5.16 T beams

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Hardly any problem attracts so much attention as the modeling of T beams;

see Fig. 5.47. Basically the interaction between a T beam and a slab is a

complex three-dimensional problem. But even though computers are becoming

more and more powerful, true 3-D solutions are much too complicated and

simply require too much effort. Engineers always tried to handle the complex

situation with simplified models, by working with an effective beam width or

other approximate models. The choice of the model depends heavily on the

accuracy to be achieved. Therefore, a multitude of possible models exist, with

differing degrees of accuracy:

X

Y

Z

X

Y

Z

5.16 T beams 463

Fig. 5.46. Hinged circular slab, bending moments (kNm/m)

• The slab and the beam are modeled as folded plates (also called the shell

model).

• The slab is modeled as a folded plate, and the T beam as an eccentric

beam or plate.

• The slab is modeled as a slab, and the T beam as an eccentric beam (with

a normal force).

• The slab is modeled as a slab, and the axis of the T beam lies in the

midsurface of the slab.

The fact that some engineers model T beams as rigid supports (EI = ∞)

suggests just how wide the range of possible models is. This approach may

be sufficient to assess the limit load of a beam, but it does not suffice to

produce an affordable solution, or to accurately predict the displacements of

the structure.

The first two models differ in how they treat the web. In the first model

the web is made up of plate elements, i.e., the distribution of the normal

stresses in the web is nonlinear. In the second model, the classical linear stress

distribution of beam theory prevails.

Both models are capable of predicting the distribution of the normal

stresses in the slab very accurately because the effective width is a result of

the analysis. On the screen it can be seen how the effective width bm increases

in the span and how it shrinks near the supports.

In all other cases the models simulate the coupling between a separate T

beam and a slab (see Fig. 5.48), where the slab is either treated as a folded

plate (mij, qi, nij) or “simply” as a slab (mij, qi).

The coupling in the finite element sense means that the movements of

the beam and the movements of the slab are synchronized at the nodes, and

464 5 Slabs

Fig. 5.47. Slab with T beams: a) system, b) bending moments in the T beams

(kNm)

Fig. 5.48. The nodes of the slab lie atop the nodes of the T beam

119.65

236.87

158.22

67.40

27.92

-27.21

262.62

-126.06

42.28

125.63

88.56

110.49 133.42

74.52

124.49

83.35

26.61

13.36

5.16 T beams 465

that the work done on both sides of the joints is the same for any virtual

displacement (energy coupling).

If u,w, and ϕ (= rotation) denote the corresponding degrees of freedom

of the slab (S) and beam (B), then

wB = wS uB = uS + ϕSe ϕB = ϕS , (5.84)

or if it is assumed that EA = ∞ in the slab, then the longitudinal displacements

uB are simply uB = ϕP · e, where e is the distance of the neutral axis

of the beam from the midsurface of the slab. That is, the slab and the beam

are imagined to be connected by a rigid bar of length e, so that rotations in

the slab lead to longitudinal displacements in the beam.

Depending on whether the eccentricity e is considered or not, we speak of

a centric or eccentric beam model. The models distinguish how the normal

forces resulting from this eccentricity are introduced into the analysis.

As a result of the coupling of various structural elements, a whole series of

incompatibilities and errors arise. Recall that at the interface of two structural

elements, the stresses are no longer pointwise opposite (see Sect. 1.34, p.

177). Only the virtual work of the resultant stresses (plus the volume forces,

etc.) is the same. Furthermore the beam and slab deflect differently, because

the deflection curve w of the beam will in general not be the same as the

deflection surface w(x, y) of the slab. Often a Reissner–Mindlin plate that

exhibits shear deformations is coupled to an Euler–Bernoulli beam that knows

no such deformations. Any idea of a pointwise match between stresses on the

two sides of the interface is misleading.

What is more, we will also see an error in the transfer of the shear stresses,

because the part of the longitudinal displacements resulting from the eccentricity

is normally of quadratic type, while the displacements due to the normal

force are most often only linear. This error decreases quadratically, but

it requires that the span be subdivided into several elements, and makes its

presence felt by a stepwise distribution of the normal force.

Even if the stiffnesses are added only at the nodes, it is helpful to think in

terms of the flexural rigidity of the whole system. The flexural rigidity of the

slab increases if the corresponding terms of the beam elements are added:

kw = bm · Eh3

12(1 − μ2)

+ EI + EA · e2 . (5.85)

This is also seen if the modified stiffness matrix of the beam is studied:

K =

⎡

⎢⎢⎢⎢⎢⎣

12EI/l3 −6EI/l2 −12EI/l3 6EI/l2

. 4EI/l + EA/l · e2 6EI/l2 2EI/l − EA/l · e2

. . 12EI/l3 6EI/l2

sym. . . 4EI/l + EA/l · e2

⎤

⎥⎥⎥⎥⎥⎦

(5.86)

466 5 Slabs

Fig. 5.49. Position of the T

beam with respect to the slab

Next most important is—besides the choice of the eccentricity e—which

moment of inertia I is attributed to the T beam. i.e., which cross-sectional

parts we apportion to the T beam; see Fig. 5.49.

If the T beam reaches the upper surface of the slab (Fig. 5.49 b) then

AB = bU d0 e = d0 − d

2 IB = bU d30

12

− bU d3

12 . (5.87)

If it is assumed that the stiffness of the T beam corresponds to the crosssectional

values, then

AB = bU dU + bmd e= bU dU

bU dU + bm d

d0

2

(5.88)

IB = bU d3

U

12

− bU dU bm d

bU dU + bm d

d20 4 . (5.89)

If only the cross-sectional area of the web is attributed to the T beam (see

Fig. 5.49 a) then

IB = bU d3

U

12

+ bU dU

d20

4 . (5.90)

Before any discussion about what the right choice might be we should

pause and recall how “virtual” the coupling is, because neither are the resultant

stresses between the beam and the slab the same, nor are the displacements

the same. In other words, the only chance we have is to think in terms

of equivalent nodal forces. But then the effects that an error in the coupling

has on the results is no longer so serious and studies have shown that the

simplest strategies often yield the best results.

Therefore, not too much effort should be put into the modeling of the

coupling of T beams and slabs. The trouble taken by the engineer is not

5.16 T beams 467

Fig. 5.50. Deformations of slab and T beam

honored by the FE program. Whether the contributions of the parallel axis

theorem are considered or not, whether cross-sectional areas are considered

twice, and whether the neutral axis runs in the midsurface or below the slab

is only important insofar as the flexural rigidity EI of the T beam and thus

its rotational and lateral flexibility is modeled more accurately; see Fig. 5.50.

The best strategy emerges if the sum of the individual stiffnesses corresponds

to the actual stiffness. By choosing a certain effective width from which

the joint stiffness of the combined system (slab + T beam) follows—if the slab

stiffness is subtracted—and an eccentricity e is chosen, the stiffness of the T

beam can be calculated. Normally the effective width has little influence on

the results; a choice of l0/3 is sufficient in most cases, [139].

In the BE method, a T beam is simply a beam (with the same bending

stiffness as the actual T beam) that supports the slab, and as in the force

method, the support reactions are determined such that the deflection of the

slab is the same as that of the T beam.

A study of various FE models—A = T beam as centric beam, B = T beam

as a centric beam, bm = ∞, C = T beam as eccentric beam, D = T beam as

rigid support—has shown that the results of the various models are not much

different.

468 5 Slabs

Model M in the beam myy (kNm/m) mxx (kNm/m) f (mm)

FE A 481 4.7 −30.0 1.8

FE B 493 4.5 −31.2 1.5

FE C 490 4.3 −30.9 1.6

FE D - 0.0 −36.4 0.0

BE 485 5.3 −31.4 1.7

The slab was a two-span slab with the T beam running down the middle of

the plate in the vertical direction (y-direction). The values f are the maximum

deflections of the T beam, M is the maximum bending moment in the T beam,

and myy and mxx are the bending moments in the slab.

If we consider how much uncertainty we must cope with in the design of

reinforced concrete, the choice of the model does not seem that important. But

in a commercial program, the limiting values of the design variables must also

yield reasonable results, and in that respect a systematic disregard of certain

effects, for example the axial displacements of the slab or the postulate that

the design value of the stiffness corresponds to the actual stiffness of the

coupled system, can lead to deviations in the results that cannot be neglected

in the extreme cases.

Recommendations

The best results are obtained if the T beam is modeled as an eccentric beam—

with an increase in the stiffness corresponding to the eccentricity e—and if

axial displacements in the slab are taken into account, because then the effective

width is a result of elasticity theory, i.e., it is automatically determined

by the FE program. In most cases however, it will be sufficient to neglect axial

displacements in the slab and work with an approximate effective width. In

that case one should choose the equivalent moment of inertia ˜I in such a way

that the sum of the flexural rigidities is equal to the flexural rigidity of the

full T beam:

EItot = bm · E · d3

12(1 − ν2)

+ b0 · du

3

12

+ E · bm · d · e2

p + E · b0 · du · e2b

,

(5.91)

EItot = bm · E · d3

12(1 − ν2)

+ E˜I . (5.92)

Here ep and eb denote the distances of the slab midsurface and neutral axis

of the beam from the center of gravity. This implies that ˜I is the moment of

inertia of the total cross section of the T beam minus the stiffness of the slab

itself.

In any case we recommend that every engineer test his model with regard

to the limits of the design variables.

5.17 Foundation slabs 469

Fig. 5.51. Winkler

load

Resultant stresses

Once the nodal displacements are determined, the resultant stresses mij, qi, nij

in the slab and MB, VB,NB in the beam, can be calculated. The next problem

is that many programs calculate the resultant stresses separately for the

slab and the beam, because no other strategy exists. If one tries to avoid this

dilemma by modifying the stiffnesses, nothing is gained in the end; rather, the

confusion increases. The result are designs in which the reinforcement lies in

the compression zone of the slab, or where it is simply ignored, which leads

either to unsafe or to wasteful designs.

A design can only be considered correct if the reinforcement is determined

for the complete cross section, and if the fact that the neutral axis lies below

the midsurface of the slab is taken into account:

M = Mbeam + Nbeam · eb +

_

slab

(myy + nxx · ep) dz , (5.93)

V = Vbeam +

_

slab

qz dz . (5.94)

With these resultant stresses, the correct reinforcement for the T beam can

be designed.

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