4.5 Column Design A non-sway column AB of 300*450 cross-section resists at ultimate limit state, an axial load of 700 KN and end moment of 90 KNM and 0 KNM in the X direction,60 KNM and 27 KNM in the Y direction causing double curvature about both axes. The column is braced with beams as shown in the figure. The concrete used Is C25/30 and rebar S500. Take eff = Step : Material data 0.85 25 f cd = = 4.6667 mpa.5 f yd = 500 = 434.7826 mpa.5 E cm = 30 Gpa E s = 200 Gpa ε yd = 2.739
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Step 2- Check slenderness limit 2. In the X direction ƛ lim = 20ABC take A = 0.7 B =. C =.7 r m n where r m = m 0 = 27 m 02 60 = 0.45 C =.7 ( 0.45) = 2.5 n = N ed 700 0 3 = A c f cd 4.667 300 450 = 0.8888 20 0.7. 2.5 ƛ lim = = 35.845 0.8888 2.. Effective length For Braced member l 0 = 0.5l [ + K 0.5 + K ] [ + K 2 0.5 + K 2 ] K = Column stiffness beam stiffness K i = (EI l )column (2 EI l )beam I column = I beam top = 450 3003 250 4003 = 0500000 mm 4 = 333333333 mm 4 K = ( 2 333333333E 4000 0500000E 4500 + 2 333333333E ) 6000 = 0.2025 I column = I beam bottom = 450 3003 250 3503 = 0500000 mm 4 = 89322966.7 mm 4 3
K 2 = ( 2 89322966.7E 4000 0500000E 4500 + 2 89322966.7E ) 6000 = 0.30227 l 0 = 0.5 4500 [ + 0.2025 0.30227 ] [ + ] = 2996.502 mm 0.5 + 0.2025 0.5 + 0.30227 ƛ = l o i i = I A = 0500000 35000 ƛ = 2996.502 86.6025 = 34.6006 ƛ < ƛ lim Short column = 86.6025 mm 2.2 In the Y direction ƛ lim = 20ABC take A = 0.7 B =. C =.7 r m n where r m = m 0 = 0 = 0. C =.7 ( 0.) =.8 m 02 90 n = N ed 700 0 3 = A c f cd 4.667 300 450 = 0.8888 20 0.7..8 ƛ lim = = 29.58299 0.8888 2.2. Effective length For Braced member l 0 = 0.5l [ + K 0.5 + K ] [ + K 2 0.5 + K 2 ] K = Column stiffness beam stiffness K i = (EI l )column (2 EI l )beam I column = 300 4503 I beam top tie beam = = 22785000 mm 4 200 3003 4 = 450000000 mm 4
K = ( 2 450000000E 5000 22785000E 9000 + 2 450000000E ) 4000 = 0.625 I column = I beam bottom = 300 4503 250 3503 = 22785000 mm 4 = 89322966.7 mm 4 K 2 = ( 2 89322966.7E 5000 22785000E 9000 + 2 89322966.7E ) 4000 = 0.3486 l 0 = 0.5 9000 [ + 0.625 0.3486 ] [ + ] = 6608.4435 mm 0.5 + 0.625 0.5 + 0.3486 ƛ = l o i i = I A = 22785000 35000 = 9.9038 mm ƛ = 6608.4435 9.9038 = 50.87 ƛ > ƛ lim Slender column Step 3. Accidental eccentricity 3. In the X direction e a = l o 400 = 376.688 = 7.9472 mm 400 3.2 In the Y direction e a = l o 400 = 6608.4435 400 Step 4. Equivalent first order eccentricity 4. In the X direction = 6.52 mm e e = max { 0.6e 02 + 0.4 e 0 0.4e 02 e 02 = M 02 60 06 = = 35.2947 mm N sd 700 03 5
e 0 = M 0 27 06 = = 5.882353 mm N sd 700 03 e e = max { 0.6e 02 + 0.4 e 0 0.4e 02 = 4.823 mm 4.2 In the Y direction e e = max { 0.6e 02 + 0.4 e 0 0.4e 02 e 02 = M 02 90 06 = = 52.947 mm N sd 700 03 e 0 = M 0 0 06 = = 5.88235 mm N sd 700 03 e e = max { 0.6e 02 + 0.4 e 0 0.4e 02 Step 5. Second order moment 5. In the X direction Because the column is short e 2 = 0 = 29.4 mm e tot = e o + e e + e 2 = 7.9472 + 4.823 + 0 = 22.76475 mm check with e = e 02 + e a = 35.2947 + 7.9472 = 43.2358 mm So take e tot = 43.2358 mm M sd,y = N sd e tot = 73.50 KNm μ sd,y = M sd,y f cd bd 2 = 73.50 0 6 4.667 450 300 2 = 0.758 5.2 In the Y direction Calculate the second order moment using either Nominal curvature or Nominal stiffness method Using Nominal curvature method l o 2 e 2 = r C C = 0 For constant cross section r = K rk r o K = + β eff β = 0.35 + f ck 200 ƛ 50 = 0.35 + 25 200 50.87 50 = 0.3586 K = + β eff = + 0.3586 =.3586 ε yd 2.7393 0 3 = = =.9282 0 5 d = effective depth r o 0.45d 0.45 405 K r = (n u n) n (n u n bal ) u = + ω 6
n = N ed 700 0 3 = A c f cd 4.66 450 300 = 0.8888 n bal = 0.4 For first iteration take e 2 = 0 e tot = e o + e e + e 2 = 6.52 + 29.4 + 0 = 45.933 mm N sd = 700 KN M sd,x = N sd e tot = 78.0862 KNm μ sd,x = M sd,x f cd bd 2 = 78.0862 0 6 4.667 300 450 2 = 0.0907 *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = 0.888 μ sd,x = 0.0907 μ sd,y = 0.758 Interpolating between v sd 0.8 and ω = 0.28 So n u = + ω =.28 K r = (n u n) (.28 0.8888) = = 0.444 (n u n bal ) (.28 0.4) r = 0.444.3586.9282 0 5 = 6.0267 0 6 e 2 = 6608.4435 2 r 0 = 26.29757 mm For Second iteration take e 2 = 26. 29757 mm e tot = e o + e e + e 2 = 6.52 + 29.4 + 26.29757 = 72.2306 mm N sd = 700 KN M sd,x = N sd e tot = 2.7922 KNm μ sd,x = M sd,x f cd bd 2 = 2.7922 0 6 4.667 300 450 2 = 0.42677 *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = 0.888 μ sd,x = 0.42677 μ sd,y = 0.758 Interpolating between v sd 0.8 and ω = 0.6 So n u = + ω =.25 K r = (n u n) (.6 0.8888) = = 0.59596 (n u n bal ) (.6 0.4) r = 0.59596.3586.9282 0 5 = 8.07466 0 6 e 2 = 6608.4435 2 r 0 = 35.26308 mm For 3rd iteration take e 2 = 35. 26308 mm e tot = e o + e e + e 2 = 6.52 + 29.4 + 35.26308 = 8.962 mm 7
N sd = 700 KN M sd,x = N sd e tot = 38.0335 KNm μ sd,x = M sd,x f cd bd 2 = 38.0335 0 6 4.667 300 450 2 = 0.604 *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = 0.888 μ sd,x = 0.604 μ sd,y = 0.758 Interpolating between v sd 0.8 and ω = 0.656 So n u = + ω =.25 K r = (n u n) (.656 0.8888) = = 0.6075 (n u n bal ) (.656 0.4) r = 0.6075.3586.9282 0 5 = 8.27506 0 6 e 2 = 6608.4435 2 r 0 = 36.3825 mm For 4th iteration take e 2 = 36. 3825 mm e tot = e o + e e + e 2 = 6.52 + 29.4 + 36.3825 = 82.0736 mm N sd = 700 KN M sd,x = N sd e tot = 39.52 KNm μ sd,x = M sd,x f cd bd 2 = 38.0335 0 6 4.667 300 450 2 = 0.62 *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = 0.888 μ sd,x = 0.62 μ sd,y = 0.758 Interpolating between v sd 0.8 and ω = 0.65 The iteration converges with similar mechanical steel ratio ω = 0. 65 A s,tot = ωf cdbd 0.65 4.66 300 450 = = 2859.87 mm 2 f yd 434.7826 A = A s,tot 4 = 2859.87 4 = 74.7968 mm 2 Check with maximum and minimum reinforcement limit 0. N ED A s,min = max { f yd = 39 mm 2 OK! 0.002A c A s,max = 0.08 A c = 0.08 300 450 = 0800 mm 2 OK! 8
Using 6 provide total of 6 6 Check using nominal stiffness method and compere the results. 9