ULTIMATE LIMIT STATE DESIGN (CEN 324)

PREPARED BY: TARTEEL AWAD HAJALI AHMED

REINFORCING STEEL CLASSIFICATION: ACCORDING

TO BRITISH STANDARDS

Designation Nominal sizes (mm)

Specified characteristic

strength fy (N/mm2)

Hot-rolled mild steel (BS 4449) All sizes 250

Hot-rolled High Yield steel (BS 4449)All sizes 460

Cold-worked high yield (BS 4461)

Hard-drawn steel wire (BS 4482) Up to and including 12 485

*For more details read chapter 1 in Mosley (page 12, 13)

STRESS-STRAIN RELATIONSHIP:

i. Concrete:

for short-term loading. Within the range of concrete mixes used in practical

design, the following general statements may be made:

a) Up to about 50% of the maximum stress, the stress/strain curve may be

approximated by a straight line.

b) The peak stress is reached at a strain of about 0.002.

c) Visible cracking and disintegration of the concrete does not occur until

the strain reaches about 0.0035.

For design purposes, BS 8110 uses the idealized curve in Fig. 1 (b), with

a maximum stress of 0.67𝑓𝑐𝑢 /𝛾𝑚 ; concrete is assumed to fail at an

ultimate strain, 𝜀𝑐𝑢 , of 0.0035.

Fig. 1 stress-strain curves for concrete in compression

The ultimate design stress is given by: 𝟎.𝟔𝟕 𝒇𝒄𝒖

𝜸𝒎

If 𝛾𝑚 = 1.5 𝑓𝑜𝑟 𝑓𝑙𝑒𝑥𝑢𝑟𝑒 (𝑙𝑒𝑐𝑡𝑢𝑟𝑒 2)

𝟎.𝟔𝟕 𝒇𝒄𝒖

𝟏.𝟓 = 0.447 fcu ≈ 0.45 fcu (1)

II. REINFORCING STEEL:

Fig. 2 stress-strain curves for reinforcement in compression & tension according to BS8110

The relation between stress & strain is linear in elastic range up to the design yield

stress of 𝑓𝑦

𝛾𝑚.

𝑓𝑦 ≈ characteristic yield strength of reinforcement

If 𝛾𝑚 = 1.05 (𝑙𝑒𝑐𝑡𝑢𝑟𝑒 2)

Stress = elastic modulus x strain = Es x 𝜀𝑠

The design yield strain= 𝜀𝑦 = 𝑓𝑦

𝛾𝑚 /𝐸𝑠 ⟶ (2)

For fy = 460 N/mm2

𝜀𝑦 =460

1.05 ×200×103= 0.00219 → (3)

For fy = 250 N/mm2

𝜀𝑦 =250

1.05 ×200×103= 0.0019 → (4)

THE DISTRIBUTION OF STRAINS AND STRESSES ACROSS A SECTION:

The theory of bending for reinforced concrete assumes that:

1. The concrete will crack in the regions of tensile strains and that, after cracking. all the

tension is carried by the reinforcement.

2. It also assumes that plane sections of a structural member remain plane after straining,

so that across the section there must be a linear distribution of strains.

• Figure 3 shows the cross-section of a member subjected to bending, and the resultant

strain diagram, together with three different types of stress distribution in the concrete.

Fig.3 section with strain diagrams and stress blocks

Three different types of stress distribution in concrete:

1. The triangular stress distribution applies when the stresses are very nearly

proportional to the strains, which generally occurs at the loading levels

encountered under working conditions and is, therefore, used at the

serviceability limit state.

2. The rectangular-parabolic stress block represents the distribution at failure

when the compressive strains are within the plastic range and it is

associated with the design for the ultimate limit state.

3. The equivalent rectangular stress block is a simplified alternative to the

rectangular-parabolic distribution.

A general theory for ultimate flexural strengths:

The following assumptions are made:

1) The strains in the concrete and the reinforcing steel are directly proportional

to the distances from the neutral axis, at which the strain is zero.

2) The ultimate limit state of collapse is reached when the concrete strain at the

extreme compression fibre reaches a specified value 𝜀𝑐𝑢 .

3) At failure, the distribution of concrete compressive stresses is defined by an

idealized stress/strain curve.

4) The tensile strength of the concrete is ignored.

5) The stresses in the reinforcement are derived from the appropriate stress/strain

curve.

Cross-Section Strains

Fig. 4

As ≡ Area of longitudinal tension reinforcement.

As\ ≡ Area of longitudinal compression reinforcement.

d≡ effective depth of section (top face to the centroid of tension reinforcement).

x ≡ neutral axis.

d\ ≡ depth of compression reinforcement.

b≡ section breadth (width).

D≡ section length (depth).

• The maximum compressive strain has a specified value 𝜀𝑐𝑢 at the collapse.

From geometry:

𝜀𝑠𝑐

𝑥−𝑑\=

𝜀𝑐𝑐

𝑥 ⇒ 𝜀𝑠𝑐 =

𝑥−𝑑\

𝑥 𝜀𝑐𝑐 → (5)

𝜀𝑠𝑡𝑑 − 𝑥

=𝜀𝑐𝑐𝑥

⇒ 𝜀𝑠𝑡 = 𝑑 − 𝑥

𝑥 𝜀𝑐𝑐 → (6)

Note: The relationship between the strain in reinforcement bar and adjacent concrete

depends on the bond strength but is accurate enough to assume that they are equal.

The depth of neutral axis x can be determined from equation (6):

𝑑

𝑥=

𝜀𝑠𝑡𝜀𝑐𝑐

+ 1 ⟹ 𝑥 = 𝑑

1 +𝜀𝑠𝑡𝜀𝑐𝑐

⟶ (7)

At ultimate limit state (ULS) the maximum compressive stress in concrete;

𝜀𝑐𝑢 = 0.0035

At fy = 460 N/mm2 ⇒ 𝜀𝑦 = 0.00219

∴ 𝑥 = 𝑑

1 +0.002190.0035

= 0.615 𝑑 ⟶ (8)

• To ensure yielding of the tension steel at the ULS: 𝑥 ≯ 0.615 𝑑

Notes:

❖ At the ultimate limit state, it is important that member sections in flexure

should be ductile and that failure should occur with the gradual yielding of

the tension steel and not by a sudden catastrophic compression failure of the

concrete.

❖ Also, yielding of the reinforcement enables the formation of plastic hinges so

that redistribution of maximum moments can occur, resulting in a safer and

more economical structure (see section 3.6 Mosley).

❖ To be very certain of the tension steel yielding, the code of practice limits the

depth of neutral axis so that: 𝑥 ≯ (𝛽𝑏 − 0.4) 𝑑

𝛽𝑏 =𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑟𝑒𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑓𝑜𝑟𝑒 𝑟𝑒𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝛽𝑏 =100 − % 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡 𝑟𝑒𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

100 → (9)

For moment redistribution ≯ 10% 𝑎𝑛𝑑 𝛽𝑏 ≥ 0.9 ⇒ 𝑥 ≯ 0.5 𝑑

Fig. 5