Department of Engineering Advanced Solid Mechanics (ENGT5258) Coursework 2 SUBJECT TUTOR– Dr.Karthikeyan Kandan Name: Dashrath Rathod Student number: P 15250471 Abstract In today’s modern word, the most of the Engineering problems are solved by the help of software there are so many designing and analysis software’s like Pro-E, Catia, Abacus, Ansys, Solid Edge, Matlab etc. In this Coursework we are doing a structural analysis (for Truss) for given figures by Finite Element Analysis using the Matlab software. The application of Matlab software is vast, we are using this software for analysis. The Matlab software solve the engineering structural problem faster and better ways which helps in saving the money and time in this coursework we are selecting the figure configuration according to the Factor of Safety after the application of forces on the members of truss 1 Introduction In this Coursework we are suggesting a firm which truss design is suitable to mounted overhead for roofing of the firm depend upon the analysis of truss structures as given in the figures. Two truss structure are made of mild steel which are rectangular shapes. The constraints and the forces are shown in the figure. We suggest the best choice among the figure by analysis and application of Finite Element Analysis and Matlab software by using a standard UK box section area to determine the stress induced in each element, displacement under the application of force and appropriate factor of safety. We using 25mm×25mm with 2mm thickness box sections for analysis which are not the smallest and nor the highest cross section according UK standard Box sizes. The Truss structure design is as follows. Figure 1: – Truss for overhead loading pulley configuration 1 Figure 2: – Truss for overhead loading pulley configuration 2 Material Properties We are using the Mild Steel material for the design of the truss configuration. The Mechanical properties of the Mild Steel are Young’s Modulus E = 210 GPAYield Strength σy = 250 MPA The Box size selected for the problem is 25mm×25mm with 2mm thickness. 2 Description of Elemental stiffness matrix 2.1 Configuration of Fig 1 Discretisation Plan for Truss Configuration of Fig 1 Truss MemberNode1Node2X1 Y1X2 Y2θC= C=S= S=lA120,0-0.8,01800-100.8B23-0.8,0-0.8,0.6900010.6C34-0.8,0.60,0.61800-100.8D410,0.60,0900010.6E420,0.6-0.8,04500.849F450,0.60.6,0.61800-100.6G510.6,0.60,04500.849 Stiffness matrix for truss element ‘A’ And the local matrix and force vector is = Stiffness Matrix for Element ‘B’ And the local matrix and force vector is = Stiffness Matrix for Element ‘C’ And the local matrix and force vector is = Stiffness Matrix for Element ‘D’ And the local matrix and force vector is = Stiffness Matrix for Element ‘E’ And the local stiffness matrix and force vector is = Stiffness Matrix for Element ‘F’ And the local stiffness matrix and force vector is = Stiffness Matrix for Element ‘G’ And the local stiffness matrix and force vector is = 2.2 Configuration of Fig 2 Discretisation Plan for Truss Configuration of Fig 2 Truss MemberNode1Node2X1 Y1X2 Y2θC= C=S= S=lA120,0-0.8,01800-100.8B23-0.8,0-0.8,0.6900010.6C34-0.8,0.60,0.61800-100.8D410,0.60,0900010.6E420,0.6-0.8,04500.849F450,0.60.6,0-4500.6G510.6,00,018000.849 Stiffness matrix for truss element ‘A’ And the local matrix and force vector is = Stiffness Matrix for Element ‘B’ And the local matrix and force vector is = Stiffness Matrix for Element ‘C’ And the local matrix and force vector is = Stiffness Matrix for Element ‘D’ And the local matrix and force vector is = Stiffness Matrix for Element ‘E’ And the local stiffness matrix and force vector is = Stiffness Matrix for Element ‘F’ And the local stiffness matrix and force vector is = Stiffness Matrix for Element ‘G And the local stiffness matrix and force vector is = 3 Assembly of global stiffness matrix Matrix: [m × n]; m = n = number of nodes × degrees of freedom. m = n = 5 × 2 = 10 Therefore, the size of the stiffness matrix is 10×10 Kglobal = KA + KB + KC + KD + KE + KF + KG Configuration of figure 1 K1= Configuration of figure 2 K2 = 4 Boundaries conditions and Forces 4.1 Box Section selected is 25mm× 25mm with 2mm thickness Boundary conditions: The only applied load is: N (Downward) Active degree of freedom: for element A for element B and for element G 4.2 Configuration of figure 1 A = (25 × 25) -(23×23) ×10-6 m2 E = 210× 109 N/ m2 Substituting the value of A and E in Global Stiffness Matrix Configuration of figure 1 K1= K1= 4.3 Configuration of figure 2 A = (25 × 25) – (23×23) ×10-6 m2 E = 210× 109 N/ m2 Substituting the value of A and E in Global Stiffness Matrix K2 = K2= 5 Governing equation 5.1 Configuration of figure 1 Equilibrium equations after applying the boundary conditions and substituting force and displacements values Subtract equation 4 from equation 3, we get And then Add Equation 2 and equation 3, we get Subtract Equation 2 from Equation 1, we get 1 in the above Equation we get 5.2 Configuration of figure 2 Equilibrium equations after applying the boundary conditions and substituting force and displacements values =0 On Solving the Equation 1 we get we get, , ( therefore From Equation 3 We will get Results obtained from Matlab analysis Nodal displacement Truss MemberConfiguration1 Nodal displacement(Configuration2 Nodal displacement (Node1Node2Node1Node2ABCDEFG Elemental strains Truss MemberConfiguration1 Elemental strainConfiguration2 Elemental strainABCDEFG The factor of safety has been calculated using the yield strength given elemental stresses Truss MembreConfiguration1 El mental stressFactor of safetyConfiguration2 El mental stressFactor of safetyA1.56251.5625B∞∞C∞∞D1.5625∞E∞∞F1.56251.5625G1.1042.4 Recommendation As per the Matlab Analysis data is clearly displaying that stresses and strains induced are less in the Figure 2 than Figure 1and also the factor of safety giving by the figure 2 is acceptable. I’ll suggest the figure 2 for truss roofing to the Firm References http://www.steelexpress.co.uk/structuralsteel/RHS.htmlDr.Karthikeyan Kandan’s class noteshttps://www.mathworks.com/products/matlab.html Matlab Codes for figure 1 clear; close all; clc; %% Set input parameter for Truss assembly % x,y coordinates of each node Nodes=[0 0; -0.8 0;-0.8 0.6;0 0.6;0.6 0.6]; % Note,since we defined E=1 and A=1, we will have same value for stress and strain % Young’s modulus of the truss member material E= 210*10^9; % Yield strength of the truss member material ys= 250*10^6; % Cross section area of the each truss member A = ((25*25)-(23-23))*10.^-6 % First node, Second node, E, A values of each element Elements = [1 2 E A;2 3 E A;3 4 E A;4 1 E A;4 2 E A;4 5 E A;5 1 E A]; % Determine number of nodes nnode = length(Nodes); % Determine number of elements nelem = size(Elements,1); %% Initialise Stiffness, Displacement, Force, Strain and Stress Vectors % Initialise Stiffness matrix, i.e size of K_global = number of nodes X number of degrees of freedom K_global = zeros(2*nnode,2*nnode); % Initialise Displacements, i.e. u1, v1, u2, v2, u3, v3, u4, v4, u5, v5, % u6, v6, u7, v7, u8, v8, u9, v9, u10 and v10 U = zeros(2*nnode, 1); % Initialise Force vector, i.e. fx1, fy1, fx2, fy2, fx3, fy3, fx4, fy4, % fx5, fy5, fx6, fy6, fx7, fy7, fx8, fy8, fx9, fy9, fx10 and fy10 F = zeros(2*nnode, 1); % Initialise Elemental Strain vector, i.e. strain1 and strain2 strain = zeros(nelem,1); % Initialise Elemental Stress vector, i,e. stress1, stress2 stress= zeros(nelem,1); %% Apply Boundary Conditions % Define number of Degrees Of Freedom i.e. u1, v1, u2, v2, u3, v3, u4, v4, % u5, v5, u6, v6, u7, v7, u8, v8, u9, v9, u10, v10 active_DOF = 1:2*nnode; % Assign zero for known Displacments i.e Degrees Of Freedom, for example, u1=v1=u3=v3=0 active_DOF([3 5 6 7 8]) = []; % Apply known forces to force vector, i.e. fxy = -10 F(10) = -10000; %% Compute elemental stiffness matrix and perform assembly for p = 1:nelem %Determine DOF for each element DOFs = [2*Elements(p, 1)-1, 2*Elements(p, 1), 2*Elements(p, 2)-1, 2*Elements(p, 2)]; % Assign x and y coordinate for each element X1 = Nodes(Elements(p,1), 1); Y1 = Nodes(Elements(p,1), 2); X2 = Nodes(Elements(p,2), 1); Y2 = Nodes(Elements(p,2), 2); % Calculate length of the element L = sqrt((X2-X1)^2+(Y2-Y1)^2); % Determine cos theta and sin theta for the transformation matrix cos_theta=(X2-X1)/L; sin_theta = (Y2-Y1)/L; % Assign Youngs modulus for each element E = Elements(p,3); % Assign Cross-sectional area of each element A = Elements(p,4); % Compute elemental stiffness matrix for each element k_element = (E*A/L)*[cos_theta^2 cos_theta*sin_theta -cos_theta^2 -cos_theta*sin_theta; cos_theta*sin_theta sin_theta^2 -cos_theta*sin_theta -sin_theta^2; -cos_theta^2 -cos_theta*sin_theta cos_theta^2 cos_theta*sin_theta; -cos_theta*sin_theta -sin_theta^2 cos_theta*sin_theta sin_theta^2]; % Assemble each element to obtain global stiffness matrix K_global(DOFs,DOFs) = K_global(DOFs,DOFs) + k_element; end %% solve for nodal displacements ONLY with active_DOFive DOFs, Ks and Forces U(active_DOF) = K_global(active_DOF,active_DOF)F(active_DOF); %% Compute elemental strain and stress for q = 1:nelem %Determine DOF for each element DOFs = [2*Elements(q, 1)-1, 2*Elements(q, 1), 2*Elements(q, 2)-1, 2*Elements(q, 2)]; % Assign x and y coordinate for each element X1 = Nodes(Elements(q,1), 1); Y1 = Nodes(Elements(q,1), 2); X2 = Nodes(Elements(q,2), 1); Y2 = Nodes(Elements(q,2), 2); % Calculate length of the element L = sqrt((X2-X1)^2+(Y2-Y1)^2); % Determine cos theta and sin theta for the transformation matrix cos_theta=(X2-X1)/L; sin_theta = (Y2-Y1)/L; % Transformation matrix to calculate axial displacement from nodal displacements t=[cos_theta sin_theta 0 0; 0 0 cos_theta sin_theta]; % calculate axial displacements for each element d = t*U(DOFs); disp(‘compute axial displacement for each element,d=’);disp(d); % Compute Strain strain(q) = (d(2) – d(1))/L; % Compute Stress stress(q)= Elements(q, 3)*strain(q); end %% Display the results disp(‘Global Stiffness Matrix, K_global=’);disp(K_global); disp(‘Computed Nodal Displacements, U=’);disp(U); disp(‘Computed Strains for all elements, strain=’);disp(strain); disp(‘Computed Stresses for all elements, stress =’);disp(stress); Matlab codes for figure 2 clear; close all; clc; %% Set input parameter for Truss assembly % x,y coordinates of each node Nodes=[0 0; -0.8 0;-0.8 0.6;0 0.6;0.6 0]; % Note,since we defined E=1 and A=1, we will have same value for stress and strain % Young’s modulus of the truss member material E= 210*10^9; % Yield strength of the truss member material ys= 250*10^6; % Cross section area of the each truss member A = ((25*25)-(23-23))*10.^-6 % First node, Second node, E, A values of each element Elements = [1 2 E A;2 3 E A;3 4 E A;4 1 E A;4 2 E A;4 5 E A;5 1 E A]; % Determine number of nodes nnode = length(Nodes); % Determine number of elements nelem = size(Elements,1); %% Initialise Stiffness, Displacement, Force, Strain and Stress Vectors % Initialise Stiffness matrix, i.e size of K_global = number of nodes X number of degrees of freedom K_global = zeros(2*nnode,2*nnode); % Initialise Displacements, i.e. u1, v1, u2, v2, u3, v3, u4, v4, u5, v5, % u6, v6, u7, v7, u8, v8, u9, v9, u10 and v10 U = zeros(2*nnode, 1); % Initialise Force vector, i.e. fx1, fy1, fx2, fy2, fx3, fy3, fx4, fy4, % fx5, fy5, fx6, fy6, fx7, fy7, fx8, fy8, fx9, fy9, fx10 and fy10 F = zeros(2*nnode, 1); % Initialise Elemental Strain vector, i.e. strain1 and strain2 strain = zeros(nelem,1); % Initialise Elemental Stress vector, i,e. stress1, stress2 stress= zeros(nelem,1); %% Apply Boundary Conditions % Define number of Degrees Of Freedom i.e. u1, v1, u2, v2, u3, v3, u4, v4, % u5, v5, u6, v6, u7, v7, u8, v8, u9, v9, u10, v10 active_DOF = 1:2*nnode; % Assign zero for known Displacments i.e Degrees Of Freedom, for example, u1=v1=u3=v3=0 active_DOF([3 5 6 7 8]) = []; % Apply known forces to force vector, i.e. fxy = -10 F(10) = -10000; %% Compute elemental stiffness matrix and perform assembly for p = 1:nelem %Determine DOF for each element DOFs = [2*Elements(p, 1)-1, 2*Elements(p, 1), 2*Elements(p, 2)-1, 2*Elements(p, 2)]; % Assign x and y coordinate for each element X1 = Nodes(Elements(p,1), 1); Y1 = Nodes(Elements(p,1), 2); X2 = Nodes(Elements(p,2), 1); Y2 = Nodes(Elements(p,2), 2); % Calculate length of the element L = sqrt((X2-X1)^2+(Y2-Y1)^2); % Determine cos theta and sin theta for the transformation matrix cos_theta=(X2-X1)/L; sin_theta = (Y2-Y1)/L; % Assign Youngs modulus for each element E = Elements(p,3); % Assign Cross-sectional area of each element A = Elements(p,4); % Compute elemental stiffness matrix for each element k_element = (E*A/L)*[cos_theta^2 cos_theta*sin_theta -cos_theta^2 -cos_theta*sin_theta; cos_theta*sin_theta sin_theta^2 -cos_theta*sin_theta -sin_theta^2; -cos_theta^2 -cos_theta*sin_theta cos_theta^2 cos_theta*sin_theta; -cos_theta*sin_theta -sin_theta^2 cos_theta*sin_theta sin_theta^2]; % Assemble each element to obtain global stiffness matrix K_global(DOFs,DOFs) = K_global(DOFs,DOFs) + k_element; end %% solve for nodal displacements ONLY with active_DOFive DOFs, Ks and Forces U(active_DOF) = K_global(active_DOF,active_DOF)F(active_DOF); %% Compute elemental strain and stress for q = 1:nelem %Determine DOF for each element DOFs = [2*Elements(q, 1)-1, 2*Elements(q, 1), 2*Elements(q, 2)-1, 2*Elements(q, 2)]; % Assign x and y coordinate for each element X1 = Nodes(Elements(q,1), 1); Y1 = Nodes(Elements(q,1), 2); X2 = Nodes(Elements(q,2), 1); Y2 = Nodes(Elements(q,2), 2); % Calculate length of the element L = sqrt((X2-X1)^2+(Y2-Y1)^2); % Determine cos theta and sin theta for the transformation matrix cos_theta=(X2-X1)/L; sin_theta = (Y2-Y1)/L; % Transformation matrix to calculate axial displacement from nodal displacements t=[cos_theta sin_theta 0 0; 0 0 cos_theta sin_theta]; % calculate axial displacements for each element d = t*U(DOFs); disp(‘compute axial displacement for each element,d=’);disp(d); % Compute Strain strain(q) = (d(2) – d(1))/L; % Compute Stress stress(q)= Elements(q, 3)*strain(q); end %% Display the results disp(‘Global Stiffness Matrix, K_global=’);disp(K_global); disp(‘Computed Nodal Displacements, U=’);disp(U); disp(‘Computed Strains for all elements, strain=’);disp(strain); disp(‘Computed Stresses for all elements, stress =’);disp(stress)

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