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Optimization of elastic spring supports for cantilever beams

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In this study,a new approach of optimization algorithm is developed. The optimum distribution of elastic springs on which a cantilever Timoshenko beam is seated and minimization of the shear force on the support of the beam is investigated.The Fourier transform is applied to the beam vibration equation in the time domain and transfer function, independent from the external influence, is used to define the structural response. For all translational modes of the beam, the optimum locations and amounts of the springs are investigated so that the transfer function amplitude of the support shear force is minimized. The stiffness coefficients of the springs placed on the nodes of the beam divided into finite elements are considered as design variables. There is an active constraint on the sum of the spring coefficients taken as design variables and passive constraints on each of them as the upper and lower bounds. Optimality criteria are derived using the Lagrange Multipliers method. The gradient information required for solving the optimization problem is analytically derived. Verification of the new approach optimization algorithm was carried out by comparing the results presented in this paper with those ones from analysis of the model of the beam without springs, with springs with uniform stiffness and with optimal distribution of springs which support a cantilever beam to minimize the tip deflection of the beam found in the literature. The numerical results show that the presented method is effective in finding the optimum spring stiffness coefficients and location of springs for all translational modes.The proposed method can give designers an idea of how to support the cantilever beams under different harmonic vibrations.

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Correspondence to Maciej Dutkiewicz.

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• The new approach of optimization algorithm is developed.

• New sensitivity-based model updating with simulated stiffness of springs with the purpose of minimization of the support shear force in Timoshenko beam.

• Application of the Lagrange Multipliers Method for optimization problem. Optimal stiffness distribution.

• Wide range of analysis of the behaviour of the beam in time and frequency domain. Development of sensitivity-based model for first sixth modes.

• Analysis of variations of first order sensitivities and transfer functions according to the step number.

• Research of influence of spring support on changes of natural frequencies of the beam for different mode control.

• Practical tool for optimal design of beam and other structures under dynamical excitation, especially for the engineering purposes.

• Proposed methodology is compared to solutions found in literature and presented model is verified with results from other models from literature.

• The problem of the beam was also analysed with the use of Gray Wolf Optimization (GWO) algorithm to compare the results of the gradient-based method. The results of the proposed gradient based method are proved to be accurate by comparing the results with a GWO method.




For a better understanding of the problem, let us do a numerical analysis and show a step on a cantilever beam model with 2 elements and 4 degrees of freedom as shown in Fig. 15. The cantilever beam (length is 6 m) is modelled as a Timoshenko beam by dividing it into 3 m finite elements and assuming vertical displacements and rotations at each node. Two vertical and rotational displacements at each node are considered and a total of 4 degrees of freedom are defined in the system. The shear modulus is G= 7,94 1010 N/m2, the correction factor κ=5/6, the cross-sectional area A=0,05 m2, the moment of inertia I= 2,08 10−4 m4, density of material ρ= 7,8 103 kg/m3, the modulus of elasticity E= 2,06 1011 N/m2. A mass of 100 kg is also added to the end of the beam. Structural damping matrix is selected as mass proportional and structural damping ratio is assumed as 0,02. The total stiffness is assumed to be \( \overline{K}=9x{10}^6N/m \) and the amount of additional stiffness in each step is calculated as \( \Delta K=\frac{9x{10}^6}{300}=30000\ N/m \) where design step number (d) is selected as 300.

Using element stiffness and mass matrices, the global stiffness and mass matrices of the 4 degree of freedom beam are found as follows

$$ K=\left[\begin{array}{ccc}3,74406\kern0.5em 0,00000& -1,87203& 2,80804\\ {}\begin{array}{cc}0,00000& 11,2807\end{array}& -2,80804& 2,78380\\ {}\begin{array}{cc}\begin{array}{c}-1,87203\\ {}2,80804\end{array}& \begin{array}{c}-2,80804\\ {}2,78380\end{array}\end{array}& \begin{array}{c}1,87203\\ {}-2,80804\end{array}& \begin{array}{c}-2,80804\\ {}5,64033\end{array}\end{array}\right]\ast {10}^7 $$
$$ M=\left[\begin{array}{ccc}\mathrm{868,698}\kern0.5em \mathrm{0,000}& \mathrm{150,651}& -\mathrm{109,277}\\ {}\begin{array}{cc}\mathrm{0,000}& \kern0.5em \mathrm{201,01}\end{array}& \mathrm{109,277}& -75,8064\\ {}\begin{array}{cc}\begin{array}{c}\mathrm{150,651}\\ {}-\mathrm{109,277}\end{array}& \begin{array}{c}\mathrm{109,277}\\ {}-\mathrm{75,806}\end{array}\end{array}& \begin{array}{c}\mathrm{534,349}\\ {}-\mathrm{183,223}\end{array}& \begin{array}{c}-\mathrm{183,223}\\ {}\mathrm{100,505}\end{array}\end{array}\right] $$

In the first step, the system’s natural circular frequencies are calculated as ωi ={29,8,881,189,4,946,366,731,694,69}rad/s from the eigenvalue-eigenvector analysis. The first eigenvector normalized to the tip degree of freedom of the system is found as ∅1 ={0,3,357,330,192,635 100,000 0,232,286}. The damping matrix of the beam model can be calculated in proportion to the mass as follows.

$$ C=2\upzeta {\upomega}_1M=2\ast 0,02\ast 29,8881\ast \left[\begin{array}{ccc}\mathrm{868,698}\kern0.5em \mathrm{0,000}& \mathrm{150,651}& -\mathrm{109,277}\\ {}\begin{array}{cc}\mathrm{0,000}& \kern0.5em \mathrm{201,01}\end{array}& \mathrm{109,277}& -75,8064\\ {}\begin{array}{cc}\begin{array}{c}\mathrm{150,651}\\ {}-\mathrm{109,277}\end{array}& \begin{array}{c}\mathrm{109,277}\\ {}-\mathrm{75,806}\end{array}\end{array}& \begin{array}{c}\mathrm{534,349}\\ {}-\mathrm{183,223}\end{array}& \begin{array}{c}-\mathrm{183,223}\\ {}\mathrm{100,505}\end{array}\end{array}\right]=\left[\begin{array}{ccc}\begin{array}{cc}1590,92& \mathrm{0,000}\end{array}& \mathrm{275,901}& -\mathrm{200,13}\\ {}\begin{array}{cc}\mathrm{0,000}& \kern0.5em \mathrm{368,127}\end{array}& \mathrm{200,130}& -\mathrm{138,831}\\ {}\begin{array}{cc}\begin{array}{c}\mathrm{275,901}\\ {}-\mathrm{200,13}\end{array}& \begin{array}{c}\mathrm{200,13}\\ {}-\mathrm{138,831}\end{array}\end{array}& \begin{array}{c}\mathrm{978,601}\\ {}-\mathrm{335,552}\end{array}& \begin{array}{c}-\mathrm{335,552}\\ {}\mathrm{184,063}\end{array}\end{array}\right] $$

The stiffness matrix of the springs added and its partial derivations with respect to k1 and k2 are given below

$$ {\boldsymbol{K}}_{\boldsymbol{S}}=\left[\begin{array}{ccc}{k}_1\kern0.5em 0& 0& 0\\ {}\begin{array}{cc}0& \kern0.5em 0\end{array}& 0& 0\\ {}\begin{array}{cc}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}& \begin{array}{c}{k}_2\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\right]\frac{\partial {\boldsymbol{K}}_{\boldsymbol{s}}}{\partial {k}_1}=\left[\begin{array}{ccc}\begin{array}{cc}1& 0\end{array}& 0& 0\\ {}\begin{array}{cc}0& 0\end{array}& 0& 0\\ {}\begin{array}{cc}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}& \begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\right]\frac{\partial {\boldsymbol{K}}_{\boldsymbol{s}}}{\partial {k}_2}=\left[\begin{array}{ccc}\begin{array}{cc}0& 0\end{array}& 0& 0\\ {}\begin{array}{cc}0& 0\end{array}& 0& 0\\ {}\begin{array}{cc}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}& \begin{array}{c}1\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\right] $$

The calculation of matrix A given in Eq. (9) is as follows

$$ \boldsymbol{A}=\left(\boldsymbol{K}+{\boldsymbol{K}}_{\boldsymbol{s}}\right)+i{\omega}_1\boldsymbol{C}-{\omega}_1^2\boldsymbol{M}=\left[\begin{array}{ccc}3,74406\kern0.5em 0,00000& -1,87203& 2,80804\\ {}\begin{array}{cc}0,00000& 11,2807\end{array}& -2,80804& 2,78380\\ {}\begin{array}{cc}\begin{array}{c}-1,87203\\ {}2,80804\end{array}& \begin{array}{c}-2,80804\\ {}2,78380\end{array}\end{array}& \begin{array}{c}1,87203\\ {}-2,80804\end{array}& \begin{array}{c}-2,80804\\ {}5,64033\end{array}\end{array}\right]\ast {10}^7+\left[\begin{array}{ccc}\begin{array}{cc}{k}_1& 0\end{array}& 0& 0\\ {}\begin{array}{cc}0& \kern0.5em 0\end{array}& 0& 0\\ {}\begin{array}{cc}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}& \begin{array}{c}{k}_2\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\right]+i\ast 29,8881\ast \left[\begin{array}{ccc}\begin{array}{cc}1590,92& \mathrm{0,000}\end{array}& \mathrm{275,901}& -\mathrm{200,13}\\ {}\begin{array}{cc}\mathrm{0,000}& \kern0.5em \mathrm{368,127}\end{array}& \mathrm{200,13}& -\mathrm{138,831}\\ {}\begin{array}{cc}\begin{array}{c}\mathrm{275,901}\\ {}-\mathrm{200,13}\end{array}& \begin{array}{c}\mathrm{200,13}\\ {}-\mathrm{138,831}\end{array}\end{array}& \begin{array}{c}\mathrm{978,601}\\ {}-\mathrm{335,552}\end{array}& \begin{array}{c}-\mathrm{335,552}\\ {}\mathrm{184,063}\end{array}\end{array}\right]-29,{8881}^2\ast \left[\begin{array}{ccc}\begin{array}{cc}\mathrm{868,698}& \mathrm{0,000}\end{array}& \mathrm{150,651}& -\mathrm{109,277}\\ {}\begin{array}{cc}\mathrm{0,000}& \kern0.5em \mathrm{201,01}\end{array}& \mathrm{109,277}& -75,8064\\ {}\begin{array}{cc}\begin{array}{c}\mathrm{150,651}\\ {}-\mathrm{109,277}\end{array}& \begin{array}{c}\mathrm{109,277}\\ {}-\mathrm{75,806}\end{array}\end{array}& \begin{array}{c}\mathrm{534,349}\\ {}-\mathrm{183,223}\end{array}& \begin{array}{c}-\mathrm{183,223}\\ {}\mathrm{100,505}\end{array}\end{array}\right]=\left[\begin{array}{ccc}\begin{array}{cc}3,66646\ast {10}^7+31040,3\ast i& 0,0000000+0,0000000\ast i\end{array}& -1,88549\ast {10}^7+5383,06\ast i& 2,8178\ast {10}^7-3904,7\ast i\\ {}\begin{array}{cc}0,0000000+0,000000\ast i\kern1.25em & \kern0.5em 1,12627\ast {10}^8+7182,47\ast i\end{array}& -2,8178\ast {10}^7+3904,7\ast i& 2,79057\ast {10}^7-2708,71\ast i\\ {}\begin{array}{cc}\begin{array}{c}-1,88549\ast {10}^7+5383,06\ast i\\ {}2,8178\ast {10}^7-3904,7\ast i\end{array}& \begin{array}{c}-2,8178\ast {10}^7+3904,7\ast i\\ {}2,79057\ast {10}^7-2708,71\ast i\end{array}\end{array}& \begin{array}{c}1,82429\ast {10}^7+19093,3\ast i\\ {}-2,79167\ast {10}^7-6546,9\ast i\end{array}& \begin{array}{c}-2,79167\ast {10}^7-6546,9\ast i\\ {}5,63135\ast {10}^7+3591,24\ast i\end{array}\end{array}\right] $$

Initial values of design variables (k1and k2) are assumed to be zero, The modal mass for 1st mode and the partial derivative of first eigenvalue with respect to k1 and k2 are obtained as

$$ {\overline{m}}_1={\boldsymbol{\varPhi}}_{\mathbf{1}}^{\boldsymbol{T}}\boldsymbol{M}{\boldsymbol{\varPhi}}_{\mathbf{1}}=\left\{\begin{array}{c}0,335733\ \\ {}0,192635\\ {}\ 1,00000\\ {}\ 0,232286\end{array}\right\}\left[\begin{array}{ccc}\mathrm{868,698}\kern0.5em \mathrm{0,000}& \mathrm{150,651}& -\mathrm{109,277}\\ {}\begin{array}{cc}\mathrm{0,000}& \kern0.5em \mathrm{201,01}\end{array}& \mathrm{109,277}& -75,8064\\ {}\begin{array}{cc}\begin{array}{c}\mathrm{150,651}\\ {}-\mathrm{109,277}\end{array}& \begin{array}{c}\mathrm{109,277}\\ {}-\mathrm{75,806}\end{array}\end{array}& \begin{array}{c}\mathrm{534,349}\\ {}-\mathrm{183,223}\end{array}& \begin{array}{c}-\mathrm{183,223}\\ {}\mathrm{100,505}\end{array}\end{array}\right]\left\{0,335733\kern0.75em 0,192635\kern1em 1,00000\kern1em 0,232286\right\}=\mathrm{679,458} $$
$$ \frac{\partial {\varOmega}_1}{\partial {k}_1}=\frac{1}{{\overline{m}}_1}\frac{\partial {\overline{k}}_1}{\partial {k}_1}=\frac{1}{\mathrm{679,458}}\left\{\begin{array}{c}0,335733\ \\ {}0,192635\\ {}\ 1,00000\\ {}\ 0,232286\end{array}\right\}\left[\begin{array}{ccc}1\kern0.5em 0& 0& 0\\ {}\begin{array}{cc}0& 0\end{array}& 0& 0\\ {}\begin{array}{cc}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}& \begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\right]\left\{0,335733\kern1em 0,192635\kern1em 1,00000\kern0.75em 0,232286\right\}=0,000165892 $$
$$ \frac{\partial {\varOmega}_1}{\partial {k}_2}=\frac{1}{{\overline{m}}_1}\frac{\partial {\overline{k}}_1}{\partial {k}_2}=\frac{1}{\mathrm{679,458}}\left\{\begin{array}{c}0,335733\ \\ {}0,192635\\ {}\ 1,00000\\ {}\ 0,232286\end{array}\right\}\left[\begin{array}{ccc}0\kern0.5em 0& 0& 0\\ {}\begin{array}{cc}0& 0\end{array}& 0& 0\\ {}\begin{array}{cc}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}& \begin{array}{c}1\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\right]\left\{0,335733\kern1em 0,192635\kern0.75em 1,00000\kern0.75em 0,232286\right\}=0,00147176 $$

The transfer function vector is calculated as

$$ \overset{\sim }{\boldsymbol{U}}\left({\omega}_1\right)=-{\boldsymbol{A}}^{-1}\boldsymbol{Mr}=\left\{\begin{array}{c}-0,0000142032+0,0135566i\\ {}0,00000110433+0,00777842i\\ {}\begin{array}{c}0,0000155147+0,040379i\\ {}0,0000140347+0,00937945i\end{array}\end{array}\right\} $$

The transfer function vector of elastic forces are obtained as follows

$$ \boldsymbol{F}\left({\omega}_1\right)=\left(\boldsymbol{K}+{\boldsymbol{K}}_{\boldsymbol{s}}\right)\overset{\sim }{\boldsymbol{U}}\left({\omega}_1\right)=\left\{\begin{array}{c}-\mathrm{428,114}+15038,9i\\ {}79,6148+4703,22i\\ {}\begin{array}{c}\mathrm{131,217}+20322,7i\\ {}-12,1448-7616,93i\end{array}\end{array}\right\} $$

where the elements of Ks are zero in the first step. The absolute value of the F(ωn) are given as

$$ \left|\boldsymbol{F}\left({\omega}_1\right)\right|=\left\{\begin{array}{c}15045,00\\ {}4703,89\\ {}\begin{array}{c}20323,10\\ {}7616,94\end{array}\end{array}\right\} $$

The transfer function amplitude of the support shear force is found as

$$ V\left({\omega}_n\right)=\sum \limits_{s=1}^2\mathit{\operatorname{sign}}\left({\varnothing}_{is}\right)\left|{F}_s\left({\omega}_1\right)\right|=\mathit{\operatorname{sign}}\left({\varnothing}_{11}\right)\left|{F}_1\left({\omega}_1\right)\right|+\mathit{\operatorname{sign}}\left({\varnothing}_{13}\right)\left|{F}_2\left({\omega}_1\right)\right|=\mathit{\operatorname{sign}}\left(0,335733\right)15045,00+\mathit{\operatorname{sign}}\left(1,00000\right)20323,10=\left(+\right)15045,00+\left(+\right)20323,10=35362,8 $$

where s is defined as the number of degree of freedom corresponding to the translational modes. The first order sensitivity with respect to the k1 and k2are given as

$$ \frac{\partial \boldsymbol{F}}{\partial {k}_1}=\left(\frac{\partial {\boldsymbol{K}}_{\boldsymbol{s}}}{\partial {k}_1}-{\boldsymbol{K}}_{\boldsymbol{s}}{\boldsymbol{A}}^{-\mathbf{1}}\frac{\partial \boldsymbol{A}}{\partial {k}_1}\right)\overset{\sim }{\boldsymbol{U}}=\left\{\begin{array}{c}-1,50914\ast {10}^{-5}-3,26491\ast {10}^{-4}i\\ {}-5,72376\ast {10}^{-8}-4,40133\ast {10}^{-6}i\\ {}\begin{array}{c}-1,70528\ast {10}^{-7}-\mathrm{1,745}\ast {10}^{-5}i\\ {}1,54417\ast {10}^{-6}+3,74729\ast {10}^{-5}i\end{array}\end{array}\right\} $$
$$ \frac{\partial \boldsymbol{F}}{\partial {k}_2}=\left(\frac{\partial {\boldsymbol{K}}_{\boldsymbol{s}}}{\partial {k}_2}-{\boldsymbol{K}}_{\boldsymbol{s}}{\boldsymbol{A}}^{-\mathbf{1}}\frac{\partial \boldsymbol{A}}{\partial {k}_2}\right)\overset{\sim }{\boldsymbol{U}}=\left\{\begin{array}{c}2,67383\ast {10}^{-5}+1,06866\ast {10}^{-3}i\\ {}3,53174\ast {10}^{-6}+5,56696\ast {10}^{-5}i\\ {}\begin{array}{c}7,5875\ast {10}^{-6}+5,10216\ast {10}^{-5}i\\ {}-5,71166\ast {10}^{-6}-1,40692\ast {10}^{-4}i\end{array}\end{array}\right\} $$

The absolute values of first order sensitivities of the elastic force in two nodes are calculated as

$$ \frac{\partial \left|{F}_1\right|}{\partial {k}_1}=\frac{1}{\left|{F}_1\right|}\left\{\mathit{\operatorname{Re}}\left[{F}_1\right]\left(\mathit{\operatorname{Re}}\left[\frac{\partial {F}_1}{\partial {k}_1}\right]\right)+\mathit{\operatorname{Im}}\left[{F}_1\right]\left(\mathit{\operatorname{Im}}\left[\frac{\partial {F}_1}{\partial {k}_1}\right]\right)\right\}=\frac{1}{15045,00}\left\{\left(-\mathrm{428,114}\right)\ast \left(-1,50914\ast {10}^{-5}\right)+\left(15038,9\right)\ast \left(-3,26491\ast {10}^{-4}\right)\right\}=-0,000325929 $$
$$ \frac{\partial \left|{F}_1\right|}{\partial {k}_2}=\frac{1}{\left|{F}_1\right|}\left\{\mathit{\operatorname{Re}}\left[{F}_1\right]\left(\mathit{\operatorname{Re}}\left[\frac{\partial {F}_1}{\partial {k}_2}\right]\right)+\mathit{\operatorname{Im}}\left[{F}_1\right]\left(\mathit{\operatorname{Im}}\left[\frac{\partial {F}_1}{\partial {k}_2}\right]\right)\right\}=\frac{1}{15045,00}\left\{\left(-\mathrm{428,114}\right)\ast \left(2,67383\ast {10}^{-5}\right)+\left(15038,9\right)\ast \left(1,06866\ast {10}^{-3}\right)\right\}=0,00106747 $$
$$ \frac{\partial \left|{F}_2\right|}{\partial {k}_1}=\frac{1}{\left|{F}_2\right|}\left\{\mathit{\operatorname{Re}}\left[{F}_2\right]\left(\mathit{\operatorname{Re}}\left[\frac{\partial {F}_2}{\partial {k}_1}\right]\right)+\mathit{\operatorname{Im}}\left[{F}_2\right]\left(\mathit{\operatorname{Im}}\left[\frac{\partial {F}_2}{\partial {k}_1}\right]\right)\right\}=\frac{1}{20323,10}\left\{\left(\mathrm{131,217}\right)\ast \left(-1,70528\ast {10}^{-7}\right)+\left(20322,7\right)\ast \left(-\mathrm{1,745}\ast {10}^{-5}\right)\right\}=-0,0000174508 $$
$$ \frac{\partial \left|{F}_2\right|}{\partial {k}_2}=\frac{1}{\left|{F}_2\right|}\left\{\mathit{\operatorname{Re}}\left[{F}_2\right]\left(\mathit{\operatorname{Re}}\left[\frac{\partial {F}_2}{\partial {k}_2}\right]\right)+\mathit{\operatorname{Im}}\left[{F}_2\right]\left(\mathit{\operatorname{Im}}\left[\frac{\partial {F}_2}{\partial {k}_2}\right]\right)\right\}=\frac{1}{20323,10}\left\{\left(\mathrm{131,217}\right)\ast \left(7,5875\ast {10}^{-6}\right)+\left(20322,7\right)\ast \left(5,10216\ast {10}^{-5}\right)\right\}=0,0000510695 $$

The first order sensitivities of transfer function amplitude of the support force with respect to design variables (k1and k2) are obtained as follows

$$ \frac{\partial V}{\partial {k}_1}=\mathit{\operatorname{sign}}\left({\varnothing}_{11}\right)\left(-0,000325929\right)+\mathit{\operatorname{sign}}\left({\varnothing}_{13}\right)\left(-0,0000174508\right)=-0,00034338 $$
$$ \frac{\partial V}{\partial {k}_2}=\mathit{\operatorname{sign}}\left({\varnothing}_{11}\right)\left(0,00106747\right)+\mathit{\operatorname{sign}}\left({\varnothing}_{13}\right)\left(0,0000510695\right)=0,00111854 $$
$$ \frac{\partial V}{\partial k}=\left\{-0,00034338\kern1.25em 0,00111854\right\} $$
$$ \mathit{\operatorname{Max}}\left(\frac{\partial V}{\partial k}\right)-\frac{\partial V}{\partial {k}_1}=0,00111854-0,00034338\ne 0 $$
$$ \mathit{\operatorname{Max}}\left(\frac{\partial V}{\partial k}\right)-\frac{\partial V}{\partial {k}_2}=0,00111854-0,00111854=0 $$

In Eq. (65), there is a z=2 index which satisfies the optimality condition. All spring stiffness coefficients are updated as follows

$$ {k}_1=0\kern1.25em {k}_2=0+\Delta K=30000\ N/m $$

According to this value, the cycle continues until the total stiffness coefficient (\( \overline{K} \)) is reached in the end of step 300. In each step, all calculations are updated such as eigenvalue, eigenvectors, objective function and sensitivities. Some basic calculations in each step are summarized briefly where the objective function is minimized.

Fig. 15
figure 15

Cantilever beam with 2 elements

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Aydin, E., Dutkiewicz, M., Öztürk, B. et al. Optimization of elastic spring supports for cantilever beams. Struct Multidisc Optim 62, 55–81 (2020).

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