## Tutorial: Catenary Cables

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

Engineering-36: Engineering Mechanics - Statics

1 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### Partial Catenary

 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.

Engineering-36: Engineering Mechanics - Statics

2 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### Eqns Used

dy dx

 sinh

µx T

0

dy dx x

x p

 sinh

µx p T

0

S

T

0

µ

  sinh

µx Q

 sinh

T

0

µx P T

0  

T p

T

0 cosh  

µx p T

0   tan 

p

 sinh

µx p T

0

Engineering-36: Engineering Mechanics - Statics

3 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

4 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

5 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

6 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

7 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### MATLAB Code

% Bruce Mayer, PE % ENGR36 * 22Jul2 % ENGR36_Tutorial_Partial_Catenary_H 13e_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) Engineering-36: Engineering Mechanics - Statics

8 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### MATLAB Results

xB = 8.2804

xA = u = 23.2804

4.9050

TO = 73.9396

TA = 181.0961

QA = 65.9026

WA = 165.3141

Engineering-36: Engineering Mechanics - Statics

9 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### Chain Lift Problem

 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

.

Engineering-36: Engineering Mechanics - Statics

10 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### Eqns Used

S

T

0

µ

  sinh

µx Q

 sinh

T

0

µx P T

0  

y

T

0    cosh 

x T

0  1  

T p

T

0 cosh  

µx p T

0   tan 

p

 sinh

µx p T

0

Engineering-36: Engineering Mechanics - Statics

11 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

12 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

13 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

Engineering-36: Engineering Mechanics - Statics

14 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### MATLAB Code

% Bruce Mayer, PE % ENGR36 * 22Jul2 % ENGR36_Tutorial_Chain_Lift_Catenar y_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) Engineering-36: Engineering Mechanics - Statics

15 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx

### MATLAB Results

Zf1 = @(q)(q/3)*sinh(75/q)-26 TO = 154.0033

h = 6.2088

Q = 26.8614

Th = 172.6297

Tup = 78 Engineering-36: Engineering Mechanics - Statics

16 Bruce Mayer, PE ENGR36_H13_Tutorial_Catenary_Cables.pptx