1 T Beams Reinforced concrete floor systems normally consist of slabs and beams that are placed monolithically As a result the two parts act together to resist loads In effect the beams have ID: 284907
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Slide1
T Beams
1Slide2
T Beams Reinforced
concrete floor systems normally consist of slabs and beams that are placed
monolithically. As
a result, the two parts act together to resist loads. In effect, the beams have extra widths at their tops, called flanges, and the resulting T-shaped beams are called T beams. The part of a T beam below the slab is referred to as the web or stem. (The beams may be L shaped if the stem is at the end of a slab.) The stirrups in the webs extend up into the slabs, as perhaps do bent-up bars, with the result that they further make the beams and slabs act together.
2
There
is a problem involved in estimating how much of the slab acts as part of the
beam. Should
the flanges of a T beam be rather stocky and compact in cross section, bending
stresses will
be fairly uniformly distributed across the compression zone.Slide3
T Beams If
, however, the flanges
are wide
and thin, bending stresses will vary quite a bit across the flange due to shear deformations. The farther a particular part of the slab or flange is away from the stem, the smaller will be its bending stress.3 Instead of considering a varying stress distribution across the full width of the flange, the ACI Code (8.12.2) calls for a smaller width with an assumed uniform stress distribution for
design purposes. The objective is to have the same total compression force in the reduced width that actually occurs in the full width with its varying stresses.
The
hatched area in Figure 5.1 shows the effective size of a T beam. For T beams
with flanges
on both sides of the web, the code states that the effective flange width may not
exceed one-fourth
of the beam span, and the overhanging width on each side may not exceed
eight times Slide4
T Beams4
the
slab thickness or one-half the clear distance to the next web. An isolated T
beam must have a flange thickness no less than one-half the web width, and its effective flange width may not be larger than four times the web width (ACI 8.12.4). If there is a flange on only one side of the web, the width of the overhanging flange cannot exceed one-twelfth the span, 6hf, or half the clear distance to the next web (ACI 8.12.3).
The analysis of T beams is quite similar to the analysis of rectangular beams in that the
specifications relating to the strains in the reinforcing are identical.
To repeat briefly, it
is desirable
to have
ϵ
t
values ≥ 0.005, and they may not be less than 0.004 unless the
member is
subjected to an axial load ≥
0.10
f’
c
A
g
. You will learn that
ϵ
t
values are almost always much larger than 0.005 in T beams because of their very large compression flanges.Slide5
T Beams5Slide6
T Beams6
For
such members, the values of c are normally very small, and calculated ϵt values very large. The neutral axis (N.A.) for T beams can fall either in the flange or in the stem, depending on the proportions of the slabs and stems. If it falls in the flange, and it almost always does for
positive moments, the rectangular beam formulas apply, as can be seen in Figure 5.2(a). The concrete below the neutral axis is assumed to be cracked, and its shape has no effect
on the
flexure calculations (other than weight). The section above the neutral axis is
rectangular. If
the neutral axis is below the flange, however, as shown for the beam of Figure 5.2(b),
the compression
concrete above the neutral axis no longer consists of a single rectangle, and
thus the
normal rectangular beam expressions do not apply.Slide7
T Beams7
If the neutral axis is assumed to fall within the flange, the value of
a can be computed as it was for rectangular beams: The distance to the neutral axis, c, equals a/β1. If the computed value of a is equal to or
less than the flange thickness, the section for all practical purposes can be assumed to be rectangular, even though the computed value of c
is actually greater than the flange thickness.
A
beam does not really have to look like a T beam to be one. This fact is shown by
the beam
cross sections shown in Figure 5.3. For these cases the compression concrete is T
shaped, and
the shape or size of the concrete on the tension side, which is assumed to be cracked,
has no
effect on the theoretical resisting moments. Slide8
T Beams8
It
is true, however, that the shapes, sizes,
and weights of the tensile concrete do affect the deflections that occur (as is described in Chapter 6), and their dead weights affect the magnitudes of the moments to be resisted.Slide9
Analysis of T Beams
The calculation of the design strengths of T beams is illustrated in
next two examples. In the
first of these problems, the neutral axis falls in the flange, while for the second, it is in the web. The procedure used for both examples involves the following steps:91. Check As min as per ACI Section 10.5.1 using bw as the web width.
2. Compute
T
=
A
s
f
y
.
3
.
Determine the area of the concrete in compression (
A
c
)
stressed
to 0.85
f’
c
.
4.
Calculate
a
,
c
, and
ϵ
r
.
5
.
Calculate
φ
M
n
.Slide10
Analysis of T Beams In the example,
where the neutral axis falls in the flange, it would be logical to
apply the
normal rectangular equations, a couple of method as a background for the solution of next example are used, where the neutral axis falls in the web. This same procedure can be used for determining the design strengths of tensilely reinforced concrete beams of any shape (Т, Г, П, triangular, circular, etc.).
10Slide11
Analysis of T Beams
Determine the design strength of the T beam shown in Figure 5.4, with
f’
c = 4000 psi and fy = 60,000 psi. The beam has a 30-ft span and is cast integrally with a floor slab that is 4 in. thick. The clear distance between webs is 50 in.11
Example 5.1
SolutionSlide12
Analysis of T Beams12Slide13
Analysis of T Beams13Slide14
Analysis of T Beams Compute the design strength for the T beam shown in
Figure,
in which
f’c = 4000 psi and fy = 60,000 psi.14Example
5.2
SolutionSlide15
Analysis of T Beams15Slide16
Analysis of T Beams16Slide17
Another Method for Analyzing T Beams
The preceding section presented
an
important method of analyzing reinforced concrete beams. It is a general method that is applicable to tensilely reinforced beams of any cross section, including T beams. T beams are so very common, however, that many designers prefer another method that is specifically designed for T beams.17 First, the value of
a is determined as previously described in this chapter. Should it be less than the flange thickness,
h
f
,
we will have a rectangular beam and the rectangular
beam formulas
will apply. Should it be greater than the flange thickness,
h
f
(as was the case
for previous example),
the special method to be described here will be very useful.Slide18
Another Method for Analyzing T Beams
The beam is divided into a set of rectangular parts consisting of the overhanging
parts of
the flange and the compression part of the web (as in next slide). The total compression, Cw, in the web rectangle, and the total compression in the overhanging flange, Cf, are computed:18
Then the nominal moment, Mn
, is determined by multiplying
C
w
and
C
f
by their
respective lever
arms from their centroids to the centroid of the steel:Slide19
Another Method for Analyzing T Beams
This procedure is illustrated in
next example.
Although it seems to offer little advantage in computing Mn, we will learn that it does simplify the design of T beams when a > hf because it permits a direct solution of an otherwise trial-and-error problem.19Slide20
Another Method for Analyzing T Beams
Repeat
previous example
using the value of a (8.19 in.) previously obtained and the alternate formulas just developed. Reference is made to Figure 5.8, the dimensions of which were taken from Figure 5.5.20Example 5.3
SolutionSlide21
Another Method for Analyzing T Beams21Slide22
22Slide23
Design of T Beams
For the design of T beams, the flange has normally already been selected in the slab design, as
it is
for the slab. The size of the web is normally not selected on the basis of moment requirements but probably is given an area based on shear requirements; that is, a sufficient area is used so as to provide a certain minimum shear capacity. It is also possible that the width of the web may be selected on the basis of the width estimated to be needed to put in the
reinforcing bars. Sizes may also have been preselected, as previously described, to
simplify formwork for architectural requirements or for deflection reasons. For the
examples that follow
the values of
h
f
,
d,
and
b
w
are given.
23Slide24
Design of T Beams
The flanges of most T beams are usually so large that the neutral axis probably
falls within
the flange, and thus the rectangular beam formulas apply. Should the neutral axis fall within the web, a trial-and-error process is often used for the design. In this process, a lever arm from the center of gravity of the compression block to the center of gravity of the steel is estimated to equal the larger of 0.9d or d − (
hf
/
2
)
,
and from this value,
called
z
, a trial
steel area
is calculated
(
A
s
=
M
n
/
f
y
z
)
. Then by the process
as used earlier,
the value of
the estimated
lever arm is
checked
. If there is much difference, the estimated value of
z
is
revised and
a new
As determined. This process is continued until the change in As
is quite small.24Slide25
Design of T Beams Often
a T beam is part of a continuous beam that spans over interior supports, such
as columns
. The bending moment over the support is negative, so the flange is in tension. Also, the magnitude of the negative moment is usually larger than that of the positive moment near midspan. This situation will control the design of the T beam because the depth and web width will be determined for this case. Then, when the beam is designed for positive moment at midspan, the width and depth are already known.
25
Example
to follow
presents a more direct approach for the case where
a >
h
f
.
This is the
case where
the beam is assumed to be divided into its rectangular parts.Slide26
Design of T Beams26
Example 5.4
SolutionSlide27
Design of T Beams27Slide28
Design of T Beams28Slide29
Design of T Beams29Slide30
Design of T Beams30Slide31
Design of T Beams31
Example
5.5
SolutionSlide32
Design of T Beams32Slide33
Design of T Beams33Slide34
Design of T Beams34Slide35
Design of T Beams35
Our
procedure for designing T beams has been to assume a value of
z, compute a trial steel area of As, determine a for that steel area assuming a rectangular section, and so on. Should a > hf, we will have a real T beam
and a trial-and-error process was used. It is easily possible, however, to determine
As
directly using the
method of
Section 5.3, where the member was broken down into its rectangular components.
The compression force provided by the overhanging flange rectangles must be
balanced by
the tensile force in part of the tensile steel,
A
sf
,
while the compression force in the web
is balanced
by the tensile force in the remaining tensile steel,
A
sw
.Slide36
Design of T Beams36
For
the overhanging flange, we have
.
from which the required area of steel,
A
sf
,
equals
The design strength of these overhanging flanges is
The remaining moment to be resisted by the web of the T beam and the steel
required to
balance that value are determined next.Slide37
Design of T Beams37
The
steel required to balance the moment in the rectangular web is obtained by the
usual rectangular beam expression. The value Muw/φbwd² is computed, and ρw is determined from the appropriate Appendix table or the expression for ρw
previously given in Section 3.4 of this book. Think of ρw
as the reinforcement ratio for the beam shown in Figure 5.7(b). ThenSlide38
Design of T Beams38
Example 5.6
Rework Example 5.5 using the rectangular component method just described.
SolutionFirst assume a ≤
hf (which is very often the case). Then the design would proceed like that of a rectangular
beam with a width equal to the effective width of the T-beam flange.
The beam acts like a T beam, not a rectangular beam, and the values for
ρ
and
a
above are
not correct. If the value of
a
had been ≤
h
f
, the value of
A
s
would have been
simply
ρbd
= 0.0072(54 in.) (24 in.) = 9.33
in.².
Now break the beam up into two parts (Figure 5.7)
and design
it as a T beam.Slide39
39Slide40
Design of T Beams40Slide41
Design of T Beams for Negative Moments
When T beams are resisting negative moments, their flanges will be in tension and the
bottom of
their stems will be in compression, as shown in Figure 5.12. Obviously, for such situations, the rectangular beam design formulas will be used. Section 10.6.6 of the ACI Code requires that part of the flexural steel in the top of the beam in the negative-moment region be distributed over the effective width of the flange or over a width equal to one-tenth of the beam span, whichever is smaller. Should the effective width be greater than one-tenth of the span length, the code requires that some additional longitudinal steel be placed in the outer portions of the flange. The intention of this part of the code is to minimize the sizes of the flexural cracks that will occur in the top surface of the flange perpendicular to the stem of a T beam
subject to negative moments.
41Slide42
Design of T Beams for Negative Moments
In Section 3.8, it was stated that if a rectangular section had a very small amount
of tensile
reinforcing, its design-resisting moment, φMn, might very well be less than its cracking moment. If this were the case, the beam might fail without warning when the first crack occurred. The same situation applies to T beams with a very small amount of tensile reinforcing.42
When the flange of a T beam is in tension, the amount of tensile reinforcing needed to make its ultimate resisting moment equal to its cracking moment is about twice that
of a
rectangular section or that of a T section with its flange in compression. As a result,
ACI
Section 10.5.1 states that the minimum amount of reinforcing required equals the larger of
the two
values that follow:Slide43
Design of T Beams for Negative Moments43
For statically determinate members with their flanges in tension,
bw in the above expression is to be replaced with either 2bw or the width of the flange, whichever is smaller.Slide44
L-Shaped Beams The author assumes for this discussion that L beams (i.e., edge T beams with a flange on
one side
only) are not free to bend laterally. Thus they will bend about their horizontal axes
and will be handled as symmetrical sections, exactly as with T beams. For L beams, the effective width of the overhanging flange may not be larger than one-twelfth the span length of the beam, six times the slab thickness, or one-half the clear distance to the next web (ACI 8.12.3).44
If an L beam is assumed to be free to deflect both vertically and horizontally, it will be necessary to analyze it as an unsymmetrical section with bending about both the
horizontal and
vertical axes. Slide45
Compression Steel
The steel that is occasionally used on the compression sides of beams is called
compression steel
, and beams with both tensile and compressive steel are referred to as doubly reinforced beams. Compression steel is not normally required in sections designed by the strength method because use of the full compressive strength of the concrete decidedly decreases the need for such reinforcement, as compared to designs made with the working-stress design method.45 Occasionally, however,
space or aesthetic requirements limit beams to such small sizes that
compression steel is needed in addition to tensile steel.
To increase the moment
capacity of
a beam beyond that of a
tensilely
reinforced beam with the maximum percentage of
steel [when
(
ϵ
t
=
0.005
)
], it is necessary to introduce another resisting couple in the beam.Slide46
Compression SteelThis
is done
by adding steel in both the compression and tensile sides of the beam. Compressive
steel increases not only the resisting moments of concrete sections but also the amount of curvature that a member can take before flexural failure. This means that the ductility of such sections will be appreciably increased. Though expensive, compression steel makes beams tough and ductile, enabling them to withstand large moments, deformations, and stress reversals such as might occur during earthquakes. As a result, many building codes for earthquake zones require that certain minimum amounts of compression steel be included in flexural members.
46Slide47
Compression Steel
Compression
steel is very effective in reducing long-term deflections due to
shrinkage and plastic flow. In this regard you should note the effect of compression steel on the long-term deflection expression in Section 9.5.2.5 of the code (to be discussed in Chapter 6 of this text). Continuous compression bars are also helpful for positioning stirrups (by tying them to the compression bars) and keeping them in place during concrete placement and vibration.47
Tests of doubly reinforced concrete beams have shown that even if the
compression concrete
crushes, the beam may very well not collapse
if the compression steel is enclosed
by stirrups
. Once the compression concrete reaches its crushing strain, the concrete cover
spalls or
splits off the bars, much as in columns (see Chapter 9). Slide48
Compression Steel48
If
the compression bars are
confined by closely spaced stirrups, the bars will not buckle until additional moment is applied. This additional moment cannot be considered in practice because beams are not practically useful after part of their concrete breaks off. (Would you like to use a building after some parts of the concrete beams have fallen on the floor?) Section 7.11.1 of the ACI Code states that compression steel in beams must be enclosed by
ties or stirrups or by welded wire fabric of equivalent area. In Section 7.10.5.1, the code states that the ties must be at least #3 in size for longitudinal bars #10 and smaller and
at least
#4 for larger longitudinal bars and bundled longitudinal bars.
The ties may not be
spaced farther
apart than 16 bar diameters, 48 tie diameters, or the least dimension of the beam
cross section
(code 7.10.5.2).Slide49
Compression Steel49
For doubly reinforced beams, an
initial assumption is made that the compression steel yields as well as the tensile steel. (The tensile steel is always assumed to yield because of the ductile requirements of the ACI Code.) If the strain at the extreme fiber of the compression concrete is assumed to equal 0.00300 and the compression steel, A’s, is located two-thirds of the distance from the neutral axis to the extreme concrete fiber, then the strain in the
compression steel equals ⅔ × 0.003 = 0.002. If this is greater than the strain in the steel at yield, as
say 50,000
/(
29 ×
10⁶
)
= 0.00172 for 50,000-psi steel, the steel has yielded.
It should be noted
that actually
the creep and shrinkage occurring in the compression concrete help the
compression steel
to yield.Slide50
Compression Steel50
Sometimes the neutral axis is quite close to the compression steel. As a matter of
fact, in some beams with low steel percentages, the neutral axis may be right at the compression steel. For such cases, the addition of compression steel adds little, if any, moment capacity to the beam. It can, however, make the beam more ductile. When compression steel is used, the nominal resisting moment of the beam is assumed
to consist of two parts: the part due to the resistance of the compression concrete and the balancing tensile reinforcing, and the part due to the nominal moment capacity of the
compression steel
and the balancing amount of the additional tensile steel. This situation is illustrated
in Figure
5.13. In the expressions developed here, the effect of the concrete in
compression, which
is replaced by the compressive steel,
A’
s
,
is neglected.Slide51
Compression Steel51
This omission will cause us
to overestimate Mn by a very small and negligible amount (less than 1%). The first of the two resisting moments is illustrated in Figure 5.13(b).Slide52
52Slide53
Compression Steel53
The second resisting moment is that produced by the additional tensile and
compressive steel (As2 and A’s), which is presented in Figure 5.13(c).
Up to this point it has been assumed that the compression steel has reached its
yield stress
. If such is the case, the values of
A
s
2
and
A’
s
will be equal because the addition to
T
of
A
s
2
f
y
must be equal to the addition to
C
of
A’
s
f
y
for equilibrium. If the compression steel
has not
yielded,
As
must be larger than
A
s
2, as will be described later in this section. Combining the two values, we obtainSlide54
Compression Steel54
The
addition of compression steel only on the compression side of a beam will have little effect on the nominal resisting moment of the section. The lever arm, z, of the internal couple is not affected very much by the presence of the compression steel, and the value of T will remain the same. Thus, the value Mn = T
z will change very little. To increase the nominal resisting moment of a section, it is necessary to add reinforcing on both the tension and
the compression
sides of the beam, thus providing another resisting moment couple.Slide55
Compression Steel55
With
the strain
obtained, the compression steel stress, f’s , is determined, and the value of As2 is computed with the following expression: In addition, it is necessary to compute the strain in the tensile steel,
ϵt , because if it is less
than 0.005, the value of the bending,
φ
, will have to be computed, inasmuch as it will
be less
than its usual 0.90 value.
The beam may not be used in the unlikely event that
ϵ
t
is
less than
0.004.
To determine the value of these strains, an equilibrium equation is written, which
upon solution
will yield the value of
c
and thus the location of the neutral axis. To write this
equation, the
nominal tensile strength of the beam is equated to its nominal compressive strength.
Only one
unknown appears in the equation, and that is
c
.Slide56
Compression Steel56
Initially
the stress in the compression steel is assumed to be at yield
(f’s = fy). From Figure 5.14, summing forces horizontally in the force diagram and substituting β1c for a leads to
Referring to the strain diagram of Figure 5.14, from similar triangles
If the strain in the compression
steel
ϵ
’
s
>
ϵ
y
=
f
y
/
E
s
, the assumption is valid and
f’
s
is
at yield
,
f
y
. If
ϵ
’
s
<
ϵ
y
, the
compression steel is not yielding, and the value of
c
calculated above is
not correct. A new equilibrium equation must be written that assumes
f’
s
<
f
y
.Slide57
Compression Steel57Slide58
Compression Steel58
The value of
c determined enables us to compute the strains in both the compression and tensile steels and thus their stresses. Even though the writing and solving of this equation are not too tedious, use of the Excel spreadsheet for beams with compression steel makes short work of the whole business. Examples 5.7 and 5.8 illustrate the computation of the design moment strength of doubly reinforced beams. In the first of these examples, the compression steel yields, while in
the second, it does not.Slide59
Compression Steel59
Example 5.7
SolutionSlide60
Compression Steel60Slide61
Compression Steel61Slide62
Compression Steel62
Example
5.8Slide63
Compression Steel63Slide64
Compression Steel64Slide65
Design of Doubly Reinforced Beams
Sufficient tensile steel can be placed in most beams so that compression steel is not
needed. But
if it is needed, the design is usually quite straightforward. Examples 5.9 and 5.10 illustrate the design of doubly reinforced beams. The solutions follow the theory used for analyzing doubly reinforced sections.65Slide66
Design of Doubly Reinforced Beams66
Example 5.9
SolutionSlide67
Design of Doubly Reinforced Beams67Slide68
Design of Doubly Reinforced Beams68
If
we had been able to select bars with exactly the same areas as calculated here,
ϵt
would have remained = 0.005 as originally assumed and
φ
= 0.90, but such was not the
case.
From
the equation for
c
in Section 5.7,
c
is found to equal 11.24 in. and
a
=
β
1
c
= 9.55
in. using
actual, not theoretical, bar areas for
A
s
and
A’
s
.Slide69
Design of Doubly Reinforced Beams69
The beam does not have sufficient capacity because of the variable
φ factor. This can be avoided
if you are careful in picking bars. Note that the actual value of As is exactly the same as the theoretical value. The actual value of
A
s
,
however, is higher than the theoretical
value by
10.12− 9.6 = 0.52
in.².
If a new bar selection for
As
is made whereby the actual value
of
A’
s
exceeds the theoretical value by about this much (0.52
in.²),
the design will be
adequate. Select
three #8 bars (
As
=
2.36
in.²)
and repeat the previous steps. Note that the actual
steel areas
are used below, not the theoretical ones. As a result, the values of
c
,
a
,
ϵ
’
s
,
and f’s must be recalculated.Slide70
Design of Doubly Reinforced Beams70Slide71
Design of Doubly Reinforced Beams71
Note that eight #10 bars will not fit in a single layer in this beam. If they were placed in
two layers
, the centroid would have to be more than 3 in. from the bottom of the section. It would be necessary to increase the beam depth, h, in order to provide for two layers or to use bundled bars (Section 7.4).Slide72
Design of Doubly Reinforced Beams72
Example
5.10
SolutionSlide73
Design of Doubly Reinforced Beams73