Limit State Method of design

Concrete

Actual curve: stress rises almost linearly for the first part, then curves, reaches a peak at strain about 0.002, then falls (descending branch) until crushing at strain about 0.0035. It has no clear yield point.

Idealised curve (IS 456, Fig. 21): a parabola from zero to strain 0.002, then a horizontal line (constant stress) up to the ultimate strain 0.0035 in bending. The peak stress is $0.67f_{ck}$ (to allow for the difference between cube and in-situ strength). The ultimate strain is 0.0035 in flexure and 0.002 in pure axial compression.

$$\sigma_c=0.446f_{ck}\left[2\frac{\varepsilon}{0.002}-\left(\frac{\varepsilon}{0.002}\right)^2\right]\quad(\varepsilon\le0.002),\qquad \sigma_c=0.446f_{ck}\quad(0.002\le\varepsilon\le0.0035)$$

Steel

Actual curve:

• Mild steel (Fe250): linear to a definite yield point, a yield plateau, then strain hardening.
• HYSD / TMT bars (Fe415, Fe500): no definite yield point; the yield stress is taken as the 0.2% proof stress.

Idealised curve (IS 456, Fig. 23): Hooke's law up to $0.8f_y$ (linear, $E_s=2\times10^5$ N/mm$^2$), then non-linear up to $f_y$ with inelastic strain of 0.002. For design, stresses are divided by $\gamma_{ms}=1.15$, giving $0.87f_y$. Yield strain for design: $\varepsilon_y=\dfrac{0.87f_y}{E_s}+0.002$ (0.00380 for Fe415).

 Concrete (IS 456 Fig. 21)        Steel (Fe415, Fig. 23)
 stress                           stress
 0.67fck |   _______________      fy   |     ______  actual
         |  /               |     0.87fy|  __/ ______ design
 (design |/                 |           |_/ /
 0.446fck)                  |           |/ /
         +------------------>          +----------------> strain
        0.002          0.0035            Es=2x10^5   0.87fy/Es+0.002

Design values of stresses

The 0.446 factor (= 0.67/1.5) is applied to the stress block. The equivalent rectangular block used in design has an average stress $0.36f_{ck}$ acting over depth $x_u$, with its centroid at $0.42x_u$ from the compression edge.

A limit state is a state beyond which the structure no longer satisfies the performance it was designed for. Limit state design makes the probability of reaching any limit state acceptably small. IS 456:2000 cl. 35 lists the following.

Limit state of strength (collapse)

The structure or a part of it must not collapse, buckle, overturn or become unstable under the worst factored loads. Checked for:

• flexure (bending), compression, shear, torsion (member level);
• overturning, sliding, loss of stability, rupture of critical sections (structure level).

It is checked with design loads ($\gamma_f\times$ characteristic load) and design material strengths ($f_{ck}/1.5$, $f_y/1.15$).

Limit state of serviceability

The structure must remain comfortable and usable under service loads. Checked for:

• Deflection: excessive sag spoils appearance, damages partitions and finishes. Limits: span/250 total, span/350 or 20 mm (whichever is less) after finishes and partitions.
• Cracking: crack width should not exceed 0.3 mm (0.2 mm in severe exposure), to protect appearance and steel.
• Durability: adequate cover, cement content and water-cement ratio for the exposure condition.
• Vibration, fire resistance, fatigue and impact where relevant.

Other limit states

Fire resistance, special impact or explosion loads, or other limit states as required for the structure.

The limit state of serviceability is the state beyond which the structure, though safe against collapse, becomes unfit for normal use because of excessive deflection, cracking, vibration or other defects. It is checked at service (unfactored) loads with material partial factor $\gamma_m=1.0$.

Requirements (IS 456:2000, cl. 35.3 and 23.2, 26.3)

1. Deflection
   • Final deflection (including creep, shrinkage and temperature) $\le$ span/250.
   • Deflection after construction of partitions and finishes $\le$ span/350 or 20 mm, whichever is less.
   • Satisfied by span/effective depth ratio: basic 20 (simply supported), 26 (continuous), 7 (cantilever), modified for steel stress and compression steel.
2. Cracking
   • Crack width limited to 0.3 mm in members (0.2 mm in severe exposure).
   • Satisfied by limiting clear spacing of bars (not more than 300 mm for tension steel, 180 mm for slabs in some cases) and by minimum steel.
3. Durability
   • Adequate nominal cover, minimum cement content, maximum water-cement ratio and minimum grade according to the exposure (mild to extreme).
4. Vibration
   • Slender members should not vibrate noticeably; limit span/depth ratios.
5. Fire resistance — cover and member dimensions as per IS 456 Table 16A.

These checks protect appearance, comfort, and the long-term life of the structure.

Idealised stress-strain diagram

The actual stress-strain curves of concrete and steel are irregular. For design they are replaced by simplified (idealised) curves given in IS 456:2000.

Concrete (Fig. 21): parabola from zero to strain 0.002, then a horizontal line up to 0.0035; peak design stress $0.67f_{ck}/1.5=0.446f_{ck}$.

Steel (Fig. 23): linear up to $0.8f_y$ (slope $E_s=2\times10^5$ N/mm$^2$), then a curve up to $f_y$ with a plastic strain of 0.002; the design curve is scaled by $1/1.15$ to peak at $0.87f_y$.

 Concrete (IS 456 Fig. 21)        Steel (Fe415, Fig. 23)
 stress                           stress
 0.67fck |   _______________      fy   |     ______  actual
         |  /               |     0.87fy|  __/ ______ design
 (design |/                 |           |_/ /
 0.446fck)                  |           |/ /
         +------------------>          +----------------> strain
        0.002          0.0035            Es=2x10^5   0.87fy/Es+0.002

Characteristic strength

• Concrete: the compressive strength $f_{ck}$ of 150 mm cubes at 28 days below which not more than 5% of test results are expected to fall. M20 means $f_{ck}=20$ N/mm$^2$.
• Steel: the yield stress (or 0.2% proof stress) $f_y$ below which not more than 5% of results fall. Fe415 means $f_y=415$ N/mm$^2$.

Both are obtained from $f_k=f_m-1.65s$.

Limit state of serviceability

It is the state beyond which the structure, although safe against collapse, is no longer fit for normal use. It is checked at service loads.

Requirements (IS 456:2000):

• Deflection: total deflection $\le$ span/250; after partitions and finishes $\le$ span/350 or 20 mm. Ensured by limiting span/effective depth ratio (20 for simply supported, 26 continuous, 7 cantilever).
• Cracking: crack width not exceeding 0.3 mm in normal exposure (0.2 mm severe); control by bar spacing and minimum steel.
• Durability: proper cover, cement content and water-cement ratio for the exposure.
• Vibration, fire resistance, fatigue where relevant.

Types of splicing of reinforcement

Splicing joins two bars when the available length is not enough. The types are:

1. Lap splice — bars overlap for a length $\ge$ development length ($L_d$, minimum $\max(15\phi, 200$ mm$)$ for flexural tension, staggered, not at points of maximum stress).
2. Welded splice — bars joined by butt welding (flash butt or arc), used for bars of weldable grade.
3. Mechanical splice — threaded couplers, sleeves with swaged joints, or cold-forged couplers; suitable for large dia bars and seismic zones.
4. End bearing splice — bar ends are cut square and held in contact by a sleeve (compression only).

Design strength

Design strength of a material is its characteristic strength divided by the partial safety factor for the material:

$$f_d=\frac{f_k}{\gamma_m}$$

For concrete $\gamma_m=1.5$ (so $f_{cd}=f_{ck}/1.5$, and in the stress block $0.446f_{ck}$), and for steel $\gamma_m=1.15$ (so $f_{yd}=f_y/1.15=0.87f_y$). The factor allows for variation in material strength, workmanship and member behaviour.

Design load

Design load is the characteristic load multiplied by the partial safety factor for load:

$$F_d=\gamma_f\,F_k$$

IS 456:2000, Table 18, gives $\gamma_f$: 1.5 for DL+LL, 1.2 for DL+LL+WL/EL, 1.5 for DL±WL/EL, and 0.9 for DL when it is favourable (0.9DL+1.5WL). The factor covers possible overloading and inaccuracies in analysis.

The member is safe when the design resistance (from design strengths) is at least the effect of the design loads: $R_d\ge S_d$.

Reinforcing steel is used in RCC to carry tension (and compression) and to control cracking. IS 1786 covers HYSD/TMT bars and IS 432 mild steel bars.

Main characteristics

1. High tensile strength: Fe250 ($f_y=250$), Fe415 ($f_y=415$), Fe500 ($f_y=500$), Fe550 N/mm$^2$.
2. Ductility: percent elongation at least 23% (Fe250), 14.5% (Fe415), 12% (Fe500); ductile steel gives warning before failure. Fe500D and Fe550D give higher ductility for seismic use.
3. Modulus of elasticity: $E_s=2\times10^5$ N/mm$^2$ for all grades.
4. Yield behaviour: mild steel has a definite yield point; HYSD/TMT bars have no sharp yield, so the 0.2% proof stress is taken.
5. Good bond: ribs/deformations on the surface give better bond with concrete (design bond stress is increased by 60% for deformed bars).
6. Thermal compatibility: coefficient of expansion about $1.2\times10^{-5}$ per $^\circ$C, nearly equal to concrete, so no internal stress develops with temperature.
7. Weldability and bendability: bars should bend without cracking (bend and rebend tests); carbon equivalent is limited for weldability.
8. Ultimate to yield ratio: $f_u/f_y\ge1.10$ (1.08 for Fe500) to ensure reserve strength.
9. Corrosion resistance: TMT bars have a tough outer surface and softer core, giving better corrosion resistance than cold twisted bars.
10. Density: 7850 kg/m$^3$; bars available in sizes 6, 8, 10, 12, 16, 20, 25, 32 mm.

Given: column 400 x 600 mm, effective height $l=3.6$ m, $DL=250$ kN, $LL=100$ kN, wind UDL $w=5$ kN/m along the height.

Wind effect on the column

• Total horizontal force: $H_w=5\times3.6=18.0$ kN.
• Column is part of a moment resisting frame, so it is assumed fixed at both ends under the lateral UDL (assumption). Fixed-end moment: $M_w=\dfrac{wl^2}{12}=\dfrac{5\times3.6^2}{12}=5.40$ kN·m (wind can act in either direction, so moment can be $\pm$).

Load combinations (IS 456:2000, Table 18 / cl. 36.4)

Working:

$$\begin{aligned} 1.5(250+100)&=525\ \text{kN}\\ 1.2(250+100)&=420\ \text{kN},\quad 1.2\times5.4=6.48\ \text{kN·m},\quad 1.2\times18=21.6\ \text{kN}\\ 1.5\times250&=375\ \text{kN},\quad 1.5\times5.4=8.10\ \text{kN·m},\quad 1.5\times18=27\ \text{kN}\\ 0.9\times250&=225\ \text{kN},\quad 1.5\times5.4=8.10\ \text{kN·m},\quad 27\ \text{kN} \end{aligned}$$

Combination 4 applies where stability against overturning or stress reversal is critical (minimum axial load with maximum moment).

Answer: The governing combinations are $P_u=525$ kN, $M_u=0$ (combination 1) and $P_u=420$ kN with $M_u=6.48$ kN·m (combination 2). The column should be designed for all four and the worst case is adopted.