1. Program Stability of a wall panel - elastic and plastic.
Stability - Which walls to choose?
Should we use the wind load or the seismic load?
Elastic design of a panel
We calculated the stresses a bottom of the wall, what if we have tension in the wall?
Shear force in a vertical connection in a wall.
Plastic design
If the eccentricity becomes to great?
Shear force in a vertical connections.
Project
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2. Intro – Stability of wall panel
The stability in a building is to be understood as its ability to withstand horizontal forces so gliding, rotation or lift wont occur
2 methods: Elastic or plastic
Lift/rotation
Glidning
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3. Wall panel – Elastic design, tension
Stresses in the bottom of the wall is found by the use of Navier:
If there is tensile stress in the wall the wall will rotate and we need to aplly some reinforcement in the foundation (anchor the wall - Danish: trækforankres).
Example Pk =10kN i LAK 6.10b:
σ = -123kN/m2 og 383kN/m2
Tension! I.e. the wall has to be anchored
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4. Wall panel – Plastic design.
Eccentric vertical reaction
By the use of static equivalent the normal force can be moved the distance “e” and distributed over a smaller area
The effective area and the compressive stress is determinate.
The eccentric is calculated:
e=M/N
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5. Plastic design– Example 1
Wind load: Pk=10kN
N is the dead load of the concrete wall and is in favorable for the stability ( in danish – virker til gunst).
Nd=0,9*24kN/m3*4,0*0,2*6=104kN
Md=1,5*10(6+3)=135kNm
e=M/N=135/104 = 1,30m
Area in compression:
Ae=(4-2*1,30)*0,2=0,280m2
Stress in compression :
σc=N/Ae=104/0,280= 0,371MPa
Okay, because the material strength is significant higher.
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6. Wall panel – Plastic and elastic design (gliding)
It has to be check that the wall panel wont glide.
For a quick design we can apply:
V < µ·N
For concrete structures µ can often be set to 0,5 (concrete against concrete – different values if plast folie or “Murpap” is used)
It is only the compressed area Ae that can transfer shear forces
i.e. when combining profiles, shear force is only obtained in the web
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7. Choice of profiles Continues walls are chosen!
Combined profiles with flange up to 1 m can be chosen. It increases the moment of inertia and the area.
Stability walls which also are load bearing are more stable because more load is favorable and it also reduce the possibility for tensile stress i the profile
Approxi-mately 1. m
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8. Horizontal load A building is check for 2 types of horizontal loads:
Wind load, i LAK 6.10b
Seismic load (mass load/vandret masselast), i LAK 6.12a/b
Ad is 1,5% of the sum of the load. Ad=0,015 (G+ψ2,i * Qi)
For grandstand 15%. (for tribuner 15%)
Remember to add the load from imperfektion i both cases
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9. Example of wind load >< mass load on the gable
The example from chapter 4.8 in Teknisk Ståbi is used in the example.
The following are selected:
The length building 40 meter
Terrain category 1 for north and south gable
Leeds to qp=0,939 kN/m2
Horizontal mass load:
Ad=0,015 (G+ψ2,i * Qi)
G1: sandwich facades mm (Concrete)
G2: Dead load of the decks including floors and sealing 3,8kN/m2
G3: Internal walls 200mm, concrete
G4: Dead load of the roof structure 1,5kN/m2 (measured horizontally)
Q1: Live load in domestic areas 1,5 kN/m2
Q2: Live load in office’s 2,5 kN/m2
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10. Example of wind load >< mass load on the gable
Ad=0,015 (G+ψ2,i * Qi)
G1=24kN/m3*0,25*(2*40*18+2*14*18)=11664kN
G2= 3,8kN/m2 *6*14*40 = 12768kN
G3=24kN/m3*0,2*18*40 = 3456kN
G4=1,5kN/m2*14*40 = 840kN
G= = 28728kN
Q1* ψ2,i =1,5 kN/m2*14*40*4*0,2= 672kN
Q2* ψ2,i =2,5 kN/m2*14*40*2*0,2=560kN
Q* ψ2,i = = 1232kN
Horizontal mass load:
Ad=0,015 ( ) = 449kN
Distributed on the gable:
Ad,gavl=449/(18*14+0,5*4*14) = 1,6kN/m2
This value is included in formula 6.12a/b
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11. Example of wind load >< mass load on the gable
Wind load on the gable:
we=qp(D-E)ρ
D og E is a formfactor for wind on the facade and is calculates to D=0,74 og E=-0,38 (Because of h/d=22/40=0,55)
correlationsfactor ρ=0,85
we=0,939*(0,74-(-0,38))*0,85=0,89kN/m2
This value is use in formula 6.10b
6.10b: 0,9*G +1,5*ψ0*Q + 1,5*V*KFI (Q=0, when it is favorable)
6.12a/b: 1,0*G +ψ2*Q + Ad (ψ2=0,2 for domestic- and office areas)
(ψ2=0,2 for både boliger og kontorer)
Which situation is the most dangerous one?
What's the contribution from geometrical imperfection Vθ?
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12. Wall panel - elastic design-Example 2:
Example Pd=15kN in LAK 6.10b:
σ = -123kN/m2 og 383kN/m2
Tension ! The wall has to be anchored.
Consider if the profile/element needs any anchoring (it need anchoring if we have tension stresses)
Calculate the necessary reinforcement
Where to place the reinforcement?
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13. Wall panel with tension - elastic design:
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14. Anchoring the wall
Stigbøjle/tophat, armering i fuge eller korrugerede rør eller spændliner:
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15. Shear force in a vartical connection in a wall Elastic design:
Connections in walls the shear stress can be calculated by the use of Grashoffs formula:
When using Grashoff the horizontal shear stress is determinate and this is equal to the vertical shear stress and can therefore be use.
See example 7.2 in the book
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16. Shear in vertical connections
Example:
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17. Anchoring of a wall- Plastic design
If the distance ”e” is out of the profile the element will rotate.
Example:
P=20kN
M=1,5*20*(6+3)=270kNm
e=M/N=270/104=2,60m
The distance ”e” from the centroid to the normal force is out side the profile, so the wall will rotate.
It needs to be anchored.
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18. Plastic design – Example 3
The wall is anchored with the force ”T” →
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19. Plastic design – Example 3
The wall is anchored with a ”stigbøjle”, type 90 from the supplier “Expan” it has a tension resistance of T= 46kN.
The “Stigbøjle” is placed 1,5m from the centroid of th½e wall.
∑Fy=0 → N =G+T → 104kN+46kN=150kN
M = 270kNm = T*1,5+N*e = 46*1,5+150*e
→ e=1,34m
Compressive stress is calculated:
σ=N/Ae=150/=(4-2*1,34)*0,2= 0,568MPa
OK, course the compressive strength in the concrete is significant higher.
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20. Plastisk dimensionering – Forskydning
(copi of example 7.4 i the book)
The connection is to be design acoording to teori about ”støbeskel” (betonkonstruktioner efter Bjarne Christian Jensen)
Vertical equilibrium:
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21. Approach in stability design
Chose/define the stable walls
The horizontal load is distributed to the walls
Elastic or plastic
Reactions are determinate according to 6.10b and 6.12a/b. remember to include contribution from geometric imperfection
The walls are checked (plastic or elastic) for rotation/lift or gliding. If rotation occurs the wall have to be anchored.
One study for each level/floor. In practice some will be left out.
Shear stresses in the connections in the walls are determinate and the connection are designed.
Deck design
Connections between walls and deck are check according to stresses and robustness.
Connections between wall and decks/støbeskel (Betonkonstruktioner af Bjarne Christian Jensen)
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