Page 34 - Handbook of Civil Engineering Calculations, Second Edition

STATICS, STRESS AND STRAIN, AND FLEXURAL ANALYSIS 1.17

3. Compute the length of the cable
2
When d/L is 1/20 or less, the cable length can be approximated from S L 8d /(3L),
2
where S cable length, ft (m). Thus, S 960 8(50) /[3(960)] 966.94 ft (294.72 m).

CATENARY CABLE SAG AND DISTANCE
BETWEEN SUPPORTS

A cable 500 ft (152.4 m) long and weighing 3 pounds per linear foot (lb/lin ft) (43.8
N/m) is supported at two points lying in the same horizontal plane. If the tension at the
supports is 1800 lb (8006 N), find the sag of the cable and the distance between the
supports.

Calculation Procedure:
1. Compute the catenary parameter
A cable of uniform cross section carrying only its own weight assumes the form of a cate-
nary. Using the notation of the previous procedure, we find the catenary parameter c from
2 0.5
2 0.5
2
d c T/w 1800/3 600 ft (182.9 m). Then c [(d c) (S/2) ] [(600) ]
2 0.5
(250) ] 545.4 ft (166.2 m).
2. Compute the cable sag
Since d c 600 ft (182.9 m) and c 545.4 ft (166.2 m), we know d 600 545.4
54.6 ft (l6.6 m).
3. Compute the span length
Use the relation L 2c ln (d c 0.5S)/c, or L 2(545.5) ln (600 250) 545.4
484.3 ft (147.6 m).

STABILITY OF A RETAINING WALL

Determine the factor of safety (FS) against sliding and overturning of the concrete retaining
3
3
wall in Fig. 10. The concrete weighs 150 lb/ft (23.56 kN/m , the earth weighs 100 lb/ft 3
3
(15.71 kN/m ), the coefficient of friction is 0.6, and the coefficient of active earth pressure
is 0.333.
Calculation Procedure:
1. Compute the vertical loads on the wall
Select a 1-ft (304.8-mm) length of wall as typical of the entire structure. The horizontal
pressure of the confined soil varies linearly with the depth and is represented by the trian-
gle BGF in Fig. 10
Resolve the wall into the elements AECD and AEB; pass the vertical plane BF through
the soil. Calculate the vertical loads, and locate their resultants with respect to the toe C.
Thus W 1 15(1)(150) 2250 lb (10,008 N); W 2 0.5(15)(5)(150) 5625; W 3
0.5(15)(5)(100) 3750. Then W 11,625 lb (51,708 N). Also, x 1 0.5 ft; x 2 1
0333(5) 2.67 ft (0.81 m); x 3 1 0.667(5) 433 ft (1.32 m).