Champlain Towers Collapse Structural Analysis Revised

Champlain Towers Collapse Structural Analysis Revisal

After the article about the collapse of Champlain Towers South structural analysis was released, the criteria of the collapse continued to be under analysis. We have received more photographic information, and more up to date plans than what we were using before, and we viewed videos of other works relating to the collapse like opinions from local professionals.

Furthermore, we detected some modifications to the original project of 1979, and have also learned of changes made to the slab because of the repairs after the inspections.

The most important aspects are:

 

1. A reinforced slab (topping) was added to the slab covering the parking lot with a welded wire mesh and paver.
2. The level of the slab covering the parking lot was elevated 12 inches from the initial level +10’ 10” to +11’ 10”. The difference of levels between the new slab and the lobby in the building (+13’ 4”) was modified from 2’ 6” to 1’ 6”.
3. The unevenness of 1’ 0” was eliminated at the K axis and the beam BMA was eliminated in axis I.
4. It was found that a typical Note about the distribution of the reinforcement in the band of columns seems to not have been applied during construction.
5. A new level was added, the penthouse, now making the building have thirteen floors.

 

It is possible that there were some changes in the distance between axes, elevations, etc.; however, it is impossible to provide all the information to be aware of other changes. If those changes do exist and they can inform us of the same, we would be very grateful.

These new situations have obliged us to revise the conclusions obtained in the first article. The following will update each aspect discussed.

Additionally, we will demonstrate the reason why the other part of the building did not suffer a collapse despite exposure to similar conditions such as the zone that collapsed.

 

Analysis of the Concrete Slab Failure in the Roof of the Parking Lot

Punching Shear Failure

The concrete slab has 9 ½ in thickness with concrete strength of fc = 4000 psi and the N columns are 12 in x 16 in. We can find the most unfavorable condition in L/11.1 as shown in Fig. 1. The following will show calculations for the failure due to punching shear, which will include all the loads, including the live load of 60 lb/ft². To this, we add the weight of 4 in. of topping and 10 lb/ft² of paver.

The tributary area:

 

.                                                                                                                                     Atrib := 19.83 • ft • 22 • ft = 436.26 ft²

 

 

Actual service loads:

.                                                                                                             VDL := Atrib • [(9.5 • in + 4 • in) • 150 • (lb/ft³) + 10 • (lb/ft²) = 77.98 • kip

.                                                                                                                                        VLL := Atrib • 60 • (lb/ft²) = 26.2 • kip

 

 

 

Fig. 1 Punching Shear Failure at column N in the parking lot area

 

Factored load:

 

.                                                                                                                                   Vu := 1.2 • VDL + 1.6 • VLL = 135.5 • kip

 

The effective height:

.                                                                                                                                       d := 9.5 • in – 0.75 • in – (Φ/2) = 8.438 • in

.                                                                                                                                       fc := 4000 • (lb/in²)

 

 

Reinforcement:

.                                                                                                                                                Φ := 5/8 • in     as := 40

 

 

Critical section perimeter:

 

.                                                                                                                           b := (12 • in +16 • in + 2 • d) • 2 = 89.75 • in

.                                                                                                                           ΦVc := 0.75 • 4 • (lb/in²)^½ • √fc • b • d = 143.7 • kip

 

 

Meaning Vu < ϕVc so the 9 ½ in. slab does not fail due to punching shear even if there is an increase of 4 in. in the slab and the paver, as well as all the live loads.

We confirm the former in the photos, if it were this kind of failure, the columns would have a cone shape in the upper part, and this was not the case.

 

Failure due to Bending

The 9 ½ in concrete slab without beams working in two directions has a positive reinforcement of #4 @ 12” o/c in both directions and a negative reinforcement of 16 #5 in one direction and between 16 #5 and 19 #5 in the perpendicular direction as shown in Fig. 2. This reinforcement corresponds to the zone of level +10’ 10”.

 

   Fig. 2 Negative Reinforcement in the zone of level +10’ 10”

 

In the Typical Notes of plans S-6 it is shown that 25% of all the reinforcement in the band of a column will be centered over the column. Of the 16 bars indicated, 25% would be 4 bars which should be placed in the center of the column. However, in the picture of axis I, it can clearly be seen that this specification was not accomplished, which leads to the conclusion that it was not done in any other case.

In the other direction, the distribution of reinforcement is approximately the same and the same observation regarding the centered reinforcement in the columns was made.

This negative reinforcement is definitely what has been affected by the corrosion due to the fact that it is in the zone in which the radial fissures were developed. This is because of the concentration of the flexural moments over the columns in both directions. Fig. 3.

First, the negative bars fail due to the corrosion, Fig. 3, then the positive bars are followed because now they do not have sufficient longitudinal anchorage and the collapse is produced from the 9 ½ in. slab. Finally, the 4 in. concrete slab topping fails due to punching shear. The previous corroborates due to the form in which the superior part of the columns of the zone in the parking lot remains.

Fig. 3 Parking lot slab failure process due to corrosion

 

Failure due to punching shear in columns A of Axis I

 If we observe Fig. 4, corresponding to axis I of the part of the building that collapsed, we can observe that the positive and negative bars of the slab protrude from the column in both directions and there are no observable effects of the corrosion in the same.

The slab of the parking lot covering 9 ½” continues at the same level of +11’ 10” penetrating within the zone of the building that withstood the collapse. This slab failed due to punching shear stress, which is evidenced in the photos showing the collapse of the slab, which show the reinforcement coming out of the columns and in some places the shape of the concrete in the form of a wedge can be seen. The following will analyze the causes of this failure.

 

 

Image by New York Times

Fig. 4 Photo of Column C of the axis I/14 after the collapse

It is clear that if the columns measure 24” x 24” and the slab is subject to similar vertical loads, the capacity of the punching shear for the vertical load would be greater than columns N of the parking lot. However, if the columns of the parking lot near this axis fail, the tributary loads increase and produce unbalanced moments which generate much greater normal stress in the critical section. This is what caused the failure due to punching shear in the slab of axis I, as shown in the photo.

A model of the slab using the program software SAFE, shows us that if one of the columns C located at I/14, when the column of the parking lot K/14.1 suffers the collapse of the slab due to corrosion, the slab over column C of I/14 cannot support the increase in vertical load and the effects of the unbalanced moments generates superior normal stress than the capacity of the 9 ½ in. slab.

We concluded that the slab of the area of the parking lot fail due to the corrosion of the negative reinforcement. This causes the rupture of the slab around the column. The slab of the columns in axis I fail due to punching shear provoked by the failure of the slab in the adjacent columns.

Failure due to punching shear in the Columns C of axis 9.1

 Lastly, we want to bring attention to the fact that in axis 9.1 the difference of level of 1’ 6” originates a beam of 12” x 27 ½ “ along that axis and there exists a beam called BMA of 12” x 15” perpendicular to each column. The presence of this beam makes it so that it is not possible for a failure to occur due to punching shear of the slab 9 ½” in these columns. Fig. 5. Later on, we will analyze the possibilities that caused the failure of the columns.

 

Fig. 5 Section of the beam along the axis 9.1

 

The Concrete Slab Acting as a Shell

Once the slab has failed in one column, the one with the worst corrosion, the remaining surrounding columns receive more loads and generate a progressive collapse.

After this moment is where the major problem commences. Supposedly, the positive reinforcement has not been affected in the same way as the negative reinforcement; therefore, the steel begins to behave in a different way. Additionally, the welded wire mesh of the topping slab adds to the positive reinforcement, giving way to a type of cable along a band as shown in the following.

In order to explain the new function of the slab and its positive reinforcement, we must choose a strip that includes half of the slab from both sides. For example, N axis is in between the pool and axis 9.1, as shown in Fig. 6.

 

 

                          Fig. 6 Strip of the slab acting as a plane in Gridline N

 

Due to the cracking of the slab over the column, the slab falls. The increase in loads for the adjacent columns causes the failure. This process repeats itself in both directions. If the slab had simple support in the edges, then the slab would have fallen to the floor and the problem would not have continued. Only the slab over the parking lot would have fallen. However, this did not occur. In actuality, the slab is anchored to the pool and the retaining wall on one side and on the lower part of the beam on gridline 9.1, about 2’ 6” under the slab at a level of +13’ 4”.

The slab starts to work like a shell anchored at both of its extremities. The positive and negative reinforcement throughout the band conforms to a cable with a sum of the areas of the rebars throughout the span of the strip analyzed.

Considering only the positive reinforcement of the 9 ½ “ slab and the topping reinforcement, for the total span of B = 19’ 0”:

 

Area of all the rebars in strip width:

 

.                                                                                                   As := ((19 • ft/12 • in) +1) • 0.20 • in² + 0.028 • (in²/ft) • 19 • ft        As = 4.532 • in²

 

All the rebars are equivalent to one cable, which we can find it supported on both extremities and in turn begins to act as a shell. After the treatment due to the slabs own weight, the slab is subject to tension and begins to fissure. In this way, the steel has the effect of a cable, similar to a cable used for handrails. The pictures show how the slab stays fixed in the retaining wall and detaches from the pool, causing the reinforcement of the anchor to fail. Fig. 7

A similar situation presents itself in a perpendicular direction, originating bands that start at the pool and in the retaining wall on one side and ends in the building at axes G.1. First, we will analyze the band in the axis N and later on we will comment on the band in axis 14.1 when the case of columns A are studied.

 

Fig. 7 The effect of the shell (working as a steel cable) along Axis N

 

 

Horizontal Force Created by the Effect of the Cable Supported on the Extremities

Effect of the slab failing over all the columns in the parking lot

 

.                                                                                          L := 60.33 • ft        l := 60.33 • ft                                                                           yc := 150 • pcf

.                                                                                          B := 19.0 • ft                                                                                                        kip := 1000 • lb

.                                                                                          h := 9.5 • in

.                                                                                          w := B • (h + 4 • in) • yc + 10 • (lb/ft²) • B                                                               w := 3396.3 lb/ft

 

 

Spc := 12⋅ in

Number of bars:

.                                                                                                                                           n := ((B/Spc) + 1)             n = 20

.                                                                                      As := 0.2 • in² • n + 0.028 • (in²/ft) • B              As := 4.532 • in²                          Es := 29000000 • psi

 

Deformation at the center:

.                                                                                                                      a := [((3 • w • L)/(64 • Es • As))^1/3] • l                a = 2.522 ft

 

Horizontal force at the extremities:

.                                                                                                                              T := (w • l²)/(8 • a)                            T = 613 • kip

 

When the steel cable is under this force, the tensions are:

.                                                                                                    Fa := T/As              Fa = 135173 • psi         > 85 ksi    strength of steel at failure

 

Surpassing the yield strength of reinforcement A615. Therefore, the maximum strength of the cable corresponds with its job of breaking tension.

The maximum strength of the steel cable will be:

.                                                                                                                      Tmax := As • 85 • ksi                          Tmax = 385.22 • kip

 

 

In the photos, we can consider that this is the most probable cause of the failure. We will suppose that the horizontal force of the cable reached a value of 385 kip which corresponds to the moment of the failure in the cable composed of positive reinforcement of the slab plus the reinforcement of the topping using steel of 85 ksi.

Structural Analysis of a Column “C”

We must consider the column located on gridline N and gridline 9.1 in the lower level. This column is 16” x 16” with 8 #11. That is about 4.88% which is high. We must consider that when we splice the bars of the foundation this number doubles to more than 8% which is the limit to the code specifications.

The column has a height of 10’ 9” and is a pin at its extremities. Which results in a conservative estimate. For the following analysis, there is an exclusivity for the effect of loads of the building self-weight components, with an overload of 20 lb/ft² and the weight of the exterior walls. When the building collapsed, the building had a minimum live load which does not influence the following analysis as shown in Fig. 8.

We suppose that there was no failure due to punching shear and the effect of the cable did not originate any failure of the slab other than that of axis 9.1.

 

 

                     Fig. 5 Maximum load on column C of gridline N and 9.1

 

The tributary area:

.                                                                                                                    Atrib := 10.67 • ft • 20 • ft + 5.67 • ft • 6.58 • ft = 250.709 ft²

 

Number of floors:

n:= 13  (11 floor slabs + roof slab + penthouse)

Dead Load DL:

 

.                                                                                                         SlabDL := [9.5 • in • 150 • (lb/ft³) + (n – 1) • 8 • in • 150 (lb/ft³)] • Atrib = 330.6 • kip

.                                                                                                         ColDL := n • 16 • in • 16 • in • 150 • (lb/ft³) • 8.83 • ft = 30.6 kip

.                                                                                                         SIDL := n • Atrib • 20 • (lb/ft²) = 65.2 • kip

.                                                                                                         WallDL := n • 8.83 • ft • 36 • (lb/ft²) • 20 • ft = 82.6 • kip

TotalDL := SlabDL + ColDL +SIDL + WallDL = 509.1 ⋅ kip

 

The vertical service load at the moment when the building collapsed was 470 kip.

In the Fig. 6, the column is the horizontal load of 350 kip acting at the level in the middle of the slab. The column is now in both extremities. This results to be the real situation and is going to be the condition that creates the least bending moment. Thus, we will be minimizing the effect of said load so that the evaluation is as safe as possible.

 

Fig. 9 Column N subjected to the effect of the horizontal load

 

.                                                                                                    P := 385 • kip              a := 8.87 • ft                  b := 1.5 • ft                    l := 10.37 • ft

 

The maximum bending moment for the service load:

.                                                                                                     Ma := (2 • P • a² • b²)/l³ = 122 • kip • ft      at the point of horizontal load

.                                                                                                     M2 := (P • a² • b)/l² = 423 • kip • ft             at the support closest to horizontal load

 

Therefore, column N is a vertical service load and a bending moment of:

 

.                                                                                                                        P := 510 • kip                                   M2 := 423 • kip • ft

 

Now we will obtain a diagram of the interaction for column N with a section of 16” x 16” reinforced with 8 #11 and concrete compressive strength of 6000 psi.

In Fig. 10 we see the diagram of the interaction obtained from the program software spColumn

 

Fig. 10 Interaction diagram of Column N

The continuous line represents the capacity of the section for the design after the combinations of the loads factored in the specifications, while the dashed line represents the nominal values, meaning the actual capacity of the section for a combination of axial load and bending moment. In other words, if the section is under a combination of P and M, the point that defines this combination should be in the interior of the nominal diagram so that it does not fail; however, if it is on the outside, failure occurs immediately.

In this case the point, which has:

 

.                                                                                                                        P := 470.3.1 • kip                              M2 := 544 • kip • ft

 

Falls exactly on the nominal diagram; thus, this column fails at the instance at which the combination starts to act simultaneously. We should remember that the unbalanced moments produced by the topping and paver have not been taken into consideration.

After this instance, the failure of the first column produces the gradual collapse of the remaining columns due to the redistribution of the loads to those adjacent columns, giving way to a type of failure similar to that of demolitions where explosives are near the lower columns.

If we now consider that the failure of the slab happened in axis 8, similar to what occurred in columns A where the failure of the slab occurred in axis G.1, column C would have an effective height being 27’ 0” – 2’ 2” – 8” = 24’ 2” which leads to a failure because of slenderness.

Another scenario that causes the failure of the structure, is the effect of the horizontal product of the work done as a shell in the slab of 9 ½ “ which fails due to the corrosion of the negative reinforcement.

Structural Analysis of a Column “A”

The three columns A of axis I did not receive the effect of the horizontal load of the shell because of the failure due to punching shear of the slab as indicated previously. The horizontal forces transmitted throughout the slab distribute the force on 4 interior columns. In other words, the effect of the  diaphragm shared the horizontal load between each column.

Column A is 24” x 24” with reinforcement of 12 #10 and fc = 6000 psi. The effective height of column A after the failure is a result of a punching shear that increased to 27’ 0” – 8” –2’ 2” = 24’ 2”.

In this case, supposing that the loads are similar to column C, we can prove that there is no failure due to slenderness when the column reaches an effective height of 24’ 2”

 

Conclusion and Observations About Building Design

The fundamental conclusion of this analysis is that the primary cause of the Champlain Towers Collapse is due to the corrosion of the bars at the top of the concrete slab over the parking lot located above the columns. The slab failure created a slab shell effect which originated the horizontal loads that caused the failure of the columns of the building and subsequently causing the progressive collapse and failure of part of the building. All because of poor maintenance.

Lastly, I would like to make observations that could help with the future design of buildings located near the ocean, with slabs at different levels on the same floor and the distribution of the shear walls of the structure.

a) If we observe the details of the slab reinforcements, we find that the net cover of the negative reinforcement bars in the slab is 3/4”. Even if the codes specify this amount, it is convenient to increase the coverage in situations where there is salt water, balconies, etc.

b) The difference in the level of the floor slab within the building and the roof of the parking lot is 2’ 6”. It is evident that the parking lot does not need elevated height, giving way to this difference in elevation. However, it is important to note that the effect of the horizontal slabs can produce horizontal loads due to the retraction of the concrete, the effects of prestressed cables within the slab, the effects of wind loads, or in this case, something unusual, the effect of slab shell.

c) In the situation of differential levels, we can achieve a discontinuity if we place a beam with a column separated at 1 or 2 in. from the column of the building or placing a support (ledge) in the beam located at the edge, avoiding the continuation of both parts. Fig. 8

d) Finally, another important factor that I noted and would like to expand on is the shear walls. If we look, the progressive collapse finalized at the moment where it met the retaining wall of gridline F.3 but it went over the retaining wall in gridline M, failing there as well. I would like to comment that they designed these walls for wind acting in one direction.

 

 

 

Fig. 8 Different connections between the slab of the parking lot and the building

On behalf of Eastern Engineering Group, we give our sincerest condolences to all affected by the Champlain Towers Collapse. This article reflects the studies of Professor Ernesto F. Valdes regarding the Champlain Towers Collapse and he wrote it specifically for Eastern Engineering Group to publish on his behalf.

 

 

Translation By: Ivet Llerena (Eastern Engineering Group)

Structural Drawings By: Eastern Engineering Group

© 2021 This article was written by Ernesto Valdes and published by Eastern Engineering Group. All rights reserved.