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)

386 Prestressed concrete continuous beams
= 5000 X 0.5 - 5000 X 0.2
= 1500 kNm
M 2 at B = (M 3 at B) - (M 1 at B) = 0
The support reactions induced by the prestressing are now calculated from
these secondary moments M 2 , and are as shown in Fig. 10.2-4(e). These
support reactions and the tendon profile e 5 completely define the shear
force VP at any section produced by the prestressing. For example at a
section between C and E,
[des]
VP = 60 kN - 120 kN - Pe dx (see eqn 9.2-22)
= 60 kN - 120 kN - 500 kN x 0.06 rad (see Fig. 10.2-4(a))
= - 360 kN
(A more convenient method of determining VP is given in Example
10.4-1.)
Example 10.2-2
With reference to the beam in Example 10.2-1, determine the prestress at
the top and bottom fibres at section E in terms of the sectional properties
of the beam.
SOLUTION
From eqns (10.1-3) and (10.1-4),
Pc Peep
bottom fibre prestress / 1 = A + 21
Pc Peep
top fibre prestress fz = A - 22
From Fig. 10.2-4(e), eP = 0.22 m at E; also Pc = 5000 kN as given.
Therefore
/ 1 = 5000 X 10 3 /A + 5000 X 10 3 X 0.22 X 10 3 /2 1 N/mm 2
fz = 5000 X 10 3 /A - 5000 X 10 3 X 0.22 X 10 3 / Z 2 N/mm 2
where A and Z are respectively in mm 2 and mm 3 units.
Example 10.2-3
If the prestressed beam in Example 10.2-1 is acted on by a uniformly
distributed load of 50 kN/m, determine the resulting line of pressure due to
the combined action of the prestressing and the imposed load.
SOLUTION
The solution is summarized in Fig. 10.2-5. The resulting line of pressure
is obtained by superposition, i.e.
Fig. 10.2-S(e) = Fig. 10.2-S(b) + Fig. 10.2-S(d)
The reader should go through the solution properly; study the solution to
Example 10.2-1 again if necessary.