Exhaustive Search Optimization: Beam Example using MathCAD - Prof. David Alciatore, Study notes of Mechanical Engineering

Download Exhaustive Search Optimization: Beam Example using MathCAD - Prof. David Alciatore and more Study notes Mechanical Engineering in PDF only on Docsity! Exhaustive Search Optimization Example Beam example: F 500:= L 10:= P 2:= σ h( ) 6 F⋅ L⋅ P 2 h−    h2⋅ := h 0.5 0.505, 0.8..:= 0.5 0.6 0.7 0.8 2 10 5 × 2.1 10 5 × 2.2 10 5 × 2.3 10 5 × 2.4 10 5 × σ h( ) h Using range variables: Nh 1000:= i 0 Nh..:= hmax P 2 0.001−:= hmin 0.001:= ∆h hmax hmin−( ) Nh := hi hmin i ∆h⋅+:= h 0 0 1 2 3 4 5 6 -31·10 -31.998·10 -32.996·10 -33.994·10 -34.992·10 -35.99·10 ... = σhi σ hi( ):= σhopt min σh( ):= σhopt 2.025 10 5 ×= hopt lookup σhopt σh, h, ( ):= hopt 0.667( )= σh 0 655 656 657 658 659 660 661 662 663 664 665 52.027·10 52.027·10 52.026·10 52.026·10 52.026·10 52.026·10 52.025·10 52.025·10 52.025·10 52.025·10 ... = Using MathCAD program: hopt σopt( ) hopt 0← σopt 10 100 ← pause "about to start loop"( ) σh σ h( )← trace "h={0} sopt={1} sh={2}" h, σopt, σh, ( ) hopt h← σopt σh← trace " new optimum found: h={0}" h, ( ) σh σopt<if h hmin hmin ∆h+( ), hmax..∈for hopt σopt( ) := large positive number necessary to initialize the minimization process (A large negative number would be used for a maximum problem) hopt 0.667= σopt 2.025 10 5 ×= wopt 1 2 P 2 hopt⋅−( ):= hopt wopt 2=