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 post, we are going to focus on 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 concrete encased 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,
*λ*(=_{EX}*L*/_{EX}*r*) and_{x}*λ*(=_{EY}*L*/_{EX}*r*)._{y} - Select suitable strut curves from Table
- Determine compressive strength, p
_{c} - Calculate compression resistance of member,
*P*= A_{c}_{g}p_{c}. - Check
*F*≤_{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 (
*λ)*using Euler’s critical load - Choose the appropriate buckling curve and calculate the corresponding reduction factor
*χ* - 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 × 380 = 15200 mm

(ii) 100b_{c}^{2}/d_{c} = (100 × 380^{2})/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*= 0.2b

_{y}_{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*= 76 mm and

_{y}*r*= 113 mm

_{x}*Slenderness ratio**λ _{EX}* =

*L*/

_{EX}*r*= (3000/113) = 26.548

_{x}*λ*=

_{EY}*L*/

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

_{y}*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*= 26.548 and

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

_{y}^{2},

*p*= 256.45 N/mm

_{c}^{2}For

*λ*= 39.47 and

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

_{y}^{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

*λ*

_{i}= (

*N*

_{pl,Rk}/

*N*

_{cr})

^{0.5}

*N*

_{cr,i}= π^{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} = 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 = *π*D^{4}/64 = (*π* 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} *= [(π

^{2}x 6.69455 x 10

^{13})

_{ }/3000

^{2}] x 10

^{-3}= 73413.955 kN

*N*[(π

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

^{13})/3000

^{2}] 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) + λ^{2}]

Φ_{y} = 0.5 [1 + 0.34 (0.310 – 0.2) + 0.310^{2}] = 0.567

Φ_{z} = 0.5 [1 + 0.49 (0.389 – 0.2) + 0.389^{2} ] = 0.622

*X* = 1/[Φ + √(Φ^{2} – λ^{2})]*X*_{y} = 1/[0.567 + √(0.567^{2} – 0.310^{2})] = 0.959*X*_{z} = 1/( 0.622 + √(0.622^{2} – 0.389^{2})) = 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