1

T Beams

2

T BeamsReinforced 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.

3

T BeamsIf, 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

4

T Beamsthe 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.

5

T Beams

6

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.

7

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.

8

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.

9

Analysis of T BeamsThe 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 .

10

Analysis of T BeamsIn 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.).

11

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.

Example 5.1

Solution

12

Analysis of T Beams

13

Analysis of T Beams

14

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.

Example 5.2

Solution

15

Analysis of T Beams

16

Analysis of T Beams

17

Another Method for Analyzing T BeamsThe 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.

18

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:

19

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.

20

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.

Example 5.3

Solution

21

Another Method for Analyzing T Beams

22

23

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.

24

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.

25

Design of T BeamsOften 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.

26

Design of T BeamsExample 5.4

Solution

27

Design of T Beams

28

Design of T Beams

29

Design of T Beams

30

Design of T Beams

31

Design of T BeamsExample 5.5

Solution

32

Design of T Beams

33

Design of T Beams

34

Design of T Beams

35

Design of T BeamsOur 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.

36

Design of T BeamsFor 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.

37

Design of T BeamsThe 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

38

Design of T BeamsExample 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 ≤ 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.

39

40

Design of T Beams

41

Design of T Beams for Negative MomentsWhen 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.

42

Design of T Beams for Negative MomentsIn 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:

43

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.

44

L-Shaped BeamsThe 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.

45

Compression SteelThe 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.

46

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.

47

Compression SteelCompression 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).

48

Compression SteelIf 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).

49

Compression SteelFor 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.

50

Compression SteelSometimes 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.

51

Compression SteelThis 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).

52

53

Compression SteelThe 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

54

Compression SteelThe 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.

55

Compression SteelWith 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.

56

Compression SteelInitially 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 .

57

Compression Steel

58

Compression SteelThe 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.

59

Compression SteelExample 5.7

Solution

60

Compression Steel

61

Compression Steel

62

Compression SteelExample 5.8

63

Compression Steel

64

Compression Steel

65

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.

66

Design of Doubly Reinforced BeamsExample 5.9

Solution

67

Design of Doubly Reinforced Beams

68

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.

69

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.

70

Design of Doubly Reinforced Beams

71

Design of Doubly Reinforced BeamsNote 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).

72

Design of Doubly Reinforced BeamsExample 5.10

Solution

73

Design of Doubly Reinforced Beams