• A continuous beam is having more than one span is carried by several supports. • It is mainly used in bridge construction. • Simple beam cannot be used for large spans, as it requires more strength and stiffness. • But continuous PSC beam not only provides adequate strength and stiffness, but also provides sufficient ductility.

1. DEPARTMENT OF CIVIL ENGINEERING
1
PRESENTATION ON PRESTRESSED CONCRETE
DESIGN OF CONTINUOUS MEMBERS IN PRESTRESSED
PRESENTED BY
LOGESH. S
M.E STRUCTURAL
19TPCV011

2. 2
DESIGN OF CONTINUOUS MEMBERS IN
PRESTRESSED

3. 3
• A continuous beam is having more than one span is carried
by several supports.
• It is mainly used in bridge construction.
• Simple beam cannot be used for large spans, as it requires
more strength and stiffness.
• But continuous PSC beam not only provides adequate
strength and stiffness, but also provides sufficient ductility.
Continuous Beam

4. 4
• An important distinction between them is that the
prestressing force does not produce any secondary moments in
simple beams, considerable secondary moments are produced
in continuous PSC beams.
• Also, the behavior of continuous PSC beams nearer to
ultimate load is different from the behavior of a corresponding
simple beam.
Continuous Beam

5. 5
Merits
• Ultimate loads carrying capacity is higher than the statically
determinate structures due to the effect of redistribution of
moments.
• An increase in stability of framed structures. This is feasible
due to the distribution of moments.
• The deflections in continuous beams are relatively small when
compared to simply supported beams. This is due to the fact
that the continuity of members makes the rotation less at the
places of supports. Thus, the deflections are less.

6. 6
• The maximum bending moment that occurs in a continuous
beam is less than that for same spans, if the beams are cut into
simply supported beams.
• In continuous beam, the bending moments are evenly
distributed throughout the span and results in reduction in size
of members. Thus, lighter weight structures made.
• An effective usage of curved cables is possible to resist the
supports and span moments.
• A reduction in cost and construction time is possible due to the
use of number of anchorages.
Merits

7. 7
• Frictional loss in continuous tendons
• Shortening of long continuous beams under prestress
• Secondary stresses
• Location of both maximum moment and shear force over
supports
• Reversal of moments
• Moment packs
• Continuity for precast elements
• Difficulty in designing.
Demerits

8. 8
The profile of the CGS for a post-tensioned beam is shown in the sketch.
Plot the pressure line due to a prestressing force Pe = 1112 kN.
Solution
1) Plot M1 diagram
The values of M1 are calculated from M1 = Pee.
e (m) M1 (kN m)
0.06 – 66.72
0.24 – 266.88
– 0.12 133.44
0.27 – 300.24
.06 C
G
C
0.08 rad
A 9 m D 6 m B 7.5 m
0.176 rad
7.5 m C
.27.12.24

9. 9
0.24
0.06
0.270.12
-66.72
-266.88 -300.24
133.44
M1 diagram (kN m)
Profile of the CGS
Plot V diagram
For AD,
V= dM1/dx
= (-266.9 - (-66.7))/9
= -22.2 kN
For DB,
V= dM1/dx
= (133.4 - (-266.9))/6
= 66.7 kN
A D B C

10. 10
For BC, to find dM1/dx, an approximate parabolic equation for the M1 diagram can
be used.
M1 = - 4Peex (L - x )/L2
V= dM1/dx
= - 4Pee(L - 2x )/L2
=- 4Pee/L
=- 4 ×(133.4 + 300.2)
15
= -115.6 kN
At B,
V =
L
M1
Pee x

11. 11
The exact value of V at B is
V = - 107.0 kN
The difference of V between C and B is given from the change in slope of the M1 diagram.
V|C - V|B = 0.176 × 1112
= 195.7 kN
Therefore, value of V at C is given as follows.
V|C = 195.7 – 107.0
= 89.0 kN
66.72
266.88 300.24
133.44
V diagram (kN)
66.7
89.0
-107.0-
22.2
M1 Diagram (kN m)

12. 12
3) Plot equivalent load (weq) diagram. Include moment 66.7 kN-m at A.
Point load at D
W|D = 66.7- (- 22.2 )
= 88.9 kN
Since B is a reaction point, the downward load at B need not be considered. Distributed load
within B and C
w BC = 89.0 - (-107.0)
15
= 13.0 kN/m
–
22.2
66.7 89.0
Equivalent load diagram
66.7 kN m 13.0
kN/m
88.9
kN
– 107.0 Kn
V diagran

13. 13
0.5 0.5
88.9x9×6^2 88.9×9^2 ×6 13.0×15^2
15^2 15^2 12
=128 = -192 = 244 –244
–194.7 244
–97 122
–38.5 –38.5
–66.7 –327.5 327.5 0
66.7
DF
88.9
FEM
Bal
CO
Bal
Total
13.0
In the previous table,
Bal = Balanced
CO = Carry Over moment
DF = Distribution Factor
FEM = Fixed End Moment
4) Plot the M2 diagram.
Calculate moment at supports by moment distribution

14. 14
M2 (kN m) ec (m)
– 66.7 0.06
327.0 0.294
0.0 0.184
Calculate values of ec at support.
The values of ec are calculated from ec= M2/Pe.
M2 diagram (kN m)
Profile of CGS
Pressure line
0.06
0.136
0.294
0.184
327.0
-66.7
66.7
knm
0
88.9 kn 13kn/
m

15. 15