From 476705d693c7122d34f9b049fa79b935405c9b49 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 14 Apr 2020 10:19:27 +0530
Subject: Initial commit

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+{
+"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
+}
-- 
cgit