T-beams are formed when reinforced concrete floor slabs, roofs, and decks are cast monolithically with their supporting beams. Generally, formworks are placed for the bottom and sides of the beams and soffit of slabs. Bent up bars and stirrups of the beam are extended up into the slab. After that, all the elements are cast at once, from the lowest point of the beam to the top of the slab.
The part of the slab around the beam, called flange, would work with the beam and resist longitudinal compression force. Interior beams have flanges on both sides and are termed as T-beams, while edge beams have flanges on one side and are called L-beams. The part of the beam extending below the slab is called a stem or web.
The design of the reinforced concrete T-beams is similar to that of a rectangular reinforced concrete beam except for flanges that need to be considered in the former type of beam.
Effective Flange Width
The effective flange width (be) of a T-beam needs to be determined in order to begin the design process. In Figure-1, the flange of the isolated T-beam is a little bit wider than the T-beam stem, and the entire flange is effective in resisting compression.
However, in Figure-2, the flange width is large; hence, parts of the flanges situated at a distance from the stem do not take their full share in resisting compression, and the stresses keep varying.
The variation of stresses leads to tedious calculations; that is why a uniform stress distribution is considered over a smaller width of the effective flange, see Figure-3.
According to ACI 318-19, the effective flange width of a T-beam can be found as follows:
1. Isolated Beams
For isolated beams, in which the flange is only used to provide an additional compression area, the flange should have a thickness greater than or equal to 1/2bw, and an effective width less than or equal to 4bw.
2. Internal T-beams
According to 318-19, the effective flange width of an internal T-beam should not exceed the smallest of:
1- One-fourth the clear span length of the beam, L/4.
2- Width of web plus 16 times slab thickness, bw +16hf .
3- Center-to-center spacing of beams.
3. Edge Beam (L-Shape)
According to 318-19, the effective flange width of an edge beam should not exceed the smallest of:
1- Effective flange width (be) equal to or smaller than (bw+(Clear span/4))
2- Effective flange width (be) equal to or smaller than (bw+(6hf)
3- Effective flange width (be) equal to or smaller than (bw+half clear distance to the next clear web beam)
T-beam Versus Rectangular Beam
If a T-shaped reinforced concrete beam is subjected to negative moments at supports, the beam is designed as a rectangular section because the concrete in tension is neglected. The width of the rectangular section is equal to the stem (web) width, see Figure-7.
However, when the T-beam is subjected to a positive moment, the flange is located in the compression zone hence the beam should be designed as a T-beam, see Figure-8.
Design of Reinforced Concrete T-beam
A T-section beam design involves calculating the dimensions (be, hf, h, and bw) of the beam and the required reinforcement area (As). The flange thickness (hf) and width (be) are usually established during the slab design.
The size of the beam web or stem is influenced by the same factors that affect a rectangular beam’s size. In the case of a continuous T-beam, the concrete compressive stresses are most critical in the negative-moment regions, where the compression zone is in the beam stem (web).
The distribution of stress in T-beam is shown in Figure-9:
- Calculate applied moment (Mu) using beam span and imposed loads.
2. Determine Effective Flange Width (be)
3. Choose the web dimensions (bw) and (h) based on either negative bending requirements at the supports, or shear requirements.
4. Assume, a=hf , then calculate (As) using the following expression:
5. Check the assumed value of (a):
In Equation 2, plug the value of (be) found in Step 2.
If a< hf, design the beam as a rectangular section and follow the design procedure of the rectangular beam.
If a> hf, design the beam a T-section and go to Step 6.
6. Compute the reinforcement area required to balance the moment of the flange use Equation 3, and then flange moment employ Equation 4:
7. Calculate moment of the web:
8. Assume a rectangular stress block depth (such as a= 100 mm), then estimate the amount of reinforcement area (Asw) required to balance the web moment:
The value of (d) should be computed using the following formula:
d= beam height-concrete cover- stirrup diameter- 0.5*longitudinal steel diameter Equation 7
Then check assumed rectangular stress block depth (a) using (Asw):
Use the new (a) and plug it into Equation 6, then compute new (Asw). Repeat this process till correct (Asw) is reached. Commonly three trials are enough.
9. Compute total As which is equal to (Asf+Asw), then determine the number of reinforcement:
No. of Bars= As/ area of single bar Equation 9
10. sketch the final design on which all necessary data are represented.
A floor system shown in Figure-10 consists of a 75 mm concrete slab supported by concrete T-beams with a 7.5 m span and 1.2 m on centers. Web dimensions, which are determined by negative moments requirements at supports, are bw= 275 mm and d= 500 mm. What is the tensile steel area required at midspan to support a factored moment of 725 KN.m? Material properties: fc’= 21 MPa and fy= 420 MPa.
1. Applied moment is provided, Mu= 725 KN.m
2. Find effective flange width (be), which is the smallest of the following:
- Span/4= 7500/4= 1875 mm
- bw +16hf= 275+16*75= 1475 mm
- Center-to-center spacing of beams= 1200 mm
Therefore, the effective flange width is equal to 1200 mm.
3. Dimensions of the web are provided.
4. Assume, a=hf= 75 mm, and assume a strength reduction factor is equal to 0.9.
As= (725*106)/(0.9*420(500-0.5*75)= 4147.004 mm2
5. Check the assumed value of (a), use (As) computed in Step 4:
a=(4147.004*420)/(0.85*21*1200)= 81.31 mm
Since a= 81.31 mm> hf=75 mm, so the beam needs to be designed as T-section.
6. Compute (Asf) and flange moment:
Asf= (0.85*21*(1200-275)*1200)/420= 2946.23 mm2
phi*Mnf= 2946.23*420*(500-0.5*75)*10-6= 572.23 KN.m
7. Calculate moment of the web:
phi*Mnw=725-572.23= 209.54 KN.m
8. Estimate the amount of reinforcement area (Asw), assume a=100 mm and phi= 0.9
Asw= (209.54*106)/(0.9*420*(500-0.5*100)= 1231.86 mm2
check (a) using the above (Asw),
a=(1231.86*420)/(0.85*21*275)= 105.4 mm
Find new (Asw) use a= 105.4 mm
Asw= (209.54*106)/(0.9*420*(500-0.5*105.4)= 1239.29 mm2
Since the new Asw is very close to the previous one, therefore further trial is not needed.
9. Compute total As which is equal to (Asf+Asw):
As= Asf+Asw= 2946.23+1239.29= 4180.29 mm2
The assumed strength reduction factor should be checked:
Choosing a single steel bar leads to a reinforcement area that is considerably higher than the total area. Therefore, no. 32 and no. 29 steel bars are selected to obtain a reinforcement area that is as close to the required reinforcement area as possible.
There are three bars with a diameter of 32 mm, and the corresponding reinforcement area is 2457 mm2
There are three bars with a diameter of 29 mm, and the corresponding reinforcement area is 1935 mm2
The total reinforcement area is equal to 4349 mm2; this is the answer to the question.
So, steel bars are arranged in two-layers and the distance between the two layers is 25 mm.
Check the strength reduction factor:
Since the compressive strength of concrete is smaller than 30 MPa, therefore B1=0.85
neutral axis depth (c)= a/B1= 105.4/0.85= 124 mm
dt: distance from the compression face of the beam to the center of the bottom layer of steel bars:
c/dt= 124/525= 0.236<0.375. Therefore, the assumption is correct.
For more details on strength reduction factor calculation, please click here
Generally, reinforced concrete floor system consist of beams and slab which are constructed monolithically. As a result, the part of the slab around the upper part of the beam work together to carry loads. In effect, the beams have extra width at the top part which are termed as flanges. The beam is called T-beam.
The effective flange width consists of web of the beam plus the width of flange on each side of the beam. The stress distribution over the width of effective flange width is uniform.
The effective depth is equal to the distance from the beam’s extreme compression fiber to the center of gravity steel bars embedded in the beam.