Structural Design of Composite Columns

Composite sections of concrete and steel have a lot of advantages, especially in the structural performance and fire resistance of a building. For columns and other compression members, they usually appear as steel-reinforced concrete columns (SRC) or as concrete-filled steel tubes. In this article, we are going to consider the structural design of concrete encased H-section subjected to concentric axial load using Eurocode 4 and BS 5950.

According to BS 5950, the steps to design composite steel columns are as follows;

1. Determine ultimate axial load F_c.
2. Select trial section and check if it is non-slender.
3. Determine r_x, r_y and A_g from steel tables.
4. Determine effective lengths, L_EX and L_EY
5. Calculate slenderness ratios, λ_EX (= L_EX/r_x) and λ_EY (= L_EX/r_y).
6. Select suitable strut curves from Table 
7. Determine compressive strength, p_c 
8. Calculate compression resistance of member, P_c = A_gp_c.
9. Check F_c ≤ P_c . If unsatisfactory return to 2.

According to Eurocode 4, the simplified steps to design encased steel columns subjected to axial load are as follows;

1. Determine the ultimate axial load on the column N_Ed
2. Select a trial section and determine its properties
3. Obtain the buckling length of the column L
4. Obtain the effective flexural stiffness (EI)_eff of the composite section
5. Calculate the plastic resistance to compression of the composite section N_pl,Rk
6. Calculate the relative slenderness of the section (λ) using Euler’s critical load
7. Choose the appropriate buckling curve and calculate the corresponding reduction factor χ
8. Multiply the plastic resistance to compression with the reduction factor to obtain the buckling resistance of the section N_b,Rd
9. Check if N_Ed < N_b,Rd else return to step 2.

Solved Example
Verify the capacity of UC 254 x 254 x 107 in grade S275 steel encased in a concrete section of 380 x 380 mm to resist a characteristic permanent axial force of 1900 kN and variable axial force of 800 kN using concrete grade C25/30. Column is 3m long and considered pinned at both ends (Area of reinforcement provided = 4H16 (804 mm² , f_yk = 500 N/ mm²).

Solution by BS 5950
The ultimate axial load on the column is given by;
Fc = 1.4Gk + 1.6Qk = 1.4(1900) + 1.6(800) = 3940 kN

Properties of the UC section from Blue Book
Area of UC section (A_g) = 13600 mm²
Radius of gyration (r_x) = 113 mm
Radius of gyration (r_y) = 65.9 mm
Design strength (p_y) = 265 N/mm² (since thickness of flange T = 20.5 mm)
Effective length (L_E) = 3.0 m

Effective length
Check that the effective length of column (L = 3000 mm) does not exceed the least of:
(i) 40b_c = 40 × 380 = 15200 mm
(ii) 100b_c²/d_c = (100 × 380²)/380 = 38000 mm
(iii) 250r_y = 250 × 65.9 = 16475 mm OK

Radii of gyration for the cased section
For the cased section r_x is the same as for UC section = 113 mm
For the cased section r_y = 0.2b_c = 0.2 × 380 = 76 mm but not greater than 0.2(B + 150) = 0.2(258.8 + 150) = 81.76 mm and not less than that for the uncased section (= 65.9 mm)
Hence r_y = 76 mm and r_x = 113 mm

Slenderness ratio
λ_EX = L_EX/r_x = (3000/113) = 26.548
λ_EY = L_EY/r_y = (3000/76) = 39.47

Compressive strength
The relevant compressive strength values for buckling about the x–x axis are obtained from Table 24(b)
of BS 5950 and from Table 24(c) of BS 5950 for bending about the y–y axis.

For λ_EX = 26.548 and p_y = 265 N/mm², p_c = 256.45 N/mm²
For λ_EY = 39.47 and p_y = 265 N/mm², p_c = 230.848 N/mm²

The compression resistance of the column is therefore given by;

P_c = (A_g + 0.45f_cuA_c/p_y)p_c

Where:
A_g = 13600 mm²
f_cu = 30 N/mm²
A_c = b_cd_c = 380 x 380 = 144400 mm²
p_c = 265 N/mm²
p_c = 230.848 N/mm²

P_c = [13600 + (0.45 x 30 x 144400)/265] x 230.848 x 10^-3 = 4837.703 kN

Check that P_c is not greater than the short strut capacity, P_cs , given by;
P_cs = (A_g + 0.25f_cuA_c /p_y)p_y = [13600 + (0.25 x 30 x 144400)/265] x 265 x 10^-3 = 4687 kN (this is less than P_c , therefore, take P_cs)

F_c /P_c = 3940 /4687 = 0.840 < 1.0 Okay

Design by Eurocode 4
At ultimate limit state;
N_Ed = 1.35Gk + 1.5Qk = 1.35(1900) + 1.5(800) = 3765 kN

Effective length of the column L = 3000 mm
Area of UC section (A_a) = 13600 mm²
Radius of gyration (i_y) = 113 mm
Radius of gyration (i_z) = 65.9 mm
Design strength (f_y) = 265 N/mm² (since thickness of flange T = 20.5 mm)
I_y = 17500 cm⁴
I_z = 5930 cm⁴
E = 210000 N/mm²

The plastic resistance to compression N_pl,Rk = A_a.f_y + 0.85A_cf_ck + A_sf_yk
A_a = 13600 mm²
f_y = 265 N/mm²
A_c = 380 x 380= 144400 mm²
f_ck = 25 N/mm²
A_s = 804 mm²
f_yk = 500 N/mm²

N_pl,Rk = [(13600 x 265) + (0.85 x 144400 x 25) + (804 x 500)] x 10^-3 = 7074.5 kN
The relative slenderness λ_i = ( N_pl,Rk / N_cr )^0.5
N_cr,i = π²(EI)_eff,i /L²

(EI)_eff,i = E_aI_a + 0.6E_cmI_c + E_sI_s

E_a = E_s = Elastic modulus of the structural steel and reinforcement respectively = 210000 N/mm²
I_a = Moment of inertia of structural steel in the relevant axis
E_cm =  Modulus of elasticity of concrete = 22(f_ck/10)^0.3 (GPa) = 28960 N/mm² (see Table 3.1 of Eurocode 2)
I_c = moment of inertia of the uncracked concrete section = bd³/12 = (380 x 380³)/12 = 17376.133 x 10⁵ mm⁴
I_s = moment of inertia of the reinforcement = πD⁴/64 = (π x 16⁴)/64 = 3216.99 mm⁴ (for four bars = 4 x 3216.99) = 12867.96 mm⁴

Hence;
(EI)_eff,y = (210000 x 17500 x 10⁴) + (0.6 x 28960 x 17376.133 x 10⁵) + (210000 x 12876.96) = 6.69455 x 10¹³ N.mm²
(EI)_eff,z = (210000 x 5930 x 10⁴) + (0.6 x 28960 x 17376.133 x 10⁵) + (210000 x 12876.96) = 4.26485 x 10¹³ N.mm²

N_cr,y = [(π² x 6.69455 x 10¹³) /3000²] x 10^-3 = 73413.955 kN
N_cr,z = [(π² x 4.26485 x 10¹³)/3000²] x 10^-3 = 46769.313 kN

λ_y = (N_pl,Rk / N_cr,y)^0.5 = (7074.5/73413.955)^0.5 = 0.310
λ_z = (N_pl,Rk / N_cr,z)^0.5 = (7074.5/46769.313)^0.5 = 0.389

Check h/b ratio = 266.7/258.8 = 1.0305 < 1.2, and t_f < 100 mm (Table 6.2 EN 1993-1-1:2005)

Therefore buckling curve b is appropriate for y-y axis, and buckling curve c for z-z axis. The imperfection factor for buckling curve b α = 0.34 and curve c = 0.49 (Table 6.1)

Φ = 0.5 [1 + α(λ – 0.2) + λ²]

Φ_y = 0.5 [1 + 0.34 (0.310 – 0.2) + 0.310²] = 0.567
Φ_z = 0.5 [1 + 0.49 (0.389 – 0.2) + 0.389² ] = 0.622

X = 1/[Φ  + √(Φ² – λ²)]
X_y = 1/[0.567 + √(0.567² – 0.310²)] = 0.959
X_z = 1/( 0.622 + √(0.622² – 0.389²)) = 0.903 

Therefore N_b,Rd = (X_z N_pl,Rk)= (0.903 x 7074.5) = 6388.685 kN

N_Ed /N_b,Rd =  3765/6388.685 = 0.589 < 1.0 kN Okay