F. K. Kong MA, MSc, PhD, CEng, FICE, FIStructE, R. H. Evans CBE, DSc, D ès Sc, DTech, PhD, CEng, FICE, FIMechE, FIStructE (auth.)-Reinforced and Prestressed Concrete-Springer US (1987)

Stresses in service: elastic theory 341
Substituting into eqns (9.2-14) and (9.2-15) (or eqns 9.2-2 and
9.2-3),
Pe = 990 kN es = 161 mm
Example 9.2-2
A Class 2 post-tensioned concrete beam is simply supported over a 10m
span. The characteristic imposed load consists of a single 100 kN force at
midspan. The characteristic concrete strength is 50 N/mm 2 and the unit
weight of concrete is 23 kN/m 3 . The beam is of uniform section having the
following properties: area A = 120000 mm 2 , Z 1 (bottom) = 19.0 x
106 mm 3 , Z 2 (top)= 21.7 x 106 mm 3 . Determine for the service condition:
(a) the minimum effective prestressing force required (Pemin) and the
corresponding midspan tendon eccentricity ( e 5);
(b) the maximum effective prestressing force (Pemax) that may safely be
used, and the midspan tendon eccentricity (es) for this force.
SOLUTION
From Table 9.2-1,
famax = 0.33 X 50 = 16.5 N/mm 2
From Table 9.2-2,
famin = -2.55 N/mm 2
(But see note (c) following Table 9.2-2.)
By inspection, the critical section is at midspan, where
Mimax = 250 kNm Mimin = 0 and Mct = 34.5 kNm
(all as in Example 9.2-1).
(a) Pemin is associated with the minimum required values of f 1 and fz.
From eqns (9.2-9) and (9.2-10),
min. reqd ft = -2 55 + 250 X 106 + 34.5 X 106 = 12 42 N/ 2
· 19.0 x 106 · mm
(b)
min. reqd !: 2 = -2 55 - 0 + 34·5 x 106 = -4 14 N/ 2
· 21.7 x 106 · mm
Substituting into eqns (9.2-14) and (9.2-15) (or eqns 9.2-2 and
9.2-3):
Pemin = 431 kN es = 390 mm
Alternatively, Pemin and es can be obtained directly from eqns
(9.2-16) and (9.2-17) without first calculatingf 1 andfz.
Pemax is associated with the maximum permissible values of ft and f 2 .
From eqns (9.2-11) and (9.2-12),
_ 0 + 34.5 X 106 _ 2
max. perm. ft - 16.5 + 19 . 0 X 106 - 18.32 N/mm