Doubly Reinforced Beams: Theory (Part - 1) - Civil Engineering (CE) PDF Download

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

• explain the situations when doubly reinforced beams are designed,  
• name three cases other than doubly reinforced beams where compression reinforcement is provided,
• state the assumptions of analysis and design of doubly reinforced beams,
• derive the governing equations of doubly reinforced beams,
• calculate the values of  f_sc from (i) d'/d and (ii) calculating the strain of the compression reinforcement,
• state the minimum and maximum amounts of  A_sc and  A_st in doubly reinforced beams,
• state the two types of numerical problems of doubly reinforced beams,
• name the two methods of solving the two types of problems, and
• write down the steps of the two methods for each of the two types of problems.  

4.8.1  Introduction 

Concrete has very good compressive strength and almost negligible tensile strength. Hence, steel reinforcement is used on the tensile side of concrete. Thus, singly reinforced beams reinforced on the tensile face are good both in compression and tension. However, these beams have their respective limiting moments of resistance with specified width, depth and grades of concrete and steel. The amount of steel reinforcement needed is known as Ast,lim. Problem will arise, therefore, if such a section is subjected to bending moment greater than its limiting moment of resistance as a singly reinforced section.

 There are two ways to solve the problem. First, we may increase the depth of the beam, which may not be feasible in many situations. In those cases, it is possible to increase both the compressive and tensile forces of the beam by providing steel reinforcement in compression face and additional reinforcement in tension face of the beam without increasing the depth (Fig. 4.8.1). The total compressive force of such beams comprises (i) force due to concrete in compression and (ii) force due to steel in compression. The tensile force also has two components: (i) the first provided by A_st,lim which is equal to the compressive force of concrete in compression. The second part is due to the additional steel in tension - its force will be equal to the compressive force of steel in compression. Such reinforced concrete beams having steel reinforcement both on tensile and compressive faces are known as doubly reinforced beams.  

Doubly reinforced beams, therefore, have moment of resistance more than the singly reinforced beams of the same depth for particular grades of steel and concrete. In many practical situations, architectural or functional requirements may restrict the overall depth of the beams. However, other than in doubly reinforced beams compression steel reinforcement is provided when:

(i) some sections of a continuous beam with moving loads undergo change of sign of the bending moment which makes compression zone as tension zone or vice versa.

(ii) the ductility requirement has to be followed.

(iii) the reduction of long term deflection is needed. It may be noted that even in so called singly reinforced beams there would be longitudinal hanger bars in compression zone for locating and fixing stirrups.

4.8.2  Assumptions

(i) The assumptions of sec. 3.4.2 of Lesson 4 are also applicable here.

(ii) Provision of compression steel ensures ductile failure and hence, the limitations of x/d ratios need not be strictly followed here. 

(iii) The stress-strain relationship of steel in compression is the same as that in tension. So, the yield stress of steel in compression is 0.87 f_y.

4.8.3  Basic Principle 

As mentioned in sec. 4.8.1, the moment of resistance M_u of the doubly reinforced beam consists of (i) M_u,lim of singly reinforced beam and (ii) M_u2 because of equal and opposite compression and tension forces (C₂ and T₂) due to additional steel reinforcement on compression and tension faces of the beam (Figs. 4.8.1 and 2). Thus, the moment of resistance M_u of a doubly reinforced beam is 

M_u  = M_u,lim + M_u2           (4.1)

The M_u,lim  is as given in Eq. 3.24 of Lesson 5, i.e., 

         (4.2)  
Also, M_u,lim  can be written from Eq. 3.22 of Lesson 5, using  x_u = x_{u, max}, i.e.,

M_{u, lim}  =  0.87 A_{st, lim} f_y (d - 0.42 x_{u, max}) 

              (4.3) 

The additional moment  M_u2  can be expressed in two ways (Fig. 4.8.2): considering (i) the compressive force C₂ due to compression steel and (ii) the tensile force T₂ due to additional steel on tension face. In both the equations, the lever arm is (d - d'). Thus, we have
M _{u 2} = A_sc ( f _sc − f _cc ) (d − d ' )        (4.4)
M _{u 2} = A_{st 2} ( 0.87 f _y ) (d − d ' )        (4.5)
where A_sc  =  area of compression steel reinforcement
f_sc   =  stress in compression steel reinforcement
f_cc  = compressive stress in concrete at the level of centroid of compression steel reinforcement
A_st2  =  area of additional steel reinforcement
Since the additional compressive force C₂ is equal to the additional tensile force T₂, we   have  A_sc (f_sc  - f_cc)  = A_st2 (0.87 f_y)  (4.6) 
Any two of the three equations (Eqs. 4.4 - 4.6) can be employed to determine A_sc and A_st2.
The total tensile reinforcement A_st  is then obtained from:  
A_st =A_st+ A_st2        (4.7) 

   (4.8) 

4.8.4  Determination of  f_sc  and f_cc  

It is seen that the values of  f_sc  and f_cc  should be known before calculating  A_sc.  The following procedure may be followed to determine the value of f_sc  and f_cc  for the design type of problems (and not for analysing a given section). For the design problem the depth of the neutral axis may be taken as  x_u,max  as shown in Fig. 4.8.2. From Fig. 4.8.2, the strain at the level of compression steel reinforcement  ε_sc may be written as 

   (4.9)

The stress in compression steel  f_sc  is corresponding to the strain  ε_sc  of Eq. 4.9 and is determined for (a) mild steel and (b) cold worked bars Fe 415 and 500 as given below: 

(a) Mild steel Fe 250  

The strain at the design yield stress of 217.39 N/mm² (f_d  =  0.87 f_y ) is 0.0010869 (= 217.39/E_s). The f_sc  is determined from the idealized stress-strain diagram of mild steel (Fig. 1.2.3 of Lesson 2 or Fig. 23B of IS 456) after computing the value of  ε_sc from Eq. 4.9 as follows:
(i) If the computed value of  ε_sc ≤  0.0010869,  f_sc  =  ε_sc E_s  =  2 (105) ε_sc
(ii) If the computed value of  ε_sc  >  0.0010869,  f_sc  = 217.39  N/mm². 

(b)  Cold worked bars Fe 415 and Fe 500  

The stress-strain diagram of these bars is given in Fig. 1.2.4 of Lesson 2 and in Fig. 23A of IS 456. It shows that stress is proportional to strain up to a stress of 0.8  f_y. The stress-strain curve for the design purpose is obtained by substituting  f_yd  for f_y  in the figure up to 0.8 f_yd. Thereafter, from 0.8 f_yd  to  f_yd, Table A of SP-16 gives the values of total strains and design stresses for Fe 415 and Fe 500. Table 4.1 presents these values as a ready reference here

Table 4.1  Values of    f_sc  and   ε_sc

Linear interpolation may be done for intermediate values.

The above procedure has been much simplified for the cold worked bars by presenting the values of  f_sc  of compression steel in doubly reinforced beams for different values of  d'/d only taking the practical aspects into consideration. In most of the doubly reinforced beams, d'/d  has been found to be between 0.05 and 0.2. Accordingly, values of f_sc can be computed from Table 4.1 after determining the value of ε_sc from Eq. 4.9 for known values of d'/d as 0.05, 0.10, 0.15 and 0.2. Table F of SP-16 presents these values of fsc for four values of d'/d (0.05, 0.10, 0.15 and 0.2) of Fe 415 and Fe 500. Table 4.2 below, however, includes Fe 250 also whose  f_sc  values  are computed  as  laid  down in sec. 4.8.4(a) (i) and (ii) along with those of Fe 415 and Fe 500. This table is very useful and easy to determine the  f_sc from the given value of d'/d. The table also includes strain values at yield which are explained below: (i)   The strain at yield of Fe 250  = 

Here, there is only elastic component of the strain without any inelastic strain.  

(ii)   The strain at yield of Fe 415  = 

(iii) The strain at yield of Fe 500  =   

Table 4.2  Values of    f_sc  for different values of  d'/d

4.8.5 Minimum and maximum steel 

4.8.5.1  In compression 

There is no stipulation in IS 456 regarding the minimum compression steel in doubly reinforced beams. However, hangers and other bars provided up to 0.2% of the whole area of cross section may be necessary for creep and shrinkage of concrete. Accordingly, these bars are not considered as compression reinforcement. From the practical aspects of consideration, therefore, the minimum steel as compression reinforcement should be at least 0.4% of the area of concrete in compression or 0.2% of the whole cross-sectional area of the beam so that the doubly reinforced beam can take care of the extra loads in addition to resisting the effects of creep and shrinkage of concrete.  

The maximum compression steel shall not exceed 4 per cent of the whole area of cross-section of the beam as given in cl. 26.5.1.2 of IS 456. 

4.8.5.2  In tension  

As stipulated in cl. 26.5.1.1(a) and (b) of IS 456, the minimum amount of tensile reinforcement shall be at least (0.85 bd/f_y) and the maximum area of tension reinforcement shall not exceed (0.04 bD).  It has been discussed in sec. 3.6.2.3 of Lesson 6 that the singly reinforced beams shall have A_st normally not exceeding 75 to 80% of A_st,lim so that x_u remains less than  x_u,max with a view to ensuring ductile failure. However, in the case of doubly reinforced beams, the ductile failure is ensured with the presence of compression steel. Thus, the depth of the neutral axis may be taken as x_{u, max} if the beam is over-reinforced. Accordingly, the A_st1  part of tension steel can go up to A_{st, lim} and the additional tension steel A_st2 is provided for the additional moment  M_u - M_{u, lim.} The quantities of A_st1  and A_st2  together form the total A_st, which shall not exceed 0.04 bD. 

4.8.6  Types of problems and steps of solution  

Similar to the singly reinforced beams, the doubly reinforced beams have two types of problems: (i) design type and (ii) analysis type. The different steps of solutions of these problems are taken up separately.  

4.8.6.1  Design type of problems  

In the design type of problems, the given data are b,  d,  D, grades of concrete and steel. The designer has to determine  A_sc and  A_st  of the beam from the given factored moment. These problems can be solved by two ways: (i) use of the equations developed for the doubly reinforced beams, named here as direct computation method, (ii) use of charts and tables of SP-16.

(a)  Direct computation method 

Step 1:  To determine  M_{u, lim}  and A_{st, lim}  from Eqs. 4.2 and 4.8, respectively. 
Step 2:  To determine M_u2, A_sc, A_st2  and A_st  from Eqs. 4.1, 4.4, 4.6 and 4.7, respectively. 
Step 3:  To check for minimum and maximum reinforcement in compression and tension as explained in sec. 4.8.5.
Step 4:  To select the number and diameter of bars from known values of A_sc and A_st. 

(b)  Use of SP table  

Tables 45 to 56 present the p_t  and p_c  of doubly reinforced sections for  d'/d = 0.05, 0.10, 0.15 and 0.2 for different f_ck  and f_y   values against  M_u /bd². The values of p_t and p_c  are obtained directly selecting the proper table with known values of M_u/bd² and d'/d.

4.8.6.2  Analysis type of problems 

 In the analysis type of problems, the data given are b,  d,  d',  D, f_ck, f_y, A_scand A_st . It is required to determine the moment of resistance M_u  of such beams. These problems can be solved: (i) by direct computation method and (ii) by using tables of SP-16.

(a)  Direct computation method 

Step 1:  To check if the beam is under-reinforced or over-reinforced. 

 First, x_u,max  is determined assuming it has reached limiting stage using  coefficients as given in cl. 38.1, Note of IS 456. The strain of tensile steel  ε_st is computed from  and is checked if   ε_st  has reached the yield strain of steel:    The beam is under-reinforced or over-reinforced if ε_st is less than or more than the yield strain. 
Step 2:  To determine M_u,lim  from Eq. 4.2 and  A_st,lim from the p_{t, lim}  given in Table 3.1 of Lesson 5.
Step 3:  To determine A_st2  and A_sc  from Eqs. 4.7 and 4.6, respectively.
Step 4:  To determine M_u2  and M_u   from Eqs. 4.4 and 4.1, respectively.

(b)  Use of tables of SP-16 

 As mentioned earlier Tables 45 to 56 are needed for the doubly reinforced beams. First, the needed parameters d'/d, p_t and p_c are calculated. Thereafter, M_u/bd² is computed in two stages: first, using  d'/d and p_t and then using  d'/d and p_c . The lower value of  M_u  is the moment of resistance of the beam.

4.8.7   Practice Questions and Problems with Answers  

Q.1: When do we go for doubly reinforced beams ?
 A.1: The depth of the beams may be restricted for architectural and/or functional requirements. Doubly reinforced beams are designed if such beams of restricted depth are required to resist moment more that its  M_{u, lim}. 

Q.2: Name three situations other than doubly reinforced beams, where the compression reinforcement is provided.
 A.2: Compression reinforcement is provided when:

(i) Some sections of a continuous beam with moving loads undergo change of sign of the bending moment which makes compression zone as tension zone, 

(ii) the ductility requirement has to be satisfied,

(iii) the reduction of long term deflection is needed.

Q.3: State the two types of problems of doubly reinforced beams specifying the given data and the values to be determined in the two types of problems.
 A.3: The two types of problems are: (i)  Design type of  problems and (ii) Analysis type of problems

(i)  Design type of problems:  The given data are  b, d, D, f_ck, f_y and  M_u .  It is required to determine  A_scand  A_st.  

(ii)  Analysis type of problems:  The given data are   b, d, D, fck, f_y,  A_sc  and  A_st.  It is required to determine the  Mu of the beam.

Q.4: Name the two methods of solving the two types of problems.
 A.4: The two methods of solving the two types of problems are:  (i)   Direct computation method, and (ii)  Use of tables of SP-16. 

Summary of this Lesson  

Lesson 8 derives the governing equations of the doubly reinforced beams explaining different assumptions and situations when they are needed. The methods of determination of compressive stress in steel are illustrated. The two types of problems and the different steps of solution of them by two different methods are explained.