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Constrained case: the welded beam design problem. HKA was compared to other<br />

metaheuristics specifically designed for solving constrained problems. In these experiments,<br />

HKA handles constraints via a new objective function which includes penalty functions (see<br />

relation (2)).<br />

A welded beam is designed for minimum cost subject to constraints on shear stress τ (q),<br />

bending stress in the beam σ (q), buckling load on the bar Pc, end deflection of the beam δ (q), and<br />

side constraints (Rao, 1996). There are four design variables as shown in Figure 10: h (denoted<br />

q1), l (denoted q2), t (denoted q3) and b (denoted q4), q = [q1, q2, q3, q4] T .<br />

Figure 10 - Welded beam design problem.<br />

The problem can be mathematically formulated as follows:<br />

Minimize 2<br />

J ( q)<br />

= ( 1+<br />

c1)<br />

q1<br />

q2<br />

+ c2q3q4<br />

( L + q2)<br />

Subject to: g ( q)<br />

= τ ( q)<br />

−τ<br />

≤ 0<br />

Where:<br />

τ ( q)<br />

=<br />

6PL<br />

σ ( q)<br />

= , 2<br />

q q<br />

1<br />

g ( q)<br />

= σ ( q)<br />

−σ<br />

2<br />

g ( q)<br />

= q − q<br />

3<br />

4<br />

5<br />

6<br />

7<br />

1<br />

1<br />

1<br />

4<br />

≤ 0<br />

2<br />

1<br />

3<br />

≤ 0<br />

g ( q)<br />

= c q + c q q ( L + q ) − 5 ≤ 0<br />

g ( q)<br />

= h<br />

min<br />

max<br />

4<br />

− q ≤ 0<br />

g ( q)<br />

= δ ( q)<br />

−δ<br />

max<br />

max<br />

≤ 0<br />

g ( q)<br />

= P − Pc(<br />

q)<br />

≤ 0<br />

2 q2<br />

2<br />

τ1<br />

+ 2τ<br />

1τ<br />

2 + τ 2 ,<br />

2R<br />

⎛ q2<br />

⎞<br />

M = P⎜<br />

L + ⎟,<br />

⎝ 2 ⎠<br />

4<br />

3<br />

R =<br />

3<br />

4PL<br />

δ ( q)<br />

= , 2<br />

Eq q<br />

τ =<br />

P<br />

2<br />

q2<br />

⎛ q1<br />

+ q3<br />

⎞<br />

+ ⎜ ⎟ ,<br />

4 ⎝ 2 ⎠<br />

4<br />

3<br />

h<br />

1<br />

t<br />

2<br />

2q<br />

q<br />

1<br />

2<br />

l<br />

2<br />

,<br />

MR<br />

τ 2 =<br />

I<br />

⎪⎧<br />

I = 2⎨<br />

⎪⎩<br />

2<br />

4.<br />

013E<br />

q3<br />

q<br />

Pc(<br />

q)<br />

=<br />

2<br />

L<br />

⎡ 2<br />

2<br />

q<br />

⎤⎪⎫<br />

2 ⎛ q1<br />

+ q3<br />

⎞<br />

2q1q2<br />

⎢ + ⎜ ⎟ ⎥⎬<br />

⎢⎣<br />

12 ⎝ 2 ⎠ ⎥⎦<br />

⎪⎭<br />

6<br />

4<br />

L<br />

/ 36<br />

P<br />

b<br />

⎛ q<br />

⎜ 3<br />

⎜<br />

1−<br />

⎝ 2L<br />

E<br />

4G<br />

⎞<br />

⎟<br />

⎠<br />

(29)<br />

(30)

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