TABLE 5. Calculations for Two-Span Continuous Beam: Part b
Section 1 2 34 B
econ, in (mm) +21.12 +26.88 +17.28 -7.68 -48.00
(536.4) (+682.8) (+438.9) (-195.1) (-1,219.2)
ftp, lb/in^2 (kPa) +2,116 +2,503 +1,857 +178 -2,535
(+14,588) (+17,256) (+12,802) (+1,227) (-17,476)
4, lb/in^2 (kPa) +37 -142 +157 +933 +2188
(+255) (-979) (+1,082) (+6,660) (+15,084)
0.85/k, lb/in^2 (kPa) +1,799 +2,128 +1,578 +151 -2,155
(+12,402) (+14,670) (+10,879) (+1,041) (-14,857)
0.85/^, lb/in^2 (kPa) +31 -121 +133 +793 +1,860
(+214) (-834) (+917) (+5,467) (+12,823)
fu, lb/in^2 (kPa) +1,157 +1,282 +1,072 +527 -355
(+7,976) (+8,838) (+7,390) (+3,633) (-2,447)
fbfe lb/in^2 (kPa) +149 -65 -51 +193 +1,305
(+1,027) (-448) (-352) (+1,331) (+8,997)
/tf, lb/in^2 (kPa) +481 +423 +520 +772 +1,180
(+3,316) (+2,916) (+3,585) (+5,322) (+8,135)
fe lb/in^2 (kPa) +794 +894 +886 +774 +260
(+5,474) (+6,163) (+6108) (+5,336) (+1,792)
nal stress at 2 and the allowable value of—60 lb/in^2 (-413.7 kPa) arises from the degree of
precision in the calculations.)
With the exception offbi at B, all stresses at the boundary sections lie within the allow-
able range.- Locate the section at which fbl = -60 lb/in
2
(-413.7 kPa)
Since fbp andfbw vary parabolically across the span, their sum fbi also varies in this man-
ner. Let x denote the distance from the interior support to a given section. Apply Eq. 66 to
find the equation for fbi using the initial-stress values at sections B 9 3, and 1. Or, -355 - 2
x 1072 + 1157 = -1342 (-9253.1 kPa); 3(-355) - 4(1072) + 1157 = -4196 (-28,931.4
kPa);/w = -2684(jt/96)^2 + 4196*/96 - 355. When/6l = -60 (-413.7), x = 7.08 ft (2.15 m).
The tensile stress at transfer is therefore excessive in an interval of only 14.16ft (4.32 m). - Design the nonprestressed reinforcement over the
interior support
As in the preceding procedures, the member must be investigated for ultimate-strength
capacity. The calculation pertaining to any quantity that varies parabolically across the
span may be readily checked by verifying that the values at uniformly spaced sections
have equal "second differences." For example, with respect to the values of/j,/ recorded in
Table 5, the verification is:
+1157 +1282 +1072 +527 -355-125 +210 +545 +882+335 +335 +337The values on the second and third lines represent the differences between successive val-
ues on the preceding line.