continuous member amenable to analysis by the theorem of three moments or moment

distribution.

Determine wL

2

/4 for each span: span AB, w^/4 = F/-0.40 - 1.20 - 1.20) = -2.80F,

in-lb (-03163F, N-m); span BC, w 2 L^22 /4 = F,.(-1.20 - 1.28 - 0.60) = -3.08F 1 in-lb

(-0.3479F,N-m).

2. 
   

   Determine the true prestress moment at B in terms of F 1
   Apply the theorem of three moments; by subtraction, find Mkb. Thus, MpJL 1 + 2M^(L 1 +
   L 2 ) + Mp 0 L 2

   
   

   ~ W 1 L]M - w 2 L2/4. Substitute the value OfL 1 and L 2 , in feet (meters), and
   divide each term by F 1 , or 0.40(60) + (2Af^ x 15OyF 1 + 0.60(90) = 2.80(60) + 3.08(90).
   Solving gives Mpb = 1 .224F, in-lb (0.1383F, N-m). Also, Mkb = Mpb - (-F^ 6 ) = F 1 (1.224 -
   1.20) = 0.024F 1 -. Thus, the continuity moment at B is positive.

   
   


3. Evaluate the prestress moment at the supports and at midspan
   Using foot-pounds (newton-meters) in the moment evaluation yields Mpa =
   0.40(96,000)712 = 3200 ft-lb (4339.2 N-m); Mpb = 1.224(96,000)712 = 9792 ft-lb (13,278
   N-m); Mpc = 0.60(96,000)712 = 4800 ft-lb (6508.0 N-m); Mpd = -Ffd + MM = Ff<-0.60 +
   Y 2 x 0.024)712 = -4704 ft-lb (-6378 N-m); Mpe = F/-0.64 + % x 0.024)712 = -5024 ft-lb
   (-6812 N-m).


4. Construct the prestress-moment diagram
   Figure 52 shows this diagram. Apply Eq. 70 to locate and evaluate the maximum negative
   moments. Thus, AF = 25.6 ft (7.80 m); BG = 49.6 ft (15.12 m); Mpf= -4947 ft-lb (-6708
   N-m); Mpg = -5151 ft-lb (-6985 N-m).

PRINCIPLE OF LINEAR TRANSFORMATION

For the beam in Fig. 50, consider that the parabolic trajectory of the prestressing force is
displaced thus: ea and ec are held constant as eb is changed to -2.0 in (-50.80 mm), the ec-
centricity at any intermediate section being decreased algebraically by an amount directly
proportional to the distance from that section to A or C. Construct the prestress-moment
diagram.

Calculation Procedure:

1. Compute the revised eccentricities
   The modification described is termed a linear transformation of the trajectory. Two meth-
   ods are presented. Steps 1 through 4 comprise method 1; the remaining steps comprise
   method 2.
   The revised eccentricities are ea = -0.40 in (-10.16 mm); ed = +0.20 in (5.08 mm);
   eb = -2.00 in (-50.8 mm); ee = + 0.24 in (6.096 mm); ec = -0.60 in (-15.24 mm).


2. Find the value of wL
   2
   /4 for each span
   Apply Eq. 71: span AB, wvL\l4 =F/-0.40 - 0.40 - 2.00) = -2.8OF 1 ; span BC, w 2 L 22 /4 =
   F,(-2.00 - 0.48 - 0.60) = -3.08F 1 ,
   These results are identical with those obtained in the previous calculation procedure.
   The change in eb is balanced by an equal change in 2ed and 2ee.


3. Determine the true prestress moment at B by applying
   the theorem of three moments; then find Mkb
   Refer to step 2 in the previous calculation procedure. Since the linear transformation of