Steel Beam Reinforcement
Steel Beam Reinforcement
On certain occasions, changes to the structure arise, causing an increase in the actual loads. Due to this, some elements are subjected to loads that may exceed that of their original design, making it necessary to increase their capacity through the use of additional reinforcement.
Therefore, we consider the case of a steel beam subjected to simple bending to which the load is increased by 60 psf. The span of the metal beam W12x58 is 14 feet.
The geometric data of the section, strength of the steel, and the total load are shown in Fig. 1.
Section W12×58 A := 17.0 · in² lx := 475 · in^4
d := 12.25 · in Sx := 78 · in³
T := 9.5 · in fy := 50000 · psi
Total load q := 9.0 · kip/ft L := 14 · ft
Fig. 1 General data
The allowable tension of the section is:
Fb := 0.6 · fy Fb = 30000 · psi
The maximum bending moment for the total load is:
M := 1/8 · q · L² M = 220.5 · kip · ft
Furthermore, the maximum stress in the outer fiber of the section is:
fb := M/Sx fb = 33923.1 · psi fb/Fb = 1.1
Check := if (fb > Fb, “N.G.” , “O.K.”) Check = “N.G.”
Results show that the section is insufficient to resist the new loads.
2. The possibility of reinforcement
There is the possibility of reinforcing the beam with steel plates, channel section, or angles welded to the lower flange, since in the upper flange there is generally no possibility of reinforcing on that side. However, if bars welded to the web are used, they can be placed above and below, keeping the centroid of the section at half the height of the beam.
Hence, we will proceed by reinforcing the metal section with four steel bars placed in the four interior corners, on each side of the web of the beam. Fig. 2 shows the 4 bars that we will be located at T/2 from the centroid.
As a result, bars will be welded at equal intervals along the element
Bars location for welding
Fig. 2 Reinforcement Distribution
The bars will have a diameter of db := 1 · in and an area of As := 0.79 · in2 , therefore, meaning that the moment of inertia and the section modulus will be:
Itotal := lx + 4 · As · (T/2)² Itotal = 546.3 · in^4
Stotal := Itotal/(d/2) Stotal = 89.2 · in³
The maximum stress in the outermost fiber will be:
fb := M/Stotal fb = 29666.5 · psi fb/Fb = 1
Check := if (fb > Fb, “N.G.” , “O.K.”) Check = “O.K.”
Then, the use of 4 #8 bars is enough for the section to reach the required capacity for the additional loads to which the section is subjected to.
Finally, we will analyze the method of fixing the bars to the metal beam to achieve the joint work between the beam and the bars. In order to do this, small lengths of welding will be placed at equal intervals to hold the bar along the beam.
3. The distribution of welding along the steel beam
We will make the following consideration, the upper bars are working under compression, then between the welding sections there must be a distance such that the bar does not fail due to stability. In the columns, the spacing between stirrups must be equal to or less than 16 times the diameter of the longitudinal bar to meet the same type of failure. An analysis of the Euler formula gives similar results. Therefore, we will use a maximum spacing of 16db.
S := 16 · db S = 16 · in
On the other hand, the welding must be capable of supporting a lateral load that is sufficient to hold the bar in position. #3 stirrups will be used in columns with #10 bars or smaller. In the corners there will be bars bent at 90 degrees, then the resulting service load will be equal to:
N := 0.11 · in² · 0.6 · 60000 · psi · 1/cos(45°) N = 5600.3 lb
The welding must support the load N. Thus, in Fig. 3 the welding of the bar to the metal section is shown, similarly, to the welding of the bars in the connection of the bars in the brackets.
Fig. 3 Welding
4. Things to take into consideration when welding
Considering that the welding is E70XX with a capacity of 70000 psi. Being that Iw is the length of the welding, we have:
Allowable shear = 0.3*70000 psi = 21000 psi = Fw
Welding dimension = db/2
N = Fw(db/2)lw
Iw := N/(21000 · psi · db/2) Iw = 0.5 · in
According to the AISC, the minimum length of the welding should be 4 times the width of the welding, thus we have:
Minimum length = 4*db/2 = 2 in
Later, we will use a weld of db/2 = ½ in with a length of 2 in.(Fig. 4) spaced at 16 in and at each extreme a length of 5 in. will be used.
Fig. 4 Welding distribution. Elevation.
Translation By: Ivet Llerena (Eastern Engineering Group)
Structural Drawings By: Eastern Engineering Group
© 2022 This article was written by Ernesto Valdes and published by Eastern Engineering Group. All rights reserved.