Transcript Chapter No 13 - WEC CIVILIANS

T Beams
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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.
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.
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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.
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
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T Beams
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 ϵt values ≥
0.005, and they may not be less than 0.004 unless the member is subjected
to an axial load ≥ 0.10f’cAg. You will learn that ϵt values are almost always
much larger than 0.005 in T beams because of their very large compression
flanges.
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T Beams
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T Beams
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.
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T Beams
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.
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T Beams
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.
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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:
1.
Check As min as per ACI Section 10.5.1 using bw as the web width.
2.
Compute T = Asfy .
3.
Determine the area of the concrete in compression (Ac) stressed
to 0.85f’c .
4.
Calculate a, c, and ϵr .
5.
Calculate φMn .
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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.).
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Analysis of T Beams
Example 5.1
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.
Solution
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Analysis of T Beams
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Analysis of T Beams
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Analysis of T Beams
Example 5.2
Compute the design strength for the T beam shown in Figure, in
which f’c = 4000 psi and fy = 60,000 psi.
Solution
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Analysis of T Beams
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Analysis of T Beams
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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.
First, the value of a is determined as previously described in this
chapter. Should it be less than the flange thickness, hf, we will have a
rectangular beam and the rectangular beam formulas will apply. Should it
be greater than the flange thickness, hf (as was the case for previous
example), the special method to be described here will be very useful.
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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:
Then the nominal moment, Mn , is determined by multiplying Cw
and Cf by their respective lever arms from their centroids to the centroid of
the steel:
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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.
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Another Method for Analyzing T Beams
Example 5.3
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.
Solution
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Another Method for Analyzing T Beams
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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 hf, d, and bw are given.
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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 (As = Mn/fyz ). 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.
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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.
Example to follow presents a more direct approach for the case
where a > hf. This is the case where the beam is assumed to be divided
into its rectangular parts.
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Design of T Beams
Example 5.4
Solution
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Design of T Beams
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Design of T Beams
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Design of T Beams
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Design of T Beams
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Design of T Beams
Example 5.5
Solution
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Design of T Beams
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Design of T Beams
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Design of T Beams
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Design of T Beams
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,
Asf, while the compression force in the web is balanced by the tensile force
in the remaining tensile steel, Asw.
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Design of T Beams
For the overhanging flange, we have.
from which the required area of steel, Asf, 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.
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Design of T Beams
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). Then
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Design of T Beams
Example 5.6
Rework Example 5.5 using the rectangular component method just described.
Solution
First 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 ≤ hf, the value of As 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.
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Design of T Beams
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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.
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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.
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:
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Design of T Beams for Negative Moments
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.
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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).
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.
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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.
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 (ϵt = 0.005)], it is necessary to introduce another resisting couple in
the beam.
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Compression Steel
This 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.
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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.
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).
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Compression Steel
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).
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Compression Steel
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 twothirds 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.
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Compression Steel
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.
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Compression Steel
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).
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Compression Steel
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 As2 and A’s will
be equal because the addition to T of As2fy must be equal to the addition to
C of A’sfy for equilibrium. If the compression steel has not yielded, As must
be larger than As2, as will be described later in this section.
Combining the two values, we obtain
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Compression Steel
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 = Tz 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.
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Compression Steel
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 ϵt 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.
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Compression Steel
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 > ϵy = fy/Es , the assumption is valid
and f’s is at yield, fy. If ϵ’s < ϵy, 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 < fy .
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Compression Steel
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Compression Steel
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.
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Compression Steel
Example 5.7
Solution
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Compression Steel
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Compression Steel
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Compression Steel
Example 5.8
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Compression Steel
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Compression Steel
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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.
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Design of Doubly Reinforced Beams
Example 5.9
Solution
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Design of Doubly Reinforced Beams
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Design of Doubly Reinforced Beams
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 =β1c = 9.55 in. using actual, not theoretical, bar areas for As and A’s.
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Design of Doubly Reinforced Beams
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 As, 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.
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Design of Doubly Reinforced Beams
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Design of Doubly Reinforced Beams
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).
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Design of Doubly Reinforced Beams
Example 5.10
Solution
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Design of Doubly Reinforced Beams
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