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

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1 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 Take d = 0.1 and section 70 mm 450 mm H Solution Step 1- Material f cd = = mpa 1.5 f yd = 460 = 400 mpa 1.15 Step -Determine the normalized axial and bending moment value v sd = N sd f cd bh = = μ sd = m sd f cd bd = = Step 3-Using d H = 0.1 read the mechanical steel ratio from uniaxial interaction chart for V sd = μ sd = From interaction chart ω = 0.3

2 A s,tot = ωf cdbd = = mm f yd 400 A = A s,tot = = mm Check with maximum and minimum reinforcement limit 0.1 N ED A s,min = max { f yd = 75 mm 0.00A c A s,max = 0.08 A c = = OK! OK! Step 4- Detailing Use 4 16 on each face

3 Example 4. [Biaxial column design] Determine the longitudinal reinforcement for corner column of size 400*400 mm and the design factored moment and axial force of P = 1360 KN M sd,h = 00 KNm M sd,b = 100 KNm Use C5/30 and S460 class 1 work take h h = b b = 0.1 Solution Step 1- Material f cd = = mpa 1.5 f yd = 460 = 400 mpa 1.15 Step - Determine the normalized axial and bending moment value v sd = μ sd,h = μ sd,b = p = = 0.6 f cd bh m sd f cd bd = = 0.06 m sd f cd db = = d Step 3- Find ω using = b = 0.1, V d B sd = 0.6, μ sd,h = 0.06, μ sd,b = From biaxial chart ω = 0.6 A s,tot = ωf cdbd = = 3400 mm f yd 400 Check with maximum and minimum reinforcement limit 0.1 N ED A s,min = max { f yd = 340 mm 0.00A c A s,max = 0.08 A c = = 1800 OK! OK!

4 Step 4- Detailing using 4 n = 3400 = 7.5 so Use

5 Example 4.3 [Column] Determine whether the column CD is slender or not, if it is subjected t loads shown below. Consider the frame to be non-sway Use C5/30 f cd = mpa Solution Step 1- Slenderness limit ƛ lim = 0ABC take A = 0.7 B = 1.1 C = 1.7 r m n

6 where r m = m 01 = 34.4 = 0.5 C = 1.7 ( 0.5) =. m n = N ed = A c f cd = ƛ lim = = Step -Find the slenderness of the column.1 Effective length For Braced member l 0 = 0.5l [1 + K K 1 ] [1 + K K ] K = Column stiffness beam stiffness K i = (EI l )column ( EI l )beam I column = I beam = = mm 4 = mm 4 K 1 = E 6 ( E ) 6 = For fixed restraint K=0 but in reality we cannot provide a fully fixed support so use K = 0.1 l 0 = [ ] [1 + ] = mm ƛ = l o i i = I A = ƛ = = mm ƛ < ƛ lim Short column = mm

7 Example 4.4 [Column Design] Design the braced column to resist an axial load of 950 KN and a moment of M sd =115 KNM at the top and M sd =-95 KNM at the bottom as shown below.length of the column is 5.5 m and cross-section of 300*300 mm use C5/30 and S460 take L e =0.66L. Solution Step 1- Material f cd = = mpa 1.5 f yd = 460 = 400 mpa 1.15 Step - Check slenderness limit Step 3-Slenderess ƛ lim = 0ABC take A = 0.7 B = 1.1 C = 1.7 r m n where r m = m 01 = 95 = 0.86 C = 1.7 ( 0.86) =.56 m n = N ed = A c f cd = ƛ lim = = ƛ = l o i i = I A = = mm

8 ƛ = = mm ƛ < ƛ lim Short column negelect second order effect Step 4- accidental eccentricity e a = l o 400 = 3630 = mm 400 Step 5- Equivalent first order eccentricity e e = max { 0.6e e e 0 e 0 = M = = mm N sd e 01 = M = = 100 mm N sd e e = max { 0.6e e e 0 = mm e tot = e o + e e + e 0 = = mm check with e = e 0 + e a = = mm So take e tot = mm Step 6- Design N sd = 950 KN M sd = N sd e tot = 13.6 KNm v sd = p f cd bh = = μ sd = m sd f cd bd = = 0.33 Using d d = 0.15 read the mechanical steel ratio from uniaxial interaction chart for V sd = μ sd = 0.33 ω = 0.79 A s,tot = ωf cdbd = = mm f yd 400 A = A s,tot = = mm

9 Check with maximum and minimum reinforcement limit 0.1 N ED A s,min = max { f yd = 37.5 mm OK! 0.00A c A s,max = 0.08 A c = = 700 mm OK! Using 0 provide 4 0 on each face Step 7- Detailing

10 Example 4.5 [Column] Design the braced column if it is subjected to the following loading.the column has total length of 6 m. L e =0.7L Use C5/30 and S460 If the column is slender.compute the total design moment using a) Nominal curvature b) Nominal stiffness Use 400*400 mm section d = 0.1 use (, t d o ) = Solution Step 1. Material data f cd = = mpa 1.5 f yd = 460 = 400 mpa 1.15 E cm = 30 Gpa E s = 00 Gpa Step - Check slenderness limit ƛ lim = 0ABC take A = 0.7 B = 1.1 C = 1.7 r m n

11 where r m = m 01 = 140 = 1 C = 1.7 (1) = 0.7 m n = N ed = A c f cd = ƛ lim = = Step 3-slenderess ƛ = l o i i = I A = = mm ƛ = = mm ƛ > ƛ lim Slender column consider second order effect Step 4- accidental eccentricity e a = l o 400 = 400 = 10.5 mm 400 Step 5- Equivalent first order eccentricity e e = max { 0.6e e e 0 e 0 = M = = mm N sd e 01 = M = = mm N sd e e = max { 0.6e e e 0 = mm Step 6- Calculate the second order moment a) Using Nominal curvature method l o e = 1 r C C = 10 For constant cross section 1 r = K 1 rk r o K = 1 + β eff (, t o ) OK eff = 0 if { ƛ 75 OK M oed h Not OK N ed

12 eff = (, t o ) M oeqp M oed where M oeqp = first order moment i nquasi permanent load(sls) M oed = First order moment in design load combination (ULS) eff = (, t o ) M oeqp M oed = = 1 β = f ck 00 ƛ 150 = = 0.35 K = 1 + β eff = = = ε yd r o 0.45d = K r = (n u n) n (n u n bal ) u = 1 + ω n = = n bal = 0.4 = d = effective depth N ed A c f cd = For first iteration take e = 0 e tot = e o + e e + e 0 = = mm N sd = 1650 KN M sd = N sd e tot = KNm v sd = N sd f cd bh = = μ sd = m sd f cd bd = = Using v sd = μ sd = So d d = 0.1 ω = 0.5 n u = 1 + ω = 1.5 K r = (n u n) ( ) = = (n u n bal ) ( ) 1 r = = e = r 10 = mm For Second iteration take e = mm e tot = e o + e e + e 0 = = mm N sd = 1650 KN M sd = N sd e tot = KNm Using v sd = and μ sd = ω = 0.35 n u = 1 + ω = 1.35 K r = (n u n) ( ) = = (n u n bal ) ( ) 1 r = =

13 e = 1 r = mm For their iteration take e = mm e tot = e o + e e + e 0 = = mm N sd = 1650 KN M sd = N sd e tot = KNm Using v sd = and μ sd = ω = 0.35 The iteration converges with similar mechanical steel ratio ω = A s,tot = ωf cdbd = = mm f yd 400 A = A s,tot = = mm Check with maximum and minimum reinforcement limit 0.1 N ED A s,min = max { f yd = 41.5 mm OK! 0.00A c A s,max = 0.08 A c = = 1800 mm OK! Using 16 provide 5 16 on each face b) Using Nominal stiffness method. EI = K c E cd I c + K s E s I s E cd = E cm = 30 = 5 Gpa γ ce 1. Initially let us assume ρ to use the simplified method K s = 0 K c = eff

14 eff = 1 K c = 0. I c = bh3 = mm4 1 EI = K c E cd I c + K s E s I s = = N b = П EI l = П o 400 = KN N b = Buckling load based on nominal stiffness Design moment M ed = M oed [1 + β = П c o c o = 8 M oed = 140 KNm β ( N ] b Ned ) 1 β M ed = M oed [1 + ( N ] = 140 [1 + b Ned ) 1 ( ) 1 ] = KNm For first iteration neglecting accidental eccentricity v sd = N sd f cd bh = = μ sd = m sd f cd bd = = 0.7 Using v sd = μ sd = 0.7 d d = 0.1 ω = 0.4 A s,tot = ωf cdbd = = mm f yd 400 A = A s,tot = = mm ρ = A s bd = = ρ < 0.01 We cannot use the simplified method for ρ > 0.00 K s = 1 K c = K 1K (1 + eff ) K 1 = f ck 0 =

15 K = n ƛ n = N ed = A c f cd K = 0. So K c = I c = bh3 = mm4 1 I s = [ (360 00) ] = mm EI = K c E cd I c + K s E s I s = EI = N b = П EI l o = П = KN β M ed = M oed [1 + ( N ] = 140 [1 + b Ned ) 1 ( ) 1 ] = KNm Design moment including accidental eccentricity is given by M = ( 10.5 ) = KNM 1000 v sd = N sd f cd bh = = μ sd = m sd f cd bd = = 0.1 Using v sd = μ sd = 0.1 d d = 0.1 ω = 0.35 A s,tot = ωf cdbd = = mm f yd 400 A = A s,tot = = mm ρ = A s bd = = ρ > 0.00 Second Iteration

16 I c = bh3 = mm4 1 I s = [ (360 00) ] = mm EI = K c E cd I c + K s E s I s = EI = N b = П EI l o = П = KN β M ed = M oed [1 + ( N ] = 140 [1 + b Ned ) 1 ( ) 1 ] = KNm Design moment including accidental eccentricity is given by M = ( 10.5 ) = KNM 1000 v sd = N sd f cd bh = = μ sd = m sd f cd bd = = 0.16 Using v sd = μ sd = 0.16 d d = 0.1 ω = 0.35 The iteration converges with similar mechanical steel ratio ω = A s,tot = ωf cdbd = = mm f yd 400 A = A s,tot = = mm Check with maximum and minimum reinforcement limit 0.1 N ED A s,min = max { f yd = 41.5 mm OK! 0.00A c A s,max = 0.08 A c = = 1800 mm OK! Using 16 provide 5 16 on each face Step 7- Detailing

Reinforced concrete structures II. 4.5 Column Design

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