Numerical Problems on Singly Reinforced Rectangular Beams (Part - 2) - Civil Engineering (CE) PDF Download

Instructional Objectives: 

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

• Apply the principles to analyse a given cross-section of a beam with specific reinforcement to determine its moment of resistance. 

3.7.1  Introduction  

This lesson explains the determination of moment of resistance of given singly reinforced rectangular beam sections with the help of illustrative analysis type of problem. The numerical problem is solved by (i) direct computation method, (ii) using charts of SP-16 and (iii) using tables of SP-16. Step by step solutions illustrate the procedure clearly.

3.7.2  Analysis Type of Problems  

It may be required to estimate the moment of resistance and hence the service load of a beam already designed earlier with specific dimensions of b, d and D and amount of steel reinforcement A_st. The grades of concrete and steel are also known. In such a situation, the designer has to find out first if the beam is under-reinforced or over-reinforced. The following are the steps to be followed for such problems.

3.7.2.1  x_u, _max

The maximum depth of the neutral axis x_{u, max} is determined from Table 3.2 of Lesson 5 using the known value of f_y.

3.7.2.2  x_u
 The depth of the neutral axis for the particular beam is determined from Eq. 3.16 of Lesson 5 employing the known values of f_y, f_ck, b and A_st. 3.7.2.3  M_u and service imposed loads  The moment of resistance M_u is calculated for the three different cases as follows: 

(a)  If  x_u < x_{u, max}, the beam is under-reinforced and Mu is obtained from Eq. 3.22 of Lesson 5.  

(b) If  x_u = x_{u, max}, the M_u is obtained from Eq. 3.24 of Lesson 5. 

(c) If  x_u > x_{u, max}, the beam is over-reinforced for which x_u is taken as x_{u, max} and then  M_u is calculated from Eq. 3.24 of Lesson 5, using  x_u = x_{u, max.} 
 

With the known value of M_u, which is the factored moment, the total factored load can be obtained from the boundary condition of the beam. The total service imposed loads is then determined dividing the total factored load by partial safety factor for loads (= 1.5). The service imposed loads are then obtained by subtracting the dead load of the beam from the total service loads. 

3.7.3  Analysis Problems 3.2 and 3.3 

Determine the service imposed loads of two simply supported beam of same effective span of 8 m (Figs. 3.7.1 and 2) and same cross-sectional dimensions, but having two different amounts of reinforcement. Both the beams are made of M 20 and Fe 415.

3.7.4  Solution by Direct Computation Method - Problem 3.2

Given data: b = 300 mm,   d = 550 mm,   D = 600 mm,   A_st = 1256 mm² (4-20 T), L_eff = 8 m and boundary condition = simply supported (Fig. 3.7.1).

3.7.4.1  x_{u, max}  

From Table 3.2 of  Lesson 5, we get  x_{u, max} = 0.479 d = 263.45 mm 

3.7.4.2  x_u

          (3.16) 

=  209.94385  mm  <  x_{u, max} = (263.45 mm)

Hence, the beam is under-reinforced. 

3.7.4.3  M_u and service imposed loads  
For x_u <   x _u, _max , we have M_u  =  0.87 f_y A_st (d - 0.42 x_u)         (3.22) 
        =  0.87 (415) (1256) {550 - 0.42(209.94385)} 
        =  209.4272  kNm Total factor load

  =  26.1784  kN/m 

Total service load  =17.452266 kN/m 

Dead load of the beam  =  0.3 (0.6) (25)  =  4.5  kN/m

Hence, service imposed loads   =  (17.452266 - 4.5)  kN/m    

=  12.952266  kN/m 

Given data:  b = 300 mm,   d = 550 mm,   D = 600 mm,   A_st = 1658 mm² (4-20 T + 2-16 T),  Leff = 8 m  and  boundary conditions = simply supported (Fig. 3.7.2)

3.7.5.1  x_u, _max  

From Table 3.2 of Lesson 5, we get  x_u, _max = 0.479 d  =  263.45  mm

3.7.5.2  x_u

 (3.16) 

=  277.13924  mm  >  x_{u, max} = (263.45 mm)

Hence, the beam is over-reinforced.

3.7.5.3  M_u and service imposed loads  

For x_u  >    x_u,max, we have 

(3.24) 

=  0.36 (0.479) {1 - 0.42(0.479)} (300) (550) (550) (20)  Nmm          

=  250.01356  kNm

If we use Eq. 3.22 using  x_u  =  x_u,max,  for M_u

M_u  =  0.87 f_y A_st (d - 0.42 x_{u, max})        (3.22) 

Then, (M_u)_steel  =  0.87 (415) (1658) {550 - 0.42 (263.45)} Nmm                

=  263.00468 kNm > (M_u)_concrete   (= 250.01356 kNm)

The higher M_u as obtained from steel is not true because the entire amount of steel (1658 mm²) cannot yield due to over-reinforcing. Prior to that, concrete fails at 250.01356 kNm. However, we can get the same of M_u as obtained from Eq. 3.24 of Lesson 5, if we can find out how much A_st is needed to have  x_u = 263.45 mm. From Eq. 3.16 of Lesson 5, we can write: 

 =  1576.1027 mm²

If we use this value for A_st in Eq. 3.22 of Lesson 5, we get ( M _u ) = 0.87 (415) (1576.1027 ) {550 − 0.42 (263.45 )}          

=  250.0135  (same as obtained from Eq. 3.24).

From the factored moment  M_u = 250.01356 kNm, we have:  

Total factored load =  kN/m 

Total service load = 

Now, Dead load of the beam  =  0.3 (0.6) (25)  =  4.5  kN/m

Hence, service imposed loads  =  20.834463 - 4.5  =  16.334463 kN/m

3.7.6  Solution by Design Chart - Problems 3.2 and 3.3  

For the two problems with known b, d, D, A_st, grade of concrete and grade of steel, chart 14 of SP-16 is applicable. From the effective depth d and percentage of reinforcement  he chart is used to find M_u per metre width. Multiplying M_u per meter width with b, we get M_u for the beam. After that, the service imposed load is calculated using the relation  Service imposed load  =    -  Dead load      (3.27) 

The results of the two problems are furnished in Table 3.7. 

Table 3.7  Results of Problems 3.2 and 3.3 (Chart of SP-16) 

3.7.7  Solution by Design Tables - Problems 3.2 and 3.3  

Table 2 of SP-16 presents the value of reinforcement percentage for different combinations of fy and (M_u/bd²) for M-20. Here, from the known values of p and f_y, the corresponding values of M_u/bd² are determined. These in turn give M_u of the beam. Then the service imposed load can be obtained using Eq. 3.27 as explained in the earlier section (sec. 3.7.6). The results of the two problems are presented in Table 3.8. 

Table 3.8  Results of Problems 3.2 and 3.3 (Table of SP-16) 

*   Linear interpolated

**  p_t = 1.2058181 is not admissible, i.e. over-reinforced. So at  p_t = 0.955, M_u/bd² =2.76.

3.7.8  Comparison of Results of Three Methods 

 The values of  M_u and service imposed loads of the under-reinforced and over-reinforced problems (Problems 3.2 and 3.3), computed by three methods, are presented in Table 3.9. 

Table 3.9  Comparison of results of Problems 3.2 and 3.3 

3.7.9  Practice Questions and Problems with Answers 

Q.1: Determine the moments of resistance M_u and service imposed loads on a simply supported beam of effective span 10.0 m with b = 300mm, d = 500 mm, D = 550 mm and grades of concrete and steel are M20 and Fe500, respectively for the two different cases employing (a) direct computation method and (b) using charts and tables of SP-16: (i) when A_st is minimum acceptable and (ii) when A_st is maximum acceptable (Fig. 3.7.3).
A.1: Given data:  b = 300 mm,  d = 500 mm,  D = 550 mm,  L_eff = 10.0 m,  f_ck = 20 N/mm² and f_y = 500 N/mm²

(a) Direct computation method: 

Case (i)  When  Ast is minimum acceptable (Eq. 3.26 of Lesson 6)  

Minimum  (3.26)

So,   =  255  mm².  Providing  4-12T    gives  A_st = 452 mm². 

Equation 3.16 of Lesson 5 gives the depth of the neutral axis  x_u: 

= 91.03 mm

Table 3.2 of Lesson 5 gives  x_{u, max}  =  0.46(500)  =  230 mm.

The beam is, therefore, under-reinforced (as  x_u < x_{u, max}).

Equation 3.22 of Lesson 5 gives the Mu as follows: M_u  =  0.87 f_y A_st (d - 0.42 x_u)            (3.22) 

=  0.87(500) (452) {500 - 0.42(91.03)}  Nmm      

=  90.79  kNm 

Total factored load  =  = 7.26  kN/m 

The dead load of the beam  =  0.3 (0.55) (25)  =  4.125  kN/m

So, the service imposed loads  =  {(Total factored load)/(Load factor)} - (Dead load) 

=  7.26/1.5 - 4.125  =  0.715  kN/m

This shows that the beam can carry maximum service imposed loads, 17 per cent of its dead load only, when the acceptable minimum tensile reinforcement is 452 mm² (4 bars of 12 mm diameter). 

Case (ii) when Ast is maximum acceptable:  

To ensure ductile failure, it is essential that the acceptable maximum tensile reinforcement should be between 75 and 80 per cent of p_{t, lim} and not as given in clause 26.5.1.1.(b), i.e. 0.04 bD. Thus, here the maximum acceptable p_t should be between 0.57 and 0.61 per cent (as  p_{t, lim} = 0.76 from Table 3.1 of  Lesson 5). However, let us start with 0.76 per cent as the span is relatively large and keeping in mind that while selecting the bar diameter, it may get reduced to some extent.

So, (A_st)_max  =  0.76 (300) (500)/100  =  1140  mm²

Selecting 4-16 and 2-12 mm diameter bars, we get A_st = 1030 mm² when pt becomes 0.67 per cent. So, the maximum acceptable tensile reinforcement is 1030 mm².

 =  207.43 mm   (Eq. 3.16 of  Lesson 5)

x_u  <  x_{u, max}  (as x_{u, max}  =  230 mm; see Case (i) of this problem).  

Therefore, M_u  is obtained from Eq. 3.22 of  Lesson 5 as,   M_u  =  0.87 f_y A_st (d - 0.42 x_u)                  (3.22) 

 =  0.87(500) (1030) {500 - 0.42(207.43)}  Nmm      

=  184.99  kNm 

Total factored load = = 14.8   kN/m

With dead load  =  4.125 kN/m  (see Case (i) of this problem), we have:  

Service imposed load  =  14.8/1.5  -  4.125  =  5.74  kN/m This beam, therefore, is in a position to carry service imposed loads of 5.74  kN/m, about 40% higher than its own dead load. 

(b) Using chart and tables of SP-16: 

Tables 3.10 and 3.11 present the results using charts and tables respectively of SP-16. For the benefit of the reader the different steps are given below separately for the use of chart and table respectively for the minimum acceptable reinforcement.

The steps using chart of SP-16 are given below:

Step 1:  With the given  f_ck, f_y and d, choose the particular chart. Here, for  f_ck = 20 N/mm², f_y  = 500 N/mm² and d = 500 mm, the needed chart no. is 17. 
Step 2: Chart 17 shows minimum  p_t = 0.13% which gives Ast = 195 mmm2
Step 3:  Provide bars of 4-12 mm diameter, which give  p_t = 0.301%
Step 4:  For  p_t = 0.301, Chart 17 shows  M_u = 300 kNm per metre width, which gives  M_u = 300 (0.3) = 90 kNm for the beam.
Step 5: The service imposed loads are calculated as follows:  Service imposed loads = - 4.125 - 0.675  kN/m, using the dead load of the beam as 4.125 kN/m (see case (i) of this problem).
Step 6: The capacity of the beam is to carry 0.675 kN/m which is (0.675/4.125) 100  =  16.36%. 
 

Table 3.10  Results of Q.1 using chart of SP-16 

Similar calculations are done for the maximum acceptable reinforcement. The steps are given below:
Step 1:  With the given    f_ck = 20 N/mm², f_y  = 500 N/mm², Table 2 of SP-16 is selected.
Step 2:  (p_t)_min = 0.07 from Table 2,  which gives Ast = 105 mmm2
Step 3:  Provide bars of 4-12 mm diameter, which give  p_t = 0.301%
Step 4:  For  (p_t)_provided, we get (M_u/bd²) from Table 2 of SP-16 by linear interpolation as follows: 

Hence, M_u  =  1  .204 (300) (500)²  =  90.3  kNm
Step 5: Same as Step 5 while using chart of SP-16.
Step 6: Same as Step 6 while using chart of SP-16. 

Table 3.11  Results of Q.1 using table of SP-16 

3.7.10  Test 7 with Solutions

Maximum Marks = 50,    
 Maximum Time = 30 minutes 

TQ.1:  Determine the moment of resistance for the beams shown in Figs. 3.7.4 and 3.7.5 using M 20 and Fe 250 by direct computation and using charts and tables of SP-16.

A.TQ.1: Case A:  TQ.1 A of Fig. 3.7.4 

(i) Direct computation method

A_st  =  1963  mm²  (4-25 φ) x_u , _max =  0.53 (450)  =  238.5 mm   (Table 3.2 of Lesson 5 gives 0.53) 

 = 197.66   (See Eq. 3.16 of Lesson 5)

So, x_u < x_{u , max} shows that it is under-reinforced section for which  M_u  is obtained from Eq. 3.22 of Lesson 5 M_u  =  0.87 f_y A_st (d - 0.42 x_u)                              (3.22) 

=  0.87(250) (1963) {450 - 0.42(197.66)}  

 =  156.68  kNm 

(ii)  Chart of SP-16 

A_st  =  1963  mm²

= 1.45,

From chart 11 of SP-16, when  p_t = 1.45,  d = 450,
we get  M_u per metre width  =  522  kNm/m
M_u   =  522 (0.3)  =  156.6  kNm 

(iii) Table of SP-16  Table 2 of SP-16 is for M-20 and Fe250.  At p_t = 1.451, we get 

 = 2.58  N/mm²

So, M_u =  2 .58 (300) (450) (450) (10^-6)  kNm  =  156.74  kNm

The three values of M_u are close to each other.  

A.TQ.1: Case B:  TQ.1 B of Fig. 3.7.5 

(i) Direct computation method 

A_st  =  1963 + 981  =  2944 mm²     (6-25 φ) 

 = 296.44

mm  (See Eq. 3.16 of  Lesson 5) 

x_{u , max} =  0.53 (450)  =  238.5 mm     (Table 3.2 of Lesson 5 gives 0.53)  

So, x_u >  x _{u, max} and the beam is over-reinforced. In such a situation, we take x_u =  x_{u, max}  =  238.5 mm. The  M_u  will be calculated from Eq. 3.24 of Lesson 5.

     (3.24) 
 =  0.36 (0.53) {1 - 0.42 (0.53)} (300) (450)2 (20)  Nmm      
=  180.22  kNm

(ii)  Chart of SP-16 

 = 2.18

.  In chart 11 (for M 20 and Fe 250), maximum admissible  p_t  is 1.75 and for this p_t when  d = 450,  M_u = 600 kNm/m. 

 So, M_u  =    600 (0.3)  =  180  kNm  

(iii) Table of SP-16  

Table 2 (for M 20 and Fe 250) has the maximum  p_t = 1.76 and at that value, (M_u/bd²) = 2.98. This gives   M_u = 2.98 (300) (450)² (10-6) = 181.03  kNm.

Here also the three values of M_u are close to each other.

3.7.11  Summary of this Lesson  

This lesson explains the use of equations derived in Lesson 5 for the analysis type of problems. The three methods  (i) direct computation, (ii) use of charts of SP-16 and (iii) use of tables of SP-16 are illustrated through the step by step solutions of numerical problems. Their results are compared to show the closeness of them.