Flanged Beams: Numerical Problems - 1 | RCC & Prestressed Concrete - Civil Engineering (CE) PDF Download

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

• identify the two types of numerical problems – analysis and design types,
• apply the formulations to analyse the capacity of a flanged beam,  
• determine the limiting moment of resistance quickly with the help of tables of SP-16.

5.11.1 Introduction  

Lesson 10 illustrates the governing equations of flanged beams. It is now necessary to apply them for the solution of numerical problems. Two types of numerical problems are possible: (i) Analysis and (ii) Design types.

This lesson explains the application of the theory of flanged beams for the analysis type of problems. Moreover, use of tables of SP-16 has been illustrated to determine the limiting moment of resistance of sections quickly for the three grades of steel. Besides mentioning the different steps of the solution, numerical examples are also taken up to explain their step-by-step solutions. 

5.11.2   Analysis Type of Problems 

The dimensions of the beam  b_f, b_w, D_f, d, D, grades of concrete and steel and the amount of steel  A_st are given. It is required to determine the moment of resistance of the beam. 

Step 1:  To determine the depth of the neutral axis  x_u  

The depth of the neutral axis is determined from the equation of equilibrium C =  T. However, the expression of C depends on the location of neutral axis, D_f /d  and D_f /x_u parameters. Therefore, it is required to assume first that the x_u is in the flange. If this is not the case, the next step is to assume x_u in the web and the computed value of x_u will indicate if the beam is underreinforced, balanced or over-reinforced. 

Other steps:  

After knowing if the section is under-reinforced, balanced or overreinforced, the respective parameter  D_f/d or D_f/X_u is computed for the underreinforced, balanced or over-reinforced beam. The respective expressions of C, T and M_u, as established in Lesson 10, are then employed to determine their values. Figure 5.11.1 illustrates the steps to be followed.

5.11.3 Numerical Problems (Analysis Type) 

Ex.1:  Determine the moment of resistance of the T-beam of Fig. 5.11.2. Given data: b_f  =  1000 mm, D_f  =  100 mm,  b_w  =  300 mm,  cover  =  50 mm, d  =  450 mm and A_st  =  1963 mm² (4- 25 T). Use M 20  and   Fe 415. 

Step 1:  To determine the depth of the neutral axis x_u 

 Assuming x_u  in  the flange and equating total compressive and tensile forces from the expressions of  C  and  T  (Eq. 3.16 of Lesson 5) as the T-beam can be treated as rectangular beam of width b_f  and effective depth  d, we get:

 = 98.44  mm  < 100 mm

So, the assumption of  x_u  in the flange is correct. x_{u, max  f}or the balanced rectangular beam  =  0.48 d  =  0.48 (450)  =  216 mm. 
It is under-reinforced since x_u  <  x_u,max. 

Step 2:  To determine  C, T  and M_u

 From Eqs. 3.9 (using b = b_f) and 3.14 of Lesson 4 for C and T and Eq. 3.23 of Lesson 5 for M_u, we have: 

C  =  0.36  b_f  x_u  f_ck          (3.9)      

=  0.36 (1000) (98.44) (20)  =  708.77  kN

T  =  0.87  f_y  A_st (3.14) 

=  0.87 (415) (1963)  =  708.74 kN 

 (3.23)     

This problem belongs to the case (i) and is explained in sec. 5.10.4.1 of Lesson 10. 

Ex.2:  Determine  A_st,lim  and  M_u,lim  of the flanged beam of Fig. 5.11.3. Given data are: b_f  =  1000 mm, D_f  =  100 mm,  b_w  =  300 mm,  cover  =  50 mm and  d =  450 mm. Use M 20  and   Fe 415.

Step 1:  To determine D_f/d  ratio  For the limiting case x_u  =  x_u,_max  =  0.48 (450)  =  216  mm  > D_f.  The ratio D_f/d  is computed. 

Df/d   =  100/450  =  0.222  >  0.2 Hence, it is a problem of case (ii b) and discussed in sec. 5.10.4.2 b of Lesson 10. 

Step 2:  Computations of  y_f , C and T 

 First, we have to compute y_f  from Eq.5.8 of Lesson 10 and then employ Eqs. 5.9, 10 and 11  of Lesson 10 to determine C, T  and  M_u, respectively.

y_f  = 0.15 x_u,max +  0.65 D_f  =  0.15 (216) + 0.65 (100)

 =  97.4 mm. (from Eq. 5.8) 

C  =  0.36  f_ck  b_w  x_u,max  +  0.45  f_ck (b_f   -  b_w) y_f (5.9)    

=  0.36 (20) (300) (216)  +  0.45 (20) (1000 - 300) (97.4) = 1,080.18 kN.

T  =  0.87  f_y  A_st    

=  0.87 (415) A_st (5.10) 

Equating  C  and  T, we have 

Provide 4-28 T (2463 mm²) +  3-16 T (603  mm²)  =  3,066  mm² 

Step 3:  Computation of  M_u

 (5.11)    

=  0.36 (0.48) {1 - 0.42 (0.48)} (20) (300) (450)2 
+ 0.45 (20) (1000 - 300) (97.4) (450 - 97.4/2) = 413.87 kNm  

Ex.3:  Determine the moment of resistance of the beam of Fig. 5.11.4 when  A_st = 2,591 mm² (4- 25 T and 2- 20 T). Other parameters are the same as those of

Ex.1: b_f  =  1,000 mm, D_f  =  100 mm,  b_w  =  300 mm,  cover  =  50 mm and d  =  450 mm. Use M 20  and   Fe 415.

Step 1:  To determine  x_u 

 Assuming x_u  t o  be in the flange and the beam is under-reinforced, we have from Eq. 3.16 of Lesson 5:  

 =  129.93  mm  > 100  mm

Since x_u > D_f, the neutral axis is in web.  Here, D_f/d  =  100/450  =  0.222  >  0.2. So, we have to substitute the term y_f  from Eq. 5.15 of Lesson 10, assuming D_f / x_u  >  0.43 in the equation of  C = T  from Eqs. 5.16 and 17 of  sec. 5.10.4.3 b of Lesson 10. Accordingly, we get:  0.36 f_ck  b_w  x_u  +  0.45  f_ck (b_f   -  b_w) y_f  =  0.87 f_y A_st

or 0.36 (20) (300) (x_u) + 0.45 (20) (1000 - 300) {0.15 x_u + 0.65 (100)}   =  0.87 (415) (2591) 

or x_u   =  169.398  mm  <  216 mm (x_u,max = 0.48 x_u = 216 mm) 

So, the section is under-reinforced.

Step 2:  To determine  M_u 

D_f /x_u  =  100/169.398  =  0.590  >  0.43 This is the problem of case (iii b) of sec. 5.10.4.3 b. The corresponding equations are Eq. 5.15 of Lesson 10 for  y_f  and Eqs. 5.16 to 18 of Lesson 10 for  C,  T and M_u, respectively. From Eq. 5.15 of Lesson 10, we have:

y_f  =  0.15 x_u +  0.65  D_f  =  0.15 (169.398) + 0.65 (100)  =  90.409  mm

From Eq. 5.18 of Lesson 10, we have

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) yf (d - y_f /2)    

or M_u  =  0.36 (169.398/450) {1 - 0.42 (169.398/450)} (20) (300) (450) (450) + 0.45 (20) (1000 - 300) (90.409) (450 - 90.409/2) 

=  138.62 + 230.56  =  369.18  kNm. 

Ex.4:  Determine the moment of resistance of the flanged beam of Fig. 5.11.5 with A_st = 4,825 mm² (6- 32 T). Other parameters and data are the same as those of Ex.1: b_f  =  1000 mm, D_f  =  100 mm,  b_w  =  300 mm,  cover  =  50 mm and d  =  450 mm. Use M 20  and   Fe 415.  

Step 1:  To determine  xu 

 Assuming x_u in  the flange of under-reinforced rectangular beam we have from Eq. 3.16 of Lesson 5:  

Here, D_f/d  =  100/450  =  0.222  >  0.2. So, we have to determine  y_f  from Eq. 5.15 and equating  C and  T from Eqs. 5.16 and 17 of Lesson 10.

y_f  =  0.15 x_u +  0.65  D_f       (5.15)

0.36 f_ck  b_w  x_u  +  0.45  f_ck (b_f   -  b_w) y_f  =  0.87 f_y A_st            (5.16 and 5.17) 

or 0.36 (20) (300) (x_u) + 0.45 (20) (1000 - 300) {0.15 x_u + 0.65 (100)}   =  0.87 (415) (4825) 

or 2160 x_u  +   945 x_u  =  -  409500  + 1742066

or x_u   =  1332566/3105  =  429.17 mm  

x_u,max  =  0.48 (450)  =  216 mm

Since  x_u   >  x_u,max, the beam is over-reinforced.

Accordingly. x_u   =  x_u, _max   =  216 mm. 

Step 2:  To determine  Mu  

This problem belongs to case (iv b), explained in sec.5.10.4.4 b of Lesson 10.

So, we can determine M_u  from Eq. 5.11 of Lesson 10.

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) y_f (d - y_f /2)             (5.11) 

where y_f  =  0.15 x_{u, max} +  0.65  D_f    =  97.4  mm     (5.8)  

From Eq. 5.11, employing the value of  y_f  = 97.4 mm, we get:  M_u  =  0.36 (0.48) {1 - 0.42 (0.48)} (20) (300) (450) (450) 

 + 0.45 (20) (1000 - 300) (97.4) (450 - 97.4/2) 

 =  167.63 + 246.24  =  413.87  kNm

It is seen that this over-reinforced beam has the same  Mu  as that of the balanced beam of Example 2. 

5.11.4   Summary of Results of Examples 1-4 

The results of four problems (Exs. 1-4) are given in Table 5.1 below. All the examples are having the common data except  A_st.  

Table 5.1  Results of Examples 1-4 (Figs. 5.11.2 – 5.11.5)

It is clear from the above table (Table 5.1), that Ex.4 is an over-reinforced flanged beam. The moment of resistance of this beam is the same as that of balanced beam of Ex.2. Additional reinforcement of 1,759 mm² (= 4,825 mm² – 3,066 mm²) does not improve the M_u of the over-reinforced beam. It rather prevents the beam from tension failure. That is why over-reinforced beams are to be avoided. However, if the M_u has to be increased beyond 413.87 kNm, the flanged beam may be doubly reinforced.

5.11.5  Use of SP-16 for the Analysis Type of Problems  

Using the two governing parameters (b_f /b_w) and (D_f /d), the M_u,lim of balanced flanged beams can be determined from Tables 57-59 of SP-16 for the three grades of steel (250, 415 and 500). The value of the moment coefficient M_u,lim /b_wd²f_ck of Ex.2, as obtained from SP-16, is presented in Table 5.2 making linear interpolation for both the parameters, wherever needed. M_u,lim  is then calculated from the moment coefficient.  

Table 5.2 M_u,lim of Example 2 using Table 58 of SP-16 

Parameters:  (i)  b_f /b_w  =  1000/300  =  3.33         (ii) D_f /d   =  100/450  =  0.222 

  by linear interpolation

So, from Table 5.2,  M_u,lim  =  0.339 b_w d² f_ck  =  0.339 (300) (450) (450) (20) 10^-6  =  411.88 kNm 

M_u,lim as obtained from SP-16 is close to the earlier computed value of M_u,lim = 413.87 kNm (see Table 5.1). 

5.11.6 Practice Questions and Problems with Answers  

Q.1:  Determine the moment of resistance of the simply supported doubly reinforced flanged beam (isolated) of span 9 m as shown in Fig. 5.11.6. Assume M 30 concrete and Fe 500 steel.
 A.1: Solution of Q.1: 
 Effective width  bf 

  = 1200  mm

Step 1:  To determine the depth of the neutral axis  

Assuming neutral axis to be in the flange and writing the equation  C = T, we have:  0.87 f_y A_st   =  0.36 f_ck b_f x_u  +  (f_sc A_sc – f_cc A_sc) Here, d ' / d  =  65/600  =  0.108 = 0.1 (say). We, therefore, have f_sc  =  353 N/mm² .  From the above equation, we have: 

So, the neutral axis is in web. D_f /d  = 120/600  =  0.2 Assuming  D_f /x_u  <  0.43, and  Equating  C = T

= 319.92 >  276 mm (  = 276 mm)

So,  x_u  =  x_u,_max  =  276 mm  (over-reinforced beam). D_f /x_u  =  120/276  =  0.4347  >  0.43

Let us assume  D_f /x_u > 0.43. Now, equating  C = T with  y_f   as the depth of flange having constant stress of 0.446 f_ck. So, we have:

y_f  =  0.15 x_u + 0.65 D_f  =  0.15 x_u + 78  0.36 f_ck b_w  x_u + 0.446 f_ck (b_f – b_w) y_f + A_sc (f_sc – f_cc) = 0.87 f_y A_st

0.36 (30) (300) x_u + 0.446 (30) (900) (0.15 x_u + 78)  =  0.87 (500) (6509) – 1030 {353 – 0.446 (30)} 

or x_u  =  305.63 mm >  x_u,max.   (x_u,max  =  276 mm)

The beam is over-reinforced. Hence,  x_u  =  x_u,max  = 276 mm. This is a problem of case (iv), and we, therefore, consider the case (ii) to find out the moment of resistance in two parts: first for the balanced singly reinforced beam and then for the additional moment due to compression steel. 

Step 2:  Determination of  xu,lim for singly reinforced flanged beam

Here, D_f /d  =  120/600  =  0.2, so  y_f  is not needed. This is a problem of case (ii a) of sec. 5.10.4.2 of Lesson 10.

Employing Eq. 5.7 of Lesson 10, we have:

M_u,lim  =  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)    

=  0.36(0.46) {1 – 0.42(0.46)} (30) (300) (600) (600) 
      +  0.45(30) (900) (120) (540) 
  =  1,220.20  kNm   

Step 3:  Determination of  M_u2  

Total A_st  =    6,509  mm²,  A_st,lim  =  5,794.62  mm²

A_st2  =  714.38  mm²  and  A_sc  =  1,030  mm²

It is important to find out how much of the total Asc  and  A_st2  are required effectively. From the equilibrium of C and T  forces due to additional steel (compressive and tensile), we have:  (A_st2) (0.87) (fy)  =  (A_sc) (f_sc) If we assume  A_sc  =  1,030 mm²   

(714.38 mm² is the total A_st2 provided). So, this is not possible.  Now, using  A_st2 = 714.38 mm² , we get Asc   from the above equation.

(880.326 ) (1,030  mm² sc 353 A == < , (1,030 mm² is the total Asc provided). M_u2  =  A_sc f _sc (d - d ' )  =  (880.326) (353) (600 - 60)  =  167.807  kNm

Total moment of resistance  =  M_u,lim + M_u2  =  1,220.20 + 167.81  =  1,388.01  kNm

Total  A_st  required  =  A_st,lim + A_st2  =  5,794.62 + 714.38  =  6,509.00  mm² , (provided A_st  = 6,509 mm²)

A_sc required  =  880.326  mm²  (provided 1,030  mm²). 
 

5.11.7  Test 11 with Solutions 

Maximum Marks = 50,    
 Maximum Time = 30 minutes
 Answer all questions. 

TQ.1: Determine  Mu,lim  of the flanged beam of Ex. 2 (Fig. 5.11.3) with the help of SP-16 using (a) M 20 and Fe 250, (b) M 20 and Fe 500 and (c) compare the results with the Mu,lim  of Ex. 2 from Table 5.2 when grades of concrete and steel are M 20 and Fe 415, respectively. Other data are: b_f  =  1000 mm, D_f  =  100 mm,  b_w  =  300 mm,  cover  =  50 mm and  d  =  450 mm.                           (10 X 3 = 30 marks)
 A.TQ.1: From the results of Ex. 2 of sec. 5.11.5 (Table 5.2), we have:

Parameters:  (i)  b_f /b_w  =  1000/300  =  3.33    (ii) Df /d   =  100/450  =  0.222

For part (a):  When Fe 250 is used, the corresponding table is Table 57 of SP16. The computations are presented in

Table 5.3 below: Table 5.3  (M_u,lim /b_w d² f_ck)  in  N/mm²   Of TQ.1 (PART a for M 20 and Fe 250) 

 by linear interpolation M_u,lim /b_w d² f_ck  = 0.354174 = 0.354 (say) 

So, M_u,lim =  (0.354) (300) (450) (450) (20) N mm   = 430.11 kNm For part (b):  When Fe 500 is used, the corresponding table is Table 59 of SP16. The computations are presented in

Table 5.4 below: Table 5.4  (M_u,lim /bw d₂ f_ck)  in  N/mm²   Of TQ.1 (PART b for M 20 and Fe 500)

* by linear interpolation

M_u,lim /b_w d² f_ck  = 0.330718 = 0.3307 (say)

So, M_u,lim =  (0.3307) (300) (450) (450) (20) mm   = 401.8  kNm

For part (c):  Comparison of results of this problem with that of Table 5.2 (M 20 and Fe    415) is given below in Table 5.5.

Table 5.5 Comparison of results of M_u,lim    

It is seen that Mu,lim of the beam decreases with higher grade of steel for a particular grade of concrete.

TQ.2: With the aid of SP-16, determine separately the limiting moments of resistance and the limiting areas of steel of the simply supported isolated, singly reinforced and balanced flanged beam of 

Q.1 as shown in Fig. 5.11.6 if the span =  9 m. Use M 30 concrete and three grades of steel, Fe 250, Fe 415 and Fe 500, respectively. Compare the results obtained above with that of

Q.1 of sec. 5.11.6, when balanced. (15 + 5  = 20 marks) 
 A.TQ.2: From the results of 

Q.1 sec. 5.11.6, we have: Parameters:  (i)  b_f /b_w  =  1200/300  =  4.0         (ii) D_f /d   =   120/600  =   0.2 

For Fe 250, Fe 415 and Fe 500, corresponding tables are Table 57, 58 and 59, respectively of SP-16. The computations are done accordingly. After computing the limiting moments of resistance, the limiting areas of steel are determined as explained below. Finally, the results are presented in Table 5.6 below:

Table 5.6 Values of M_u,lim  in  N/m_m2   Of TQ.2 

The maximum area of steel allowed is .04 b D = (.04) (300) (660) = 7,920 mm² . Hence, Fe 250 is not possible in this case.

Summary of this Lesson  

This lesson mentions about the two types of numerical problems (i) analysis and (ii) design types. In addition to explaining the steps involved in solving the analysis type of numerical problems, several examples of analysis type of problems are illustrated explaining all steps of the solutions both by direct computation method and employing SP-16. Solutions of practice and test problems will give readers the confidence in applying the theory explained in Lesson 10  in solving the numerical problems.