Analysis of Singly Reinforced Rectangular Sections

Analysis of a given reinforced concrete section at the ultimate limit state of flexure implies the determination of the ultimate moment of resistance of the section.

This is easily obtained from the couple resulting from the flexural stresses [Fig. 4.17(c)]: where and T are the resultant (ultimate) forces in compression and tension respectively, and z is the lever arm. and the line of action of T corresponds to the level of the centroid of the tension steel.

u

Concrete Stress Block in Compression

In order to determine the magnitude of and its line of action, it is necessary to analyse the concrete stress block in compression. As ultimate failure of a reinforced concrete beam in flexure occurs by the crushing of concrete, for both under- and over-reinforced beams, the shape of the compressive stress distribution (‘stress block’) at failure will be, in both cases, as shown in Fig. 4.18. [also refer assumptions (b) and (c) in Section 4.7.1]. The value of can be computed knowing that the compressive stress in concrete is uniform at 0.447 f

C_u

Fig. 4.18 Concrete stress-block parameters in compression

For a rectangular section of width b,

C f b x x Also, the line of action of is determined by the centroid of the stress block, located at a distance

C_u

x from the concrete fibres subjected to the maximum compressive strain. Accordingly, considering moments of compressive forces Cu, C1

and C2 [Fig. 4.18] about the maximum compressive strain location,

(

^{0 362. f bx}ck u

)

× =^x

(

^{0.447 f bx}ck u

)

satisfying equilibrium of forces. Equating = , with expressions for and

, given by Eq. 4.53 and Eq. 4.52 respectively:

C_{u u} limit state, the steel would not have ‘yielded’ (as per the proof stress definition

x_u>x_u,max ε_st < ε^*_st

† for fy) and the steel stress cannot be taken as f_y γ = 0.87_s fy. Hence Eq. 4.52 and therefore Eq. 4.55 are not applicable. When the steel has not yielded, the true location of the neutral axis is obtained by a trial-and-error method, called strain compatibility method, involving the following steps:

1) Assume a suitable initial (trial) value of x_u

2) Determine

ε

_st by considering strain compatibility [Eq. 4.49]:

ε

st = 0.0035 (

d x

_u

− 1

) (4.56) 3) Determine the design stress fst corresponding to

ε

_st using the design

stress-strain curve [Fig. 3.7, Table 3.2].

4) Derive the value of x_u corresponding to fst by considering Tu = fst Ast

and applying the force equilibrium condition Cu = Tu, whereby

† In the case of low grade mild steel (Fe 250), which has a sharply defined yield point, the steel would have yielded and reached fy even at a strain slightly lower than . In such cases, one may find that f

ε^*_st

st = 0.87fy even for values of xu slightly in excess of xu,max.

x f A

difference between the two values is acceptably small, accept the value given by step (4); otherwise, repeat steps (2) to (5) with an improved (say, average) value of , until convergence.

x_u

x_u

Ultimate Moment of Resistance

The ultimate moment of resistance MuR of a given beam section is obtainable from Eq. 4.51. The lever arm z, for the case of the singly reinforced rectangular section [Fig. 4.17(d), Fig. 4.18] is given by

z = d− 0.416x_u (4.58) Accordingly, in terms of the concrete compressive strength,

M_uR =0 362. f bx_{ck u}(d−0 416. x_u) for all x_u (4.59) Alternatively, in terms of the steel tensile stress,

( )

M_uR = f A_{st st} d− 0 416. x_u for all xu (4.60) with fst = 0.87 fy for x_u ≤x_u,max Limiting Moment of Resistance

The limiting moment of resistance of a given (singly reinforced, rectangular) section, according to the Code (Cl. G−1.1), corresponds to the condition , defined by Eq. 4.50. From Eq. 4.59, it follows that:

M_u,lim

The values of the non-dimensional parameter K for different grades of steel [refer Table 4.3] are obtained as 0.1498, 0.1389 and 0.1338 for Fe 250, Fe 415 and Fe 500 respectively.

Limiting Percentage Tensile Steel

Corresponding to the limiting moment of resistance , there is a limiting percentage tensile steel = 100×

M_u,lim

p_t,lim A_st,lim bd. An expression for is obtainable from Eq. 4.55 with :

p_t,lim x_u=x_u,max

x d

f f

u y p

ck t

,max . ,lim

= 0 87. ×

0 362 100

⇒

p f

f x

t d

ck y

u ,lim

. ,max

= ⎛

⎝⎜⎜ ⎞

⎠⎟⎟⎛

⎝⎜ ⎞

⎠⎟

41 61 (4.62) The values of p_t,lim and M_{u ,lim} bd² (in MPa units) for, different combinations of steel and concrete grades are listed in Table 4.4. These values correspond to the so-called ‘balanced’ section [refer Section 4.5.3] for a singly reinforced rectangular section.

Table 4.4 Limiting values of p_t,lim and _{Mu,lim bd}² for singly reinforced rectangular beam sections for various grades of steel and concrete.

(a) pt,lim values

M 20 M 25 M 30 M 35 M 40 M 45 M 50

Fe 250 1.769 2.211 2.653 3.095 3.537 3.979 4.421

Fe 415 0.961 1.201 1.441 1.681 1.921 2.162 2.402

Fe 500 0.759 0.949 1.138 1.328 1.518 1.708 1.897

(b) Mu,lim/bd² values (MPa)

M 20 M 25 M 30 M 35 M 40 M 45 M 50

Fe 250 2.996 3.746 4.495 5.244 5.993 6.742 7.491

Fe 415 2.777 3.472 4.166 4.860 5.555 6.249 6.943

Fe 500 2.675 3.444 4.013 4.682 5.350 6.019 6.688

Safety at Ultimate Limit State in Flexure

The bending moment expected at a beam section at the ultimate limit state due to the factored loads is called the factored moment Mu. For the consideration of various combinations of loads (dead loads, live loads, wind loads, etc.), appropriate load factors should be applied to the specified ‘characteristic’ loads (as explained in Chapter 3), and the factored moment Mu is determined by structural analysis.

The beam section will be considered to be ‘safe’, according to the Code, if its ultimate moment of resistance MuR is greater than or equal to the factored moment Mu. In other words, for such a design, the probability of failure is acceptably low.

It is also the intention of the Code to ensure that at ultimate failure in flexure, the

type of failure should be a tension (ductile) failure ⎯ as explained earlier. For this reason, the Code requires the designer to ensure that x_u≤x_u,max[Table 4.3], whereby it follows that, for a singly reinforced rectangular section, the tensile reinforcement percentage p_t should not exceed p_{t ,lim} and the ultimate moment of resistance M_uR should not exceed M_u,lim [Table 4.4]. The Code (Cl. G−1.1d) clearly states:

“If xu /d is greater than the limiting value, the section shall be redesigned.”

The topic of design is covered in detail in Chapter 5. The present chapter deals with analysis ⎯ and, in analysis, it is not unlikely to encounter beam sections

(already constructed) in which , whereby and .

Evidently, in such ‘over-reinforced’ sections, the strength requirement may be satisfied, but not the ductility

p_t>p_t,lim x_u>x_u,max M_uR>M_u,lim

† requirement. The question arises: are such sections acceptable ? The answer, in general, would be in the negative, except in certain special situations where the section itself is not ‘critical’ in terms of ductility, and will not lead to a brittle failure of the structure under the given factored loads. In such exceptional cases, where MuR > Mu, and inelastic flexural response^‡ is never expected to occur under the given factored loads, over-reinforced sections cannot be strictly objected to.

It may be noted that the exact determination of MuR of an over-reinforced section generally involves considerable computational effort, as explained in the next section. An approximate (but conservative) estimate of the ultimate moment capacity of such a section is given by the limiting moment of resistance, Mu,lim, which can be easily computed.

Variation of M uR with p_t (for singly reinforced rectangular sections) pt ≤ pt,lim

For . it is possible to arrive at a simple closed−form expression for the ultimate moment of resistance of a given section with a specified

x_u≤x_u,max and then substituting in Eq. 4.55,

x f Further substituting Eq. 4.63 and Eq. 4.64 in Eq. 4.60,

† The ductility requirement may be partly satisfied in the case of mild steel (Fe 250), even if xu slightly exceeds xu,max; this is explained later with reference to Fig. 4.19. [See also footnote on p 136.]

‡ for details on ‘plastic hinge’ formation at the ultimate limit state, refer Chapter 9.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 percentage tension reinforcement (pt )

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

M uR/ bd2 (MPa)

Fe 250 Fe 415

0.961

pt= p t,lim 1.201

2.211

1.769

M 20 M 25

(a )

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 100

150 200 250 300 350 400

0.961 1.201

1.769 2.211

pt= pt,lim

Fe250 Fe415

M 20 (0.87fy) M 25

(0.87fy)

fst (MPa)

p e rc e n ta g e te n s io n re in fo rc e m e n t (pt)

(b )

Fig. 4.19 Variation of (a) MuR /bd² and (b) fst with pt

M f p value f which in general is not a constant, and depends on the value of [Eq. 4.56].

x_u>x_u,max

st x_u

To determine fst, the (trial-and-error) strain compatibility method (described earlier) has to be employed. The final expression, comparable to Eq. 4.65, takes the following form: strain

ε

_st corresponding to [Fig. 3.6, 3.7] must satisfy the strain compatibility condition [Eq. 4.56]. For convenience, Eq. 4.57 is re-arranged as follows:

f_st

The steps involved in the ‘strain-compatibility method for determining M_uR bd² for a given p_t, are as follows: within acceptable tolerance, proceed to step (6); otherwise repeat steps (2) to (5) until convergence is attained.

6) Apply Eq. 4.66 and determine M_uR bd².

A quick solution can be obtained by means of a computer program. The relationship between M_uR bd² (expressed in MPa units) and pt is plotted in Fig. 4.19(a) for two typical grades of steel (Fe 250 and Fe 415) combined with the commonly used grades of concrete (M 20 and M 25). The corresponding relationship between stress f_st (at the ultimate limit state) and pt is depicted in

† This condition p_t >p_t,lim is not permitted in design. Its only relevance is in analysis.

Fig. 4.19(b). The relatively thick lines represent the under-reinforced condition p_t ≤ p_t,lim (i.e., M_uR ≤ M_uR,lim), whereas the thin lines denote the over-reinforced condition p_t > p_t,lim (i.e., M_uR > M_uR,lim), and the transition points are marked by thin vertical lines.

It can be seen that these curves in Fig. 4.19(a) (for the ultimate limit state) bear resemblance with the corresponding curves in Fig. 4.13 (for the service load state).

The gain in MuR with increase in ptfollows a nearly linear relationship almost up to the

‘balanced’ point. Also the gain in MuR with higher grades of concrete is marginal for low values of pt and becomes pronounced only when pt exceeds pt,lim.

It may be noted from Fig. 4.19(b) that with the steel percentage limited to pt,lim, as ultimate moment of resistance MuR is reached, the steel would have already ‘yielded’

(fst = 0.87fy) and gone into the domain of large inelastic strains, thus ensuring a ductile response. For pt > pt,lim, the tension steel would not have ‘yielded’ at the ultimate limit state, with the definition of steel strain at balanced condition as

ε_st^* =0 002. +( .0 87f_y) E_s [refer Section 4.7.1]. However, in the case of Fe 250 steel [Fig. 3.6], it can be seen that ‘yielding’ will actually take place even with a steel strain less than ε^*_st.

Hence, in the case of Fe 250 steel, the actual ‘under-reinforced’ (ductile failure) behaviour extends to some pt values greater than pt,lim [Fig. 4.19]. However, as explained earlier, the use of pt values in this range is not permitted by the Code for design purpose.

It can also be seen that the gain in M_uR with p_t falls off significantly, and somewhat exponentially, beyond the point where f_st drops below 0.87 f_y. Beyond the

‘balanced’ point, there is a stage when the ultimate moment capacity is dictated entirely by the compressive strength of concrete, and hence does not depend on the grade of steel; in this range, ε_st << ε_y, whereby the steel stress is given by f_st = Esε_st, regardless of the grade of steel, and the same Tu = Ast fst is obtained whether Fe 250 or Fe 415 steel is used. This is indicated by the merging together of the thin lines (for a given concrete grade, and for Fe 250 and Fe 415 steel grades) in Fig. 4.19(a).

It is thus evident that over-reinforced sections are undesirable not only from the Code perspective of the loss in ductility, but also from the practical viewpoint of economy.

Analysis Aids

The variation of M_u/bd² with p_t for different grades of concrete and steel (depicted in Fig. 4.19) is expressed in tabular form and presented in Tables A.2(a), (b) in Appendix A of this book. As with analysis by WSM [Tables A.1(a), (b)], these Tables serve as useful analysis aids. They enable rapid determination of the ultimate

moment capacity of any given singly reinforced rectangular beam section. The use of these Tables is demonstrated in Example 4.11.

EXAMPLE 4.9

Determine the neutral axis depth (at the ultimate limit state) for the beam section in Example 4.2.

• As , steel would not have ‘yielded’; accordingly, the ‘strain compatibility method’ is adopted to obtain the correct value of .

x_u>x_u,max

4) ^xu=349 0. ×

(

0 9038.

)

= 315.4 mm.

The final value of x_u may be taken as: x_u = 315 mm.

EXAMPLE 4.10

Repeat the problem in Example 4.9, considering Fe 250 grade steel in lieu of Fe 415.

SOLUTION

• Given: b = 300 mm, d = 550 mm, Ast = 1963 mm², fy = 250 MPa , fck = 20 MPa

• For Fe 250 steel, x_u,max d = 0.5313 [Table 4.3, Eq. 4.50]

⇒

x_u,max = 0 5313. ×550 =292.2 mm.

• Assuming x_u≤x_u,max, and applying the force equilibrium condition

x_u = × ×

× ×

0 87 250 1963 0 362 20 300

.

. = 196 6. mm< x_u,max = 292.2 mm.

Therefore, x_u=196.6 mm.

EXAMPLE 4.11

Determine the ultimate moments of resistance for the beam sections in (a) Example 4.9 and (b) Example 4.10.

SOLUTION

(a)

• Given: b = 300 mm, d = 550 mm, Ast = 1963 mm², fy = 415 MPa , fck = 20 MPa.

• x_u=315 mm> x_u,max = 263.5 mm (from Example 4.9).

• Taking moments about the tension steel centroid,

M_uR=0 362. f bx_{ck u}(d−0.416x_u)=0.362 × 20 × 300 × 315

× (550 – 0.416 × 315) = 286.6 × 10⁶ Nmm = 287 kNm.

• Note that M_uR can also be calculated in terms of the steel tensile stress fst , which is less than 0.87fy , as xu> xu,max. From the last cycle of iteration in Example 4.9, the value of fst is obtained as 349 MPa.

⇒ M_uR= f A_{st st}(d−0.416x_u) = 287 kNm (as before).

Evidently, it is easier to evaluate MuR in terms of the concrete strength.

Alternative (using analysis aids) p_t = ×

× 100 1963

300 550 = 1.190

Referring to Table A.2(a) — for M 20 concrete and Fe 415 steel, for pt = 1.190,

M bd

uR

2 = ( .3 145+3 170 2. ) = 3.158 MPa

⇒ M_uR= 3.158 × 300× 550² = 286.6 × 10⁶ Nmm

= 287 kNm (exactly as obtained earlier).

(b)

• Given: b = 300 mm, d = 550 mm, Ast = 1963 mm², fy = 250 MPa , fck = 20 MPa

• x_u = 196.6 mm < x_u,max = 292.2 mm (from Example 4.10)

• Taking moments about the tension steel centroid,

M_uR=0 362. ×20 300 196 6× × . ×(550−0.416×196 6. )

x

= 199.9 × 10⁶ Nmm = 200 kNm.

• Alternatively, as x_u < xu,max , it follows that fst = 0.87fy, and M_uR =0 87. f A_{y st}(d−0.416 _u)

= 0.87 × 250 × 1963 × (550 – 0.416 × 196.6) = 199.6 × 10⁶ Nmm = 200 kNm.

Alternative (using analysis aids) pt= 1.190 (as in the previous case).

Referring to Table A.2(a) — for M 20 concrete and Fe 250 steel, M

bd

uR

2 = ( .2 188+2 219 2. ) = 2.204 MPa

⇒ M_uR= 2.204 × 300× 550² = 200.0 × 10⁶ Nmm

= 200 kNm (exactly as obtained earlier).

4.7.4 Analysis of Singly Reinforced Flanged Sections

Flanged beams (T-beams and L−beams) were introduced in Section 4.6.4, where the analysis at service loads was discussed. The present section deals with the analysis of these beam sections at the ultimate limit state.

The procedure for analysing flanged beams at ultimate loads depends on whether the neutral axis is located in the flange region [Fig. 4.20(a)] or in the web region [Fig. 4.20(b)].

If the neutral axis lies within the flange (i.e., xu ≤ D_f ), then ⎯ as in the analysis at service loads [refer Section 4.6.4] ⎯ all the concrete on the tension side of the neutral axis is assumed ineffective, and the T-section may be analysed as a rectangular section of width bf and effective depth d [Fig. 4.20(a)]. Accordingly, Eq. 4.55 and Eq. 4.59 are applicable with b replaced by bf .

If the neutral axis lies in the web region (i.e., xu > Df ), then the compressive stress is carried by the concrete in the flange and a portion of the web, as shown in Fig. 4.20(b). It is convenient to consider the contributions to the resultant compressive force Cu , from the web^† portion (bw × xu) and the flange portion (width

† In the computation of C_uw, the ‘web’ is construed to comprise the portion of the flanged beam (under compression) other than the overhanging parts of the flange.

bf − bw) separately, and to sum up these effects. Estimating the compressive force Cuw in the ‘web’ and its moment contribution Muw is easy, as the full stress block is operative:

C_uw = 0.362f_ckb x_{w u} (4.68a)

( )

M_uw =C_uw d− 0.416 x (4.68b) _u However, estimating the compressive force Cuf in the flange is rendered difficult by the fact that the stress block for the flange portions may comprise a rectangular area plus a truncated parabolic area [Fig. 4.20(b)]. A general expression for the total area of the stress block operative in the flange, as well as an expression for the centroidal location of the stress block, is evidently not convenient to derive for such a case. However, when the stress block over the flange depth contains only a rectangular area (having a uniform stress 0.447 f_ck), which occurs when

3x_u 7≥D_f,

Fig. 4.20 Behaviour of flanged beam section at ultimate limit state

an expression for C_ufand its moment contribution M_uf can easily be formulated. For the case, 1<x_u D_f <7 3, an equivalent rectangular stress block (of area 0.447f_cky_f) can be conceived, for convenience, with an equivalent depth y_f ≤ D_f , as shown in Fig. 4.20(c). The expression for yf given in the Code (Cl. G − 2.2.1) is necessarily an approximation, because it cannot satisfy the two conditions of ‘equivalence’, in terms of area^† of stress block as well as centroidal location. A general expression for yf

may be specified for any xu > Df:

† It may be noted that the equivalence in terms of area is approximately satisfied at the limiting conditions xu / Df = 1, and exactly satisfied at xu / Df = 7/3.

y x D x

D x D

f

u f u

f u f

= + < <

≥

⎧⎨

⎪

⎩⎪

0 15 0 65 1

73

. . for D 7

3

for ^f (4.69) The expressions for Cufand Muf are accordingly obtained as:

( )

C_uf = 0.447f_ck b_f −b_w y_f for x_u >D_f (4.70a)

( )

M_uf = C_uf d−y_f 2 (4.70b) The location of the neutral axis is fixed by the force equilibrium condition (with yf

expressed in terms of xu [Eq. 4.70]).

C_uw +C_uf = f A_{st st} (4.71) where fst = 0.87 fyfor xu ≤ xu,max. Where xu > xu,max, the strain compatibility method has to be employed to determine xu.

Substituting Eq. 4.68a and Eq. 4.70a in Eq. 4.71, and solving for x_u,

( )

x f A f b b y

u f b

st st ck f w f

ck w

= −0.447 −

0.362 for x_u>D_f (4.72) The final expression for the ultimate moment of resistance MuRis obtained as:

M_uR=M_uw +M_uf (4.73)

( ) ( ) ( )

⇒ M_uR = 0 362. f b x_{ck w u} d−0.416x_u +0.447f_ck b_f −b_w y_f d−y_f 2 (4.74)

Limiting Moment of Resistance

The limiting moment of resistance Mu,lim is obtained for the condition xu = xu,max, where xu,max takes the values of 0.531d, 0.479d and 0.456d for Fe 250, Fe 415 and Fe 500 grades of tensile steel reinforcement [refer Table 4.3]. The condition xu /Df≥7/3 in Eq. 4.69, for the typical case of Fe 415, works out, for xu= xu,max, as

0.479d D_f ≥ 37 , i.e., D_f d≤ 0 205. . The Code (Cl. G−2.2) suggests a simplified condition of D_f d≤ 0 2. for all grades of steel — to represent the condition xu /Df≥ 7/3.

Eq. 4.74 and Eq. 4.69 take the following forms:

( )

M_u,lim = 0.362f b x_{ck w u,max} d −0.416x_u,max +

( ) ( )

0.447f_ck b_f −b_w 2y_f d − y_f for x_u,max > D_f (4.75) where

y x D As mentioned earlier, when it is found by analysis of a given T-section that

x

u

> x

u,max , then the strain compatibility method has to be applied. As an approximate and conservative estimate, MuR may be taken as Mu,lim, given by Eq. 4.76 / 4.77. From the point of view of design (to be discussed in Chapter 5), Mu,lim provides a measure of the ultimate moment capacity that can be expected from a T-section of given proportions. If the section has to be designed for a factored moment Mu > Mu,lim , then this calls for the provision of compression reinforcement in addition to extra tension reinforcement.

EXAMPLE 4.12

Determine the ultimate moment of resistance for the T−section in Example 4.4 SOLUTION

= 233.3 mm, the compression in the ‘flange’ is given by

( )

C_uf =0 447. f_ck b_f −b_{w f}

= 0.447 ×20×

(

850 −250

)

×100 = 536400 N.

• Also assuming x_u≤x_u,max = 276.1 mm, T_u=0 87. ×250×3695 = 803662 N.

• Applying the force equilibrium condition

(

^{Cuw +Cuf =Tu}

)

^,

1810x_u+ 536400 = 803662.

⇒

x_u = 147.7 mm< 73D_f = 233.3 mm.

Hence, this calculated value of x_u is also not correct.

• As _{Df <xu< 7 D}

3 ^f )

f =0 15. u+0 65. f

, the depth of the equivalent concrete stress block is obtained as:

Repeat the T−section problem in Example 4.12, considering 8 - 28φ bars instead of

= 850 mm, D = 100 mm, b = 250 mm, d = 520 mm, fy = 250 MPa and

Hence this calculated value of is not correct.

•

( )

C_uf = 0 447. ×20× 850 250− ×100 = 536400 N.

• = 276.1 mm,

= 10714

• Applying the force e tion Further assuming x_u ≤x_u,max

Exact Solution (considering strain compatibility)

(

520295.6 1

)

0035

• Corresponding to x_u = 295.6 mm, ε_st =0. − = 0.00266 ield for Fe 250,

[Eq. 4.56]

which is clearly greater than the strain at y i.e., 0 87. ×250 2 0( . ×10⁵) = 0.00109.

This is the correct estimate of the of the section; as the Nmm

steel strain is beyond the yield strain a limited amount of ductile behaviour can also be expected. However, as per the Code, this will not qualify as an admissible under-reinforced section since xu > xu,max. [Note that if the ε_st computed had turned out to be less than ε_y, fst < 0.87fy and a trial-and-error procedure has to be resorted to.]

Approximate Solution

• An approximate and conservative solution for M_uR can be obtained by limiting x_u to x_u,max = 276.1 mm, and taking moments of _uw and C_uf about the centroid of the tension steel (Note that, following the Code procedure,

C

† This is a case where, being Fe 250 grade steel with a sharp yield point, the strain at first yield, ε_y = f_y E_s, is lower than the strain for the ‘balanced’ condition specified by the Code. Hence, even though x

ε^*_st

u > x_u,max, the steel has yielded. See also footnote on p. 136 and Fig. 4.19(b).

= = 455 kNm.

In document reinforced concrete structures (Page 151-168)