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Solution to Problem 4.35: Minimum LAC Value for a 2-Cable Supported Load, Slides of Computational Methods

The solution to problem 4.35 in engr225, which involves finding the minimum value of lac (lateral offset) for a 2-cable supported load while ensuring that both tab (tension in the first cable) and tac (tension in the second cable) do not exceed 2000 lbs. The solution is presented through a matlab code snippet, which plots the tension in tab and tac against lac and identifies the minimum value of lac and the corresponding tensions.

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2012/2013

Uploaded on 04/30/2013

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Download Solution to Problem 4.35: Minimum LAC Value for a 2-Cable Supported Load and more Slides Computational Methods in PDF only on Docsity! Chp4 Tutorial: Prob 4.35 Solution Docsity.com Problem 4.35 • The Situation – Cable Supported Weight = 6ft 3ft = = 2000 lbs = 2 kip  Find Minimum Value for LAC subject to Constraint • TAB & TAC Both ≤2000 lbs Docsity.com When in Doubt → Plot • Zoom-In on Answer • Locate LACmin more Precisely 3.5 3.6 3.7 3.8 3.9 4 4.1 4.2 4.3 4.4 4.5 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 LAC (ft) Te ns io n: T A B & T A C (K ip ) P4.35 - 2-Cable Supported Load TAB TAC Max-T TAB Limits ≈4.025 ft Docsity.com ZO Ce wt is 2S Bruce Mayer, PE { 2 paert | a a Engineering instiuctor Chabot College 25665 Hesperian Blvd Hayward, CA 94545 eMail: [email protected] CABLE SUOPPORTERP WIG GFT pees A c BY EvVQ2~26 METHOPS PKRAW FREE te Trea ® _ BOPY PIAGRAN FOR STATIC EGUILIBRIOM NEEP ZFxEa ¢ ZhH=O MOUSsT USE PROPER. S/GNS Docsity com fs | ENGRES | 2 Z Fp =O = -~Tra coe © + The coed =a Fy SF OD> Tag dn©® +7Tac wed ~w HAVE TWO GOONS Ww 2 OrtKos<eS Tre CRO + 7TraccRP =0 TaadnS + TAcdnd =v OR eo MAFRIC FORA “ce © arg wee — [ ° | US aon ln pe wv A a TT = P KO Tre LAW! OF Sores ® val Tt Pf = @ = & a ~ P np a tiny AND FIRE LAD OF COSfivelt f= £407 ZRO ce pe (® THIS CADE r~-s PP Lae e-Laa = 2 EP OG P=GFr rod ®@ SOLVE FaXtr~S Docsity| com Mm et e@e2s |S S) OS& poe-hoor 4 Cameron” BuceeT Fort save, aie ) The Chee, mew) 3 Tw (Lac, acre) Line = AISPACE (hig, Luz, $00) Tae + TAw = S008 5 We 2000 Fok K= /, S00 crrce © chica FIND YEecrot T BY Bkek DIV TEST Cov srTRAre, /F TOI) & 20cm & Tl2)<ee0e rhe Tee Forel CHECK 1F Lae. 1S 2685 THaat LET VALUE, JF SO COLAECT hac mv, Tre , Tae THEN 7 kre Ce) < “Aegan hac, mine HhypelK) Tag = TOS TAe = 7C2) END EMP cNe @).., “hae pean » Tp) Tre Docsity. om The Command Session >> run Prob4_35_Loaded_Cables Minimum value for LAC in ft = 4.0260 TAB at LACminimum in kip => 1.9993 TAC at LACminimum in kip => 1.7878 Support Angle at LACminimum in ° => 36.8932 index of LACmin => 139 Docsity.com % Bruce Mayer, PE % ENGR225 * 20Feb12 % Prob 4.35 * file Prob4_35_Loaded_Cables_1202.m % % Plot L_AC for 0° to 90° using 500pts by LinSpace LAC0 = 3; LAC90 = 6.71; % length in Feet; n = 500; % number of plotted points LAC = linspace(LAC0, LAC90, n); % in ft % % the Fixed Parameters LAB = 3; % in feet D = 6; % in feet W = 2; % wt in kip (kilo-lbs) % % Calc angle vectors theta = acos((D^2 + LAB^2 - LAC.^2)/(2*D*LAB)); phi = asin(LAB.*sin(theta)./LAC); % % Calc TAB & TAC plotting Vectors for k = 1: length(LAC) % Define for THIS VALUE of k the Matrix Mult terms A = [-cos(theta(k)), cos(phi(k)); sin(theta(k)), sin(phi(k))]; P = [0; W]; % For THIS VALUE of k Solve for 2-unknowns in 2-eqns using BackDivision T = A\P; % Assign temporary tension solns to permanent vectors TAB(k) = T(1); TAC(k) = T(2); end % % Plot the tension % make Horizontal line vectors at 2000 lbs (2 Kip) Xln = [LAC(1), LAC(length(LAC))]; Yln = [2, 2]; % in Kip plot(LAC, TAB, 'g--', LAC, TAC,'b:', Xln, Yln, 'r-', 'LineWidth', 3),... xlabel('LAC (ft)'), ylabel('Tension: TAB & TAC (Kip)'),... title('P4.35 - 2-Cable Supported Load '), grid,... legend('TAB', 'TAC', 'Max-T'), axis([3 6.7 0 3]) % axis([3 6.7 0 3]) ** axis([3.5 4.5 1.5 2.2]) % % Find LACmin and Corresponding Tensions % use FOR Loop to test every Case, Collect Mins in Bucket % that depends on LAC % Initialize colletion buckets LACmin = 7; % ft TABmin = 3; % kip TACmin = 3; % kip Mmin = n + 9; % make larger than the no. items in LAC-vector for m = 1:length(LAC) Test_No = m; if (TAB(m)<=2)&(TAC(m)<=2) % OK to here; now test if current LAC < LACmin % m_ok = m if LAC(m) < LACmin % if OK here => replace OLD LACmin with current LAC LACmin = LAC(m); TABmin = TAB(m); TACmin = TAC(m); Mmin = m; end end Docsity.com