- Compute the stresses at midspan due to the beam weight
Thus, Mw = (^1 / 8 )(83)(20)^2 (12) = 49,800 in-lb (5626.4 N-m); fbw = -49,800/133 = -374
lb/in^2 (-2578.7 kPa),/^ = +374 lb/in^2 (2578.7 kPa). - Set the critical stresses equal to their allowable values to
secure the allowable unit superimposed load
Use Fig. 32 or 33 as a guide. At support:^ = +2400 lb/in^2 (+16,548 kPa);/ft- = -190 lb/in^2
(-1310.1 kPa); at midspan, fbf = 0.85(2400) - 374 + fbs = -425 lb/in^2 (-2930.4 kPa);
ftf = 0.85(-19O) + 374 + fts = +2250 lb/in
2
(+15,513.8 kPa). Also,/fe = -2091 lb/in
2
(-14,417.4 kPa);/ft = +2038 lb/in^2 (+14,052 kPa).
Since the superimposed-load stresses at top and bottom will be numerically equal, the
latter value governs the beam capacity. Or ws = w^fjf^ = 83(2038/374) = 452 Ib/lin ft
(6596.4 N/m). - Find Firmax and its eccentricity
The value of ws was found by setting the critical value Of^ 1 and offtf equal to their re-
spective allowable values. However, since Sb is excessive for the load ws, there is flexibil-
ity with respect to the stresses at the bottom. The designer may set the critical value of ei-
ther fbi or fbf equal to its allowable value or produce some intermediate condition. As
shown by the calculations in step 3,^may vary within a range of 2091 - 2038 = 53 lb/in^2
(365.4 kPa). Refer to Fig. 34, where the lines represent the stresses indicated.
Points B and F are fixed, but points A and E may be placed anywhere within the 53-
lb/in^2 (365.4-kPa) range. To maximize F 1 -, place A at its limiting position to the right; i.e.,
set the critical value offbi rather than that offbf equal to the allowable value. Thenj£fl/ =
FimJA = Y 2 (TAQQ - 190) = + 1105 lb/in^2 (+7619.0 kPa); F/max = 1105(80) = 88,400 Ib
(393,203.2 N);/^ = 1105 + 88,400e/133 = + 2400; e=l.95 in (49.53 mm).
5. Find Fifmin and its eccentricity
For this purpose, place A at its limiting position to the left. Then/^ = 2,400 - (53/0.85) =
+2338 lb/in^2 (+ 16,120.5 kPa);/cfl/ = + 1074 lb/in^2 (+7405.2 kPa); F,min = 85,920 Ib
(382,172.2 N); e = 1.96 in (49.78 mm). - Verify the value of F 1 - max by checking the critical stresses
At support: fbi = + 2400 lb/in^2 (+16,548.0 kPa); fti = -190 lb/in^2 (-1310.1 kPa). At
midspan:/^ = +2040 - 374 - 2038 = -372 lb/in^2 (-2564.9 kPa);/^= -162 + 374 + 2038
= +2250 lb/in^2 (+15,513.8 kPa).
FIGURE 34. Stresses at midspan under maximum prestressing force.