Module
5
Flanged Beams – Theory
and Numerical
Problems
Version 2 CE IIT, Kharagpur

Lesson
12
Flanged Beams –
Numerical Problems
(Continued)
Version 2 CE IIT, Kharagpur

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
flanged beams, spacing of
analysis before the design.
Df, bw, D, effective span,
imposed loads.

some preliminary dimensions of width and depth of
the beams and span for performing the structural
Thus, the assumed data known for the design are:
effective depth, grades of concrete and steel and

There are four equations: (i) expressions of compressive force C, (ii)
expression of the tension force T, (iii) C = T and (iv) expression of Mu in terms
of C or T and the lever arm {M = (C or T ) (lever arm)}. However, the relative
dimensions of Df, D and xu 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 Mu
Step 2: To determine the Mu,lim of the given or the assumed section
The beam shall be designed as under-reinforced, balanced or doubly
reinforced if the value of Mu is less than, equal to or more than Mu,lim. The
design of over-reinforced beam is to be avoided as it does not increase the
bending moment carrying capacity beyond Mu,lim either by increasing the depth
or designing a doubly reinforced beam.
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Step 3: To determine xu, the distance of the neutral axis, from the
expression of Mu
Here, it is necessary to assume first that xu is in the flange. Later on, it
may be necessary to calculate xu if the value is found to be more than Df . This is
to be done assuming first that Df /xu < 0.43 and then Df /xu > 0.43 separately.
Step 4: To determine the area(s) of steel
For doubly reinforced beams Ast = Ast,lim + Ast2 and Asc are to be
obtained, while only Ast is required to be computed for under-reinforced and
balanced beams. These are calculated employing C = T (for Ast and Ast, lim) and
the expression of Mu2 to calculate Ast2 and Asc.
Step 5:
4

It may be necessary to check the xu and Ast once again after Step

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 Mu. For such beams and
for under-reinforced beams, it may be necessary to estimate the Ast
approximately immediately after Step 2. This value of Ast will facilitate to
determine xu.

5.12.3

Numerical Problems

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

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Ex.5: Design the simply supported flanged beam of Fig. 5.12.1, given the
following: Df = 100 mm, D = 750 mm, bw = 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 = (1.5)

(35.6875) (12) (12)
= 963.5625 kNm
8

Step 2: Computation of xu,lim
Effective width of flange =(lo/6) + bw+ 6 Df = (12000/6) + 350 + 600 = 2,950
mm.
xu,max = 0.48 d = 0.48 (660) = 316.80 mm. This shows that the neutral
axis is in the web of this beam.
Df /d = 100/660 = 0.1515 < 0.2, and
Df /xu = 100/316.8 = 0.316 < 0.43
The expression of Mu,lim is obtained from Eq. 5.7 of Lesson 10 (case ii a of sec.
5.10.4.2) and is as follows:
Mu,lim = 0.36(xu,max /d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d Df /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 Mu = 963.5625 kNm is less than Mu,lim. Hence, one underreinforced beam can be designed.

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Step 3: Determination of xu
Since the design moment Mu is almost 50% of Mu,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.

= 0.87 f y

Mu

Ast

⎧
d ⎨1 −
⎩

Ast f y ⎫
⎬
f ck b d ⎭

(3.23)
Here, all but Ast are known. However, this will give a quadratic equation of Ast
and the lower one of the two values will be provided in the beam. The above
equation gives:
Ast2 - 93831.3253 Ast + 379416711.3 = 0

which gives the lower value of Ast as:
Ast = 4,234.722097 mm2. The reason of selecting the lower value of Ast
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
xu

=

0.87 f y Ast
0.36 b f ck

(3.16)
or

xu = 71.98 mm.

Again, employing Eq. 3.24 of Lesson 5, we can determine xu first and then Ast
from Eq. 3.16 or 17 of Lesson 5, as explained in the next step.
Eq. 3.24 of Lesson 5 gives:
Mu = 0.36(xu /d) {1 - 0.42(xu /d)} fck bf d2
= 0.36 (xu) {1 - 0.42 (xu /d)} fck bf d
963.5625 (106) = 0.36 (xu) {1 - 0.42 (xu /660)} (20) (2950) (660)
or

xu = 72.03 mm.

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The two values of xu are the same. It is thus seen that, the value of xu can be
determined either first finding the value of Ast, from Eq. 3.23 of Lesson 5 or
directly from Eq. 3.24 of Lesson 5 first and then the value of Ast can be
determined.
Step 4: Determination of Ast
Equating C = T, we have from Eq. 3.17 of Lesson 5:
0.87 f y Ast
xu
=
d
0.36 f ck b f d
Ast

=

0.36 f ck b f xu
0.87 f y

=

0.36 (20) (2950) (72.03)
0.87 (415)

= 4,237.41 mm 2

Minimum Ast = (0.85/fy) bw d = (0.85/415) (350) (660) = 473.13 mm2
Maximum Ast = 0.04 bw D = (0.04) (350) (660) = 9,240 mm2
Hence,

Ast = 4,237.41 mm2 is o.k.

Provide 6 - 28 T (= 3694 mm2) + 2-20 T (= 628 mm2) to have total Ast = 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: Df = 100 mm, bw = 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, bf = 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

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(Mu)factored =

1.5 (63.6875) (12) (12)
8

= 1,719.5625 kNm

Step 2: Determination of Mu,lim
Mu,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 Mu,lim of the section.
Step 3: Determination of d
Assuming Df /d < 0.2, we have from Eq. 5.7 of Lesson 10,
Mu = 0.36(xu,max /d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d - Df
/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

( M u )factored

=

1.5 (63.38125) (12) (12)
8

= 1,711.29 kNm

Approximate value of Ast:

Ast

=

Mu
0.87 f y (d -

Df
2

=
)

1711.29 (106 )
0.87 (415) (625 - 50)

= 8,243.06 mm 2

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Step 4: Determination of Ast (Fig. 5.12.2)
xu = xu,max = 0.48 (625) = 300 mm
Equating T and C (Eq. 5.5 of Lesson 10), we have:
0.87 fy Ast = 0.36 xu,max bw fck + 0.45 fck (bf - bw) Df
or

Ast

=

0.36 (300) (350) (20) + 0.45 (20) (2600) (100)
0.87 (415)

= 8,574.98 mm 2

Maximum Ast = 0.04 b D = 0.04 (350) (715) = 10,010.00 mm2
Minimum Ast = (0.85/fy) bw d = (0.85/415) (350) (625) = 448.05 mm2
Hence, Ast = 8,574.98 mm2 is o.k.
So, provide 8-36 T + 2-18 T = 8143 + 508 = 8,651 mm2
Step 5: Determination of xu
Using Ast = 8,651 mm2 in the expression of T = C (Eq. 5.5 of Lesson 10),
we have:
0.87 fy Ast = 0.36 xu bw fck + 0.45 fck (bf - bw) Df
or

xu =

0.87 f y Ast - 0.45 f ck (b f - bw )
0.36 bw f ck

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=

0.87 (415) (8651) - 0.45 (20) (2600) (100)
= 310.89 > xu ,max (= 300 mm)
0.36 (350) (20)

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

xu =

0.87 (415) (8545) - 0.45 (20) (2600) (100)
= 295.703 mm < xu ,max (= 300 mm)
0.36 (350) (20)

Step 6: Checking of Mu
Df /d = 100/625 = 0.16 < 0.2
Df /xu = 100/215.7 = 0.33 < 0.43. Hence, it is a problem of case (iii a)
and Mu can be obtained from Eq. 5.14 of Lesson 10.
So,

Mu = 0.36(xu /d) {1 - 0.42(xu /d)} fck bf d2 + 0.45 fck (bf - bw) (Df) (d - Df /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 > (Mu)design (= 1,711.29 kNm)

Hence, the design is o.k.
Ex.7: Determine the tensile reinforcement Ast 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: Df = 100 mm, D = 750 mm, bw = 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 Mu
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

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Total loads = 63.6875 kN/m
Factored Mu = (1.5) (63.6875) (12) (12)/8 = 1,719.5625 kNm.
Step 2: Determination of Mu,lim
From Ex.5, the Mu,lim of this beam = 1,835.43 kNm. Hence, this beam
shall be designed as under-reinforced.
Step 3: Determination of xu
Assuming xu to be in the flange, we have from Eq. 3.24 of Lesson 5 and
considering b = bf,
Mu = 0.36xu {1 - 0.42(xu /d)} fck bf d
1719.5625 (106) = 0.36 xu {1 - 0.42 (xu /660)} (20) (2950) (550)
Solving, we get xu = 134.1 > 100 mm
So, let us assume that the neutral axis is in the web and Df /xu < 0.43, from Eq.
5.14 of Lesson 10 (case iii a of sec. 5.10.4.3), we have:
Mu = 0.36(xu /d) {1 - 0.42(xu /d)} fck bw d2 + 0.45 fck (bf - bw) (Df) (d - Df /2)
= 0.36 xu {1 - 0.42 (xu /660)} (20) (350) (660)
+ 0.45 (20) (2600) (100) (660 - 50)
Substituting the value of Mu = 1,719.5625 kNm in the above equation and
simplifying,
xu2 - 1571.43 xu + 276042 = 0
Solving, we have xu = 201.5 mm
Df /xu = 100/201.5 = 0.496 > 0.43.
So, we have to use Eq. 5.15 and 5.18 of Lesson 10 for yf and Mu (case iii b of
sec. 5.10.4.3). Thus, we have:
Mu = 0.36 xu {1 - 0.42( xu /d)} fck bw d + 0.45 fck (bf - bw) yf (d - yf /2)
where,
So,

yf = (0.15 xu + 0.65 Df)
Mu = 0.36 xu {1 - 0.42 (xu/660)} (20) (350) (660)

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+ 0.45 (20) (2600) (0.15 xu + 65) (660 - 0.075 xu - 32.5)
or

1719.5625 (106) = 3.75165 (106) xu - 795.15 xu2 + 954.4275 (106)

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

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 fy Ast = 0.36 fck bw xu + 0.45 fck (bf – bw) yf
where,

yf = 0.15 xu + 0.65 Df

Here, using xu = 213.63 mm, Df = 100 mm, we get
yf = 0.15 (213.63) + 0.65 (100) = 97.04 mm

So,

Ast

Minimum

=

0.36 (20) (350) (213.63) + 0.45 (20) (2600) (97.04)
= 7,780.32 mm2
0.87 (415)

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.

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Provide 6-36 T + 3-28 T (6107 + 1847 = 7,954 mm2). Please refer to Fig.
5.12.3.
Step 5: Checking of xu and Mu using Ast = 7,954 mm2
From T = C (Eqs. 5.16 and 5.17 of Lesson 10), we have
0.87 fy Ast = 0.36 fck bw xu + 0.45 fck (bf – bw) yf
where,

yf = 0.15 xu + 0.65 Df

or
Df)

0.87 (415) (7954) = 0.36 (20) (350) xu + 0.45 (20) (2600) (0.15 xu + 0.65

or

xu = 224.01 mm

Df /xu = 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,

Mu = 0.36 xu {1 - 0.42( xu /d)} fck bw d + 0.45 fck (bf - bw) yf (d - yf /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.

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Ex.8: Design the flanged beam of Fig. 5.12.4, given in following: Df = 100 mm, D
= 675 mm, bw = 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, Mu
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 Mu = (1.5) (63.032) (12) (12)/8 = 1,701.87 kNm
Step 2: Determination of Mu,lim
Assuming the neutral axis to be in the web, Df /xu < 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:
Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2
+ 0.45 fck (bf – bw) Df (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 Mu > Mu,lim, the beam is designed as doubly reinforced.
Mu2 = Mu - Mu,lim = 1701.87 - 1582.4 = 119.47 kNm

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Step 3: Determination of area of steel

Ast,lim is obtained equating T = C (Eqs. 5.5 and 6 of Lesson 10).
0.87 fy (Ast,lim) = 0.36 bw (xu,max /d) d fck + 0.45 fck (bf - bw) Df
or

Ast ,lim =

0.36 bw ( xu, max / d ) d f ck + 0.45 f ck (b f - bw ) D f

=

0.87 f y

0.36 (350) (0.48) (585) (20) + 0.45 (20) (2600) (100)
0.87 (415)

=

8,440.98

mm2

Asc =
where

M u2
(Eq. 4.4 of Lesson 8).
( f sc - f cc ) (d - d ' )
fsc

fcc

= 353 N/mm2 for d'/d = 0.1

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

Mu2 = 119.47 (106) Nmm
d'

= 58.5 mm

d

= 585 mm

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

Ast 2 =

Asc ( f sc - f cc )
(Eqs. 4.4 and 4.5 of Lesson 8).
0.87 f y

Substituting the values of Asc, fsc, fcc and fy we get
Ast2 = 628.48 mm2
Total Ast = Ast,lim + Ast2 = 8,440.98 + 628.48 = 9,069.46 mm2
Maximum Ast = 0.04 bw D = 0.04 (350) (675) = 9,450 mm2
and minimum Ast = (0.85/fy) bw d = (0.85/415) (350) (585) = 419.37 mm2
Hence, Ast = 9, 069.46 mm2 is o.k.
Provide 8-36 T + 3-20 T = 8143 + 942 = 9,085 mm2 for Ast and 1-20 + 2-16
= 314 + 402 = 716 mm2 for Asc (Fig. 5.12.5).
Step 4: To check for xu and Mu (Fig. 5.12.5)
Assuming xu in the web and Df /xu < 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 fy Ast = 0.36 bw xu fck + 0.45 (bf - bw) fck Df + Asc (fsc - fcc)
or

0.87 (415) (9085) = 0.36 (350) xu (20) + 0.45 (2600) (20) (100)
+ 716 {353 - 0.45 (20)}

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

xu < xu,max, Df /xu = 100/275.33 = 0.363 < 0.43

and

Df /d = 100/585 = 0.17 < 0.2.

The assumptions, therefore, are correct. So, Mu 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:

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So,

Mu = 0.36 bw xu fck (d - 0.42 xu) + 0.45 (bf - bw) fck Df (d - Df /2)
+ Asc (fsc - fcc) (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 Df = 100 mm, D = 700 mm, cover = 50 mm, d = 650
mm, bw = 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 (Mu)factored
Weight of slab = (0.1) (25) = 2.5 kN/m2
Imposed loads =
10.0 kN/m2
_________
12.5 kN/m2

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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 Mu = (1.5) (54.50) (10) (10)/8 = 1,021.87 kNm
Step 2: Determination of Mu,lim
Effective width of the flange bf = lo/6 + bw + 6 Df = (10,000/6) + 300 + 600 =
2,567 mm.
xu,max = 0.48 d = 0.48 (650) = 312 mm
Hence, the balanced neutral axis is in the web of the beam.
Df /d = 100/650 = 0.154 < 0.2
Df /xu = 100/312 = 0.32 < 0.43
So, the full depth of flange is having a stress of 0.446 fck. From Eq. 5.7 of Lesson
10 (case ii a of sec. 5.10.4.2), we have,
Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2
+ 0.45 fck (bf – bw) Df (d – Df /2)
= 0.36(0.48) {1 – 0.42(0.48)} (20) (300) (650) (650)
+ 0.45(20) (2267) (100) (650 – 50)
= 1573.92 kNm > Mu (= 1021.87 kNm)
So, the beam will be under-reinforced one.
Step 3: Determination of xu
Assuming xu is in the flange, we have from Eq. 3.24 of Lesson 5
(rectangular beam when b = bf ).
Mu = 0.36 (xu /d) {1 – 0.42 (xu /d)} fck bf d2
= 0.36 xu {1 – 0.42 (xu /d)} fck bf d
1021.87 (106) = 0.36 xu {1 – 0.42(xu/650)} (20) (2567) (650)

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xu = 89.55 mm2 < 100 mm (Hence, the neutral axis is in the flange.)
Step 4: Determination of Ast
Equating C = T, we have from Eq. 3.17 of Lesson 5:

or

xu
d

=

Ast

=

0.87 f y Ast
0.36 f ck b f d
0.36 f ck b f xu
0.87 f y

=

0.36 (20) (2567) (89.55)
0.87 (415)

Minimum Ast = (0.85/f y ) (bw ) d =

= 4,584.12 mm 2

0.85 (300) (650)
= 399.39 mm 2
415

Maximum Ast = 0.04 bw D = (0.04) (300) (700) = 8,400 mm2
So,
Provide

5.12.5

Ast = 4,584.12 mm2 is o.k.
6-28 T + 2-25 T = 3694 + 981 = 4,675 mm2 (Fig. 5.12.6).

References

1. Reinforced Concrete Limit State Design, 6th Edition, by Ashok K. Jain,
Nem Chand & Bros, Roorkee, 2002.
2. Limit State Design of Reinforced Concrete, 2nd Edition, by P.C.Varghese,
Prentice-Hall of India Pvt. Ltd., New Delhi, 2002.
3. Advanced Reinforced Concrete Design, by P.C.Varghese, Prentice-Hall of
India Pvt. Ltd., New Delhi, 2001.
4. Reinforced Concrete Design, 2nd Edition, by S.Unnikrishna Pillai and
Devdas Menon, Tata McGraw-Hill Publishing Company Limited, New
Delhi, 2003.
5. Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam,
Oxford & I.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004.
6. Reinforced Concrete Design, 1st Revised Edition, by S.N.Sinha, Tata
McGraw-Hill Publishing Company. New Delhi, 1990.
7. Reinforced Concrete, 6th Edition, by S.K.Mallick and A.P.Gupta, Oxford &
IBH Publishing Co. Pvt. Ltd. New Delhi, 1996.
8. Behaviour, Analysis & Design of Reinforced Concrete Structural Elements,
by I.C.Syal and R.K.Ummat, A.H.Wheeler & Co. Ltd., Allahabad, 1989.
9. Reinforced Concrete Structures, 3rd Edition, by I.C.Syal and A.K.Goel,
A.H.Wheeler & Co. Ltd., Allahabad, 1992.
10. Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited,
New Delhi, 1993.

Version 2 CE IIT, Kharagpur

11. Design of Concrete Structures, 13th Edition, by Arthur H. Nilson, David
Darwin and Charles W. Dolan, Tata McGraw-Hill Publishing Company
Limited, New Delhi, 2004.
12. Concrete Technology, by A.M.Neville and J.J.Brooks, ELBS with
Longman, 1994.
13. Properties of Concrete, 4th Edition, 1st Indian reprint, by A.M.Neville,
Longman, 2000.
14. Reinforced Concrete Designer’s Handbook, 10th Edition, by C.E.Reynolds
and J.C.Steedman, E & FN SPON, London, 1997.
15. Indian Standard Plain and Reinforced Concrete – Code of Practice (4th
Revision), IS 456: 2000, BIS, New Delhi.
16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi.

5.12.6

Test 12 with Solutions

Maximum Marks = 50,

Maximum Time = 30 minutes

Answer all questions.

TQ.1: Determine the steel reinforcement Ast of the simply supported flanged
beam of Q.1 (Fig. 5.12.6) having Df = 100 mm, D = 700 mm, cover = 50
mm, d = 650 mm, bw = 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.

Version 2 CE IIT, Kharagpur

A.TQ.1: Solution:
Step 1: Computation of (Mu)factored
Total loads from Q.1 of sec. 5.12.4 = 54.50 kN/m
Factored Mu = (1.5) (54.50) (12) (12)/8 = 1,471.5 kNm
Step 2: Determination of Mu,lim
Effective width of flange = lo/6 + bw + 6 Df
= (12000/6) + 300 + 600 = 2,900 mm (Fig. 5.12.7)
xu,max = 0.48 d = 0.48 (650) = 312 mm
Hence, the balanced neutral axis is in the web.
Df /d = 100/650 = 0.154 < 0.2
Df /xu = 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
Mu,lim = 0.36 (xu,max /d) {1 – 0.42 (xu,max /d)} fck bw d2
+ 0.45 fck (bf – bw) Df (d – Df /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 xu
Assuming xu to be in the flange, we have from Eq. 3.24 of Lesson 5 (singly
reinforced rectangular beam when b = bf ):
Mu = 0.36 xu {1 – 0.42 (xu /d)} fck bf d
or

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

or

xu2 – 1547.49 xu + 167.81 (103) = 0

Version 2 CE IIT, Kharagpur

Solving, we have xu = 117.34 mm > 100 mm
So, neutral axis is in the web.
Assuming Df /xu < 0.43, we have from Eq. 5.14 of Lesson 10 (case iii a of sec.
5.10.4.3),
Mu = 0.36 xu {1 – 0.42 (xu /d)} fck bw d + 0.45 fck (bf – bw) Df (d – Df /2)
= 0.36 xu {1 – 0.42 (xu/650)} (20) (300) (650)
+ 0.45(20) (2600) (100) (650 – 50)
or

xu2 – 1547.62 xu + 74404.7 = 0

Solving, we have xu = 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 xu 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 yf from Eq. 5.15 of Lesson 10.
Assuming Df /xu > 0.43, from Eqs. 5.15 and 5.18 of Lesson 10 (case iii b
of sec. 5.10.4.3), we have:
Mu = 0.36 xu {1 – 0.42 (xu /d)} fck bw d + 0.45 fck (bf – bw) yf (d – yf /2)
where,
So,

yf = (0.15 xu + 0.65 Df)
Mu = 0.36 xu {1 – 0.42 (xu/650)} (20) (300) (650)
+ 0.45(20) (2600) (0.15 xu + 0.65) (650 – 0.075 xu – 0.325 xu)

or,

1471.5 (106) = - 1170.45 xu2 + 3.45735 xu + 939.2175 (106)

Solving, we get xu = 162.9454 mm. This shows that the assumption of Df /xu >
0.43 is correct as Df /xu = 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

Version 2 CE IIT, Kharagpur

0.87 fy Ast = 0.36 fck bw xu + 0.45 fck (bf – bw) yf
or

Ast =

0.36 (20) (30) (162.9454) + 0.45 (20) (2600) {0.15 (162.9454) + 65}
0.87 (415)

= 974.829 + 5,796.81 = 6,771.639 mm2
Minimum

Ast = (0.85/fy) (bw) (d) = 0.85 (300) (650)/415 = 399.39 mm2

Maximum Ast = 0.04 (bw) (D) = 0.04 (300) (700) = 8,400 mm2
So,

Ast = 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).

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.

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