Reinforcement of a Concrete Beam: Everything you Need to Know

August 12, 2022 | Ernesto F. Valdes | Publications | No Comments

During the process of extending a home, a contractor will demolish a load-bearing wall. The reinforcing beam does not have sufficient reinforcement to withstand the new loads, so it is necessary to reinforce the beam in a way that allows it to withstand these loads.

There are several solutions to the problem: demolish it and make it new, increase the height, increase the width, etc.; however, this complicates the construction process or affects the architectural side. That is why we consider steel plate reinforcements on both sides, connected with screws to the existing beam so that it is not necessary to prop up or increase the height of the beam.

Fig. 1 shows the dimensions of the extension and the existing beam.

Detail-of-the-extension-and-existing-beam

Detail-of-the-extension-and-existing-beam-P.2

Detail-of-the-extension-and-existing-beam-P.3

Fig. 1 Detail of the extension and existing beam

In the first place, it will be verified if the reinforcement of the existing beam is or is not sufficient to resist the total loads.

Load Components and Structural Demands

The loads will be:

Own weight of the beam = (8 × 12) × 150/144 = 100 lb/ft

Dead load of the deck = 25 psf × (12/2 + 30/2) = 525 lb/ft

Total dead load = 100 lb/ft + 525 lb/ft = 625 lb/ft

Live load of the deck = 30 psf × (12/2 + 30/2) = 630 lb/ft

Factored load pu = 1.2 × 625 + 630 × 1.6 = 1758 lb/ft

Conservatively considering the simply supported beam, we have:

Mu ∶= 1/8 ∙ 1758 ∙ lb/ft ∙ (13.33 ∙ ft)² = 39047 lb ∙ ft

The steel area will be:

As ∶= (Mu/4∙d) As = 3.9 kip.ft/ (4 × 9.81 in) = 0.996 in²

Therefore, a greater reinforcement is required (0.996 in²) than the existing 0.62 in².

Note that even if the reinforcement would have been sufficient, the beam in question has a shear reinforcement lower than the minimum for beams, since it was a cladding supported on a load-bearing wall that is now eliminated and converted into a beam. Only if it had complied with both reinforcements would an additional reinforcement not have been necessary.

Reinforcement Design and Connector Analysis

Two steel plates measuring 1⁄2 in × 10 in will be used (one on each side of the concrete beam). In Fig. 2 the elevation and the cross section of the beam reinforced with the two steel plates are shown.

Elevation-and-cross-section-of-the-reinforced-concrete-beam-P.1

Elevation-and-cross-section-of-the-reinforced-concrete-beam-P.2

Fig. 2 Elevation and cross section of the reinforced concrete beam

Structural Verification with Steel Plates

Assuming t = 1⁄2 in hp = 10 in Fy = 36000 psi

The service load will be:

. p = 625 lb/ft + 630 lb/ft = 1255 lb/ft

The bending moment will be:

. M = (1/8) × 1255 lb/ft × (13.33 ft)² = 27875 lb.ft

The module of the section:

. S = 2.(0.5 × 10²) /6 = 16.67 in³ for both steel plates

Extreme fiber stress:

. fb = M / S = 27875 × 12/16.67 = 20066 psi

. < 0.6 × 36000 psi = 21600 psi O.K.

The shear force:

. V = (1/2) × 1255 lb/ft × 13.33 ft = 8365 lb

. fvmax = 1.5 V / (2. hp × t) = (1.5 × 8365) / (2 × 0.5 × 10) = 1255 psi

. < 0.4 x 36000 psi = 14400 psi O.K.

Then the 2 sections measuring 1⁄2 in × 10 in are enough to withstand the new loads.

Another variant would be to consider plates measuring 3/8 in × 12 in, thinner but wider plates.
In this case:

. S = 2.(3/8 × 12²) / 6 = 18 in³

. fb = M / S = 27875 × 12/18 = 18583 psi

. < 0.6 × 36000 psi = 21600 psi O.K.

The result is satisfactory; however, since it is the same height as the beam, it would possibly cause problems in locating it, since it could affect the ceiling. Therefore, the best solution will be accepted in this case, which would be 1⁄2 in × 10 in.

Connector Placement and Shear Transmission

Continuing the calculation, the connectors along the beam will be calculated. It will be considered that in a section of the beam Spc, the connector must transmit the load received by the beam to the steel plates, that is, each connector will receive a shear equal to: Fig. 3

. V = 1225 lb/ft x Spc

Fig. 3 Connectors along the beam

Using KWIK BOLT 3 Expansion Anchor HITLI

3/8 in diam. 3 1⁄2 embedment Allowable shear = 1315 lb for fc = 3000 psi

. fAN = fAv = 1 Spacing > 6 in

. fRV1 = 0.62 Distance to edge = 3 in

. Allowable shear = 1315 × 1 × 0.62 = 815 lb

Then, 2 connectors (one on each side) would give:

. 2 × 815 = 1225 × Spc

. where Spc = 15 1⁄2 in

Fig. 3 shows the distribution of the alternating connectors along the beam at a distance of 15 1⁄2 in.

Now it remains to calculate the connectors at both ends of the steel plates. The end reaction, considering the simply supported steel plates, will be:

. Reaction = V = 8365 lb

We will consider (4) 1⁄2 in diam. KWIK BOLT 3 Exp. Anchors from HILTI (on each side) with 3 1⁄2 in embedment

Allowable shear = 2415 lb

Spacing = 4 in fAV = 0.92

Distance to edge = 3 in fRV = 0.50

Allowable shear = 8 × (2415 × 0.92 × 0.5) = 8887 lb > 8365 lb O.K.

Fig. 4 shows the location of the connectors at the ends of the beams. These connectors will be on both sides of the indicated position.

Fig. 4 Connectors at the ends of the beam

In this way, the reinforcement of the concrete beam is completed with two steel plates placed on both sides, connected to the beam with uniformly spaced anchors to transmit the acting load and with anchors at the ends to transmit the load to the supports.

It is not always possible to place the plates as indicated in Fig. 4, with equal spacing and one after the other. In case that it is not possible to extend the inner plate to the end of the other one, a variant solution is shown in Fig. 5.

Additionally, if the length is not very long, it is possible for a plate only on one side to solve the problem. The calculation is similar to the one shown in the previous example.

Fig. 5 Connectors in ends that are not the same

Final Design Considerations

Finally, there is an aspect that can modify the design developed so far. This consists of the upper area of the slab being subjected to compression along its entire span so that between two connectors at that level the separation is 15.5 × 2 = 31 in. and a failure due to slenderness could occur.

We consider a column with a width of 4 in. in the upper band with a thickness of 1⁄2 in. subjected to compression.

The code recommends that the slenderness

. (k ∙L/r) be less than 200

Assuming that the column (top of the slab) has

. b ∶= 4 ∙ in t ∶= 0.5 ∙ in k ∶= 0.8

. I ∶= (b∙t³/12) I = 0.042 in^4 A ∶= b ∙ t A = 2 in²

. r ∶= √(I/A) r = 0.144 in

Considering a safety factor of 2, the maximum spacing of the connectors at the top must not exceed:

. FS ∶= 2 L ∶= (200/FS) ∙ (r/k) = 18.042 in

This means that the connectors of the upper compressed zone will be placed at a distance of 15 1/2 in. and the lower ones will be placed at 31 in., giving rise to the final design of the additional reinforcement plates for the concrete beam as shown in Fig. 6.