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Concrete Beams With Lateral Reinforcement

Enhancing Strength: Concrete Beams with Lateral Reinforcement

Beams or bending elements are normally subjected to bending moments in relation to their horizontal axis, which determines the upper and lower reinforcement of the section.

If the beam is connected horizontally through a slab, reinforcement, prefabricated elements, etc., the horizontal effect of the loads does not cause the beam to bend; in other words, the moment in relation to the vertical axis is dissipated in the diaphragm that is generated per the reasons mentioned above.

When the beam is acting independently of any type of diaphragm, then the effect of horizontal wind loads, vehicle impacts, earth pressures, etc. can generate horizontal bending, so the beam requires reinforcement at the lateral faces.

The big problem is that design programs do not consider this situation. When the structural element is specified as a beam, the program only designs the upper and lower reinforcement but does not calculate the lateral reinforcement. The engineer must make the necessary calculations to obtain the reinforcement that is needed in each situation.

A beam with bending in the two principal planes constitutes the so-called skewed bending, which is very difficult to calculate by hand because a trial-and-error process is required. Here we will present a simplified procedure that is easy to apply and sufficiently safe.

Concrete Beams
Fig. 1 Facade Beam.

Fig. 1 shows a case of a double strut facade in which the beam to be analyzed supports a block wall on its upper part and below there is an opening with a sliding door.

As a result of the vertical loads and the support conditions, reinforcement is required at the top and the base of the beam, Ast and Asb, respectively.

Due to the effect of the wind, pressure and suction, reinforcement will be required on both lateral faces. It will be considered that this reinforcement is the same on each lateral face and the reinforcement on each face will be called Asl. The lateral reinforcement will be distributed in the section with n areas on each side, so that one of these areas is located at the level of the upper reinforcement and another at the level of the lower reinforcement, leaving n-2 intermediate areas. Fig. 2

Lateral Reinforcement
Fig. 2 Lateral reinforcement distribution.

The procedure to follow will be the following:

  • It is assumed that the upper and lower reinforcement due to bending due to vertical loads has been calculated or obtained from some program. (Ast, Asb).
  • The moment originating from lateral loads, wind, earth or water pressure, etc. is calculated, or obtained from the program, and is factored to obtain the design moment Mu (kip. Ft). 

The horizontal effective depth is obtained

dh = b – cover – фv – фt/2    (inches)

where cover = 1.5 inches in beams

фv = diameter of the transverse reinforcement (stirrups)

фt = diameter of the lateral reinforcement (intermediate bars)

The lateral reinforcement is obtained by the simplified formula:

Asl := Mu4∙dh        (in2)

but never less than the minimum bending reinforcement.

where:   Mu is in kip.ft, dh is in inches, and the result Ast is in2

  1. The distribution of the lateral reinforcement is carried out in such a way that it complies with the regulations of the skin reinforcement and that, in turn, can work together with the vertical flex reinforcement. The latter is only applicable when the depth is greater than 36 inches.

Said lateral reinforcement Asl will be distributed in at least three parts, one area is in the center and the remaining two form part of the top and the base, adding each of said parts to the reinforcement Ast and Asb. Now, the three parts are not necessarily the same, the central part is chosen so that it has an area equal to one bar or more bars (m) and the rest is distributed above and below.

Asl :=m∙Ab+Aslt+Aslb

where m is the number of intermediate bars and Ab is the area of ​​the bar used. Aslt and Aslb are the areas located at the top and bottom respectively.

The intermediate bars will have a separation equal to or less than 15” -2.5. Cc but no greater than 12 inches.

  1. Lastly, the final upper and lower reinforcement will be the sum of the reinforcement calculated in vertical bending plus twice Aslt or Aslb as the case may be.

The number of bars required and the spacing between bars that complies with code regulations are chosen.

Next, a beam with a depth of less than 36 inches will be analyzed, and then a case of a beam with a depth greater than 36 inches.

Example No. 1

Facade beam B-1 in Fig. 1 has a 16-foot clear span with 12″ columns on each side of a sliding door entry. The beam is 8″ x 16″ and supports 8″ block wall. The factored wind pressure is 50 lb/ft2. A simply supported section will be conservatively considered.

Design of the top and bottom reinforcement for vertical loads

        Section        b :=8∙in            h :=16∙in            fc :=3000∙psi

Clear span calculation    L :=16∙ft+12∙in           L :=17 ft

Vertical loads

Beam         BeamDL :=b∙h∙150∙lbft3                   BeamDL=133.333 lbft

Wall        WallDL :=72∙lbft210∙ft +12∙ft2          WallDL=792 lbft

Factored load    pu :=BeamDL+WallDL∙1.4       pu=1295.5lbft

Bending moment    Mu := 18∙pu∙L2         Mu=46.8∙kip∙ft

Assuming #5 bars and

#3 stirrup        d :=h-1.5∙in-38∙in-58∙in∙12     d=13.8∙in

Steel area at the bottom     Asb :=46.84∙13.8in2          Asb=0.848∙in2

Bending moment     Mu := 112∙pu∙L2              Mu=31.2∙kip∙ft

Steel area at the top     Ast :=31.24∙13.8in2          Ast=0.565∙in2

Lateral reinforcement design (equal on each side)


Factored wind load        ωu :=50∙lbft2(10∙ft+12∙ft)2     ωu=550 lbft

Bending moment     Mu := 18∙ωu∙L2              Mu=19.9∙kip∙ft


Assuming #5 bars and

#3 stirrup          d :=b-1.5∙in-38∙in-58∙in∙12     d=5.8∙in


Lateral steel area    Asl := 19.94∙5.8in2        Asl=0.858∙in2

Variation 1    Use 2 #4 intermediate

#4    Ab :=0.20∙in2      m :=2


Asl-2∙0.2∙in2=0.458∙in2  then     Aslt=Aslb=0.458.in22=0.229.in2


Reinforcement at the top

Ast :=0.565∙in2+0.229∙in2∙2=1.023∙in2    2 #7     Ast=1.2. in2

Reinforcement at the bottom


Asb :=0.848∙in2+0.229∙in2∙2=1.306∙in2     2 #8     Asb=1.58. in2

Total :=2∙0.6∙in2+2∙0.79∙in2+4∙0.2∙in2=3.58∙in2

Variation 2    Use 2 #5 intermediate

#5     Ab :=0.31∙in2      m :=2

Asl-2∙0.31∙in2=0.238∙in2     then   Aslt=Aslb=0.238in22=0.119.in2

Reinforcement at the top

Ast :=0.565∙in2+0.119∙in2∙2=0.803∙in2    2 #6     Ast=0.88. in2


Reinforcement at the bottom


Asb :=0.848∙in2+0.119∙in2∙2=1.086in2     2 #7     Asb=1.2. in2

Total :=2∙0.44∙in2+2∙0.6∙in2+4∙0.31∙in2=3.32∙in2


Therefore, it is more economical to use variation 2.


Translation By: Ivet Llerena (Eastern Engineering Group).

Structural Drawings By: Sebastian Paz (Eastern Engineering Group).

© 2022 This article was written by Ernesto Valdes and published by Eastern Engineering Group. All rights reserved.



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