Flanged Beams: Theory - Civil Engineering (CE) PDF Download

Instructional Objectives: At the end of this lesson, the student should be able to:

• identify the regions where the beam shall be designed as a flanged and where it will be rectangular in normal slab beam construction,  
• define the effective and actual widths of flanged beams,
• state the requirements so that the slab part is effectively coupled with the flanged beam,
• write the expressions of effective widths of T and L-beams both for continuous and isolated cases,
• derive the expressions of  C, T and M_u for four different cases depending on the location of the neutral axis and depth of the flange.

5.10.1 Introduction

Reinforced concrete slabs used in floors, roofs and decks are mostly cast monolithic from the bottom of the beam to the top of the slab. Such rectangular beams having slab on top are different from others having either no slab (bracings of elevated tanks, lintels etc.) or having disconnected slabs as in some pre-cast systems (Figs. 5.10.1 a, b and c). Due to monolithic casting, beams and a part of the slab act together. Under the action of positive bending moment, i.e., between the supports of a continuous beam, the slab, up to a certain width greater than the width of the beam, forms the top part of the beam. Such beams having slab on top of the rectangular rib are designated as the flanged beams - either T or L type depending on whether the slab is on both sides or on one side of the beam (Figs. 5.10.2 a to e). Over the supports of a continuous beam, the bending moment is negative and the slab, therefore, is in tension while a part of the rectangular beam (rib) is in compression. The continuous beam at support is thus equivalent to a rectangular beam (Figs. 5.10.2 a, c, f and g). 

The actual width of the flange is the spacing of the beam, which is the same as the distance between the middle points of the adjacent spans of the slab, as shown in Fig. 5.10.2 b. However, in a flanged beam, a part of the width less than the actual width, is effective to be considered as a part of the beam. This width of the slab is designated as the effective width of the flange. 

5.10.2  Effective Width 

5.10.2.1  IS code requirements  

The following requirements (cl. 23.1.1 of IS 456) are to be satisfied to ensure the combined action of the part of the slab and the rib (rectangular part of the beam). 

(a)  The slab and the rectangular beam shall be cast integrally or they shall be effectively bonded in any other manner.

(b)  Slabs must be provided with the transverse reinforcement of at least 60 per cent of the main reinforcement at the mid span of the slab if the main reinforcement of the slab is parallel to the transverse beam (Figs. 5.10.3 a and b). 

The variation of compressive stress (Fig. 5.10.4) along the actual width of the flange shows that the compressive stress is more in the flange just above the rib than the same at some distance away from it. The nature of variation is complex and, therefore, the concept of effective width has been introduced. The effective width is a convenient hypothetical width of the flange over which the compressive stress is assumed to be uniform to give the same compressive force as it would have been in case of the actual width with the true variation of compressive stress.

5.10.2.2  IS code specifications  

Clause 23.1.2 of IS 456 specifies the following effective widths of T  and  L-beams:

(a) For T-beams, the lesser of  
(i)    b_f  =  l_o/6  +  b_w  +  6 D_f
(ii) b_f  =   A ctual width of the flange

(b) For isolated T-beams, the lesser of 

 (ii) b_f  =  A ctual width of the flange

(c)  For  L-beams, the lesser of  

(i)    b_f  =  l_o/12  +  b_w  +  3 D_f
(ii) b_f  =   A ctual width of the flange

(d) For isolated L-beams, the lesser of 

(ii) b_f  =  A ctual width of the flange
where b_f  =  effective width of the flange,
l_o  =  distance between points of zero moments in the beam, which is the effective span for simply supported beams and 0.7 times the effective span for continuous beams and frames,
b_w =  beadth of the web,
D_f =  thickness of the flange, and  
b  =  actual width of the flange. 

5.10.3  Four Different Cases  

The neutral axis of a flanged beam may be either in the flange or in the web depending on the physical dimensions of the effective width of flange  b_f, effective width of web  b_w, thickness of flange D_f and effective depth of flanged beam d (Fig. 5.10.4). The flanged beam may be considered as a rectangular beam of width  b_f  and effective depth  d  if the neutral axis is in the flange as the concrete in tension is ignored. However, if the neutral axis is in the web, the compression is taken by the flange and a part of the web.

All the assumptions made in sec. 3.4.2 of Lesson 4 are also applicable for the flanged beams. As explained in Lesson 4, the compressive stress remains constant between the strains of 0.002 and 0.0035. It is important to find the depth h of the beam where the strain is 0.002 (Fig. 5.10.5 b). If it is located in the web, the whole of flange will be under the constant stress level of 0.446 f_ck. The following gives the relation of  D_f  and  d  to facilitate the determination of the depth h where the strain will be 0.002.  From the strain diagram of Fig. 5.10.5 b: 

or (5.1) 

when x_u = x_{u, max} ,  we get 

h=  = 0.227d ,  0.205d and  0.197 7d  , for  Fe 250, Fe 415 and Fe 500, respectively. In general, we can adopt, say

h/d  =  0.2            (5.2)

The same relation is obtained below from the values of strains of concrete and steel of Fig. 5.10.5 b. 

or                            (5.3) 

Dividing Eq. 5.1 by Eq. 5.3 

                (5.4) 

Using      + 0.002   in Eq. 5.4, we get  h/d = 0.227, 0.205 and 0.197 for Fe 250, Fe 415 and Fe 500 respectively, and we can adopt   h/d = 0.2 (as in Eq. 5.2).

 Thus, we get the same Eq. 5.2 from Eq. 5.4,   h/d  =  0.2            (5.2) 

It is now clear that the three values of  h  are around 0.2 d  for the three grades of steel. The maximum value of  h  may be  D_f, at the bottom of the flange where the strain will be 0.002, if  D_f /d = 0.2. This reveals that the thickness of the flange may be considered small if D_f /d does not exceed 0.2 and in that case, the position of the fibre of 0.002 strain will be in the web and the entire flange will be under a constant compressive stress of 0.446  f_ck .  

On the other hand, if D_f   is >  0.2 d, the position of the fibre of 0.002 strain will be in the flange. In that case, a part of the slab will have the constant stress of 0.446  f_ck where the strain will be more than 0.002.  

Thus, in the balanced and over-reinforced flanged beams (when  x_u = x_{u , max} ), the ratio of D_f /d  is important to determine if the rectangular stress block is for the full depth of the flange (when D_f /d does not exceed 0.2) of for a part of the flange (when D_f /d > 0.2). Similarly, for the under-reinforced flanged beams, the ratio of D_f /x_u  is considered in place of D_f /d. If D_f /x_u does not exceed 0.43 (see Eq. 5.1), the constant stress block is for the full depth of the flange. If D_f /x_u > 0.43, the constant stress block is for a part of the depth of the flange.  Based on the above discussion, the four cases of flanged beams are as follows: 

(i) Neutral axis is in the flange (x_u < D_f ), (Fig. 5.10.6 a to c) 

(ii) Neutral axis is in the web and the section is balanced (x_u = x_u,max > D_f), (Figs. 5.10.7 and 8 a to e)  

It has two situations: (a) when D_f /d does not exceed 0.2, the constant stress block is for the entire depth of the flange (Fig. 5.10.7), and (b) when D_f /d > 0.2, the constant stress block is for a part of the depth of flange (Fig. 5.10.8). 

(iii) Neutral axis is in the web and the section is under-reinforced (x_u,max > x_u > D_f), (Figs. 5.10.9 and 10 a to e) 

This has two situations: (a) when D_f /x_u does not exceed 0.43, the full depth of flange is having the constant stress (Fig. 5.10.9), and (b) when D_f /x_u > 0.43, the constant stress is for a part of the depth of flange (Fig. 5.10.10). 

(iv) Neutral axis is in the web and the section is over-reinforced (x_u > x_u,_max> D_f), (Figs. 5.10.7 and 8 a to e)  

As mentioned earlier, the value of x_u is then taken as x_u,max when x_u> x_u,max. Therefore, this case also will have two situations depending on D_f /d not exceeding 0.2  or  >  0.2 as in (ii) above.

The governing equations of the four different cases are now taken up. 

5.10.4  Governing Equations  

The following equations are only for the singly reinforced T-beams. Additional terms involving  M_u,lim, M_u2,  A_sc ,  A_st1  and  A_st2 are to be included from Eqs. 4.1 to 4.8 of  sec. 4.8.3 of Lesson 8 depending on the particular case. Applications of these terms are explained through the solutions of numerical problems of doubly reinforced T-beams in Lessons 11 and 12. 

5.10.4.1  Case (i): When the neutral axis is in the flange (x_u < D_f ), (Figs. 5.10.6 a to c)  
Concrete below the neutral axis is in tension and is ignored. The steel reinforcement takes the tensile force (Fig. 5.10.6). Therefore, T and L-beams are considered as rectangular beams of width  b_f  and effective depth  d. All the equations of singly and doubly reinforced rectangular beams derived in Lessons 4 to 5 and 8 respectively, are also applicable here.

5.10.4.2  Case (ii): When the neutral axis is in the web and the section is balanced (x_u,max > D_f ), (Figs. 5.10.7 and 8 a to e) 

(a) When  D_f /d  does not exceed  0.2, (Figs. 5.10.7 a to e)  

As explained in sec. 5.10.3, the depth of the rectangular portion of the stress block (of constant stress = 0.446 f_ck) in this case is greater than D_f (Figs. 5.10.7 a, b and c). The section is split into two parts: (i) rectangular web of width b_w and effective depth d, and (ii) flange of width (b_f - b_w) and depth D_f  (Figs. 5.10.7 d and e).

Total compressive force  =  Compressive force of rectangular beam of width  b_w and depth  d  +  Compressive force of rectangular flange of width (b_f - b_w) and depth  D_f . Thus, total compressive force

C  =  0.36 f_ck b_w x_{u, max}  +  0.45  f_ck (b_f - b_w) D_f    (5.5)    

(Assuming the constant stress of concrete in the flange as  0.45  f_ck in place of  0.446 fck ,as per G-2.2 of IS 456), and the tensile force T  =  0.87  f_y A_st           (5.6)

The lever arm of the rectangular beam (web part) is (d - 0.42 x_{u, max}) and the same for the flanged part is (d - 0.5  D_f ).

So, the total moment =  Moment due to rectangular web part + Moment due to rectangular flange part

or M_u  =  0.36  f_ck b_w x_{u, max} (d - 0.42 x_{u, max} ) + 0.45  f_ck (b_f - b_w) D_f (d - D_f /2)

or M_u  =  0.36(x_{u, max} /d){1 - 0.42( x_{u, max}/d)} f_ck b_w d² + 0.45  f_ck(b_f - b_w) D_f(d - D_f /2)               (5.7) 

Equation 5.7 is given in G-2.2 of IS 456. 

(b)  When Df /d  >  0.2, (Figs. 5.10.8 a to e)  

In this case, the depth of rectangular portion of stress block is within the flange (Figs. 5.10.8 a, b and c).

It is assumed that this depth of constant stress (0.45  f_ck) is y_f, where y_f  =  0.15 x_{u, max} + 0.65  D_f,  but not greater than  D_f (5.8) 

The above expression of y_f  is derived in sec. 5.10.4.5.  As in the previous case (ii a), when D_f /d  does not exceed 0.2, equations of  C, T  and M_u are obtained from

Eqs. 5.5, 6 and 7 by changing D_f  to  y_f. Thus, we have (Figs. 5.10.8 d and e) C  =  0.36 f_ck b_w x_u, _max  +  0.45  f_ck (b_f - b_w) y_f    (5.9)    

T  =  0.87  f_y A_st (5.10) 

The lever arm of the rectangular beam (web part) is (d - 0.42 x_{u, max} ) and the same for the flange part is (d - 0.5  y_f ). Accordingly, the expression of  M_u is as follows: 

Mu  =  0.36(x_{u, max} /d){1 - 0.42( x_{u, max/}d)} f_ck b_w d² + 0.45  f_ck(b_f - b_w) y_f(d - y_f /2)              (5.11) 

5.10.4.3  Case (iii): When the neutral axis is in the web and the section is under-reinforced (x_u > D_f ), (Figs. 5.10.9 and 10 a to e) 

 (a) When D_f / x_u does not exceed  0.43, (Figs. 5.10.9 a to e) 

Since D_f doe s not exceed  0.43 x_u   and  h (depth of fibre where the strain is 0.002) is at a depth of 0.43 x_u, the entire flange will be under a constant stress of 0.45 f_ck (Figs. 5.10.9 a, b and c). The equations of  C, T  and M_u  can be written in the same manner as in sec. 5.10.4.2, case (ii a). The final forms of the equations are obtained from Eqs. 5.5, 6 and 7 by replacing  x_{u, max} by x_u. Thus, we have (Figs. 5.10.9 d and e) C  =  0.36 f_ck b_w x_u  +  0.45  f_ck (b_f - b_w) D_f                         (5.12) 
T  =  0.87  f_y A_st                                (5.13) 
Mu  =  0.36(x_u /d){1 - 0.42( x_u /d)} f_ck b_w d² + 0.45  f_ck(b_f - b_w) D_f (d - D_f /2)                   (5.14) 

(b) When Df / xu >  0.43, (Figs. 5.10.10 a to e) 

 Since D_f  >    0.43 x_u   and  h (depth of fibre where the strain is 0.002) is at a depth of   0.43 x_u, the part of the flange having the constant stress of 0.45  f_ck is assumed as  y_f  (Fig. 5.10.10 a, b and c). The expressions of y_f , C, T  and  M_u can be written from Eqs. 5.8, 9, 10 and 11 of sec. 5.10.4.2, case (ii b), by replacing  x_u,_max by x_u. Thus, we have (Fig. 5.10.10 d and e) y_f  =  0.15 x_u +  0.65  D_f,  but not greater than  D_f                        (5.15) 
C  =  0.36 f_ck b_w x_u  +  0.45  f_ck (b_f - b_w) y_f                         (5.16) 
T  =  0.87  f_y A_st                                      (5.17) 
M_u  =  0.36(x_u /d){1 - 0.42( x_u /d)} f_ck b_w d² + 0.45  f_ck(b_f - b_w) y_f (d - y_f /2)                      (5.18)

5.10.4.4  Case (iv):  When the neutral axis is in the web and the section is over-reinforced (x_u > D_f ), (Figs. 5.10.7 and 8 a to e) 

 For the over-reinforced beam, the depth of neutral axis  x_u  is more than   x_{u, max}  as in rectangular beams. However, x_u  is restricted up to  x_u,_max. Therefore, the corresponding expressions of   C, T  and  M_u  for the two situations (a) when D_f / d does not exceed 0.2  and (b)  when D_f / d >  0.2  are written from Eqs. 5.5 to 5.7 and 5.9 to 5.11, respectively of sec. 5.10.4.2 (Figs. 5.10.7 and 8). The expression of y_f for (b) is the same as that of Eq. 5.8.

(a)  When  D_f /d  does not  exceed 0.2 (Figs. 5.10.7 a to e)

The equations are:  C  =  0.36 f_ck b_w x_{u, max}  +  0.45  f_ck (b_f - b_w) D_f    (5.5)    T  =  0.87  f_y A_st           (5.6)

M_u  =  0.36(x_u, _max /d){1 - 0.42( x_u, _max/d)} f_ck b_w d² + 0.45  f_ck(b_f - b_w) D_f(d - D_f /2)               (5.7) 

(b)  When D_f /d  >  0.2 (Figs. 5.10.8 a to e) 

y_f  =  0.15 x_u, _max + 0.65  D_f,  but not greater than  Df (5.8) 
C  =  0.36 f_ck b_w x_u, _max  +  0.45  f_ck (b_f - b_w) y_f    (5.9)    T  =  0.87  f_y A_st (5.10) 
M_u  =  0.36(x_{u, max} /d){1 - 0.42( x_{u, max}/d)} fck bw d2 + 0.45  f_ck(b_f - b_w) y_f(d - y_f /2)  (5.11) 

It is clear from the above that the over-reinforced beam will not have additional moment of resistance beyond that of the balanced one. Moreover, it will prevent steel failure. It is, therefore, recommended either to re-design or to go for doubly reinforced flanged beam than designing over-reinforced flanged beam. 

5.10.4.5 Derivation of the equation to determine y_f , Eq. 5.8, Fig. 5.10.11 

Whitney's stress block has been considered to derive Eq. 5.8. Figure 5.10.11 shows the two stress blocks of IS code and of Whitney. y_f  =  Depth of constant portion of the stress block when D_f /d  >  0.2.  As y_f is a function of   x_u  and  D_f  and let us assume y_f   =  A x_u  +  B  D_f                    (5.19) 
where  A  and  B  are to be determined from the following two conditions:  (i) y_f   =      0.43  x_u ,  when  D_f  =  0.43  x_u               (5.20) 
 (ii) y_f   =     0.8 x_u ,     when  D_f  =  x_u                                (5.21) 
Using the conditions of Eqs. 5.20 and 21 in Eq. 5.19, we get  A  =  0.15  and  B  =  0.65. Thus, we have y_f   =   0.15 x_u  +  0.65  D_f                 (5.8) 

 
Summary of this Lesson  

This lesson illustrates the practical situations when slabs are cast integrally with the beams to form either T and L-beams or rectangular beams.  The concept of effective width of the slab to form a part of the beam has been explained. The requirements as per IS 456 have been illustrated so that the considered part of the slab may become effective as a beam. Expressions of effective widths for different cases of T and L-beams are given. Four sets of governing equations for determining C, T and M_u are derived for four different cases. These equations form the basis of analysis and design of singly and doubly reinforced T and L- beams.