Handbook of Civil Engineering Calculations

FIGURE 17

^F 7 P 1 +P 2 + P 3 - 20,000 = O, or P 1 +P 2 + P 3 = 20,000, Eq. a; also, ^Mc = 16P 1 + 1OP 2

• 20,000(12) = O, or 16P 1 + 1OP 2 = 240,000, Eq. b.

2. Establish the relations between the deformations
   Selecting an arbitrary center of rotation O 9 show the bar in its deflected position (Fig.
   lie). Establish the relationships among the three deformations. Thus, by similar triangles,
   (AZ 1 - A/ 2 )/(A/ 2 - A/ 3 ) = 6/10, or 1OA/! - 16A/ 2 + 6A/ 3 = O, Eq. c.


3. Transform the deformation equation to an axial-force equation
   By substituting axial-force relations in Eq. c, the following equation is obtained:
   10Pi(5)/(1.25£) - 16P 2 (9)/(1.20E) + 6P 3 (I.S)IE = O, or 40Pi - 12OP 2 + 45P 3 = O, Eq. c'.


4. Solve the simultaneous equations developed
   Solve the simultaneous equations a, b, and c' to obtain PI = 11,810 Ib (52,530 N); P 2 =
   5100 Ib (22,684 N); P 3 = 3090 Ib (13,744 N).
   5. Locate the center of rotation
   To locate the center of rotation, compute the relative deformation of rods 1 and 2. Thus
   A/! = 11,810(5)/(1.25£) = 47,240/£; A/ 2 = 5100(9)7(1.2OF) = 38,250/£.'
   In Fig. 17c, by similar triangles, x/(x - 6) = A/!/A/ 2 = 1.235; x = 31.5 ft (9.6 m).
   6. Verify the computed values of the tensile forces
   Calculate the moment with respect to A of the applied and resisting forces. Thus MAa =
   20,000(4) - 80,000 lb-ft (108,400 N-m); MAr = 5100(6) + 3090(16) = 80,000 lb-ft
   (108,400 N-m). Since the moments are equal, the results are verified.

ANALYSIS OF CABLE SUPPORTING

A CONCENTRATED LOAD

A cold-drawn steel wire % in (6.35 mm) in diameter is stretched tightly between two

points lying on the same horizontal plane 80 ft (24.4 m) apart. The stress in the wire is

Deflected position

•Initial position

Assumed center

of rotation