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2. Prestressed Concrete:
Beam in bending at
working load
Introduction
Bending stresses in an uncracked section
Jacking force Po, initial force Pi and
effective force Pe.
Two important properties of a prestressed
section
Stress distribution across an uncracked section
University of Western AustraliaSchool of Civil and Resource Engineering 2004
INTRODUCTION
Basic assumptions:
Bernoulli/Navier postulate ( plane sections remain plane)
Linear performance for concrete
Tendon stress unaltered through range from transfer to working load
sc
ec
f c
Approx.
linear to
0.5 f c
NB Demonstrating that stresses are
acceptable at working load does not
ensure that section has adequate
safety. This is treated later in the
course.
Lecture 5
So how to estimate the bending stresses at working load? . . .
BENDING STRESSES IN AN
UNCRACKED SECTION
Start by assuming that the section is uncracked and linear, then demonstrate
that each is true.
Limiting stresses:
Concrete is linear up to about 0.5 fc where f c = current strength
Two conditions of concern: At transfer, when f c = f cp and
At maturity, when f c = f c
At transfer to ensure linearity : sb = - 0.25 (f cp)0.5
At maturity to ensure linearity : sa = - 0.25 (f c)0.5
sc
ec
f c
0.5 f cCONCRETE
Consider this example . . .
centroidal
axis
Tendon stressed
to force P
applied loading
Simple Beam - stresses in working load range
Examine this typical section, where
the applied bending moment = M
This section may be anywhere along the beam,
EXCEPT very close to the supports . . .
applied loading
axis
Examine this section, whereTendon stressed
to force P
centroidal
applied moment = M.
Simple Beam - actions at a typical section
Stresses due to prestress only:
e
qP R
Components of
force on concrete
Resultants at
section
P Pe
Pq
From these actions we can work out the stresses . . .
C = Rcosq = P
V = Rsinq= Pq
Prestress force
sa
sb
P
Pe
P/A Pe / Z
Pe / Z
= P/A - Pe/Z
= P/A + Pe/Z
(Pq does not cause
any bending at the
section)
So top fibre stress sa = P/A - Pe/Z and
bottom fibre stress sb = P/A + Pe/Z
NOTE:
Position of neutral axis is high in the section.
Some tension may occur high in the section.
Stresses depend on the eccentricity of prestress at section.
Now consider stresses due to applied loading . . .
Simple Beam - stresses in working load range
Stresses due to prestress only:
applied loading
reaction
MM / Z
M / Z
These stresses are now added
to stresses due to prestress to
give the total stresses . . . .
Simple Beam - stresses in working load range
= P/A + Pe/Z - M/Z
applied loading
reaction
P
M
V
M / Z
M / Z
sa
sb
Stresses due to: Prestress Applied loading Combined
So top fibre stress
s
a = P/A - Pe/Z + M/Z and
bottom fibre stress
s
b
Provided that section is not cracked in tension, AND . . .
. . . concrete is not too highly stressed in compression.
Simple Beam - stresses in working load range
JACKING FORCE PoINITIAL FORCE PiEFFECTIVE FORCE Pe Jack applies force to
tendon at the live end.
The maximum value
of this force is the
Jacking Force Po
Immediately after jacking
and lock-off are completed,
tendon force at typical
location is less than Po due to
friction losses and anchorage
losses. This tendon force is
called Initial Force Pi
Tendon force progressively
diminishes with time due to
shrinkage and creep of concrete, and
relaxation of tendon. Ultimately the
force settles at a value called the
Effective Force Pe
Important for
anchorage design
Important at transfer
Important for maximum loads
Pe = 0.75 - 0.85 Pi approx.
Now we know what were doing!
Ok!
Lets try an example!
10.000
Study of stresses at mid-span of beam :
Applied working live load w = 25 kN/m (+ s/wt )
parabolically draped
tendon, stressed to
Pi of 1250 kN,
Pe 1000 kN, at mid-span.
e = 0 e = 0e = 225
BMD due to LL of 25 kN/m
312.5 kNm
400
750
e varies 0 to 225 mm
Note that the maximum eccentricity e maxis governed by the need to provide
enough space below the duct to place
concrete, and any secondary
reinforcement which may be needed.
Example 2.1
400
750
Stresses due to prestress only :
sa = P/A - Pe/Z
-3.33
sb = P/A + Pe/Z +11.67
NOTE:
Neutral
axis . . .
Pi = 1250 kN
. . . does not coincide
with centroidal axis
This is fictitious case, since self-weight of beam is engaged
when the beam is stressed and curves upwards . . .
So . . . .
Stresses due to p/s + s/wt of beam :
sa = P/A - Pe/Z + Mswt / Z400
750
-0.83
sb = P/A + Pe/Z - Mswt / Z
+9.17
NOTE:
both top and bottom fibre stresses are reduced by self-weight.
neutral axis is even higher.
Now what about stresses due to applied load? . . . .
Bending stresses
prestress
only
Stresses due to p/s, s/wt and applied load :
Mtot = Mswt + Mapplied = 93.75 + 312.5 = 406.25 kNm
sa = P/A - Pe/Z + Mtot /Z+8.16
sb = P/A + Pe/Z - Mtot /Z
-1.50
400
750
NOTE:
stress reversals at extreme fibres, so
movement of neutral axis.
Now check stresses to ensure our assumptions are o.k. . .
Bending stresses
prestress only
p/s and s/wt
Our stress estimates will not be valid if, at any time j,
Compressive stress anywhere exceeds 0.5 f cj , since above that stress the stress /strain curve for concrete is not linear, or
Tensile stress anywhere exceeds 0.25 (fcj ) 0.5, since above that tensile
stress the section may crack.
where fcj is compressive strength at that time j.
So we must check stresses at
Time of transfer, when f cp is probably less than f c, and
At full load, when concrete strength = f c.
If excessive stress is indicated anywhere,
then a different approach will be required
Lecture 4
Now consider stresses across the entire section . . .
Bending stresses Checking of stresses :
STRESS DISTRIBUTION ACROSS
AN UNCRACKED SECTION
Consider again Example 2.1. The applied load is all live load, (except for the
self-weight). Remember that live load may or may not be present. So in the
mature life of the beam, there are a range of extreme fibre stresses to be
considered.
The bending stress at any distance y below the
centroidal axis is:
sy = P/A + Pey/I - My/I
So the stress varies linearly with depth y
The range of stresses to be considered (for a simply
supported beam) is:
From MQ = 0 to MQ = MQ max
400
750
-0.17 +8.16
-1.50 +6.83
Life-time range of stresses :
Live load may, or may not, be present. So stresses will vary
throughout the beams life. The full range is displayed by a
diagram which pivots about the centroidal axis.
Note that there is a particular value of w, somewhere
between 0 and 25 kN/m for which the stress will be
uniformly compressive. More on this next lecture.
Example 2.1
total
p/s + s/wt
TWO IMPORTANT PROPERTIES
OF A PRESTRESSED SECTION
Decompression moment Mo or Mdec :
The total moment which is just enough to
eliminate tension in the bottom fibre:
sa
sb = 0
sb = 0 so P/A + Pe/Z - Mdec/Z = 0
then M dec = Z [P/A + Pe/Z]
Cracking moment Mcr :
The total moment which is just enough to
cause tensile cracking in the bottom fibre
sa
sb = - f cf
sb = - f cf so P/A + Pe/Z - Mcr/Z = - f cf
then Mcr = Z [P/A + Pe/Z + f cf ]
Rectangular section T section
I sectionBox section
Some Typical Prestressed Beam Sections :
y
Top fibre stress : sa = P/A - Pe/Ztop + M/Ztop , and
Bottom fibre stress : sb = P/A + Pe/Zbott - M/Zbott
Z is section modulus.
For sections which
are not symmetrical
about the horizontal
axis, Z is different
for top and bottom.
Z top = I / y top, and
Z bott = I / y bott.
Also, stress at any level y below centroidal axis :
sy = P/A + Pey/I - My/I
Enough for
today !
Bending stresses
y
y
Bending stresses must be checked for the transfer condition, and the maximum load condition.
Stresses must be checked to ensure assumptions of linearity and non-cracking are valid.
Simple beam theory is appropriate, noting that the action caused by prestress may be modelled in terms
of an axial load applied at the centroidal axis, and a
bending action.
Mdec and Mcr are important properties of a prestessed section.
This theory may be applied to any prismatic member.
SUMMARY