Design of Flange Beams to Eurocode 2: an Overview

A flange section is that which has a rectangular extension from the main rectangular body. The extension is called a flange while the rectangular body is called web. This type of section mostly occurs when a beam is cast monolithically with a slab. 

The design of a flange beam is similar to that of a rectangular beam only that the flange gives additional area of concrete which adds to the overall compressive strength of the section. So instead of using the breadth of the rectangular section only in estimating the compressive strength of the section, the breath of the rectangular section plus that of the flange is used thereby augmenting the compressive strength of the beam. This increase in compressive strength means flange beams, when the flanges are in compression throughout the member as in the case of a simply supported beam, rarely require to be designed as doubly reinforced.

However, it is very pertinent to note that this increase in compressive strength due to the flange area will be valid only when the flange is in compression. Whenever the flange is in tension as in the case above columns in a continuous beam, the section should be designed as a rectangular beam as more concrete area in tension will actually serve little or no purpose.

Types of Flange Section

Flange sections are however categorized based on their geometry. The two basic types are:

1. L section
2. T section

The L sections are mostly found in perimeter beams of building as the slab is only spanning from the beam on one side while the T section on the other hand is mostly found in buildings intermediate beams as slab spans from both sides of the beam.

EFFECTIVE WIDTH OF FLANGE

When the flange of a beam is exploited in designing a beam, the total span of the adjoining slab when large does not often act with the supporting beam to resist the loads on the beam, only some portion of the slab does. To calculate the portion of the slab that act together with the beam to resist loads results to the concept of effective width of flange.

The effective flange width according to Eurocode 2 depends on the web and flange dimensions, the type of loading, the span, the support conditions and the transverse reinforcement.

The effective width (b_eff) of flange according to Eurocode 2 can be calculated as:

$
b_{eff}\,\,=\,\,b_{effj\,\,}+\,\,b_{w\,\,}\leqslant \,\,b
$

Where:

$
b_{effj\,\,}=0.2b_i+0.1l_o\leqslant 0.2l_o
$

b_effj <= b_i

l_o is defined as the distance between point where moment is zero. The value of for different support configurations is defined in fig (5) as shown below. Other parameters in the formular are defined in fig (4)

 

Click to also Read: Design of Reinforced Concrete Flange beam to Eurocode 2 – Worked Example

Design of Flange Beam

Flexural design of a Flange beam

For the design of a flange in flexure either of these two cases are expected:

1. When the portion of concrete to resist compression lies within the flange
2. When the portion of concrete to resist compression lies below the flange.

In designing a flange section, when the area of concrete within the flange section is enough to develop the strength required to resist the compressive force, then the beam is designed as a rectangular section with breath b_eff (effective flange width). Therefore, the moment of resistance of the section can be calculated using:

M_u = Kb_effd²f_ck (k = 0.167)

When M_u < M, then the stress block is truly within the flange and the beam is designed as singly reinforced.

The area of tensile reinforcement in this case can be determined using:

$
A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{ck}Z}}
$

Where Z which is the lever arm is:

$
Z\,\,=\,\,d\left( \text{0.5}+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) \,\,\,\,\leqslant \,\,0.95d
$

Although Eurocode 2 (EC2) does not limit the lever arm to 0.95d, it is a good practice nonetheless to limit it as such in accordance to British standard.

When the stress block to resist compressive force extends below the flange, then these can be designed by either:

1. Calculating the exact depth of the web below the flange that is harnessed to resist the compressive force.
2. By conservatively assuming that the neutral axis depth is equals to maximum value allowed by the code which 0.45d.

To calculate the actual depth of the web that partakes in resisting compression can be very tedious, as a work around, the author prefers conservatively taking the depth of the web that resist compression to be 0.45d.

Consequently, the moment of resistance of the section will be the moment of resistance of the flange plus the moment of resistance of the web.

ie: M_u = M_f + M_w

0.156f_ck(b_efff – b_w)(d – h_f/2)h_f + 0.167f_ckbd²

If the design moment (M) is greater than the moment obtained after evaluating the above formula, then the section shall be designed as doubly reinforced.

After analyzing a T-beam section In Reinforced Concrete Design to Eurocode 2 by Mosley, Bungey, and Hulse; the formula for calculating the area of tensile reinforcement for singly reinforced section when the depth of the neutral axis is assumed as 0.45d is rendered as:

$
A_s\,\,=\,\,\frac{M\,\,+\,\,0.1f_{ck}b_wd\left( 0.36d\,\,-\,\,h_f \right)}{0.87f_yk\left( d\,\,-\,\,0.5h_f \right)}
$

 

Vertical Shear Design

Shear force is an internal force that tends to make the part of a structural member slide against another. It is often significant close to supports and under point loads. BS EN 1992-1-1: 2004 adopts the approach of variable strut inclination method for shear design; this method allows for flexibility in the angle of inclination of the compressive strut between 22 and 45.

Shear resistance of section without Shear reinforcement

Before shear reinforcement are designed, it is important that the concrete section is checked whether it has sufficient shear capacity (V_Rd.c) to resist the maximum design shear force (V_Ed) that will act on it without shear reinforcement. The shear capacity of the concrete section without shear reinforcement is often sufficient for lightly loaded beams of minor importance, however minimum area of shear reinforcement is always provided.

For member which shear reinforcement is not required, the shear capacity is given in section 6.2.2 (1) as:

V_Rdc = (C_RdcK(100_ρLf_ck)^1/3 + K1σ_cp) b_wd

 σ_cp = N_Ed/A_{c, hence where there is no axial force the equation is reduced to:}

_{VRdc = {CRdcK(100ρLfck   ^1/3}bwd}

From EC2 and Table NA.1 from of the UK National Annex to the code _CRdc =  0.18/ϒ, hence the formular becomes:

_{VRdc = {0.12K(100ρLfck)^1/3}bwd}

This section resistance must not, however, be less than the permitted minimum resistance which is:  

_{VRdc = {0.035K3/2fck^1/2}bwd}

where:

K = (1 +√200/d)  ≤   2.0

ρL   =  Asl/bwd  ≤  0.02

The Variable Strut Inclination Method

As for section which has the member shear capacity (V_Rdc) less than the design shear force (V_Rdc) the shear reinforcement is required and designed using the variable strut inclination method. This method is based on an imaginary truss model where concrete acts as the top chord and also acts as the diagonal strut members inclined at an angle ϴ, the bottom chords are the main tensile reinforcement while the designed links serve as the transverse tension members. The angle ϴ changes in value proportionately to the shear force acting on the member. The angle ranges between the lower and upper limit of 22 and 45 degrees respectively. It should be noted that in this imaginary truss, the contribution of the concrete to shear resistance is ignored.

Before the vertical stirrup is designed, it is important to check whether the design shear force is not too large enough to cause the crushing of the inclined compressive strut. To ensure this does not happen, the maximum resistance of the section must be greater than the design shear force. 

ie: V_Rdmax    > V_Ed

The formula for V_Rdmax is given in equation (6.9) of the code as:

V_Rdmax = ∝_cwb_wZv₁f_cd/(cotϴ + tanϴ)

When the value of V1 in accordance with equation (6.6N) of the code and ∝cw = 1.0 are entered into the formula, it can be further simplified as:

V_Rdmax  =  (0.36bwd(1-fck/250)fck)/(cotϴ + tanϴ)

The value of ϴ to be used to check whether the section is adequate such that the compressive strut is not crushed is 45 degrees as this is the highest value of ϴ allowed by EC2 which gives the maximum resistance. When 45 is substituted for ϴ then the equation becomes:

 V_Rdmax(45) = 0.81bwd(1 – fck/250)fck

Should V_Rdmax be less than  V_Ed, the compressive strut will fail by crushing so the section is inadequate and have to be resized. However, if V_Rdmax is greater than V_Ed then we can proceed to designing the vertical links that will be required to resist the shear in the member. Different values of ϴ can be tried to achieve a shear resistance greater than the design shear force. The first value to be tried is 22 degrees as this angle gives the least resistance. If substituting 22 for ϴ does not give sufficient resistances, then a higher value of ϴ will be tried.

 As an alternative and easier workaround to trying different values for ϴ until a suitable one is found, the design shear force should be equated to V_Rdmax  so that ϴ is made the subject of the formular and a suitable value of  ϴ  is obtained in one swoop. Thus:

V_Ed = (0.36b_wd(1 – f_ck/250)f_ck)/(cotϴ + tanϴ)

Once the perfect value of ϴ is obtained having modified the above equation to make ϴ the subject of the formular, this value can be plugged into the equation below: To get the shear resistance of a vertical stirrup of area A_sw and spacing s, the code gives the equation to be used as:

V_Rd,s = (A_sw/s)Zf_ywdCotϴ 

Where; f_ywd = design yield strength of shear reinforcement.

The ratio of the Area and spacing of the link should not however be less than the minimum ratio as given in equation (9.5N) thus:

Asw/s = 0.08b_w(√f_ck/f_ywk)

Additional Longitudinal tension reinforcement

There’s always need to provide additional longitudinal tension reinforcement at the bottom face of the section to resist the horizontal component of the force in the inclined strut. The required force to be resisted is given in equation (6.18) of EC2 which goes:

F_td:  0.5V_Ed(Cotϴ – Cotα)

For all practical purpose, when curtailment length of longitudinal bars is increased beyond the position where they are not needed, then required force to counteract the longitudinal component of the force on the strut is unwittingly provided, hence separate calculation for additional longitudinal steel is often superfluous.

 

Longitudinal Shear Design

Longitudinal shear stress (v_ED) occurs along the interface between the flange and the web of a flange beam. This longitudinal shear stress is a result of the change in the longitudinal force acting at different positions along the beam.  The variation of this longitudinal force is as a result of the deformation of the beam under bending.

Transverse reinforcement over full effective flange width should be provided at the top face of the flange to resist this longitudinal shear stress. These reinforcement act as tension ties while the concrete flange act as compressive structs in an assumption similar to that in vertical shear design of a rectangular beam.

The shear stress in the flange can be evaluated using:

$
v_{ED\,\,=\,\,}\frac{\varDelta F_d}{h_f\,\,X\,\,\varDelta x}
$

where:

∆x is the length under consideration (The maximum value of ∆x is half of the distance between point where moment is 0 and where moment is maximum for simply supported beam)

$
\varDelta F_d\,\,=\,\,\frac{\varDelta M}{d\,\,-\,\,h_f/2}\,\,X\,\,\frac{b_{eff}\,\,-\,\,\frac{b_w}{2}}{b_{eff}}
$

∆M is the change in moment over the distance ∆x.

As with vertical shear, it is also necessary to check whether the compressive struct in the flange will not crush under the shear stress. To verify this, the equation below must be satisfied:

$
v_{ED}\,\,=\,\,vf_{cd}\sin \theta _f\cos \theta _f
$

EC2 gives a range of angle of inclination of the struct whether the flange is in tension or compression as shown below:

26.5⁰ ≤ ϴ_f ≤ 45⁰  OR   1.0 ≤ cotϴ_f ≤ 2.0 (for flanges in compression)

38.6⁰ ≤ ϴ_f ≤ 45⁰  OR   1.0 ≤ cotϴ_f ≤ 1.25 (for flanges in tension)

The transverse reinforcement required by the flange beam can be evaluated using:

$
\frac{A_{sf}\,\,X\,\,f_{yd}}{s_f}\,\,=\,\,\frac{V_{Ed}X\,\,h_f\,\,}{\cot \theta _f}
$

If the longitudinal shear acts together with transverse bending of the flange as in a cantilever, the transverse reinforcement will be the greater of:

a) $
\frac{A_{sf}\,\,X\,\,f_{yd}}{s_f}\,\,=\,\,\frac{V_{Ed}X\,\,h_f\,\,}{\cot \theta _f}
$

b) $
\frac{A_{sf}\,\,X\,\,f_{yd}}{s_f}\,\,=\,\,\frac{V_{Ed}X\,\,h_f\,\,}{\cot \theta _f}
$ + A_sf X b/2_sf X b

 

 

 

On no account should the area of transverse reinforcement be less than A_smin given by:

$
\frac{A_{s\min}}{b\,\,X\,\,d_f}\,\,>\,\,\text{0.26}X\frac{f_{ctm}\,\,}{f_{yk}}\,\,>\,\,\text{0.0013}
$

It is however worthy of note that should v_Ed is less than or equal to 40percent of the cracking tensile strength of the concrete (f_ctd then no transverse reinforcement is required.)

Ie:  v_Ed ≤ 0.4 f_ctd; no transverse shear reinforcement required.

Where: 0.4 f_ctd = 0.4 X f_ctk/1.5 = 0.27f_ctk

 

Deflection Check

Deflection is an important serviceability limit state that must be checked whether its limit is not exceeded. Excessive deflection can compromise the aesthetic of the structure, give impression of unsafe structure, cause damage to brittle finishes, cause problem to fixtures etc. So ensuring deflection limit is not exceeded is one of the important criteria of beam design.

The factors affecting deflection of beams are numerous likewise is calculating deflection for heterogenic material like reinforced concrete very tedious. Quasi-permanent load combination is considered for deflection; this entails considering the permanent load plus some percentage of the variable load. A fraction of the variable load is considered because the element is not expected to be exposed to the full extent of variable load consistently for long term.

The code suggests that the deflection of a beam can be estimated by calculating the curvature of the beam when un-cracked and when cracked. But a less tedious approach is to limit the span-depth ratio as this in turn limit the possible curvature of the beam. The limiting span-depth ratio can be estimated using equations below:

l/d = K[11 + 1.5√f_ck ρ0/ρ + 3.2√f_ck (ρ^o/ρ – 1)3/2]   if ρ ≤ ρ^o

l/d = K[11 + 1.5√fck ρ^o/ρ + 3.2√fck √ρ^o/ρ ]  if ρ > ρ^o

The definition of the symbols in the formulas can be found under clause 7.4.2 (2) of the code.

When the ratio of the flange breath the web breadth is greater than 3, the limiting value for deflection should be multiplied by 0.8. It should also be put into consideration that the code assumes the steel stress at serviceability limit state at critical section of an element is 310MPa. This stress corresponds with yield stress of 500MPa. If other serviceability stress is used, then the equation for the limiting span-depth ratio should be modified accordingly by multiplying it by 310MPa/rows.

When the span-depth ratio of the member falls under the limiting value, it is inferred that the member has passed deflection criteria and has conformed to the code’s objective, which is to limit the deflection of the member to Span/250 or span/500 for members with brittle finishes or vulnerable adjacent part. Should the actual span-depth ratio of the member exceed the limiting span-depth ratio, the member is said to have failed deflection criteria and should therefore be redesigned.

References:

BS EN 1992-1-1: 2004 (Eurocode 2: Design of Concrete Structures)

Renforced Concrete Design to Eurocode 2 by W.H. Mosley, J.H.Bungey, and Ray Hulse