Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

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.

Typology: Slides

2012/2013

1 / 13

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