Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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Engineering 25

Tutorial: CatenaryCables

Bruce Mayer, PE

Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

- The cable has a mass of 0.5 kg/m and is 25 m long.
- Determine the vertical and horizontal components of force it exerts on the top of the tower.

% Bruce Mayer, PE

% ENGR36 * 22Jul2

% ENGR36_Tutorial_Partial_Catenary_H13e_P7_119_1207.m

%

K1 = tand(30)

K2 = asinh(K1)

S0 = @(z) (z/K2)*(sinh(K2*(z+15)/z)-K1) - 25

xB = fzero(S0, 10)

xA = xB+15

u = 0.5*9.81

TO = u*xB/K2

TA = TO*cosh(u*xA/TO)

QA = atand(sinh(u*xA/TO))

WA = TO*tand(QA)

xB =

8.2804

xA =

23.2804

u =

4.9050

TO =

73.9396

TA =

181.0961

QA =

65.9026

WA =

165.3141

- The man picks up the 52-ft chain and holds it just high enough so it is completely off the ground. The chain has points of attachment A and B that are 50 ft apart. If the chain has a weight of 3 lb/ft, and the man weighs 150 lb, determine the force he exerts on the ground. Also, how high h must he lift the chain? Hint: The slopes at A and B are zero.

% Bruce Mayer, PE

% ENGR36 * 22Jul2

% ENGR36_Tutorial_Chain_Lift_Catenary_H13e_P7_124_1207.m

%

Zf1 = @(q) (q/3)*sinh(75/q) - 26

TO = fzero(Zf1, 150)

h = (TO/3)*(cosh(75/TO) - 1)

Q = atand(78/TO)

Th = 3*h + TO

Tup = Th*sind(Q)

Zf1 = @(q)(q/3)*sinh(75/q)-26

TO =

154.0033

h =

6.2088

Q =

26.8614

Th =

172.6297

Tup =

78