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## Wooden Beams Reinforced with Steel Plates

To achieve the objective of this work, the legalization of a covered terrace will be used as an example, in which it was detected that one of the beams did not have the necessary resistance to support the actual loads on it.

Fig. 1 shows the floor plan of the terrace, named as B-1, the wooden beam with a 2×8 section will be reinforced. It will first be analyzed as is to show that the section cannot resist the actual loads and immediately the section will be reinforced with several variants. One of the variants would be to increase the depth of the beam; however, architectural reasons do not allow it. Therefore, the variants that will be chosen will have a maximum depth of 8 in.

##### Fig. 1 Plan view of covered terrace

The actual loads on the beam B-1 will be:

Joists (2) 2×8 spaced at 16 in (SP No. 2). Own weight 36 lb/ft³

36 ∙ lb/ft³ ∙ 1.5 ∙ in ∙ 2 ∙ 7.5 ∙ in ∙ (1/16 ∙ in) = 4.1 lb/ft²

Plywood 3⁄4 in          3.0 psf

Insulation DL          1.0 psf

. Shingles DL          3.0 psf

.       Total DL =       11.1 psf

##### Fig. 2

The live load on the deck is LL = 30 psf, then the total load will be 41.1 psf.

The tributary width will be equal to:                       (12 ft/2) + (16 in/2) =6.67 ft

The actual loads on the beam                                  ω ∶= 41.1 ∙ lb/ft² ∙ 6.67 ∙ ft = 274 lb/ft

For a 2×8 SP No. 2 wooden section, the AWC NDS 2018 code provides the following admissible values:

Reference: NDS Supplements 2018 edition Table 4B, page 39 and 40

Adjustment factors           Cr := 1.15                         Cfu := 1.0

Allowable stresses:          Fb := 925∙psi∙Cr∙Cfu         Fv := 175∙psi       Fc := 565∙psi

For the height of               L := 10∙ft            and tributary width          S := 6.67∙ft

The total load                   ω ∶= 41.1∙psf∙S                 ω ∶= 274.137∙plf

Maximum bending moment

.                                        M ∶= 1/8 ∙ ω ∙ L²                M = 3426.7 lb ∙ ft

Maximum shear

.                                        V ∶= (ω∙L/2)                       V = 1370.7 lb

##### Fig. 3

The section should have S, A, and ∆ values greater than the following:

.                                        Sreq ∶= M/Fb                     Sreq = 38.656 ∙ in³

.                                        Areq ∶= (1.5∙V/Fv)              Areq = 11.749 ∙ in²

.                                        Allow∆ ∶= L/360                 Allow∆ = 0.33 ∙ in

The 2×8 section that actually has: b := 1.5∙in h := 7.25∙in only reaches the following values:

.                                        Sprov ∶= (b∙h²/6)               Sprov = 13.141 ∙ in³

.                                        Aprov ∶= b ∙ h                    Aprov = 10.875 ∙ in²

.                                        ∆prov ∶= (5/384) ∙ ω ∙ (L^4/((b∙h³/12 )∙1400000∙psi))          ∆prov = 0.925 ∙ in

which shows that the section is not capable of resisting all the design loads and therefore requires reinforcement by means of wooden sections or reinforcement with a steel plate.

If (2) 2×8 were to be placed, the section module would be doubled

.                                        Sprov ∶= (2∙b∙h²/6)            Sprov = 26.281 ∙ in³                    less than Sreq

If (3) 2×8 were to be placed, the section module would triple

.                                        Sprov ∶= (3∙b∙h²/6)            Sprov = 39.422 ∙ in³                    greater than Sreq

then a possible solution would be to add two 2×8 sections to the existing one. It is likely that the solution will not be accepted by the owner, especially if it is exposed to view.

##### Fig. 4 Solution with three sections (3) 2×8

Another possible solution is to reinforce with a steel plate located between two 2×8 sections of wood. This solution would hide the steel plates. Fig. 5

##### Fig. 5 Solution with interleaved steel plate

We will now try with a 1⁄4 in. × 7 in. steel plate. This plate would increase the total load.

Weight of steel plate per foot                   1 ∙ ft ∙ ¼ ∙ in ∙ 7 ∙ 490 ∙ pcf = 6 lb

Being the total load equal to                    ω ∶= 274.1 ∙ plf + 6 ∙ plf = 280.1 lb/ft

The bending moment would be               M ∶= 1/8 ∙ ω ∙ L 2 = 3501.2 ft ∙ lb

The section modulus of the steel plate would be:

.                                                                Spl ∶= (¼∙in∙(7∙in)²)/6           Spl = 2.042 ∙ in3

The maximum tension will be:

.                                                                fb ∶= M/Spl                           fb = 20578.8 psi

less than           Fy ∶= 36000 ∙ psi          Fb ∶= 0.6 ∙ Fy                       Fb = 21600 ∙ psi

for which the section with (2) 2×8 and a 1⁄4 in. × 7 in. steel plate is sufficient to resist the actual loads.

##### Fig. 6 Reinforcement with 1/4 in. × 7 in. steel plate

It remains for us to define the necessary screws to transfer the load between the elements, along the beam and at the supports. Fig. 6 shows how the loads on the joists are laterally transmitted to the existing section. These loads must be supported jointly between the two sections, the wooden ones and the steel plate. To do this, screws will be placed that go through all of the members.

In the central section, the screws will be in charge of transmitting the actual loads located in the midway point between screws. S is the spacing between screws, the shear in each screw will be:

.                                                                V ∶= ω ∙ S                               S ≔ V/ω

Table 12-B indicates the allowable shear depending on the type of connection, the type of wood, size of the wooden element, thickness of the steel plate, and the direction of the load, parallel or perpendicular to the grain of the wood.

###### .

Reference: AWC NDS 2018 Table 12-B page 96

For SP No. 2 G = 0.55

Main member = 1 1⁄2 in.

Lateral member = 1⁄4 in. steel plate

Diameter of screw                        Zperp (allowable shear perpendicular to grain)

.                            1/2………………….350 lb
.                            5/8………………….400 lb
.                            3/4………………… 450 lb                           Minimum distance to the end, to the
.                            7/8………………….510 lb                           edges, and minimum spacing of screws
.                            1   ………………… 550 lb                           = 4D, where D = ф screw diameter.

For                      ф ∶= ½ ∙ in                       V = Zperp = 350 lb                       4 ∙ ф = 2 ∙ in

.                           S ∶= (350∙lb)/ω                S = 15 in           L/S = 8           spacing at 15 in

To ensure that the loads are transmitted to the supports by all the members and not only by the plate, avoiding the crushing of the concrete, we will place screws that will be able to transmit the entire reaction.

.                           R ∶= ω ∙ L/2                     R = 1400.5 lb

Number of 1⁄2 in. screws

.                           n ∶= R/(350∙lb)                n = 4        ½in. screws

Fig. 7 shows the distribution of the screws in the supports and in the central area. Note that the minimum spacing is met, minimum of 2 in. distance to the edges and extremities.

##### Fig. 7 Screw distribution in the central and support areas

Translation By: Ivet Llerena (Eastern Engineering Group).

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