- Alternatively, evaluate F/ by assigning an arbitrary depth to
the member
Thus, set A = 10 in (254 mm); yb = SJil(Sb + St) = 4.815 in (122.301 mm);/cfl/ =fbp - (fbp -
ftp)yblh = 2973 - (2973 + 807)0.4815 = + 1153 lb/in^2 (+7949.9 kPa); F 1 = 1153(500) =
576,500 Ib (2,564,272.0 N).
EFFECT Of INCREASE IN BEAM SPAN
Consider that the span of the beam in the previous calculation procedure increases by 10
percent, thereby causing the midspan moment due to superimposed load to increase by 21
percent. Show that the member will be adequate with respect to flexure if all cross-sec-
tional dimensions are increased by 7.2 percent. Compute the new eccentricity in the cen-
ter interval, and compare this with the original value.
Calculation Procedure:
- Calculate the new section properties and bending moments
Thus A = 500(1.072)
2
= 575 in
2
(3709.9 cm
2
); Sb = 4584(1.072)
3
= 5647 in
3
(92,554.3
cm
3
); S 1 = 4257(1.072)
3
= 5244 in
3
(85,949.2 cm
3
); M 5 = 9000(1.21) = 10,890 in-kips
(1230.4 kN-m); Mw = 3500(1.072)^2 (1.21) = 4867 in-kips (549.9 kN-m).
- Compute the required section moduli, prestresses, prestressing
force, and its eccentricity in the central interval, using the same
sequence as in the previous calculation procedure
Thus Sb = 5649 in^3 (92,587.1 cm^3 ); St = 5246 in^3 (85,981.9 cm^3 ). Both these values are ac-
ceptable. Then4, = + 3046 lb/in^2 (+21,002.2 kPa);^ = -886 lb/in^2 (-6108.9 kPa); F 1 =
662,800 Ib (2,948,134.4 N); e = 16.13 in (409.7 mm). The eccentricity has increased by
11.5 percent.
In practice, it would be more efficient to increase the vertical dimensions more than
the horizontal dimensions. Nevertheless, as the span increases, the eccentricity increases
more rapidly than the depth.
EFFECT OF BEAM OVERLOAD
The beam in the second previous calculation procedure is subjected to a 10 percent over-
load. How does the final stress in the bottom fiber compare with that corresponding to the
design load?
Calculation Procedure:
- Compute the value of fbs under design load
Thus,/fe = -MJSb = -9,000,000/4584 = -1963 lb/in^2 (-13,534.8 kPa).
- Compute the increment or fbs caused by overload and the
revised value of fbf
Thus, A/fa = 0.10(-1963) = -196 lb/in^2 (-1351.4 kPa); ^ 7 = -200 - 196 = -396 lb/in^2
(-2730.4 kPa). Therefore, a 10 percent overload virtually doubles the tensile stress in the
member.
