Example_on_Seismic_including_Torsion

Example_on_Seismic_including_Torsion - Example For the...

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Unformatted text preview: Example: For the shown plan: it’s required to get the lateral forces due to seismic loads according to NBCC 2005 requirements. A 1 hs=180mm All columns (400*400)mm 1.0m 0.2 2m 0.25 5m 3 5m 5m 5m 5m 5m 5m B C D E F 2 2.0m 0.25 1.0m 0.2 Given: floor cover=1.5KPa, partitions=1 KPa, L.L=1.9KPa Storey height (h) =3.0m, No. of stories (n) = 8, γc = 24KN/m3 Snow load = 2 kPa , the building is located in Montréal. The soil class is (C), ductile shear walls are constructed. Solution: V= S (Ta ) M v I E W Rd Ro a) Estimation of W: W=dead load +25% of snow loads =(0.18×24+1.0+1.5) ×(10×25) ×8 + 18×0.4×0.4×3×8×24 column weight +2×(0.25×2+0.2×1) ×3×8×24+0.25×2×(25×10) Shear walls =16287.9KN b) Fundamental period (Ta): Cl .4.1.8.11(3) Ta = 0.05(hn )3 / 4 = 0.05(24)3/ 4 = 0.542sec For Montreal: from Table (C-2) Appendix C Sa (0.2) =0.69, Sa (0.5) =0.34 Sa (1.0) =0.14 Sa (2.0) =0.048 For soil class (C) : from Table 4.1.8.4 (B&C) Fa=1, Fv=1 To get Mv: Table 4.1.8.11 Sa (0.2) 0.69 = = 14.38 > 8 Sa (2.0) 0.048 To get J: from linear interpolation J=0.9832. Mv =1.0 To get Ro ,Rd : Table 4.1.8.9 Rd=3.5, Ro=1.6 (for ductile shear walls). To get S(Ta): S (Ta)=0.3232 Therefore, V = S (Ta ) M v I EW 0.3232 × 1×1 W = 0.05771W = Ro Rd 1.6 × 3.5 Vmin = Vmax = S (2.0) M v (2.0) I EW 0.048 × 2.5 ×1 = W = 0.0214W Ro Rd 1.6 × 3.5 2 S (0.2) I EW 2 0.69 ×1 W = 0.0821W =× 3 Ro Rd 3 1.6 × 3.5 OK Therefore V=939.97KN Vmin < V < Vmax Torsion effect on the building: Wall Thickness (m) A 0.25 (assume B ≤ 1.70) C 0.2 E 0.2 F 0.25 I (m4) A (m2) 0.1667 0.5 0.03255 0.25 0.03255 0.25 0.1667 0.5 h = 3m, E = 27×106 KN/m2, G = 11.25×106KN/ m2 Wall stiffness: ΔT = Δf + Δs = A A h3 h = , A* = + * 3EI GA α s 1.2 KA =KF = for rectangular section 1 1 = = 378788KN / m ΔT 2.64 ×10−6 Kc=KE=86805.4 KN/m Relative stiffness KA: KC: KE: KF = 0.407 : 0.093 : 0.093 : 0.407 Location of center of rigidity (CR) is: x= ∑ kixi = 378788(−12.5 + 12.5) + 86805.4(−2.5 + 7.5) = 0.466m 931186.8 ∑ ki ex=0.466m Design eccentricities: Dnx=25m ex=0.466m , (i) e = 0.466 + 0.1 × 25 = 2.966 m (ii) e = 0.466 - 0.1 × 25 = - 2.034m T(i)= 2.966 F T(ii)= - 2.034 F Calculation of torsional shears: ΔA=12.966 θ ΔC=2.966 θ ΔE=-7.034 θ ΔF=-12.034 θ NBCC 4.1.8.11(10) FA=KA* ΔA= KA(12.966 θ ) & the same for C, E & F 2 Moment @ C.R due to FA = (12.966) KA θ FC = (2.966) KC θ FE = (7.034) KE θ FF = (12.034) KF θ For moment equilibrium: 2 T = ⎡(12.966 ) K A + (2.966) 2 K C + (7.034) 2 K E + (12.034) 2 K F ⎤ θ = K Rθ ⎣ ⎦ Solve for θ 2 2 2 Case(i): T=T(i) = 2.966F θ =2.4× 10-8(radians) FA = 0.1179 F, FC = 6.18×10-3 F FE= - 0.01465 F, Case(ii): FF = -0.1094 F θ = -1.6457×10-8 F (radians) FA = -0.0808 F, FC = -4.24×10-3 F FE = -0.010 F, FF = 0.075 F Ti=2.966F Tii=2.034F 0.1179F 6.18*10E−3 F 0.01465F 0.1094F 0.0808F 4.24*10E−3F 0.01F 0.075F case(i) case(ii) CR lateral resisting shears. assuming no torsion 0.407F 0.093F 0.093F 0.407F Wall Design Shear Control Case A 0.5249F (i) C 0.0992F (i) E 0.103F (ii) F 0.482F (ii) T 0.7 Ft=0 Fx= Fx = VWx hx ∑W h i =1 n ii Shear and moments for the structure without torsion effect Shear and moments for the walls taking torsion effect Wall moments after reduction by Jx ...
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This note was uploaded on 03/10/2009 for the course ENGR BCEE 345 taught by Professor Drgala during the Winter '07 term at Concordia Canada.

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