L01_Unit Info & Design Revision(2).pdf

1. 1

2. ADVANCED CONCRETE DESIGN AND CONSTRUCTION
2
STEN4005
Thong Pham
Unit coordinator and lecturer
Contact details:
Building 204 Room 420
Email: thong.pham@curtin.edu.au

3. Assessments
• Assignments 60%
– 30% Assignment 1 Bridge Design
– 30% Assignment 2 Behaviour and Construction
• EOS Exam 40%
• Pass requirements:
– 50% overall
3

4. Text Books and References
• Concrete Structures, Warner et al., Longman, 1999
• Australia Standards AS 3600‐2018,
AS 5100.2, AS 5100.5, and more
• Prestressed Concrete Design by Edward G. Nawy, Prentice
Hall, 5th ed. 2006 (updated 2010)
• Notes for Prestressed Concrete Design by Joe Wyche, Wyche
Consulting, 2008.
4

5. Content
Section Design – Reinforced and Prestressed Concrete
Loading and Load path
Hyperstatic actions
Prestress Losses
Anchorage Zone
Box Girde (T‐roff) Design
Bridge Type and Construction
Other Design and Construction aspects
5

6. 6

7. 7
DESIGN REVISION
Advanced Concrete Design and Construction

8. • Review of RC Design Concept
• Review of PC Design Concept
• Calculation of fibre stresses
• Transfer Limit State
• Cracking Limit State
• Flexural Strength
Objectives
8

9. Concrete Structures
9
Any form and shape
Prestressed concrete
Reinforced concrete
Precast
Cast in‐situ
taken from various internet sources

10. Concrete Structures
10
World's Largest 3D-Printed Concrete Pedestrian Bridge
Completed in China
©Prof Xu Weiguo, ArchDaily
Acciona built the world’s first 3D-printed pedestrian bridge in
Madrid, Spain
https://www.constructionweekonline.com/projects‐tenders/168913‐meet‐the‐builder‐behind‐
the‐worlds‐first‐3d‐printed‐bridge‐in‐spain

11. Examples of Prestressed Concrete Structures
www.prestasi‐concrete.com
http://www.dormanlongtechnology.com/images/Sutong_PC_01.jpg 17 June2010
Cast in‐situ or Precast
11
Post‐tensioned Pre‐tensioned
Incremental Launch
Beam/Span Bridge

12. Examples of Prestressed Concrete Structures
Two‐way post‐tensioned slab
https://deltacorp.com.au/precast‐products/major‐infrastructure/teeroff‐bridge‐beams/
12
T‐roff bridge
https://theconstructor.org/concrete/post‐tension‐slab/27708/

13. Basic Concepts: Reinforced Concrete
• Concrete is good in compression
but very poor in tension
• Steel bars are inserted to provide
tension in a concrete section,
called a reinforced concrete (RC)
section
• The moment capacity of the RC
section is calculated from the
coupling between compression in
concrete and tension in steel r/f.
RC beam
13
RC slab
RC sections
What happens when this section is
subjected to bending?
What are the main assumptions
used when calculating the
moment capacity?

14. Basic Concepts: Prestressed Concrete
• A prestressed concrete section has
prestressing steel as the main
reinforcement.
• The prestressing is performed by
pulling the steel, either before or
after the concrete is cast.
• The prestressing of steel is also
called tensioning of steel
• The force used is called the
prestressing force (P).
• Steel used for prestressing is either
strand, tendon or cable
P
Section
What happens to the concrete member
when we apply the prestressing force (to
pull the tendon)?
14

15. Basic Concept Review
• Pre‐tensioned member is when
the tendon is stressed (by
tensioning) before casting the
concrete.
• Post‐tensioned member is when
the tendon is stressed after
casting the concrete.
• Tendon can be concentric or
eccentric.
cgc
e
Denote e as the distance between the
centroid of the area of the concrete gross
section (cgc) and the centroid of the area
of the prestressing steel (cgs).
Note that e can be below or above the
cgc, i.e. the steel can be either at the top,
or bottom, or both.
Section
cgs
15

16. Pre-tensioned vs Post-tensioned Concrete

17. Forces on Prestressed Concrete Section
• Prestressing Force, P,and
eccentricity, e, measured from
the centroid, cgc, of the section
• Other loads on the members
• Fibre stresses calculation
P
cgc
e
P
Pe
P
cgc
e
Longitudinal View
Longitudinal View Section
More forces due to
selfweight, SDL & LL
When are these forces applied?
Transfer & Service States
May need to consider more than one service state!
P
Pe
Forces
17

18. Calculation of Fibre Stresses
• Stresses due to P and e
P
cgc
e
Longitudinal View Section
Pe
P
A I
P Pey

 
Stresses
c (+ve)
t
(-ve)
c
(+ve)
Stress at any distance y from the cgc line
Notes:
• σ is the elastic stress at distance y from the cgc.
• A is the cross sectional area
• I is the moment of inertia about the same axis of
the moment considered
y σ
18
σmax
(in compression)
min
(either in tension or compression,
controlled by P and e)

19. 19
Comparison between RC and PC
RC section PC section
Reo does not exert any force on the member Member is preloaded (due to steel tensioning)
Reo resists tension Prestressing steel acts as r/f
Cracks are irrecoverable Cracks are recoverable
Member depth greater than PC Require less concrete and steel
More material cost Less material cost but more equipment and
operation cost
Heavier members and foundations Lighter members and foundations
Not suitable for large span Suitable for large span members

20. Comparison between RC and PC
RC member PC member
P P
P
Pe
P
Pe
Deflection due to selfweight
20
Deflection due to total load

21. Example 1.1 – Fibre stress due to prestressing
cgc
Section
800
 
P

Pey
A I
P cgc
Section
e
P = 1500 kN
800
Section
e P = 1500 kN
Based on the calculation of the fibre stresses, draw the stress
diagram of the PC section below. Consider the following cases:
1. e = 0, i.e., the prestressing is at the cgc line (concentric tendon)
2. e = 300 mm below cgc (eccentric tendon)
3. e = 350 mm above cgc (eccentric tendon)
500 500 500
800 cgc
21

22. Example 1.1 (ctd.)
1. When e = 0, the stress is uniform
= 1500X1000/(500*800) = 3.75 MPa (in compression)
2. When e = 300 mm below cgc.
21
cgc
Section
300
c (+ve)
t (-ve)
c (+ve)
max
= 3.75 + 8.43 = 12.18 MPa
P = 1500 kN
500
800
P
A
Pey
I
= 3.75 = ±8.43 (max)
t = elastic stress at top fibre
b = elastic stress at bottom fibre
yt = distance from cgc to top fibre
yb = distance from cgc to bottom fibre
min= 3.75 - 8.43 = -4.68 MPa
top
= P
- P.e.yt

bot
A I
= P
+ P.e.yb
A I
 = P
A

23. t (-ve)
Example 1.1 (ctd.)
3. When e = 350 mm above cgc
Section
c (+ve)
max
= 3.75 + 9.84 = 13.59MPa
min= 3.75 - 9.84 = -6.09MPa
500
P = 1500 kN
350
800 cgc.
I
Pey
c (+ve)
P
= 3.75
A
= ±9.84(max)
top
= P
+ P.e.yt

bot
A I
= P
- P.e.yb
A I
23

24. Acting loads on prestressed concrete member
• Self weight, SW
• Superimposed dead load, SDL, & Live load, LL
FD
a
L
L
SW
SDL
Loads are either distributed
load (uniform or non‐uniform),
concentrated load, moment
FL
b
L
LL
24

25. Acting loads on prestressed concrete member
• Other loads e.g., Traffic load, TL
• Horizontal load, e.g., Wind load, Earthquake, Vehicle collision,
moving debris
25

26. Example 1.2 – Fibre stress due to prestressing and
self weight
• A simply supported beam has the cross section as shown below. The beam is
prestressed with P and e as given.
• Calculate the maximum fibre stresses due to prestressing and selfweight of
the beam. Use concrete density = 25 kN/m3
Solution:
• Calculate SW = 25*0.5*0.8 = 10 kN/m
• Max moment due to SW is at midspan,
MSW = SW*L2/8 = 10*202/8 = 500kN‐m
• Recall: elastic stress due to Moment,
L=20m
SW
P
cgc
I
  
My
P
P =
1500 kN
Section
500
26
800
cgc
350

27. Pey
min
= 3.75 - 9.84 + 9.375
= 3.28MPa (c)
max
= 3.75 + 9.84 - 9.375
= 4.22MPa(c)
c (+ve)
= ±500E6*400/(2.13E10)
= ± 9.375 MPa(max)
I
t (-ve)
M SWy
Example 1.2 ctd.
• Calculate fibre stresses
total stress = stress due to P + stress due to P*e + stress due to MSW
26
Section
cgc
0
P =
1500 kN
500
80035
t (-ve)
c (+ve)
P
A
= 3.75
c (+ve)
I
= ±9.84(max)
The max stresses are to be checked against the permissible
stresses specified in design standards
top A
= P
- P.e
+ MG
Ztop Ztop
bot A
= P
+ P.e
- MG
Zbot Zbot
y
 = +
A I
- G
P P.e.y M .y
I

28. Design Concept of PC members
cgc
e
Section Forces
Fibre stresses calculation
L
SW + SDL + LL
P
cgc
P
P
Pe MDL M
Pey
I
t
t
c
MDL y
t
LL c (+ve) c t
c
I I
Stresses due to bending
MLL y
P
A
Uniform stress
b
Total stress
Question:
when to use + , when – ?
28
I
y
M
I
y
M
I
Pey
A
P LL
DL





(check both comp and tens)

29. Design Concept of PC members
Notes:
This method is based on the elastic
stress check of the section.
This design method is aka “Limit State
of Decompression”. Any load beyond
this state will cause cracking in the
section (the moment of rupture of
concrete is reached due to cracking
moment).
Ultimate strength design method will
be covered in Section 4 of the lecture.
29
• If stresses are greater than
allowable, i.e.,   allow
• Need to redesign by (either one
or combination of the following):
• Change P and/or e
• Change concrete section
• Change material properties (e.g.
select stronger strength grade)
• Then, recheck fibre stresses until
  allow

30. 30
Prestressing forces
• Initial prestressing Pi: the force
after initial losses.
At this state, the PC member is
usually subjected to its own weight
only.
This state is called “transfer state”.
• Thus, the prestressing force should
be Pi in Example 1.2.
• Jacking force Pj: the total force • Effective prestressing Pe: the force
applied to stress the tendon when the member is at
serviceability, i.e., under all the
superimposed loadings.
All losses must be taken into
account at this state.
This state is called “service state”
• More details on how to calculate
prestress loss will be covered later.

31. Kern Limit
• Kern limit is the limit of the
prestressing zone in the cross‐
section.
• It is an envelope within which the
applied prestressing force will
not cause any tension in concrete
• How to calculate: Recall
• Lower kern limit means that if
the prestressing force is applied
further from this limit, there will
be tension in concrete (in this
case tension is at the top fibre).
• Upper kern limit is as seen
below: (in this case bottom fibre
is in tension and must be zero)
cgc
e
Section
Pe
P
cgc
e
Section
Pe
P
31
0



I
Pey
A
P t
t

t
b
Ay
I
e
e 

lower kern limit
I
Pey
A
P



0



I
Pey
A
P b
b

b
t
Ay
I
e
e 

upper kern limit

32. Tendon Profile (Envelopes)
• Tendon envelopes are the limiting eccentricities along the span
such that there is no tension in the extreme fibre
• Calculated as the shift of kern limits when the section is
subjected to dead load and/or live load moments
• The envelopes depend on the applied moments
Examples: the shift of lower and upper kern limits due to dead
load moment
31
cgc
e
Section
P
Pe MDL
cgc
Section
e P
Pe MDL
0




I
y
M
I
Pey
A
P t
DL
t
t

P
M
e
P
M
Ay
I
e
e DL
b
DL
t
b 



 *
0




I
y
M
I
Pey
A
P b
DL
b
b

P
M
e
P
M
Ay
I
e
e DL
t
DL
b
t 



 *

33. Design Concept of PC members
Examples: the shift of lower and upper kern limits
due to hogging moment
cgc
e Pe
P cgce
Section
Pe
P
MDL
Based on this knowledge, tendon envelopes along the span can be
drawn
Question: What is the shift if there is an additional live load moment?
+ve for upper kern
MDL +ve for lowerkern
Sign convention ?
33
0




I
y
M
I
Pey
A
P t
DL
t
t

P
M
e
P
M
Ay
I
e
e DL
b
DL
t
b 



 *
0




I
y
M
I
Pey
A
P b
DL
b
b

P
M
e
P
M
Ay
I
e
e DL
t
DL
b
t 



 *

34. Example 1.3
L = 20m
450
34
1000
150
150
150
100
Section
(not‐to‐scale)
Draw the tendon envelopes (the shift of upper and lower kern limits) along the span of
the beam shown. The beam is subjected to dead load and live load, note that the
load combinations are 1) DL only, and 2) DL+LL. The elastic stress in concrete must
be in compression at all time (tensile stress is zero)
DL and LL
SDL = 1.5kN/m
LL = 15.0kN/m

35. 3
4

36. 35
Allowable tensile stress

37. 37
Limit States from AS 3600‐2018
Prestressed concrete members, like all other concrete structures in Australia, are
designed to comply with AS3600, which lays down certain limit states and minimum
requirements. The limit states include:
• Strength at transfer of prestress
• Strength in bending
• Serviceability – cracking or crack control
• Serviceability – deflection
• Shear strength
In addition, prestressed beams must comply with rules for fire resistance and
durability, and may even be affected by concerns for vibration stability and fatigue.

38. 38
Stresses Calculation
General Design Assumptions
• Strains vary linearly acros.s the section
• Cross sectional properties may be
taken as gross properties ignoring the
transformed area of steel.
• There is complete adhesion of the
tendons to the concrete.
• No sudden changes in cross section
Design Situations are:
• Transfer state – stresses limit,
hogging/sagging deflection limit
• Service state ‐ when under service load
and prestress after long‐term loss
• Partial prestressing – limited control of
cracking, deflection control
• Full prestressing – full control of
cracking, deflection control

39. 39
Transfer Limit State
• AS 3600‐2018 [8.1.6.2] permits transfer to be checked by limiting the maximum
compressive stress to 0.5 fcp.
• Clause 2.5.2 noted the importance of the load at transfer, wtr, is the larger of:
• 1.15G + 1.15P
• 0.9G + 1.15P
• Where the prestress is the initial prestress Pi.
• Take note of Clause [6.2.6] on the Secondary Bending Moments and Shears
resulting from prestress.
• Where applicable, prestressing effect shall be included with a load factor of
unity (1) in all load combinations for both ultimate and serviceability design
(load combinations in accordance with AS/NZS 1170.0).

40. Transfer Limit State
300
10000 mm
Prestress = 4  12.7 mm super strands at 350 mm spacing:
From Table [3.3.1]: Pu = 184 kN  4 / 0.350 m = 2103 kN/m
Given: Initial Prestressed force, Pi = 1694 kN/m
Effective Prestressed force, Pe = 1337 kN/m
e = 120.5 mm
Example 1.4: Slab under prestressed action.
12 kN/m2 Imposedload
40
c.g.

42. 42
Transfer Limit State
= 300 mm
= 270.5 mm (depth of centre of cable)
Example 1.4:
Dimensions:
D
dp
e = dp –yC.G. = 120.5 mm (from C.G. to cable)
Section Properties:
= 0.3 m2 per metre width ofslab
= bD3/12 = 1000  3003/12 = 2250  106 mm4 per meter width
A
I
Z = I/y = 2250  106 /(300/2) = 15  106 mm4 per meter width

43. 43
Transfer Limit State
Example 1.4:
Loading:
WG
WQ
WTR
= 1  0.3  25 = 7.5 kN/m per meterwidth
= 1  12 = 12 kN/m per meter width
= Larger of [1.15G + 1.15P; 0.9G + 1.15P]
Transfer unfactored moment, MTR = WGL2/8 = 7.5  102/8
= 93.75 kNm per meter width
Materials: f’c = 32MPa
fcp = 27MPa
(compressive strength at transfer depends on the age of concrete at transfer)

44. Transfer Limit State
Example 1.4:
At transfer, for the load combination of 1.15G + 1.15P:
Stress at bottom, bot. A Zbot.
=
1.15Pi
+
1.15Pi.e
-
1.15MG
=
1.15X1694X1000
0.3X10002
+
Zbot.
1.15X1694X1000X120.5
15X106
-
1.15X93.75X106
15X106
= 6.49 +15.65 - 7.19
= 14.95 MPa
44

45. Transfer Limit State
Example 1.4:
At transfer, for the load combination of 0.9G + 1.15P:
Stress at bottom, bot. A Zbot.
=
1.15Pi
+
1.15Pi.e
-
0.9MG
=
1.15X1694X1000
0.3X10002
Zbot.
+
1.15X1694X1000X120.5
15X106
-
0.9X93.75X106
15X106
= 6.49 + 15.65 - 5.63
=16.51 MPa
From Clause 2.4.2, maximum stress at transfer is larger of [14.95 MPa or 16.51
MPa] = 16.51 MPa in compression.
45
max = 16.51 Mpa > 0.5 fcp = 13.5MPa
Stress at Transfer not OK. What to do?

46. Cracking Limit State
Cracking is checked under the short term service load, Wsev = WG+ sWQ,
and the simplified requirements are that the limiting tensile stress should
be:
 0.25 f , ‐ no requirement [8.6.3 & 9.5.2.3]
C
Otherwise provide reinforcement or bonded tendons near the tensile face
with centre‐to‐centre spacing not exceeding 300 mm and limiting either:
  0.60 f , for beams [8.6.3] or 0.50 f , for slabs[9.5.2.3]
C C
 Increment steel stress near the tension face to Table [8.6.3] for beams
and Table [9.5.2.3] for slabs.
46

47. Cracking Limit State
300
10000 mm
Prestress = 4  12.7 mm super strands at 350 mm spacing:
From Table [3.3.1]: Pu = 184 kN  4 /0.350 m = 2103 kN/m
Given: Initial Prestressed force, Pi = 1694 kN/m
Effective Prestressed force, Pe = 1337 kN/m
e = 120.5 mm
Example 1.5: From previous example
12 kN/m2 Imposedload
47
c.g.

48. Cracking Limit State
Example 1.5: From previous example
Wsev = WG+ sWQ = 7.5 + 0.7  12 = 15.9 kN/m2 (kN/m per meter width)
Note: s is the short-term factor, Table 4.1 (AS 1170.0)
Msev =WsevL2/8 = 15.9  102/8 = 198.8 kNm per meterwidth
Stress at bottom, bot. A
= Pe
+ Pe.e
- Msev
Zbot. Zbot.
= 0.3X10002
+ -
1337X1000 1337X1000X120.5 198.8X106
15X106 15X106
= 4.46 + 10.74 - 13.25
C
48
= 1.95 Mpa < 0.50 f , = 2.83MPa
Therefore, the section is uncracked.
At service load,

49. 47
Flexural Strength Design Provision in AS3600‐2018
• Strength of beams in bending [8.1]
• Strength of slabs in bending [9.1]
• Points to note:
– Rectangular stress block
– Modification factors, 2 and 
– Min strength requirement (e.g., min Ast)
– Crack control
– Durability requirements, environmental exposure, fire rating

50. Equilibrium: C = T
α2.f’c. bdn + Asc.sc = Ast.st + Ap.py
Taking moment about top edge:
Mu = (Ast.st) × dst + (Ap.py) × dp – (α2.f’c. bdn) × (dn/2) – (Asc.sc) × dsc
Flexural Strength of RC & PC Concrete Section
50
• Rectangular Stress Block

51. Total Strain in Prestressing Tendon
Total Strain of the Prestressing Tendon, p
• Strain in tendon due to prestressing force (in tension), pe =
𝜎𝑝𝑒
𝐸𝑝
• Strain in concrete due to prestressing force (in compression), ce =
𝜎𝑐𝑒
𝐸𝑐
; 𝜎𝑐𝑒 =
𝑃𝑒
𝐴𝑔
+
𝑃𝑒.𝑒2
𝐼𝑔
• Strain in concrete due to prestressing force and moment (in tension), cp =
𝜀0.(𝑑𝑝−𝑑𝑛)
𝑑
𝑛
• Total strain in tendon (in tension), p = pe + ce + cp
51

52. Shear Design Provision in AS3600‐2018
eq
• Strength of beams in shear [8.2]
• Strength of slabs in shear [9.3]
• Points to note:
– Torsional effects ( ) – find
equivalent shear force V∗
– Prestress Pv contributes to shear
resistance
– Shear due to concrete strength Vuc
– Shear strength limited by web crushing
Vu.max
– Shear due to transverse reinforcement
Vus
• Hanging Reinforcement – hanging
load to transfer to top chord [8.2.6]
• Additional longitudinal tension
forces caused by shear, Ftd [8.2.7]
• Proportioning longitudinal
reinforcement [8.2.8] on tension
side Ttd and compression sideTcd
50

53. Serviceability – Deflection Limit
• Refined calculation – iterative
process
• Simplified – short term and long
term deflection, Icr, Ief, kcs, sus,
• Deemed to conform span‐to‐
depth ratios
Deflection of beams [8.5] Deflection of slabs [9.4]
• Refined calculation
• Simplified – equivalent
beam/strip
• Deemed to conform span‐to‐
depth ratios [9.4.4.1]

54. Crack Control
300
(RC Detailing Handbook)
Crack control of Beams [8.6]
• Limit max crack width
• Min tensile steel reinforcement
• Limit max tensile steel stress
• Spacing of reinforcement from edge
or soffit  100 mm. Centre to centre
spacing of reinforcement must not
exceed 300 mm.
Other requirements: vibration [8.7],
flange width [8.8], slenderness [8.9]

55. Crack Control
Crack control of Slabs [9.5]
• Limited crack width
• Min reinforcement and spacing of reinforcement must not exceed 2Ds or 300mm.
• Limited tensile steel stress
• Shrinkage and temperature effects
Other requirements: vibration [9.6], slab width supporting point loads [9.7], longitudinal shear
in slab [9.8]
Fig 14:3 RC Detailing Handbook

56. Summary
• Review of RC Design
• Review of PC Design, Limit State of
Decompression
• Calculation of fibre stresses, Kern
Limit, Tendon profile
• Transfer Limit State
• Cracking Limit State
• Strength Limit State
• Serviceability
54
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sand that can
control humidity
and deodorise
the air.