Reinforced concrete structures II. 4.5 Column Design

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Emory Logan
6 years ago
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1 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 f cd = = mpa.5 f yd = 500 = mpa.5 E cm = 30 Gpa E s = 200 Gpa ε yd = 2.739

2 2

3 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 = 0.45 C =.7 ( 0.45) = 2.5 n = N ed = A c f cd = ƛ lim = = Effective length For Braced member l 0 = 0.5l [ + K K ] [ + K K 2 ] K = Column stiffness beam stiffness K i = (EI l )column (2 EI l )beam I column = I beam top = = mm 4 = mm 4 K = ( E E E ) 6000 = I column = I beam bottom = = mm 4 = mm 4 3

4 K 2 = ( E E E ) 6000 = l 0 = [ ] [ + ] = mm ƛ = l o i i = I A = ƛ = = ƛ < ƛ lim Short column = 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 n = N ed = A c f cd = ƛ lim = = Effective length For Braced member l 0 = 0.5l [ + K K ] [ + K K 2 ] K = Column stiffness beam stiffness K i = (EI l )column (2 EI l )beam I column = I beam top tie beam = = mm = mm 4

5 K = ( E E E ) 4000 = I column = I beam bottom = = mm 4 = mm 4 K 2 = ( E E E ) 4000 = l 0 = [ ] [ + ] = mm ƛ = l o i i = I A = = mm ƛ = = ƛ > ƛ lim Slender column Step 3. Accidental eccentricity 3. In the X direction e a = l o 400 = = mm In the Y direction e a = l o 400 = Step 4. Equivalent first order eccentricity 4. In the X direction = 6.52 mm e e = max { 0.6e e 0 0.4e 02 e 02 = M = = mm N sd

6 e 0 = M = = mm N sd e e = max { 0.6e e 0 0.4e 02 = mm 4.2 In the Y direction e e = max { 0.6e e 0 0.4e 02 e 02 = M = = mm N sd e 0 = M = = mm N sd e e = max { 0.6e 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 = = mm check with e = e 02 + e a = = mm So take e tot = mm M sd,y = N sd e tot = KNm μ sd,y = M sd,y f cd bd 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 = f ck 200 ƛ 50 = = K = + β eff = =.3586 ε yd = = = d = effective depth r o 0.45d K r = (n u n) n (n u n bal ) u = + ω 6

7 n = N ed = A c f cd = n bal = 0.4 For first iteration take e 2 = 0 e tot = e o + e e + e 2 = = mm N sd = 700 KN M sd,x = N sd e tot = KNm μ sd,x = M sd,x f cd bd 2 = = *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = μ sd,x = μ sd,y = Interpolating between v sd 0.8 and ω = 0.28 So n u = + ω =.28 K r = (n u n) ( ) = = (n u n bal ) ( ) r = = e 2 = r 0 = mm For Second iteration take e 2 = mm e tot = e o + e e + e 2 = = mm N sd = 700 KN M sd,x = N sd e tot = KNm μ sd,x = M sd,x f cd bd 2 = = *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = μ sd,x = μ sd,y = Interpolating between v sd 0.8 and ω = 0.6 So n u = + ω =.25 K r = (n u n) ( ) = = (n u n bal ) (.6 0.4) r = = e 2 = r 0 = mm For 3rd iteration take e 2 = mm e tot = e o + e e + e 2 = = mm 7

8 N sd = 700 KN M sd,x = N sd e tot = KNm μ sd,x = M sd,x f cd bd 2 = = *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = μ sd,x = μ sd,y = Interpolating between v sd 0.8 and ω = So n u = + ω =.25 K r = (n u n) ( ) = = (n u n bal ) ( ) r = = e 2 = r 0 = mm For 4th iteration take e 2 = mm e tot = e o + e e + e 2 = = mm N sd = 700 KN M sd,x = N sd e tot = KNm μ sd,x = M sd,x f cd bd 2 = = 0.62 *Using h = b = 0.0 read the mechanical steel ratio from biaxial interaction chart for h b v sd = μ sd,x = 0.62 μ sd,y = Interpolating between v sd 0.8 and ω = 0.65 The iteration converges with similar mechanical steel ratio ω = A s,tot = ωf cdbd = = mm 2 f yd A = A s,tot 4 = = 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 = = 0800 mm 2 OK! 8

9 Using 6 provide total of 6 6 Check using nominal stiffness method and compere the results. 9

Example 4.1 [Uni-axial Column Design] Solution. Step 1- Material Step 2-Determine the normalized axial and bending moment value

Example 4.1 [Uni-axial Column Design] 1. Design the braced short column to sustain a design load of 1100 KN and a design moment of 160KNm which include all other effects.use C5/30 and S460 class 1 works