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)

340 Prestressed concrete simple beams
To obtain Pemin' and the es to be used with Pemin' it is only necessary
to insert in these equations the minimum / 1 and fz. Substituting eqns
(9.2-9) and (9.2-10) into eqns (9.2-14) and (9.2-15),
p . = Uamin(Zt + Zz) + Mr]A
emm Zt + Zz
es ZzMimax + ZtMimin + (Zt + Zz)Md
(for Pemin) = [famin(Zt + Zz) + Mr]A
Example 9.2-1
(9.2-16)
(9.2-17)
A Class 1 pre-tensioned concrete beam is simply supported over a 10 m
span. The characteristic imposed load Qk is a 100 kN force at midspan. The
concrete characteristic strength is 50 N/mm 2 and the unit weight of
concrete is 23 kN/m 3 .
(a) Determine the minimum required sectional moduli for the service
condition.
(b) If the section adopted is of area 120 000 mm 2 and exactly the
minimum required moduli, determine the effective prestressing force
SOLUTION
Pe required under service condition and the tendon eccentricity es at
midspan.
For Class 1 members, famin = 0. From Table 9.2-1,
famax = 0.33 X 50 = 16.5 N/mm 2
design imposed load for the service condition = 1.0 Qk = 100 kN
Therefore
Mimax = ! X 100 X 10 = 250 kNm;
(a) From eqn (9.2-8),
Mimin = 0
. 250 X 106 06 3
mm. reqd Z = 16 _ 5 _ 0 = 15.15 x 1 mm
(b) Adopted section: A = 120000 mm 2 ; Z 1 = Z 2 = 15.15 x 106 mm 3 .
. 120 X 103
destgn dead load = 106 x 23 = 2. 76 kN/m
Md = i x 2.76 x 10Z = 34.5 kNm
Since exactly the minimum required Z's have been used, eqns
(9.2-4) to (9.2-7) become identities, as explained in statement (h)
above. Equations (9.2-4) and (9.2-5) will give the same value for / 1;
similarly eqns (9.2-6) and (9.2-7) will give the same f2 . Use, say,
eqns (9.2-4) and (9.2-6):
f 250 X 10 6 + 34.5 x 106 - 0 therefore f 1 = 18.78 N/mm2
1 - 15.15x106 -
f + 250 X 10 6 + 34.5 x 106 = 16 5 therefore f 2 = -2.28 N/mm2
2 15.15 X 106 .