Mechanical engineering - LIGHTWEIGHT DESIGN OF MECHANICAL STRUCTURES

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Lightweight design of mechanical structures Homework report Optimum design of a channel beam Chiara Moreschini ID 10601144 Points Correctness of the formulation of the optimum design problem / 7 Correctness of the solution / 8 Ability of discussing critically results / 10 Quality of the presentation /5 Chiara Moreschini – ID 10601144 1 1. Problem description The structure to be optimize d is a C - section cantilever beam constrained at one end , with length L=2100mm and a vertical load P applied to the free end equal to 1,0 kN. The beam has the following characteristics: Figure 1: Geometric characteristics of the beam ! = # − 2 & ( ) ! = # 2 ) " = 2 ℎ # & + ! & # 2ℎ # − ,! ( ℎ − & ) ⎩ ⎨ ⎧ 2 ! = 2 & ℎ $ + ! & $ 3 − ( 2 & ℎ + !& ) ) " # 2 " = ℎ # $ − ! $ ( ℎ − & ) 12 Where (C x , C y ) are the centroi d’s coordinates with respect to the initial reference system (X c , Y c ), while I x and I y are the moments of inertia of the cross - section about the new (X,Y) reference system centered in C . 2. Data and information collection Table 1: Mechanical properties of S355J0 The beam i s made of steel S355J0 , which is a structural steel with the mechanical properties reported in Table 1. C - section beams produce d with this kind of steel are usually manufactured by hot rolling . 3. Definition of design variables The design variables are the dimensions of the section of the beam: t hickness (t) , w idth (w) and h eight (h). 4. Optimization criterion The objective function that has been chosen is the volume: 5 = [ # ℎ − ! ( ℎ − & ) ] ∙ 9 [ :: $ ] Since the material has not been considered as a design variable, it is equivalent to use the volume or the mass as objective function, but since the mass requires a n additional parameter to be computed (the density), volume has been used as objective function. Also the area of t he cross - section could have been chosen as objective function, since the length L of the beam is not a design variable. Density ( r ) 7850 kg/m 3 Young modulus (E) 210000 MPa Shear modulus (G) 81000 MPa Poisson’s ratio ( n ) 0.3 Yield strength ( s y ) 355 MPa Tensile strength (R m ) 490 MPa Chiara Moreschini – ID 10601144 2 5. Formulation of constraints The following constraints have been considered : a) Aspect ratio constraint : the aspect ratio should not be lower than the prescribed value # ℎ ≥ 2 .5 → 1 − # 2 .5 ∙ ℎ ≤ 0 b) Deflection constraint : the maximum deflection A of the beam should not exceed the limit value A ≤ A %&' = 9 200 = 10 ,5 :: → A ∙ 10 $ A %&' − 1 ≤ 0 The maximum deflection A of the beam is located at its free end and can be computed as follows 1 : A = C 9 $ 3 D 2 ! c) Normal stress constraint : the maximum normal stress E developed in the structure should not exceed the yield strength of the material : E ≤ E %&' = 355 FCG → E ∙ 10 $ E %&' − 1 ≤ 0 The maximum normal stres s E developed inside the structure is located at the constrained end of the beam and can be computed as: E = C F ( ,'*! 2 ! H '*! = C9 2 ! ( ℎ − ) " ) Where H '*! = ( ℎ − ) " ) is the maximum distance of a material’s fiber from the neutral axis, which passes through the centroid of the beam at y=C y . Figure 2 : Normal stress distribution on the cross - section 1 Note that the displacement ! is in [m], so in the formulation of the constraint it has been added a conversion factor of 10 ! to have both ! and ! "#$ in [mm]. As for the displacement, a conversion factor of 10 ! will be used also for the following constraints on the normal stres s and on the shear stress (to convert all parameters in MPa). Chiara Moreschini – ID 10601144 3 d) Shear stress constraint : the maximum shear stress I developed in the structure should not exceed the limit shear stress of the material : I ≤ I %&' ≅ E %&' √ 3 → I ∙ 10 $ I %&' − 1 ≤ 0 The maximum shear stress I is in correspondence of the neutral axis ( NA ) of the section and can be computed with the Jourawsky approach : I = 5 L 2 ! & +, = C [ M +, NO ] 2 ! 2 & = C [ 2 ( ℎ − ) " ) # & ] 2 ! 2 & = C ( ℎ − ) " ) # 2 2 ! Figure 3 : On the left, reference geometry used to compute the shear stress on the beam ’ s section, on the right shear stress distribution on the beam section e) Design variables’ bounds : t he design variables should not exceed the ir prescribed dimensions : P 50 ≤ # ≤ 300 [ :: ] 5 ≤ & ≤ 15 [ :: ] 10 ≤ ℎ ≤ 100 [ :: ] O nly a vertical deflection is supposed to be present in the beam, due to the load P. However there is a critical value of P that could lead to lateral torsional buckling , so the force applied should be lower than th is critical value if we want to consider o nly the presence of a vertical deflection : C -.&/ = 4 ,2 9 # R D 2 " S T / Where T / is the torsional stiffness of the beam and can be computed with the formula referred to open sections made of i=1,…,n plates of thickness & & and length U & : T / = 1 3 V ( W 1 ,& & & $ U & ) 2 & 3 1 = 1 3 & $ ( 2ℎ + ! ) #X& ℎ W 1 ,& = 1 ,XYZ[ & & ≪ U & The condition C ≤ C -.&/ will be checked once the solution has been obtained, otherwise also a horizontal deflection and the stresses related to lateral torsional buckling should be considered. There exist also some additional manufacturing constraints : t his kind of beam is usually manufactured by hot rolling and even if not all sizes are commercially available, we can assume that reasonable measures for the variables t, w and h are integer millimetres , so the design variables are discrete . Chiara Moreschini – ID 10601144 4 Moreover, since some of the constraints are non - linear functions of the design variables, the present problem is a discrete non - linear programming problem . The simplest procedure to solve discrete problems is assuming the variables continuous and solve the associated continuous problem ; t hen the nearest discrete values are assigned to the variables and the design is checked again for feasibility. Therefore , the manufacturing constraints are not formulated as mathematical constraints insid e the analytical model at this stage. 6. Description of the solution method The optimization problem has been solved in Excel , that uses the non - linear GRG optimization algorithm . 6.a. T he Excel file and its structure Figure 4 : Structure of the Excel file used to solve the optimiz ation problem Chiara Moreschini – ID 10601144 5 6.b. Results of the optimization in Excel The minimum point found by the Excel solver is the following: ] & = 5 :: # = 220 ,2751432 :: ℎ = 88 ,11005986 :: The procedure has been repeated for different initial guesses, to ensure that the optimum point found was not a local minimum depending on the initial conditions chosen, but it is a global minimum of the problem. The sensitivity and the answer reports related to this solution are shown in the table s below: Table 2 : R esults from the answer and sensitivity reports Constraint Slack Status t lower bound 0 Binding t upper bound 10 Not binding w lower bound 170,2751 Not binding w upper bound 79,7248 Not binding h lower bound 78,1100 Not binding h upper bound 11,8899 Not binding The lower bound of the thickness, the aspect ratio and the deflection are the active constraint s of the problem , since their slack s are equal to zero . The constraints on the stresses instead are very far from the ir limit and they remain in the inactive state, with slack variables close to - 1 (in particular the shear stresses). Exploiting the constraint variation sensitivity theorem , we can say that the constraint s with largest Lagrange multiplier s will be the ones with the largest impact on the cost function : in this case the aspect ratio and the deflection constraints . In order to study the effect of a modification of those constraint, we can try relaxing the aspect ratio constraint, imposing it equal to # / ℎ ≥ 2 instead of # / ℎ ≥ 2 ,5 (relaxation of 20%) ; in this case we obtain a slightly different optimum point, but the active constrains remain the aspect ratio and the deflection, and the Lagrange multiplier of the aspect ratio decreases of 21% (from - 1929058 to - 1511659 ) as expected, since the constraint has been relaxed and has now a lower impact on the cost function. Same observation s could be done for the deflection constraint, even if its effect on the cost function will be lower : by relaxing the deflection constraint of 20 % , we obtain as Lagrange multiplier - 1271749 (decrease of 5,8%) . It is also worth of noticing that the proposed constraints do not consider safety factors, that instead are usually taken into account (especially for what concerns the stresses), so the actual constrains should be tightened , resulting in an increase of the related Lagrange multipliers. The thickness is the only integer solution, since the lower bound of this variable is an active constraint. The other two variables instead have non - integer solutions, but as explained above there are some manufacturing constraints tha t have to be take in into account : to do so the solution has been rounded to the closest integers . The final solution is therefore: ] & = 5 :: # = 220 :: ℎ = 88 :: After this approximation , o n e of the constrain ts , i.e. the deflection one , is not more strictly satisfied, but we obtain A = 10 ,54 :: ≥ A %&' = 10 ,5 :: . However, the deflection exceeds the limit of only 0,04 mm, so the Constraint Lagrange Multiplier Slack Status Aspect ratio - 1929058 0 Binding Deflection - 1351293 0 Binding Normal stress 0 0,7183 Not binding Shear stress 0 0,9922 Not binding Chiara Moreschini – ID 10601144 6 proposed design can be considered feasible ; if instead the deflection constraint is assumed to be inviolable , then a possible design solution could be w=224 mm, h=88 mm and t=5 mm, that is a point which satisfies strictly all the constraints , but is farer from the continuous optimum solution (so it will be heavier) . In this case, we assume a small violation of 0,4% of the maximum deflection to be acceptable. With the proposed solution it has also been verified that the lateral torsional buckling does not occur: C -.&/ = 59 !c ≫ C = 1 !c 7. Graphical solution in Matlab Since the lower bound constraint of the thickness reaches the active state and the thickness becomes integer , we can fix its value in order to obtain a 2 - variables problem, that can be solved graphically in Matlab. Figure 5 : Graphical solution in Matlab for a fixed value of thickness (t=5mm). The result obtained from the graphical solution, shown in figure 5, are consistent with the previous one, displaying a unique minimum point around w=220 mm and h=88 mm. The active constraints are the deflection and the aspect ratio constraints, while the constraints on the normal and shear stresses remain inactive (in particular the shear stress es are so low that they do not appear in the proposed graphical solution). 8. FEA of the optimum solution previously found 8.a. FEA model setup The finite element model of the beam has been created as a 3D solid extruded starting from the C - section with w=220 mm, h=88 mm and t=5 mm ; the geometry was not created as a shell, in order to be able to observe also the stress components in the axial direction of the beam on the different sections . w lower bound w upper bound h upper bound h lower bound OPTIMUM DESIGN Chiara Moreschini – ID 10601144 7 Th e n the material S355 J 0 has been assigned to the part and a new static general step has been created. The end section ’s surface of the bea m has been constrained with an encastre boundary condition, while on the other end the load has been applied by creating a refence point in the middle of the section and coupling it with the section. Finally, the part has been meshed using quadratic hexahedral elements with global size of 20 mm. 8 . b . Results of FEA model and comparison with analytical results The results of the FE A mode l are reported below in terms of deflection, normal stress and shear stress. The deflection of the beam is only in the vertical direction, confirming the absence of buckling 2 : the vertical deflection is zero at the constrain ed end of the beam and increases along its axis, reaching the m aximum value at the free end of the beam, as expected . The maximum deflection registered is 10,58 mm . Figure 6 : Results of FEA model in terms of deflection of the beam On the other end, the normal stress increases moving from the free end to the constrained one, w here it reaches the maximum value, that is 118,9 MPa (in compression) ; however this maximum value is in correspondence of the constrained nodes ( encastre ) and it could be affected by boundary effects, while the adjacent nodes show a normal stress equal to 96, 52 MPa , which is a more reliable maximum value ; this value has been computed as mean value of the nodes adjacent to the final ones, on both sides of the s ection . Moreover, the normal stress distribution in each section of the beam is consistent with the expect ed b utterfly distribution (Figure 2) , with fibers in tension on the upper part and in compression in the bottomer part. Figure 7 : Results of FEA model in terms of normal stresses The maximum shear stress instead is reached at the neutral axis , that is theoretically located at a distance of 21,44 mm from the top surface of the beam ; the closest nodes available where to compute the shears are instead 2 There is o nly a small transverse displacement near the end of the beam , due to its deformation mode, not to buckling. Mean value : 96, 52 MPa Chiara Moreschini – ID 10601144 8 at 22 mm from the to p surface of the beam. The maximum shear stress has been obtained as average value between the se nodes (on both sides of t he section) , that are the closest to the neutral axis , and it is equal to 1, 66 MPa . Moreover the overall distribution of the shear stress is consistent with the theoretical one (Figure 3) . Figure 8 : Results of FEA model in terms of shear stresses; on the left component S23, on the right S13 Finally t he results of the FE A mode l are compared to the ones of the analytical model; the latter are referred to the final configuration chosen (t=5 mm, w=220 mm, h=88 mm) and not to the continuou s optimum solution . Table 3 : Comparison between the results of the FEA and the analytical results Parameter FEA model Analytical model Relative error Maximum deflection 10,58 mm 10 , 54 mm 0,3 % Maximum normal stress 96, 52 MPa 100 , 25 MPa 3, 7 % Maximum shear stress 1, 66 MPa 1 , 5 9 MPa 4 , 4 % The relative errors between the analytical and the numerical solutions for all the parameters considered (computed with respect to the analytical solution) are lower than 5%, showing consistency between the two models. The error with respect to the shear stress is the highest , but this value could be related to the impossibility of evaluating the numerical s olution exactly on the neutral axis , where the analytical solution is co nsidered ; in order to obtain a better comparison of the shear stress es , a mesh refinement should be performed , or the reference analytical solution should be considered in correspondence to the available nodes’ position . 9. Conclusions The proposed problem consisted in the optimum design of a C - section cantilever beam and it has been solved in Excel, that allowed to identify all the active constraints of the problem , their effects on the beam ’s volume (considered as cost function) and the continuous optimum solution. The final solution has been obtained by discretizing the lat ter one, due to manufacturing constraints , and checking again its feasibility. T he final optimum design has been analysed numerically in Abaqus, in order to check the consistency of the analytical and numerical model s . The errors between the two solutions resulted to be acceptable with respect to all the parameters considered , i.e. lower than 5% , so the two models seem to be consistent and reliable. However it should be remembered that the proposed design is not the o nly possible solution , since the discretization of the continuous solution (which was unique) should have been performed in many ways, depending on which constraint is considered to be the most limiting one. 1, 66 MPa