Handbook of Civil Engineering Calculations

50,000 lb/in^2 (344,700 kPa). A load of 200

Ib (889.6 N) is suspended at the center of

the cable. Determine the sag of the cable

and the final stress in the cable. Verify that

the results obtained are compatible.

FIGURE 18
Calculation Procedure:

1. Derive the stress and strain
   relations for the cable
   With reference to Fig. 18,Z = distance between supports, ft (m); P = load applied at cen-
   ter of cable span, Ib (N); d = deflection of cable center, ft (m); e = strain of cable caused
   by P; S 1 and S 2 = initial and final tensile stress in cable, respectively, lb/in^2 (kPa).
   Refer to the geometry of the deflection diagram. Taking into account that dlL is ex-
   tremely small, derive the following approximations: S 2 = PL/(4Ad), Eq. a; e = 2(d/L)^2 , Eq.
   b.


2. Relate stress and strain
   Express the increase in stress caused by P in terms of e, and apply the above two equa-
   tions to derive 2E(dlLf + s^dlL) = P/(4A), Eq. c.


3. Compute the deflection at the center of the cable
   Using Eq. c, we get 2(30)(10)^6 (d/I)^3 + 50,000^/1 = 200/[4(0.049)], so dlL = 0.0157 and /.
   d= 0.0157(80) = 1.256 ft (0.382 m).


4. Compute the final tensile stress
   Write Eq. a as S 2 = [P/(4A)]/(d/L) = 1020/0.0157 = 65,000 lb/in^2 (448,110 kPa).


5. Verify the results computed
   To demonstrate that the results are compatible, accept the computed value of dlL as cor-
   rect. Then apply Eq. b to find the strain, and compute the corresponding stress. Thus e =
   2(0.0157)2 - 4.93 x IQ-^4 ; S 2 = S 1 + Ee = 50,000 + 30 x 106 x 4.93 x 10^ = 64,800 lb/in^2
   (446,731 kPa). This agrees closely with the previously calculated stress of 65,000 lb/in^2
   (448,11OkPa).

DISPLACEMENT OF TRUSS JOINT

In Fig. 19a, the steel members AC and BC both have a cross-sectional area of 1.2 in^2 (7.7
cm^2 ). If a load of 20 kips (89.0 IcN) is suspended at C, how much is joint C displaced?

Calculation Procedure:

1. Compute the length of each member and the tensile forces
   Consider joint C as a free body to find the tensile force in each member. Thus LAC =192
   in (487.7 cm); LBC= 169.7 in (431.0 cm); PAC= 14,640 Ib (65,118.7 N); PBC= 17,930 Ib
   (79,752.6 N).


2. Determine the elongation of each member
   Use the relation A/ = PLI(AE). Thus MAC = 14,640(192)/[1.2(30 x 106 )] = 0.0781 in
   (1.983 mm); MBC= 17,930(169.7)/[1.2(30 x 106 )] - 0.0845 in (2.146 mm).


3. Construct the Williott displacement diagram
   Selecting a suitable scale, construct the Williott displacement diagram as follows: Draw