf('\nThe Average Normal Stress for section b-b = %.2f kPa',avg_normal_stress);\n", +"printf('\nThe Average Shear Stress for section b-b = %.2f kPa',ceil(avg_shear_stress));\n", +"\n", +"//--------------------------------------------------------------------------END--------------------------------------------------------------------------\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11: S11.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.11 : ')\n", +"\n", +"//Given :\n", +"f = 5000; //N\n", +"d_rod = 10;//Diameter of steel rod in mm.\n", +"l_bc = 20; //Length of side bc in mm.\n", +"l_bd = 40; //Length of side bd in mm.\n", +"a_rod = (%pi/4)* (d_rod^2); //Area of cross section of the rod in mm^2.\n", +"a_strut = l_bc*l_bd ; //Area of strut in mm^2.\n", +"\n", +"\n", +"//Average shear stress\n", +"\n", +"avg_shear_rod = f/a_rod; //for rod in Mpa\n", +"avg_shear_strut = (f/2)/a_strut; //for strut\n", +"\n", +"//Display:\n", +"\n", +"printf('\n\nThe average shear stress for the rod = %.2f MPa',avg_shear_rod);\n", +"printf('\nThe average shear stress for the strut = %.2f MPa',avg_shear_strut);\n", +"\n", +"\n", +"\n", +"//--------------------------------------------------------------END----------------------------------------------------------------------------" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.12: S12.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.12 : ')\n", +"\n", +"//Given:\n", +"l_bc = 50; //Length of BC in mm.\n", +"l_db = 75; // mm.\n", +"l_ed = 40; // mm.\n", +"l_ab = 25; // mm.\n", +"f_diagonal = 3000; //N\n", +"a1 = l_ab*l_ed; //Area of face AB in mm^2.\n", +"a2 = l_bc*l_ed ; //mm^2.\n", +"a3 = l_db*l_ed ; // mm^2.\n", +"\n", +"//Internal loadings - The free body diagram of the inclined member is shown in 1-26b. \n", +"\n", +"//Equilibrium Equations\n", +"\n", +"//Balancing forces along the x- direction.\n", +"f_ab = f_diagonal*(3/5); //Force on segment AB in N\n", +"V = f_ab; //Shear force acting on the sectioned horizontal plane EDB in N\n", +"\n", +"//Balancing forces along the Y direction.\n", +"f_bc = f_diagonal*(4/5); //Force on segment BC in N.\n", +"\n", +"//Average compressive stresses along the horizontal and vertical planes:\n", +"\n", +"avg_comp_ab = f_ab/a1; // N/mm^2\n", +"avg_comp_bc = f_bc/a2; // N/mm^2\n", +"\n", +"//Average shear stress acting on the horizontal plane defined by EDB :\n", +"\n", +"avg_shear = f_ab/a3; // N/mm^2\n", +"\n", +"//Display:\n", +"\n", +"\n", +"printf('\n\nThe Force on segment AB = %.2f N',f_ab);\n", +"printf('\nThe Shear Force on sectioned plane EDB = %.2f N',V);\n", +"printf('\nThe Force on segment BC = %.2f N',f_bc);\n", +"printf('\nThe average compressive stress along AB = %.2f N/mm^2',avg_comp_ab);\n", +"printf('\nThe average compressive stress along BC = %.2f N/mm^2',avg_comp_bc);\n", +"printf('\nThe average shear stress along EDB = %.2f N/mm^2',avg_shear);\n", +"\n", +"//-------------------------------------------------------------------------------END---------------------------------------------------------------------------\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.13: S13.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"\n", +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.13 : ')\n", +"\n", +"//Given:\n", +"shear_allow = 90; //MPa\n", +"tensile_allow = 115; //MPa\n", +"\n", +"l_AP = 2; //m\n", +"l_PB = 1; //m\n", +"resultant_A = 5.68; //kN\n", +"resultant_B = 6.67; //kN\n", +"v_a = 2.84; //kN\n", +"v_b = 6.67; //kN\n", +"\n", +"\n", +"//Diameter of the Pins:\n", +"A_A = (v_a*10^3)/(shear_allow*10^6); //Area of pin A\n", +"da = (sqrt((4*A_A)/%pi))*10^3 // d = (square root of(area*4/pi)) in mm\n", +"A_B = (v_b*10^3)/(shear_allow*10^6) ; //Area of pin B\n", +"db = (sqrt((4*A_B)/%pi))*10^3 // Area = (%pi\4)d^2 in mm^2\n", +"\n", +"chosen_da = ceil(da);\n", +"chosen_db = ceil(db);\n", +"\n", +"//Diameter of Rod:\n", +"A_bc = (resultant_B*10^3)/(tensile_allow*10^6); //Area of BC\n", +"dbc = (sqrt((4*A_bc)/%pi)*10^3); // Area = %pi\4)d^2\n", +"chosen_dbc = ceil(dbc);\n", +"\n", +"//Displaying Results:\n", +"\n", +"printf ('\n\n The diameter of pin A = %.3f mm',da);\n", +"printf ('\n The diameter of pin B = %.3f mm',db);\n", +"printf ('\n The diameter of rod BC = %.2f mm',dbc);\n", +"printf ('\n\n\nThe chosen diameters are: ');\n", +"printf ('\n The diameter of pin A = %.3f mm',chosen_da);\n", +"printf ('\n The diameter of pin B = %.3f mm',chosen_db);\n", +"printf ('\n The diameter of rod BC = %.2f mm',chosen_dbc);\n", +"\n", +"//-----------------------------------------------------------------------END--------------------------------------------------------------------\n", +"\n", +"\n", +"\n", +"\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.14: S14.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.14 : ')\n", +"\n", +"//Given:\n", +"shear_allow = 55; //MPa\n", +"l_ac = 200; //mm\n", +"l_cd= 75; //mm\n", +"l_de = 50; //mm\n", +"l_ce = l_cd + l_de;\n", +"load_d =15; //kN\n", +"load_e = 25; //kN\n", +"\n", +"//Internal Shear Force:\n", +"//summation Mc = 0\n", +"\n", +"f_ab = ((load_d*l_cd +load_e*(3/5)*l_ce)/l_ac);\n", +"c_x =-load_d + (load_e*(4/5)); //resolving C in x dir\n", +"c_y = load_d + (load_e*(3/5)); //resolving C in y dir\n", +"\n", +"f_c = sqrt(c_x^2 + c_y^2); //kN\n", +"V = f_c/2;\n", +"\n", +"//Required Area\n", +"A = ((V*10^3)/(shear_allow)); //A = V/Allowable shear in mm^2\n", +"d = ((sqrt((4*A)/%pi))) // Area = (%pi\4)d^2 in mm^2\n", +"\n", +"chosen_d = ceil(ceil(d))+1;\n", +"\n", +"//Displaying Results:\n", +"\n", +"\n", +"printf('\n\nThe force at AB = %.2f kN',f_ab);\n", +"printf('\nThe resultant force at C = %.2f kN',f_c);\n", +"printf('\nThe area of pin = %.2f mm^2',A);\n", +"printf('\nThe diameter of pin = %.2f mm',chosen_d);\n", +"\n", +"//---------------------------------------------------------------END--------------------------------------------------------------------------------------\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.15: S15.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.15 : ')\n", +"\n", +"//Given:\n", +"P= 20; //kN\n", +"d_hole = 40; //mm\n", +"normal_allow = 60; //MPa\n", +"shear_allow = 35; //MPa\n", +"\n", +"\n", +"//Diameter of Rod:\n", +"area1 = (P*10^3)/(normal_allow*10^6); //Area in m^2\n", +"d = ((sqrt((4*area1)/%pi))*1000); // Area = (%pi\4)d^2\n", +"\n", +"\n", +"//Thickness of disc:\n", +"V = P;\n", +"area2 = (V*10^3)/(shear_allow*10^6); //Area in m^2\n", +"thickness = (area2*10^6)/(d_hole*%pi);// A = pi*d*t\n", +" \n", +"\n", +"printf('\n\nThe cross sectional area of disc = %.8f m^2',area1);\n", +"printf('\nThe diameter of rode = %.2f mm',d);\n", +"printf('\nThe thickness of disc = %.2f mm',thickness);\n", +"\n", +"//------------------------------------------------------------------------END------------------------------------------------------------------------------------\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.16: S16.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.16 : ')\n", +"\n", +"//Given:\n", +"bearing_allow = 75; //MPa\n", +"tensile_allow = 55; //MPa\n", +"d_shaft = 60; //mm\n", +"r_shaft = d_shaft/2; //mm\n", +"area_shaft = %pi*(r_shaft^2); //Area = pi*r^2\n", +"d_collar = 80; //mm\n", +"r_collar = d_collar/2; //mm\n", +"area_collar = %pi*(r_collar^2); //Area = pi*r^2\n", +"thick_collar = 20; //mm\n", +"\n", +"//Normal Stress:\n", +"P1 = (tensile_allow* area_shaft)/3; //Tensile stress = 3P/A.\n", +"P1_kN = P1/1000;\n", +"\n", +"\n", +"//Bearing Stress:\n", +"bearing_area = area_collar-area_shaft; //mm^2\n", +"P2 = (bearing_allow*bearing_area)/3; //Bearing stress = 3P/A.\n", +"P2_kN= P2/1000;\n", +"\n", +"if(P2_kN<P1_kN)\n", +" big = P2_kN;\n", +"else big = P1_kN;\n", +" end\n", +" \n", +"//Displaying Results:\n", +"\n", +"printf('\n\nThe load calculated by Normal Stress = %.1f kN',P1_kN);\n", +"printf('\nThe load calculated by Bearing Stress = %.1f kN',P2_kN);\n", +"printf('\nThe largest load that can be applied to the shaft = %.1f kN',big);\n", +"\n", +"//----------------------------------------------------------------------------END----------------------------------------" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.17: S17.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.17 : ')\n", +"\n", +"//Given:\n", +"d_ac= 20; //mm\n", +"area_ac = %pi*(d_ac/2)^2; //Area = (%pi\4)d^2\n", +"area_al = 1800; //mm^2\n", +"d_pins = 18; //mm\n", +"area_pins = %pi*(d_pins/2)^2;\n", +"st_fail_stress = 680; //MPa\n", +"al_fail_stress = 70; //MPa\n", +"shear_fail_pin = 900; //MPa\n", +"fos = 2; //Factor of safety\n", +"l_ab = 2; //m\n", +"l_ap = 0.75; //m\n", +"\n", +"\n", +"st_allow= st_fail_stress /fos; //MPa\n", +"al_allow = al_fail_stress/fos; //MPa\n", +"pin_allow_shear = shear_fail_pin/fos; //MPa\n", +"\n", +"//Rod AC\n", +"f_ac = (st_allow*area_ac)/1000;\n", +"P1 = ((f_ac*l_ab)/(l_ab-l_ap));\n", +"\n", +"//Block B\n", +"f_b =(al_allow*area_al)/1000;\n", +"P2 = ((f_b*l_ab)/l_ap);\n", +"\n", +"//Pin A or C:\n", +"V = (pin_allow_shear*area_pins)/1000;\n", +"P3 = (V*l_ab)/(l_ab-l_ap);\n", +"\n", +"if(P1<P2 & P1<P3)\n", +" big = P1;\n", +"else if(P2<P1 & P2<P3)\n", +" big = P2;\n", +"else big = P3;\n", +"end\n", +"\n", +"//Displaying Results:\n", +"\n", +"printf('\n\nThe load allowed on rod AC = %.1f kN',round(P1));\n", +"printf('\nThe load allowed on block B = %.1f kN',P2);\n", +"printf('\nThe load allowed on pins A or C = %.1f kN',P3);\n", +"printf('\nThe largest load that can be applied to the bar = %.1f kN ',big);\n", +"\n", +"//----------------------------------------------------------------------------------END----------------------------------------------------------------------------\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1: S1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.1 :')\n", +"\n", +"w_varying = 270;\n", +"l_crossection = 9;\n", +"l_cb = 6;\n", +"l_ac = 2;\n", +"w_c = (w_varying/l_crossection) * l_cb //By proportion, load at C is found.\n", +"f_resultant_c = 0.5* w_c *l_cb \n", +"// Equations of Equilibrium\n", +"\n", +"//Balancing forces in the x direction:\n", +"n_c = 0\n", +"\n", +"//Balncing forces in the y direction:\n", +"v_c = f_resultant_c\n", +"\n", +"// Balncing the moments about C:\n", +"m_c = - (f_resultant_c*l_ac)\n", +"\n", +"\n", +"// Displaying results:\n", +"\n", +"printf('\n\nThe resultant force at C = %.2f N',f_resultant_c);\n", +"printf('\nThe horizontal force at C = %.2f N',n_c);\n", +"printf('\nThe vertical force at C = %.2f N',v_c);\n", +"printf('\nThe moment about C = %.2f Nm',m_c);\n", +"\n", +"\n", +"// ---------------------------------------------------------END-------------------------------------------------" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2: S2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.2 : ')\n", +"\n", +"f_d = 225; //N\n", +"w_uniform = 800; // N/m\n", +"l_ac = 0.200; //m\n", +"l_cb = 0.05+0.1; //m\n", +"l_bd = 0.100; //m\n", +"l_bearing = 0.05; //m\n", +"f_resultant = w_uniform*l_cb //120N\n", +"l_f_resultant_b = (l_cb/2)+ l_bearing; //0.125m\n", +"l = l_ac + l_cb + l_bearing + l_bd \n", +"\n", +"\n", +"// This problem is solved by considering segment AC of the shaft.\n", +"\n", +"//Support Reactions:\n", +"\n", +"m_b = 0; // Net moment about B is zero for equilibrium . Sum Mb = 0.\n", +"a_y = -((f_d*l_bd) - (f_resultant*l_f_resultant_b))/ (l - l_bd) // finding the reaction force at A\n", +"\n", +"// Refer to the free body diagram in Fig.1-5c.\n", +"f_c = 40 //N\n", +"//Balancing forces in the x direction:\n", +"n_c = 0\n", +"\n", +"//Balncing forces in the y direction:\n", +"v_c = a_y - f_c //-18.75N - 40N-Vc = 0\n", +"\n", +"// Balncing the moments about C:\n", +"m_c = ((a_y * (l_ac + 0.05)) - f_c*(0.025) ) // Mc+40N(0.025m)+ 18.75N(0.250m) = 0\n", +"\n", +"\n", +"// Displaying results:\n", +"\n", +"printf('\n\nThe resultant force = %.2f N',f_resultant);\n", +"printf('\nThe reaction force at A = %.2f N',a_y);\n", +"printf('\nThe horizontal force at C = %.2f N',n_c);\n", +"printf('\nThe vertical force at C = %.2f N',v_c);\n", +"printf('\nThe moment about C = %.2f Nm',m_c);\n", +"\n", +"//-------------------------------------------------------------------END-----------------------------------------------------------------------------------------\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3: S3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.3 :')\n", +"\n", +"// Given:\n", +"l_ac = 1; //m.\n", +"l_cd = 1.5 ; //m.\n", +"l_bd = 0.5; //m.\n", +"r_a = 0.125; //m.\n", +"r_d = 0.125; //m.\n", +"W = 2000; // N\n", +"\n", +"\n", +"// Equations of equilibrium:\n", +"\n", +"//Balancing forces in the x direction:\n", +"n_c = -W; // N\n", +"\n", +"//Balncing forces in the y direction:\n", +"v_c = -W; //N\n", +"\n", +"// Balncing the moments about C:\n", +"m_c = - (W*(r_a +l_ac)- W*r_a)\n", +"\n", +"\n", +"// Displaying results:\n", +"\n", +"printf('\n\nThe horizontal force at C = %.2f N',n_c);\n", +"printf('\nThe vertical force at C = %.2f N',v_c);\n", +"printf('\nThe moment about C = %.2f Nm',m_c);\n", +"\n", +"//----------------------------------------------------------------------------END--------------------------------------------------------------------------------------------\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4: S4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.4 :')\n", +"\n", +"// Given:\n", +"l_ag = 1; //Length of AG is 1m.\n", +"l_gd = 1; //Length of GD is 1m.\n", +"l_de = 3; //Length of DE is 1m.\n", +"f_a = 1500; //Force at A is 1500N.\n", +"l_ec = 1.5; //Length of EC is 1m.\n", +"l = l_ag +l_gd +l_de;\n", +"w_uniform_varying = 600; //Nm.\n", +"\n", +"w_resultant = 0.5*l_de*w_uniform_varying;\n", +"// calling point of action of resultant as P\n", +"l_ep = (2/3)*l_de; //Distance between points P and E.\n", +"l_ap = l - l_ep; // Distance between points A and P.\n", +"\n", +"\n", +"f_ba = 7750; //N\n", +"f_bc = 6200; //N\n", +"f_bd = 4650; //N\n", +"\n", +"//Free Body Diagram: Using the result for Fba, the left section AG of the beam is shown in Fig 1-7d.\n", +"\n", +"// Equations of equilibrium:\n", +"\n", +"//Balancing forces in the x direction:\n", +"n_g = -f_ba * (4/5); // N\n", +"\n", +"//Balncing forces in the y direction:\n", +"v_g = -f_a + f_ba*(3/5); //N\n", +"\n", +"// Balncing the moments about C:\n", +"m_g = (f_ba * (3/5)*l_ag) - (f_a * l_ag); //Nm\n", +"\n", +"\n", +"\n", +"// Displaying results:\n", +"\n", +"\n", +"printf('\n\nThe horizontal force at G = %.2f N',n_g);\n", +"printf('\nThe vertical force at G = %.2f N',v_g);\n", +"printf('\nThe moment about G = %.2f Nm',m_g);\n", +"\n", +"\n", +"//-------------------------------------------------------------------END----------------------------------------------------------------------------------------" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5: S5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.5 :')\n", +"\n", +"// Given:\n", +"f_a = 50; //N\n", +"m_a = 70; // Moment at A in Nm\n", +"l_ad = 1.25; //Length of AD in m.\n", +"l_bd = 0.5; //Length of BD in m.\n", +"l_cb = 0.75; //Length of BC in m.\n", +"w_l = 2; //Kg/m\n", +"g = 9.81; //N/kg- acceleration due to gravity\n", +"\n", +"\n", +"\n", +"//Free Body Diagram :\n", +"\n", +"w_bd = w_l*l_bd*g; //in N. Weight of each segment of pipe that acts through the centre of gravity of each segment.\n", +"w_ad = w_l*l_ad*g;\n", +"\n", +"// Equations of Equilibrium\n", +"\n", +"//Balancing forces in the x direction:\n", +"f_b_x = 0; // N\n", +"\n", +"//Balncing forces in the y direction:\n", +"f_b_y = 0; //N\n", +"\n", +"//Balncing forces in the z direction:\n", +"f_b_z = g + w_ad + f_a; //N\n", +"\n", +"// Balancing Moments in the x direction:\n", +"m_b_x = - m_a + (f_a*l_bd) + (w_ad*l_bd) + (l_bd/2)*g; //Nm\n", +"\n", +"// Balancing Moments in the y direction:\n", +"m_b_y = - (w_ad*(l_ad/2)) - (f_a*l_ad); //Nm\n", +"\n", +"// Balancing Moments in the z direction:\n", +"m_b_z = 0; //Nm\n", +"\n", +"v_b_shear = sqrt(f_b_z ^2 + 0); //Shear Force in N\n", +"t_b = - m_b_y; //Torsional Moment in Nm\n", +"m_b = sqrt(m_b_x ^2+ 0); // Bending moment in Nm\n", +"\n", +"\n", +"//Display\n", +"\n", +"// Displaying results:\n", +"\n", +"\n", +"printf('\n\n The weight of segment BD = %.1f N',w_bd);\n", +"printf('\n The weight of segment AD = %.1f N',w_ad);\n", +"printf('\n The force at B in the Z direction = %.1f N',f_b_z);\n", +"printf('\n The moment about B in the X direction = %.1f Nm',m_b_x);\n", +"printf('\n The moment about G in the Y direction = %.1f Nm',m_b_y);\n", +"printf('\n The Shear Force at B = %.1f N',v_b_shear);\n", +"printf('\n The Torsional Moment at B = %.1f Nm',t_b);\n", +"printf('\n The Bending Moment at B = %.1f Nm',m_b);\n", +"\n", +"\n", +"\n", +"//-----------------------------------------------------END-----------------------------------------------------------------------------\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6: S6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.6 :')\n", +"\n", +"//Given:\n", +"netf_b = 18*(10 ^3); //N Net force at B.\n", +"netf_c = 8*(10^3); //N Net force at C.\n", +"f_a = 12 *(10^3); //N Force at A.\n", +"f_d = 22* (10^3); //N Force at D.\n", +"w = 35; //mm Width.\n", +"t = 10; //mm Thickness.\n", +"\n", +"//calculations:\n", +"p_bc = netf_b + f_a; //N Net force in region BC.\n", +"a = w*t; //m^2 The area of the cross section.\n", +"avg_normal_stress = p_bc/a; //Average Normal Stress.\n", +"\n", +"\n", +"\n", +"// Displaying results:\n", +"\n", +"printf('\n\n Net force in the region BC = %.2f N',p_bc);\n", +"printf('\nThe Area of cross section = %.2f m^2',a);\n", +"printf('\nThe Average Normal Stress in the bar when subjected to load = %.2f MPa',avg_normal_stress);\n", +"\n", +"//---------------------------------------------------------END----------------------------------------------------------------------------------------" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7: S7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.7 :')\n", +"\n", +"//Given :\n", +"m_lamp = 80; //Mass of lamp in Kg.\n", +"d_ab = 10; // Diameter of AB in mm.\n", +"d_bc = 8; // Diameter of BC in mm.\n", +"ab_h = 60 *(%pi/180); // In degrees - Angle made by AB with the horizontal.\n", +"w = m_lamp*9.81; //N\n", +"a_bc = (%pi/4)*(d_bc^2); //m^2 Area of cross section of rod BC\n", +"a_ab = (%pi/4)*(d_ab^2); //m^2 Area of cross section of rod AB\n", +"\n", +"\n", +"\n", +"// Equations of equilibrium: Solving equilibrium equations simultaneously ,using matrices ,in the x and y directions to obtain force in BC and force in BA.\n", +"\n", +"\n", +"a = [(4/5) -(cos(ab_h)) ; (3/5) (sin(ab_h))];\n", +"b = [0 ; w];\n", +"f = zeros(1)\n", +"\n", +"f = a\b;\n", +"f_bc = f(1); // Force in BC in N.\n", +"f_ba = f(2); //Force in BA in N.\n", +"avg_normal_stress_a = f_ba / a_ab; //Mpa Average Normal Stress in AB\n", +"avg_normal_stress_c = f_bc/ a_bc;// Mpa Average Normal Stress in BC\n", +"\n", +"\n", +"// Displaying results:\n", +"\n", +"\n", +"printf('\n\nThe Weight of lamp = %.2f N',w);\n", +"printf('\nThe Net force in BC = %.2f N',f_bc);\n", +"printf('\nTheNet force in BA = %.2f N',f_ba);\n", +"printf('\nThe Average Normal Stress in AB when subjected to load = %.2f MPa',avg_normal_stress_a);\n", +"printf('\nThe Average Normal Stress in BC when subjected to load = %.2f MPa',avg_normal_stress_c);\n", +"\n", +"//------------------------------------------------------------------END----------------------------------------------------------------------------------\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.8: S8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"\n", +"disp('Scilab Code Ex 1.8 :')\n", +"\n", +"//Given:\n", +"h_above_ab = 0.8; \n", +"h_below_ab = 0.2; \n", +"d_a = 0.2; \n", +"d_b = 0.1; \n", +"sp_w = 80; \n", +"\n", +"// Equation of Equilibrium:\n", +"\n", +"\n", +"a = %pi* (d_a^2); // Area of cross section in m^2\n", +"p = sp_w * h_above_ab * a;\n", +"avg_comp_stress = p/a; // The average compressive stress in kN/m^2\n", +"\n", +"//Display:\n", +"\n", +"printf('\nThe internal Axial force P = %.2f kN',p);\n", +"printf('\nThe average compressive stress = %.2f kN/m^2',avg_comp_stress);\n", +"\n", +"\n", +"//--------------------------------------------------------------------------------END------------------------------------------------------------------------------\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.9: S9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear all; clc;\n", +"\n", +"disp('Scilab Code Ex 1.9 : ')\n", +"\n", +"//Given :\n", +"f = 3000; //N Force acting at distance x from AB.\n", +"l_ac = 200; //Length of AC in mm.\n", +"a_ab = 400; //Cross sectional area of AB in mm^2.\n", +"a_c = 650; // area of C in mm^2.\n", +"\n", +"\n", +"f_ans = zeros(3)\n", +"\n", +"k = [1 1 0;0 l_ac -f; 1.625 -1 0]\n", +"l = [f ; 0 ; 0 ]\n", +"f_ans = k\l;\n", +"\n", +"f_ab = f_ans(1)\n", +"f_c = f_ans(2)\n", +"x = f_ans(3)\n", +"\n", +"//Display:\n", +"\n", +"printf('\n\nThe Net force on AB = %.2f N',ceil(f_ab));\n", +"printf('\nNet force on C = %.2f N',f_c);\n", +"printf('\nDistance of force from AB = %.2f mm',ceil(x));\n", +"\n", +"\n", +"//------------------------------------------------------------------------------END------------------------------------------------------\n", +"" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |