C 6.1 Ta Clause 1 (1) able 3.1 E T a s 1 R P R A a . = = F p Example of The column and has no steel and d 1. SELECTIO Required A Partial facto Resistance Assume the advance U .: Required =3500 x 103 =98.60 cm2 From Table provided 16 Pinned Co n shown in f o intermedia esign comp N OF SECTI rea, A = NEd or for partic of cross-sec e nominal th KB S355 to Gross Area / (355 / 1.0 of Propertie 64 cm2 of A olumn w/o I figure below ate restraint pression for ON d / (fy / γMO cular resista ctions what hk. of flange be used, .: a , A = NEd 0) es, select 3 Area ntermediat w is pin-end t. Column w rce, NEd = 3 ) ance, γM tever the c e and web yield streng / (fy / γMO ) 56 x 368 x 1 te Restrain ded about was design 3500 kN lass, γMO = 1 b is less than gth, fy = 355 129 UKC in S both axes n using S355 1.0 n 40mm and 5N/mm2 S355 steel R 5 d NEd = 356 x UKC Remarks = 3500 kN 368 x 129
Table 3.1 Table 5.2(1) Table 3.1 Table 5.2(2)
From section property tables, section properties for 356 x 368 x 129 UKC: Depth ,h = 355.6 mm Width, b = 368.6 mm Web thickness, tw = 10.4 mm Flange thickness, tf = 17.5 mm Root radius, r = 15.2 mm
Depth between fillets, d = 290.2 mm Ratio for Local Buckling (flange), cf/tf = 9.4
Ratio for Local Buckling (web), cw/tw = 27.9 Radius of gyration y axis, iy = 15.6 cm Radius of gyration z axis, iz = 9.43 cm
Torsional constant, IT = 153 cm4 Warping constant, Iw = 4.18 dm6 Area, A = 164 cm2 Modulus of elasticity, E = 210 000 N/mm2 Shear modulus, G ≈ 81 000 N/mm2 2. CLASIFICATION OF SECTION
Take steel grade as S355 and tw = 10.4 mm < 40 mm .: fyw = 355 N/mm2
Web
ε = 0.81 and cw/tw = 27.9 < 38ε .: Web section was Class 2
Take steel grade as S355 and tf = 17.5 mm < 40 mm .: fyf = 355 N/mm2
Flange
ε = 0.81 and cf/tf = 9.4 < 14ε .: Flange section was Class 3
fyw = 355N/mm2
6.2.4 6.2.4(1) 6.2.4(2) 6.3 6.3.1 6.3.1.1 6.3.1.1(1) 6.3.1.1(3)
3.RESISTANCE OF CROSS- SECTION SUBJECT TO COMPRESSION Compression
The design value of the compression force NEd at each cross-section shall satisfy: NEd /Nc,Rd ≤ 1 (eq. 6.9) Nc,Rd = A fy / γMO (eq 6.10) = 164 x 102 (355) / 1.0 = 5,822kN NEd /Nc,Rd = 3,500kN/5,822kN = 0.6 <1
.: Cross-section is adequate for compression resistance Buckling resistance of members
Uniform Members in Compression Buckling resistance
A compression member should be verified against buckling as follows:
NEd /Nb,Rd ≤ 1 (eq. 6.46)
The design buckling resistance of a compression member should be taken as:
Nb,Rd = (X A fy ) / γM1 (eq. 6.47)
where X is the reduction factor for the relevant buckling mode
6.3.1.2 6.3.1.2(1) 6.3.1.3(1) 6.3.1.2(2) Table 6.2 Table 6.1 Buckling Curves
For axial compression in members the value of X for the appropriate non-dimensional slenderness λ should be determined from the relevant buckling curve according to:
Reduction factor, X = 1 / (Φ + √Φ2 - λ2) but X ≤ 1 (eq.6.49) Where:
λ
= √Afy / Ncr = (Lcr / i) x (1 /λ
1) (eq. 6.50) for class 1,2,3 cross sectionλ
1 = 93.9ε= 93.9 (0.81)
= 76.06
Lcr = 6000mm (see additional notes)
.: λ
z = (Lcr / iz) x (1 /λ
1)= (6000/94.3) x (1/76.06) = 0.84
Φ = 0.5 ( 1 + α (
λ
– 0.2) +λ
2)
Where imperfection factor, α corresponding to the appropriate buckling curve should be obtained from
Table 6.1 and Table 6.2. h/b = 355.6/368.6
= 0.96
For rolled section grade S355 with h/b < 1.2 and tf ≤ 100mm, buckling about z-z axis .: buckling curve ‘c’
For buckling curve ‘c’ .: Imperfection factor , α = 0.49
Lcr = 6000mm
.: Buckling resistance Nb,Rd = (X A fy ) / γM1 (eq. 6.47) = (0.64 x 16400 x 355) / 1.0 = 3726.08 kN
NEd /Nb,Rd ≤ 1 (eq. 6.46)
3500 / 3726.08 = 0.94 < 1 .: section is satisfactory for buckling resistance
6.1 (1)
Table 3.1
Example of Pinned Column with Intermediate Restrained
The column shown in figure below is pin-ended about both axes and has a tie at mid-height providing restraint about the z-z axis. Column was used S355 steel and design compression force, NEd = 3,500 kN
1. SELECTION OF SECTION
Required Area, A = NEd / (fy / γMO ) Partial factor for particular resistance, γM
Resistance of cross-sections whatever the class, γMO = 1.0
Assume the nominal thk. of flange and web is less than 40mm and
Table 3.1 Table 5.2(1) Table 3.1 Table 5.2(2)
From section property tables, section properties for 356 x 368 x 129 UKC: Depth ,h = 355.6 mm Width, b = 368.6 mm Web thickness, tw = 10.4 mm Flange thickness, tf = 17.5 mm Root radius, r = 15.2 mm
Depth between fillets, d = 290.2 mm Ratio for Local Buckling (flange), cf/tf = 9.4
Ratio for Local Buckling (web), cw/tw = 27.9 Radius of gyration y axis, iy = 15.6 cm Radius of gyration z axis, iz = 9.43 cm
Torsional constant, IT = 153 cm4 Warping constant, Iw = 4.18 dm6 Area, A = 164 cm2 Modulus of elasticity, E = 210 000 N/mm2 Shear modulus, G ≈ 81 000 N/mm2 2. CLASIFICATION OF SECTION
Take steel grade as S355 and tw = 10.4 mm < 40 mm .: fyw = 355 N/mm2
Web
ε = 0.81 and cw/tw = 27.9 < 38ε .: Web section was Class 2
Take steel grade as S355 and tf = 17.5 mm < 40 mm .: fyf = 355 N/mm2
Flange
ε = 0.81 and cf/tf = 9.4 < 14ε .: Flange section was Class 3
.: Section was Class 3 under compression
fyw = 355N/mm2
fyf = 355 N/mm2
Section was Class 3 under
6.2.4 6.2.4(1) 6.2.4(2) 6.3 6.3.1 6.3.1.1 6.3.1.1(1) 6.3.1.1(3)
3.RESISTANCE OF CROSS- SECTION SUBJECT TO COMPRESSION Compression
The design value of the compression force NEd at each cross-section shall satisfy: NEd /Nc,Rd ≤ 1 (eq. 6.9) Nc,Rd = A fy / γMO (eq 6.10) = 164 x 102 (355) / 1.0 = 5,822kN NEd /Nc,Rd = 3,500kN/5,822kN = 0.6 <1
.: Cross-section is adequate for compression resistance Buckling resistance of members
Uniform Members in Compression Buckling resistance
A compression member should be verified against buckling as follows:
NEd /Nb,Rd ≤ 1 (eq. 6.46)
The design buckling resistance of a compression member should be taken as:
Nb,Rd = (X A fy ) / γM1 (eq. 6.47)
where X is the reduction factor for the relevant buckling mode
6.3.1.2 6.3.1.2(1) 6.3.1.3(1) 6.3.1.2(2) Table 6.2 Buckling Curves
For axial compression in members the value of X for the appropriate non-dimensional slenderness λ should be determined from the relevant buckling curve according to:
Reduction factor, X = 1 / (Φ + √Φ2 - λ2) but X ≤ 1 (eq.6.49) Where:
λ
= √Afy / Ncr = (Lcr / i) x (1 /λ
1) (eq. 6.50) for class 1,2,3 cross sectionλ
1 = 93.9ε= 93.9 (0.81)
= 76.06
The member is effectively held in position at both ends, but not restrained in direction at either end. The tie provides restraint in position only for buckling about the z-z axis (i.e. the member is not restrained in direction by the tie. Therefore the buckling lengths are: About the y-y axis Lcr,y= 6000 mm
About the z-z axis Lcr,z = 3000 mm
.: λ
y = (Lcr,y / iy) x (1 /λ
1) = (6000/156) x (1/76.06) = 0.51.: λ
z = (Lcr,z/ iz) x (1 /λ
1) = (3000/94.3) x (1/76.06) = 0.42 Φ = 0.5 ( 1 + α (λ
– 0.2) +λ
2)
Where imperfection factor, α corresponding to the appropriate buckling curve should be obtained from
Table 6.1 and Table 6.2. h/b = 355.6/368.6
= 0.96
For rolled section grade S355 with h/b < 1.2 and tf ≤ 100mm, buckling about y-y axis .: buckling curve ‘b’
For rolled section grade S355 with h/b < 1.2 and tf ≤ 100mm, buckling about z-z axis .: buckling curve ‘c’
λ
1= 76.06
Lcr,y= 6000 mm Lcr,z = 3000 mmλ
y = 0.51λ
z = 0.42 h/b = 0.96Table 6.1 For buckling curve ‘b’ .: Imperfection factor , αy = 0.34 For buckling curve ‘c’ .: Imperfection factor , αz = 0.49
Φy= 0.5 (1 + α (
λ
– 0.2) +λ
2)
= 0.5(1+0.34(0.51-0.2)+0.51
2) = 0.68 Φz= 0.5 (1 + α (λ
– 0.2) +λ
2)
= 0.5(1+0.49(0.42-0.2)+0.42
2) = 0.64 .: Reduction factor, X = 1 / (Φ + √Φ2 - λ2 ) Xy= 1 / (Φ + √Φ2 - λ2 ) = 1 / (0.68 + √0.682 - 0.512) = 0.8851 Xz = 1 / (Φ + √Φ2 - λ2 ) = 1 / (0.64 + √0.642 - 0.422) = 0.8905In case of different value of reduction factor, take the more onerous effect.
.: Buckling resistance Nb,Rd = (X A fy ) / γM1 (eq. 6.47) = (0.8851 x 16400 x 355) / 1.0 = 5,153.05 kN
NEd /Nb,Rd ≤ 1 (eq. 6.46)
3,500 / 5,153.05 = 0.68 < 1 .: section is satisfactory for buckling resistance αy = 0.34 αz = 0.49 Φy = 0.68 Φz = 0.64 Xy= 0.8851 Xz = 0.8905 Nb,Rd = 5,153.05 kN
