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 Det Tekniske Fakultet

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 Det Tekniske Fakultet

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 Det Tekniske Fakultet

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 Det Tekniske Fakultet

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. Det Tekniske Fakultet

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 Det Tekniske Fakultet

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 Det Tekniske Fakultet

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 Det Tekniske Fakultet

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 180+70mm (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 Det Tekniske Fakultet

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=11664+12768+3456+840 = 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 = 672+560 = 1232kN Horizontal mass load: Ad=0,015 (28728+1232) = 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 Det Tekniske Fakultet

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? Det Tekniske Fakultet

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? Det Tekniske Fakultet

Wall panel with tension - elastic design: Det Tekniske Fakultet

Anchoring the wall Stigbøjle/tophat, armering i fuge eller korrugerede rør eller spændliner: http://expan.dk/ http://www.spaendbeton.dk http://www.peikko.dk/products-dk Det Tekniske Fakultet

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 Det Tekniske Fakultet

Shear in vertical connections Example: Det Tekniske Fakultet

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. Det Tekniske Fakultet

Plastic design – Example 3 The wall is anchored with the force ”T” → Det Tekniske Fakultet

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. Det Tekniske Fakultet

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: Det Tekniske Fakultet

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) Det Tekniske Fakultet