Limit State of Collapse
Flexure (Theories and
Examples)
Version 2 CE IIT, Kharagpur

Determination

of

Neutral Axis Depth and
Computation of Moment
of Resistance
Version 2 CE IIT, Kharagpur

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

-

determine the depth of the neutral axis for a given cross-section with known
value of As, and grades of steel and concrete,

o write and derive the expressionsof xmmax,
pm,-m,
Mu,Iin)/bd2
and state the
influences of grades of steel and concrete on them separately,
0 derive the corresponding expression of M, when x,, < x,,,,,,_,,,,,
x,, = xmmaxand x,,
> Xu,maxs

o finally take a decision if x., > x.,,,,,_,,,,.

3.5.1

Introduction

After learning the basic assumptions, the three equations of equilibrium
and the computations of the total compressive and tensile forces in Lesson 4, it is
now required to determine the depth of neutral axis (NA) and then to estimate the
moment of resistance of the beams. These two are determined using the two
equations of equilibrium (Eqs. 3.1 and 3.3). It has been explained that the depth
of neutral axis has important role to estimate the moment of resistance.
Accordingly, three different cases are illustrated in this lesson.

3.5.2 Computation of the Depth of Neutral Axis xu
From Eqs. 3.1, 9 and 14, we have
0.87 r, As, = 0.36 b x,, re,
(3.15)

or

f A
x 2 0.87
0.36
byfck

(3.16)
We can also write:

xu _ 0.87 fy A,,
d

0.36 bd fck

(3.17)
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Substituting
theexpression
of :7"fromEq.3.17intoEq.3.13of Lesson
4, we

have
_
_o.42[036fckbdj}
0.87
fy
Ax,
lever
arm
_(1%
= d[11.015
Af]
fckbd

(3.18)
Ignoring multiplying factor 1.015 in Eq. 3.18, we have

lever
arm
= d[1 AMY]
fckbd

(3.19)

3.5.3 Limiting Value of X (= xusmax)
Should there be a limiting or maximum value of x,,?

Equation
3.17reveals
that 7"increaseswiththeincrease
of percentage

ofsteelreinforcement
;_Z
forfixedvalues
of fyandfck.Thus,thedepthofthe
neutral axis x,, will tend to reach the depth of the tensile steel. But, that should
not be allowed. However, let us first find out that value of x., which will satisfy
assumptions (ii) and (vi) of sec. 3.4.2 and designate that by x.,, max for the
present, till we confirm that x., should have a limiting value.

Ei§g.3.5.l: Etxrainof

for thsreecases

Figure 3.5.1 presents the strain diagrams for the three cases: (i) when x., = x.,,max
; (ii) when x., is less than xu,max and (iii) when x., is greater than xu,max. The
following discussion for the three cases has the reference to Fig. 3.5.1.
Xu = Xulmax

The compressive strain at the top concrete fibre = 0.0035 and the tensile

strain
atthe
level
ofsteel
= [0-3;?
+0.002].
Thus,
itsatisfies
the
assumptions
(ii)
and (vi) of sec. 3.4.2.
(ii) When x,, is less than xu,max
There are two possibilities here:

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(a) If the compressive strain at the top fibre = 0.0035, the tensile strain is

more
than
[°&#39;87fy
+0002}.
Thus,
this
possibility
satisfies
the
two
assumptions
(ii)
and (vi) of sec. 3.4.2.

(b)When
thesteel
tensile
strain
is[°&#39;8;fy
+0002],
thecompressive
concrete strain is less than 0.0035. Here also, both the assumptions (ii) and (vi)
of sec. 3.4.2

are satisfied.

(iii) When xu is more than x,,,,,,_,,,,
There are two possibilities here:
(a) When the top compressive strain reaches 0.0035, the tensile steel

strain
isless
than[°-:5+0.002].
Itviolets
theassumption
(vi)though
assumption (ii) of sec. 3.4.2 is satisfied.

(b)When
the
steel
tensile
strain
is[032
ft +0002],
the
compressive
strain
of concrete exceeds 0.0035. Thus, this possibility violets assumption (ii) though
assumption (vi) is satisfied.
The above discussion clearly indicates that the depth of x., should not
become more than xu,max.Therefore, the depth of the neutral axis has a limiting
or maximum value = x.,,max.Accordingly, if the Ast Provided yields x., > x.,,max,the
section has to be redesigned.
Since xu depends on the area of steel, we can calculate A5,,,,-m
from Eq.
3.17.

From Eq. 3.17 (using x,,= xmmaxand

As, = As,-,,,),we have

xu,max_ 0&#39;87fy
A.rt,lJ&#39;m
d

0|

0.36 bd fck

Ast,lim
fck xu,max
(400) = ptlim = L
(100)
bd
&#39; 0.87fy d

(3.20)

Intheaboveequationx-3canbeobtainedfromthestraindiagram
of Fig.
3.5.1

as follows:

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xu,max

d

0.0035

0.87fy

+ 0.0055

(3.21)

3.5.4

Values of "-m
and puim
d
Equation 3.20 shows that the values of pi, Iimdepend on both the grades

ofsteelandconcrete,
whileEq.3.21reveals
that"&#39;;"
depends
onthegrade
of
steel alone and not on the grade of concrete at all. The respective values of pi, ;,-m
for the three grades of steel and the three grades of concrete are presented in

Table3.1.Similarly,
therespective
values
of ""-:1for three
grades
ofsteelare
presented in Table 3.2.
Table 3.1 Values of p,, ,,m
fy=250 N/mm

fy=415 N/mm

fy=500 N/mm

0.479 = 0.48 (say)

0.456 = 0.46 (say)

0%:
I0

Table 3.2

Values of ""-W
d

0.531 = 0.53 (say)
d

A careful study of Tables 3.1 and 3.2 reveals the following:
(i) The pt, 1,-mincreases with lowering the grade of steel for a particular
grade of concrete. The pt, 1,-m,
however, increases with increasing the grade of
concrete for a specific grade of steel.
(ii) The maximum depth of the neutral axis xmmaxincreases with lowering
the grade of steel. That is more area of the section will be utilized in taking the
compression with lower grade of steel.

3.5.5 Computation of M,
Equation 3.3 of Lesson 4 explains that M, can be obtained by multiplying
the tensile force T or the compressive force C with the lever arm. The
expressions of C, lever arm and T are given in Eqs. 3.9, 3.13 (also 3.19) and
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3.14 respectively of Lesson 4. Section 3.5.3 discusses that there are three
possible cases depending on the location of x.,. The corresponding expressions
of Mu are given below for the three cases:
(i) When X,_.<xmmax
Figure 3.5.1 shows that when concrete reaches 0.0035, steel has started

flowing
showing
ductility
(Strain
> 0&#39;2
+ 0.002).
So,thecomputation
ofM,isto
be done using the tensile force of steel in this case. Therefore, using Eqs. 3.13
and 3.14 of Lesson 4, we have

Mu = T(|ever arm) = 0.87 fy As;(d- 0.42 xu)
(3.22)
Substituting the expressions of Tand lever arm from Eqs. 3.14 of Lesson
4 and 3.19 respectively we get,

M=0/87fyAs,d[1
2:13]
(3.23)

Xu=

Xu!maX

From
Fig.3.5.1,
it isseen
thatsteel
justreaches
thevalue
of 0&#39;2
+ 0.002
and

concrete

also

reaches

its maximum

value.

The

strain

of steel

can

further

increase but the reaching of limiting strain of concrete should be taken into
consideration to determine the limiting M, as Mu,,,-m
here. So, we have
M,,,,,-m= C(|ever arm)
Substituting the expressions of C and lever arm from Eqs. 3.9 of Lesson 4 and
3.19 respectively, we have

Mu,m
=0.36
x";*"[1_0.42
x;*"jbdz
fck

(3.24)
(iii)
When
x,,>x.,,,,,,,,,
In this case, it is seen from Fig. 3.5.1 that when concrete reaches the

strain
of0.0035,
tensile
strain
ofsteelismuch
lessthan(asgfy
+ 0.002)
andany
further increase of strain of steel will mean failure of concrete, which is to be

avoided. On the other hand,

when steel reaches

0.87fy

+ 0.002,the strain of

concrete far exceeds 0.0035. Hence, it is not possible. Therefore, such design is
avoided and the section should be redesigned.
However, in case of any existing reinforced concrete beam where x., > x,,,
max,the moment of resistance M, for such existing beam is calculated by
restricting x., to x.,,maxonly and the corresponding M, will be as per the case when
Xu = Xu, max-

3.5.6 Computation of Limiting Moment of Resistance Factor
Equation 3.24 shows that a particular rectangular beam of given
dimensions of b and d has a limiting capacity of Mu,,,-mfor a specified grade of

concrete.The limitingmomentof resistancefactor F?,i.-m
(= M.,,i,-m/bdz)
can be
established from Eq. 3.24 as follows:

R_1,m
= Jim=0.36
x";"[1 0.42
x";*"J
fck
M

.

(3.25)
It is seen that the limiting moment of resistance factor F?,i,-m
depends on

x&#39;;
andfck.Since"&#39;;"
depends
onthegradeofsteelfy,wecansaythat
F?,iim
depends on fckand fy. Table 3.3 furnishes the values of Ry,-m
for three grades
of concrete and three grades of steel.

Table3.3 Limitingvaluesof R. =
Jim

fy=250 N/mm

M

""&#39;
factors(in N/mm2)

b d2

fy=415 N/mm

fy=500 N/mm

A study of Table 3.3 reveals that the limiting moment of resistance factor
F?,,,-m
increases with higher grade of concrete for a particular grade of steel. It is
also seen that this factor increases with lowering the grade of steel for a
particular grade of concrete. The increase of this factor due to higher grade of
concrete is understandable. However, such increase of the factor with lowering
the grade of steel is explained below:

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Lowering
thegradeof steelincreases
the "&#39;;
(videTable3.2)andthis
enhanced
x&#39;;
increases
M,asseenfromEq.3.24.However,
onemayargue

that
Eq.
3.24
has
two
terms:
"&#39;;
and[1_0.42
"-3}.

With
the
increase
ofx"-;
, [10.42 isdecreasing.
Then
how
doweconfirm
thattheproduct
isincreasing
withtheincrease
of "&#39;;
? Actual
computation will reveal the fact. Otherwise, it can be further explained from Table
3.1 that as the grade of steel is lowered for a particular grade of concrete, the pi,
1,-mgets increased. Therefore, amount of steel needed to have Mu,,,-mwith lower

grade
ofsteelishigher.
Thus,higher
amount
ofsteelandhigher
values
of "&#39;;

show
higher factor
with
thelowering
ofgrade
ofsteel
foraparticular
grade
of concrete (see Table 3.3).

3.5.7

Practice

Questions

and Problems

with Answers

Q.1:

Which equation is needed to determine the depth of the neutral axis?

A.1:

Eq. 3.16 or 3.17.

Q.2:

How to find the lever arm?

A.2:

Eq. 3.13 of Lesson 4 or Eq. 3.19.

Q.3:

Should there be limiting or maximum value of up? If so, why? What is the

(eggglt-?ion
tofind
themaximum
value
ofx.,?
How
tofind
A5,,
,,-m
forsuch

A.3:

Sec. 3.5.3 is the complete answer.

Q.4:

State the effects of grades of concrete and steel separately on pi, Iimand
d

A.4:

(i) p,, ,,-m
(see Eq. 3.20 and Table 3.1)
(a) pt, ,,-mincreases with lowering the grade of steel for a particular grade of
concrete

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(b) Pt, Iim increases with increasing the grade of concrete for a particular
grade of steel

Q.5:

(ii)

"&#39;;"
(seeEq.3.21andTable
3.2)

(a)

xu,m-ax

increases with lowering the grade of steel

(b)

xu,m-ax

is independent on the grade of concrete.

Write the corresponding expression of M, when (i) x., < x.,,max;and (ii) x., =

Xu, max
A.5:

(i) Eq. 3.23, when x., < x.,,max
Eq. 3.24,

Xu= Xuimax

Q.6:

What is to be done if x., > x.,,maxand why ?

A.6:

When x., > x.,, max,the section has to be redesigned as this does not
ensure

Q.7:

failure

of the beam.

Write
theexpression
oflimiting
moment
ofresistance
factor
F?,,,-m=
A2"!-1;
.
What

A.7:

ductile

is its unit?

Rljm= Mu,m
= 0.36
bdz

xu,max

[1 0.42 Jm.(Eq.3.25).
Itsunitis
xu,max

N/mm2.
Q.8:

State separately the effects of grades of concrete and steel on the limiting

moment
ofresistance
factorF?,,,-m=
.
bd
A.8:

increases with increasing the grade of concrete for a particular
(i) F?,,,-m
grade of steel (see Table 3.3).
(ii) F?,,,-m
increases with lowering the grade of steel for a particular grade of
concrete (see Table 3.3).

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3.5.8

References

1. ReinforcedConcreteLimitState Design,6" Edition,by Ashok K. Jain,
Nem Chand & Bros, Roorkee, 2002.

2. LimitStateDesignof ReinforcedConcrete,2 Edition,by P.C.Varghese,
PrenticeHa|I 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. ReinforcedConcreteDesign,2 Edition,by S.UnnikrishnaPillai 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. ReinforcedConcreteDesign, 13 RevisedEdition,by S.N.Sinha,Tata
McGraw-Hill Publishing Company. New Delhi, 1990.

7. ReinforcedConcrete,6" Edition,by S.K.Ma|Iickand A.P.Gupta,Oxford &
IBH Publishing Co. Pvt. Ltd. New Delhi, 1996.
8. Behaviour, Analysis & Design of Reinforced Concrete Structural Elements,
by |.C.Sya| and R.K.Ummat, A.H.Whee|er & Co. Ltd., Allahabad, 1989.

9. ReinforcedConcreteStructures,3 Edition,by |.C.Sya|and A.K.Goe|,
A.H.Whee|er & Co. Ltd., Allahabad, 1992.

10.Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited,
New Delhi, 1993.

11.Designof ConcreteStructures,13" 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.Propertiesof Concrete,4" Edition, 13 Indian reprint, by A.M.Neville,
Longman, 2000.

14.Reinforced
ConcreteDesignersHandbook,10"Edition,by C.E.Reyno|ds
and J.C.Steedman, E & FN SPON, London, 1997.

15.IndianStandardPlain and ReinforcedConcrete Codeof Practice(4"
Revision), IS 456: 2000, BIS, New Delhi.
16. Design Aids for Reinforced Concrete to IS: 456

3.5.9

Test

5 with

Maximum Marks = 50,

1978, BIS, New Delhi.

Solutions
Maximum Time = 30 minutes

Answer all questions.
TQ.1:

Tickthe correct answer:
20)

(4 x 5 =
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(i) The depth of the neutral axis is calculated from the known area of steel and it
should

be

(a) less than 0.5 times the full depth of the beam
(b) more than 0.5 times the effective depth of the beam
(c) less than or equal to limiting value of the neutral axis depth
(d) less than 0.43 times the effective depth of the beam
A.TQ.1: (i):

(c)

(ii) For a particular grade of concrete and with lowering the grade of steel, the
Pt,Iim

(a) increases
(b) decreases
(c) sometimes increases and sometimes decreases
(d) remains constant
A.TQ.1: (ii): (a)
(iii) For a particular grade of steel and with increasing the grade of concrete, the
Pt,Iim

(a) decreases
(b) increases
(c) remains constant
(d) sometimes increases and sometimes decreases
A.TQ.1: (iii): (b)
(iv)

Which of the statements is correct?

(a) xmmax
/d is independent of grades of concrete and steel
(b) xmax /d is independent of grade of steel but changes with grade of
steel

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(c) xmmax
/d changes with the grade of concrete and steel
(d) xmax /d is independent of the grade of concrete and changes with
the grade of steel
A.TQ.1: (iv): (d)
TQ.2:

Derive the expression of determining the depth of neutral axis and lever
arm of a singly reinforced rectangular beam with known quantity of tension
steel.
(10)

A.TQ.2: Section 3.5.2 is the complete answer.
TQ.3:

Establish the expressions of the moment of resistance of a singly
reinforced rectangular beam when (i) x,, <x,,,,,,_,,,,,
(ii) x,, = x,,,,,,_,,,,
and (iii)
Xu>Xu,maX.

(8+7=15)
A.TQ.3: Section 3.5.5 is the complete answer.
TQ.4:

Derive the expression of limiting moment of resistance factor and explain
how it is influenced by the grades of concrete and steel.
(5)

A.TQ.4:

Section 3.5.6 is the full answer.

3.5.10 Summary of this Lesson
Understanding the various assumptions of the design, stress-strain
diagrams of concrete and steel, computations of the total compressive and
tensile forces and the equations of equilibrium in Lesson 4, this lesson illustrates
the applications of the two equations of equilibrium. Accordingly, the depth of
neutral axis and the moment of resistance of the beam can be computed with the
expressions derived in this lesson. Different cases that arise due to different
values of the depth of neutral axis are discussed to select the particular
expression of the moment of resistance. Further, the expressions of determining
the limiting values of percentage of steel, depth of neutral axis, moment of
resistance

and

moment

of resistance

factor

are

established.

The

influences

of

grades of concrete and steel on them are also illustrated.

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