A Comparative Study of Recent Non-traditional Methods for Mechanical Design Optimization

Abstract

Solving practical mechanical problems is considered as a real challenge for evaluating the efficiency of newly developed algorithms. The present article introduces a comparative study on the application of ten recent meta-heuristic approaches to optimize the design of six mechanical engineering optimization problems. The algorithms are: the artificial bee colony (ABC), particle swarm optimization (PSO) algorithm, moth-flame optimization (MFO), ant lion optimizer (ALO), water cycle algorithm (WCA), evaporation rate WCA (ER-WCA), grey wolf optimizer (GWO), mine blast algorithm (MBA), whale optimization algorithm (WOA) and salp swarm algorithm (SSA). The performances of the algorithms are tested quantitatively and qualitatively using convergence speed, solution quality, and the robustness. The experimental results  on the six mechanical problems demonstrate the efficiency and the ability of the algorithms used in this article.

Appendix

• 1. Coupling with a bolted rim

The problem can be mathematically formulated as follows:

Objective function: $f\left( x \right) = \beta_{1} \left( {\frac{N}{{N_{m} }}} \right) + \beta_{2} \left( {\frac{{R_{B} + \phi_{4} (d) + c}}{{R_{m} }}} \right) + \beta_{3} \left( {\frac{M}{{M_{T} }}} \right)$

Subject to:

$$\begin{aligned} g_{1} (x) & = \frac{\alpha M}{{NR_{B} K(d)}} - 1 \le 0, \, g_{2} (x) = 1 - \frac{{2\pi R_{B} }}{{\phi_{5} (d)N}} \le 0 \\ g_{3} (x) & = 1 - \frac{{R_{B} }}{{\phi_{4} (d)}} + R_{M} \le 0, \, g_{4} (x) = N - N_{\hbox{max} } \le 0 \\ g_{5} (x) & = R_{B} - R_{\hbox{max} } \le 0, \, g_{6} (x) = N_{M} - N \le 0 \\ g_{7} (x) & = R_{M} - R_{B} \le 0, \, g_{8} (x) = M - M_{\hbox{max} } \le 0 \\ g_{9} (x) & = M_{T} - M \le 0, \, g_{10} (x) = d - 24 \le 0 \\ g_{11} (x) & = 6 - d \le 0 \\ \end{aligned}$$

where $K(d) = \frac{{0.9f_{m} R_{e} \pi \left( {\phi_{1} (d)} \right)^{2} }}{{4\sqrt {1 + 3(0.16\phi_{3} (d)f_{1} /\phi_{1} (d))^{2} } }}, \, M_{T} = 40\,{\text{Nm}}, \, M_{\hbox{max} } = 1000\,{\text{Nm}}, \, f_{m} = 0.15, \, f_{1} = 0.15$

$$\begin{aligned} \alpha & = 1.5, \, R_{e} = 627\,{\text{MPa}}, \, N_{M} = 8, \, N_{\hbox{max} } = 100, \, R_{M} = 50\,{\text{mm}}, \, R_{\hbox{max} } = 1000\,{\text{mm}}, \, c = 5\,{\text{mm}}, \, \beta_{1} = \beta_{2} = \beta_{3} = 1. \\ & \quad 6 \le d \le 24 , { }8 \le N \le 100, \, 50 \le R_{B} \le 100,40 \le M \le 100. \\ \end{aligned}$$

See Table 15.

Table 15 Discrete values for bolts

• 2. Car side impact design

The problem can be mathematically formulated as follows:Objective function: $f\left( x \right) = 1.98 + 4.90x_{1} + 6.67x_{2} + 6.98x_{3} + 4.01x_{4} + 1.78x_{5} + 2.73x_{7}$

Subject to:

$$g_{1} (x) = 1.16 - 0.3717x_{2} x_{4} - 0.00931x_{2} x_{10} - 0.484x_{3} x_{9} + 0.01343x_{6} x_{10} \le 1$$

$$\begin{aligned} g_{2} (x) = 0.261 - 0.0159x_{1} x_{2} - 0.188x_{1} x_{8} - 0.019x_{2} x_{7} + 0.0144x_{3} x_{5} + 0.0008757x_{5} x_{10} + 0.080405x_{6} x_{9} \hfill \\ \, +\, 0.00139x_{8} x_{11} + 0.00001575x_{10} x_{11} \le 0.32 \hfill \\ \end{aligned}$$

$$\begin{aligned} g_{3} (x) & = 0.214 + 0.00817x_{5} - 0.131x_{1} x_{8} - 0.0704x_{1} x_{9} + 0.03099x_{2} x_{6} - 0.018x_{2} x_{7} + 0.0208x_{3} x_{8} \\ & \quad + \,0.121x_{3} x_{9} - 0.00364x_{5} x_{6} + 0.0007715x_{5} x_{10} - 0.0005354x_{6} x_{10} + 0.00121x_{8} x_{11} \le 0.32 \\ \end{aligned}$$

$$g_{4} (x) = 0.074 - 0.061x_{2} - 0.163x_{3} x_{8} + 0.001232x_{3} x_{10} - 0.166x_{7} x_{9} + 0.227x_{2 \, }^{2} \le 0.32$$

$$g_{5} (x) = 28.98 + 3.818x_{3} - 4.2x_{1} x_{2} + 0.0207x_{5} x_{10} + 6.63x_{6} x_{9} - 7.7x_{7} x_{8} + 0.32x_{9} x_{10} \le 32$$

$$g_{6} (x) = 33.86 + 2.95x_{3} + 0.1792x_{10} - 5.057x_{1} x_{2} - 11.0x_{2} x_{8} - 0.0215x_{5} x_{10} - 9.98x_{7} x_{8} + 22.0x_{8} x_{9} \le 32$$

$g_{7} (x) = 46.36 - 9.9x_{2} - 12.9x_{1} x{}_{8} + 0.1107x_{3} x_{10} \le 32$

$$g_{8} (x) = 4.72 - 0.5x_{4} - 0.19x_{2} x_{3} - 0.0122x_{4} x_{10} + 0.009325x_{6} x_{10} + 0.000191x_{11}^{2} \le 4$$

$$g_{9} (x) = 10.58 - 0.674x_{1} x_{2} - 1.95x_{2} x_{8} + 0.02054x_{3} x_{10} - 0.0198x_{4} x_{10} + 0.028x_{6} x_{10} \le 9.9$$

$$g_{10} (x) = 16.45 - 0.489x_{3} x_{7} - 0.843x_{5} x_{6} + 0.0432x_{9} x_{10} - 0.0556x_{9} x_{11} - 0.000786x_{11 \, }^{2} \le 15.7$$

where $0.5 \le x_{1} - x_{7} \le 1.5,x_{8} ,x_{9} \in (0.192,0.345)$ and $- 30 \le x_{10} ,x_{11} \le 30.$

• 3. Rolling element bearing

The problem can be mathematically formulated as follows:

Objective function:

$$f(x) = \left\{ \begin{aligned} C_{d}& = f_{c} Z^{{\frac{2}{3}}} D_{{^{b} }}^{1.8} {\text{ if }}D_{b} \le 25.4\,{\text{mm}} \hfill \\ C_{d} & = 3.647f_{c} Z^{{\frac{2}{3}}} D_{{^{b} }}^{1.4} {\text{ if }}D_{b} > 25.4\,{\text{mm}} \hfill \\ \end{aligned} \right.$$

Subject to:

$$\begin{aligned} g_{1} (x) & = \frac{{\phi_{0} }}{{2\sin^{ - 1} \left( {\frac{{D_{b} }}{{D_{m} }}} \right)}} - Z + 1 \ge 0, \, g_{2} (x) = 2D_{b} - K_{D\hbox{min} } \left( {D - d} \right) \ge 0 \\ g_{3} (x) & = K_{D\hbox{max} } \left( {D - d} \right) - 2D_{b} \ge 0, \, g_{4} (x) = \zeta B_{\omega } - D_{b} \ge 0 \\ g_{5} (x) & = D_{m} - 0.5\left( {D + d} \right) \ge 0, \, g_{6} (x) = \left( {0.5 + e} \right)\left( {D + d} \right) - D_{m} \ge 0 \\ g_{7} (x) & = 0.5\left( {D - D_{m} - D_{b} } \right) - \varepsilon D_{b} \ge 0, \, g_{8} (x) = f_{i} - 0.515 \ge 0 \\ \end{aligned}$$

$$g_{9} (x) = f_{o} - 0.515 \ge 0$$

where

$$f_{c} = 37.91\left[ {1 + \left\{ {1.04\left( {\frac{1 - \gamma }{1 + \gamma }} \right)^{1.72} \left( {\frac{{f_{i} \left( {2f_{o} - 1} \right)}}{{f_{o} \left( {2f_{i} - 1} \right)}}} \right)^{0.41} } \right\}^{10/3} } \right]^{ - 0.3} , \, \gamma = \frac{{D_{b} \cos \alpha }}{{D_{m} }}, \, f_{i} = \frac{{r_{i} }}{{D_{b} }}, \, f_{o} = \frac{{r_{o} }}{{D_{b} }}$$

$$\phi_{0} = 2\pi - 2\cos^{ - 1} \left( {\frac{{\left\{ {\left( {D - d} \right)/2 - 3\left( {T/4} \right)} \right\}^{2} + \left\{ {D/2 - T/4 - D_{b} } \right\}^{2} - \left\{ {d/2 - T/4} \right\}^{2} }}{{2\left\{ {\left( {D - d} \right)/2 - 3\left( {T/4} \right)} \right\}\left\{ {D/2 - T/4 - D_{b} } \right\}}}} \right)$$

where $T = D - d - 2D_{b} , \, D = 160, \, d = 90, \, \beta_{\omega } = 30,0.5\left( {D + d} \right) \le D_{m} \le 0.6\left( {D + d} \right),0.15\left( {D - d} \right) \le D_{b} \le 0.45\left( {D - d} \right),$$4 \le Z \le 50,0.515 \le f_{i} \le 0.6,0.515 \le f_{o} \le 0.6,0.6 \le K_{D\hbox{max} } \le 0.7,0.3 \le \varepsilon \le 0.4,0.02 \le e \le 0.1,0.6 \le f_{i} \le 0.85.$

• 4. Step-cone pulley

The problem can be mathematically formulated as follows:

Objective function:

$$f(x) = \rho \omega \left[ {d_{1}^{2} \left\{ {1 + \left( {\frac{{N_{1} }}{N}} \right)^{2} } \right\} + d_{2}^{2} \left\{ {1 + \left( {\frac{{N_{2} }}{N}} \right)^{2} } \right\} + \,d_{3}^{2} \left\{ {1 + \left( {\frac{{N_{3} }}{N}} \right)^{2} } \right\} +\, d_{4}^{2} \left\{ {1 + \left( {\frac{{N_{4} }}{N}} \right)^{2} } \right\}} \right]$$

Subject to:

$$\begin{aligned} h_{1} (x) & = c_{1} - c_{2} = 0,\quad \quad h_{2} (x) & = c_{1} - c_{3} = 0 \\ h_{3} (x) & = c_{1} - c_{4} = 0,\quad \quad \, g_{1,2,3,4} (x) & = R_{i} \ge 2 \\ g_{5,6,7,8} (x) & = P_{i} \ge \left( {0.75 \times 745.6998} \right) \, \\ \end{aligned}$$

where:

• $C_{i}$ indicates the length of the belt to obtain speed $N_{i}$ and is given by

  $$C_{i} = \frac{{\pi d_{i} }}{2}\left( {1 + \frac{{N_{i} }}{N}} \right) + \frac{{\left( {\frac{{N_{i} }}{N} - 1} \right)^{2} }}{4a} + 2a,\quad \, i = 1, \ldots ,4$$

• $R_{i}$ is the tension ratio and is given by

  $$R_{i} = \exp \left[ {\mu \left\{ {\pi - 2\sin^{ - 1} \left\{ {\left( {\frac{{N_{i} }}{N} - 1} \right)\frac{{d_{i} }}{2a}} \right\}} \right\}} \right], \, i = 1, \ldots ,4$$

• $P_{i}$ is the power transmitted at each step

  $$R_{i} = st\omega \left[ {1 - \exp \left[ { - \mu \left\{ {\pi - 2\sin^{ - 1} \left\{ {\left( {\frac{{N_{i} }}{N} - 1} \right)\frac{{d_{i} }}{2a}} \right\}} \right\}} \right]} \right]\frac{{\pi d_{i} N_{i} }}{60}, \, i = 1, \ldots ,4$$
  $$\rho_{1} = 7200\,{\text{kg/m}}^{3} , \, a = 3\,{\text{m}}, \, \mu = 0.35, \, s = 1.75\,{\text{MPa}}, \, t = 8\,{\text{mm}},40 \le d_{i} \le 100,\,16 \le \omega \le 100.$$

• 5. Belleville spring

The problem can be mathematically formulated as follows:

Objective function:$f(x) = 0.07075\pi \left( {D_{e}^{2} - D_{i}^{2} } \right)t$

Subject to:

$$\begin{aligned} g_{1} (x) & = S - \frac{{4E\delta_{\hbox{max} } }}{{\left( {1 - \mu^{2} } \right)\alpha D_{e}^{2} }}\left[ {\beta \left( {h - \frac{{\delta_{\hbox{max} } }}{2}} \right) + \gamma t} \right] \ge 0 \\ g_{2} (x) & = \left( {\frac{4E\delta }{{\left( {1 - \mu^{2} } \right)\alpha D_{e}^{2} }}\left[ {\left( {h - \frac{\delta }{2}} \right)\left( {h - \delta } \right)t + t^{3} } \right]} \right) - P_{\hbox{max} } \ge 0 \\ \end{aligned}$$

$$\begin{aligned} g_{3} (x) & = \delta_{l} - \delta_{\hbox{max} } \ge 0,\quad \quad \, g_{4} (x) = H - h - t \ge 0 \\ g_{5} (x) & = D_{\hbox{max} } - D_{e} - \ge 0,\quad \quad \, g_{6} (x) & = D_{e} - D_{i} \ge 0 \\ g_{7} (x) & = 0.3 - \left( {\frac{h}{{D_{e} - D_{i} }}} \right) \ge 0 \\ \end{aligned}$$

where $\alpha = \left( {\frac{6}{\pi \ln \left( K \right)}} \right)\left( {\frac{K - 1}{K}} \right)^{2} ,\beta = \left( {\frac{6}{\pi \ln \left( K \right)}} \right)\left( {\frac{K - 1}{K} - 1} \right),\gamma = \left( {\frac{6}{\pi \ln \left( K \right)}} \right)\left( {\frac{K - 1}{2}} \right)$$P = \frac{{\log_{10} \log_{10} \left( {8.122{\text{e}}6\mu + 0.8} \right) - C_{1} }}{n},h = \left( {\frac{2\pi N}{60}} \right)^{2} \left( {\frac{2\pi \mu }{{E_{f} }}} \right)\left( {\frac{{R^{4} }}{4} - \frac{{R_{o}^{4} }}{4}} \right),P_{\hbox{max} } = 1000\,{\text{lb}},\;\delta_{\hbox{max} } = 0.2\,{\text{in}} .,\;S = 200\,{\text{KPsi}},$$E = 30e6\,psi,\;\mu = 0.3,\;H = 2\,{\text{in}} .,\;D_{\hbox{max} } = 12.01\,{\text{in}} .,\;K = \frac{{D_{e} }}{{D_{i} }},\;\delta \left( l \right) = f\left( a \right)a,\;a = \frac{h}{t}.$

$$5 \le D_{e} \le 15,5 \le D_{i} \le 15,0.2 \le t \le 0.25,0.2 \le h \le 0.25.$$

See Table 16.

Table 16 Variation of $f\left( a \right)$ with a

• 6. Speed Reducer

The problem can be mathematically formulated as follows:Objective function: $f\left( x \right) = 0.7854bm^{2} \left( {3.3333z^{2} + 14.9334z - 43.0934} \right) - 1.508b\left( {d_{1}^{2} + d_{2}^{2} } \right) + 7.4777\left( {d_{1}^{3} + d_{2}^{3} } \right) + 0.7854\left( {l_{1} d_{1}^{2} + l_{2} d_{2}^{2} } \right)$

Subject to:

$$\begin{aligned} g_{1} (x) & = \frac{27}{{b \, m^{2} z}} - 1 \le 0,\quad \quad g_{2} (x) = \frac{397.5}{{bm^{2} z^{2} }} - 1 \le 0 \\ g_{3} (x) & = \frac{{1.93l_{1}^{3} }}{{mzd_{1}^{4} }} - 1 \le 0,\quad \quad g_{4} (x) = \frac{{1.93l_{2}^{3} }}{{mzd_{2}^{4} }} - 1 \le 0 \\ g_{5} (x) & = \frac{{\sqrt {\left( {\frac{{745l_{1} }}{mz}} \right)^{2} + 16.9 \times 10^{6} } }}{{\left( {110d_{1}^{3} } \right)}} - 1 \le 0,\quad \quad g_{6} (x) = \frac{{\sqrt {\left( {\frac{{745l_{2} }}{mz}} \right)^{2} + 157.5 \times 10^{6} } }}{{\left( {85d_{2}^{3} } \right)}} - 1 \le 0 \\ g_{7} (x) & = \frac{mz}{40} - 1\le 0,\quad \quad g_{8} (x) = \frac{5m}{b} - 1 \le 0 \\ g_{9} (x) & = \frac{b}{ 1 2m} - 1\le 0,\quad \quad g_{10} (x) = \frac{{1.5d_{1} + 1.9}}{{l_{1} }} - 1 \le 0 \\ g_{11} (x) & = \frac{{1.1d_{2} + 1.9}}{{l_{2} }} - 1 \le 0 \\ \end{aligned}$$

where $2.6 \le b \le 3.6,0.7 \le m \le 0.8,17 \le z \le 28,7.3 \le l_{1} \le 8.3,7.3 \le l_{2} \le 8.3,2.9 \le d_{1} \le 3.9$ and $5 \le d_{2} \le 5.5.$