PRESTRESS chapter 4 design of beams and joist using cables

prestress concrete design

1. Prestressed Concrete Design
Short Term Analysis of Prestressed Concrete Beams

2. SHORT TERM ANALYSIS OF PRESTRESSED CONCRETE BEAMS
INTRODUCTION TO
SHORT TERM ANALYSIS OF PRESTRESSED CONCRETE
BEAMS

3. Review: Hooke’s Law
✓Hooke’s Law (law of elasticity)
discovered by the English Scientist
Robert Hooke in 1660.
✓States that for relatively small
deformations of an object, the
displacement or size of the
deformation is directly proportional
to the deforming force or load.
https://phys.org/news/2015-02-law.html

4. Review: Hooke’s Law
✓Under these conditions the object
returns to its original shape and size
upon removal of the load.
✓Elastic behavior of solids according
to Hooke’s Law can be explained by
the fact that small displacements of
their constituents molecules, atoms
or ions from normal positions is
also proportional to the force that
causes the displacement.
https://phys.org/news/2015-02-law.html

5. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
✓Due to the well-established relationships between strain and stress in
the linear elastic range, strain monitoring is regarded as the closest to
direct monitoring of stress.
✓Recent advances in the development of durable and stable strain
sensors have led to the increasing popularity and use of strain-based
methods for Prestress force monitoring.
Abdel-Jabel and Glisic, 2019

6. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
✓If structures are instrumented with sensors during construction, either
through attaching them to prestressing strands or embedding them in
the concrete, long-term strain changes can be monitored.
✓Additionally, instrumentation during construction allows for
measuring strain changes during prestress force transfer, thereby
enabling calculation of the initial stress in the prestressing strands.
Abdel-Jabel and Glisic, 2019

7. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
✓If the strain change at the location of the prestressing strands is
determined, the stress change can be calculated using HOOKE’S
LAW given by the equation below:Ds
P
=EDe
P
Where:
-is the average stress change in the prestressing strand
-is the modulus of elasticity of the prestressing steel.
-is the strain change at the location of the center of gravity of
the prestressing strands.Ds
P E De
P
Abdel-Jabel and Glisic, 2019

8. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
✓In cases where the location of the prestressing strand is not known
with accuracy or where there is concerns regarding separating the
effects of bending due operational load on the structure from the
effects of prestress loss, strain changes at the centroid of stiffness of
the concrete cross-section can be monitored instead of strain changes
at the center of gravity of prestressing strands.
Abdel-Jabel and Glisic, 2019

9. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
✓Because strain at the centroid of stiffness is minimally affected by
bending due to load and temperature changes, the effect of long-term
prestress losses can be isolated.
✓In applications where the strain sensors are embedded in the concrete,
at least two strain sensors are commonly embedded in every
instrumented cross section.
Abdel-Jabel and Glisic, 2019

10. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
✓In applications where the strain sensors are embedded in the concrete,
at least two strain sensors are commonly embedded in every
instrumented cross section.
✓Assuming a linear strain distribution, the strain at the center of
gravity of the prestressing strand can be determined.
✓Assuming a perfect bond between the prestressing strands and
surrounding concrete, the strain changes measured in the concrete
can be assumed to be equal to the strain changes in the prestressing
strands at the same location. Abdel-Jabel and Glisic, 2019

11. Review: Hooke’s Law in Prestressed Concrete
STRAIN – BASED METHODS
Abdel-Jabel and Glisic, 2019

12. Short Term Analysis of Cross Section
✓The short term behavior of a prestressed concrete cross section can be
determined by transforming the bonded reinforcement into equivalent
areas of concrete and performing a simple elastic analysis on the
equivalent concrete section.
Section Transformed Section

13. Short Term Analysis of Cross Section
✓The strain at a depth “y” below the top of the cross section is defined
in terms of the top fiber strain “??????
oi” and the initial curvature x
i:e
i=e
oi+yx
i()
✓If the short term behavior of concrete is assumed
to be linear elastic, the initial concrete stress at
“y” below the top fiber is:s
i=E
ce
i=E
ce
oi+yx
i()é
ë
ù
û

14. Short Term Analysis of Cross Section
✓The resultant axial force on the section (Ni) can be determined by:
✓Note:
For a prestressed section in pure bending, the
initial axial force on the transferred concrete
section immediately after transfer is compressive
and equal in magnitude of the prestressing force. N
i
=E
c
e
oi
A+E
c
x
i
B N
i
=-P
i

15. Short Term Analysis of Cross Section
✓Resultant moment about the top surface (Mi)
✓Note:
For a prestressed beam sectionM
i
=E
c
e
oi
B+E
c
x
i
I M
i=M
o-P
id
p

16. Short Term Analysis of Cross Section
✓Derived Formula:e
oi
=
BM
i()-IN
i()
E
cB
2
-AIé
ë
ù
û x
i
=
BN
i()-AM
i()
E
cB
2
-AIé
ë
ù
û Where:
n = short term modular ration=
E
s
E
c
??????
oi = top fiber strain
??????
i = strain at depth “y” below the top
cross section
y = depth below the top
x
i = initial curvature

17. Short Term Analysis of Cross Section
✓Derived Formula:e
oi
=
BM
i()-IN
i()
E
cB
2
-AIé
ë
ù
û x
i
=
BN
i()-AM
i()
E
cB
2
-AIé
ë
ù
û Where:
n = short term modular ration=
E
s
E
c
??????
oi = top fiber strain
??????
i = strain at depth “y” below the top
cross section
y = depth below the top
x
i = initial curvature

18. Short Term Analysis of Cross Section
✓Derived Formula:e
oi
=
BM
i()-IN
i()
E
cB
2
-AIé
ë
ù
û x
i
=
BN
i()-AM
i()
E
cB
2
-AIé
ë
ù
û Where:
??????
?????? = stress at depth “y” below the top fiber
N
i = resultant axial force
P
i = prestressing force
A = area of the transformed section
B = first moment of the transformed area
about the top surface of the section
M
i = resultant moment about the top surface
M
o = external moment at the section at transfer

19. Short Term Analysis of Cross Section
✓Derived Formula:e
oi
=
BM
i()-IN
i()
E
cB
2
-AIé
ë
ù
û x
i
=
BN
i()-AM
i()
E
cB
2
-AIé
ë
ù
û Where:
d
p = depth of the prestressing steel
I = second moment of area about the top
surface of the transformed section.

20. SHORT TERM ANALYSIS OF PRESTRESSED CONCRETE BEAMS
SAMPLE PROBLEMS IN
SHORT TERM ANALYSIS OF PRESTRESSED CONCRETE
BEAMS

21. Example 1:
The short term behavior of the post – tensioned cross section shown
in the figure below is to be determined immediately after transfer. The
section contains a single unbonded cable containing 10-12.7mm dia. Strands
(Ap=1000 mm
2
) loaded within a 60mm dia duct and two layers of non-
prestressed reinforcements as shown. The force in the prestressing steel is Pi
= 1350 kN and the applied moment is Mo=100 kN.m. The elastic moduli for
concrete and steel are Ec=30,000MPa and E=200,000MPa.
1.What is the stress at the top when y=0?
2.What is the stress at the bottom when y=800mm?
3.What is the stress at the top layer on non-prestressed reinforcement?
4.What is the stress at the bottom layer of the non-prestressed reinforcement?

22. Example 1:
Section Transformed Sectionp
4
60()
2 =2827.433mm
2
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.mn=
Es
Ec
=
200,000
30,000
=6.667

23. Example 1:
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.m, n = 6.667
A. Solve for the basic components of ??????
oi

and x
ip
4
60()
2 =2827.433mm
2
i. AreaA=300800()+5100+10200-2827.433 A=252.473x10
3
mm
2
ii. 1
st
Moment of the Transformed Area about
the Top Surface of the BeamB=
300800()
800
2
æ
è
ç
ö
ø
÷+510060()[phone]()
-2827.43600()
é
ë
ê
ê
ê
ù
û
ú
ú
ú B=102.158x10
6
mm
3

24. Example 1:
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.m, n = 6.667
A. Solve for the basic components of ??????
oi

and x
ip
4
60()
2 =2827.433mm
2
iii. 2
nd
Moment of Area about the top surface
of the transformed section
I=
300800()
3
3
+510060()
2
[phone]()
2
-2827.433600()
2
é
ë
ê
ê
ê
ù
û
ú
ú
ú I=55.786x10
9
mm
4
iv. Resultant Moment about the Top SurfaceM
i=M
o-P
id
p M
i
=100-1350()
600
1000
æ
è
ç
ö
ø
÷=-710kN.m

25. Example 1:
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.m, n = 6.667
A. Solve for the basic components of ??????
oi

and x
ip
4
60()
2 =2827.433mm
2
v. Resultant Axial Force
Since it is assumed that pure bending occurs,N
i
=P
i
=-1350kN

26. Example 1:
Given:
Ec = 30,000 MPa
Es = 200,000 MPa
Pi = 1350 kN
Mo = 100kN.m
n = 6.667
A = 252.473x10
3
mm
2
B = 102.158x10
6
mm
3
I = 55.786x10
9
mm
4
Mi = -710 kN.m
Ni = -1350kN
A. Solve for ??????
oi
e
oi=
BM
i
-IN
i
E
c
B
2
-AIé
ë
ù
û e
oi=
102.158x10
6
-710x10
6
( )-55.786x10
9
( )-1350x10
3
( )
30000102.158x10
6
( )
2
-252.473x10
3
( )55.786x10
9
( )
é
ë
ù
û e
oi=-25.391x10
-6
B. Solve for x
ix
i=
BN
i
-AM
i
E
c
B
2
-AIé
ë
ù
û x
i=
102.158x10
6
( )-1350x10
3
( )-252.473x10
3
( )-710x10
6
( )
30000102.158x10
6
( )
2
-252.473x10
3
( )55.786x10
9
( )
é
ë
ù
û x
i=
-0.378x10
-6
mm

27. Example 1:
Given:
Ec = 30,000 MPa
Es = 200,000 MPa
Pi = 1350 kN
Mo = 100kN.m
n = 6.667
A = 252.473x10
3
mm
2
B = 102.158x10
6
mm
3
I = 55.786x10
9
mm
4
Mi = -710 kN.m
Ni = -1350kN
??????
oi = -25.391x10
-6
X
i = -0.378x10
-6
/mm
A. Solve for the stress at the top (y=0).
B. Solve for the stress at the bottom (y=800).s
y=0
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=0=30000-25.391x10
-6
( )+0()-0.378x10
-6
( )é
ë
ù
û
s
y=0
=-0.762MPa s
y=800
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=800=30000-25.391x10
-6
( )+800()-0.378x10
-6
( )é
ë
ù
û
s
y=800
=-9.834MPa

28. Example 1:
Given:
Ec = 30,000 MPa
Es = 200,000 MPa
Pi = 1350 kN
Mo = 100kN.m
n = 6.667
A = 252.473x10
3
mm
2
B = 102.158x10
6
mm
3
I = 55.786x10
9
mm
4
Mi = -710 kN.m
Ni = -1350kN
??????
oi = -25.391x10
-6
X
i = -0.378x10
-6
/mm
C. Solve for the stress at the top layer of non-PS reinforcement (y=60).
D. Solve for the stress at the bottom layer of the non-PS reinf. (y=740).s
y=60
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=60=200000-25.391x10
-6
( )+60()-0.378x10
-6
( )é
ë
ù
û
s
y=60
=-9.614MPa s
y=740
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=740=200000-25.391x10
-6
( )+740()-0.378x10
-6
( )é
ë
ù
û
s
y=740
=-61.022MPa

29. Example 2:
The short term behavior of the post – tensioned cross section shown
in the figure below is to be determined immediately after transfer. The
section contains a single bonded cable containing 10-12.7mm dia. Strands
(Ap=1000 mm
2
) loaded within a 60mm dia duct and two layers of non-
prestressed reinforcements as shown. The force in the prestressing steel is Pi
= 1350 kN and the applied moment is Mo=100 kN.m. The elastic moduli for
concrete and steel are Ec=30,000MPa and E=200,000MPa.
1.What is the stress at the top when y=0?
2.What is the stress at the bottom when y=800mm?
3.What is the stress at the top layer on non-prestressed reinforcement?
4.What is the stress at the bottom layer of the non-prestressed reinforcement?

30. Example 2:
Section Transformed Section
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.mn=
Es
Ec
=
200,000
30,000
=6.667

31. Example 2:
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.m, n = 6.667
A. Solve for the basic components of ??????
oi

and x
i
i. Area
ii. 1
st
Moment of the Transformed Area about
the Top Surface of the Beam

32. Example 2:
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.m, n = 6.667
A. Solve for the basic components of ??????
oi

and x
i
iii. Solve for I
iv. Solve for MiM
i=M
o-P
id
p M
i
=100-1350()
600
1000
æ
è
ç
ö
ø
÷=-710kN.m

33. Example 2:
Given:
Ec = 30,000 MPa, Es = 200,000 MPa, Pi = 1350 kN, Mo = 100kN.m, n = 6.667
A. Solve for the basic components of ??????
oi

and x
i
v. Solve for Ni
Since it is assumed that pure bending occurs,N
i
=P
i
=-1350kN

34. Example 2:
Given:
Ec = 30,000 MPa
Es = 200,000 MPa
Pi = 1350 kN
Mo = 100kN.m
n = 6.667
A = 260.967x10
3
mm
2
B = 107.254x10
6
mm
3
I = 58.840x10
9
mm
4
Mi = -710 kN.m
Ni = -1350kN
A. Solve for ??????
oi
e
oi=
BM
i
-IN
i
E
c
B
2
-AIé
ë
ù
û
B. Solve for x
ix
i=
BN
i
-AM
i
E
c
B
2
-AIé
ë
ù
û

35. Example 2:
Given:
Ec = 30,000 MPa
Es = 200,000 MPa
Pi = 1350 kN
Mo = 100kN.m
n = 6.667
A = 260.967x10
3
mm
2
B = 107.254x10
6
mm
3
I = 58.840x10
9
mm
4
Mi = -710 kN.m
Ni = -1350kN
A. Solve for the stress at the top (y=0).s
y=0
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=0=30000-25.391x10
-6
( )+0()-0.378x10
-6
( )é
ë
ù
û
s
y=0
=-0.762MPa
B. Solve for the stress at the bottom (y=800).s
y=800
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=800=30000-25.391x10
-6
( )+800()-0.378x10
-6
( )é
ë
ù
û
s
y=800
=-9.834MPa

36. Example 2:
Given:
Ec = 30,000 MPa
Es = 200,000 MPa
Pi = 1350 kN
Mo = 100kN.m
n = 6.667
A = 260.967x10
3
mm
2
B = 107.254x10
6
mm
3
I = 58.840x10
9
mm
4
Mi = -710 kN.m
Ni = -1350kN
C. Stress at y = 60 (top layer of non-PS Reinforcement)
D. Stress at y=740 (bottom layer of non-PS reinforcements
y=60
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=60=200000-25.391x10
-6
( )+60()-0.378x10
-6
( )é
ë
ù
û
s
y=60
=-9.614MPa s
y=740
=E
c
e
oi
+yx
i()é
ë
ù
û
s
y=740=200000-25.391x10
-6
( )+740()-0.378x10
-6
( )é
ë
ù
û
s
y=740
=-61.022MPa

37. Reference:
Raju, N. K. (2006). Prestressed concrete. Tata McGraw-Hill Education.
https://www.britannica.com/science/Hookes-law
Abdel‐Jaber, H., & Glisic, B. (2019). Monitoring of prestressing forces in prestressed concrete
structures—An overview.Structural Control and Health Monitoring,26(8), e2374.