LINEAR ELASTIC ANALYSIS OF COMPOSITE SECTIONS

The analysis described is not directly applicable to non-homogeneous materials like reinforced concrete. For
‘composite’ materials, made of two (or more) different (linear elastic) materials [Fig. 4(a)], the theory has to
be suitably modified.

Distribution of Strains and Stresses

The fundamental assumption in flexural theory that initially plane cross-sections remain plane, while subject
to bending, is valid — provided the two materials are bonded together to act as an integral unit, without any
‘slip’ at their interface.

Fig. 4 Concept of ‘transformed section’

Accordingly, the strain variation in the section will be linear [Fig. 4(b)]. When two different (but bonded)
materials are located at the same distance y from the ‘neutral axis’, both materials will have exactly the
same strain ε y. The corresponding stresses will be f1y = E1ε y in the case of ‘material 1’ and f2y = E2ε y in the
case of ‘material 2’, where E1 and E2 represent the elastic moduli of materials 1 and 2 respectively [Fig.
4(c)].
The stress f2y in the ‘material 2’ can be expressed in terms of the corresponding stress f1y in the ‘material 1’
(at points located at the same distance y from the neutral axis) as follows:
𝑓2𝑦 = 𝑚 × 𝑓1𝑦 (5)
𝐸
𝑚 = 𝐸2 (6)
1

The ratio of the two moduli of elasticity, m, is called the modular ratio.

Compiled By: Dr. U. K. Nath Professor, CED, AEC

Reference Book---Reinforced Concrete Design: S Unnikrishna Pillai & Devdas Menon
Page 1 of 6
Concept of ‘Transformed Section’
The concept of ‘modular ratio’ makes it possible, for the purpose of analysis, to transform the composite
section into an equivalent homogeneous section made up entirely of one material (say, ‘material 1’).
Evidently, this transformation must not alter the magnitude, direction and line of action of the resultant
forces in the ‘material 2’ due to the flexural stresses f2y.
Considering the resultant force dF2 in an infinitesimal element of ‘material 2’ having thickness dy (and
corresponding breadth b2), located at a distance y from the neutral axis [Fig. 4 (a),(c)].
𝒅𝑭𝟐 = 𝒇𝟐𝒚 (𝒃𝟐 𝒅𝒚)
Substituting equation 5, dF2 can be expressed in terms of f1y as follows.
𝒅𝑭𝟐 = 𝒎𝒇𝟏𝒚 (𝒃𝟐 𝒅𝒚) = 𝒇𝟏𝒚 (𝒎𝒃𝟐 )𝒅𝒚 (7)
Equation 7 indicates that ‘material 2’ may be transformed into an equivalent ‘material 1’ simply by
multiplying the original breadth b2 (dimension parallel to the neutral axis at the depth y) with the modular
ratio m.
In the transformed section [Fig. 4 (d)], as the material is homogeneous (all of ‘material 1’) and ‘linear elastic’.
The use of the ‘transformed section’ concept may be limited to determining the neutral axis as the
‘centroidal-axis’ of the transformed section. The stresses induced in the two materials due to a given
moment can then be determined by applying the basic equations of static equilibrium [Eq. 1, 2].
Alternatively, the stresses can be computed with the ‘transformed section’ itself, by applying the flexure
formula [Eq. 3]; in this case the second moment of area Ig of the ‘transformed section’ has to be considered.
The stresses thus computed with reference to ‘material 1’ can be converted to the equivalent stresses in
‘material 2’ by involving the ‘modular ratio’ concept [Eq. 5].
MODULAR RATIO AND CRACKING MOMENT
Modular Ratio in Reinforced Concrete
In the case of the working stress analysis of reinforced concrete sections, it is usual to transform the
composite section into an equivalent concrete section. Accordingly, for reinforced concrete, the ‘modular
ratio’ m [Eq. 6] is defined as the ratio of the elastic modulus of steel to that of concrete.
Transformed Area of Reinforcing Steel
Tension Steel
Applying the concept of ‘transformed section’, the area of tension reinforcement steel Ast is transformed
into equivalent concrete area as mAst. This transformation is valid in reinforced concrete not only for flexural
members but also for members subjected to direct tension.
The stress in the tension steel, fst, is obtained from the corresponding stress fcs in the equivalent
‘transformed’ concrete (at the level of the steel) as fst = mfcs.
This concept is used subsequently in some derivations. Hence, for consistency in calculations, the value of
m should not be rounded off to an integer (as done in the traditional WSM).

Compiled By: Dr. U. K. Nath Professor, CED, AEC

Reference Book---Reinforced Concrete Design: S Unnikrishna Pillai & Devdas Menon
Page 2 of 6
Compression Steel
When reinforcing steel is provided in compression in reinforced concrete beams or columns, the modular
ratio to be considered for transformation is generally greater than that used for tension steel. This is because
the long-term effects of creep and shrinkage of concrete, as well as the nonlinearity at higher stresses,
result in much larger compressive strains in the compression steel than those indicated by the linear elastic
theory using the normally specified value of m. Accordingly, the Code recommends that the transformed
area of compression steel Asc be taken as 1.5mAsc, rather than mAsc.
The stress in the compression steel, fsc, is obtained from the corresponding stress fcsc in the equivalent
‘transformed’ concrete (at the level of the compression steel) as fsc = 1.5mfcsc

Cracking Moment
Concrete in the extreme tension fibre of a beam section is expected to crack (for the first time) when the

stress reaches the value of the modulus of rupture fcr (Flexural strength= 0.7√𝑓𝑐𝑘 𝑁/𝑚𝑚2 (Clause 6.22 of
IS 456-2000).
At this stage, the maximum strains in compression and tension are of a low order. Hence, assuming a linear
stress-strain relation for concrete in both tension and compression, with same elastic modulus, the following
𝑓
formula is obtained [applying Eq. .4 i.e. 𝑀 = 𝑦𝑚𝑎𝑥 𝐼] for the ‘moment at first crack’ or cracking moment Mcr.
𝑚𝑎𝑥

𝑓𝑐𝑟
𝑀𝑐𝑟 = 𝐼𝑇 (10)
𝑦𝑡

where yt is the distance between the neutral axis and the extreme tension fibre,
IT is the second moment of area of the transformed reinforced concrete section with reference to the NA.

If the contribution of the transformed area of reinforcing steel is not significant, an approximate value Mcr is
obtainable by considering the ‘gross (concrete) section’, i.e., treating the beam section as a plain concrete
section.
If the beam is very lightly loaded (or designed to be crack-free), the maximum applied bending moment may
be less than Mcr. In such a case of ‘uncracked section’, the concrete and steel both participate in resisting
tension. The computation of stresses for such a situation is described in Example below.

Example 1: A reinforced concrete beam of rectangular section has the cross-sectional dimensions shown in
Fig. 4.5(a). Assuming M 20 grade concrete and Fe 415 grade steel, compute (i) the cracking moment and
(ii) the stresses due to an applied moment of 50 kNm.

Compiled By: Dr. U. K. Nath Professor, CED, AEC

Reference Book---Reinforced Concrete Design: S Unnikrishna Pillai & Devdas Menon
Page 3 of 6
Solution

Fig. 5: Uncracked section

Material Properties: For M 20 concrete
σcbc=7 N/mm2 (Table 21 of IS 456-2000)
280
modular ratio 𝑚 = 3𝜎 = 13.33
𝑐𝑏𝑐

Flexural strength (Modulus of rupture) =0.7√𝑓𝑐𝑘 𝑁/𝑚𝑚2 (Clause 6.22 of IS 456-2000).

= 3.13 N/mm2
Approximate Cracking Moment (assuming gross section):
𝑏𝐷 2 300×6002
Section Modulus 𝑍 = = = 18 × 106 mm3
6 6

Thus, cracking moment,

𝑀𝑐𝑟 ≈ 𝑓𝑐𝑟 𝑍 = 3.13 × 18 × 106 = 56.34 × 106 Nmm = 56.34 kNm
Transformed section properties
252
Area of steel, 𝐴𝑠𝑡 = 4 × 𝜋 × = 1963 mm2
4

The transformed area, 𝐴𝑇 = 𝑏𝐷 + (𝑚 − 1)𝐴𝑠𝑡

Depth of neutral axis (𝑦̅)
𝐷
𝑏𝐷 × 2 + (𝑚 − 1)𝐴𝑠𝑡 𝑑
̅=
𝒚
𝑏𝐷 + (𝑚 − 1)𝐴𝑠𝑡
Thus,
(300 × 600) × 600⁄2 + (13.33 − 1) × 1963 × 550
𝑦̅ = = 329.6 𝑚𝑚
(300 × 600) + (13.33 − 1) × 1963
Thus, the distance from the neutral axis to extreme compression fibre is, yc =329.6 mm
the distance from the neutral axis to extreme tension fibre is, yt =600-329.6 = 270.4 mm
the distance from the neutral axis to reinforcement steel is, ys =550-329.6 = 220.4 mm
Transformed second moment of area:
𝒃𝒚𝟑𝒄 𝒚𝒄 𝟐 𝒃𝒚𝟑𝒕 𝒚𝒕 𝟐
𝑰𝑻 = + 𝒃𝒚𝒄 × ( ) + + 𝒃𝒚𝒕 × ( ) + (𝒎 − 𝟏) × 𝑨𝒔𝒕 × 𝒚𝟐𝒔
𝟏𝟐 𝟐 𝟏𝟐 𝟐

Compiled By: Dr. U. K. Nath Professor, CED, AEC

Reference Book---Reinforced Concrete Design: S Unnikrishna Pillai & Devdas Menon
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 𝒃𝒚𝟑𝒄 𝒃𝒚𝟑𝒕
𝑰𝑻 = + + (𝒎 − 𝟏) × 𝑨𝒔𝒕 × 𝒚𝟐𝒔
𝟑 𝟑
𝟑𝟎𝟎
𝑰𝑻 = (𝟑𝟐𝟗. 𝟔𝟑 + 𝟐𝟕𝟎. 𝟒𝟑 ) + (𝟏𝟑. 𝟑𝟑 − 𝟏) × 𝟏𝟗𝟔𝟑 × 𝟐𝟐𝟎. 𝟒𝟐 =6.733 × 109 mm4
𝟑

Cracking moment
𝒇𝒄𝒓 𝟑.𝟏𝟑×𝟔.𝟕𝟑𝟑×𝟏𝟎𝟗
𝑴𝒄𝒓 = 𝑰𝑻 = =77.9 kNm > Applied Moment => Uncracking (Thus, the whole section is
𝒚𝒕 𝟐𝟕𝟎.𝟒

considered for calculation of the stresses)

Stresses due to applied moment M= 50 kNm:

(As M < Mcr ,the assumption of ‘uncracked section’ is valid.)

(1) Maximum Compressive Stress in Concrete:

𝑴𝒚𝒄 𝟓𝟎×𝟏𝟎𝟔 ×𝟑𝟐𝟗.𝟔
𝒇𝒄 = = = 𝟐. 𝟒𝟓 N/mm
𝑰𝑻 𝟔.𝟕𝟑𝟑×𝟏𝟎𝟗

(2) Maximum Tensile Stress in Concrete:

𝒚 𝟐𝟕𝟎.𝟒
𝒇𝒕 = 𝒚𝒕 𝒇𝒄 = 𝟑𝟐𝟗.𝟔 × 𝟐. 𝟒𝟓 = 𝟐. 𝟎𝟏 N/mm2
𝒄

(3) Maximum Stress in Steel:

From the stress distribution diagram, stresses in concrete at the level of steel is
𝒚 𝟐𝟐𝟎.𝟒
𝒇𝒔 = 𝒚𝒔 𝒇𝒄 = 𝟑𝟐𝟗.𝟔 × 𝟐. 𝟒𝟓 = 𝟏. 𝟔𝟑 N/mm2
𝒄

Thus, stress in steel is = m × fs =13.33 × 1.63 = 21.73 N/mm2

Example 2: A reinforced concrete beam of rectangular section has the cross-sectional dimensions shown in
Fig. 4.5(a). Assuming M 20 grade concrete and Fe 415 grade steel, compute (i) the cracking moment and
(ii) the stresses due to an applied moment of 140 kNm.

Compiled By: Dr. U. K. Nath Professor, CED, AEC

Reference Book---Reinforced Concrete Design: S Unnikrishna Pillai & Devdas Menon
Page 5 of 6
Solution

Fig.6: Cracked section

Material Properties: For M 20 concrete

σcbc=7 N/mm2 (Table 21 of IS 456-2000)
280
modular ratio 𝑚 = 3𝜎 = 13.33
𝑐𝑏𝑐

Flexural strength (Modulus of rupture) =0.7√𝑓𝑐𝑘 𝑁/𝑚𝑚2 (Clause 6.22 of IS 456-2000).

= 3.13 N/mm2
Since Mcr = 77.9 kNm < Applied moment, M =140 kNm => hence the section would have cracked. (Concrete
of tension zone is neglected in calculation of transformed section)
Transformed section properties
252
Area of steel, 𝐴𝑠𝑡 = 4 × 𝜋 × = 1963 mm2
4

Transformed steel area 𝒎𝑨𝒔𝒕 = 𝟏𝟑. 𝟑𝟑 × 𝟏𝟗𝟔𝟑 = 𝟐𝟔𝟏𝟔𝟕 mm2

Equating moments of areas about the neutral axis
𝟏 𝟐 𝟏
𝒃𝒏 = 𝒎𝑨𝒔𝒕 (𝒅 − 𝒏) => 𝟑𝟎𝟎𝒏𝟐 = 𝟐𝟔𝟏𝟔𝟕(𝟓𝟓𝟎 − 𝒏) => 𝒏 = 𝟐𝟑𝟒. 𝟔
𝟐 𝟐
𝟑𝟎𝟎×𝟐𝟑𝟒.𝟔𝟑
𝑰𝒄𝒓 = + 𝟐𝟔𝟏𝟔𝟕 × (𝟓𝟓𝟎 − 𝟐𝟑𝟒. 𝟔)𝟐 = 𝟑. 𝟖𝟗𝟒 × 𝟏𝟎𝟗 mm4
𝟑

𝑴𝒚𝒄 𝟏𝟒𝟎 × 𝟏𝟎𝟔 × 𝟐𝟑𝟒. 𝟔

𝒇𝒄 = = = 𝟖. 𝟒𝟑 𝑴𝑷𝒂
𝑰𝒄𝒓 𝟑. 𝟖𝟗𝟒 × 𝟏𝟎𝟗
𝒇𝒔𝒕⁄
𝒎 𝒅−𝒏 𝒅−𝒏 𝟓𝟓𝟎 − 𝟐𝟑𝟒. 𝟔
= => 𝒇𝒔𝒕 = 𝒎 × 𝒇𝒄 × => 𝒇𝒔𝒕 = 𝟏𝟑. 𝟑𝟑 × 𝟖. 𝟒𝟑 × = 𝟏𝟓𝟏 𝑴𝑷𝒂
𝒇𝒄 𝒏 𝒏 𝟐𝟑𝟒. 𝟔

Compiled By: Dr. U. K. Nath Professor, CED, AEC

Reference Book---Reinforced Concrete Design: S Unnikrishna Pillai & Devdas Menon
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