Stability II report - Lastendaling van een rijwoning PDF

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Lastendaling van een rijwoning...

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2018-2019 A

UNIVERSITEIT ANTWERPEN FACULTEIT TOEGEPASTE INGENIEURSWETENSCHAPPEN

Stability II Group 6 Benramdane Yassine Celik Bedirgan Ghorbani Hassan Robertson Nikolai

Professor: Uwe Muehlich Serge Vandemeulebroecke

Introduction Before starting the construction of a building or a series of buildings, the stability of all aspects related to this building should be checked. A general stability check includes a load reduction from roof to foundation by means of load cases, the related settlement, safety checks through design calculations, etc. The assignment contains nine row houses in Boechout that need to be inspected. In order to do this in an efficient way, a detailed calculation is made for the first house (lot 70) and afterwards the differences with the other houses are discussed followed by the possible influences and consequences. All this is discussed briefly and concisely, with the necessary results, in the report. The detailed discussions and calculations can be found in the appendix.

Roof All the row houses have a gable roof with an angle of 45 degrees. The roof consists of rafters and purlins made of solid timber. Two approaches have been applied for the roof modelling. The first approach only takes vertical loads into account. These vertical loads consist of the weight of the roof construction and the snow load. Due to the symmetrical shape of the roof construction and the loads, one can make a simplification and only one half of the roof is considered. The supports are chosen in such a way that the entire structure remains stable and that the reality is approached as closely as possible. At the bottom left corner, a hinge has been placed so that the roof construction cannot perform any translation movement. Figure 1 shows the selected supports. The second approach also takes into account the horizontal forces generated by the wind. Because the wind load is not symmetrical, it is no longer possible to simplify the construction by using symmetry (see figure 2).

Figure 1: roof model with symmetric load Figure 2: roof model with asymmetric load

First approach The first approach results in a 2-times hyperstatic system that is calculated by hand. Figure 3 shows the selected systems used in the manual calculations. The compatibility equations are derived from this figure.

Figure 3: selected systems with compatibility equations

From this, the two unknown reaction forces are calculated. After the manual calculation, the construction is calculated in diamonds so that a comparison can be made between the manual calculations and diamonds. Table 1 shows that the two calculation methods show similar results, which means that the hand calculations were correct, and that Diamonds can be used for the second approach. Table 1: hand calculation vs Diamonds calculation for the first approach

Reaction force

Manual calculation

Diamonds

Ax

31,05 kN

30,58 kN

Ay

33,26 kN

32,22 kN

By

25,91 kN

26,94 kN

X1

26,60 kN

25,69 kN

X2

4,45 kN

4,89 kN

Second approach

The second approach takes into account all the vertical loads as well as the wind load. The wind load is calculated according to Eurocode 1. The magnitude and direction of the wind load depend on the location, orientation and height of the building. This approach also results in a 2-times hyperstatic system. Figure 4 shows the “0”-system, “1”-system and “2”-system that can be used to solve this hyperstatic construction manually. However, Diamonds is used since the correctness of this program was proven in the first approach.

Figure 4: systems that can be used for the hand calculation

Table 2 shows the reaction forces obtained from Diamonds. From this table it can be concluded that the wind has a limited influence on the reaction forces. This is because the height of the building is small. Table 2: Reaction forces generated by Diamonds for the second approach

Reaction force

Diamonds

Ax

28,55 kN

Ay

29.62 kN

X1

30.92 kN

X2

24.54 kN

Cx

32.45 kN

Cy

34.54 kN

Load reduction After obtaining the reaction forces of the roof construction, it was possible to begin with the load reduction. The roof is attached to the attic floor, so it does not rest on the masonry. For the construction of the attic floor, the choice was made for concrete hollow core slabs. This simplified the calculations, because the floor was carrying in one direction. There is been opted for the shortest span because of the fact that oversizing had to be avoided and also due to economic reasons. Afterwards the floor had to be divided in several zones, because each zone confronts a different type of load case. There are also zones with additional supports due to a bearing wall. Also, the span varies. In addition, there are also beams and lintels added if necessary due to an interruption of a bearing wall, so the core slabs could rest on these beams and lintels. The same assumptions and approaches were also made for the first floor. For defining the supporting walls, the ground floor had to be checked. For the acting forces, it was necessary to look at the upper floor. In this case most of the core slabs were placed horizontally. So, a vertical bearing wall, for example, was defined as a point load. This point load equals to the reaction force of this bearing wall on the upper floor including its selfweight. Horizontal walls are defined as a distributed load. The following figure... illustrates the implementation of the thinking for the first floor. The attic floor can be found in the appendix. The reasoning is that the first floor contains more varied load cases. Very important to know is that in this theoretical explanation, the hinges of the resting floors always find themselves at the outer walls.

Figure 5: analysis load cases first floor

Table 3: legende analysis load cases first floor

Legende Bearing Direction Supporting Wall

Beam Concrete hollow core slabs

•

Load case 1

Figure 6: load case 1

This load case defines a flat roof. It is calculated as an isostatic imposed floor plate as one is dealing with a single-family house. The apparent clamping can therefore be replaced by a support. In the case of higher buildings, this should be adapted to a clamping. The outer wall of the rear side is also supported by the flat roof, which generates a perpendicular and axial force and a moment at that point. •

Load case 2

Figure 7: load case 2

At first this load case seems like a simple imposed floor but in addition to the deadweight and permanent loads, there is also a point load caused by the deadweight of the vertical separation wall on top of the floor. There is also a point load above the support point B (roller) because of the stairs. This is guided directly down, so it does not affect the concrete hollow core slab. But this point load above support B has to be taken in account for the load reduction. •

Load case 3

Figure 8: load case 3

This load case is almost identical to load case 2. This means that there is also a point load caused by the deadweight of the vertical separation wall. The only difference is that the stairs don’t have to be taken in account. •

Load case 3’

Figure 9: load case 3’

This load case is more complex than most of the other load cases. Besides the vertical separation wall, there is also a horizontal separation wall over a certain length (not over the total length of the floor). The weight of this horizontal wall is taken as the second distributed load. However, this wall only has a thickness of 9 cm. The deadweight is spread over a width of 1 m (unit width), as the reinforcement is calculated using a beam with the same width. Distribution is done as follows:

• Wall thickness: 9 cm • Height masonry: 3,5 m The self-weight per running meter has to be determined! • Total volume/m: 0.09*3.5*1 (unit length) = 0,315 m3/m • Density of masonry: 1800 kg/m3 • Wall weight/m: 567 kg/m àWall weight/m: 567*9,81 = 5,562 kN/m However, this must be spread over 1 m width instead of 0,09 m. This multiplies the weight of the wall by 0,09. The second distributed load will then be: 5,562 * 0,09 = 0,5 kN/m •

Load case 4

Figure 10: load case 4

In this load case an additional support is placed in point B by the load-bearing wall on the ground floor. This makes the load case hyperstatic. Again, there is a second distributed load through a horizontal separation wall. This load case also supports the stairs. In support C, we obtain tension, so that the masonry is theoretically raised. But in support C of load case 4, support B of load case 6' comes together. Support B of load case 6' creates a high pressure, so that the masonry still undergoes a compressive stress. •

Load case 4’

Figure 11: load case 4’

In this load case the dead weight and a point load, because of the vertical wall on top, are taken into account. The support in point B also makes this load case hyperstatic. •

Load case 5

Figure 12: load case 5

This load case is the same as load case 4, except for the place of support and point load. Also, at support point B there is a point load from the load-bearing wall on the first floor. This is not taken into account here. However, this must be taken into account when determining the load reduction.

•

Load case 6

Figure 13: load case 6

For load case 6 both sides are seemingly trapped by the masonry on the floor. However, as with the previous load cases, these apparent entrapments are replaced by a simple imposed floor. • Load case 6’

Figure 14: load case 6’

This load case is the same as load case 6. However, there are also two horizontal dividing walls that act as distributed load. These are calculated the same way as in load case 3’. •

Pillar (Load case 6):

Figure 15: load case 6

Instead of a hollow core slab, this is a load case which contains a concrete beam at the end of a hollow core slab. The beam supports point A of load case 6. To simplify the calculation, the pillar is defined as a point load. To calculate the point load By, which represents the reaction force on top of the pillar, we calculate the concrete beam on top as an isostatic calculation. This is essential for the load reduction.

Figure 16: load case pillar

Horizontal wind load on façade The wind load does not only affect the load reduction via the roof structure, but also via the masonry. First the wind pressures have to be generated. This is done with the BBRI program based on two factors: the location and the orientation of the house. After the correct data has been implemented in the program, both pression and suction values are generated per surface zone. The values can be found in the appendix. The wind load on a certain façade is assumed to be carried from the face brick on to the inner masonry through cavity anchors that are placed every 75 cm. There are two methods used in the calculations for the load reduction. The first method uses the theory of plates where the shortest span is seen as the weakest direction. This means that, according to the placement of the bricks in the brickwork, the long horizontal strips of mortar are the weakest components. In this method the wind load is seen as a vertical line load on the façade. For the calculation the cavity anchors are assumed as clamped on to the inner masonry, which makes the calculation x times hyperstatic. The reaction forces obtained by Diamonds have to be summed up from top to bottom (on figure… this means from left to right because it is shown horizontally) to carry the loads down to the foundation. In figure … the calculation of the reaction forces is shown for the front façade. The calculation of the other façades can be found in the appendix.

Figure 17: load case example front façade

The second method takes the wind load into account as a horizontal load, which shows that the horizontal length of the masonry is the bending direction. This calculation was done to compare the results with the first assumption. If the wind load is seen as a horizontal load, the pression or suction value (in kN/m2) has to be multiplied with the height of the masonry. The maximum reaction force obtained with Diamonds, gets carried down to the foundation (worst case). In figure … the calculation of the reaction forces is shown for the front façade. The calculation of the other façades can be found in the appendix.

Figure 18: example front façade

After generating the results, table 4 shows that both methods generate similar results. The results of the other façades can be found in the appendix. Table 4: results front façade

Vertical [kN/m]

Horizontal [kN/m]

8,305

7,12

Reinforcement After the analysis of each load case, the moment diagram was drawn to determine the maximum bending moment and its corresponding required reinforcing bars. Of course, at the end a reinforcing bar suitable for each load case was chosen. This reinforcement was provided from the bottom as well as from the top. This choice was made on purpose because there are also hyperstatic cases. To illustrate with an example, reference is made to load case 5, where additional support is provided. If the deflection is drawn, it can be concluded that in point 1 there is compression from above and tension from bottom. Due to the load. In point 2 there is tension from top and compression from bottom due to support B.

Figure 19: load case 5

According to the calculations, it is sufficient to provide the following reinforcement for the core slabs: Table 5: reinforcement

Diameter bar: Number of bars from top: Number of bars from top: Distance between bars

10 mm 9 9 10,6 mm

Steel beam On the ground floor plan, it can be clearly seen that there is no bearing wall between the kitchen and the living room. This implies that there is a need for a beam, on which the concrete hollow core slabs of the kitchen can be placed vertically. This beam is supported by the vertical outer walls and horizontal inner wall. For the calculation of this beam, the horizontal wall is been replaced by a distributed load, acting in the opposite direction.

Figure 20: steel beam

It's clear that this is a hyperstatic case. With “0”-system the support is neglected, and the load is kept, while with “1”-system the load is neglected, and the support is kept. Ultimately, the sum of the deflections of both systems at the level of support "X" must be zero to obtain a value for X. Afterwards X can be divided by the length of the horizontal wall to get supporting force per running meter.

Figure 21: “0”-system & “1”-system for the steel beam

Service limit state In addition to the ULS check, a Service Limit State (SLS) check must be performed. It is very important that the maximum deflection of the floor does not exceed L/250. The most extreme load case has been chosen for both floors (Load case 4’ for first floor and load case 2 for attic floor). Of course, this depends on the crosssection of the slab and the stiffness of the material. The following dimensions have been taken into account:

Table 6: stiffness of the slab

b, slab

1000

h, slab

180

d, hollow core

100

Number of hollow cores

7

Moment of inertia, Total

= I, rect – 7 * I, hollow core

Stiffness

17500 kN/m2 Figure 22: determining I with Steiner

According to the calculations of both cases, load case 4' has the biggest deflection at x=2.65 with w(x=2.65) = -5.11 mm. The maximum allowable deflection is 16.76 mm, so all load cases meet the requirements.

Settlement The settlement of the building is calculated using two methods in service limit state (SLS). In the first method the settlement of plate and sole is calculated separately (see Figure 23) because in practice, the floor is completely supported by the ground. Based on Terzaghi's bearing capacity theory the setting of the sole is calculated by hand according to NBN EN 1997. The sole support outer bearing walls. In this case, a settlement of 10 mm is obtained, see Table 7. The setting of the plate is determined by Diamonds because the forces from inner bearing walls are not uniformly distributed over the plate. Diamonds gives a maximum settlement of 12 mm.

Figure 23:The sole and plate in method 1

Table 7: Settlement from two methods

Method 1

Method 2

Hand calculation

10.12 mm (sole)

4.59 mm (plate)

Diamonds

12.25 mm (plate)

14.40 mm (plate)

Method 1 is a theoretical calculation of the settlement, but in practice method 2 is often used. All forces are spread over the entire surface of the plate. This method is more unfavourable and has a bigger influence on the settlement, as shown on Figure 24. Foundation with lager surface has a greater influence line which reaches more underlying ground layers.

Figure 24:Influence line of foundation

In method 2 the settlement is calculated manually and by Diamonds. The difference is due to the compression coefficient of the soil and other factor which Diamonds takes into account, e.g. soil consolidation. In both methods the settlements are acceptable because the maximum allowed settlement of the sole and plate are 15mm and 50 mm. [1]

Design safety checks (ULS) Timber roof Once all the forces have been determined, all roof elements can be checked on design safety according to NBN EN 1995. Based on the direction of the forces, it is possible to know what can be checked per element, e.g. bending and compression for element 1 (see Figure 25).

Figure 25: Timber roof construction

Table 8 shows the results from design safety checks of roof elements. All elements are safe because the ratios are smaller than 1. Table 8: Results from safety checks roof

Element

Formula

Ratio

Check

nr. 1

Bending:

smyd/fm,y,d & Km*smyd/fm,y,d

0.88 & 0.62

OK

Compression:

sc,a,d/fc,a,d

0.08

OK

2

Compression:

sc,0,d/fc,0,d

0.16

OK

Buckling:

sc,0,d/ kcy*fc,0,d & kcz*sc,0,d/fc,0,d

0.18 &0.79

OK

3

Compression: sc,0,d/fc,0,d Buckling: sc,0,d/ kcy*fc,0,d & kcz*sc,0,d/fc,0,d

0.16 0.16 & 0.24

OK OK

When the ratio is too small, an optimisation can be applied, e.g. for element 3 a smaller cross section or another wood class can be chosen.

Masonry Different types of masonry (face brick, inner masonry, pillar) can be checked on design safety according to NBN EN 1996. Compression and shear are here most important to check. Table 9 shows the results for different types of masonry. The most loaded element is checked per type of masonry. The results show that the different types of masonry fulfil the safety check. Table 9: Results from safety check masonry

Type of check

Formula...