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;

- Determine ultimate axial load
*F*._{c} - Select trial section and check if it is non-slender.
- Determine
*r*,_{x}*r*and_{y}*A*from steel tables._{g} - Determine effective lengths,
*L*and_{EX}*L*_{EY} - Calculate slenderness ratios,
*Î»*(=_{EX}*L*/_{EX}*r*) and_{x}*Î»*(=_{EY}*L*/_{EX}*r*)._{y} - Select suitable strut curves from TableÂ
- Determine compressive strength, p
_{c}Â - Calculate compression resistance of member,
*P*= A_{c}_{g}p_{c}. - Check
*F*â‰¤_{c }*P*. If unsatisfactory return to 2._{c}

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

- Determine the ultimate axial load on the column
*N*_{Ed} - Select a trial section and determine its properties
- Obtain the buckling length of the column
*L* - Obtain the effective flexural stiffness
*(EI)*of the composite section_{eff} - Calculate the plastic resistance to compression of the composite section
*N*_{pl,Rk} - Calculate the relative slenderness of the section (
*Î»)*using Eulerâ€™s critical load - Choose the appropriate buckling curve and calculate the corresponding reduction factor
*Ï‡* - Multiply the plastic resistance to compression with the reduction factor to obtain the buckling resistance of the section
*N*_{b,Rd} - 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 mm^{2} , *f*_{yk} = 500 N/ mm^{2}).

* 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*) = 13600 mm

_{g}^{2}

Radius of gyration (

*r*) = 113 mm

_{x}Radius of gyration (

*r*) = 65.9 mm

_{y}Design strength (p

_{y}) = 265 N/mm

^{2}(since thickness of flange T = 20.5 mm)

Effective length (

*L*) = 3.0 m

_{E}*Effective length*

Check that the effective length of column (L = 3000 mm) does not exceed the least of:

(i) 40b_{c} = 40 Ã— 380 = 15200 mm

(ii) 100b_{c}^{2}/d_{c} = (100 Ã— 380^{2})/380 = 38000 mm

(iii) 250r_{y} = 250 Ã— 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*= 0.2b

_{y}_{c}= 0.2 Ã— 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*= 76 mm and

_{y}*r*= 113 mm

_{x}*Slenderness ratio**Î» _{EX}* =

*L*/

_{EX}*r*Â = (3000/113) = 26.548

_{x}*Î»*=

_{EY}*L*/

_{EY}*r*= (3000/76) = 39.47

_{y}*Compressive strength*

The relevant compressive strength values for buckling about the xâ€“x axis are obtained from Table 24(b)

of BS 5950 and from Table 24(c) of BS 5950 for bending about the yâ€“y axis.

For

*Î»*= 26.548 and

_{EX}*p*= 265 N/mm

_{y}Â^{2},

*p*= 256.45 N/mm

_{c}^{2}For

*Î»*= 39.47 and

_{EY}*p*= 265 N/mm

_{y}Â^{2},

*p*= 230.848 N/mm

_{c}^{2}

^{ }

The compression resistance of the column is therefore given by;

*P*

_{c}

*= (A*

_{g}

*+ 0.45f*

_{cu}

*A*

_{c}

*/p*

_{y}

*)p*

_{c}

Where:

*A*= 13600 mm

_{g}^{2}

f

_{cu}= 30 N/mm

^{2}

A

_{c}= b

_{c}d

_{c}= 380 x 380 = 144400 mm

^{2}

*p*= 265 N/mm

_{c}^{2}

*p*= 230.848 N/mm

_{c}^{2}

*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*_{cu}*A*_{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 mm

^{2}

Radius of gyration (

*i*) = 113 mm

_{y}Radius of gyration (

*i*) = 65.9 mm

_{z}Design strength (

*f*

_{y}) = 265 N/mm

^{2}(since thickness of flange T = 20.5 mm)

I

_{y}= 17500 cm

^{4}

I

_{z}= 5930 cm

^{4}

E = 210000 N/mm

^{2}

The plastic resistance to compression *N*_{pl,Rk}* = A*_{a}*.f*_{y}* + 0.85A*_{c}*f*_{ck}* + A*_{s}*f*_{yk} *A _{a}* = 13600 mm

^{2}

*f*

_{y}= 265 N/mm

^{2}

*A*

_{c}= 380 x 380= 144400 mm

^{2}

*f*

_{ck}= 25 N/mm

^{2}

*A*

_{s}= 804 mm

^{2}

*f*

_{yk}= 500 N/mm

^{2}

*N*= [(13600 x 265) + (0.85 x 144400 x 25) + (804 x 500)] x 10

_{pl,Rk}^{-3}= 7074.5 kN

The relative slenderness

*Î»*

_{i}= (

*N*

_{pl,Rk}/

*N*

_{cr})

^{0.5}

*N*

_{cr,i}= Ï€^{2}(EI)_{eff,i }/L^{2}(EI)_{eff,i} = E_{a}I_{a} + 0.6E_{cm}I_{c} + E_{s}I_{s}

E_{a} = E_{s} = Elastic modulus of the structural steel and reinforcement respectively = 210000 N/mm^{2}

I_{a} = Moment of inertia of structural steel in the relevant axis

E_{cm} =Â Modulus of elasticity of concrete = 22(f_{ck}/10)^{0.3} (GPa) = 28960 N/mm^{2} (see Table 3.1 of Eurocode 2)

I_{c} = moment of inertia of the uncracked concrete section = bd^{3}/12 = (380 x 380^{3})/12 = 17376.133 x 10^{5} mm^{4}

I_{s} = moment of inertia of the reinforcement = *Ï€*D^{4}/64 = (*Ï€* x 16^{4})/64 = 3216.99 mm^{4} (for four bars = 4 x 3216.99) = 12867.96 mm^{4}

Hence;

(EI)_{eff,y} = (210000 x 17500 x 10^{4}) + (0.6 x 28960 x 17376.133 x 10^{5}) + (210000 x 12876.96) = 6.69455 x 10^{13} N.mm^{2}

(EI)_{eff,z} = (210000 x 5930 x 10^{4}) + (0.6 x 28960 x 17376.133 x 10^{5}) + (210000 x 12876.96) = 4.26485 x 10^{13} N.mm^{2}

*N _{cr,y} *= [(Ï€

^{2}x 6.69455 x 10

^{13})

_{ }/3000

^{2}] x 10

^{-3}= 73413.955 kN

*N*[(Ï€

_{cr,z}=^{2}x 4.26485 x 10

^{13})/3000

^{2}] x 10

^{-3}= 46769.313 kN

*Î»*

_{y}= (

*N*

_{pl,Rk}/

*N*

_{cr,y})

^{0.5}= (7074.5/73413.955)

^{0.5}= 0.310

*Î»*

_{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}Â < 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 Î± = 0.34 and curve c = 0.49 (Table 6.1)

Î¦ = 0.5 [1 + Î±(Î» â€“ 0.2) + Î»^{2}]

Î¦_{y} = 0.5 [1 + 0.34 (0.310 â€“ 0.2) + 0.310^{2}] = 0.567

Î¦_{z} = 0.5 [1 + 0.49 (0.389 â€“ 0.2) + 0.389^{2} ] = 0.622

*X* = 1/[Î¦Â + âˆš(Î¦^{2} â€“ Î»^{2})]*X*_{y} = 1/[0.567 + âˆš(0.567^{2} â€“ 0.310^{2})] = 0.959*X*_{z} = 1/( 0.622 + âˆš(0.622^{2} â€“ 0.389^{2})) = 0.903Â

Therefore N_{b,Rd}Â = (X_{z }*N _{pl,Rk}*)= (0.903 x 7074.5) = 6388.685 kN

N_{Ed} /N_{b,Rd} =Â 3765/6388.685 = 0.589 < 1.0 kN Okay