Page 294 - Handbook of Civil Engineering Calculations, Second Edition

PRESTRESSED CONCRETE 2.79

procedure, since this renders the continuous member amenable to analysis by the theorem
of three moments or moment distribution.
2 2
Determine wL /4 for each span: span AB, w 1 L 1 /4 F i ( 0.40 1.20 1.20) 2.80F i
2
in.·lb ( 0.3163F i N·m); span BC, w 2 L 2 /4 F i ( 1.20 1.28 0.60) 3.08F i in.·lb
( 0.3479F i N·m).
2. Determine the true prestress moment at B in terms of F 1
Apply the theorem of three moments; by subtraction, find M kb . Thus, M pa L 1 2M pb (L 1
3
3
L 2 ) M pc L 2 w 1 L 1 /4 w 2 L 2 /4. Substitute the value of L 1 and L 2 , in feet (meters), and
divide each term by F i , or 0.40(60) (2M pb 150)/F i 0.60(90) 2.80(60)
3.08(90). Solving gives M pb 1.224F i in.·lb (0.1383F i N·m). Also, M kb M pb ( F i e b )
F i (1.224 1.20) 0.024F i . 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 M pa
0.40(96,000)/12 3200 ft·lb (4339.2 N·m); M pb 1.224(96,000)/12 9792 ft·lb
(13,278 N·m); M pc 0.60(96,000)/12 4800 ft·lb (6508.0 N·m); M pd F j e d M kd
F i ( 0.60 /2 0.024)/12 4704 ft·lb ( 6378 N·m); M pe F i ( 0.64 /2
1
1
0.024)/12 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); M pf 4947 ft·lb
( 6708 N·m); M pg 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: e a and e c are held constant as e b is changed to 2.0 in. ( 50.80 mm), the
eccentricity at any intermediate section being decreased algebraically by an amount di-
rectly 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 e a 0.40 in. ( 10.16 mm); e d 0.20 in. (5.08 mm);
e b 2.00 in. ( 50.8 mm); e e 0.24 in. (6.096 mm); e c 0.60 in. ( 15.24 mm).
2
2. Find the value of wL /4 for each span
2
2
Apply Eq. 71: span AB, w 1 L 1 /4 F i ( 0.40 0.40 2.00) 2.80F i ; span BC, w 2 L 2 /4
F i ( 2.00 0.48 0.60) 3.08F i .
These results are identical with those obtained in the previous calculation procedure.
The change in e b is balanced by an equal change in 2e d and 2e e .
3. Determine the true prestress moment at B by applying
the theorem of three moments; then find M kb
Refer to step 2 in the previous calculation procedure. Since the linear transformation of
the trajectory has not affected the value of w 1 and w 2 , the value of M pb remains constant.
Thus, M kb M pb ( F i e b ) F i (1.224 2.0) 0.776F i .