Instructional Objectives: At the end of this lesson, the student should be able to:
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 SP16 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 stepbystep 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 overreinforced.
Other steps:
After knowing if the section is underreinforced, balanced or overreinforced, the respective parameter D_{f}/d or D_{f}/X_{u }is computed for the underreinforced, balanced or overreinforced 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 Tbeam 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 mm^{2} (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 Tbeam 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 underreinforced 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 428 T (2463 mm^{2}) + 316 T (603 mm^{2}) = 3,066 mm^{2 }
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 mm^{2} (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 underreinforced, 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 underreinforced.
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} d^{2} + 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 mm^{2 }(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 underreinforced 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 overreinforced.
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} d^{2} + 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 overreinforced beam has the same Mu as that of the balanced beam of Example 2.
5.11.4 Summary of Results of Examples 14
The results of four problems (Exs. 14) are given in Table 5.1 below. All the examples are having the common data except A_{st}.
Table 5.1 Results of Examples 14 (Figs. 5.11.2 – 5.11.5)
It is clear from the above table (Table 5.1), that Ex.4 is an overreinforced 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 mm^{2} (= 4,825 mm^{2} – 3,066 mm^{2}) does not improve the M_{u }of the overreinforced beam. It rather prevents the beam from tension failure. That is why overreinforced 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 SP16 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 5759 of SP16 for the three grades of steel (250, 415 and 500). The value of the moment coefficient M_{u,lim} /b_{w}d^{2}f_{ck} of Ex.2, as obtained from SP16, 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 SP16
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 }d^{2} f_{ck} = 0.339 (300) (450) (450) (20) 10^{6} = 411.88 kNm
M_{u,lim} as obtained from SP16 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/mm^{2 }. 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 (overreinforced 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 overreinforced. 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} d^{2} + 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 mm^{2}, A_{st,lim }= 5,794.62 mm^{2}
A_{st2} = 714.38 mm^{2} and A_{sc} = 1,030 mm^{2}
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 mm^{2}
(714.38 mm^{2} is the total A_{st2} provided). So, this is not possible. Now, using A_{st2} = 714.38 mm^{2} , we get Asc from the above equation.
(880.326 ) (1,030 mm^{2} sc 353 A == < , (1,030 mm^{2} 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 mm^{2} , (provided A_{st } = 6,509 mm^{2})
A_{sc} required = 880.326 mm^{2 } (provided 1,030 mm^{2}).
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 SP16 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} d^{2} f_{ck}) in N/mm^{2} Of TQ.1 (PART a for M 20 and Fe 250)
by linear interpolation M_{u,lim} /b_{w} d^{2 }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 d_{2} f_{ck}) in N/mm^{2} Of TQ.1 (PART b for M 20 and Fe 500)
* by linear interpolation
M_{u,lim }/b_{w} d^{2} 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 SP16, 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 SP16. 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 mm^{2} . 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 SP16. Solutions of practice and test problems will give readers the confidence in applying the theory explained in Lesson 10 in solving the numerical problems.
13 videos42 docs34 tests

1. What is a flanged beam and how is it used in civil engineering? 
2. How is the moment of inertia of a flanged beam calculated? 
3. What are the advantages of using flanged beams in construction? 
4. How can flanged beams be reinforced to enhance their strength? 
5. What are the limitations of using flanged beams in construction? 

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