Flanged Beams: Numerical Problems - 2 | 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 problems – analysis and design types,
• apply the formulations to design the flanged beams. 

5.12.1 Introduction   

Lesson 10 illustrates the governing equations of flanged beams and Lesson 11 explains their applications for the solution of analysis type of numerical problems. It is now necessary to apply them for the solution of design type, the second type of the numerical problems. This lesson mentions the different steps of the solution and solves several numerical examples to explain their step-by-step solutions. 

5.12.2 Design Type of Problems  

We need to assume some preliminary dimensions of width and depth of flanged beams, spacing of the beams and span for performing the structural analysis before the design. Thus, the assumed data known for the design are:  D_f, b_w, D,  effective span, effective depth, grades of concrete and steel and imposed loads.  There are four equations: (i) expressions of compressive force C, (ii) expression of the tension force T,  (iii)  C = T  and (iv) expression of  M_u  in terms of  C or T  and the lever arm {M  =  (C or T ) (lever arm)}. However, the relative dimensions of  D_f, D and x_u and the amount of steel (under-reinforced, balanced or over-reinforced) influence the expressions. Accordingly, the respective equations are to be employed assuming a particular situation and, if necessary, they need to be changed if the assumed parameters are found to be not satisfactory. The steps of the design problems are as given below. 

Step 1: To  determine the factored bending moment  M_u

Step 2: To determine the  M_u,lim  of the given or the assumed section  

The beam shall be designed as under-reinforced, balanced or doubly reinforced if the value of  M_u  is less than, equal to or more than  M_u,lim. The design of over-reinforced beam is to be avoided as it does not increase the bending moment carrying capacity beyond  M_u,lim  either by increasing the depth or designing a doubly reinforced beam. 

Step 3: To determine x_u, the distance of the neutral axis, from the expression of M_u  

Here, it is necessary to assume first that x_u is in the flange. Later on, it may be necessary to calculate x_u  if the value is found to be more than D_f . This is to be done assuming first that D_f /x_u  < 0.43 and then D_f /x_u  > 0.43 separately. 

Step 4:  To determine the area(s) of steel   

For doubly reinforced beams A_st =  A_st,lim + A_st2   and  A_sc   are to be obtained, while only  A_st  is required to be computed for under-reinforced and balanced beams. These are calculated employing  C = T (for A_st  and A_{st, lim}) and the expression of  M_u2  to calculate A_st2 and A_sc. 

Step 5:    It may be necessary to check the xu  and Ast  once again after Step 4 
 It is difficult to prescribe all the relevant steps of design problems. Decisions are to be taken judiciously depending on the type of problem. For the design of a balanced beam, it is necessary to determine the effective depth in Step 3 employing the expression of bending moment M_u.  For such beams and for under-reinforced beams, it may be necessary to estimate the A_st approximately immediately after Step 2. This value of A_st  will facilitate to determine x_u. 

5.12.3 Numerical Problems  

Four numerical examples are solved below explaining the steps involved in the design problems.

Ex.5:  Design the simply supported flanged beam of Fig. 5.12.1, given the following: D_f  =  100 mm,  D  =  750 mm,  b_w  =  350 mm, spacing of beams  =  4000 mm c/c,  effective span  =  12 m,  cover  =  90 mm,  d  =  660 mm and imposed loads  =  5 kN/m².  Fe 415 and M 20 are used.

Solution:

Step 1:  Computation of factored bending moment  

Weight of slab per m2  =  (0.1) (1) (1) (25)  =  2.5  kN/m2            

So, Weight of slab per m    = (4) (2.5) = 10.00 kN/m            

Dead loads of web part of the beam  =  (0.35) (0.65) (1) (25)  =  5.6875  kN/m  

Imposed loads  =  (4) (5) = 20  kN/m  Total loads  =  30 + 5.6875  =  35.6875  kN/m  

Factored Bending moment =         963.5625  kNm

Step 2:  Computation of x_u,lim  

Effective width of flange =(l_o/6) + b_w+ 6 D_f = (12000/6) + 350 + 600 = 2,950 mm. 
x_u,max  =  0.48 d  =  0.48 (660)  =  316.80 mm. This shows that the neutral axis is in the web of this beam.

D_f /d   =  100/660  =  0.1515  <  0.2, and  

D_f /x_u   =  100/316.8  =  0.316 <  0.43

The expression of  M_u,lim  is obtained from Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2) and is as follows:

M_u,lim = 0.36(xu,max /d){1 - 0.42 (x_u,max /d)} f_ck b_w d² + 0.45 fck (b_f - b_w) D_f (d - D_f /2) 
           = 0.36(0.48) {1 - 0.42(0.48)} (20) (350) (650) (650) 
           + 0.45 (20) (2950 - 350) (100) (660 - 50)  =  1,835.43 kNm

The design moment  M_u = 963.5625 kNm is less than  M_u,lim. Hence, one underreinforced beam can be designed. 

Step 3: Determination of  x_u 

Since the design moment M_u is almost 50% of M_u,lim, let us assume the neutral axis to be in the flange. The area of steel is to be calculated from the moment equation (Eq. 3.23 of Lesson 5), when steel is ensured to reach the design stress fd = 0.87 (415) = 361.05 N/mm².  It is worth mentioning that the term b of Eq. 3.23 of Lesson 5 is here bf  as the T- beam is treated as a rectangular beam when the neutral axis is in the flange. 

 (3.23) 
Here, all but Ast are known. However, this will give a quadratic equation of A_st and the lower one of the two values will be provided in the beam. The above equation gives: 

which gives the lower value of A_st as: A_st  =  4,234.722097 mm². The reason of selecting the lower value of A_st is explained in sec 3.6.4.8 of Lesson 6 in the solution of  Design Problem 3.1. Then, employing Eq. 3.16 of Lesson 5, we get  

(3.16) 

or x_u  =  71.98 mm. Again, employing Eq. 3.24 of Lesson 5, we can determine x_u first and then A_st from Eq. 3.16 or 17 of Lesson 5, as explained in the next step. Eq. 3.24 of Lesson 5 gives:

M_u  =  0.36(x_u /d) {1 - 0.42(x_u /d)} f_ck b_f d²        

=  0.36 (x_u) {1 - 0.42 (x_u /d)} f_ck b_f d

963.5625 (10⁶) =  0.36 (x_u) {1 - 0.42 (x_u /660)} (20) (2950) (660)

or x_u  =  72.03 mm. 

The two values of x_u are the same. It is thus seen that, the value of  x_u  can be determined either first finding the value of  A_st, from Eq. 3.23 of Lesson 5 or directly from Eq. 3.24 of Lesson 5 first and then  the value of  A_st can be determined.

Step 4: Determination of  A_st  

Equating C = T  , we have from Eq. 3.17 of Lesson 5: 

Minimum  A_st  =  (0.85/f_y) b_w d  =  (0.85/415) (350) (660)  =  473.13  mm²

Maximum  A_st  =  0.04 b_w D  =  (0.04) (350) (660)  =  9,240  mm²

Hence, A_st  =   4,237.41 mm² is o.k.

Provide  6 - 28 T (= 3694 mm²) + 2-20 T (= 628 mm²)  to have total  A_st = 4,322 mm²
 

Ex.6:  Design a beam in place of the beam of Ex.5 (Fig. 5.12.1) if the imposed loads are increased to 12 kN/m². Other data are: D_f = 100 mm, b_w = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported and cover = 90 mm. Use Fe 415 and M 20.  

Solution:  As in Ex.5,  b_f = 2,950 mm.

Step 1:  Computation of factored bending moment  

Weight of slab/m²  =  2.5 kN/m² (as in Ex.1)  

Imposed loads  =  12.0  kN/m² (given)  

Total loads  =  14.5 kN/m²

Total weight of slab and imposed loads  =  14.5 (4)  =  58.0  kN/m  

Dead loads of the beam  =  0.65 (0.35) (25)  =  5.6875 kN/m  

Total loads  =  63.6875  kN/m 

Step 2: Determination of  M_u,lim

M_u,lim  of the beam of Ex.5 = 1,835.43 kNm. The factored moment of this problem (1,719.5625 kNm) is close to the value of M_u,lim of the section.   

Step 3: Determination of  d  

Assuming D_f /d    <  0.2, we have from Eq. 5.7 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) D_f (d - D_f /2) 
1719.5625 (10⁶) = 0.36(0.48) {1 - 0.42(0.48)} (20) (350) d²   + 0.45 (2600) (20) (100) (d - 50)  

Solving the above equation, we get  d  =  624.09 mm,  giving total depth  =  624.09 + 90 = 715 mm (say).  

Since the dead load of the beam is reduced due to decreasing the depth of the beam, the revised loads are calculated below:  

Loads from the slab  =  58.0  kN/m  

Dead loads (revised)  =  0.615 (0.35) (25)  =  5.38125  kN/m  

Total loads  =  63.38125  kN/m 

Approximate value of  A_st: 

Step 4: Determination of Ast (Fig. 5.12.2)

x_u  =  x_u,max  =  0.48 (625)  =  300 mm

Equating T and C  (Eq. 5.5 of Lesson 10), we have:  0.87 f_y A_st   =  0.36 x_u,max b_w f_ck + 0.45 f_ck (b_f - b_w) D_f

Maximum   A_st  =  0.04 b D  =  0.04 (350) (715)  =  10,010.00 mm²

Minimum  A_st  =  (0.85/fy) b_w d  =  (0.85/415) (350) (625)  =  448.05 mm²

Hence, A_st  =  8,574.98 mm²    is o.k.

So, provide 8-36 T + 2-18 T  =  8143 + 508  =  8,651 mm²

Step 5: Determination of x_u  

Using A_st =  8 ,651 mm2 in the expression of  T = C (Eq. 5.5 of Lesson 10), we have: 

 0.87 f_y A_st   =  0.36 x_u b_w f_ck +  0.45 f_ck (b_f - b_w) D_f

So, A_st  provided is reduced to 8-36 + 2-16  =  8143 + 402  =  8,545 mm².  Accordingly, 

Step 6: Checking of  M_u 

D_f /d  =  100/625  =  0.16  <  0.2 D_f /x_u  =  100/215.7  =  0.33  <  0.43.

Hence, it is a problem of case (iii a) and M_u can be obtained from Eq. 5.14 of Lesson 10.

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

= 0.36 (295.703/625) {1 - 0.42 (295.703/625)} (20) (350) (625) (625) 
         + 0.45 (20) (2600) (100) (625 - 50)   
  =  1,718.68  kNm  >  (M_u)design  (= 1,711.29 kNm)

Hence, the design is o.k. 
 

Ex.7:  Determine the tensile reinforcement A_st of the flanged beam of Ex.5 (Fig. 5.12.1) when the imposed loads = 12 kN/m². All other parameters are the same as those of Ex.5: D_f = 100 mm, D = 750 mm, b_w = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, simply supported, cover = 90 mm and d = 660 mm. Use Fe 415 and M 20. 

Solution: 

Step 1: Computation of factored bending moment  M_u  

Dead loads of the slab (see Ex.5)  =  2.5  kN/m²

Imposed loads  =  12.0  kN/m² Total loads  =  14.5  kN/m²

Loads/m  = 14.5 (4)  =  58.0  kN/m  

Dead loads of beam  =  0.65 (0.35) (25)  =  5.6875  kN/m

Total loads  =  63.6875  kN/m  

Factored M_u  =    (1.5) (63.6875) (12) (12)/8  =  1,719.5625 kNm.
 

Step 2: Determination of  M_u,lim  

From Ex.5, the  M_u,lim  of this beam  =  1,835.43  kNm. Hence, this beam shall be designed as under-reinforced.
 

Step 3: Determination of  x_u  

Assuming x_u  to   be in the flange, we have from Eq. 3.24 of Lesson 5 and considering b = b_f,

M_u  =  0.36x_u  {1 - 0.42(x_u /d)} f_ck b_f d  1719.5625 (10⁶)  

=  0.36 x_u {1 - 0.42 (xu /660)} (20) (2950) (550)

Solving, we get  x_u  =  134.1  >  100 mm

So, let us assume that the neutral axis is in the web and D_f /x_u   <  0.43, from Eq. 5.14 of Lesson 10 (case iii a of sec. 5.10.4.3), 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) (D_f) (d - D_f /2)      

= 0.36 x_u {1 - 0.42 (x_u /660)} (20) (350) (660)          

+ 0.45 (20) (2600) (100) (660 - 50)   
Substituting the value of  

M_u  =  1,719.5625  kNm

 in the above equation and simplifying, x_u²  - 1571.43 x_u + 276042  =  0

Solving, we have

x_u  =  201.5 mm D_f /x_u  =  100/201.5  =  0.496  >  0.43.

So, we have to use Eq. 5.15 and 5.18 of Lesson 10 for y_f  and M_u (case iii b of sec. 5.10.4.3).

Thus, we have: M_u  =  0.36 x_u {1 - 0.42( x_u /d)} f_ck b_w d + 0.45  f_ck (b_f - b_w) y_f (d - y_f /2)    

where, y_f =  (0.15 x_u + 0.65 D_f)

So, M_u  =  0.36 x_u {1 - 0.42 (xu/660)} (20) (350) (660) 

+ 0.45 (20) (2600) (0.15 x_u + 65) (660 - 0.075 x_u - 32.5) or 1719.5625 (10⁶ )  

=  3.75165 (106) x_u - 795.15 x_u²   +  954.4275 (10⁶)

Solving, we get  x_u  =  213.63 mm. D_f /x_u  =  100/213.63  =  0.468  >  0.43. Hence, o.k. 

Step 4: Determination of  A_st  

Equating C = T    from Eqs. 5.16 and 5.17 of Lesson 10 (case iii b of sec. 5.10.4.3), we have:

0.87 f_y A_st   =  0.36 f_ck b_w x_u + 0.45 f_ck (b_f – b_w) y_f where, y_f =  0.15 x_u + 0.65 D_f

Here, using   x_u  =  213.63 mm,  D_f =  100 mm,

we get y_f =  0.15 (213.63)  +  0.65 (100)  =  97.04 mm 
So,

Minimum    Ast  =  (0.85/fy) (bw) (d)  =  0.85 (350) (660)/(415)  =  473.13  mm2 Maximum   Ast  =  0.04 bw D  =  0.04 (350) (750)  =  10,500  mm2 Hence,       Ast  =  7,780.32  mm2  is o.k.

Provide  6-36 T + 3-28 T  (6107 + 1847  =  7,954  mm²). Please refer to Fig. 5.12.3.
 

Step 5: Checking of  x_u  and  M_u  using  A_st  =  7,954  mm² 

From  T = C (Eqs. 5.16 and 5.17 of Lesson 10), we have

0.87 f_y A_st  =  0.36 f_ck b_w x_u + 0.45 f_ck (b_f – b_w) y_f where, y_f =  0.15 x_u + 0.65 D_f

or      0.87 (415) (7954)  =  0.36 (20) (350) x_u + 0.45 (20) (2600) (0.15 x_u + 0.65 D_f)

or x_u  =  224.01 mm D_f /x_u  =  100/224.01  =  0.446  >  0.43.

Accordingly, employing Eq. 5.18 of Lesson 10 (case iii b of sec. 5.10.4.3), we have:

So, M_u  =  0.36 x_u {1 - 0.42( x_u /d)} f_ck b_w d + 0.45  f_ck (b_f - b_w) y_f (d - y_f /2)      

=  0.36 (224.01){1 - 0.42 (224.01/660)} (20) (350) (660) 
            + 0.45 (20) (2600) {(0.15) 224.01 + 65} {(660) - 0.15 (112) - 32.5} 
     =  1,779.439  kNm  >  1,719.5625  kNm

Hence, o.k. 

Ex.8: Design the flanged beam of Fig. 5.12.4, given in following:  D_f = 100 mm, D = 675 mm, b_w = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported, cover = 90 mm, d = 585 mm and imposed loads = 12 kN/m².  Use Fe 415 and M 20. 

Step 1: Computation of factored bending moment, M_u  

Weight of slab/m²  =  (0.1) (25)  =  2.5  kN/m²

Imposed loads  =  12.0  kN/m² Total loads  =  14.5  kN/m²

Total weight of slab + imposed loads/m  =  14.5 (4)  =  58  kN/m  

Dead loads of beam  =  0.575 (0.35) (25)  =  5.032  kN/m  

Total loads  =  63.032  kN/m  

Factored M_u  =    (1.5) (63.032) (12) (12)/8  =  1,701.87  kNm 

Step 2: Determination of  M_u,lim  

Assuming the neutral axis to be in the web, D_f /x_u <  0.43 and Df /d = 100 / 585 = 0.17  < 0.2, we consider the case (ii a) of sec. 5.10.4.2 of Lesson 10 to get the following:

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 – Df /2)    

=  0.36(0.48) {1 – 0.42 (0.48)} (20) (350) (585) (585) 
      +  0.45(20) (2600) (100) (585 – 50) =  1,582.4  kNm  

Since, factored  M_u  >  M_u,lim, the beam is designed as doubly reinforced.

M_u²  =  M_u - M_u,lim  =  1701.87 - 1582.4  =  119.47 kNm   

Step 3: Determination of area of steel 

A_st,lim  is obtained equating  T = C  (Eqs. 5.5 and 6 of Lesson 10).  

0.87 f_y (A_st, l_im)  =  0.36 b_w (x_u,max /d) d f_ck  +  0.45 f_ck (b_f - b_w) D_f 
or 

mm²

where   f_sc     =  353  N/mm²  for  d'/d  =  0.1 f_cc    

=  0.446 f_ck  =  0.446 (20)  =  8.92  N/mm² 

M_u²   =  119.47 (10⁶)  Nmm

d'     =  58.5  mm

d     =  585  mm 

Using the above values in the expression of  A_sc (Eq. 4.4 of Lesson 8), we get A_sc  =  659.63  mm²

Substituting the values of  A_sc,  f_sc,  f_cc  and  f_y  we get A_st²  =  628.48  mm²

Total A_st  = A_st,lim  + A_st²  =  8,440.98 + 628.48  =  9,069.46  mm²  

Maximum A_st  =  0.04 bw D  =  0.04 (350) (675)  =  9,450 mm²and minimum  A_st  =  (0.85/fy) b_w d  =  (0.85/415) (350) (585)  =  419.37 mm²

Hence,   A_st  =  9, 069.46 mm² is o.k.

Provide  8-36 T + 3-20 T  =  8143 + 942  =  9,085  mm²  for  A_st and 1-20 + 2-16 = 314 + 402 = 716 mm²  for  A_sc  (Fig. 5.12.5). 
 

Step 4: To check for  x_u  and  M_u  (Fig. 5.12.5)

Assuming  x_u  in the web and  D_f /x_u  <  0.43  and using  T = C (case ii a of sec. 5.10.4.2 of Lesson 10 with additional compression force due to compression steel), we have:

0.87 f_y A_st  =  0.36 b_w x_u  f_ck  +  0.45 (b_f - b_w) f_ck D_f + A_sc (f_sc - f_cc)

or 0.87 (415 ) (908 5)  =  0.36 (350) x_u (20) + 0.45 (2600) (20) (100)      

+ 716 {353 - 0.45 (20)} 

This gives  x_u  =  275.33 mm. x_u,max  =  0.48 (d)  =  0.48 (585)  =  280.8 mm.

So, x_u  <  x_u,max,  D_f /x_u  =  100/275.33  =  0.363  <  0.43   and   D_f /d =  100/585  =  0.17  <  0.2.

The assumptions, therefore, are correct. So, M_u  can be obtained from Eq. 5.14 of sec. 5.10.4.3 of Lesson 10 with additional moment due to compression steel, as given below:
So, M_u  =  0.36 b_w x_u f_ck (d - 0.42 x_u) + 0.45 (b_f - b_w) f_ck D_f (d - D_f /2)

+  A_sc (f_sc - f_cc) (d - d')      

 =  0.36 (350) (275.33) (20) {585 - 0.42 (275.33)} 
           + 0.45 (2600) (20) (100) (585 - 50) + 716 (344) (585 - 58.5) 
      =  325.66 + 1251.9 + 129.67  =  1,707.23  kNm  

Factored moment  =  1,701.87 kNm < 1,707.23 kNm.  Hence, o.k. 

5.12.4 Practice Questions and Problems with Answers  

Q.1:  Determine the steel reinforcement of a simply supported flanged beam (Fig. 5.12.6) of  D_f = 100 mm, D  =  700 mm, cover = 50 mm,  d  =  650 mm,  b_w  =  300 mm, spacing of the beams = 4,000 mm c/c, effective span = 10 m and imposed loads = 10 kN/m².  Use M 20 and Fe 415. 
 A.1: Solution: 

Step 1:  Computation of (M_u)_factored  

The design moment  M_u = 963.5625 kNm is less than  M_u,lim.

Hence, one underreinforced beam can be designed. 

Total loads per m = (12.5) (4) = 50 kN/m  

Dead loads of beam = (0.3) (0.6) (25)  =   4.50 kN/m  

Total loads  =  54.50 kN/m  

Factored M_u =  ( 1.5) (54.50) (10) (10)/8  =  1,021.87 kNm 

Step 2:  Determination of  M_u,lim  

Effective width of the flange b_f = l_o/6 + b_w + 6 D_f = (10,000/6) + 300 + 600 = 2,567 mm.

x_u,max  =  0.48 d  =  0.48 (650)  =  312 mm

Hence, the balanced neutral axis is in the web of the beam.

D_f /d  =  100/650  =  0.154  <  0.2 D_f /x_u  =  100/312  =  0.32  <  0.43

So, the full depth of flange is having a stress of 0.446 f_ck.

From Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2), 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 fck (b_f – b_w) D_f (d – D_f /2)    

=  0.36(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650) 
      +  0.45(20) (2267) (100) (650 – 50) 
  =  1573.92 kNm  >  M_u (= 1021.87 kNm) So, the beam will be under-reinforced one. 

Step 3:  Determination of  x_u  

Assuming x_u is  in the flange, we have from Eq. 3.24 of Lesson 5 (rectangular beam when b = b_f ).

M_u  =  0.36 (x_u /d) {1 – 0.42 (x_u /d)} f_ck b_f d²        

=  0.36 x_u {1 – 0.42 (x_u /d)} f_ck b_f d 1021.87 (106)  

=  0.36 x_u {1 – 0.42(x_u/650)} (20) (2567) (650)

x_u  =  89.55  mm²  <  100 mm (Hence, the neutral axis is in the flange.) 
 

Step 4:  Determination of  A_st  

Equating C = T  , we have from Eq. 3.17 of Lesson 5: 

Minimum

Maximum  A_st  =  0.04 b_w D  =  (0.04) (300) (700)  =  8,400  mm²

So,     A_st  =  4,584.12  mm²  is o.k.

Provide    6-28 T + 2-25 T  =  3694 + 981  =  4,675  mm²  (Fig. 5.12.6

  
5.12.6 Test 12 with Solutions 

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

TQ.1: Determine the steel reinforcement A_st of the simply supported flanged beam of
 Q.1 (Fig. 5.12.6) having D_f = 100 mm, D = 700 mm, cover = 50 mm,  d  =  650 mm,  b_w  =  300 mm, spacing of the beams = 4,000 mm c/c, effective span = 12 m and imposed loads = 10 kN/m². Use M 20 and Fe 415.
A.TQ.1: 

Solution: Step 1:  Computation of (M_u)_factored 

Total loads from Q.1 of sec. 5.12.4 = 54.50  kN/m  

Factored M_u  =    (1.5) (54.50) (12) (12)/8  =  1,471.5  kNm 

Step 2:  Determination of  M_u,lim 

Effective width of flange  =  l_o/6 + b_w + 6

D_f = (12000/6) + 300 + 600  =  2,900 mm  (Fig. 5.12.7)

x_u,max  =  0.48 d  =  0.48 (650)  =  312 mm

Hence, the balanced neutral axis is in the web.

D_f /d  =  100/650  =  0.154 < 0.2

D_f /x_u  =  100/312  =  0.32 < 0.43

So, the full depth of flange is having constant stress of 0.446 fck. From Eq. 5.7 of Lesson 10 (case ii a of sec. 5.10.4.2), 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(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650) +  0.45(20) (2600) (100) (650 – 50) =  1,753.74  kNm  >  1,471.5  kNm 

So, the beam will be under-reinforced.

Step 3:  Determination of  x_u  

Assuming x_u to b e in the flange, we have from Eq. 3.24 of Lesson 5 (singly reinforced rectangular beam when b =  b_f ):

M_u  =  0.36 x_u {1 – 0.42 (x_u /d)} f_ck b_f d

or 1471.5 (106 )  =  0.36 (x_u) {1 – 0.42 (x_u /650)} (20) (2900) (650)

or x_u² – 1547.49 x_u + 167.81 (103)  =  0 

Solving, we have  x_u  =  117.34 mm > 100 mm

So, neutral axis is in the web. Assuming  D_f /x_u < 0.43, we have from Eq. 5.14 of Lesson 10 (case iii a of sec. 5.10.4.3), 

M_u  =  0.36 x_u {1 – 0.42 (x_u /d)} f_ck b_w d  + 0.45 f_ck (b_f – b_w) D_f (d – D_f /2)  

=  0.36 xu {1 – 0.42 (xu/650)} (20) (300) (650)   +  0.45(20) (2600) (100) (650 – 50) 

or x_u² – 1547.62  x_u + 74404.7  =  0

Solving, we have  x_u  =  49.67  <  100 mm 

However, in the above when it is assumed that the neutral axis is in the flange xu is found to be 117.34 mm and in the second trial when x_u is assumed in the web  xu is seen to be 49.67 mm. This indicates that the full depth of the flange will not have the strain of 0.002, neutral axis is in the web and  Df /xu is more than 0.43. So, we have to use Eq. 5.18 of Lesson 10, with the introduction of  y_f  from Eq. 5.15 of Lesson 10.  

Assuming D_f /x_u  > 0.43, from Eqs. 5.15 and 5.18 of Lesson 10 (case iii b of sec. 5.10.4.3), we have:

M_u  =  0.36 x_u {1 – 0.42 (x_u /d)} f_ck b_w d  + 0.45 f_ck (b_f – b_w) y_f (d – y_f /2)  

where, y_f  = (0.15 x_u + 0.65 D_f) So,

M_u =  0.36 x_u {1 – 0.42 (x_u/650)} (20) (300) (650)   +  0.45(20) (2600) (0.15 x_u + 0.65) (650 – 0.075 x_u – 0.325 x_u)

or, 1471.5 (10 6)  =  - 1170.45 x_u² + 3.45735 xu + 939.2175 (106)

Solving, we get  x_u  =  162.9454  mm.

This shows that the assumption of  D_f /x_u > 0.43 is correct as  D_f /x_u  = 100 / 162.9454 = 0.614.

Step 4:  Determination of  Ast  

Equating C = T    from Eqs. 5.16 and 5.17 of Lesson 10 (case iii b of sec. 5.10.4.3), we have

 0.87 f_y A_st   =  0.36 f_ck b_w x_u + 0.45 f_ck (b_f – b_w) y_f

=  974.829 + 5,796.81  =  6,771.639  mm²Minimum    

A_st  =  (0.85/f_y) (b_w) (d)  =  0.85 (300) (650)/415  =  399.39  mm²

Maximum   A_st  =  0.04 (b_w) (D)  =  0.04 (300) (700)  =  8,400  mm²

So,       A_st  =  6,771.639  is o.k. Provide  2-36 T + 6-32 T  =  2035 + 4825  =  6,860  mm²  >  6,771.639  mm² (Fig. 5.12.7).  

Summary of this Lesson  

This lesson explains the steps involved in solving the design type of numerical problems. Further, several examples of design type of numerical problems are illustrated explaining the steps of their solutions. Solutions of practice problems and test problems will give the readers confidence in applying the theory explained in Lesson 10 in solving the numerical problems.