Beams - 2

(b) Doubly reinforced section

When the available depth or width of a beam section is restricted or when the bending moment to be resisted exceeds the moment of resistance of a balanced singly reinforced section, two choices are possible:

• Provide an over-reinforced section (not recommended because it leads to brittle failure), or
• Provide a doubly reinforced section (preferred), i.e. reinforcement provided in both compression and tension zones.

Main features and practical notes

• Reinforcement is provided on both sides of the neutral axis (NA): tension steel at the soffit and compression steel near the top.
• In modern practice the Limit State Method (LSM) is normally used for design; earlier Working Stress Method (WSM) treatments exist and are useful for understanding but have limitations (see later section).
• For analysis of forces, steel is usually taken with design stress 0.87fy in tension; compressive stress in compression steel is obtained from the steel stress-strain curve corresponding to its strain (denoted fsc).
• Equivalent concrete area occupied by steel is sometimes accounted for; the correction term 0.446 fck Asc arises when the loss of concrete area by compressive bars is included. This term is small and often neglected for simplicity.

Equilibrium and neutral axis

Consider a rectangular section of breadth B and overall effective depth d, with compression reinforcement area Asc at an effective distance dc from the top fibre and tension reinforcement area Ast at depth d from the top fibre.

Compressive forces:

• Compressive force in concrete, C1 = 0.36 fck B Xu
• Compressive force in compression steel, C2 = fsc Asc

Total compressive force, C = C1 + C2 = 0.36 fck B Xu + fsc Asc (neglecting loss of concrete area due to steel).

Tensile force, T = 0.87 fy Ast.

Equilibrium (to locate NA) gives:

0.36 fck B Xu + fsc Asc = 0.87 fy Ast

Here Xu is the depth of neutral axis from the extreme compression fibre. The compressive stress in compression steel fsc must be obtained from the stress-strain relation of steel using the strain at the location of compressive reinforcement:

εsc = 0.0035 × (Xu - dc) / Xu

Because fsc depends on εsc which depends on Xu, the equation for Xu is implicit and an iterative solution is generally required.

Ultimate moment of resistance (Mu) for a doubly reinforced rectangular section

The internal moment of resistance about the neutral axis is the sum of the moment of the concrete compressive resultant and the moment of the compressive steel force (or simpler forms if the small concrete area under compression steel is accounted):

Mu = 0.36 fck B Xu (d - 0.42 Xu) + (fsc - 0.446 fck) Asc (d - dc)

When the 0.446 fck Asc term is neglected (because it is normally small), a useful approximation is

Mu ≈ 0.36 fck B Xu (d - 0.42 Xu) + fsc Asc (d - dc)

Iterative procedure to compute Xu and Mu

1. Compute Xu(lim) (depth of NA for the balanced section) using strain compatibility (see Limit State Method section for formulae) and take an initial guess Xu = Xu(lim).
2. Compute εsc from εsc = 0.0035 × (Xu - dc) / Xu.
3. From the steel stress-strain curve find fsc corresponding to εsc (use 0.87fy only for tension design value; for compression steel use the actual steel curve or design stress corresponding to εsc).
4. Substitute fsc into the equilibrium equation 0.36 fck B Xu + fsc Asc = 0.87 fy Ast and solve for a new Xu.
5. Repeat steps 2-4 until Xu converges to an acceptable tolerance.
6. Calculate Mu using the converged Xu and the Mu expression above.

Design concept - view as two imaginary beams (A and B)

A doubly reinforced beam may be conceptually split into two parts for design:

• Beam A: a singly reinforced beam (concrete compression + tension steel Ast1) that provides the moment equal to the balanced moment Mu(lim). Ast1 is the tension reinforcement required for the balanced portion.
• Beam B: an imaginary portion in which compression steel provides additional compressive force C2 and additional tension steel Ast2 provides corresponding tensile force to resist the excess moment (M - Mu(lim)).

Thus total tension steel Ast = Ast1 + Ast2.

If the factored moment M exceeds Mu(lim), the additional moment ΔM = M - Mu(lim) is resisted by the internal couple formed by compression steel C2 and additional tension steel T2:

ΔM = (fsc - 0.446 fck) Asc (d - dc) ≈ fsc Asc (d - dc)

From equilibrium of forces for the additional part:

fsc Asc - 0.446 fck Asc = 0.87 fy Ast2 (or approximately fsc Asc = 0.87 fy Ast2).

Working Stress Method (brief) - limitations

• The WSM assumes linear elastic stress-strain relations and uses allowable stresses in materials. It does not account well for time-dependent effects.
• Shrinkage and creep of concrete significantly influence stresses under service loads. Typical magnitudes: drying shrinkage strain may be around 0.03% and elastic strain under service loads about 0.035% - these are comparable, so omission of shrinkage/creep causes error.
• Creep depends on sustained stress and age at loading; creep coefficient cc (ratio of creep strain to initial elastic strain) typically ranges 1.1-2.2; a reference value of about 1.6 is often used for approximate estimates.
• Cracks render elastic theories inaccurate for shear design; shear should be based on ultimate behaviour or limit state concepts.
• Because stress-strain behaviour is not linear up to collapse, the safety margin under WSM is not reliably represented by a single factor - this motivates the Limit State Method.

(II) Limit State Method (LSM)

The Limit State Method designs structures so that the probability of reaching an unfavourable limit state during the design life is acceptably low. A limit state is the condition beyond which the structure is unfit for use.

Classification of limit states

• Limit state of serviceability - relates to satisfactory performance under normal service loads (deflection, cracking, vibration, durability, etc.). When a structure reaches a serviceability limit state it generally recovers on removal of loads.
• Limit state of collapse (safety) - relates to failure modes (loss of load Bearing capacity, overturning, sliding, buckling, collapse of sections). When collapse limit state is reached the structure does not recover.

Assumptions commonly used for flexural design in LSM

• Plane sections before bending remain plane after bending (strain varies linearly over the section depth).
• Maximum compressive strain in concrete at the extreme compressive fibre at failure = 0.0035.
• Tensile strength of concrete is ignored (cracked section), so tension is assumed to be carried entirely by reinforcement.
• Stress in steel is obtained from its stress-strain curve and reduced by a partial safety factor; for design the permissible stress in reinforcement is commonly taken as 0.87 fy.
• Partial safety factor for steel γs = 1.15 (used in some contexts as per codal provisions); partial safety factor for concrete γm = 1.5 (used when converting characteristic strength to design strength: fcd = fck/γm).

Recommended stress block for concrete (IS 456 procedure)

The parabolic-rectangular stress block (equivalent) used in LSM is derived from the actual nonlinear stress distribution. Using the simplified block leads to:

• Equivalent resultant compressive force in concrete: C = 0.36 fck B Xu
• Location of resultant from the top fibre: 0.42 Xu

These simplified parameters arise from dividing the actual stress diagram into a rectangular part and a parabolic part and combining their resultant effects.

Limiting depth of neutral axis Xu(lim)

The limiting depth of neutral axis Xu(lim) (expressed as ratio Xu/d) corresponds to the balanced failure condition where concrete crushing and steel yielding occur simultaneously. Typical values based on steel grade are:

• fy = 250 N/mm²: Xu(lim)/d = 0.53
• fy = 415 N/mm²: Xu(lim)/d = 0.48
• fy = 500 N/mm²: Xu(lim)/d = 0.46

Equilibrium for rectangular section (actual Xu)

Balance of forces gives (for singly reinforced section):

0.36 fck B Xu = 0.87 fy Ast

From this Xu can be found directly if Ast is known, or Ast can be found if Xu (from moment equation) is known.

Ultimate moment of resistance (Mu(lim)) for rectangular section

With Xu(lim) known, the limiting moment of resistance with respect to concrete is:

Mu(lim) = 0.36 fck B Xu(lim) × (d - 0.42 Xu(lim))

Equivalently (introducing coefficient Q), for a balanced singly reinforced rectangular section:

Mu(lim) = Q B d²

Approximate values of Q (moment-of-resistance coefficient) commonly used are:

• Q ≈ 0.148 fck for Fe 250
• Q ≈ 0.138 fck for Fe 415
• Q ≈ 0.133 fck for Fe 500

Design of a rectangular beam (LSM)

The general objective of design is to determine cross-sectional dimensions B and d and the required area of reinforcement Ast to develop a given ultimate moment Mu.

Note on economy: a balanced section gives the smallest concrete section but the maximum steel area. Because steel is costlier than concrete, a balanced design may not be economical. Under-reinforced sections are generally preferred in limit state design because they provide ductile failure.

Design procedure - determine B, d and Ast when Mu is given

1. Determine Xu(lim) for the chosen grade of steel.
2. Choose a suitable proportion of d and B; typical d/B range 1.5 to 3.0.
3. From Mu(lim) = 0.36 fck B Xu(lim) (d - 0.42 Xu(lim)) = Q B d² compute d (or check if chosen section can resist Mu).
4. Knowing B and d, compute the required steel area from Mu = 0.87 fy Ast (d - 0.42 Xu) (for under-reinforced Xu found from equilibrium) or use the doubled reinforcement procedure if Mu > Mu(lim).

Design procedure - determine Ast when B and d are known

1. Compute Mu(lim) = 0.36 fck B Xu(lim) (d - 0.42 Xu(lim)).
2. Compare Mu (required) with Mu(lim):
   • If Mu = Mu(lim), section is balanced - find Ast from 0.87 fy Ast = 0.36 fck B Xu(lim).
   • If Mu > Mu(lim), a doubly reinforced section is required - use procedures given in the doubly reinforced section above.
   • If Mu < Mu(lim), an under-reinforced section results - find Xu from Mu = 0.36 fck B Xu (d - 0.42 Xu) (a quadratic in Xu) and then find Ast from 0.87 fy Ast = 0.36 fck B Xu.
3. For under-reinforced sections an approximate relation often used is Mu ≈ 0.87 fy Ast d (when Xu is small), from which Ast can be directly estimated and refined if needed.

Doubly reinforced section (summary of design steps)

1. Find Mu(lim) for the given cross-section assuming a singly reinforced balanced section using Xu(lim).
2. Compute Ast1 required for Mu(lim) from 0.87 fy Ast1 = 0.36 fck B Xu(lim).
3. If factored moment M exceeds Mu(lim), compute the additional moment ΔM = M - Mu(lim).
4. Compute required compression steel Asc from ΔM ≈ fsc Asc (d - dc) (or more precisely ΔM = (fsc - 0.446 fck) Asc (d - dc)).
5. Compute additional tension steel Ast2 from equilibrium fsc Asc - 0.446 fck Asc = 0.87 fy Ast2 (approx. Ast2 = fsc Asc / 0.87 fy).
6. Total tension steel Ast = Ast1 + Ast2.

T-beams - comments and cases

For T-beams the flange contributes to compression. Effective flange width and behaviour depend on whether the neutral axis lies within the flange or in the web. The following cases are relevant:

• Case 1: Neutral axis in flange (Xu < Df)
  • Compressing area is flange only: Mu = 0.36 fck Bf Xu (d - 0.42 Xu) = 0.87 fy Ast (d - 0.42 Xu).
• Case 2: Neutral axis in web (Xu > Df)
  • Concrete compression comprises part of web and full flange. Equilibrium and moment expressions become:
  • 0.36 fck bw Xu + 0.446 fck (Bf - bw) Df = 0.87 fy Ast (force equilibrium)
  • Mu = 0.36 fck bw Xu (d - 0.42 Xu) + 0.446 fck (Bf - bw) Df (d - Df/2) (moment of concrete compressive resultant about tension steel)

Designers commonly convert the flange area outside the web into an equivalent rectangular area (Bf - bw) yf, with an equivalent stress of 0.45 fck for the converted part as per IS 456 provisions. The depth yf is calculated as yf = 0.15 Xu + 0.65 Df but must satisfy yf < Df.

Analysis of stress block parameters (review)

The simplified stress block used in LSM is obtained by approximating the actual nonlinear compressive stress distribution by a rectangular block plus a small parabolic correction. Combining these portions yields the compact expressions used earlier:

• Total compressive resultant C = 0.36 fck B Xu
• Resultant centroid at 0.42 Xu from the top fibre

Limiting neutral axis depth and related values

From strain compatibility and the assumed maximum compressive strain 0.0035, the limiting neutral axis depths Xu(lim) are derived and tabulated for common steel grades. Representative values (Xu(lim)/d): 0.53 for Fe 250, 0.48 for Fe 415, 0.46 for Fe 500.

Limitations of working stress method (reiterated)

• Shrinkage and creep produce strains of the same order as elastic strains under service loads; these must be addressed in long-term behaviour calculations.
• Cracking and nonlinearity make elastic shear design unreliable; design should be based on ultimate behaviour.
• Limit State Method provides a rational framework accounting for material strengths, partial safety factors and ultimate behaviour - it is the preferred approach for modern reinforced concrete design.