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Mechanical Engineering Problem Solving

by: Lionel Hansen

Mechanical Engineering Problem Solving MECH 102

Marketplace > Colorado State University > Mechanical Engineering > MECH 102 > Mechanical Engineering Problem Solving
Lionel Hansen
GPA 3.87

David Alciatore

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David Alciatore
Class Notes
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This 12 page Class Notes was uploaded by Lionel Hansen on Tuesday September 22, 2015. The Class Notes belongs to MECH 102 at Colorado State University taught by David Alciatore in Fall. Since its upload, it has received 59 views. For similar materials see /class/210252/mech-102-colorado-state-university in Mechanical Engineering at Colorado State University.

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Date Created: 09/22/15
Optimization Example find the maximum strength minimum stress rectangular h by w fixedlength L fixedperimeter P cantilevered beam supporting an end load F The highest stress in the beam is 6FL 3 hh2 2 J 0h The minimum should occur when the derivative wrt h is 0 6FL F solving for h substituting perimeter equation and solving for h 1 h 2h 2w h 2w Using MathCAD39s Symbolic commands 6FL Gal112 0h P d oh o solveh db 3 1 h 2h 2w solveh gt 2w Temperature Distribution FiniteDifference BoundaryValue Example Bar and heat transfer info A1 10 length TA 4o temperature at left end k o001 conductionconvection constant TB 200 temperature at right end 11 20 number of intervals Ta 20 ambient temperature ODE 2 d ZT kTai T 0 dx BC39sr W TA TltLgt TB Theoretical solution using methods of Differential Eguations 2 d ZT 7 kT ikTa dx homogeneous solution 52 7 k 0 s JEjE Thx Cl39e x f C2e7 kix particular solution Tpx Ta general solution Tx Thx Tpx Cle liiX C2e7 ix Ta applying the 80s T0TAC1 C2 Ta TL T3 Cle 39L Cze 39L Ta using algebra TB 7 Ta7 TA 7 TaequotTlt39L C 7 1 e L 7 ei lTCL C1 257246 c2 7 TA 7 Ta7 c1 c2 7 7237246 using MathCAD Given TA C1 C2 Ta T3 Cle 39L Cze 39L Ta C 25724642704932558079 quotCm 7 FindC1C2 7 N 723724642704932558079 true analytical solution Ttx Cle IE39X C2e7 ix Ta MathCAD numerical solution Given d2 2Tx kTa 7 Tx 0 dx T0 TA TL TB TM OdesolveXL ODE finite difference form L AX interval Size H 2 2 0c 2 kAX 3 kAX Ta 7T17109T17Ti1i3 i1n71 Matrix form of finite difference equations AT B i0n72 j0n72 W1n72 0 A a A 71 A 71 J 11 marl m71m B03TA Bm3 Bn723TB 200 Linearized Regression Example fitting a theoretical function to experimental stressstrain data Provided exgerimental data 6 2250 3575 4250 4350 4250T 6 6 6 6 T 25001o 1000107 1500107 2000107 23751076 11 lengths n 5 i 01171 Theoretical nonlinearfunction to be fit with linearized regression Taking the natural log of both sides 1116 111a 7 1112 7 bs 1116 7 1112 111a 7 bs 111a 7 bs Y a0 alX 5 X Si Y 1 1 1 1 Si a1 slopeXY a1 7492333 a0 interceptXY a0 15577 a ae0 a5823gtlt106 b 7 7a1 b 7 492333 Comguting the standard error and correlation coefficient X s y cs fX aXei b39X 2X1 Zyi Xibar 1475 X 10 3 yibar yibar 3735 X 103 St 3139 gtlt106 Sr 8815 x103 syix 76468 r 0999 Plotting the results 0 A XmaX 7 Xmin X X maX X X mm maX Xf XminXmn AX quotXmax 5000 I I I I 4000 yi 3000 ooo fXD 2000 1000 0 I 0 5104 0001 00015 0002 00025 Xi Linear Regression Example Trending vehicle milespergallon versus vehicle weight Data grovided mpg 35 30 28 22 18 17 15T y mpg weight 2000 2200 2600 2900 3400 3800 4100T X weight 11 1engthmpg n 7 1 0n 7 1 Form of regression eguation ya0a1x Using regression eguations yibar yibar 23571 11 a1 a1 79188 X107 1 1 i 2 N2 392 Xi 2 xi 1 J a0 yibar 7 a1 X7bar a0 51137 Using MathCAD39s builtin functions a1 slopexy a1 79188 X 107 3 a0 interceptxy a0 51137 Com utin the standard error and the correlation coefficient St 101 7 yibar2 st 341714 1 Sr 2B 7 a0 a1 xi2 Sr 19199 1 7X 196 Sy r 0972 Plottinq the oriqinal data and the reqression line fx a0 alX I 2000 2500 3000 3500 4000 4500 Xi 10 Now we can estimate a vehicles mpg given its weight For example weight 3100 mpg fweight mpg 22653 Solution to Linear Algebraic Equations Matrix form A X B Truss exalee from class initialize all elements in the A matrix to 0 i05 j05 A0 M 000000 00 000 000000 A 0000 000000 000000 copy the matrix above fill in the nonzero elements and redefine A 030 cos 30deg 530 sin30deg 060 cos 60deg 560 sin60deg 7030 0 c60 0 0 0 0 7530 0 7560 0 0 0 1000 c30 1 0 1 0 0 0 A 13 s30 0 0 01 0 0 0 71 7060 0 0 0 0 0 0 560 0 0 1 0 coefficient matrix righthand side constants Matrix Inverse 7500 433013 1 7866025 X A B X 0 250 L 750 j check inverse results i 0 1 1gtlt103 0 AX 0 AA l 0 72842gtlt 10 14 0 0 0 0 j 0 example equation row 2 row 1 in 0based MathCADD 3 isSOXO 7 s60X2 1 gtlt 10 Cramer39s Rule to solve for the 2nd unknown x2 X1 in MathCAD replace the 2nd column with B in the numerator 7030 0 c60 0 0 0 7530 1000 7560 0 0 0 do 0 0 1 0 0 530 0 0 0 1 0 0 0 7060 0 0 0 0 0 560 0 0 1 lei X2 shortcut way to do this in MathCAD OOOOHO OOOHOO start with the coefficient matrix ie initialize the numerator matrix a A replace the 2nd column with the constant vector 70866 0 05 lt1gt 3 N B 705 1 X 10 70866 N 0866 0 0 05 0 0 0 0 705 0 0 0866 OOHOO 2 3 to solve for all 6 unknowns 7500 433013 7866025 for is 05 0 N e A l 250 N lt B 750 j X e m 1 IAI


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