Introduction to T-Beams Reinforced concrete structural systems such as floors, roofs, decks, etc., are almost monolithic, except for precast systems. Forms are built for beam sides the underside of slabs, and the entire construction is poured at once, from the bottom of the deepest beam to the top of the slab.

Beam and Girder System This system is composed of slab on supporting reinforced concrete beams and girder.. The beam and girder framework is, isupported by columns. In such a system, tare placed monolithically with the slab. The typical monolithic structshown in Fig. 1.

Beam and Girder Floor System

Introduction to T-BeamsPositive Bending Moment In the analysis and design of floor and roof systems, it is common practice to assume that the monolithically placed slab and supporting beam interact as a unit in resisting the positive bending moment.

Occurrence and Configuration of TBeams Common construction type.- used in conjunction with either on-way or two-way slabs. Sections consists of the flange and web or stem; the slab forms the beam flange, while the part of the beam projecting below the slab forms is what is called web or stem.

T-Beam as Part of a Floor System

The Effective Flange Width (be) When reinforced concrete slabs are cast integrally with the supporting beams they may contribute to the compressive strength of the beams during flexure. When subject to sagging moments the resulting beam cross section is either a T-section or a L- section, as shown in figure 1 and 2, where the top surface, i.e the slab is in compression. Both type of beam are referred to as Flanged beams When subject to hogging moments the top surface is subject to tension and hence the beams are designed as rectangular sections.

The question arises of what width of the slab is to be taken as the effective width?

Where b = effective width bw = web width

Clause 3.4.1.5 BS 8110-1: 1997 gives the followingrecommendation for a T-beam, the effective width b should be taken as: Web width + 0.2 lz or Actual flange width if less

Where lz is the distance between points of zero moment (which, for a continuous beam, may be taken as 0.7 times the

effective span).

Be for single span

BS8110:Part 1:1997, Clause 3.4.1.5 Effective flange width = be < (bw + lz/5) 350 + 6000/5 1550 mm Actual flange width = 4000 mm Therefore, be = 1550 mm

be For Continuous span

BS8110:Part 1:1997,Clause 3.4.1.5 Effective flange width = be < (bw + lz/5) 300 + ( 0.7x6000/5 ) 1140 mm Actual flange width = 4000 mm Therefore, be = 1140 mm

Example 1

Under Reinforced

Balanced Reinforced

Over Reinforced

ANALYSIS OF FLANGE SECTION. Determine the ultimate moment of resistance of the T-section as shown in figure. Bf=800 Given, fy =460N/mm^2FLANGE =150

fcu=30N/mm^2 D =420mm As =1470mm^2

Bf=800 hf=150

Bf=800=150

X

S

0.45fcu

Depth,d Fst

Z=d-(s/2)

As =1470mm^2

CALCULATION. Using concept of equilibrium of force. Force compression, Fcu= Force tension, Fy Therefore, 0.45 fcubfs=0.95fyAs Solve the equation to find s, s= (0.95x460x1470) / (0.45x30x800) s=59mm x=s/0.9 =66mm

Since x < height of flange beam,hf ,then theneutal axis lies on the flange. Then we need to find the moment resist by the beam. First we need to find the lever arm, z z=d- (s/2) z=420-(59/2)= 390mm

Now by taking the moment at the centroid of the reinforcement, the moment of resistance,M M=Fcu x z =0.45fcubfsz =0.45x30x800x59x390x10^-6 =249 kN m

FLANGE BEAM WITH COMPRESSION REINFORCEMENT. From the equation for solving t beam with stress block lies within the web.

Compression=Tension 0.45fcu[(bf-bw)hf +bws]=0.95fyAs And by taking the moment at the tensile steel, M= 0.45fcu[(bf-bw)hf (d-(hf /2)) + bws(d(s/2))] However, the limit of the depth of the neutral axis is x=d/2.or s=0.9(d/2) By substituting s=0.45d into moment equation

M= 0.45fcu[(bf-bw)hf (d-(hf /2)) + bw0.45d(d-(0.45d/2))]

Simplify it, M= 0.45fcu (bf-bw)hf (d-(hf /2)) + 0.156fcubd^2

Therefore if the applied moment is greater than this maximum moment, compression reinforcement is required.

In order to find the area of steel compression reinforcement, As= Ma M/ (0.95fy(d-d) Where Ma= apply moment M =maximum moment d =distance from surface of flange to center of steel compression reinforcement.

Alternative 1: Using an Exact method to determine the depth of N.A

Alternative 2: Designing for the Conservative Condition of x = 0.5d