FIGURE 57. Composite Magnel diagram.- Plot the concordant trajectory
Do this by applying the values of C 3 appearing in Table 4; for example, ev = +21.87
(0.0550)70.0700 = +17.19 in (436.626 mm). At midspan, em = +21.87(0.0625)70.0700 =
+19.53 in (496.062 mm).
Record the eccentricities on line 10 of the table. It is apparent that this concordant tra-
jectory is satisfactory in the respect that it may be linearly transformed to one falling
within the confines of the section; this is proved in step 14. - Apply Eq. 56 to find fbp and ftp
Record the results in Table 4. For example, at section 1, fbp = 1,160,000(10.32 +
17.19)714,860 - +2148 lb/in^2 (+14,810.5 kPa);j^ - 1,160,000(22.32 - 17.19)732,140 = +
185 lb/in
2
(+1275.6 kPa). - Multiply the values of fbp and ftp by 0.85, and record
the results
These results appear in Table 4. - Investigate the stresses at every boundary section
In calculating the final stresses, apply the live-load stress that produces a more critical
condition. Thus, at section \Jbi = -959 + 2148 = + 1189 lb/in^2 (+8198.2 kPa);/ft/ = -959
- 691 + 1826 = + 176 lb/in^2 (+1213.6 kPa);/,, = + 444 + 185 = + 629 lb/in^2 (+4337.0
kPa);/0-= + 444 + 319 + 157 = +920 lb/in^2 (+6343.4 kPa). At section 2:fbi = -1221 +
2513 = + 1292 lb/in^2 (+8908.3 kPa);/fc/ = -1221 - 972 + 2136 = -57 lb/in^2 (-393.0 kPa);
fa = + 565 + 16 = + 581 lb/in^2 (+4006.0 kPa);/^= +565 + 450 + 14 = + 1029 lb/in^2
(+7095.0 kPa). At section 3:fbi = -785 + 1903 = + 1118 lb/in^2 (+7706.8 kPa);/^= -785 -
844 + 1618 = - 11 lb/in^2 (-75.8 kPa);/ft- = + 363 + 298 = + 661 lb/in^2 (+4558.0 kPa);