T Beams - PowerPoint Presentation

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